Applied Partial Di erential Equations...

Applied Partial Differential Equations

MAT1508/APM446 ∗

I. M. Sigal †

Dept of Mathematics, Univ of TorontoWinter Thursdays 3-6pm (BA 6183)

Contents

1 Introduction 3

1.1 Important examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Abstract evolution equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Local Existence for Evolution Equations 7

2.1 Reduction to a fixed point problem . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 The contraction mapping principle . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3 Local existence for evolution partial differential equations . . . . . . . . . . . . . 9

2.4 Local existence for Hartree and Schrodinger equations . . . . . . . . . . . . . . . 12

2.5 Classical solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.6 A priori estimates and global existence . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Key classes of solutions of PDEs and symmetries 16

3.1 Key special solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Symmetries and solutions of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Gateaux and Frechet derivatives 27

5 The implicit and inverse function theorems 30

5.1 Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5.2 Applications: existence of breathers and constant mean curvature surfaces . . . . 33

5.3 Appendix I: Direct proof of Theorem 5.2. . . . . . . . . . . . . . . . . . . . . . . 36

5.4 Appendix II: Derivation of the inverse function theorem from the implicit functiontheorem (Theorem 5.2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.5 Minimal surfaces (to be done) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.6 Harmonic maps (to be done) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6 Existence of bubbles and Lyapunov - Schmidt decomposition 37

6.1 Appendix. Details of the Lyapunov-Schmidt reduction . . . . . . . . . . . . . . . 43

∗January 10, 2015†Thanks to Jordan Bell for careful reading these notes and flagging typos and sloppiness

1

2 Lectures on Applied PDEs, January 10, 2015

7 Theory of bifurcation 47

8 Bifurcation of surfaces of constant mean curvature 54

8.1 Bifurcation of new surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

8.2 Proof of the existence part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

8.3 Proof of the linearized stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

9 Stability of static solutions (and solitons) 58

9.1 Stability: generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

9.2 Turing instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

9.3 Appendix: Derivation of Theorem 9.1 from Theorem 9.2 . . . . . . . . . . . . . . 62

9.4 Pattern formation in reaction-diffusion equations . . . . . . . . . . . . . . . . . . 63

9.5 Asymptotic stability of homogeneous and bifurcating soltions . . . . . . . . . . . 64

10 Abrikosov lattices 64

10.1 Fixing the gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

10.2 The linear problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

10.3 Rescaling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

10.4 Reformulation of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

10.5 Reduction to a finite-dimensional problem . . . . . . . . . . . . . . . . . . . . . . 71

10.6 Proof of Theorem 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

10.7 Appendix: Proof of Proposition 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . 75

10.8 Appendix: Solving the equation (10.25b) . . . . . . . . . . . . . . . . . . . . . . . 76

11 Keller-Segel Equations of Chemotaxis 76

11.1 M > 8π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

11.2 Blowup criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

11.3 Appendix: Gradient formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

11.4 Appendix: Criterion of break-down in the dimension d ≥ 3 . . . . . . . . . . . . 84

12 Mean curvature flow 85

12.1 Definition and general properties of the mean curvature flow . . . . . . . . . . . . 85

12.2 Solitons: self-similar surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

12.3 Linearized stability of self-similar surfaces . . . . . . . . . . . . . . . . . . . . . . 90

12.4 F−stability of self-similar surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 94

13 Asymptotic stability of kinks in the Allen-Cahn equation 95

A Banach and Hilbert Spaces 96

A.1 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

A.2 Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

A.3 Lp–spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

A.4 Proofs of Holder’s and Minkowski’s inequalities . . . . . . . . . . . . . . . . . . 100

A.5 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

A.6 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

Lectures on Applied PDEs, January 10, 2015 3

B Fourier transform 103

B.1 Definitions and properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

B.2 Applications of Fourier transform to partial differential equations . . . . . . . . . 107

B.3 Estimates on propagators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

C Linear operators 110

D Elements of spectral theory 116

D.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

D.2 Location of the essential spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . 118

D.3 Perron-Frobenius Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

E Linear evolution and semigroups 120

1 Introduction

In this course we will consider several key partial differential equations arising in physics, materialscience, biology and geometry. We develop an existence theory for these equations, describe theirkey properties, isolate their most important solutions and study stability or instability of thesesolutions.

1.1 Important examples

Here are some examples of the equations we will be studying in the course:

The Allen-Cahn equation. This equation describes among other things motion of interfacesbetween different compounds and is given by

∂u

∂t= ε2∆u+ u− u3. (1.1)

This simple looking equation has a rich structure and is an object of an intensive investigationin the last 50 years. It has an important set of static solutions, called kinks

χε(x) = tahn

((x− x0) · e√

2ε

), (1.2)

for any e, x0 ∈ Rd. It turns out that solutions of this equation change as kinks across interfacesmoving by the mean curvature flow.

The Ginzburg-Landau equations. These describe equilibrium configurations of supercon-ductors, and of the U(1) Higgs model from particle physics, are described by a system of non-linear PDE called the Ginzburg-Landau equations:

∆AΨ = κ2(|Ψ|2 − 1)Ψ, (1.3a)

curl∗ curlA = Im(Ψ∇AΨ), (1.3b)


for (Ψ, A) : R2 → C× R2, ∇A = ∇− iA, and ∆A = ∇2A, the covariant derivative and covariant

Laplacian, respectively. These equations have remarkable special solutions - the magnetic vor-tices, which are labeled by the equivalence classes of the homomorphisms of S1 into U(1), i.e.by integers n, and for d = 2 are given by

Ψ(n)(x) = f (n)(r)einθ, A(n)(x) = a(n)(r)∇(nθ), (1.4)

where (r, θ) are the polar coordinates of x ∈ R2. Even more interestingly, they solutions whichare lattices of vortices, for which A.A. Abrikosov received a few years back a Nobel prize inphysics.

The Keller - Segel equations. The next system of equations models the chemotaxis, whichis the directed movement of organisms in response to the concentration gradient of an externalchemical signal and is common in biology. The chemical signals either come from externalsources or are secreted by the organisms themselves. The latter situation leads to aggregationof organisms and to the formation of patterns.

Chemotaxis underlies many social activities of micro-organisms, e.g. social motility, fruitingbody development, quorum sensing and biofilm formation.

Consider organisms moving and interacting in a domain Ω ⊆ Rd, d = 1, 2 or 3. Assumingthat the organism population is large and the individuals are small relative to the domain Ω,Keller and Segel derived a system of reaction-diffusion equations governing the organism densityρ : Ω× R+ → R+ and chemical concentration c : Ω× R+ → R+. The equations are of the form

∂tρ = Dρ∆ρ−∇ (f(ρ)∇c)∂tc = Dc∆c+ αρ− βc.

(1.5)

Here Dρ, Dc, α, β are positive functions of x and t, ρ and c, and f(ρ) is a positive functionmodelling chemotaxis (positive chemotaxis).

Assuming α and β are constants, this system has the simple homogeneous static solution c =const and ρ = β/α =const. However this solution is unstable under small perturbations. Forinitial conditions close to this static solution, ρ starts growing at some point and becomes highlyconcentrated around this point. This is a chemotactic aggregation which leads to formation offruition bodies (colonies formed by Escherichia coli under starvation conditions, or multicellularstructures of ∼ 105 cells, by single cell bacterivores, when challenged by adverse conditions).Similar phenomenon occurs in dynamics of tumours.

A simimplified version of these equations is used to model stellar collapse and crime patterns.

The Fisher-Kolmogorov-Petrovsky-Piskunov equation. This equation describes prop-agation of mutations in population biology and flames in combustion theory. It is given by

∂u

∂t= ∆u− (u− 1)u. (1.6)

It has interesting solutions, called traveling waves, which are of the form

u(x, t) = φ(x− vt), (1.7)

where ϕ(y) and v are independent of t and are called the traveling wave profile and the travelingwave velocity.


1.2 Abstract evolution equations

We would like to address a general theory of evolution PDEs, i.e. equations of the form

∂tu = F (u), u|t=0 = u0, (1.8)

where t → u(t) is a path in some space, F is a map defined on the same space, ∂tu = u = ∂u∂t

and we understand (1.8) as ∂tu(t) = F (u(t)). u0 is called the initial condition and (1.8), theinitial value problem.

The situation we will be mostly interested in is when F is a partial differential operator,linear or nonlinear. An example of such an equation is the equation:

∂u

∂t= ∆u+ g(u). (1.9)

In this example, F (u) = ∆u+ g(u). Here ∆ is the Laplace operator (the Laplacian):

∆u :=n∑j=1

∂2u

∂x2j

.

This is the celebrated nonlinear heat or reaction-diffusion equation. The term ∆ is responsiblefor diffusion and the term, g(u), describes the reaction in the system modelled. Let us considerfirst several other examples of maps F :

1) F (u) = ∆u,

2) F (u) = f u for a given function f ,

3) F (u) = div( ∇u√1+|∇u|2

).

4) F (u) = ϕ ∗ u for given ϕ ∈ L1(Rn).

Note that the map in 1) is linear, while the map in 2) is linear if f is linear, and nonlinearotherwise.

Sometimes (1.8) is called the dynamical system and F , the vector field (finite or infinitedimensional). Though we define F on a vector space, it can be also defined on a manifold(again, finite or infinite dimensional).

To solve equation (1.8), we have to choose a space, say Y , to which the vector-functiont → u(t) belongs for all t considered. We have to make sure that F is defined on this space.Depending on the problem at hand, we choose different spaces. For instance, for the examples1) and 3) above, F maps the space Y := Ck(Ω) into the space Ck−2(Ω), and in the examples4), Lp(Rn) into itself. More abstractly, Y could be some space with a norm, or a Banach space.For the definition of Banach spaces see Appendix A.

We also have to choose a space, say X, to which the function t→ u(t) belongs. If u(t) ∈ Yfor t in some interval I, then we can choose

• X = C(I, Y ), where, say, I := [0, T ].

Here X = C(I, Y ) is the space of continuous vector - functions u : t→ u(t) on I with values inY and with the norm

‖u‖X := supt∈I‖u(t)‖Y .


Flows. If the initial problem (1.8) has a unique solution, u(t), for every initial conditionu0 ∈ Ω ⊂ Y , we can define the map Φt : u0 → u(t), where u(t) is the solution to (1.8) at time t.This map is called the flow generated by (1.8) (or by F ). It has the following properties

• Φ0 = 1; Φt Φs = Φt+s; ∂tΦt(u0) = F (Φt(u0)), t ≥ 0.

The first and third properties follow from the definition and the second one, from the uniquenessof solutions of the initial problem (1.8). The families satisfying the first two properties are calledthe semi-groups and F is called the generating vector field.

If the map F (u) is linear, F (u) = Au, i.e. if we consider the linear equation

∂u

∂t= Au and u|t=0 = u0, (1.10)

then so is the flow, Φt(u). In the linear case, we denote the flow as Φt(u) = eAtu. The linearflow is also called the semi-group. An important example of the linear evolution equation is thelinear (free) heat equation

∂u

∂t= ∆u and u|t=0 = u0, (1.11)

where u : Rnx × R+t → R, R+

t := t ∈ R : t ≥ 0, is an unknown function, and u0 : Rn → R is agiven initial condition. The linear flow for this equation is called the heat kernel.

For linear flows, there are general situations where we can show the existence of the global(all t’s) flow:

• A is a bounded operator;

• A is either self-adjoint (A∗ = A) and bounded above, A ≤ C for some C <∞ (〈u,Au〉 ≤C‖u‖2) or anti-self-adjoint (A∗ = −A);

• A is a ‘constant coefficient pseudo-differential’ operator; more precisely, A = a(−i∇x),where a is some decent function.

For a bounded operator A the flow always exists since eAt can be defined by

etA :=

∞∑n=0

(tA)n

n!. (1.12)

The series on the r.h.s. converges absolutely since∥∥∥∥∥∞∑n=0

(tA)n

n!

∥∥∥∥∥ ≤∞∑n=0

‖(tA)n‖n!

≤∞∑n=0

tn

n!‖A‖n = et‖A‖ <∞.

For the second case we refer to [30]. In the third case A and eAt are defined using the Fourier

transform as (Au)(k) = a(k)u(k) and (eAtu)(k) = ea(k)tu(k), respectively (see Appendix B and[30] for more on the Fourier transform). Applying the inverse Fourier transform F−1 : f 7→ f ,so that (f ) = f = (f ) , and using that F−1 : gf 7→ (2π)−n/2g ∗ f , we obtain

u = gt ∗ u0, (1.13)

where gt(x) is the inverse Fourier transform of the function ea(k)t. As an illustrative example,we consider


The operator et∆ (heat kernel). Since (e−|k|2t) = (4πt)−n/2e−|x|

2/(4t) (see Appendix B),the equation (1.13), in this case, gives

u = pt ∗ u0, (1.14)

where pt(x) = (4πt)−n/2e−|x|2/4t. In particular, u→ u0 as t→ 0, as it should be.

Thus the operator et∆ has the integral kernel pt(x, y) = pt(x − y) := (4πt)−n/2e−|x−y|2/4t,

t > 0, which satisfies pt(x, y) > 0 and∫pt(x, y) dy = 1.

Discussion. The semi-group Pt = et∆ has the following properties

(a) Pt is positivity improving, i.e. if f ≥ 0, then Ptf > 0,

(b) Pt1 = 1.

Semi-groups with such properties are called stochastic semigroups. (The conditions f ≥ 0and f > 0 can be stated for abstract vector spaces in terms of closed and open cones.)

Remark. We can arrive at the formula (eAtu)(k) = ea(k)tu(k), by first applying the Fouriertransform the equation (1.10), whilch we write as ∂u

∂t = a(−i∇x)u and u|t=0 = u0, treating t

as a parameter and using the property F : a(−i∇x)f 7→ a(k)f . As a result, we obtain

∂u(k, t)

∂t= a(k)u(k, t) and u(k, t)|t=0 = u0(k).

We solve the latter equation (treating k as a parameter) to find u(k, t) = ea(k)tu0(k).

2 Local Existence for Evolution Equations

2.1 Reduction to a fixed point problem

Canonical form of evolution equations. We say the initial problem (1.8) is written in thecanonical form if the map F (u) is presented as F (u) = Au+ f(u), where A is a linear operatorand f(u) satisfies f(u) = o(‖u‖). Then (1.8) can be rewritten as

∂tu = Au+ f(u), u|t=0 = u0. (2.1)

where A is a linear operator on Y and f is a nonlinear map, i.e. f(u) = o(‖u‖) in some norm, orf(0) = f ′(0) = 0. We will call f the nonlinearity. We will see later that, if F (0) = 0 and F (u) isonce continuously (Gateaux) differentiable, it can be always split into linear and nonlinear parts,F (u) = Au+ f(u), with A and f as above. (This is a general result, valid for all differentiablemaps. It generalizes the Taylor theorem of Calculus.)

Duhamel principle. Consider the inhomogeneous initial value problem

∂tu = Au+ f, u|t=0 = u0, (2.2)

where f = f(t) is a given function f : [0, T ) → Y . Assuming the homogeneous equation∂tu = Au, u|t=0 = u0, has a unique solution, the solution, u(t), of (2.2), is given by

u(t) = etAu0 +

∫ t

0e(t−s)Af(s) ds. (2.3)


Weak solutions. Consider the initial value problemanonical form (2.1). We apply (2.3) to(2.1) to obtain

u(t) = etAu0 +

∫ t

0e(t−s)Af(u(s)) ds. (2.4)

If u(t) solves (2.1), then it also solves the equation (2.4). Conversely, if u(t) solves (2.4) and isdifferentiable in t, then it solves the equation (2.1).

If u(t) solves (2.4), but we do not know whether it is differentiable or not, we call u(t) aweak solution to (2.1).

Fixed point problem. Eq (2.4) can be written as the equation u = H(u), where

H(u)(t) := etAu0 +

∫ t

0e(t−s)Af(u)(s) ds, (2.5)

called the fixed point equation or the fixed point problem. A solution of such an equation is calleda fixed point. In our next step we learn how to solve fixed point equations.

2.2 The contraction mapping principle

A key to dealing with a large class of equations is to reduce them to a fixed point problem,

H(u) = u, (2.6)

for some map H and then use one of several fixed point theorem stating existence and uniquenessof solutions of the latter problem. (The equation (2.6) is called the fixed point equation and itssolution, a fixed point of the map H.) The most useful among these theorems is the Banachcontraction mapping principle which we now formulate and prove.

Let X be a Banach space. Denote by d(u, v) = ‖u − v‖ the distance between the vectors uand v. We remark that actually all we need for the next theorem is that X is a complete metricspace (i. e. it does not have to have a norm). Let B be a closed set in X. A map H : B → Bis called a strict contraction if and only if there is a number α ∈ (0, 1) s.t.

d(H(u), H(v)) ≤ αd(u, v), ∀u, v ∈ B.

Theorem 1 (the contraction mapping principle). If H is a strict contraction in B, then H hasa unique fixed point in B.

Proof. We use the method of successive approximations to solve the equation u = H(u). Picksome u0 ∈ B and define u1 = H(u0), . . . , un = H(un−1). Since H is a contraction, un ∈ B.

We claim that un is a Cauchy sequence in X. Indeed, let n ≥ m, then

d(un, um) ≤ αmd(un−m, u0).

Taking here n = m + 1, we find d(um+1, um) ≤ αmd(u1, u0). Next, by the triangle inequality(i.e. d(v, u) ≤ d(v, w) + d(w, u), ∀w ∈ X), we obtain

d(uk, u0) ≤ d(uk, uk−1) + d(uk−1, uk−2) + · · ·+ d(u1, u0).

Applying d(um+1, um) ≤ αmd(u1, u0) to each term on the r.h.s. gives

d(uk, u0) ≤(αk−1 + αk−2 + . . .+ 1

)d(u1, u0)

≤ 1

1− αd(u1, u0).


(If B is a bounded set in X, then we do not need the step above, since in this case ‖uk‖ areuniformly bounded.) The last two inequalities imply that

d(un, um) ≤ αm

1− αd(u1, u0)→ 0 as m,n→∞.

Thus un is a Cauchy sequence in X. Now since X is complete, un ∈ B and B is closed,there is a u∗ ∈ B s.t. un → u∗. Since d(H(un), H(u∗)) ≤ αd(un, u∗) → 0, we have alsothat H(un) → H(u∗). This and the equation un = H(un−1) imply that the limit u∗ satisfiesthe equation u∗ = H(u∗). This demonstrates existence of a fixed point in B, and we finishthe proof by showing its uniqueness. Suppose that H(u∗) = u∗, and H(v∗) = v∗. Then wehave d(v∗, u∗) = d(H(v∗), H(u∗)) ≤ αd(v∗, u∗). Hence, since α ∈ (0, 1), d(v∗, u∗) = 0 and sov∗ = u∗.

Remark. As was mentioned above, there are many fixed point theorems and the Banachcontraction mapping principle is one of these, albeit the most useful one.

2.3 Local existence for evolution partial differential equations

In this subsection, we use the contraction mapping principle to prove local existence of solu-tions for several evolution PDEs, the reaction-diffusion (nonlinear heat), Hartree and nonlinearSchrodinger equations.

Local existence for reaction-diffusion (nonlinear heat) equations. We show local exis-tence to the initial value problem for the reaction-diffusion (nonlinear heat) equation (or reaction-diffusion equation):

∂u

∂t= ∆u+ λ|u|p−1u, u|t=0 = u0, (2.7)

on Rn, with p > 1 and bounded initial conditions u0 on Rn. The special case of equation(2.7) without nonlinearity first appeared in the theory of heat diffusion. In that case, u0(x)is a given distribution of temperature in a body at time t = 0, and u(x, t) is the unknowntemperature–distribution at time t. Presently, this equation appears in various fields of science,including the theory of chemical reactions and mathematical modeling of stock markets. Similarequations appear in the motion by mean curvature flow, vortex dynamics in superconductors,surface diffusion and chemotaxis.

We note that (2.7) is the initial value problem (1.8), ∂tu = F (u), u|t=0 = u0, in the canonicalform, (2.1), with A = ∆, a linear operator acting on H2 and f(u) = λ|u|p−1u, a nonlinear map.

We will look for week solutions in space X := C([0, T ], Y ), with T specified later and Y :=L∞(Rn). The space L∞ is one of the standard Lp-spaces and is defined as (see Appendix A.3for more details)

L∞(Ω) := f : Ω→ C | f is measurable, and ess sup |f | <∞, (2.8)

where, recall that ess sup |f | := infsup |g| : g = f a.e., with the norm defined as

‖f‖∞ := ess sup|f |. (2.9)

(Strictly speaking, elements of Lp(Ω) are equivalence classes of measurable functions: two func-tions define the same elements of Lp(Ω) if they differ only on a set of measure 0.) We often usethe abbreviations Lp and ‖v‖p for Lp(Ω) and ‖v‖Lp .


Theorem 2. Let u0 ∈ L∞(Rn) and R > ‖u0‖∞. Then, for T <(pRp−1

)−1, (R−‖u0‖∞) (Rp)−1,

the nonlinear heat equation (2.7) has a weak solution u ∈ C([0, T ], L∞), satisfying ‖u‖C([0,T ],L∞) ≤R and unique in the ball ‖u‖C([0,T ],L∞) ≤ R.

Proof. Using Duhamel’s principle, Eq (2.7) can be written as the fixed point equation u = H(u),where

H(u)(t) := et∆u0 +

∫ t

0e(t−s)∆f(u)(s) ds (2.10)

and we have written f(u)(s) for f(u(s)). Let Y := L∞(Rn) and X := C([0, T ], Y ), with Tspecified later. The proof of existence and uniqueness will follow if we can show that the mapH has a unique fixed point in the ball

BR := u ∈ X, ‖u‖X ≤ R,

for some R > 0. We prove this statement via the contraction mapping principle.We begin by proving that there is R > 0 s.t. H is a well-defined map from BR to BR. First,

we show the estimate ∥∥et∆u∥∥Y≤ ‖u‖Y (2.11)

We have shown above that the operator et∆ has the integral kernel pt(x, y), t > 0, i.e. (et∆u)(x) =∫pt(x, y)u(y) dy, with the following properties: pt(x, y) > 0 and

∫pt(x, y) dy = 1. Using these

properties, we obtain the estimate (2.11).Next, the elementary bound ||u|p−1u| ≤ Rp, for |u| ≤ R, shows that, if t < T and u ∈ BR,

thensup

‖w‖Y ≤R‖f(w)‖Y = sup

|w|≤R|f(w)| ≤ Rp (2.12)

(remember that Y := L∞(Rn)), which, together with the estimate (2.11), gives∥∥∥∥∫ t

0e(t−s)Af(u)(s) ds

∥∥∥∥X

≤ supt≤T

∫ t

0‖f(u)(s)‖Y ds ≤ TRp. (2.13)

Estimates (2.11) and (2.13) and the definition (2.10) of the map H imply that H : BR → BR,provided ‖u0‖Y + TRp ≤ R.

Now, we prove that H : BR → BR is a strict contraction. Recalling the definition f andusing the elementary bound

||u1|p−1u1 − |u2|p−1u2| ≤ pRp−1|u1 − u2|,

for |u|, |u1|, |u2| ≤ R, we obtain, for u1, u2 ∈ BR,

sup‖w1‖Y ,‖w2‖Y ≤R

‖f(w1)− f(w2)‖Y / ‖w1 − w2‖Y ≤ pRp−1, (2.14)

which, together with the definitions of H and ‖u‖X and estimate (2.11), gives

‖H(u1)−H(u2)‖X ≤ supt≤T

∫ t

0‖f(u1)(s)− f(u2)(s)‖Y ds

≤ TpRp−1‖u1 − u2‖X .

Therefore, if pRp−1T < 1, then H is a strict contraction in BR. We see that the inequalities‖u0‖Y + TRp ≤ R and pRp−1T < 1 are satisfied if R > ‖u0‖Y and T < (pRp−1)−1, (R −‖u0‖Y )/Rp. This gives the existence and uniqueness for the stated conditions on R and T .

This completes the proof of existence and uniqueness of the solution u and the estimate onit.


Abstract result on local existence. Before proceeding to other equations we prove anabstract result on local existence of evolution equations in canonical form. We prove existenceand uniqueness of weak solutions for the initial value problem (1.8), ∂tu = F (u), u|t=0 = u0, inthe canonical form, (2.1), which we reproduce here

∂tu = Au+ f(u), (2.15)

with the initial condition u|t=0 = u0. Recall, that here A is a linear operator on Y and f isa nonlinear map, i.e. f(u) = o(‖u‖). For simplicity we assumed that f does not depend on texplicitly (and only through u).

Recall, that we think of u as a path u : t ∈ I → u(t) ∈ Y in Y and we are looking for weaksolutions to (2.15). Specifically, we will consider solutions in the space C([0, T ], Y ), for someT > 0. We have the following result:

Theorem 3. Consider the abstract nonlinear equation (2.15) and assume that

• A generates a semi-group eAt, satisfying, for some constant K,

supt≥0

∥∥etAw∥∥Y≤ K‖w‖Y , (2.16)

• the nonlinearity f is a locally Lipschitz map, f : Y → Y , obeying, for some MR, LR <∞,

sup‖w‖Y ≤R

‖f(w)‖Y ≤MR, (2.17)

andsup

‖w1‖Y ,‖w2‖Y ≤R‖f(w1)− f(w2)‖Y / ‖w1 − w2‖Y ≤ LR. (2.18)

Let u0 ∈ Y and R > K‖u0‖Y . Then the equation (2.15) has a unique weak solution u ∈C([0, T ], Y ), for

T < (KLR)−1, (R−K‖u0‖Y )/KMR,

in the ball ‖u‖C([0,T ],Y ) ≤ R. The solution u depends continuously on the initial condition u0.Furthermore, either the solution is global in time or blows up in Y in a finite time (i.e. either‖u(t)‖Y <∞, ∀t, or ‖u(t)‖Y <∞ for t < t∗ and ‖u(t)‖Y →∞ as t→ t∗ for some t∗ <∞).

Proof. Using Duhamel’s principle, Eq (2.15) can be written as the fixed point equation u = H(u),where

H(u)(t) := etAu0 +

∫ t

0e(t−s)Af(u)(s) ds (2.19)

and we have written f(u)(s) for f(u(s)). Let X := C([0, T ], Y ), with T < (KLR)−1, (R −K‖u0‖Y )/KMR and R > K‖u0‖Y . The proof of existence and uniqueness will follow if we canshow that the map H has a unique fixed point in the ball

BR := u ∈ X, ‖u‖X ≤ R.

We prove this statement via the contraction mapping principle.We begin by proving that H is a well-defined map from BR to BR. Using the assumptions

(2.16) and (2.17), we obtain the estimates∥∥etAu0

∥∥X≤ K‖u0‖Y (2.20)


and, if t < T and u ∈ BR,∥∥∥∥∫ t

0e(t−s)Af(u)(s) ds

∥∥∥∥X

≤ supt≤T

∫ t

0K ‖f(u)(s)‖Y ds ≤ TKMR. (2.21)

Estimates (2.20) and (2.21) imply that H : BR → BR, provided K‖u0‖Y +KTMR ≤ R.Now, we prove that H : BR → BR is a strict contraction. Recalling the definition of ‖u‖X

and using (2.16) and (2.18), we obtain, for u1, u2 ∈ BR,

‖H(u1)−H(u2)‖X ≤ supt≤T

∫ t

0K ‖f(u1)(s)− f(u2)(s)‖Y ds

≤ KTLR‖u1 − u2‖X .

Therefore, if LRKT < 1, then H is a strict contraction in BR. We see that the inequalitiesK‖u0‖Y + KTMR ≤ R and LRKT < 1 are satisfied if R > K‖u0‖Y and T < (KLR)−1, (R −K‖u0‖Y )/KMR. This completes the proof of existence and uniqueness of the solution u andthe estimate on it.

Now, we prove that the solution to the initial value problem is continuous with respect tochanges in the initial condition u0. Let u and v be the solutions with initial conditions u0 andv0. We estimate

‖u− v‖X ≤ ‖etA(u0 − v0)‖X + ‖∫ t

0e(t−s)A(f(u)(s)− f(v)(s)) ds‖X .

The estimate of these terms proceeds as above (take u1 = u and u2 = v) and if u, v ∈ BR, then

‖u− v‖X ≤ K‖u0 − v0‖Y +KTLR‖u− v‖X .

Thus, if T is as above, then ‖u − v‖X ≤ K(1 − KTLR)−1‖u0 − v0‖Y completing the proof ofcontinuity.

(expand) Finally, assume [0, t∗) is the maximal interval of existence of u and sup0≤t<t∗‖u(t)‖Y =: U <∞. Take R = 2KU and let

τ := min((3KLR)−1 , U (3MR)−1).

Then taking u(t∗ − τ) as a new initial condition, we see that the solution exists in the interval[0, t∗ + τ), a contradiction. This proves the dichotomy claimed in the theorem.

Discussion: Generalize (2.16) to

supt≥0

ρ(t)∥∥etAw∥∥

Y≤ K‖w‖Y , (2.22)

for some constant K and an appropriate positive function ρ(t).

2.4 Local existence for Hartree and Schrodinger equations

In this subsection we prove the local existence of solutions for Hartree equations. To this endwe will use Theorem 3 proven in the previous subsection, whose conditions we will verify for thespecific equations we consider.


Recall that under the conditions (2.16) and (2.18) for some Banach space Y , the abstractnonlinear equation (2.15) has a weak unique solution u ∈ C([0, T ], Y ), for T < (KLR)−1, (R−K‖u0‖Y )/KMR, in the ball ‖u‖C([0,T ],Y ) ≤ R, with R > K‖u0‖Y . For the space Y , we will useone of the Lp-spaces:

Lp(Ω) := f : Ω→ C | f is measurable, and

∫|f |pdx <∞. (2.23)

(Strictly speaking, elements of Lp(Ω) are equivalence classes of measurable functions: two func-tions define the same elements of Lp(Ω) if they differ only on a set of measure 0.) The normson these spaces are defined as

‖f‖p :=

(∫|f |p

)1/p

if 1 ≤ p <∞, (2.24)

We often use the abbreviations Lp and ‖v‖p for Lp(Ω) and ‖v‖Lp .

Local existence for the Hartree equation. In this subsection we show local existence inL2(Rn) to the initial value problem for the Hartree equation

i∂u

∂t= −∆u+ (v ∗ |u|2)u, u|t=0 = u0, (2.25)

on Rn, where, as usual v ∗f is the convolution of two functions, v ∗g(x) :=∫v(x−y)g(y)dy. We

assume that v is a ’nice’ function, i.e. sufficiently smooth and fast decaying at infinity. Equation(2.25) arises in the problem in quantum physics of many-body systems. In the treatment of thisequation we will use Lp− spaces defined and discussed in Appendix A.3.

Theorem 4. Assume v ∈ L∞. Let u0 ∈ L2(Rn) and R > ‖u0‖2. Then, for T <(3R2‖v‖∞

)−1, (R−

‖u0‖2)(R3‖v‖∞

)−1, the Hartree equation (2.25) has a solution u ∈ C([0, T ], L2), satisfying

‖u‖C([0,T ],L2) ≤ R and unique in the ball ‖u‖C([0,T ],L2) ≤ R. The solution u depends continu-ously on the initial condition u0. Furthermore, either the solution is global in time or blows upin L2 in a finite time.

Proof. We check the conditions of Theorem 3 for (2.7) with p > 1 and Y = L2(Rn). Letwt := eit∆w. Passing to the Fourier transform and using the Plancherel theorem, we obtain‖wt‖2 = ‖wt‖2. Next, since wt = e−i|k|

2tw, we have ‖wt‖2 = ‖w‖2 = ‖w‖2, which implies theestimate ∥∥eit∆w∥∥

2= ‖w‖2, (2.26)

uniformly in t. This gives the condition (2.16) with K = 1. Next, using the elementary inequality‖v ∗ g‖p ≤ ‖v‖p‖g‖1, ∀p ≥ 1, we obtain∥∥(v ∗ |u|2)u

∥∥2≤ ‖v‖∞‖u‖32.

Moreover, using the triangle inequality |(v ∗ |u1|2)u1 − (v ∗ |u2|2)u2| ≤ |(v ∗ (|u1|2 − |u2|2))u1|+|(v ∗ |u2|2)(u1 − u2)| and again the above inequality, we find∥∥(v ∗ |u1|2)u1 − (v ∗ |u2|2)u2

∥∥2≤ 3‖v‖∞(max

i‖ui‖2)2‖u1 − u2‖2.

The conditions (2.17) and (2.18) hold, with MR = R3‖v‖∞ and LR = 3‖v‖∞R2. Applying theabstract result, Theorem 2, we arrive at the statement of the theorem.

Problem. Extend the above theorem from L2(Rn) to arbitrary Sobolev spaces Hs(Rn), s > 0,and (separately) from −∆ to an arbitrary self-adjoint operator A on L2(Rn). (For the definitionof the Sobolev spaces Hs(Rn), s > 0, see Appendix A.6.)


Local existence for nonlinear Schrodinger equation. Consider the initial value problemfor the nonlinear Schrodinger equation (or reaction-diffusion equation):

i∂u

∂t= −∆u+ λ|u|p−1u, u|t=0 = u0, (2.27)

with an initial condition u0 ∈ Hs(Rn). We assume p > 1 and s > 0. Equation (2.27) arises innonlinear optics, plasma physics, theory of water waves and in condensed matter physics.

Problem. Show local existence for (2.27) in the spaces Hs(Rn) with s > n/2 and p, an oddinteger. (Hint: Use Sobolev embedding theorems, e.g. Hs(Rn) ⊂ L∞ for s > n/2, so that Hs isan algebra and so MR ≤ |λ|Rp and LR ≤ |λ|pRp−1.)

Discussion. Notice the difference in behaviour between the heat kernel Pt := et∆ and thepropagator Ut := eit∆. The family et∆ is a one parameter semi-group, which is well defined onthe entire space, say Lp, only for t ≥ 0, while eit∆ is a one parameter group, defined for all t.

While∥∥et∆u∥∥

Lp< ‖u‖Lp , for u not identical 1, the characteristic property of the propagator

Ut is its unitarity. In particular it preserves the L2− inner product, 〈u, v〉 :=∫uvdx: 〈u, Utv〉 =

〈u, v〉.

2.5 Classical solutions

Now we show that weak solutions of the abstract nonlinear equation, (2.15),

∂tu = Au+ f(u), (2.28)

with the initial condition u|t=0 = u0, are differentiable in time, if Au0 ∈ Y . Recall that a weaksolution solves the fixed point problem u = H(u), where, recall,

H(u)(t) := etAu0 +

∫ t

0e(t−s)Af(u)(s) ds. (2.29)

Using thatAu0 ∈ Y and ∂tetAu0 = etAAu0 and ∂t

∫ t0 e

(t−s)Af(u)(s) = f(u)(t)+∫ t

0 ∂te(t−s)Af(u)(s)

and ∂te(t−s)A = −∂se(t−s)A and integrating by parts, we obtain

∂tH(u)(t) = ∂tetAu0 + f(u)(t)− e(t−s)Af(u)(s)|t0 (2.30)

+

∫ t

0e(t−s)A∂sf(u)(s) ds, (2.31)

= etAAu0 + etAf(u)(0) +

∫ t

0e(t−s)A∂sf(u)(s) ds. (2.32)

using this relation we can extend Theorem 3 to the space C1((0, T ], Y ).

Exercise. Do this.

Now, we consider the nonlinear heat equation (2.7) and show that weak solutions u ∈C([0, T ], L∞), found in Theorem 2 are, in fact, smooth. First we show that they have onemore derivative than the initial condition u0(x). To this end we consider this equation in thefixed point form u = H(u), where

H(u)(t) = et∆u0 +

∫ t

0e(t−s)∆f(u)(s) ds. (2.33)


First, we recall that the heat kernel et∆ is given by

(et∆u)(x) = (4πt)−n/2∫e−|x−y|

2/4tu(y)dy, (2.34)

Differentiating this expression w.r.to x, we obtain

∂x(et∆u)(x) = −(4πt)−n/2∫x− y

2te−|x−y|

2/4tu(y)dy. (2.35)

Using that |x−y|2√te−|x−y|

2/4t ≤ Ce−|x−y|2/5t, we obtain, as before, that

‖∂x(et∆u)‖∞ ≤ CAt−1/2‖u‖∞, (2.36)

where

A = (4πt)−n/2∫e−|x−y|

2/5tdy = (4πt)−n/2∫e−|y|

2/5tdy

= (4π)−n/2∫e−|y|

2/5dy. (2.37)

Using this, we find

‖∂xH(u)(t)‖∞ ≤ At−1/2‖u0‖∞ (2.38)

+A

∫ t

0(t− s)−1/2‖f(u)(s)‖∞ ds. (2.39)

This as the relations u = H(u) and sup‖w‖Y ≤R ‖f(w)‖Y ≤ MR imply that ‖∂xu(t)‖∞ . t−1/2.We can iterate this argument, using that

∂x(et∆u)(x) = (4πt)−n/2∫e−|x−y|

2/4t∂yu(y)dy, (2.40)

to obtain

‖∂αxu(t)‖∞ . t−|α|/2,

where, recall, α = (α1, ..., αn) with αj non-negative integers, ∂α :=∏nj=1 ∂

αjxj and |α| =

∑ni=1 αj .

2.6 A priori estimates and global existence

Let us take R = 2K‖u0‖Y in Theorem 3. This leads to the following conditions of the existenceinterval

T < (KLR)−1, ‖u0‖Y /MR.

Hence if we have the a priori estimate

‖u(t)‖Y ≤ C

on the solution u(t), then we can iterate the local existence argument and obtain the existenceon the infinite time interval, i.e. the global existence. (Remember that Theorem 3 also statesthat either the solution is global in time or blows up in Y in a finite time, i.e. ‖u(t)‖Y ≤ R ast→ T for some T <∞).


Assume there is a map E(u) : M → R from a subset M of a vector space X (not necessarilyopen) into R (such maps are called functionals; we describe them in detail later), which does notincrease under the evolution (1.8), i.e. ∂tE(u) ≤ 0. (E(u) is called sometimes energy or entropy,or in general a Lyapunov functional.) This implies that

E(u) ≤ E(u0), (2.41)

where u0 is the initial condition. It might happen that E(u) ≥ c‖u(t)‖2Y , for some Banach spaceY . This would give that ‖u(t)‖Y ≤ c−1E(u0), which would imply the global existence in Y .

Examples:.

1) Under the Hartree and Scrodinger evolutions (a) the L2−norm of solutions and (b) theenergy,

E(ψ) =

∫(1

2|∇ψ|2 +G(ψ)), (2.42)

where G(ψ) := 14(v ∗ |ψ|2)|ψ|2, of solutions are conserved, ∂tE(ψ) = 0.

2) Under the nonlinear heat evolution, (2.7), the energy

E(u) =

∫(1

2|∇u|2 +G(u)), (2.43)

where G(u) := 1p |u|

p, decreases (∂tE(u) < 0), as t increases.

Problem. Show this.One can use the above fact 1a) to iterate the local existence result in L2 for the Hartree

equation to global existence result. For the Sobolev space H1(Rn), one can use in addition theconservation of the energy, E(ψ) = E(ψ0), where ψ0 is the initial condition, in 1b) or decreasingof the energy to prove that ‖ψ(t)‖H1 ≤ C.

For the nonlinear heat equation (or reaction-diffusion equation) (2.7), one can use that theenergy (2.43) decreases as t increases (∂tE(u) < 0) and therefore in particular (2.41) holds, whereu0 is the initial condition, to prove that for certain nonlinearities G, ‖u(t)‖H1 ≤ C.

3 Key classes of solutions of PDEs and symmetries

3.1 Key special solutions

We consider key solutions for the dynamical system on some space Y of functions on Rn, given,as usual, by

∂u

∂t= F (u). (3.1)

where F is a map on this space Y .

Static solutions. Static solutions are the solutions independent of time t. As the result theysatisfy the equation F (u) = 0. Here are some examples.

Allen-Cahn equation. First, we consider the Allen-Cahn equation, which plays a central role inmaterial science. It presents a basic model with many generalizations and extensions. It is areaction diffusion equation of the form

∂u∂t = ε2∆u− g(u), x ∈ Ω, t > 0,∂u∂n = 0, x ∈ ∂Ω,

(3.2)


where Ω ⊆ Rn, u : Ω→ R, ε is a small parameter and g(u) = u3 − u, with the initial conditionu|t=0 = u0(x), x ∈ Ω. More generally, g : R → R is the derivative, g = G′, of a double-wellpotential: G(u) ≥ 0 and has two non-degenerate global minima with the minimum value 0.G(u) is also called a bistable potential (see Figure 1). In our case, G(u) = 1

2(u2 − 1)2.

ïG

ba

a

b

x

u

Figure 1: Hill-valley-hill structure for −G, and the kink for u.

We consider static solutions of Allen-Cahn equation. They satisfy the static Allen-Cahnequation,

−ε2∆u+ g(u) = 0. (3.3)

For complex u and g(u) = |u|2u−u, (3.3) is the Ginzburg-Landau equation. It appeared first incondensed matter physics. For this equation G is of the form G1(|u|2), where G1 is the “half”of the double well potential.

Equation (3.3) has two trivial solutions u = ±1, which descibe pure phases.We are interested in the phase separation phenomena when the solutions are close to +1 or

−1 (pure phases) in most of the space with sharp transitional interfaces separating +1 regionsfrom −1 regions. We consider the simplest of planar interface.

We look for static solutions which very only in a direction, e, transversal to a plane, x ·e = 0,i.e. of the form

χε(x) = χ

(x · eε

), (3.4)

for any e ∈ Rd, where χ is a function of single variable. By rotational symmetry and scalingproperies of (3.3), the function χ(s) satisfies the equation

χ′′ = g(χ). (3.5)

This is the Newton’s equation with the potential −G(φ), where G′(φ) = g(φ). The potential−G(φ) has equilibria at φ = ±φ0 and at φ = 0. Hence the equation (3.3) has the homogeneoussolutions, φ = ±φ0 and φ = 0.

It is clear from this mechanical analogy that (3.3), besides these homogeneous solutions, hasthe solutions which go asymptotically to ±φ0 as s→ ±∞ (see Figure 1). To show the existenceof these solutions, we observe that by the conservation of mechanical energy, we have

1

2(φ′)2 −G(φ) = 0 (3.6)

(we take the mechanical energy to be 0), which can be solved for φ′ as φ′ = ±√

2G(φ), and thenintegrated to obtain φ. The solution with the + sign is called the kink and, with the − sign, theanti-kink.

In our case, G(φ) = 12(φ2 − 1)2 and the integration can be performed explicitly and gives

χε(x) = tahn

(x · e√

2ε

), (3.7)


for any e, x0 ∈ Rd.We investigate (3.5) also from the point of view of elementary dynamical system theory.

Rewrite (3.5) asχ′ = ψ, ψ′ = g(χ). (3.8)

The linearized vector field at an equilibrium (ϕ, 0), where ϕ solves g(φ) = 0, i.e. ϕ = ±φ0, 0, is(0 1

g′(ϕ) 0

). (3.9)

The eigenvalue equation, λ2 − g′(ϕ) = 0, gives λ = ±√g′(ϕ). At ϕ = 0 we have g′(0) < 0,

which gives two purely imaginary eigenvalues. Moreover, g′(±φ0) > 0, which gives one positiveand one negative eigenvalue. Hence the stationary point (0, 0) is a stable equilibrium (a stablefocus) and (±φ0, 0) are unstable equilibria (saddle points). There is one trajectory going from(−φ0, 0) to (φ0, 0) and one from (φ0, 0) to (−φ0, 0). These are exactly the kinks and anti-kinks(or fronts) mentioned above (see Figure 1).

Example 3.1.

G(ϕ) =

ωo(ϕ− ϕo)2, ϕ ≥ 0,ωw(ϕ− ϕw)2, ϕ ≤ 0,

and ωoϕ2o = ωwϕ

2w. (3.10)

Then equation (3.3) is piecewise linear, and we get

ϕk(z) =

ϕw(1− e−

√ωwz), z ≤ 0,

ϕo(1− e√ωoz), z ≥ 0.

(3.11)

ϕ is a function consisting of a kink and an antikink glued together at a distance R.

Exercise 1. Show (3.11).

One can show existence of kink (and anti-kink) solutions by variational techniques (see e.g.[43, 50]).

Lamellar phase. In this situation, layers of +1 and −1 phases (substances) coexist in aperiodic array. We conjecture that one can construct a periodic solution corresponding to anarray of kinks and antikinks (see below for a discussion of an approach to this problem).

Stationary waves. Stationary or standing waves (or breathers) are solutions of the form

Ψ(x, t) = e−iλtϕ(x), (3.12)

where ϕ and λ are time-independent. The function ϕ is called the profile of the stationary wave(3.12). As an example, consider the nonlinear Schrodinger or Gross - Pitaevskii equation

i∂tΨ = −∆Ψ + κ|Ψ|2Ψ, (3.13)

for Ψ : Rd → C and Ψ|t=0 = Ψ0. It has solutions of the form (3.12), where ϕ and λ ∈ R satisfythe equation

−∆ϕ+ κ|ϕ|2ϕ = λϕ (3.14)

(called the nonlinear eigenvalue problem).


Traveling waves. Traveling waves are solutions of the form

u(x, t) = φ(x− vt), (3.15)

where ϕ(y) and v are independent of t and are called the traveling wave profile and the travelingwave velocity. Consider the reaction-diffusion equation

∂u

∂t= ∆u− g(u). (3.16)

Substituting (3.48) into (3.16), we obtain the following equation for ϕ(y) and v

∆φ− g(φ) + v · ∇φ = 0. (3.17)

Consider the one-dimensional case. Then this equation can be rewritten as

φ′′ = g(φ)− vφ′. (3.18)

This is the Newton’s equation with the potential −G(φ), where G′(φ) = g(φ), and the frictionterm vφ′.

Compute the change in mechanical energy:

∂x(1

2(φ′)2 −G(φ)

)= (φ′′ − g(φ))φ′ = −v(φ′)2 < 0, for v > 0. (3.19)

This implies that for v > 0 the particle descends to a (local) minimum of −G.To be specific we consider the Fisher-Kolmogorov-Petrovsky-Piscunov (FKPP) equation,

which appears in population biology and combustion theory. This is Eqn. (3.16) with g(u) =u(u− 1):

∂u

∂t= ∆u− (u− 1)u. (3.20)

For this equation, G = 13u

3− 12u

2, see Figure 2 (where the label G should be replaced by −G).

g

G

p vn

vm

vm

o oo ph

Figure 2: Functions −G, g and φ.

Using the mechanical analogy described above, we can solve the equation (3.18) for thenonlinearities described. At the remote past (time y = −∞), the particle leaves the top of thehill in −G(φ) at φ = 1 and moves to the left toward the wall loosing the altitude due to thedissipation. Hitting the wall particle turns around and moves toward the opposite wall and soforth until it relaxes to the bottom of the well at φ = 0. The front of the wave is at y at whichφ(y) hits 0 the first time.


We investigate (3.18) from the point of view of elementary dynamical system theory. Rewrite(3.18) as

φ′ = ψ, ψ′ = g(φ)− vψ. (3.21)

The equilibria (or static solutions or stationary points) of this system are (φ0, 0) where φ0 solvesg(φ) = 0. For g(φ) = φ(φ−1) we have the equilibria (0, 0) and (1, 0). To determine whether theseequilibria are stable or not, we compute the linearized vector field (Jacobi derivative of the vector

field on the r.h.s.) at (φ0, 0). It is

(0 1

g′(φ0) −v

). The eigenvalue equation, λ(λ+v)−g′(φ0) = 0,

gives

λ =−1

2v ±

√v2

4+ g′(φ0). (3.22)

At φ0 = 0 we have g′(0) = −1, which gives two negative eigenvalues for v ≥ 2 and twocomplex eigenvalues with negative real parts for v < 2. Hence the equilibrium (0, 0) is a stableequilibrium. (0,0) is a stable focus for v > 2 and stable spiral for v < 2. For φ0 = 1 we findg′(1) = 1 and the point (1, 0) is a saddle point. For v > 2 there is a separatrix. This is a bitsubtle, see Figure 3 below.

−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4

−0.25

−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25

X

Y

Separatrix

Figure 3: v > 2 there exists a separatrix. Plotted is arrows with direction dψ/dφ, at X-coordinate φ and Y -coordinate ψ, note that dψ/dφ = ψ−1φ(φ− 1)− v.

More generally we assume g(u) satisfies the conditions g(0) = g(1) = 0 together with g′(0) <0 and g′(1) > 0, see Figure 2 (where the label G should be replaced by −G).

Another type of the the nonlinearity g(u), which is used in combustion theory is shown inFig. 4.

Exercise 3.1. Consider the reaction-diffusion equation with 1) g(u) = (u − a)(u − 1)u with0 < a < 1/2 (the a = 1/2 case gives the Allen-Cahn equation) 2) g(u) as in Figure 4 (flamepropagation for the ignition nonlinearity).


u1

T T

g −G

1

Figure 4: Ignition case

As another example we consider the celebrated KdV equation originated in the descriptionof waves in shallow channels:

∂tu+ u∂xu+ ∂3xu = 0. (3.23)

The traveling waves u(x, t) = φ(x−vt) for this equation satisfy the equation φ∂yφ+∂3yφ−v∂yφ =

0. Integrate this equation, to obtain 12φ

2 +∂2yφ−vφ = C. To solve this equation, we interpret it

as a Newton’s equation (in time y), we can write out the conservation of energy, or in differentword multiply it by ∂yφ and integrate the result to obtain 1

6φ3 + 1

2(∂yφ)2 − 12vφ

2 − Cφ = C ′.This is the first order ODE, which can be solved to obtain (not finished).

3.2 Symmetries and solutions of PDEs

It turns out that special solutions we considered above come from specific symmetries of theequation we consider. We explore the relation between symmetries of evolution equations andsolutions of such equations in more depth. Transformations which map solutions of an equationinto solutions are called symmetries of this equation. Clearly, all symmetries of a given equationform a group. This group is called the symmetry group of the given equation. It is often arepresentation of an abstract group.

Let F be a map on some space Y . We consider the dynamical system (3.1), ∂u∂t = F (u). If

the map F is invariant under an invertible transformation T , i.e.

F (Tu) = TF (u), (3.24)

then T is a symmetry: if u is a solution, then so is Tu. We distinguish space - time symmetries,i.e. symmetries of the form

u(x, t) 7→ u(g(x, t)), (3.25)

where g is a transformation of the space time Rn+1, and gauge symmetries, i.e. symmetries ofthe form

u(x, t) 7→ gu(x, t), (3.26)

where g is a transformation of the target space of u(x, t). Here are examples of space - timesymmetries:

Translation symmetry: Assume the additive group Rn acts on the space of u’s as: for anyh ∈ Rn,

Th : u(x, t) 7→ u(x+ h, t); (3.27)

Rotation and reflection symmetry: for any R ∈ O(n) (including the reflectionsf(x)→ f(−x))

TR : u(x, t) 7→ u(Rx, t). (3.28)


Translations and rotations are called rigid motions. They are a part of Galilean group:

(R, h, v, τ)(R′, h′, v′, τ ′) = (R′R,R′h+ h′, R′v + v′, τ + τ ′), (3.29)

with the representation

T(R,h,v,τ) : u(x, t) 7→ u(Rx+ h+ vt, t+ τ). (3.30)

Examples of gauge symmetries will be special cases of the following general situation. AssumeV is a vector space and G, a group acting on V . Then for ψ : Rn × R+ → V , we define

ψ(x, t) 7→ gψ(x, t), g ∈ G. (3.31)

The group G is called the gauge group and (3.31), the gauge transformation with the corre-

sponding gauge group G. Specifically, consider V = Cm, or = Rm. Then G = U(m) or O(m),the unitary or orthogonal (rotation) group. In particular, for the gauge group U(1), we have

ψ(x, t) 7→ eiαψ(x, t), α ∈ R. (3.32)

Example. Consider the nonlinear Schrodinger or Gross - Pitaevskii equation (3.13). Show thatit has translational, rotational and gauge symmetries. What about NLH and NLW?

Consider first solutions of the dynamical system (3.1), invariant under some subgroup of itssymmetry group:

Static solutions. Solutions invariant under time translations are called static solutions. Theyare independent of time and form the most important class of special solutions. They are alsocalled equilibria and sometimes stationary solutions. Thus u∗ is an equilibrium or static solutioniff u∗ is independent of time and satisfy the equation F (u∗) = 0.

Examples: NLH, NLS, NLW.

Homogeneous solutions. Solutions invariant under spatial translations are independent of xand are called the homogeneous solutions.

Spherically symmetric solutions. Solutions invariant under spatial rotations are calledspherically symmetric solutions. They are independent of angular variables and depend only ontwo variables, |x| and t.

Covariance. We generalize the above definition a bit and say that a map F is covariant underan invertible transformation T , iff there is a real number a s.t.

F (Tϕ) = aTF (ϕ). (3.33)

In this case we can upgrade T to a symmetry by defining: if u is a solution, then so is

T ut := Tuat. (3.34)

Space-time symmetries. An example of a space-time symmetry is the Galilean symmetryof the nonlinear Schrodinger equation (NLS):

ψ(x, t) 7→ ei(12v·x− 1

4|v|2t)ψ(x− vt, t). (3.35)


Equivariant solutions. We begin with auxiliary definitions. We say two solutions are equiv-alent if one can be obtained from the other by a gauge transformation. The group of rigidmotions is defined as a semi-direct product of the groups of translations and rotations. Wedenote by Tg, g ∈ Grm, the action of the group, Grm, of rigid motions on space of solutions. Forthe groups of translations, and rotations, is given (3.27) and (3.28), respectively.

We say a solution u is equivariant (under a subgroup, G, of the group of rigid motions) iffthe action of G on u takes it into the (gauge-) equivalent function, i.e., for any g ∈ G, there isγ = γ(g) s.t.

Tgu = Γγu,

where Γγ is the action of for the gauge group, given in (3.32). Here are two important examplescoming from the Gross - Pitaevski equation, mentioned above in (3.36), and the Ginzburg-Landau equations studied in detail later.

Vortices. We consider the static Gross - Pitaevskii (or Ginzburg-Landau) equation, appearingin the condensed matter physics (the theory of superfluidity and Bose - Einstein condensation).It is given by

∆Ψ = κ2(|Ψ|2 − 1)Ψ, (3.36)

for Ψ : Rd → C. In the dimension d = 2 and for G, the group of rotations, O(2), these equationshave O(2)− equivariant solutions, which are labeled by integers n, which are given by

Ψ(n)(x) = f (n)(r)einθ, (3.37)

where (r, θ) are the polar coordinates of x ∈ R2. These solutions are called vortices. (n labelsthe equivalence classes of the homomorphisms of S1 into U(1).)

Abrikosov lattices. If G is the subgroup of the group of lattice translations for a lattice L, thenwe call the corresponding equivariant solution a lattice, or L-gauge-periodic, solution. Explicitly,

TsΨ(x) = Γgs(x)Ψ(x), ∀s ∈ L, (3.38)

where gs : R2 → R is, in general, a multi-valued differentiable function, with differences of valuesat the same point ∈ 2πZ, and satisfying

gs+t(x)− gs(x+ t)− gt(x) ∈ 2πZ. (3.39)

The latter condition on gs can be derived by computing Ψ(x+ s+ t) in two different ways.Note that for Abrikosov lattices, the physical quantity, |Ψ|2, (the probability density (of

superfluid or Bose-Einstein condensed atoms) is doubly-periodic with respect to some lattice L.One can also show the converse: a state Ψ ∈ H1

loc(R2;C) for which the physical quantity,|Ψ|2, is doubly-periodic with respect to some lattice L (we call such a state the (generalized)Abrikosov (vortex) lattice) is a L-gauge-periodic state.

Solitons. Symmetries allow us to introduce special classes of solutions: stationary, rotatingand traveling waves, which sometimes grouped as solitons.

Consider the evolution equation (3.1). Let Tλ be a one-parameter generalized symmetrygroup of the equation (3.1) in the sense of (3.33). This means that for each λ, Tλ is the symmetryin the sense of

F (Tλϕ) = b(λ)TλF (ϕ), (3.40)

and the family Tλ is a one-parameter group, as defined above, i.e.


• T0 = 1; Tt Ts = Tt+s.

The latter means that b(λ) satisfies b(s+ t) = b(s)b(t). We assume that Tλ is strongly differen-tiable at λ = 0 and define the generator, A, of the group Tλ as Aϕ = ∂λTλϕ|λ=0. Then we have∂tTtu0 = ATtu0, t ≥ 0. We consider solutions of the form

u = Tλϕ, (3.41)

where λ = λ(t), while ϕ is t−independent. We call such solutions, if they exist, solitary waves.Substituting this into (3.1) and using (3.40), we obtain the equation for ϕ and h:

∂λ

∂tAϕ = b(λ)F (ϕ), (3.42)

where A is the generator of the group Tλ. Since the r.h.s. of (3.42) is independent of t, wesee that ∂λ

∂t b(λ)−1 must be independent of t as well, say = −v, i.e. λ must solve the equation∫ λ0 b(s)ds = −vt−λ0, for some constant λ0. In the simplest case of b(λ) = 1, it should be of the

formλ = −λ0 − vt, (3.43)

for some constants λ0 and v, so that (3.42) becomes

F (ϕ) + vAϕ = 0. (3.44)

This is a time-independent equation for ϕ and v. The function ϕ is called the solitary waveprofile. We summarize this as

Theorem 3.1. Let Tλ be a one-parameter symmetry group of the equation (3.1) with a gen-erator A. Then (3.1) has a solitary wave solution of the form u = Tλϕ, where λ depends on t,while ϕ is a t−independent function iff λ = −λ0 − vt, for some constants λ0 and v, and ϕ andv satisfy (3.44).

Moving reference frame. We transform the equation to the moving frame by setting u =Tλw. Then w satisfies the equation

∂w

∂t= F (w) + vAw. (3.45)

Now, if u = Tλϕ is a solitary wave for (3.1), then the solitary wave profile ϕ is a stationarysolution to the new equation (3.45). Thus one can apply general definitions and results for thestatic solutions to solitary waves as well.

Assume that the space, Y , of solutions is a space of functions on Rn and let the additivegroup Rm acts on the space of u’s: h → Thu, ∀h ∈ Rn and we assume that the map F iscovariant w.r.to this group, i.e. (3.34) holds. We consider several realizations of the action Thof Rm.

Stationary waves. The notion of stationary waves is related to the U(1)−gauge symmetry.Let Y be a space of complex functions. Let Y be a complex space and consider one parametergroup of transformations Tαf := eiαf, α ∈ R, on Y . The generator of this transformation isAf = if , the solitary waves in this case are of the form

u(x, t) = ei(α0+λt)ϕ(x), (3.46)


where ϕ(y) and λ satisfy the equation

F (ϕ) + iλϕ = 0, (3.47)

the nonlinear eigenvalue problem. Such solutions are called the stationary or standing waves orbreathers.

Traveling waves. The traveling waves arise from the translational symmetry. Let Th be thegroup of translations,

(Thf)(x) := f(x+ h).

Exercise. Show that the generator Aj for translations along the coordinate xj is Aj = ∂xj .For the translation invariance, the equations (3.41), (3.43) and (3.44) become

u(x, t) = ϕ(x− vt), (3.48)

where ϕ(y) and v satisfy the equation

F (ϕ) + v · ∇ϕ = 0. (3.49)

They are called the traveling waves. The function ϕ(y) is called the traveling wave profile.

Exercise. Show that the map F (u) := ∆u+ |u|p−1u is covariant under the shifts (we say thatF is translation invariant).

Rotating waves. The above construction can be generalized to other Lie groups. For instance,consider an action of the group O(m) of rotations, or the group U(m). If Y be a space of complexfunctions on Rn with values in Rm or Cm, then we can define an action of O(n) on Y by

(TRf)(x) := f(R−1x), R ∈ O(n),

or an action of O(m) on Y by

(TRf)(x) := Rf(x), R ∈ O(m).

Similarly for an action of U(m). Note that the generator of TRf in the first case is −i times theangular momentum of Quantum Mechanics. One can also combine various transformations.

Shrinkers and expanders. The group R can be also realized on functions of n variables as ascaling transformation:

(Tθf)(x) := eαθf(eθx)

for some α.Exercise. Show that the generator A is A = x · ∇+ α.Exercise. Show that the map F (u) := ∆u+|u|p−1u is covariant under the scaling transformationabove with α = 2

p−1 .

Taking λ = eθ and we write the solitary waves in this case in the form

u(x, t) = λ(t)αϕ(λ(t)x), (3.50)

where v := ∂λ∂t satisfy the equation

F (ϕ) + v · (y · ∇+ α)ϕ = 0. (3.51)


They are called variously the self-similar solution or shrinker or expanders, depending onwhether λ(t) grows or decreases. The function ϕ(y) is called the self-similar wave profile.

Passing to the variables y = λ(t)x is the same as using the frame of reference expanding orshrinking together with the traveling wave. In this variable, the traveling wave is a stationarysolution to the new equation

∂w

∂t= F (w) + v · (y · ∇+ α)w. (3.52)

As an example we consider the mean curvature flow (MCF) equation of a surface S. If Sis given by a graph, S = graph f , of function f : U → R, where U is an open set in Rn+1 i.e.S := x = (x

′, f(x

′)) x

′ ∈ U, then the MCF equation reduces to (see Subsection 12.1)

∂f

∂t= −

√|∇f |2 + 1H(f), (3.53)

where H(f) := −div

(∇f√|∇f |2+1

), the mean curvature of S.

‘Spherically symmetric’ (equivariant) solutions:

a) Sphere: upper hemisphere can be given as the graph of the function f =√R2 − |x′ |2.

Using the formula H(f) := −div

(∇f√|∇f |2+1

), it is easy to compute H(f) = n

R and

therefore we get R = − nR which implies R =

√R2

0 − 2nt. So this solution shrinks to apoint.

b) Cylinder: it is a sphere in in the variables (x1, . . . , xn−1) H(x) = n−1R and R = −n−1

R

which implies R =√R2

0 − 2(n− 1)t.

The MCF equation ((3.53)) is invariant under the rescaling transformation

Tλ : f(u, t)→ λf(λ−1u, λ−2t), (3.54)

which corresponds to surfaces re-scaled as S(t) ≡ Sλ(t) := λ(t)S (standing waves). This leadsto the soliton

f(u, t) = λφ(λ−1u), λ depends on t. (3.55)

Substituting this into (3.53) and setting y = uλ we find√

1 + |∇yφ|2H(φ) = a(y∂y − 1)φ, (3.56)

where a = −λλ. Since the l.h.s. is independent of time, we conclude that

a is time-independent. (3.57)

In the non-trivial case, a is a non-zero constant, which implies λ =√λ0 − 2at. So

i) a > 0⇒ λ→ 0 as t→ T :=λ202a ⇒ (3.55) (Sλ) is a shrinker.

ii) a < 0⇒ λ→∞ as t→∞⇒ (3.55) (Sλ) is an expander.

Solution to (3.56): Sphere φ(y) =√R2 − |y|2, with R =

√na .


Combined symmetries. We consider solitons generated by two groups, one a space group,say the group of translations, Th, and another, a gauge group, say the U(1)−gauge group,Γα = eiα. (In other words, we allow the soliton to move inside the gauge-equivalent class ofsolutions.) Hence we look solutions of the form

u = ΓαThϕ, (3.58)

where h = h(t) and α = α(t), while ϕ is t−independent. To be specific, we consider the nonlinearSchrodinger or Gross - Pitaevskii equation (3.13) and look for solution of the form (3.58). Asimple computation gives α(t) = 1

2v · x −14 |v|

2t and h(t) = x − vt − x0, which shows that hasthe solutions of the form

u(x, t) := ei(12v·x− 1

4|v|2t−λt)φ(x− vt− x0) (3.59)

where e−iλtφ(x) is a stationary solution to (3.13). This solution is obviously a traveling wave.Put differently, the solution (3.59) comes from applying the Galilean symmetry transforma-

tion (3.35) to the stationary solution e−iλtφ(x).

Manifolds of static solutions and traveling waves If the dynamical system (3.1) withthe symmetry group G has a static solution u∗, then it has a manifold of static solutions

M = Tgu∗ : g ∈ G.

Indeed, if u is a static solution to (3.1), then, due to (3.34), T = Tg, so is Tgu∗.For example, since the Allen-Cahn equation (3.2) is translational invariant and has the kink

and anti-kink solutions, (3.4) and its negative, the functions χ±ε ( (x−x0)·e√2

) ≡ ±χ( (x−x0)·e√2ε

), are

also solutions for any x0 ∈ R. These are kink and antikink solutions centered at a. Thus theequation (3.2) has a family (manifold) of stationary solutions, χ±ε ( (x−x0)·e√

2) ≡ ±χ( (x−x0)·e√

2ε), x0 ∈

Rn, e ∈ Sn−1.

Exercise 3.2. Check the translation invariance of (3.2).

The same of course is true for traveling wave, as they can be reduced to static solution whenthe equation is considered in the moving reference frame.

4 Gateaux and Frechet derivatives

Our goal is is to develop a differential calculus of maps between Banach spaces. Let X and Ybe Banach spaces and M , an open subset of X. We consider a map F : M → Y . The mapF is called Gateaux differentiable at u ∈ M if and only if there exists a bounded, linear mapdF (u) : X → Y , s.t. for any ξ ∈ X:

dF (u)ξ :=∂

∂λ

∣∣∣∣λ=0

F (u+ λξ). (4.1)

The map dF (u) is called the Gateaux derivative or sometimes, the variational derivative.The map F is called continuously differentiable at u ∈ X if and only if it is Gateaux differ-

entiable for u in a neighborhood of u, and moreover, u 7→ dF (u) is a continuous map from Xto L(X,Y ) at the point u; i.e. if un → u in X, then dF (un) → dF (u) in L(X,Y ). The mapF is called continuously differentiable, or C1 (written F ∈ C1) if and only if it is continuouslydifferentiable for every u ∈M .


Example 4.1 (formal). 1) If F (u) = Lu, where L is a linear map, then dF (u) = L (inde-pendently of u). Indeed, dF (u)ξ = ∂

∂λL(u+ λξ)|λ=0 = ∂∂λ(Lu+ λLξ)|λ=0 = Lξ. Thus if L

is bounded, then F is C1.

2) If F (u) = f u (composition map), for a fixed C1–function f : R → R, and u : Rn → R,then dF (u) is the multiplication operator by f ′(u). Indeed, dF (u)ξ = ∂

∂λF (u + λξ)|λ=0 =∂∂λf(u(x) + λξ(x))|λ=0 = f ′(u)ξ. So if f ′(u) is a bounded function, say for some u ∈Lp(Rn), then F : Lp(Rn)→ Lp(Rn) is differentiable at u.

Exercise 4.1. Compute dF (u) for F : Rn → Rm, and for

F (u) = div

(∇u√

1 + |∇u|2

).

Exercise 2. Let F (u) = f u, and let Ω be a domain in Rn with a smooth boundary. Show thatif f ∈ Ck+1(R), then F : Ck(Ω)→ Ck(Ω), and F is C1 with dF (u)ξ = f ′(u)ξ.

As usual, the symbol o(‖ξ‖) stands for a map R : X → Y s.t.

‖R‖‖ξ‖

→ 0, as ‖ξ‖ → 0.

The following properties of the Gateaux derivative play an important role in applications.

Theorem 4.1. (a) (The fundamental theorem of calculus) If F is Gateaux differentiable atu ∈M , then we have

F (u+ ξ)− F (u) =

∫ 1

0dsdF (u+ sξ)ξ.

• (b) (The chain rule) If F and G are Gateaux differentiable at u ∈ M and F (u) ∈ F (M),then we have d(G F )(u) = dG(F (u))dF (u).

(c) Let K be a convex subset of a Banach space X (i.e. if u, v ∈ K, then su+ (1− s)v ∈ K,for all s ∈ [0, 1]). Show that if F : K → K satisfies ‖dF (ψ)‖ ≤ α, ∀ψ ∈ K, then F isLipschitz: ‖F (ψ)− F (ϕ)‖ ≤ α‖ψ − ϕ‖, ∀ψ,ϕ ∈ K.

(d) If F is C1 at u ∈ X, then as ‖ξ‖ → 0

F (u+ ξ) = F (u) + dF (u)ξ + o(‖ξ‖). (4.2)

Proof. Define the function g : [0, 1]→ Y by

g(t) = F (u+ tξ),

for u, ξ ∈ X fixed. According to the definition of the Gateaux derivative (4.1), we have

g′(t) = dF (u+ tξ)ξ.

By Fundamental Theorem of Calculus,

g(1)− g(0) =

∫ 1

0g′(t)dt. (4.3)

Since g′(t) = dF (u+ tξ)ξ, this gives the first statement.


Exercise 4.2. Show the properties (c) and (d).

Using (4.3), we find g(1)− g(0) = g′(0) +∫ 1

0 (g′(t)− g′(0))dt, which in turn gives

‖F (u+ ξ)− F (u)− dF (u)ξ‖= ‖g(1)− g(0)− g′(0)‖≤ sup

0<t<1‖g′(t)− g′(0)‖

≤ sup0<t<1

‖dF (u+ tξ)− dF (u)‖ ‖ξ‖

= o(‖ξ‖).

In the last step, we used the continuity of dF (u).

Discussion. The Frechet derivative. Though the Gateaux derivative is straightforwardto compute, for theoretical considerations, one needs often a stronger notion of derivative: theFrechet derivative. Before we define the Frechet derivative, let us remark that equation (4.1) isequivalent to

F (u+ λξ)− F (u) = λdF (u)ξ + o(λ), (4.4)

were o(λ) is a vector in Y satisfying limλ→0 ‖o(λ)‖/λ = 0. Notice that in general, o(λ) dependson ξ.

The map F is called Frechet differentiable at u ∈ X if and only if there exists a boundedlinear map dF (u) ∈ L(X,Y ) s.t. (7.7) holds as ‖ξ‖ → 0. The operator dF (u) satisfying (7.7) iscalled the Frechet derivative of F at the point u.

From this definition and equation (4.4), it is clear that if F is Frechet differentiable at u,with Frechet derivative dF (u), then F is Gateaux differentiable at u with Gateaux derivativegiven by the same operator dF (u). In the opposite direction, if F ∈ C1, then the precedingtheorem implies

Theorem 5. If F is continuously Gateaux differentiable at u ∈ X, with Gateaux derivativedF (u), then F is Frechet differentiable at u, and the Frechet derivative is given by the sameoperator dF (u).

For a detailed discussion of Frechet and Gateaux derivatives, we refer to [54].

In everything that follows, by the derivative dF (u) we understand the Gateaux derivative.We point out that in most of our applications, we deal with C1 maps, in which case the Frechetand Gateaux derivatives coincide, according to the last theorem.

We conclude this section with some useful rigorous results about Gateaux derivatives ofcomposition operators F (u) = f u, where f is a fixed function and u belongs to the space ofdifferentiable functions. Such operators appear often in applications. The statements below areuseful in this context. An important result in this direction is the following.

Proposition 4.2. Let Ω ⊂ Rn. Let F (u) = f u with f ∈ C1(Ω) and obeying the estimates

|f (k)(u)| ≤ c|u|p+1−k for k = 0, 1, 2 (4.5)

for some p ≥ 1. Then F : Hr(Ω) → L2(Ω) and is C1, provided the indices p and r satisfy therelation p < 2r

(n−2r)+and r > 0.


Proof. Let u ∈ Hr(Ω). Then by the Sobelev embedding theorem (see Section A.6) u ∈ L2(p+1)(Ω)since n

p >n2 − r. Hence

|f(u)| ≤ c|u|p+1 ∈ L2.

This shows that F : Hr(Ω)→ L2(Ω). To show that F ∈ C1, we compute for h > 0

1

h(f(u+ hξ)− f(u)) = f ′(u)ξ +Rh(ξ)

where Rh(ξ) = 1h

∫ h0 (f ′(u+ tξ)− f ′(u))ξ dt. Now we estimate by the mean value theorem

||Rh(ξ)||2 ≤1

h

∫ h

0‖f ′′(u+ tξ)ξ2‖2t dt

for some 0 ≤ t ≤ t. Using the estimate |f ′′(u)| ≤ c|u|p−1 and the triangle and Holder inequalitieswe derive furthermore

‖Rh(ξ)‖2 ≤ c

h

∫ h

0‖|u+ tξ|p−1ξ2‖2t dt

≤ hc(‖|u|p−1ξ2‖2 + ‖ξp+1‖2

)≤ ch

(‖u‖p−1

2(p+1)‖ξ‖22(p+1) + ‖ξ‖p+1

2(p+1)

).

Hence ‖Rh(ξ)‖2 → 0 and therefore ‖ 1h(f(u + hξ) − f(u)) − f ′(u)ξ‖2 → 0 as h → 0. Thus F is

C1.

Corollary 4.3. Assume f : C→ C satisfies estimates (4.5) with 1 ≤ p < 2(n−2r)+

, r > 0. Define

F (u) = −∆ + f(u). Then F : Hr(Ω)→ L2(Ω) and is C1.

Furthermore, we have

Theorem 4.4. Let F (u) = f u, Ω ⊂ Rn be a bounded domain with smooth boundary, and letk > n/2. If f ∈ Ck+1(R), then F : Hk(Ω)→ Hk(Ω), and F is C1.

Exercise 3. Prove this theorem for n2 < k ≤ 2 and l = 0, k.

For a complete proof of the Theorem, see [43], page 221.

5 The implicit and inverse function theorems

5.1 Theorems

The implicit and inverse function theorems are two key theorems in analysis. One easily followsfrom the other. We begin with the simpler one - the implicit function theorem.


Inverse function theorem. Let X and Y be two Banach spaces and consider a map F :Y → X. We wish to solve the equation

F (y) = x

for y. Recall that a map G : X → Y is called the inverse of the map F : Y → X if and only ifG F = 1lX and F G = 1lY . Here, 1lZ denotes the identity on the space Z. We write G = F−1.Recall that a linear map L is called invertible if and only if it has a bounded inverse.

If the map F is linear, F (y) = Ly, where L is a linear operator on Y , then the problem wewant to solve reads Ly = x, which is reduced to inverting the operator L.

The following theorem is a generalization of the corresponding theorem in multivariablecalculus:

Theorem 5.1 (The inverse function theorem). Let U be an open neighborhood of 0 ∈ X, andlet F : U → Y be a C1 map s.t. dF (0) : Y → X has a bounded inverse (i.e. dF (0) : Y → Xis bijective). Then there is a neighborhood W of F (0) in Y and a unique map G : V → X s.t.F (G(y)) = y, for all y ∈W .

G

F

. .0 F(0)V

Y

U

X

Figure 5: Maps F and G

Proof. First we explain the idea of the proof. We want to solve F (y) = x for y near y = 0.Expand F in y around 0: F (y) = F (0)+dyF (0)y+R(y), with R(y) = o(‖y‖). So we can rewritethis equation we want to solve as the following equation for y:

x0 + dyF (0)y +R(y) = x,

where x0 = F (0). By the assumptions, the operator L := dyF (0) is invertible. Hence theequation above can be rewritten as the fixed point equation Hx(y) = y, where Hx(y) is the mapgiven by

Hx(y) := L−1 (x− x0 −R(y)) . (5.1)

If we neglect the remainder term R(y), then equation (5.7) yields for each given x the corre-sponding y = G(x) = L−1(x − x0). In the general case, the remainder is not zero, but o(‖y‖)for y small, and we can use the fixed point argument to show existence of a solution to (5.7).To do so, denote by BZ(z, r) the open ball in Z of radius r centered at z ∈ Z.

Claim 5.1. ∃ε > 0 and δ > 0 such that ∀x ∈ BX(x0, δ)

(i) Hx : BY (0, ε)→ BY (0, ε),

(ii) ‖dHx(y)‖ ≤ 1/2 ∀y ∈ BY (0, ε).


Given (i) and (ii), we see that for any x ∈ BX(0, δ), Hx is a contraction map and thereforeit has a unique fixed point in BY (0, ε). Call this fixed point y = G(x). It solves the equationy = Hx(y) which is equivalent to the equation F (y) = x. Thus, the theorem follows withW = BX(0, δ).

It remains to prove the claim. (i) Let M = ‖L−1‖ and pick δ = ε/2M . Next, by the propertyof R(y) and the continuity of dyF (y), we can pick ε > 0 so that

(a) ‖R(y)‖ = o(‖y‖) ≤ ε/2M ;

(b) ‖dyF (y)− dyF (0)‖ ≤ 1/2M .

Now, inequality (a) and the definition of Hx imply

‖Hx(y)‖ ≤M (ε/2M + ε/2M) = ε

∀x ∈ BX(0, δ) and y ∈ BY (0, ε). So (i) follows.Next, the definition, R(y) := F (y)− F (0)− dyF (0)y, of the remainder R(y) gives dyR(y) =

dyF (y)− dyF (0). Using this and definition (5.6) of the map Hx, we compute

dyHx(y) = −L−1dyR(y)

= −L−1 [dyF (y)− dyF (0)] .

Applying inequaly (b), we conclude that ‖dyHx(y)‖ ≤ 1/2 for any y ∈ BY (0, ε). This proves (ii)and therefore completes the proof of the claim and with it, the theorem.

Implicit function theorem. Now we derive the implicit function theorem. Consider threeBanach spaces X,Y and Z, and a map F : X × Y → Z. We wish to solve the equation

F (x, y) = 0

for y, i.e. we want to find a function y = g(x) s.t. F (x, g(x)) = 0. For a map F (x, y) dependingon two arguments, x and y, we introduce the notion of partial Gateaux derivative, say dyF (x, y),by fixing y and taking the Gateaux derivative in y.

Theorem 5.2 (The implicit function theorem). Let U and V be neighborhoods of 0 ∈ X and0 ∈ Y respectively. Assume

(1) F : U × V → Z is C1 in y and C in x;

(2) F (0, 0) = 0;

(3) dyF (0, 0) has a bounded inverse.

Then there is a neighborhood W of 0 ∈ X and a map g : W → Y such that F (x, g(x)) = 0,∀x ∈W , and g(0) = 0.

We derive the implicit function theorem from the inverse function theorem under strongercondition that F is C1 in y and x. In Appendix I we give the proof under the original conditions.

Proof of Theorem 5.2 under the assumption that F is C1 in y and x. Let U := U×V and X :=X × Z. Define the map F (x, y) := (x, F (x, y)) : U → X. Then the implicit function equationF (x, y) = 0 is equivalent to inverse function one, F (y) = x, where y = (x, y) and x = (x, 0).By the assumption that F is C1, we have that F is C1 and, as can be easily verified, the


assumpions of the inverse function theorem are satisfied. The latter theorem gives the mapG : W → Y , where W ⊂ X a neighbourhood of 0 and Y := X × Y , satisfying F (G(x)) = x, orG(x, 0) = (x, y). We write G(x, 0) =: (h(x), g(x)). Then h(x) = x, g(x) = y. Collecting theserelations and using the definition of F give (x, F (x, g(x))) = (x, 0), or F (x, g(x)) = 0.

Remark 1. Of course, we can replace the point (0, 0) in Theorem 5.2 by an arbitrary point(x0, y0). We can think of Theorem 5.2 as a perturbation result in around the point (x0, y0). y0

is called the unperturbed solution and y = g(x), its pertrbation.

Remark 2. The derivation of the inverse function theorem from the implicit function theoremis given in Appendix II.

5.2 Applications: existence of breathers and constant mean curvature sur-faces

Existence of breathers. Consider the discrete nonlinear Schrodinger equation (DNLS)

i∂ψ

∂t= −ε∆ψ − |ψ|2ψ

in l2(Zd). Here Zd = α = (α1, . . . , αd) |αj integers, l2(Zd) = ψ : Zd → C |∑

x∈Zd |ψ(α)|2 <∞ and ∆ is the discrete Laplacian

(∆f)(α) =∑

|α−β|=1

f(β)− 2df(α). (5.2)

Remark. The discrete Laplacian can be also defined as follows. Let E := ej = (0, ..., 1, ..., 0)|j =1...d be the basis of the elementary cell at the origin in Zd. Define ∆ := −div ·∇ where

(∇f)(α) =∑e∈E

(f(α+ e)− f(α))e

and div (the divergence) is the adjoint negative of operator, div = −∇∗, i.e., 〈∇f,~g〉 = −〈f, div ~g〉.Hence

(div ~g)(α) =∑e∈E

(ge(α)− ge(α− e))

Example 5.1. For d = 1, (∇f)(α) = f(α + 1) − f(α), (div g)(α) = g(α) − g(α − 1) whichimplies ∆f(α) = f(α+ 1) + f(α− 1)− 2f(α).

Exercise 4. 1) Derive (5.2) from the formula ∆ := −div · ∇ and, separately, from the resultof Example 5.1 (applying the latter in each coordinate). 2) Show ∆ is a bounded operator onl2(Zd).

Breathers are time-periodic solutions to DLNS of the form

ψ(α, t) = eiλtφ(α)

This implies that φ(α) satisfies

−ε∆φ− |φ|2φ+ λφ = 0 (5.3)

Theorem 5.3. If ε is sufficiently small and λ > 0, then (5.3) has a solution in l2(Zd).


Proof. We look for real solutions φ to (5.3). Define F (ε, φ) := −ε∆φ + λφ − |φ|2φ. Note that(5.3) ⇐⇒ F (ε, φ) = 0. Therefore, we have to solve F (ε, φ) = 0 for φ, provided ε sufficientlysmall. We want to use IFT. Note that 1) F (ε, φ) : R × l2(Zd) → l2(Zd) is continuous in ε andφ (since l2 ⊂ lp ∀p ≥ 2); 2) F (0, φ0) = 0, where φ0 =

√λ δα0 for any α0; 3) F (ε, φ) is C1 in φ:

∂φF (ε, φ) = −ε∆ + λ + 3φ2 is a bounded operator on l2(Zd), and is continuous in ε and φ; 4)∂φF (0, φ0) = λ(1− 3δα0) is invertible if λ > 0. By 1)− 4), IFT is applicable and hence (5.3) hasa unique solution for ε sufficiently small.

Remark. One can extend the above theorem to large ε: let F (ε, φε) = 0 for ε small; ∂φF (ε, φε) =−ε∆ + λ− 3φ2

ε is invertible if

λ /∈ Ran 3φε and ε <minα |λ− 3φ2

ε |‖∆‖

. (5.4)

Exercise 5. Show that if (5.4) holds, then dφF (ε, φε) = −ε∆ + λ − 3φ2ε is invertible. Hint:

Write

−ε∆ + λ− 3φ2ε =

(λ− 3φ2

ε

) (1l− ε(λ− 3φ2

ε )−1∆

).

If λ /∈ Ran 3φε, then the first factor on the r.h.s (the multiplication operator λ−3φ2ε) is invertible.

Show that the second factor is invertible as well.

Existence of surfaces with prescribed mean curvature. Now, we consider some appli-cations the inverse function theorem. Assume S is a hypersurface in Rn+1, given as a graph ofa function ψ : Ω ⊂ Rn → R, S = graph ψ (see Fig. 6). We assume Ω is a bounded domain withsmooth boundary.

The mean curvature of S at a point x′ = (x, ψ(x)) ∈ S is given by H(x) = −div

(∇ψ(x′)√

1+|∇ψ(x)|2

)

,,

,,

xn+1

S = graph

1

Figure 6: Graph of ψ

for x ∈ Ω.

Given a function h(x′) on Ω, is there a surface S = graph ψ which has mean curvatureh(x) at points x′ = (x, ψ(x))? This problem can be reformulated as finding a solution ψ to the


equation H = h, i.e.

−div

(∇ψ(x)√

1 + |∇ψ(x)|2

)= h(x), (5.5)

on Ω. We sketch a proof that

• for any h ∈ Hk−2(Ω) sufficiently small, there is a unique ψ ∈ Hk(Ω) solving this equation.

First, we reformulate the problem (5.5) as F (ψ) = h, where

F (ψ) = div

(∇ψ√

1 + |∇ψ|2

).

To solve the latter equation, we use the inverse function theorem, so we need to show that forsome spaces X and Y we have

1) F : U → Y , where U is a neighborhood of 0 ∈ X,

2) F is C1,

3) dF (0) has a bounded inverse.

In a first step, let X, Y be the Sobolev spaces X = Hk(Ω), Y = Hk−2(Ω). In order to show 1),we define for p ∈ Rn the smooth function:

G(p) =p√

1 + |p|2,

then F (ψ)(x′) = div G(∇ψ(x′)). We know that ∇ : Hk(Ω)→ Hk−1(Ω).In Theorem 4.4 of Section 4, we saw that if k − 1 > n/2, then, since G is smooth, the

composition with G leaves Hk−1(Ω) invariant: G : Hk−1(Ω)→ Hk−1(Ω).Finally, div : Hk−1(Ω) → Hk−2(Ω), and therefore the composition of these three maps

satisfies:F = div G ∇ : Hk(Ω)→ Hk−2(Ω),

which shows 1).In order to check 2), i.e. F ∈ C1, we remember from Exercise 4.1 #1 that

dF (ψ)ξ = div

(∇ξ

(1 + |∇ψ|2)1/2− (∇ψ · ∇ξ)∇ψ

(1 + |∇ψ|2)3/2

).

Exercise 6. Show dF (ψ) : Hk(Ω)→ Hk−2(Ω) is bounded and continuous in ψ (i.e. ‖dF (ψ′)−dF (ψ)‖ → 0, as ‖ψ′ − ψ‖ → 0).

Finally, we have to verify that 3) is satisfied, i.e. that dF (0) has a bounded inverse. NowdF (0) = ∆, and we have discussed the existence of ∆−1 in Appendix B, where it is shown thatthe inverse Laplacian exists on the orthogonal complement of a constant (0−eigenfunction).Hence we have to take X = Hk

0 (Ω).Again, we have not shown yet boundedness of ∆−1 : Hk−2(Ω) → Hk

0 (Ω). This can be doneusing variational calculus. (We might do this in a later section.) Modulo the proof of thisfact, we have thus shown that the conditions of the inverse function theorem are satisfied, andtherefore, for any sufficiently small h ∈ Hk−2(Ω), the equation F (ψ) = h has a unique solutionψ ∈ Hk

0 (Ω). In other words, there exists a surface S = graph ψ with prescribed small meancurvature h.

If h is constant the corresponding surface is called the constant mean curvature surface. Forh = 0, this is a minimal surface.


5.3 Appendix I: Direct proof of Theorem 5.2.

First we explain the idea of the proof. We want to solve F (x, y) = 0 for y near (x, y) = (0, 0).Expand F in y around 0 (see Theorem 4.1): F (x, y) = F (x, 0) + dyF (x, 0)y + R(x, y), withR(x, y) = o(||y||). So we can rewrite this equation we want to solve as the following equationfor y:

F (x, 0) + dyF (x, 0)y +R(x, y) = 0.

We will show in a moment that the operator Lx = dyF (x, 0) is invertible for sufficiently small‖x‖. Hence the equation above can be rewritten as the fixed point equation Hx(y) = y, whereHx(y) is the map given by

Hx(y) := −L−1x (F (x, 0) +R(x, y)) , (5.6)

y = −L−1x (F (x, 0) +R(x, y)) . (5.7)

If we neglect the remainder term R(x, y), then equation (5.7) yields for each given x the corre-sponding y = G(x) = −L−1

x F (x, 0). In the general case, the remainder is not zero, but o(‖y‖)for y small, and we can use the fixed point argument to show existence of a solution to (5.7).To do so,

We now show that the map Hx is well defined and has a fixed point y = y(x). First weshow that there is a δ1 such that Lx is invertible for ‖x‖ ≤ δ1. Write Lx = L0 + Vx, whereVx := Lx − L0. Since Lx is continuous in x by the conditions of the theorem we can choose δ1

such that ‖Vx‖ ≤ (2‖L−10 ‖)−1 if ‖x‖ ≤ δ1. Hence Lx is invertible if ‖x‖ ≤ δ1 by Theorem 23 in

Section C.Denote by BZ(z, r) the open ball in Z of radius r centered at z ∈ Z.

Claim 5.2. ∃ε > 0 and δ > 0 such that ∀x ∈ BX(0, δ)

(i) Hx : BY (0, ε)→ BY (0, ε),

(ii) ‖dHx(y)‖ ≤ 1/2 ∀y ∈ BY (0, ε).

Given (i) and (ii), we see that for any x ∈ BX(0, δ), Hx is a contraction map and thereforeit has a unique fixed point in BY (0, ε). Call this fixed point y = g(x). It solves the equationy = Hx(y) which is equivalent to the equation F (x, y) = 0. Thus, the theorem follows withW = BX(0, δ).

It remains to prove the claim. (i) First, we pick δ > 0 so that for M = 2‖L−10 ‖ and given

ε > 0 and for any x ∈ BX(0, δ), we have

(a) ‖L−1x ‖ ≤M ;

(b) ‖F (x, 0)‖ ≤ ε/2M .

This is possible by the continuity of ∂yF (x, 0)−1 (due to the continuity of ∂yF (x, 0), see above)and of F (x, 0) in x and the relation F (0, 0) = 0. Next, we pick ε > 0 so that, for any x ∈ BX(0, δ)and any y ∈ BY (0, ε),

(c) ‖R(x, y)‖ = o(‖y‖) ≤ ε/2M ;

(d) ‖∂yF (x, y)− ∂yF (x, 0)‖ ≤ 1/2M .


Again this is possible by a property of R(x, y) and the continuity of ∂yF (x, y).Now, inequalities (a)-(c) and the definition of Hx imply

‖Hx(y)‖ ≤M (ε/2M + ε/2M) = ε

∀x ∈ BX(0, δ) and y ∈ BY (0, ε). So (i) follows.(ii) Using the definition, R(x, y) := F (x, y)− F (x, 0)− dyF (x, 0)y, of the remainder R(x, y)

and definition (5.6) of the map Hx, we compute

dyHx(y) = −dyF (x, 0)−1dyR(x, y)

= −dyF (x, 0)−1 [dyF (x, y)− dyF (x, 0)] .

Applying inequalities (a) and (d), we conclude that ‖dyHx(y)‖ ≤ 1/2 for any x ∈ BX(0, δ) andany y ∈ BY (0, ε). This proves (ii) and therefore completes the proof of the claim and with it,the theorem.

5.4 Appendix II: Derivation of the inverse function theorem from the implicitfunction theorem (Theorem 5.2).

Proof. Finding the inverse function G is equivalent to solving the equation

φ(v, u) := F (u)− v = 0

for u, given v ∈ V . Applying to this equation the implicit function theorem yields the result.

5.5 Minimal surfaces (to be done)

5.6 Harmonic maps (to be done)

6 Existence of bubbles and Lyapunov - Schmidt decomposition

For a field equation describing dynamics of interfaces, by a bubble we mean a smooth sphericallysymmetric static solution, u(x) = v(|x|), with a single interface (or more precisely a boundarylayer) across which the solution changes from approximately constant value to another.

We will consider two field equations describing interfaces, the Allen-Cahn equation, (3.2),and the Cahn-Hilliard equation, introduced below. In this case, a bubble is a smooth sphericallysymmetric static solution, u(x) = v(|x|), with the property that

v(r)→ +1, r → 0, v(r)→ −1, r →∞. (6.1)

In additions, we make a technical assumption that v′(r)→ −1, r →∞.First, we consider the Allen-Cahn equation, (3.2), which describes the phase separation

phenomena. Unlike in (3.2), in this section we consider the Allen-Cahn equation in the entirespace. We reproduce it here

∂u

∂t= ε2∆u− g(u), (6.2)

where u : Rd×R+ → R and ε is a small parameter, with the initial condition u|t=0 = u0(x), x ∈Rd. Here g : R→ R is the derivative, g = G′, of a double-well potential: G(u) ≥ 0 and has twonon-degenerate global minima with the minimum value 0 (see Figure 1). Specifically, we takeg(u) = u3 − u and G(u) = 1

2(u2 − 1)2.We consider static solutions of this equation in the entire space, Rd. They satisfy the static

Allen-Cahn equation,−ε2∆u+ g(u) = 0. (6.3)


Theorem 6. Let dimensions d ≥ 2. The Allen-Cahn equation (6.3) has no bubble solutions.

Proof. We use the well-known approach which goes under names of virial relation, Derrick’stheorem or Pohozaev identity.

By rescaling, we reduce to the case of ε = 1. Let r = |x| and u = v(r) be a sphericallysymmetric, C1 stationary solution to the Allen-Cahn equation:

−∂2rv −

d− 1

r∂rv + g(v) = 0, (6.4)

with v|r=0 = 1, and v|r=∞ = −1. We multiply (6.4) by ∂rv = x ·∇v and integrate in r to obtain

−1

2(∂rv)2|∞0 − (d− 1)

∫ ∞0

(∂v

∂r

)2 1

rdr +G(v)|∞0 = 0.

Since the last term is zero, this equation implies

1

2(∂rv)2|r=0 = (d− 1)

∫ ∞0

(∂v

∂r

)2 1

rdr.

If ∂v∂r |r=0,∞ = 0,, then

∫∞0

(∂v∂r

)2 1r dr = 0 and therefore v must be a constant; this however

contradicts the boundary conditions on v. If ∂v∂r |r=0 6= 0 (by the continuity, this condition must

hold in a neighborhood of the origin), then the above relation leads to a contradiction since theright side is infinite and the left side is finite. This completes the proof of the first part of thetheorem.

The reason the Allen-Cahn equation has no bubbles, or that bubble solutions collapse to apoint, is that the mass is not conserved and the surface tension shrinks the interface to a point.In a variant of the Allen-Cahn equation with the mass conservation - the Cahn-Hilliard equation- the bubbles do exist as shown below.

Now, we consider the Cahn-Hilliard equation, which play a central role in material science.In a sense, this is a version of the Allen-Cahn equation with mass conservation. As the Allen-Cahn equation, (6.2), it presents a key model with many generalizations and extensions. In Rd,it is of the form

ut = −∆(ε2∆u− g(u)), (6.5)

where g is the same as for the Allen-Cahn equation. In particular, we can take g(u) = u3 − u.The equation (6.5) is derived from the conservation law of mass

∂u

∂t= −div J, (6.6)

where J flux of the material and Fick’s law,

J = −D∇µ, (6.7)

connecting J to the chemical potential µ coming from thermodynamic consideration and theexpression of the latter in terms of the free energy, E ,

µ = dE(u). (6.8)

(Recall, that dE(u) is the Gateaux derivative of E(u).) If we take the standard expression

E(u) :=

∫Rd

1

2(ε2|∇u|2 +G(u)), (6.9)


with G′ = g, for the free energy E and D constant, say D = 1, then the above implies theCahn-Hilliard equation, (6.5).

It follows from Gauss’ theorem that∫Rdu dx = constant

along solutions to (6.5), in agreement with the conservation of the average mass fraction of thecomponents of the alloy.

We consider static solutions of the Cahn-Hilliard equation in Rd. They satisfy the staticCahn-Hilliard equation, which, after rescaling, is given by

−∆u+ g(u) = µ, (6.10)

It has a (homogeneous) solution u ≡ u for any constant u ∈ g−1(µ). This gives a static solutionof (6.5).

Theorem 6.1. [Alikakos and Fusco] Let dimensions d ≥ 2. The Cahn-Hilliard equation, (6.5),has a one-parameter family of static bubble solutions parametrized by their radii.

First, we present our main strategy. To solve the equation (6.10) equation for u we

(i) Construct a family, (φR(r), µR) of approximate solutions to (6.10) parametrized by R > 0(radii of the bubbles). (Here we reparameterized our problem from µ to R.)

(ii) Solve the equation (6.10) near (φR(r), µR) by using the Lyapunov - Schmidt decomposition.

Ideas and sketch of the proof of Theorem 6.1. Recall that the static equation (6.10) with µ = 0,i.e. the Allen-Cahn equation, has the kink solutions:

χx0e(x) = χ((x− x0) · e

), where x0, e ∈ Rn, (6.11)

where χ(s) → ±1 as s → ±∞ (see Section 3.1 and Figure 1). Recall that for the key g(u) =u3 − u, the function χ is given explicitly as

χ(s) = tahn(s√2

). (6.12)

Denote F (u, µ) := −∆u+ g(u)− µ. Then the equation (6.10) can be rewritten as

F (u, µ) = 0. (6.13)

Let r = |x|. As an approximate solution, we try the shifted kink

χR(r) := χ(r −R) (6.14)

We run into two problems. First, using that χR is a spherically symmetric function and therefore

∆χR = ∂2rχR +

d− 1

r∂rχR,

we obtain F (χR, µ) = −∂2rχR − d−1

r ∂rχR + g(χR)− µ. Using that χR satisfies

−∂2rχR + g(χR) = 0, (6.15)


we find furthermore

F (χR, µ) = −d− 1

r∂rχR − µ.

The r.h.s. is small (if µ is small and R is large) but is not L2: as r → ∞, we have ∂rχR → 0and therefore F (χR, µ)→ −µ. Hence we have to modify χR(r) at infinity.

We ignore this for now and look for a solution u(x) of (6.13) in the form u(x) = χR(r)+α(r),expecting α to be small (the perturbation theory). We rewrite the equation (6.13) as an equationfor α, by expanding F (u, µ) in (6.13) as

F (χR + α, µ) = F (χR, µ) + LRα+N(α, µ), (6.16)

where LR = duF (u, µ)|u=χR and N(α, µ) is defined by this equation, explicitly, N(α) := g(χR +α)− g(χR)− g′(χR)α, so that the equation F (χR + α, µ) = 0 becomes

LRα = −(F (χR, µ) +N(α, µ)).

We try to solve this equation for the fluctuation term α, by inverting the operator LR andreducing it to the fixed point problem α = H(α), where H(α) := −L−1

R (F (χR, µ) +N(α, µ)).To show that H is a contraction, we need that the linearized map LR = duF (u, µ)|u=χR is

invertible and its inverse is not too large. The second problem is that, as we indicate below,

• LR has an eigenvalue of the size O(

1R

)and therefore it is either non-invertible or invertible

with the inverse of the size O (R).

Indeed, using that the shifted kink χR satisfies the equation −∂2rχR + g(χR) = 0 and differenti-

ating this equation w.r.to R to obtain

(−∂2r + g′(χR))χ′R = 0, (6.17)

where χ′R ≡ ∂RχR. Thus we conclude that χ′R is the zero eigenfunction of the operator L0 :=−∂2

r + g′(χR). Using this and LR = L0 − d−1r ∂r, we compute

LR(−χ′R) =d− 1

rχ′′R. (6.18)

Now we use that χ′′R is concentrated near r = R, more precisely, that

|χ(k)R (r)| ≤ Ce−2|r−R| (6.19)

to obtain the estimate

LR(−χ′R) = O

(1

R

)χ′′R. (6.20)

(It is easy to check (6.19) for χ for the specific nonlinearity given in (6.12), but it holds for thegeneral nonlinearities described in (6.11), though with a different exponent.) By the spectraltheory (see Appendix D (to be done) and [30, 31]), this implies this implies the statementabove.

Thus we have to devise a way to overcome these problems. To tackle the first one, wemodify the shifted kink χR(r) := χ(r−R) to find a better approximate solution. To handle thesecond problem, we use the Lyapunov - Schmidt reduction and contraction mapping techniques.We ignore for the moment the first problem, treating χR as a good approximate solution, andconcentrate on the second one.


To begin with, the Lyapunov-Schmidt reduction consists of the following steps:

(a) Parameterization of solutions u(x). We parameterize our solution u(x) by a pair (R, ξ),where R > 0 and the function α(r), so that u(x) can be represented uniquely as

u(x) = χR(r) + α(r), α ⊥ χ′R. (6.21)

The condition α⊥χ′R ≡ ∂RχR (in L2(Rd)) determines R. Indeed, applying the implicit functiontheorem to the equation f(R, v) := 〈v − χR, χ′R〉 = 0, we obtain the unique solution R = R(v).

To clarify the geometric meaning of th decomposition (6.21), we define the kink manifoldMkink := χR(r)| R > 0 and and observe that (6.21) is equivalent to projecting u(x) to thismanifold, which gives the point χR(r) on Mkink and the function α(r) which is orthogonal tothis manifold, α ⊥ TχRMkink. Since the tangent space to Mkink at χR(r) is spanned by χ′R(r),this leads to (6.35). The function α(r) is called the fluctuation of u(x) around χR(r).

(b) Decomposition of the equation. Plug the decomposition (6.21) into (6.13) and project theresulting equation onto the tangent space TχRMkink and its orthogonal complement, (TχRMkink)⊥

to obtain the equations for two unknowns R and α:

PRF (χR + α, µ) = 0, (6.22)

P⊥R F (χR + α, µ) = 0, (6.23)

where PR be the orthogonal projector onto χ′R, namely, PRf := χ′R∫χ′Rf/norm, and P⊥R =

1l− PR. These are two equations for two unknowns ξ and h.

(c) Solution of the second equation (6.23) for α. We plug the expansion (6.16) into (6.23)and use the relation α = P⊥R α and the notation L⊥R := P⊥RLRP

⊥R , restricted to Ran P⊥R , to

obtain the equation

P⊥R F (χR, µ) + L⊥Rα+ P⊥RN(α, µ) = 0. (6.24)

Now, we use the key fact (see Corollary 6.3, below) that the operator LR, restricted to the or-thogonal complement, Ran P⊥R = (TχRMkink)⊥, of χ′R, is invertible, with the uniformly boundedinverse. To prove this one needs a fair amount of spectral theory, see [30, 31]. (For the relevantdefinitions and facts, see Appendix D (under construction).) As a result, we can rewrite(6.24) as

α = Φ(α, µ), where Φ(α, µ) := −(L⊥R)−1P⊥R [F (χR, µ) +N(α, µ)]. (6.25)

Now, it not hard to show that it is a contraction for R sufficiently large, which implies that

• For R−1 and µ sufficiently small, the equation (6.23) has a unique solution for α = α(R,µ)and this solution satisfies the estimate

α(R,µ) = O(Rd−12 (R−2 + µ)). (6.26)

We prove a similar result (for the correct approximate solution) later.

(d) Derivation of the reduced equation. Plug the solution α = α(R,µ) of (6.23) into (6.22)to obtain the new equation for R,

f(R,µ) := 〈χ′R, F (χR + α(R,µ), µ)〉 = 0. (6.27)

This is a scalar equation for a single unknown R (the reduced equation). Solve the reducedequation for R. Then χR+α(R,µ), where R is a solution to (6.27), is a solution to the stationaryCahn-Hilliard.


(e) Solution of the reduced equation for R. Since PRf := χ′R∫χ′Rf/norm, we can rewrite

(6.27) as 〈χ′R, F (χR + α, µ)〉 = 0. Remembering the expansion (6.16), this gives

〈χ′R, F (χR, µ)〉+ 〈χ′R, LRα(R,µ)〉+ 〈χ′R, N(α(R,µ)〉 = 0. (6.28)

To be specific, in what follows we take g(u) = u3−u. Then the function χ is given explicitlyby (6.12) and one can easily compute the terms 〈χ′R, F (χR, µ)〉, 〈χ′R, LRα〉 and χ′R, N(α)〉 (seeAppendix 6.1, below) and use the estimate (6.26), to derive from (6.28) the following equationfor R

−d− 1

2|Sd−1|R−1 + O(R−4 +R−2µ+ µ2) = 0. (6.29)

This equation has the solution of the form R = O(µ−2).(f) Summing up, we expect that if R is large enough, there exists a unique spherically

symmetric solution to the stationary Cahn-Hilliard equation of the form χR + α(R,µ), where

χR is defined in (6.14) and α(R,µ) = O(Rd−12 (R−2 + µ)).

Finally, we explain how to construct the family, φR(r), R > 0, of approximate solutions. Weassume R is sufficiently large and µ > 0, sufficiently small. To construct the family, φR(r), R >0, of approximate solutions to (6.13), we look for a spherically symmetric solution to (6.10),satisfying (approximately) the boundary conditions v(0) = −1 and v(∞) = 1.

As was discussed above, the shifted kink χR(r) := χ(r − R), where r = |x|, is not a suit-able approximate solution and we have to modify it. We write the family, φR(r), R > 0, ofapproximate solutions to (6.13) as follows

φR(r) = χR(r) + ηR(r), (6.30)

where ηR(r) := η(r − R). To be specific, we take g(u) = u3 − u. We show that ηR(r) can bechoosen so that

F (φR, µ)→ 0, as r →∞, (6.31)

F (φR, µ) is bounded, as r → 0. (6.32)

Namely, we take η(∞) = ηR(∞) = δ∞, with the number δ∞ satisfying δ3∞ − 3δ2

∞ + 2δ∞ = µ,which gives δ∞ = O(µ), and η′(−R) = η′R(0) = χ′R(0) = χ′(−R). Indeed, using that χR satisfiesthe equation (6.15), we compute

F (φR, µ) = −d− 1

r∂rχR − ∂2

rηR −d− 1

r∂rηR − ηR + 3ηRχ

2R + 3η2

RχR + η3 − µ. (6.33)

Since ∂rχR → 0 and χR(r)→ −1, as r →∞, the first relation requires that

−ηR + 3ηR − 3η2R + η3

R − µ→ 0,

as r → ∞. Hence, it suffices to choose η(∞) = ηR(∞) = δ∞, with the number δ∞ satisfyingδ3∞ − 3δ2

∞ + 2δ∞ = µ. This gives δ∞ = O(µ) and F (φR, µ)→ 0, as r →∞.For the second relation, (6.32), we require that ∂rχR +∂rηR = O(r), as r → 0, and therefore

chooseη′(−R) = η′R(0) = χ′R(0) = χ′(−R),

which gives (6.32). Finally, for simplicity, we take ηR(r) = 0, for R/2 ≤ r ≤ 3R/2.


Using the above expression, the estimate (6.19) and a convenient choice of ηR, we obtain thesimple estimates (this requires a fair amount of work not displayed here) |F (φR, µ)| . e−δR, forr ≤ R/2 or r ≥ 3R/2, and |F (φR, µ)| . (R−1 + µ), for R/2 ≤ r ≤ 3R/2. Using these estimates,we find

‖F (φR, µ)‖Hr . Rd−12 (R−1 + µ). (6.34)

(The term Rd−12 comes from the volume of the shell of radius R and thickness 1.)

This completes the construction of the approximate solution φR. The Lyapunov-Schmidtreduction based on this solution follows the outline above for the decomposition (6.21), i.e. forthe rough approximate solution χR and is done in Appendix to this section.

6.1 Appendix. Details of the Lyapunov-Schmidt reduction

Our goal is to solve the equation (6.10) near φR(r), for some R depending on µ to be chosenlater. We look for a solution u(x) of (6.13) in the form u(x) = φR(r) + ξ(r), and try to solvefor the fluctuation term ξ. We follow the approach sketched above for the decomposition (6.21),i.e. for the rough approximate solution χR. We begin with the following proposition

Proposition 6.2. The operator LR = −∆r + g′(χR) has the following properties:

1. The smallest point, inf σ(LR), of the spectrum of LR is a non-degenerate, isolated eigen-value of size O

(1R

)and with an approximate eigenfunction χ′R(r);

2. For large enough R, the gap between the smallest eigenvalue and the rest of the spectrumis O(1).

Proving this proposition, as well as the corollary below, requires a fair amount of the spectraltheory, see e.g. [30, 31] and Appendix D (to be expanded), and the proof is given below. Theproperty 2 shows that

Corollary 6.3. The operator LR is invertible on the orthogonal complement of the subspacespanned by the vector χ′R(r) and its inverse on this orthogonal complement is uniformly boundedin R.

Now, we follow closely the analysis in the main text. It consists of the following steps:(a) Parameterization of solutions u(x). We parameterize our solution u(x) by a pair (R, ξ),

where R > 0 and the function ξ(r), so that u(x) can be represented uniquely as

u(x) = φR(r) + ξ(r), ξ ⊥ φ′R. (6.35)

where φ′R ≡ ∂RφR. The condition ξ⊥φ′R (in L2(Rd)) determines R. Indeed, observing that∂RφR ≈ −φ′R and applying the implicit function theorem to the equation f(R, v) := 〈v −φR, φ

′R〉 = 0, we obtain the unique solution R = R(v).

To clarify the geometric meaning of th decomposition (6.35), we define the kink manifoldM := φR(r)| R > 0 and and observe that (6.35) is equivalent to projecting u(x) to thismanifold, which gives the point φR(r) on M and the function ξ(r) which is orthogonal to thismanifold, ξ ⊥ TφRM. Since the tangent space toM at φR(r) is spanned by φ′R(r), this leads to(6.35). The function ξ(r) is called the fluctuation of u(x) around φR(r).

(b) Decomposition of the equation. It is convenient to replace M := φR(r)| R > 0 by itsapproximate Mkink := χR(r)| R > 0. Plug the decomposition (6.35) into (6.13) and project


the resulting equation onto the tangent space TχRM and its orthogonal complement, (TχRM)⊥

to obtain the equations for two unknowns R and ξ:

PRF (φR + ξ, µ) = 0, (6.36)

P⊥R F (φR + ξ, µ) = 0, (6.37)

where PR be the orthogonal projector onto χ′R, namely, PRf := χ′R∫χ′Rf/norm, and P⊥R =

1l− PR. These are two equations for two unknowns ξ and h.

(c) Solution of the second equation (6.37) for ξ. (Here the choice of the family φR(r), R > 0,plays a crucial role.)

Lemma 6.4. For R−1 and µ sufficiently small, the equation (6.37) has a unique solution for

ξ = ξ(R,µ) and this solution satisfies the estimates ξ(R,µ) = O(Rd−12 (R−2 + µ)).

We prove this lemma later. Meantime we explain the idea of the proof. Expand F (φR+ξ, µ)in ξ to obtain

F (φR + ξ, µ) = F (φR, µ) + duF (φR, µ)︸︷︷︸LR,µ

ξ +N(ξ, µ), (6.38)

where N(ξ, µ) is defined by this equation, explicitly,

N(ξ) := g(φR + ξ)− g(φR)− g′(φR)ξ.

Plug this into (6.37) and use the notation L⊥ ≡ L⊥R,µ := P⊥R duF (φR, µ)P⊥R , restricted to

Ran P⊥R , to obtain the equation

P⊥R F (φR, µ) + L⊥ξ + P⊥RN(ξ, µ) = 0. (6.39)

By Corollary 6.3, the operator L⊥R,µ is invertible, with the uniformly bounded inverse. Hencewe can rewrite (6.39) as

ξ = Φ(ξ, µ), where Φ(ξ, µ) := −(L⊥)−1P⊥R [F (φR, µ) +N(ξ, µ)]. (6.40)

The estimation of the map Φ(ξ, µ) is done below and shows that it is a contraction for Rsufficiently large, which implies that that (6.40) has a unique solution, ξ = ξ(R,µ), satisfying

the estimate ξ(R,µ) = O(Rd−12 (R−2 + µ)).

(d) Derivation of the reduced equation. Plug the solution ξ = ξ(R,µ) of (6.37) into (6.36)to obtain the new equation for R,

f(R,µ) := 〈χ′R, F (φR + ξ(R,µ), µ)〉 = 0. (6.41)

This is a scalar equation for a single unknown R (the reduced equation). Solve the reducedequation for R. Then φR+ξ(R,µ), where R is a solution to (6.42), is a solution to the stationaryCahn-Hilliard.

(e) Solution of the reduced equation for R. Since PRf := χ′R∫χ′Rf/norm, we can rewrite

(6.41) as 〈χ′R, F (φR + ξ, µ)〉 = 0. Remembering the expansion (8.9), this gives

〈χ′R, F (φR, µ)〉+ 〈χ′R, Lξ(R,µ)〉+ 〈χ′R, N(ξ(R,µ)〉 = 0. (6.42)


Using (6.33) and (10.20), we compute

〈χ′R, F (φR, µ)〉 ≈ −d− 1

2|Sd−1|

∫ ∞0

1

rχ′R

2rd−1 dr +O

(Rd−1µ

),

〈χ′R, Lξ〉 = 〈Lχ′R, ξ(R,µ)〉 = O(Rd−12−2‖ξ‖L2

),

〈χ′R, N(ξ)〉 = O(‖ξ‖2H1 + ‖ξ‖3H1

).

For ξ = ξ(R,µ), the last two estimates give by Lemma 6.4, 〈χ′R, Lξ〉 = O(Rd−3(R−2 + µ)) and〈χ′R, N(ξ)〉 = O(Rd−1(R−4 + µ2)). Inserting the estimates above into (6.42) and dividing byRd−1, gives

−d− 1

2|Sd−1|R−1 + O(R−4 +R−2µ+ µ2) = 0. (6.43)

This equation has a solution R = O(µ−2).

Summing up, we have shown that if R is large enough, there exists a unique sphericallysymmetric solution to the stationary Cahn-Hilliard equation of the form φR + ξ(R,µ), where

φR is defined in (6.30) and ξ(R,µ) = O(Rd−12 (R−2 + µ)). Thus, the Cahn-Hilliard equation has

bubble solutions.

Proof of Lemma 6.4. We estimate the map Φ(ξ, µ). To keep things specific we consider thecanonical nonlinearity g(u) = u3−u and d = 3. Then N(ξ) = 3φRξ

2 +ξ3. Using this expression,the Sobolev embedding theorems and the simple inequality ‖ξm−ηm‖L2 ≤ m

(‖ξ‖m−1

H1 + ‖η‖m−1H1

)‖ξ−

η‖H1 , we obtain

‖N(ξ)‖L2 . ρ2, ‖N(ξ)−N(η)‖L2 . ρ‖ξ − η‖H1 , (6.44)

provided ‖ξ‖H1 , ‖η‖H1 ≤ ρ ≤ 1. Next, by (6.34) and the boundedness of P⊥R , we have

‖P⊥R F (φR, µ)‖Hr . Rd−12 (R−2 + µ). (6.45)

Using the last two relations, the uniform boundedness of L−1 and (6.40), we obtain

‖Φ(ξ, µ)‖H1 . Rd−12 (R−2 + µ) + ρ2,

‖Φ(ξ, µ)− Φ(η, µ)‖H1 . ρ‖ξ − η‖H1

Hence for ρ < 1 s.t. Rd−12 (R−2+µ)+ρ2 ρ, which is easy to satisfy if R

d−12 (R−2+µ) sufficiently

small (which means that R is large for d < 5), we have that Φ is a strict contraction on the ballin H1 of radius ρ, centered at the origin, and it has a unique fixed point in this ball.

(Fill in details!)

Proof of Proposition 6.2. The proof below uses the definition and properties of the essentialspectrum (see [30, 31] and Appendix D (to be expanded)). The first statement follows fromthe spectral fact that σess(L) = σ(L∞), where L∞ is the evaluation of L at infinity: L∞ :=−∆r + g′(χR(∞)) = −∆r + g′(1), and the simple computation, which uses the definition of thespectrum (see [30, 31]) and the Fourier transform,

σ(L∞) = σ(−∆r + g′(1)) = [g′(1),∞).


To prove the second statement, we observe that the equation (6.20) implies that 〈χ′R, LRχ′R〉 =O(

1R

)which shows that the operator LR has an eigenvalue of the order O

(1R

)below its essential

spectrum. Using the fact that −χ′R is a positive function by the Perron - Frobenius theory, weconclude that, for R large enough, the smallest eigenvalue of LR is non-degenerate and of theorder O

(1R

), with the approximate eigenvector χ′R.

To prove the third statement, we perform a geometric analysis of L. Let (j0, j1) be apartition of unity, normalized as j2

0 + j21 = 1, and with supp j0 ⊂ x ∈ Rd||x| ≤ R/2 and

supp j1 ⊂ x ∈ Rd||x| ≥ R/3 and |∂mji| . R−m. We use the IMS formula

L :=∑i

jiLji −∑i

|∇ji|2, (6.46)

which holds for any operator L of the form L = −∆ + V (x). To prove it, we use the relationsL =

∑1i=0 j

2i L = −

∑1i=0 ji[ji,∆]+

∑1i=0 jiV ji,

∑1i=0 j

2i ∆ =

∑1i=0 ji∆ji+ji[ji,∆] and ji[ji,∆] =

−ji(2∇ji · ∇+ ∆ji), to obtain

L = −1∑i=0

ji∆ji + ji(2∇ji · ∇+ ∆ji) +

1∑i=0

jiV ji.

Similarly, we have

L = −1∑i=0

ji∆ji + (2∇ji · ∇+ ∆ji)ji +1∑i=0

jiV ji.

Subtracting the second relation from the first and dividing the result by 2, we obtain L =−∑1

i=0 ji∆ji +∑1

i=0 jiV ji + [ji,∇ji · ∇]. Since [ji,∇ji · ∇] = −∑

i |∇ji|2, this gives (6.46).Since, by the choice of ji,

∑i |∇ji|2 = O

(1R2

), (6.46) implies that

L :=

1∑i=0

jiLji +O( 1

R2

). (6.47)

We estimate each term jiLji separately, using the properties of supp ji.

First we observe that χR ≈ 1 and therefore L ≈ −∆ + g′(1) ≥ g′(1) > 0 on x ∈ Rd||x| ≥R/3. Hence j1Lj1 ≥ j1g′(1)j1 = g′(1)j2

1 .

To show that the operator j0Lj0 is bounded below by cj20 on the subspace (χ′R)⊥, for some

c > 0 of the order O(1), we notice that the operator

L = −∂2r −

d− 1

r∂r + g′(χR(r))

on L2([0,∞), rd−1dr) is unitarily equivalent to the operator

−∂2r +

d− 1

4

1

r2+ g′(χR(r)) (6.48)

on L2([0,∞), dr). Indeed, the transformation u 7→ rd−12 u from L2(rd−1dr) to L2(r

d−12 dr) is

unitary and maps L into (6.48) as follows from the computation

(∂2r +

d− 1

r∂r)(r

− d−12 v) = r−

d−12 (∂r −

d2 − 1

4r2)v.


Next, under the unitary shift ξ(r)→ ξ(r −R), the operator (6.48) is unitarily equivalent to

L := −∂2r +

d− 1

4(r +R)2+ g′(χ(r))

on L2([−R,∞], dr).Now, under the unitary shift ξ(r)→ ξ(r−R), supp j0 is mapped into a subset of r ≥ −2R/3

and therefore L = L∗ + O(1/R) on r ≥ −2R/3, where L∗ := −∂2r + g′(χ(r)). The operator

L∗ is independent of R, has a non-degenerate eigenvalue 0 with the eigenfuction χ′R and theessential spectrum σess(L∗) = [g′(−1),∞). Hence L∗ ≥ c on the subspace (χ′R)⊥, for some c > 0of the order O(1), and therefore, by the unitary equivalence, this implies that j0Lj0 ≥ c0j

20 on

the subspace (χ′R)⊥.The estimates j0Lj0 ≥ c0j

20 on the subspace (χ′R)⊥ and j1Lj1 ≥ g′(1)j2

1 , together with therelation (6.47), imply that L ≥ c = min(c0, g

′(1)) on the subspace (χ′R)⊥, for large enough R.This gives the assertion four.

7 Theory of bifurcation

Consider a C1–map F : R× Y → Z, where Y and Z are Banach spaces. We would like to finda function u = u(µ), implicitly defined by the equation

F (µ, u) = 0. (7.1)

Assume F satisfies F (µ, 0) = 0 for all µ, i.e. (µ, 0) satisfies (7.1) for all µ. The branch ofsolutions (µ, 0) : µ ∈ R is called the trivial branch. The corresponding solutions are calledtrivial solutions. Our task is to find nontrivial solutions to (7.1) in a vicinity of the trivialbranch.

The “curve” (µ, u(µ)), µ ∈ [−ε, ε], is called a branch of solutions if F (µ, u(µ)) = 0. A point(µ0, 0) at which a branch of nontrivial solutions appears is called a bifurcation point.

bifurcation point trivial branch

nontrivial branch

µ

u

Figure 7: Bifurcation

An important example is given by the case of a linear (in u) map F : R× Y → Y :

F (µ, u) = µLu− u,

where L is a linear operator on Y . Then (µ, 0) is the trivial branch of solutions. Let us findthe bifurcation points. The candidates for bifurcation points are the points where duF (µ, 0) is


not invertible. We have dF (µ, 0) = µL− 1l. Assume here for simplicity that L has purely pointspectrum (i.e. only eigenvalues, no continuous spectrum), then µL − 1l is not invertible if andonly if 0 is an eigenvalue of µL− 1l (indeed, recall 0 /∈ σ(A) if and only if A is invertible). Now(µL − 1l)u0 = 0 ⇔ Lu0 = 1

µu0, i.e. 1/µ is an eigenvalue of L. If 1/µ is an eigenvalue of Lwith an eigenfunction u0, then (µ, au0), for any a ∈ C, is the bifurcating solution: F (µ, au0) =µa(Lu0 − µ−1u0) = 0. (If 1/µ is an eigenvalue of L, then we call µ a characteristic value of L.)

µ

Ru1

Ru2

(λ−1,0 ) (λ−1

21,0 ) . . . . . . .

. . . . . . .

. . .

u

Figure 8: Bifurcation:linear case

Example 7.1. Let Y = Z = L2([0, 2π]), L = −∆ with Dirichlet boundary conditions: u(0) =u(2π) = 0. Recall that the domain of L satisfies D(L) = H2([0, 2π]) ⊂ C([0, 2π]) by the Sobolevembedding theorem, and therefore the boundary conditions u(0) = u(2π) = 0 make sense.

To find the eigenvalues of L, we need to solve the equation −∆u = λu, i.e. u′′ = −λu. Thesolutions satisfying the Dirichlet boundary conditions are un = a sin(n2x), where a ∈ R, and theeigenvalues are given by λn = (n2 )2, n = 1, 2, 3, . . .. Thus besides the trivial branch (µ, 0), µ ∈ R,the equation µ(−∆u)− u = 0 has the branches of solutions (( 2

n)2,Run), for n = 1, 2, 3, . . ..

Observe that in the the above example, F is linear in u, and as a result, the bifurcatingbranches are straight lines. In general, if F is nonlinear, we expect the bifurcating branches tobe bent, as in the following example.

Example 7.2. Let Y = Z = R, and F (µ, u) = µu−u3. Clearly we have F (µ, 0) = 0 ∀µ ∈ R, so(µ, 0) is the trivial branch. We calculate the derivative duF (µ, u) = µ−3u2, so duF (µ, 0) = µ = 0has the solution µ0 = 0, hence (0, 0) is a candidate for a bifurcation point. On the other hand,we can solve the equation F (µ, u) = 0 explicitly, obtaining the solutions (µ, 0) and u = ±√µ.This shows that (0, 0) is indeed a bifurcation point (the bifurcation here is called a pitchforkbifurcation because of the shape of its bifurcating branch).

Below we will learn how to find out the qualitative behavior of the bifurcating branch withoutactually solving for it. But before, let us find a necessary condition for a bifurcation to happenat a point (µ0, 0). By the implicit function theorem, we know that if duF (µ0, 0) has a boundedinverse, then equation (7.1) has a unique solution in a neighborhood of (µ0, 0), which must bethe trivial branch (µ0, 0). Hence we have the following

Proposition 7.1. If (µ0, 0) is a bifurcation point, then duF (µ0, 0) does not have a boundedinverse.

Now, let us find a sufficient condition for a bifurcation to happen at a point (µ0, 0). Webegin with following cautionary example.

Example 7.3. For F : R× R2 → R2 given by F (µ, u1, u2) = (u1, u2)− µ(u1 − u32, u2 + u3

1), wefind

duF (µ, u)(ξ1, ξ2) = (ξ1, ξ2)− µ(ξ1 − 3u22ξ2, ξ2 + 3u2

1ξ1),


and therefore DuF (µ, 0) = (1 − µ)1l. Thus duF (1, 0) is not invertible. However, (1, 0) is not abifurcation point! Indeed, look at the two components of the equation F (µ, u) = 0. Multiplyingthe first one by −u2, the second one by u1, we obtain

−(1− µ)u1u2 − µu42 = 0

(1− µ)u1u2 − µu41 = 0.

Adding the above two equations yields −µ(u41 + u4

2) = 0, so u1 = u2 = 0 (for µ 6= 0), whichshows that F (µ, u1, u2) = 0 has only the trivial solution (u1, u2) = (0, 0), ∀µ ∈ R (if µ = 0, thenthis follows directly from the definition of F ). (1, 0) is therefore not a bifurcation point.

We now give a sufficient condition for a bifurcation to take place. Let (µ0, 0) be a candidatefor a bifurcation point for the equation (7.1), i.e. we assume duF (µ, u) is not invertible at (µ0, 0).The non-invertibility of duF (µ0, 0) is equivalent to the statement that

0 ∈ σ(duF (µ0, 0)).

Here σ(A) denotes the spectrum of a linear operatorA, i.e. σ(A) := z ∈ C : A−z1l is not invertible.Clearly, eigenvalues of A belong to σ(A) (see Appendix C). In general, the spectrum can alsocontain continuous pieces and it can take very peculiar forms. The multiplicity of an eigenvalueλ of A is defined as the number of linearly independent eigenvectors with eigenvalue λ, i.e.

• the multiplicity of an eigenvalue λ of L is dim Null(L− λ1l).

For details on spectral theory see Appendix C. To minimize the technicalities, we formulate andprove the next result for the special case of when duF (µ0, 0) is a bounded, symmetric operator.(An operator A on an inner product space is said to be symmetric it satisfies 〈Au, v〉 = 〈u,Av〉.)

Theorem 7.2 (Krasnoselski). Assume that Y is dense in a Hilbert space Z and that F : R×Y →Z is a C1 map such that

(i) duF (µ0, 0) is a self–adjoint operator;

(ii) 0 is an isolated eigenvalue of duF (µ0, 0) of odd multiplicity;

(iii) there exists a v0 ∈ Null duF (µ0, 0) such that 〈v0, ∂µduF (µ0, 0)v0〉 6= 0.

(In (iii), we assumed that duF is C1 in µ at the point (µ0, 0).) Then (µ0, 0) is a bifurcationpoint.

We conduct the proof in two steps. One the first step, under very general conditions, wereduce the problem of solving the equation (7.1) to the problem of solving an equation in a fewdimensions (the reduced or effective, or bifurcation equation). On the second step, we solve thelatter equation.

Theorem 7.3 (Reduction to effective equation). Assume that Y is dense in a Hilbert space Zand that F : R× Y → Z is a C1 map such that duF is C1 in µ at the point (µ0, 0) and

(i) duF (µ0, 0) is a self–adjoint operator;

(ii) duF (µ0, 0) has an isolated eigenvalue at 0.


Then any solution, u, to (7.1) in a neighbourhood of (0, µ0) satisfies

‖u− v‖ = o(‖v‖) +O(|µ− µ0|‖v‖),

where v ∈ null duF (µ0, 0) and solves a certain finite - dimensional equation.

The equation (7.4) is called the bifurcation equation or branching equation. It describes thebifurcating branches, and usually, it is a system of n = dim nullL(µ) algebraic equations forn+ 1 variables µ and v.

Proof. To prove this theorem we use the powerful technique, called the Lyapunov-Schmidt de-composition. We will use this technique repeatedly below. Let L(µ) := duF (µ, 0), and denoteby P the orthogonal projection onto the subspace nullL(µ0) and L(µ0)P = PL(µ0) = 0. LetP⊥ := 1l − P . We project u ∈ Y and the equation F (µ, u) = 0 onto the subspaces Ran P andRanP⊥: u = v + w, where v ∈ Ran P and w ∈ RanP⊥, and

PF (µ, v + w) = 0, (7.2)

P⊥F (µ, v + w) = 0. (7.3)

We have thus two equations, (7.2) and (7.3), for two variables v and w. Observe that sincedim Ran P <∞, v is a finite–dimensional variable.

We solve (7.3) for w, and substitute the solution, w = w(µ, v), into (7.2) to obtain theequation

f(v, µ) = 0, (7.4)

with f(v, µ) := PF (µ, v + w(µ, v)).We now show equation (7.3) has a unique solution w. Define

F1(µ, v, w) :=P⊥F (µ, v + w) : X ×P⊥Y →P⊥Z.

where X := R × PY . Then the problem (7.3) can be reformulated as solving the equationF1(µ, v, w) = 0 for w. Now observe that

(α) F1 is C1,

(β) F1(µ, 0, 0) = 0 for any µ,

(γ) dwF1(µ0, 0, 0) is invertible.

Indeed, (α) follows from the condition that F is C1, (β) results from the relation F1(µ, 0, 0) =P⊥F (µ, 0) = 0, and (γ) is due to the relation

dwF1(µ0, 0, 0) =P⊥duF (µ0, 0)P⊥

plus the fact that the r.h.s. is invertible as an operator fromP⊥Y toP⊥Z.Hence we can apply the implicit function theorem to the equation F1(µ, v, w) = 0, which

shows thus that for any (µ, v) sufficiently close to (µ0, 0), the equation F1(µ, v, w) = 0, andtherefore (7.3), has a unique solution. We denote this solution by w = w(µ, v).

Below, we show that w = o(‖v‖)+O(|µ−µ0|‖v‖). This together with the relation u = v+wgives that the solution of the original problem has the form

u = v + w(µ, v), with w(µ, v) = o(‖v‖) +O(|µ− µ0|‖v‖). (7.5)

This completes the first step of the proof.


Proof of the Krasnoselski theorem. We now proceed to the second step, i.e., we solve equation(7.4). For this, we use conditions (i) and (ii) of the theorem. Assume for simplicity thatdim NullL(µ0) = 1, i.e. that the eigenvalue 0 of L(µ0) is simple and show that equation (7.4)has a unique solution for µ as a function of v ∈ R. Let v0 the zero eigenvector of the operatorL(µ0), normalized as 〈v0, v0〉 = 1. Then Pu = 〈v0, u〉v0. Since v = sv0 for some s ∈ R, equation(7.4) is equivalent to the equation f(s, µ) = 0, where

f(s, µ) :=

1s 〈v0, F (µ, sv0 + w(µ, sv0))〉 for s 6= 0,

〈v0, L(µ)v0〉 for s = 0.(7.6)

Let u1 := s−1u. Next, using the fact that F (µ, 0) = 0 for any µ, we expand the map F (µ, u)around u = 0 as

F (µ, u) = L(µ)u+R(µ, u), (7.7)

with R(µ, u) = o(‖u‖) and, recall, L(µ) := duF (µ, 0). Using this expansion, we rewrite f(s, µ)as

f(s, µ) = 〈v0, L(µ)u1〉+ 〈v0, s−1R(µ, su1)〉, (7.8)

Since s−1‖R(µ, su1)‖ → 0 and u1 − v0 = s−1w → 0, as s → 0, we have that f(s, µ) →〈v0, L(µ)v0〉, as s→ 0. We compute the µ−derivative of f . Hence f(s, µ) is continuous at s = 0.Next, we have

∂f

∂µ(s, µ) = 〈v0, ∂µL(µ)u1〉+ 〈v0, L(µ)∂µu1〉+ 〈v0, s

−1(∂µR(µ, su1) + sduR(µ, su1))∂µu1〉.

Write u1 := s−1u = v0 + w1, where w1 := s−1w. Since ‖w1‖, ‖∂µw1‖ → 0, as s → 0 and|µ − µ0| → 0 (see (7.11) below), and s−1‖∂µR(µ, su1)‖, ‖duR(µ, su1)‖ → 0, as s → 0, we havethat

∂f

∂µ(0, µ) = 〈v0, ∂µduF (µ, 0)v0〉, (7.9)

which shows that f(s, µ) is differentiable in µ also at s = 0. Moreover, by condition (iii) ofthe Krasnoselski theorem, ∂f

∂µ(0, µ0) = 〈v0, ∂µduF (µ0, 0)v0〉 6= 0. Therefore equation (7.6) has aunique solution µ = µ(s), for µ as a function of s, if s is in a neighbourhood of s = 0. Thiscompletes the second step.

We have shown above that the solution of the original problem has the form (7.5), where vsatisfies the equation (7.4). In the case when 0 is a simple eigenvalue of L(µ0), we have shownthat the equation (7.4) can be solved for µ as a function of v, µ = µ(v). This give the bifurcationbranch of solutions

(u = v + w(µ, v), µ = µ(v)), (7.10)

parametrized by v ∈ nullL(µ).

Proof of the estimate w, ∂µw = o(‖v‖) +O(|µ− µ0|‖v‖). We show the following important prop-erty of the solution w(µ, v):

w, ∂µw = o(‖v‖) +O(|µ− µ0|‖v‖). (7.11)

To show (7.11), we use the expansion (7.7), to rewrite the equation (7.3), as

L⊥(µ)w +P⊥L(µ)v +P⊥R(µ, u) = 0, (7.12)


where L⊥(µ) :=P⊥µ L(µ)P⊥µ . Since the operator L⊥(µ) :P⊥µ Y →P⊥µ Z is invertible, we derive

w = −L⊥(µ)−1P⊥µ (R(µ, u) + L(µ)v) .

The relations R(µ, u) = o(‖u‖) and L(µ)v = (L(µ) − L(µ0))v = O(|µ − µ0| ‖v‖) imply w =o(‖u‖) + O(|µ − µ0| ‖v‖). Since u = v + w, this shows (7.11) for w. To prove (7.11) for ∂µw,we differentiate (7.12) w.r.to µ and then proceed with the resulting equation as we did with(7.12).

Example. As in an example above, let F (µ, u) = µu − u3. We have duF (µ, u) = µ − 3u2, so0 is an eigenvalue of duF (0, 0) of multiplicity 1. Next, ∂µduF (0, 0) = 1, so the condition (ii) issatisfied as well. Therefore (0, 0) is a bifurcation point.

Corollary 7.4. Let F be a C1 map satisfying duF (µ, 0) = µL − 1l, where L is a linear, self–adjoint operator. If µ0 is a characteristic value of L of odd multiplicity, then (µ0, 0) is a bifur-cation point.

Proof. Clearly the first two conditions of the Krasnoselski’s theorem are satisfied. To checkthe third condition, we compute ∂µduF(µ, 0) = L, and 〈u0, ∂µduF(µ, 0)u0〉 = 〈u0, Lu0〉 =µ−1

0 ‖u0‖2 6= 0 ∀u0 ∈ Null (µ0L− 1l). Hencethe third condition is satisfied as well.

To illustrate the corollary, let us consider the nonlinear eigenvalue problem

Lu+ f(u) = λu, with f ∈ C1 and f(0) = f ′(0) = 0. (7.13)

Then the corollary implies that if λ0 is an eigenvalue of a self–adjoint operator L of odd mul-tiplicity, then equation (7.13) has a nontrivial branch of solutions near the bifurcation point(λ−1

0 , 0).We specify this example further by considering the following nonlinear eigenvalue problem

(−∆ + V )u(x) + f(u(x)) = λu(x)

where u ∈ H2(Rn,R), ∆ is the Laplacian on Rn and the function f : Rn → R satisfies f(0) = 0,f ′(0) = 0 and |f(u)| ≤ c|u|p with n

2 −n2p < 2.

Exercise 7. Find the bifurcation points for the equation ∆u + λu + u3 = 0 on [−L,L]n withDirichlet boundary conditions (i.e. u = 0 on the boundary), and where u ∈ H2

sym([−L,L]n).Here L is a bifurcation parameter and H2

sym([−L,L]n) denotes the subspace of H2([−L,L]n)consisting of functions symmetric with respect to permutations, i.e., ψ(x1, . . . , xn) = ψ(xπ(1), . . . ,xπ(n)) for all permutations π.

Example 7.4. Let L be a very large fixed number, and take the family of spaces [−α/2, α/2]×[−L/2, L/2], where α ∈ R is a parameter. On these spaces, consider the equation

−∆u+ g(u) = 0, (7.14)

with periodic boundary conditions in x and Dirichlet boundary conditions in y:

u(−α/2, y) = u(α/2, y) ∀y,u(x,−L/2) = u(x, L/2) = 0 ∀x. (7.15)

Here g(u) satisfies g(u) = G′(u), where G(u) is a function of the form


u0

u1

G

−G

Figure 9: Function G and solution u1

Let a be the first positive zero of G(u) (see figure). Assume G′′(a) < 0.Equation (7.14) with boundary conditions (7.15) has the following branches of solutions

which are independent of x: (α, u0), ∀α, and (α, u1), ∀α, where u0 is a constant function equalto the local minimizer of G(u) (see figure) and u1 satisfies 0 ≤ u1(y) ≤ a, u1(∞) = 0 andu1(0) = a (see figure). To prove the existence of the solution u1, we observe that it satisfies theequation

−∂2u

∂y2+ g(u) = 0 on [−L/2, L/2], (7.16)

with the Dirichlet boundary conditions. The last equation is just Newton’s equation if weinterpret y as being the time-variable, and −G as a potential (whose derivative is a force). Thenthe existence of u1 is obvious from the picture above.

The problem here is to find solutions bifurcating from the branches above as α varies. Thisproblem arises in the theory of tunneling of line vortices in superconductors. There α is inversetemperature.

Exercise 8. (a) Check whether any solutions bifurcate from (α, u0).

(b) Find the bifurcation points from the branch (α, u1).

Lamellar phase. Show the existence of lamellar phase of Section ??. Recall that in lamellarphase, the layers of +1 and −1 phases (substances) coexist in a periodic array. We conjecturethat a periodic solution corresponding to an array of kinks and antikinks bifurcates from thetrivial solution of the static Allen-Cahn equation.


8 Bifurcation of surfaces of constant mean curvature

Let S be a hypersurface in Rn+2. Denote by H(x) its mean curvature at a point x ∈ S. IfS is described by an immersion θ : U → Rn+2 (where U , an open set in Ω), then we writeH(x) = H(θ). We consider the equation

H(θ) = h (8.1)

for some constant h. Surfaces satisfying this equation are called the constant mean curvaturesurface. One of the reasons they are interesting is that they solve the isoperimetric problem:

• minimize surface area for a fixed enclosed volume.

Critical points for this variational problem are surfaces of constant mean curvature.

There is a vast number of constant mean curvature surfaces, e.g. rotationally symmetricsurfaces in R3, known as Delaunay surfaces, and various surfaces obtained by perturbing severalknown surfaces, glued together ([?, ?]).

However, it is probably only spheres which solve the isoperimetric minimization problem.

Now, given a > 0, we consider the domain

Ωa := x ∈ Rn+2 : 0 ≤ xn+2 ≤ a.

We assume ∂S ⊂ ∂Ωλ are non-empty and ∂S intersects ∂Ωa transversally. In this case

• Every round cylinder in Ωa, with the axis transversal to the plane x ∈ Rn+2 : xn+2 = 0and every half sphere are static solutions of (8.1).

Obvious solutions of the equation (8.1) are half-spheres of radius n+1h and cylinders of radius n

h .To determine which of them are stable and which are not, one can compare the surface area ofthe cylinder and half-sphere of the same volume, namely which one is smaller. For a = 1, thesurface area of a cylinder with volume V between the two planes is Acyl = 2

√πV and the surface

area of a sphere with volume V is Asph = (36πV 2)1/3. Acyl ≤ Asph is equivalent to V ≤ 814π . In

particular, one expects that cylinders of radius R are stable when R is large and as R decreasescylinders become unstable.

We are interested in rotationally symmetric static solutions in Ωa, i.e. solutions of (8.1),with non-empty boundaries which are subsets of ∂Ωa and are transversal to the latter, theirdependence on the parameter a and their stability.

At the moment we give an ad hoc definition of stability, which we will explain later. (This isthe stability w.r.to perturbation of initial conditions of the volume preserving mean curvatureflow.) We say that a CMC surface θ is stable (in the sense of geometric analysis) iff dH(θ) ≥ 0on the subspace

Vθ = ξ : S → Rn+1 :

∫Sξ · ν = 0 (8.2)

and unstable, otherwise.

8.1 Bifurcation of new surfaces

We consider solutions of Eq. (8.1) in the form of graphs, θρ : (ω, x) 7→ (ρ(ω, x)ω, x), over theunit round cylinder Cn+1

a = Sn× [0, a] = (ω, x) : ω ∈ Sn, x ∈ [0, a], with the graph functions ρ.


Theorem 7. • For 0 < a < π nh , the cylindrical static solution, ρ = nh , of Eq. (8.1) is

unique in Uδ and stable.

• At a = π nh , it looses its stability and a new stable branch of solutions bifurcates at thispoint. The new solutions are graphs over the cylinder, i.e. their immersions are of theform θs(ω, x) = θρs(ω, x) = (ρs(ω, x)ω, x), with the graph function ρ of the form ρs =nh + s cos(πx) +O(s2).

• The latter branch is stable for nh < a < 2π nh , at a = 2π nh , it looses its stability and a new

stable branch of solutions bifurcates at this point and so forth.

In the next section we prove existence of bifurcating solutions. After that we prove theirstability. Before proceeding to the proofs, we give the expression of the mean curvature H, ofsurfaces of revolution.

First, we rewrite out the expression for the mean curvature for the level set representationof S. Below, all differential operations, e.g. ∇,∆, are defined in the corresponding Euclidianspace (Rn+2). First, we recall that, if a surface S given in the level set representation, S = x′ ∈Rn+2 : ϕ(x′) = 0, then we have

ν(x′) =∇ϕ|∇ϕ|

, H(x′) = div ν(x′) = div

(∇ϕ|∇ϕ|

). (8.3)

Using this expression, we can obtain an expression for the mean curvature of surfaces of revolu-tion. Let x′ = (x⊥, x) ∈ Rn+2, with x⊥ ∈ Rn+1. The the mean curvature H(x′), of a surface of

revolution S = r⊥ = ρ(x), where r⊥ = (∑n+1

i=1 x2i )

12 , are given by

H(ρ) ≡ H(x′) =1√

1 + (∂xρ)2

(− ∂2

xρ

1 + (∂xρ)2+n

ρ

). (8.4)

We define the map F (ρ, h, a) := H(θρ)− h. Then the equation (8.1) can rewritten as

F (ρ, h, a) = 0. (8.5)

Note that the round cylinder of the radius R is given by the equation ρ = R.

8.2 Proof of the existence part

We consider solutions to (8.1) which are surfaces of revolution. In other words they are definedby immersions θ(ω, x), which are graphs over the cylinder, θ(ω, x) = (ρ(ω, x)ω, x), with the graphfunction ρ independent of ω: ρ = ρ(x). Then the mean curvature is given by the expression(??)nd the equation (8.1) can rewritten as F (ρ, h, a) = 0, with

F (ρ, h, a) :=1√

1 + (∂xρ)2

(− ∂2

xρ

1 + (∂xρ)2+n

ρ

)− h, (8.6)

where ρ is defined on the interval [0, a] with the Neumann boundary conditions. Note thefollowing properties of F :

• cylinder of radius R = R(h) := n−1h solves F (R, h, a) = 0, ∀a, h.

• F (ρλ, λ−1h, λa)(x) = λ−1F (ρ, h, a)(λ−1x), where ρλ(x) := λρ(x/λ).


• dρF (ρ, h, a)∂ρλ∂λ∣∣λ=1

= −h.

Proof. The first and second properties are obvious from the expression (8.6). They can be alsodeduced from the general facts that H(θcyl) = n

R and (see e.g. [?])

H(λθ) = λ−1H(θ). (8.7)

Because of θ(ω, x) = (ρ(ω, x)ω, x), the rescaling θ → λθ induces the rescaling ρ→ λρ(x/λ) of ρ.By the definition F (ρ, h, a) := H(θρ)− h, we have the desire property.

The third property is related to the fact that ρ ≡ R breaks the scaling covariance of theequation: if ρλ(x) := λρ(x/λ), then F (ρλ, h, a)(x) = 1

λF (ρ, λh, a)(x/λ). Therefore, if ρ solvesF (ρ, h, a) = 0, then ρλ satisfies F (ρλ,

1λh, a)(x) = 0. Differentiating the latter equation w.r.to λ

at λ = 1, we find dρF (ρ, h, a)∂ρλ∂λ∣∣λ=1

= −h.

Since Rλ = λR and therefore ∂λRλ = R, the last property implies that

• dρF (ρ, h, a)∣∣ρ=R

has the eigenvalue − hR , with the eigenfunction 1.

The latter fact and the Perron-Frobenius argument imply that −h/R is the smallest eigenvalueand is simple. Moreover, for the round cylinder Cn+1

R of the radius R to solve the CMC equation

(8.1), we have to take R = nh and therefore this eigenvalue becomes −h2

n .Using the second property we rescale the equation F (ρ, h, a) = 0, to eliminate one of

the parameters, e.g. by taking λ = h, or λ = a−1, in the equation F (ρλ, λ−1h, λa)(x) =

λ−1F (ρ, h, a)(λ−1x). (This might depend on which parameter physically we want to keepfixed and which to vary.) This gives the new equation F (ρ′, a′)(x) = 0, 0 ≤ x ≤ ha, whereF (ρ′, a′)(x) := h−1F (ρ′, 1, a′)(x), ρ′(x) := hρ(x/h) and a′ := ha, or F (ρ′, h′)(x) = 0, 0 ≤ x ≤ 1,where F (ρ′, h′)(x) := (h′)−1F (ρ′, h′, 1)(x), ρ′(x) = a−1ρ(ax) and h′ := ah. We choose the secondrescaling and drop the primes. Consequently, we have the equation

F (ρ, h) :=1√

1 + (∂xρ)2

(− ∂2

xρ

1 + (∂xρ)2+n

ρ

)− h, (8.8)

on the interval [0, 1] with the Neumann boundary conditions. The cylindrical (homogeneous)

solution to this equation is solutions are ρ = n−1h , which in the old variables is ρ = (n)a

h .We linearize F (ρ, h) at ρ = R and let LR = −dρF (R, h), then by direct computation

LR = −∂2x −

n

R2,

acting on the space L2([0, 1]), with the Neumann boundary conditions.

Proposition 1. The operator LR is self-adjoint and its spectrum, σ(LR), is purely discrete withthe eigenvalues − n

R2 , (πk)2 − nR2 , (π(k + 1

2))2 − nR2 , k = 1, 2, · · · , of the multiplicity 1, with the

eigenfunctions 1, cos(πkx) and sin((k + 12)πx).

The self-adjointness and the form of the spectrum are obvious. We mention that the factthat LR has the eigenvalue −n−1

R2 is not accidental. As shown above, it is related to the factthat ρR ≡ R breaks the scaling covariance of the equation.

Now, we fix R = nh , the cylindrical solution to (8.8) and write Lh = LR

∣∣R=n

h. If h satisfies

either (πk)2 − nR2 = 0 or (π(k + 1

2))2 − nR2 = 0, k = 1, 2, · · · , where R = n

h , then the operatorLh has a zero eigenvalue. These values of h are the candidates for the bifurcation points. Letus check the first of these values, π2 = n

R2 or h = π√n.

To prove the bifurcation of the new solutions, we use the bifurcation theory, specifically, theKrasnoselski theorem. We check the conditions of this theorem (below R(h) = n

h ):


(a) F is C1;

(b) F (R(h), h) = 0, ∀h;

(c) dρF (R(h), h) has the eigenvalue 0 for h =√nπk, k = 1, 2, . . . and this eigenvalue is simple;

(d) ∂hdρF (ρ, h)∣∣ρ=R(h)

= −h−2 nR(h) = −h−1 6= 0.

Hence the Krasnoselski theorem is applicable and implies that the equation (9.10)as non-trivial solution branches bifurcating at from the trivial branch (R(h), h), ∀h. The first non-trivialsolution branch is of the form ρ = n

h + s cos(πx) + ws, with ws = O(s2) and h depending on s.To obtain the information about ρs we spell out the proof of the Krasnoselski theorem.We use Lyapunov-Schmit decomposition. We write ρ = R+u. Let P be orthogonal projection

onto eigenfunctions corresponding to eigenvalue π2− n−1R2 , i.e. span cos(πx) and P⊥ = 1−P .

Apply P and P⊥ to F (ρ, h) = 0:

• PF (R+ v + w, h) = 0,

• P⊥F (R+ v + w, h) = 0,

where v = Pu and w = P⊥u. We write v = s cos(πx), where s = 〈cos(πx), u〉. s is a newparameter. We can express it in terms of the parameter h, but it is more convenient to expressh in terms of s, which is the new bifurcation parameter.

We expand F (ρ, h) around the solution ρ = R = R(h) (to F = 0) and use that LR cos(πx) =(π2 − n−1

R2 ) cos(πx) to obtain

F (ρ, h) = −(π2 − n− 1

R2)s cos(πx)− LRw +Nh(s cos(πx) + w), (8.9)

where Nh(u) := F (R+ u, h)− F (R, h)− LRu. Hence the equation P⊥F (R+ v + w, h) = 0 canbe rewritten as

L⊥Rw = P⊥Nh(s cos(πx) + w),

where L⊥R := P⊥LRP⊥|Ran P⊥ . Notice that w ⊥ 1, cos(πx), and σ(L⊥R) is (kπ)2 − 1

R2 : k =2, 3, · · · ⊂ (0,∞). So L⊥R is invertible and therefore we can rewrite the last equation as

w = (L⊥R)−1P⊥NR(s cos(πx) + w).

By a fixed point argument there exists a unique solution w = ws satisfying ws|s=0 = 0. Pluggingthis into PF (v + w, h) = 0, we obtain the following equation for h and s:

〈cos(πx), F (R+ s cos(πx) + ws, h)〉 = 0, (8.10)

which, due to (8.9) and and∫

cos2(πx)dx = 1/2, can be also rewritten as

−1

2(π2 − n− 1

R2)s+

∫ 1

0cos(πx)NR(s cos(πx) + ws(x))dx = 0. (8.11)

This is an equation for the unknowns, h and s. Since ws|s=0 = 0 and NR(0) = 0, this equationalways has solutions s = 0 and arbitrary h. But there are also nontrivial solutions to (8.11). Tosee this, we first notice that, since NR(ξ) = O(ξ2), one can show, as usual, that ws = O(s2).Now, we expand the l.h.s. of (8.11) into powers of s, observe that the lowest power of s is 2 andcheck that there are higher powers, to conclude that, after factoring s2, the remaining equationhas a non-trivial solution for h as a function of s. To sum up, we obtain the solution of the formρ = R(h) + s cos(πx) + ws, with ws = O(s2) and h depending on s.


8.3 Proof of the linearized stability

UNDER CONSTRUCTIONIn this section we show that (a) for 0 < a < π nh , the cylindrical static solution, ρ = n

h , of Eq.(8.1) is stable, (b) at a = π nh , it looses its stability and (c) for n

h < a < 2π nh the new solutionthat bifurcated at a = π nh is stable.

The notion of stability defined in at the beginning of this section translate in the presentcontext as follows. Let dρF (ρ∗, h, a) denote the Gateaux derivative of the map F (ρ, h, a) atρ = ρ∗. In the present context the subspace (8.2) becomes

V∗ = ξ : Cn+1a → R :

∫Cn+1a

ξ = 0. (8.12)

We say that a static solution ρ∗ of (8.5), i.e. a solution of F (ρ, h, a) = 0, is linearly stable iff thespectrum of dρF (ρ∗, h, a) restricted to V∗ lies in the half-plane Re z > 0 and we say that ρ∗ islinearly unstable iff σ(duF (ρ∗, h, a))

∣∣V∗∩Re z < 0 6= ∅. (Since in our case, dρF (ρ∗, h, a) is self-

adjoint, the above relations can be replaced by duF (ρ∗, h, a))∣∣V∗≤ 0 and duF (ρ∗, h, a))

∣∣V∗≥ 0,

respectively.)Now, duF (R, h, a))

∣∣V∗

= LRa and therefore the static solution ρR is linearly stable iff

σ(LRa∣∣1⊥

) ⊂ Re(z) > 0, and unstable iff σ(LR∣∣1⊥

) ∩ Re(z) < 0 6= ∅.Proposition 1 implies that on the subspace 1⊥, the eigenvalues of LR are (πk)2− n

R2 , (π(k+12))2− n

R2 , k = 1, 2, · · · , of the multiplicity 1, with the eigenfunctions cos(πkx) and sin((k+ 12)πx).

If π√n > h, then σ(Lh) on Vθ is positive and for h = π

√n, the lowest eigenvalue vanishes.

This suggests that a new stable stationary solution bifurcates at this point from the old one.Moreover, for π

√n < h, the operator LR has a negative eigenvalue π2 − 1

R2 ) and therefore,while the old solution, the cylinder ρR, is stable for π

√n > h, it is unstable for π

√n < h. We

expect that the new bifurcating solution is stable, at least for R not too large, more precisely,for π

√n < h < π 3

2

√n.

If R > 1π , σ(LR) is positive and hence cylinder ρR is stable. If 1

2π < R < 1π , LR has a

negative eigenvalue π2 − 1R2 and hence cylinder ρR is unstable.

9 Stability of static solutions (and solitons)

9.1 Stability: generalities

So far we studied mainly existence of static solutions. (Remember that at the same time thisgives also the existence of of stationary or standing waves and traveling wave solutions.) Thenext key question is, starting with initial conditions close to a static solution, how the solutionsof the dynamical equation in question behave? This leads to the question of stability of staticsolutions of general dynamical systems,

∂u

∂t= F (u). (9.1)

Assume F : U → Y , where U is an open set X and X and Y are Hilbert spaces. Let (9.1) have astatic solution, u∗, i.e., time independent solutions. (Such solutions are also called equilibria andsometimes stationary solutions.) Thus u∗ is an equilibrium or static solution iff u∗ is independentof time and satisfy the equation F (u∗) = 0.

We would like to understand the behavior of solutions to equation (9.1) for initial conditionsu0 near an equilibrium one u∗. There are the following general scenarios


• The solutions stay in a neighborhood of u∗ (Lyapunov stability);

• The solutions converge to u∗ as t→ +∞ (asymptotic stability);

• The solutions move away from u∗ as t→∞ (instability).

More precisely, we say that a static solution of (9.1) is Lyapunov stable if for any neighbor-hood, U of u∗ have another neighborhood, V ⊂ U , of u∗ such that if an initial condition, u0, isin V , then the solution, u, stays in U . Otherwise, u∗ is said to be unstable.

We say a static solution, u∗, is asymptotically stable if ∃δ > 0 such that for any initialcondition u0 satisfying ‖u0 − u∗‖ < δ, we have limt→∞ ‖φt(u0)− u∗‖ = 0.

For systems with symmetry, the notion of stability/instability should be modified. Recall,that, if the dynamical system (9.1) with the symmetry group G has a static solution u∗, then ithas the manifold of static solutions

M∗ = Tgu∗ : g ∈ G.

We say that a static solution u∗ (or more precisely, the manifold of static solutions M∗) isorbitally stable if any solution to the equation (9.1), starting in a small neighborhood of u∗ staysin a small neighborhood of the manifoldM∗. In other words the solution ut sticks very close toa possibly moving static solution Tg(t)u∗.

Linear stability. We also want to give a weaker notion of stability which is usually is thefirst step in proving asymptotic stability and often gives the necessary condition. We say that astatic solution u∗ of the evolution equation ∂tu = F (u) is linearly stable iff the spectrum of theGateaux derivative dF (u∗) at u∗ of the map F (u) on the r.h.s. of (9.1), lies in the half-planeRe z < 0 and we say that u∗ is linearly unstable iff σ(dF (u∗)) ∩ Re z > 0 6= ∅.

To exhibit the origin of this definition, we write the equation (9.1) in the canonical form asfollows. First, we expand F (u) around u∗ and useg that F (u∗) = 0 to obtain

F (u∗ + ξ) = dF (u∗)ξ + f(ξ), (9.2)

where the term f(ξ) is defined by this expansion. According to (7.7) of Theorem 4.1, it is of ahigher order, f(ξ) = o(‖ξ‖). Substituting u = u∗ + ξ and this expansion into (9.1), we find

∂ξ

∂t= L∗ξ + f(ξ), (9.3)

where L∗ := dF (u∗). If the initial condition, u0, for (9.1) is close to u∗, then the initial condition,ξ0 = u0 − u∗, for (9.4) is close to 0. If u∗ is stable, then ξ remains small for all times. Thissuggests that f(ξ) which is of a higher order in ξ does not have significant effect on the behaviourof the solution and therefore can be omitted on the first step. This leads to the linear equation

∂ξ

∂t= L∗ξ. (9.4)

In a very large variety of situations, the spectrum of the linear operator L∗ determines the longtime behaviour of the latter equation (see Appendix E), which leads to the definition above.


9.2 Turing instability

Let f be a continuously differentiable function f : Rn → Rn, n ≥ 2, and let Ω ⊆ Rn be abounded open set with smooth boundary. We consider the following reaction-diffusion systemwith Neumann boundary conditions:

∂u∂t (x, t) = δ∆u(x, t) + f(u(x, t)) (x, t) ∈ Ω× (0,∞)∂u∂ν (x, t) = 0 (x, t) ∈ ∂Ω× (0,∞),

(9.5)

where δ is a positive definite n× n matrix (the diffusion tensor) and u : Ω× (0,∞)→ Rn.

Consider the associated kinetic system (the reaction equation)

∂u(t)

∂t= f(u(t)). (9.6)

Notice that static solutions to (9.6) are also static, homogeneous (i.e. x−independent) solutionsof (9.5) and, if a static, homogeneous solution to (9.6) is stable, then it is also a stable solutionof (9.5) under homogeneous perturbations.

Turing proposed that pattern formation in biological systems could arise through a processwhere a stable static solution of (9.6) became unstable in the presence of diffusion, i.e., the staticsolution is not stable as a solution of (9.5). This is called Turing or diffusion-driven instability.

Let u ≡ u∗ be a static homogeneous solution to (9.5), i.e. u∗ solves f(u∗) = 0. u = u∗ isalso a static solution of (9.6). If u(t) = u∗ is a stable equilibrium (static solution) of (9.6), thenu(x, t) ≡ u∗ is also a stable solution of (9.5) under homogeneous perturbations.

The problem then is: if u = u∗ is a stable solution of (9.6), what conditions on f , d, and Ωdetermine the stability of u = u∗ as a solution of (9.5), and in the case where u = u∗ becomesunstable, what can be said about the evolution of solutions which are initially close to u = u∗.We address this problem in this notes.

To simplify the notation, we take in what follows u∗ = 0. In this section we prove thefollowing

Theorem 9.1. Suppose n = 2 and δ is a diagonal matrix δ = diag (d1, d2). Let fij be the ij

entry of f ′(u∗), i.e. fij = ∂fi∂xj

(u∗).

1. If f11 ≤ 0, then u∗ is a linearly stable solution of (9.5).

2. If f11 > 0, then there exists ε such that u = 0 is linearly stable for d2 < εd1 and linearlystable for d2 > εd1, provided σ(−∆) ∩ (αδ, βδ) 6= ∅, for some 0 < αδ < βδ .

First we observe that −∆ is a self-adjoint operator on L2(Ω,C) (with the appropriate bound-ary conditions) and, since, Ω is bounded, it has a purely discrete spectrum with non-negativeeigenvalues, which we denote by 0 ≤ λ1 ≤ λ2 ≤ . . . , marching off to ∞, λn → ∞, as n → ∞.We define the n× n matrix

Lδ(λ) = −δλ+ f ′(u∗). (9.7)

We derive Theorem 9.1 from the following more general result

Theorem 9.2. Suppose n is even (n = 2) and and u(t) = u∗ is a stable equilibrium (staticsolution) of (9.6). Then the homogeneous static solution u∗ of (9.5) is a linearly stable, if andonly if detLδ(λ) > 0 for all λ ∈ σ(−∆).


Proof of Theorem 9.2. For the proof of the theorem, we will require a number of lemmas. Webegin investigating the linearization of the r.h.s. of (9.5) around u∗, which is

Lδ = δ∆ + f ′(u∗). (9.8)

We prove the following result on the spectrum of Ld.

Lemma 9.3.

σ(Lδ) =⋃

λ∈σ(−∆)

σ(Lδ(λ)). (9.9)

Proof. Let φλ λ∈σ(−∆) be a orthonormal basis of eigenfunctions of −∆. Then for any u : Ω→Rn, there exist vectors cλ ∈ Rn such that

u =∑λ

cλφλ.

We then have

(Lδ − z)u =∑λ

(Lδ(λ)− z) cλφλ.

Now, since Ω ⊆ Rm is a bounded open set with smooth boundary, Lδ has purely pointspectrum. Suppose that z ∈ σ(Ld). Then there is non-zero u such that (Lδ − z)u = 0. For eachµ ∈ σ(−∆), we multiply by φµ and integrate, to obtain, using the orthonormality of the basis,

0 =∑λ

(Lδ(λ)− z) cλ∫

Ωφµφλ = (Lδ(µ)− z) cµ,

which implies that either cµ = 0 or z is an eigenvalue of Lδ(λ). Since u is non-zero, not all cµare zero, and therefore z ∈ σ(Ld(µ)) for some µ.

Conversely, suppose that z ∈ σ(Lδ(λ)) for some λ. Fix non-zero c ∈ Rn such that Lδ(λ)c = 0.Then if we set u = cφλ, (Lδ − z)u = 0. Therefore z ∈ σ(Ld), and that completes the proof.

Lemma 9.4. Suppose that λ ≥ 0. Then TrLδ(λ) < 0.

Proof. We then have

TrLδ(λ) = −λTr δ + Tr f ′(0).

Since f ′(u∗) is negative definite, we know that Tr f ′(u∗) < 0. Since δ > 0 and therefore Tr δ > 0,this implies the result.

For n = 2, the equation TrLδ(λ) < 0 implies

Corollary 9.5. One eigenvalue of Lδ(λ) is always negative. For n = 2, the other eigenvalue isnegative iff detLδ(λ) > 0.

This corollary implies Theorem 9.2.


9.3 Appendix: Derivation of Theorem 9.1 from Theorem 9.2

Suppose n = 2 and δ is a diagonal matrix δ = diag (d1, d2). We have

detLδ(λ) = d1d2λ2 − (d1f22 + d2f11)λ+ det f ′(u∗) =: hδ(λ). (9.10)

This means that detLd(λ) > 0 if and only if hδ(λ) > 0. Hence Lδ(λ) is negative definite if andonly if hδ(λ) > 0.

We we begin with studying the function hδ(λ) and prove the following result.

Lemma 9.6.

1. If f11 ≤ 0, then for all d > 0, hδ(λ) > 0 for all λ ≥ 0.

2. If f11 > 0, then there exists δ > d1 with the following properties:

(a) If d2 < δ, hd(λ) > 0 for all λ ≥ 0.

(b) If d2 = δ, there exists λ0 > 0, such that hd(λ) = d1d2(λ− λ0)2.

(c) If d2 > δ, there exist λ2 > λ1 > 0, such that hd(λ) = d1d2(λ− λ1)(λ− λ2).

Proof. It is clear from the definition of hδ(λ), (9.10), that hδ(λ) has a unique global minimumpoint, λmin, which can be easily calculated to be

λmin =f11

2d1+f22

2d2. (9.11)

We let hmin be the global minimum value. A number of properties of hδ(λ) can also be deter-mined using its discriminant, D, which is

D = (d1f22 + d2f11)2 − 4d1d2 det f ′(0). (9.12)

Now suppose f11 ≤ 0. Since f11 + f22 < 0, we have f22 < −f11. If f11 = 0, then λmin < 0 forall d > 0. Now hd(0) = det f ′(u∗) > 0, so it follows that for all λ > 0 > λmin, hδ(λ) ≥ hδ(0) > 0.So suppose f11 < 0. Fix d1 and set ε = −f22

f11d1. Then if d2 > ε, then again λmin < 0 and the

same argument gives hδ(λ) > 0 for all λ ≥ 0. Suppose then that d2 < ε. We will show thathmin > 0, or equivalently that D < 0. Since d2 < ε, d2

2f211 < d2

1f222. Also, since det f ′(0) > 0, we

have f12f21 < f11f22. Therefore,

D = d21f

222 + d2

2f211 − 2d1d2f11f22 + 4d1d2f12f21

< 2d21f

222 + 2d1d2f11f22

< 2d21f

222 − 2d2

1f222 = 0.

That establishes (1).

Now suppose f11 > 0. Then f22 < −f11 < 0. Set ε = −f22f11d1, so ε > d1. For d2 ≤ ε, one can

check that λmin ≤ 0, and therefore the same argument as before shows that hδ(λ) > 0 for allλ ≥ 0.

For d2 > ε, on the other hand, λmin > 0. As can be seen from above, the discriminant Discis a quadratic polynomial in d2 and D → ∞ as d2 → ∞. Now for d2 = ε, a simple calculationshows that

D = −4d1d2 det f ′(0) < 0.


This means there exist a α > ε such that D < 0 for ε ≤ δ < α, D = 0 at d2 = α, and D > 0 ford2 > α. One can calculate that

α =1

f211

[2√−f12f21 Tr f ′(u∗) + Tr f ′(u∗)− f12f21

]d1. (9.13)

This then implies (2), except for the assertion that λ0, λ1, λ2 > 0, but this follows from the factthat λmin > 0 and hδ(0) > 0.

We now turn to the proof of Theorem 9.1.

Suppose first that f11 ≤ 0. Then we have seen that for all δ, hδ(λ) > 0 for all λ ≥ 0. Thismeans that for all λ ≥ 0, Lδ(λ) is negative definite. Now if λ ∈ σ(−∆), then λ ≥ 0, so thismeans that the eigenvalues of Ld lie in the plane Re z < 0 , and therefore u∗ is linearly stable.

Now suppose that f11 > 0 and let ε be the following (from (9.13)):

ε =1

f211

[2√−f12f21 Tr f ′(u∗) + Tr f ′(u∗)− f12f21

].

Then if d2 < εd1, an argument similar to the one above, shows that u = 0 is stable. If d2 > εd1,on the other hand, then there are λ ≥ 0, for which hδ(λ) < 0. If there is λ ∈ σ(−∆) such thathδ(λ) < 0, then Lδ(λ) has an eigenvalue with positive real part, and therefore so does Lδ, whichmeans that u∗ is not linearly stable.

9.4 Pattern formation in reaction-diffusion equations

(UNDER CONSTRUCTION) Recall that for n is even (n = 2), the homogeneous staticsolution u∗ of (9.5) is a linearly stable, if and only if detLδ(λ) > 0 for all λ ∈ σ(−∆).

Theorem 9.7. Suppose n is even (n = 2) and and u(t) = u∗ is a stable equilibrium (staticsolution) of (9.6). Let δ∗ be the smallest δ at which detLδ(λ1), where λ1 is the smallest eigen-value of −∆, vanishes. Then a non-homogeneous static solution to (9.6) bifurcates at from thehomogeneous one. At the bifurcation point, the solution u∗ looses its stability, while the new non-homogeneous static solution is stable (the exchange of the stability). The bifurcating solution isof the form ???.

Proof. Define F (δ, u) = δ∆u+ f(u). Then the static equation for (9.5) can be written as

F (δ, u) = 0. (9.14)

Suppose n = 2 and δ is a diagonal matrix δ = diag (d1, d2). Let d1 be fixed for simplicity’s sake.Write d2 = αd1. We have

1. (9.14) has a trivial branch: F (δ, 0) = 0 for all δ > 0.

2. There exists α ≥ ε, such that Lδ := duF (δ, 0)∣∣d2=αd1

has a 0 eigenvalue for some δ. The

latter is s.t. 0 is an eigenvalue of Lδ(λ) for some λ ∈ σ(−∆).

3. The multiplicity of the zero eigenvalue of Lδ = the multiplicity of λ as an eigenvalue of−∆.

4. Fix any non-zero u in the nullspace of Lδ and non-zero v in the nullspace of L∗. Then〈v, dd2,uF (d1, αd1, 0)u〉 6= 0.

We prove these properties. We know that for d2 > εd1, λmin > 0 and hmin < 0. We also knowthat λmin →∞ as d2 →∞, so therefore there is a smallest d2 such that detLδ(λ) = hδ(λ) = 0


for some λ ∈ σ(−∆). Since hδ(λ) has two roots, it is possible for there to be two such λ, but inthe generic case, there will only be one and we assume that from now on.

This means that 0 is an eigenvalue of Lδ(λ), which, since n = 2, has at most one eigenvaluein Re z ≥ 0, so it has multiplicity 1.

Now 0 is an eigenvalue of Lδ if and only if it is an eigenvalue of Lδ(λ) for some λ ∈ σ(−∆).To show that the multiplicity of the 0 eigenvalue of Lδ is equal to the multiplicity of λ as aneigenvalue of −∆, let φ1, . . . , φn be the n eigenfunctions corresponding to λ. Then the nullspaceof Lδ is generated by cφ1, . . . , cφn , where c is any non-zero vector in the nullspace of Lδ(λ).

First we show that if c ∈ R2 is such that Lδ(λ)c = 0, then c2 6= 0. Suppose, on the contrary,that c2 = 0. Then we have f21c1 = 0, so f21 = 0. This implies that 0 < det f ′(u∗) = f11f22, andsince f11 > 0, this implies f22 > 0. But that would mean Tr f ′(u∗) = f11 + f22 > 0 and thatcontradiction proves c2 6= 0.

A similar proof shows that if c ∈ R2 is such that Lδ(λ)∗c = 0, then d2 6= 0.

Now, a simple calculation gives dd2,uF (δ, u) =

(0 00 d2∆

), so dd2,uF (d1, αd1, u) =

(0 00 αd1∆

).

Next, u = a1cφ1 + · · ·+ ancφn and v = b1dφ1 + · · ·+ bndφn. So

〈v, dd2,uF (d1, αd1, 0)u〉 = c2d2〈a1φ1 + · · ·+ anφn, b1φ1 + · · ·+ bnφn〉= c2d2(a1b1 + · · ·+ anbn),

which can at least be made to not be zero. The four properties above imply the theorem.

Literature. [51] found conditions on f ′(0) and d for u = 0 to be unstable or a linearly stable.[48] have Ω ⊂ R. They find conditions for when Turing instability can and cannot occur, seealso [52]. For a textbook exposition see [44].

9.5 Asymptotic stability of homogeneous and bifurcating soltions

(under constriction)

10 Abrikosov lattices

In this section we prove existence of the Abrikosov lattices defined in Section 3.2. Recallthat equilibrium configurations of superconductors, and of the U(1) Higgs model from parti-cle physics, are described by a system of nonlinear PDE called the Ginzburg-Landau equations:

∆AΨ = κ2(|Ψ|2 − 1)Ψ, (10.1a)

curl∗ curlA = Im(Ψ∇AΨ). (10.1b)

for (Ψ, A) : R2 → C× R2, ∇A = ∇− iA, and ∆A = ∇2A, the covariant derivative and covariant

Laplacian, respectively. We discussed the origin of these equations and the physical meaning ofthe unknowns (Ψ, A) in Section 3.2.

After Abrikosov, we look for solutions (Ψ, A) defined on all of R2, whose all physical prop-erties, namely the density of superconducting pairs of electrons, ns := |Ψ|2, the magnetic field,B := curlA, and the current density, J := Im(Ψ∇AΨ), are doubly-periodic with respect to somelattice L. We call such states Abrikosov lattice states.

Below we use the bifurcation theory developed above to show existence of such solutions.The branch of ‘trivial’ solutions they bifurcate from are the normal (or non-superconducting)


solutions, uN ≡ ub := (ΨN ≡ 0, AN ≡ Ab), where Ab is a vector potential with constantmagnetic field curlAb =: b = constant. (The bifurcation parameter b will come through thespace on which the equations are defined, not through the equations themselves.) (There isanother simple, homogeneous solution, describing the perfect superconductor solution, namelyuS := (ΨS ≡ 1, AS ≡ 0).)

Results. Here we formulate a simplified version of the result. The general result can be foundin [?].

Theorem 10.1. For every lattice L satisfying 0 < κ2 − b 1, where b = 2π|Ω| , the equations

(10.1) have an Abrikosov lattice solution with this lattice in a neighbourhood of the branch ofnormal solutions.

Before proceeding to the proof we present some general discussion of properties of the equa-tions (10.1) and of some general notions.

Symmetries. As we already mentioned in Section 3.2, the Ginzburg-Landau equations (10.1)admit several symmetries, that is, transformations which map solutions to solutions.

Gauge symmetry: for any sufficiently regular function η : R2 → R,

Γγ : (Ψ(x), A(x)) 7→ (eiη(x)Ψ(x), A(x) +∇η(x)); (10.2)

Translation symmetry: for any h ∈ R2,

Th : (Ψ(x), A(x)) 7→ (Ψ(x+ h), A(x+ h)); (10.3)

Rotation and reflection symmetry: for any R ∈ O(2) (including the reflections

f(x)→ f(−x))

TR : (Ψ(x), A(x)) 7→ (Ψ(Rx), R−1A(Rx)). (10.4)

Equivalence classes of solutions, gauge conditions.

L-equivariant states. One can show that a pair (Ψ, A) is an Abrikosov (vortex) lattice iff itis mapped by lattice traslations into a gauge equivalent pair: there are, in general, multi-valueddifferentiable functions, gs : R2 → R, s ∈ L, s.t.

Ψ(x+ t) = eigt(x)Ψ(x) and A(x+ t) = A(x) +∇gt(x). (10.5)

Indeed, if state (Ψ, A) satisfies (10.5), then all associated physical quantities are L−periodic, i.e.(Ψ, A) is an Abrikosov lattice. In the opposite direction, if (Ψ, A) is an Abrikosov lattice, thencurlA(x) is periodic w.r.to L, and therefore A(x+s) = A(x)+∇gs(x), for some functions gs(x).Next, we write Ψ(x) = |Ψ(x)|eiφ(x). Since |Ψ(x)| and J(x) = |Ψ(x)|2(∇φ(x)−A(x)) are periodicw.r.to L, we have that ∇φ(x+s) = ∇φ(x)+∇gs(x), which implies that φ(x+s) = φ(x)+gs(x),where gs(x) = gs(x) + cs, for some constants cs.

We call a pair satisfying (10.5) the lattice, or L-gauge-periodic state. In terminology ofSection 3.2 it is an equivariant state w.r. to the group of lattice translations for a lattice L.

Since T transs is a commutative group, we see that the family of functions gs has the important

cocycle property

gs+t(x)− gs(x+ t)− gt(x) ∈ 2πZ. (10.6)


This can be seen by evaluating the effect of translation by s + t in two different ways. We callgs(x) the gauge exponent.

Remark. Relation (10.6) for Abrikosov lattices was isolated in [?], where it played an importantrole. This condition is well known in algebraic geometry and number theory where eigs(x) is calledthe automorphy factor (see e.g. [?]). However, there the associated vector potential (connectionon the corresponding principal bundle) A is not considered there.

Flux quantization. For a lattice L ⊂ R2, we denote by ΩL and |ΩL| the basic lattice cell andits area, respectively. The important property of lattice states is that the magnetic flux througha lattice cell is quantized: ∫

ΩcurlA = 2πn (10.7)

for some integer n. Indeed, if |Ψ| > 0 on the boundary of the cell, we can write Ψ = |Ψ|eiθand 0 ≤ θ < 2π. The periodicity of ns and J ensure the periodicity of ∇θ − A and thereforeby Green’s theorem,

∫Ω curlA =

∮∂ΩA =

∮∂Ω∇θ and this function is equal to 2πn since Ψ is

single-valued.

Equation (10.7) then implies the relation between the average magnetic flux, b, per latticecell, b = 1

|Ω|∫

Ω curlA, and the area of a fundamental cell

b =2πn

|Ω|. (10.8)

We note that due to the reflection symmetry of the problem we can assume that b ≥ 0.

10.1 Fixing the gauge

The gauge symmetry allows one to fix solutions to be of a desired form. Let Ab(x) = b2Jx, where

J is the symplectic matrix

J =

(0 −11 0

). (10.9)

We will use the following proposition

Proposition 10.2. Let (Ψ, A′) be an L-lattice state, i.e. it satisfies (10.5) and let b be theaverage magnetic flux per cell. Then there is a L-lattice state (Ψ, A), that is gauge-equivalent toa translation of (Ψ, A), such that

Ψ(x+ s) = eib2x·JsΨ(x) and A(x+ s) = A(x) +

b

2Js, ∀s ∈ L. (10.10)

Moreover, we have for α := A−Ab,

(i) α(x+ s) = α(x) for all s ∈ L;

(ii) α has mean zero:∫

Ω α = 0;

(iii) α is divergence-free: divα = 0.

A proof of this proposition is given in an appendix at the end of this section.

From now on we assume that (Ψ, A) satisfy (10.10).


By identifying R2 with C, any lattice L can be given a basis ν1, ν2 such that the complexnumber τ = ν2

ν1satisfies Imτ > 0. τ will be called the shape parameter of the lattice. By rotating

L, if necessary, we can bring it to the form

Lω = r(Z + τZ), where ω = (τ, r), with τ ∈ C, Imτ > 0, and r > 0. (10.11)

By the quantization condition (10.8), r :=√

2πnImτb . We will say that a gauge-periodic pair

(Ψ, A) is of type ω = (τ, r), if the underlying lattice is Lω.

10.2 The linear problem

To find candidates for bifurcation points we have solve the linear problem:

−∆Abψ0 = λψ0, (10.12)

for ψ satisfying the gauge - periodic boundary condition

ψ0(x+ s) = eib2s·Jxψ0(x), ∀s ∈ Lω. (10.13)

This quasiperiodic boundary condition is consistent with the fact that ψ0 is a single valuedfunction if and only if the magnetic flux, b|Ω|, through the fundamental cell Ω is quantized:b|Ω| = 2πn, for some integer n.

We consider this eigenvalue problem above on the Sobolev space of order two, H2(Ω,C),whose elements satisfy the quasiperiodic boundary condition (10.13). We identify R2 with C inthe usual way as x = (x1, x2)↔ z = x1 + x2. The key result here is

Theorem 10.3. The smallest λ for which the problem (10.12) - (10.13) with the quantizationcondition b|Ω| = 2πn has a non-trivial solution is λ = b. In this case the solutions are of theform

ψ0 (z) = e−b4

(|z|2−z2)θ(z′), z′ :=1

rz, (10.14)

where Θ is an n− theta function, i.e. an entire function satisfying

θ(z + 1) = θ(z), (10.15a)

θ(z + τ) = e−2πinze−iπnτθ(z) (10.15b)

(θ is the theta function, see [42]), and therefore span an n−dimensional vector space. Here n isthe integer entering the quantization of the flux condition, b|Ω| = 2πn.

Proof. Let L denote the operator −∆Ab on the space L2(Ω,C), with the domain consisting offunctions from H2(Ω,C) satisfying the boundary conditions (10.13). By a standard result, itis self-adjoint. Spectral information about L can be obtained by introducing the complexifiedcovariant derivatives (harmonic oscillator annihilation and creation operators), ∂Ab and ∂∗

Ab=

−∂Ab , with

∂Ab := (∇Ab)1 + i(∇Ab)2 = ∂x1 + i∂x2 +1

2bx1 +

1

2ibx2. (10.16)

One can verify that these operators satisfy the following relations:

1. [∂Ab , ∂∗Ab

] = 2 curlAb = 2b;

2. −∆Ab − b = ∂∗Ab∂Ab .


(As for the harmonic oscillator (see citeGS), this gives σ(L) = (2k + 1)b : k = 0, 1, 2, . . . .)The second property implies

null(L− b) = null ∂Ab . (10.17)

We find null ∂Ab . A simple calculation gives the following operator equation

eb2

(ix1x2−x22)∂Abe− b

2(ix1x2−x22) = ∂x1 + i∂x2 .

(The transformation on the l.h.s. is highly non-unique.) This immediately proves that Ψ ∈null ∂Ab if and only if ξ(x) = e

b2

(ix1x2−x22)ψ(x) satisfies ∂x1ξ + i∂x2ξ = 0.

For z′ = 1r (x1 + ix2), we define

θ(z′) = eb2

(ix1x2−x22)φ(x). (10.18)

By the above, the function θ is entire and, due to the periodicity conditions on φ, satisfies(10.15).

To complete the proof, we now need to show that the space of the analytic functions whichsatisfy these relations form a vector space of dimension n. By the first relation, θ has theabsolutely convergent Fourier expansion

θ(z) =

∞∑k=−∞

cke2πkiz. (10.19)

The second relation, on the other hand, leads to a relation for the coefficients of the expansion.Namely, we have ck+n = einπτe2kiπτck and that means such functions are determined solely bythe values of c0, . . . , cn−1 and therefore form an n-dimensional vector space. This completes theproof of Theorem 10.4.

Abrikosov considered the case n = 1. In this case, the space (10.14) is one-dimensional andspanned by the function

ψ := ei2x2(x1+ix2)

∞∑k=−∞

ckeik√

2πImτ(x1+ix2), ck = ceikπτk−1∏m=1

ei2mπτ . (10.20)

This is the leading approximation to the Abrikosov lattice solution. The normalization coefficientc cannot be found from the linear theory and is obtained by taking into account nonlinear termsby perturbation theory.

10.3 Rescaling.

Suppose, that we have a Lω-lattice state (Ψ, A), where ω = (τ, r). We now define the rescaledfields (ψ, a) to be

(ψ(x), a(x)) := (rΨ(rx), rA(rx)).

Let Lτ be the lattice spanned by 1 and τ , i. e. Lτ := Z + τZ, with Ωτ being a primitive cell ofthat lattice. We note that |Ωτ | = 2πn. We summarize the effects of the rescaling above:

(A) (ψ, a) is a Lτ -lattice state.


(B) Ψ and A solve the Ginzburg-Landau equations if and only if ψ and a solve

(−∆a − λ)ψ = −κ2|ψ|2ψ, (10.21a)

curl∗ curl a = Im(ψ∇aψ) (10.21b)

for λ = κ2r2. The latter equations are valid on Ωτ with the boundary conditions given inthe next statement.

(C) If (Ψ, A) is of the form described in Proposition 10.2, then (ψ, a) satisfies

ψ(x+ t) = eikn2x·Jtψ(x), (10.22)

where k = k(τ) := r2b/n = 2πImτ , and, with a = an + α, an(x) := kn

2 Jx,

α(x+ t) = α(x), ∀t ∈ Lτ , and

∫Ωτα = 0, divα = 0. (10.23)

Our problem then is, for each n = 1, 2, . . ., find (ψ, a), solving the rescaled Ginzburg-Landauequations (10.21) and satisfying (i).

10.4 Reformulation of the problem

In this section we reduce two equations (10.21) for ψ and a to a single equation for ψ. Weintroduce the spaces Ln(τ) := L2(Ωτ ,C) and ~L (τ) := a ∈ L2(Ωτ ,R2) | 〈a〉 = 0, div a = 0, inthe distributional sense and the Sobolev spaces of order two: Hn(τ) and ~H (τ), whose elements,ψ and α, satisfy the quasiperiodic boundary condition (10.22) and (10.23), respectively. Definethe operators

Ln := −∆an and M := curl∗ curl, (10.24)

on the spaces Ln(τ) and ~L (τ), with the domains being . Their properties that will be usedbelow are summarized in the following propositions:

Proposition 10.4. Ln is a self-adjoint operator on Hn(τ) with spectrum σ(Ln) = (2m+1)kn :m = 0, 1, 2, . . . and dimC null(Ln − kn) = n, where recall k = k(τ) = 2π

Imτ .

Proposition 10.5. M is a strictly positive operator on ~H (τ) with discrete spectrum.

The proofs of these results are standard and, for the convenience of the reader, are givenbelow.

Proof of Proposition 10.5. The fact that M is positive follows immediately from its definition.We note that its being strictly positive is the result of restricting its domain to elements havingmean zero.

Proof of Proposition 10.4. First, we note that Ln is clearly a positive self-adjoint operator. Tosee that it has discrete spectrum, we first note that the inclusion H2 → L2 is compact forbounded domains in R2 with Lipschitz boundary (which certainly includes lattice cells). Thenfor any z in the resolvent set of Ln, (Ln− z)−1 : L2 → H2 is bounded and therefore (Ln− z)−1 :L2 → L2 is compact. In fact, the operator Ln is unitary equivalent (by rescaling) to k times theoperator L, whose the spectrum was found explicitly in the previous section. This completesthe proof of Proposition 10.4.


Our goal now is to solve the equation (10.21b) for α and, substituting the solution intothe equation (10.21a), find an equation containing only ψ. First we reformulate the equations(10.21), by substituting a = an + α, to obtain

(Ln − λ)ψ + 2iα · ∇anψ + |α|2ψ + κ2|ψ|2ψ = 0, (10.25a)

(M + |ψ|2)α− Im(ψ∇anψ) = 0. (10.25b)

In Appendix 10.8 at the end of this section, we solve the second equation, (10.25b), for α interms of ψ. We write the result as α = α(ψ), where

α(ψ) = (M + |ψ|2)−1Im(ψ∇anψ). (10.26)

We collect the elementary properties of the map α in the following proposition, where we identify

Hn(τ) with a real Banach space using ψ ↔−→ψ := (Reψ, Imψ).

Proposition 10.6. The unique solution, α(ψ), of (10.25b) maps Hn(τ) to ~H (τ) and has thefollowing properties:

(a) α(·) is analytic as a map between real Banach spaces.

(b) α(0) = 0.

(c) For any δ ∈ R, α(eiδψ) = α(ψ).

Proof. The only statement that does not follow immediately from the definition of α is (a). Itis clear that Im(ψ∇anψ) is real-analytic as it is a polynomial in ψ and ∇ψ, and their complexconjugates. We also note that (M − z)−1 is complex-analytic in z on the resolvent set of M ,and therefore, (M + |ψ|2)−1 is analytic. (a) now follows.

Now we substitute the expression (10.26) for α into (10.25a) to get a single equation

F (λ, ψ) = 0, (10.27)

where the map F : R×Hn(τ)→ Ln(τ) is defined as

F (λ, ψ) = (Ln − λ)ψ + 2iα(ψ) · ∇anψ + |α(ψ)|2ψ + κ2|ψ|2ψ. (10.28)

For a map F (λ, ψ), we denote by ∂ψF (λ, φ) its Gateaux derivative in ψ at φ. The followingproposition lists some properties of F .

Proposition 10.7.

(a) F is analytic as a map between real Banach spaces,

(b) for all λ, F (λ, 0) = 0,

(c) for all λ, ∂ψF (λ, 0) = Ln − λ,

(d) for all δ ∈ R, F (λ, eiδψ) = eiδF (λ, ψ).

(e) for all ψ, 〈ψ,F (λ, ψ)〉 ∈ R.


Proof. The first property follows from the definition of F and the corresponding analyticity ofa(ψ). (b) through (d) are straightforward calculations. For (e), we calculate that

〈ψ,F (λ, ψ)〉 = 〈ψ, (Ln − λ)ψ〉+ 2i

∫Ωτψα(ψ) · ∇ψ

+ 2

∫Ωτ

(α(ψ) · an)|ψ|2 +

∫Ωτ|α(ψ)|2|ψ|2 + κ2

∫Ωτ|ψ|4.

The final three terms are clearly real and so is the first because Ln − λ is self-adjoint. For thesecond term we calculate the complex conjugate and see that

2i

∫Ωτψα(ψ) · ∇ψ = −2i

∫Ωτψα(ψ) · ∇ψ = 2i

∫Ωτ

(∇ψ · α(ψ))ψ,

where we have integrated by parts and used the fact that the boundary terms vanish due tothe periodicity of the integrand and that divα(ψ) = 0. Thus this term is also real and (e) isestablished.

10.5 Reduction to a finite-dimensional problem

In this section we reduce the problem of solving the equation F (λ, ψ) = 0 to a finite dimensionalproblem. We address the latter in the next section. We use the standard method of Lyapunov-Schmidt reduction. Let X := Hn(τ) and Y := Ln(τ) and let K = null(Ln − kn). We let P bethe Riesz projection onto K, that is,

P := − 1

2πi

∮γ(Ln − z)−1 dz, (10.29)

where γ ⊆ C is a contour around n that contains no other points of the spectrum of Ln. Thisis possible since n is an isolated eigenvalue of Ln. P is a bounded, orthogonal projection, and ifwe let Z := nullP , then Y = K ⊕ Z. We also let Q := I − P , and so Q is a projection onto Z.

The equation F (λ, ψ) = 0 is therefore equivalent to the pair of equations

PF (λ, Pψ +Qψ) = 0, (10.30)

QF (λ, Pψ +Qψ) = 0. (10.31)

We will now solve (10.31) for w = Qψ in terms of λ and v = Pψ. To do this, we introducethe map G : R×K ×Z → Z to be G(λ, v, w) := QF (λ, v +w). Applying the Implicit FunctionTheorem to G, we obtain a function w : R×K → Z, defined on a neighbourhood of (n, 0), suchthat w = w(λ, v) is a unique solution to G(λ, v, w) = 0, for (λ, v) in that neighbourhood. Thissolution has the following properties

w(λ, v) real-analytic in (λ, v); (10.32)

‖w‖ = O(‖v‖3) and ‖∂λw‖ = O(‖v‖3), (10.33)

where the norms are in the space Hn(τ). The last property follows from (10.26) and (10.28)together with (10.31). (Indeed, by (10.26), we have ‖α(ψ)‖H2 . ‖ψ‖2H2 and therefore combining(10.28) with (10.31) and using (as above) that the product of Hn(τ) functions is again a Hn(τ)function (and the norms are bounded correspondingly), one concludes that (10.33) holds.)


We substitute the solution w = w(λ, v) into (10.30) and see that the latter equation in aneighbourhood of (n, 0) is equivalent to the equation (the bifurcation equation)

γ(λ, v) := PF (λ, v + w(λ, v)) = 0. (10.34)

Note that γ : R ×K → C. We have shown that in a neighbourhood of (n, 0) in R ×X, (λ, ψ)solves F (λ, ψ) = 0 if and only if (λ, v), with v = Pψ, solves (10.34). Moreover, the solution ψof F (λ, ψ) = 0 can be reconstructed from the solution v of (10.34) according to the formula

ψ = v + w(λ, v). (10.35)

Finally we note that w and γ inherit the symmetry of the original equation:

Lemma 10.8. For every δ ∈ R, w(λ, eiδv) = eiδw(λ, v) and γ(λ, eiδv) = eiδγ(λ, v).

Proof. We first check that w(λ, eiδv) = eiδw(λ, v). We note that by definition of w,

G(λ, eiδv, w(λ, eiδv)) = 0,

but by the symmetry of F , we also have G(λ, eiδv, eiδw(λ, v)) = eiδG(λ, v, w(λ, v)) = 0. Theuniqueness of w then implies that w(λ, eiδv) = eiδw(λ, v). We can now verify that

γ(λ, eiδv) = PF (λ, eiδv + w(λ, eiδv))

= eiδPF (λ, v + w(λ, v))〉 = eiδγ(λ, v).

Solving the bifurcation equation (10.34) is a subtle problem, unless n = 1. In the latter case,this is done in the next section.

10.6 Proof of Theorem 10.1

In this section we look at the case n = 1, and look for solutions near the trivial solution. Recallthat by Theorem 10.3, the nullspace of the operator Ln−kn, where k = 2π

Imτ , acting on Hn(τ) isa one-dimensional complex subspace for n = 1. We denote a0 = an=1 and drop the (super)indexn = 1 from the notation Ln. We begin with the following result which gives the existence anduniqueness of the Abrikosov lattices.

Theorem 10.9. For every τ there exist ε > 0 and a branch, (λs, ψs, αs), s ∈ [0,√ε), of nontrivial

solutions of the rescaled Ginzburg-Landau equations (10.21), unique modulo the global gaugesymmetry (apart from the trivial solution (1, 0, a0)) in a sufficiently small neighbourhood of(1, 0, a0) in R×H (τ)× ~H (τ), and such that

λs = k + gλ(s2),

ψs = sψ0 + sgψ(s2),

as = a0 + ga(s2),

(10.36)

where k = 2πImτ , (L−k)ψ0 = 0, gψ is orthogonal to null(L−k), gλ : [0, ε)→ R, gψ : [0, ε)→H (τ),

and gα : [0, ε)→ ~H (τ) are real-analytic functions such that gλ(0) = 0, gψ(0) = 0, gα(0) = 0.


Proof. The proof of this theorem is a slight modification of a standard result from bifurcationtheory. Our goal is to solve the equation (10.34) for λ. Since the projection P , defined there, isrank one and self-adjoint, we have

Pψ =1

‖ψ0‖2〈ψ0, ψ〉ψ0, with ψ0 ∈ null ∂ψF (λ0, 0). (10.37)

We can therefore view the function γ in the bifurcation equation (10.34) as a map γ : R×C→ C,where

γ(λ, s) = 〈ψ0, F (λ, sψ0 + w(λ, sψ0)〉. (10.38)

We now show that γ(λ, s) ∈ R. Since the projection Q is self-adjoint, Qw(λ, v) = w(λ, v),w(λ, v) solves QF (λ, v + w) = 0 and v = sψ0, we have

〈w(λ, sψ0), F (λ, sψ0 + w(λ, sψ0))〉 = 〈w(λ, sψ0), QF (λ, sψ0 + w(λ, sψ0))〉 = 0.

Therefore, for s 6= 0,

〈ψ0, F (λ, sψ0 + Φ(λ, sψ0))〉 = s−1〈sψ0 + w(λ, sψ0), F (λ, sψ0 + w(λ, sψ0))〉,

and this is real by property (e) of Proposition 10.7. Thus, since by Lemma 10.8, γ(λ, s) =ei arg sγ(λ, |s|), it therefore suffices to solve the equation

γ0(λ, s) = 0 (10.39)

for the restriction γ0 : R × R → R of the function γ to R × R, i.e. for real s. Since by (10.33),w(λ, sψ0) = O(s2) and therefore (10.39) has the trivial branch of solutions s ≡ 0 for all λ. Hencewe factorize γ0(λ, s) as γ0(λ, s) = sγ1(λ, s), i.e. we define the function

γ1(λ, s) := s−1γ0(λ, s), if s > 0, and = 0 if s = 0, (10.40)

and solve the equation γ1(λ, s) = 0 for λ. The definition of the function γ1(λ, s) and (10.32) implythat it has the following properties: γ1(λ, s) is real-analytic, γ1(λ,−s) = γ1(λ, s), γ1(1, 0) = 0and, by (10.28) and (10.33), ∂λγ1(1, 0) = −‖ψ0‖2 6= 0. Hence by a standard application of theImplicit Function Theorem, there is ε > 0 and a real-analytic function φλ : (−

√ε,√ε)→ R such

that φλ(0) = 1 and λ = φλ(|s|) solves the equation γ(λ, s) = 0 with |s| <√ε.

We also note that because of the symmetry, φλ(−|s|) = φλ(|s|), φλ is an even real-analyticfunction, and therefore must in fact be a function solely of s2. We therefore set φλ(s) = φλ(

√s)

for s ∈ [0, ε), and so φλ is real-analytic.

We now define gψ : [0, ε)→H (τ) to be

gψ(s) =

1√sw(φλ(s),

√sψ0) s 6= 0,

0 s = 0,(10.41)

It is easily check that gψ is real-analytic and satisfies sgψ(s2) = w(φλ(s), sψ0) for any s ∈ [0,√ε).

Now, we know that there is a neighbourhood of (1, 0) in R×H (τ) such that in this neighbour-hood F (λ, ψ) = 0 if and only if γ(λ, s) = 0 where Pψ = sψ0. By taking a smaller neighbourhoodif necessary, we have proven that F (λ, ψ) = 0 in this neighbourhood if and only if either s = 0or λ = φλ(s2). If s = 0, we have ψ = sψ0 + w(φλ(s), sψ0) = 0, which gives the trivial solution.In the other case, ψ = sψ0 + w(φλ(s), sψ0) = sψ0 + sgψ(s2).


If we now also define gλ(s) = 1 − φλ(s), then the above gives us a neighbourhood of (1, 0)in R ×H (τ) such that the only non-trivial solutions of the equation (10.27) are given by thefirst two equations in (10.36). We now define the function ga(s) = α(ψs), where, recall, α(ψ)is defined in (10.26) and ψs = sψ0 + sgψ(s2) was found above. This function is real-analyticand satisfies ga(−s) = α(−ψs) = ga(s), and therefore is really a function of s2, ga(s

2). Defineas = a0 + ga(s

2). Then (λs, ψs, αs), s ∈ [0,√ε), solves the rescaled Ginzburg-Landau equations

(10.21).

For ψ0 a non-zero element in the nullspace null(L− k), we define the function of τ as

β(τ) :=〈|ψ0|4〉〈|ψ0|2〉2

, (10.42)

which we call this function the Abrikosov function.

Simple computations give the following expression for the derivative g′λ(0) at 0 of the functiongλ(s2) defined in (10.36)

g′λ(0) =2π

Imτ

[(κ2 − 1

2

)β(τ) +

1

2

]〈|ψ0|2〉. (10.43)

Note that the definition λ = κ2r2 (n = 1), the first equation (10.36) and the relation (10.43)imply that for (κ2 − 1

2)β(τ) + 12 ≥ 0, the bifurcated solution exists for b ≤ κ2, and for (κ2 −

12)β(τ)+ 1

2 < 0, it exists for b > κ2. Thus Theorem 10.9, after rescaling to the original variables,implies Theorem 10.1.

Remark. The proof of Theorem 10.9 gives in fact the following abstract result.

Theorem 10.10. Let X and Y be complex Hilbert spaces, with X a dense subset of Y , andconsider a map F : R×X → Y that is analytic as a map between real Banach spaces. Supposethat for some λ0 ∈ R, the following conditions are satisfied:

(1) F (λ, 0) = 0 for all λ ∈ R,

(2) ∂ψF (λ0, 0) is self-adjoint and has an isolated eigenvalue at 0 of (geometric) multiplicity 1,

(3) For non-zero ψ0 ∈ null ∂ψF (λ0, 0), 〈ψ0, ∂λ,ψF (λ0, 0)ψ0〉 6= 0,

(4) For all α ∈ R, F (λ, eiαψ) = eiαF (λ, ψ).

(5) For all ψ ∈ X, 〈ψ,F (λ, ψ)〉 ∈ R.

Then (λ0, 0) is a bifurcation point of the equation F (λ, ψ) = 0, in the sense that there is a familyof non-trivial solutions, (λs, ψs), for s ∈ [0,

√ε), unique modulo the global gauge symmetry (apart

from the trivial solution (1, 0)) in a neighbourhood of (λ0, 0) in R × X. Moreover, this familyhas the form

λ = φλ(s2),

ψ = sψ0 + sφψ(s2).

Here ψ0 ∈ null ∂ψF (λ0, 0), and φλ : [0, ε) → R and φψ : [0, ε) → X are unique real-analyticfunctions, such that φλ(0) = λ0, φψ(0) = 0.


10.7 Appendix: Proof of Proposition 10.2

Proof of Proposition 10.2. We begin by defining the function B : R→ R to be

B(ζ) =1

r

∫ r

0curlA′(ξ, ζ) dξ.

It is clear that b = 1rτ2

∫ rτ20 B(ζ) dζ. A calculation shows that B(ζ + rτ2) = B(ζ).

We now define P = (P1, P2) : R2 → R2 to be

P (x) = (bx2 −∫ x2

0B(ζ) dζ,

∫ x1

τ1τ2x2

curlA′(ξ, x2) dξ +τ ∧ xτ2

B(x2)).

A calculation shows that P is doubly-periodic with respect to L.

We now define η′ : R2 → R to be

η′(x) =b

2x1x2 −

∫ x1

0A′1(ξ, 0) dξ −

∫ x2

0A′2(x1, ζ)− P2(x1, ζ) dζ.

η′ satisfies ∇η′ = −A′ + A0 + P and let η′′ be a doubly-periodic solution of the equation∆η′′ = −divP . Also let C = (C1, C2) be given by

C = − 1

|Ω|

∫Ω

(P +∇η′′) dx,

where Ω is any fundamental cell, and set η′′′ = C1x1 + C2x2 and define η = η′ + η′′ + η′′′.

We claim that the pair (eηΨ′, A′+∇η) satisfies (10.10) and (i) - (iii) of the proposition. Letα := A − Ab = A′ + ∇η − Ab. We first note that α = P + ∇η′′ + C and by the above, it isperiodic. We also calculate that divα = divP + ∆η′′ = 0. Finally

∫Ω α =

∫Ω(P +∇η −C) = 0.

We note that, since α(x) and L−periodic, Ab(x) satisfies Ab(x + s) = Ab(x) + b2

(−s2

s1

)and

(−s2

s1

)= ∇(s ∧ x), we have that A = Ab + α satisfies

A(x+ s) = A(x) +b

2∇(s ∧ x), ∀s ∈ L.

Next, if (Ψ′, A′) satisfies condition (10.5) with the exponent g′s(x), then (eηΨ′, A′ + ∇η)satisfies condition (10.5) with the exponent gs(x) := g′s(x) + η(x + s) − η(x). We have shownabove that for our choice of η, ∇gs(x) = b

2∇(t ∧ x). Therefore gs(x) = b2s ∧ x + cs for some

constant cs. To establish (10.10), we need to have it so that cs = 0 for s = r, rτ . First let l besuch that r∧ l = − cr

b and rτ ∧ l = − crτb . This l exists as it is the solution to the matrix equation(

0 r−rτ2 rτ1

)(l1l2

)=

(− cr

b− crτ

b

),

and the determinant of the matrix is just r2τ2, which is non-zero because (r, 0) and rτ form abasis of the lattice. Let ζ(x) = b

2 l ∧ x. A straightforward calculation then shows that Ψ(x) :=

eiζ(x)Ψ′(x+ l) satisfies (10.10) and that α(x) := α′(x+ l) +∇ζ(x) still satisfies (i) through (iii).This proves the proposition.


10.8 Appendix: Solving the equation (10.25b)

(under construction)

11 Keller-Segel Equations of Chemotaxis

Chemotaxis is the directed movement of organisms in response to the concentration gradientof an external chemical signal and is common in biology. The chemical signals can come fromexternal sources or they can be secreted by the organisms themselves. The latter situation leadsto aggregation of organisms and to the formation of patterns.

Chemotaxis is believed to underly many social activities of micro-organisms, e.g. social motil-ity, fruiting body development, quorum sensing and biofilm formation. A classical example isthe dynamics and the aggregation of Escherichia coli colonies under starvation conditions [14].Another example is the Dictyostelium amoeba , where single cell bacterivores, when challengedby adverse conditions, form multicellular structures of ∼ 105 cells [12, 18]. Also, endothelial cellsof humans react to vascular endothelial growth factor to form blood vessels through aggregation[17].

Consider organisms moving and interacting in a domain Ω ⊆ Rd, d = 1, 2 or 3. Assumingthat the organism population is large and the individuals are small relative to the domain Ω,Keller and Segel derived a system of reaction-diffusion equations governing the organism densityρ : Ω× R+ → R+ and chemical concentration c : Ω× R+ → R+. The equations are of the form

∂tρ = Dρ∆ρ−∇ (f(ρ)∇c)∂tc = Dc∆c+ αρ− βc.

(11.1)

Here Dρ, Dc, α, β are positive functions of x and t, ρ and c, and f(ρ) is a positive functionmodelling chemotaxis (positive chemotaxis). Assuming a closed system, one is led to impose noflux boundary conditions on ρ and c:

∂νρ = 0 and ∂νc = 0 on ∂Ω, (11.2)

where ∂νg is the normal derivative of g.The equations (11.1) have the family of homogeneous static solutions, (ρ∗, c∗) : αρ∗−βc∗ = 0.

A simple calculation shows that these solutions are linearly unstable: the linearized operator hastwo spectral bands, one positive and the other negative (they are separated by the gap). Dueto the positive chemotaxis in the system, one expects that the system evolves to a non-uniformstate describing organism aggregation (the organisms secrete the chemical and move towardsareas of higher chemical concentration). One refers to this process as (chemotactic) collapse.

In this case, the density may become infinite and form a Dirac delta singularity. Moreprecisely, we say that a solution ρ(x, t) undergoes collapse at a point x0 ∈ Rd in finite timeT <∞ if it exists for 0 ≤ t < T and

limt↑T

∫|x−x0|≥ε

ρ(x, t) dx = 0, ∀ε > 0.

Since the total mass is conserved, this implies that limt↑T ρ(x0, t) =∞.The common approximations made in the literature for system (11.1) is based on the fact

that, in practically all situation, the coefficients in (11.1) are constant and satisfy

ε :=Dρ

Dc 1, α :=

α

Dc= O (1) and β :=

β

Dc 1. (11.3)


The first of these conditions says that the chemical diffuses much faster than the organisms do.As a result of this relation, one drops the ∂tc term in (11.1) (after rescaling time t→ t/Dρ, thisterm becomes ε∂tc). Furthermore, one takes f(ρ) to be a linear function f(ρ) = Kρ and neglectsthe term βc in (11.1) compared with αρ, as one expects that it would not effect the blow-upprocess where ρ 1 (it is also small due to the last relation in (11.3)). These approximations,after rescaling, lead to the system

∂ρ

∂t= ∆ρ−∇ · (ρ∇c) ,

0 = ∆c+ ρ,(11.4)

with ρ and c satisfying the no-flux Neumann boundary conditions. Equation (11.4) is thesimplest model of positive chemotaxis considered in the literature. This is the equation studiedin these lectures. We note that Eq. (11.4) in three dimensions also appear in the context of stellarcollapse (see [33, 53, 19, ?]); similar equations—the Smoluchowski or nonlinear Fokker-Planckequations—model non-Newtonian complex fluids (see [24, ?, 21, 22].

Properties Keller-Segel equations. Equations (11.4) have the following properties

• (11.4) preserves positivity: if the initial condition ρ0(x) is positive, then so is ρ(x, t) (bythe maximum principle).

• (11.4) is scaling invariant, that is, if a pair ρ(x, t) and c(x, t) is a solution to (11.4), thenfor any λ > 0 so is the pair

1

λ2ρ

(1

λx,

1

λ2t

)and c

(1

λx,

1

λ2t

). (11.5)

• With the no flux boundary conditions, the total number of organisms in Ω is conserved:∫Ωρ(x, t) dx =

∫Ωρ(x, 0) dx.

• (11.4) is a gradient system, ∂tρ = −gradF(ρ), with the energy

F(ρ) =

∫R2

[− 1

2ρ(−∆)−1ρ+ ρ ln ρ

]dx. (11.6)

and the metric 〈v, w〉J := −⟨v, J−1w

⟩L2 , where J := ∇ · ρ∇ < 0. In particular, the

functional F(ρ) decreases under the evolution and its critical points are static solutions of(11.4).

(The first term of F(ρ) can be thought of as the internal energy of the system and theremaining terms are the entropy.)

• For d = 2, the system (11.4) has the radially symmetric static solution, R(r), where r = |x|,with the total mass, M =

∫R(|x|)d2x = 8π, it is given explicitly by

R(r) :=8

(1 + r2)2. (11.7)


Under this scaling, the total mass changes as∫1

λ2ρ

(1

λx, t

)= λ(d−2)

∫ρ (x, t) .

Thus one does not expect collapse for d = 1 and that collapse is possible for d ≥ 2 with criticalcollapse for d = 2 and supercritical collapse for d > 2. (Equation (11.4) in d = 2 is said to beL1−critical, etc.)

By the scaling invariance, for d = 2, the system (11.4) has the one-parameter family ofradially symmetric static solutions, Rλ(r) = 1

λ2R( 1

λr).

Existence vs blowup dichotomy. The dimension of the interest for us is the critical dimen-sion d = 2. Recall that for d = 2, system (11.4) has a radially symmetric static solution (11.7).We note that

∫R2 Rdx = 8π. This mass turns out to be the threshold separating a regular

behavior and a breakdown of the solution. It is shown that for the initial condition ρ0 ≥ 0,

• (Blanchet, Dolbeault, Perthame) If the initial total mass satisfies M :=∫R2 ρ0 dx ≤ 8π,

then the solution to (11.4) exists globally and, for M :=∫R2 ρ0 dx < 8π, converges to 0, as

t→∞;

• (Biler) If the initial total mass satisfies M :=∫R2 ρ0 dx > 8π, then the solution to (11.4)

blows up in finite time.

We sketch some key ideas in proving the first statement. To prove the global existence, wehave to control some appropriate positive quantity, like a Sobolev norm. In the present case,this is the entropy,

∫ρ ln ρ.

If the blowup takes place along the family, Rλ(r) = λ2R(λr), of static solutions with λ→∞,then the entropy would blow up,

∫ρλ ln ρλ =

∫ρ ln ρ + 2 lnλ

∫ρ → ∞ as λ → ∞, for any ρ

with finite entropy and the total mass.Hence, if the entropy stays bounded during the evolution, this would indicate that ρ does

not blow up by the compression as in ρλ, λ → ∞. To check this, we start with computing thechange in the entropy

∂t

∫ρ ln ρ = −4

∫|∇√ρ|2︸︷︷︸

entropy dissipation

+

∫ρ2︸︷︷︸

entropy production

. (11.8)

Depending on whether the entropy dissipation or the entropy production wins we expect eitherdissipation of the solution or the collapse (blowup). The Nirenberg - Gagliardo inequality,

‖f‖24 ≤ cgn‖∇f‖2‖f‖2

shows that the dissipation wins if Mc2gn ≤ 4.

To sharpen this result one uses that the free energy (the free energy production or generalizedFisher information) decreases as per

∂tF(ρ) = −∫ρ|∇ ln ρ−∇c|2 (11.9)

(this can be thought of as an entropy monotonicity formula) and that, due to the logarithmicHardy-Littlewood-Sobolev inequality, On the other hand, the logarithmic Hardy-Littlewood-Sobolev inequality ∫

f ln f ≥ 1

(M/8π)

1

2

∫f(−∆)−1f − C(M),


where M :=∫f (the dimension n = 2) and C(M) := M(1+log π−logM), we have the following

lower bound on the free energy,

F(ρ) ≥ (1

(M/8π)− 1)

1

2

∫ρ(−∆)−1ρ− C(M).

Combining these two relations one finds the bound on the entropy

(1−M/8π)

∫ρ ln ρ ≤ F(ρ0)− 1

4πC(M),

which is used to prove the global existence for M ≤ 8π. (Cf. the standard Hardy-Littlewood-

Sobolev inequality,∫ ∫ f(x)f(y)

|x−y| dxdy ≤ ‖f‖2p.)

Virial relation. To see how the critical mass M∗ = 8π enters here, we consider the secondmoment of mass

W :=

∫R2

x2ρ(x) dx.

We have the following virial relation

∂tW = 4M(1− 1

8πM).

If M > 8π, then the right hand side is constant and negative, and hence, W becomes negative infinite time. When this happens we have a contradiction since W is by definition always positive(recall that if ρ0 ≥ 0 then ρ(t) ≥ 0 by the maximum principle). Thus, if M > 8π, then thesolution ρ exists only for a finite time (t < t∗, where t∗ is the point of time when W vanishes).

11.1 M > 8π

Rescaling. We are interested in behaviour of solutions for initial condition with M > 8π. Weoutline some ideas of the approach used in [?]. Recall that (11.4) has the manifold of staticsolutions M0 := R(r/λ) | λ > 0. Assuming this manifold is stable, one can slide along iteither in the direction λ → ∞ (dissipation) or in the direction λ → 0 (collapse). This suggestsa likely scenario of collapse: sliding along M0 in the direction of λ→ 0.

To understand which scenario takes place, we pass to the reference frame collapsing with thesolution, by introducing the adaptive (blowup) variables,

u(y, τ) = λ−2ρ(x, t), c(x, t) = v(y, τ), where y = λx and τ =

∫ t

0λ2(s) ds,

where λ : [0, T )→ [0,∞), T > 0, is a positive differentiable function (compression or dilatationparameter). The advantage of passing to blowup variables is that, if the solution ρ blows up ata finite time T , we expect λ adjusts in such a way that λ(t) → ∞, while u stays bounded, ast ↑ T , and similarly for the dispersion. Moreover, as τ → ∞, as t ↑ T , the blowup time, T , iseliminated from consideration (it is mapped to ∞).

Writing (11.4) in blowup variables, we find the equation for the rescaled mass function

∂τu = ∆yu−∇ · (u∇v)− ay · ∇yu, (11.10)

where a := −λλ. Now, the blowup problem for (11.4) is mapped into the problem of asymptoticdynamics of solitons for the equation (11.10).


Now, one can forget about λ and consider (11.10) as an equation for u and a and then findλ, given λ(0) = λ0, according to the formula

λ2(t) = λ20 − 2

∫ t

0a(s) ds. (11.11)

Stability analysis of R(|y|). Equation (11.10) has the static solution (R(|y|), a = 0). Toinvestigate stability properties of the rescaled stationary solution R(|y|), we consider the lin-earization, La, of the r.h.s. on R(|y|). One can show that this operator is self-adjoint on acertain weighted space. Since R(|y|) breaks the translational and scaling symmetries, La hasthe following eigenfunctions

???

Since La commutes with the rotations, it can be decomposed into spherical harmonics (theFourier series in the polar angular variable θ) as

La = ⊕m≥0Lam, with Lam = La0 +m2

r2.

The radially symmetric part, La0, was investigated in [23], where it was shown that it has

• one negative eigenvalue −2a+ aln 1a

+ O(a ln−2 1

a

)(corresponding to the scaling mode—for

a fixed parabolic scaling it is connected to possible variation of the blowup time) and

• one near zero eigenvalue, and while the third eigenvalue, 2a+ 2aln 1a

+O(a ln−2 1

a

), is positive,

but vanishing as a→ 0.

It also isolates the correct perturbation (adiabatic) parameter— 1ln 1a

. Namely, we have

λn =

µ+a

ln 1a

+K+γ+ O

(a ln−3 1

a

)n = 0

2na+ 2aln 1a

+K+γ−Hn−1−µ+a2an

+ O(a ln−3 1

a

)n ≥ 1,

(11.12)

where K := ln 2− 1− 2γ (here γ = −Ψ(1) = 0.577216 . . . is the Euler-Mascheroni constant) andHn :=

∑nk=1 1/k.

Modification of the leading term. Hence we have to construct a one-parameter deformationof R(|y|) (besides the parameter λ, or a). For technical reasons it is convenient to use a two-parameter family, Rbc(|y|)

Rbc(|y|) :=8b

(c+ |y|2)2, (11.13)

with b > 1 and both parameters b and c are close to 1, with an extra relation between theparameters a, b and c. The family Rbc(|y|) gives approximate solutions to (11.10) (see (11.7))and forms the deformation (or almost center-unstable) manifold M := Rbc(|y|/λ) | λ > 0, p.

We expect that the solution to (11.10) approaches this manifold as τ →∞, and therefore wedecompose the solution u(y, τ) to (11.10) as the leading term, Rb(τ)c(τ)(|y|), and the fluctuation,φ(y, τ),

u(y, τ) = Rb(τ)c(τ)(|y|) + φ(y, τ), (11.14)

and require that the fluctuation φ(y, τ) is orthogonal to the tangent space ofM at Rb(τ)c(τ)(|y|),〈∂pRp(τ)(·), φ(·, τ)〉 = 0, where p := (b, c).

Note that this family evolves on a different spatial scale than φ(y, τ) in (11.14), as it can

rewritten as Rbc(|y|) = R bc4,1( |y|√

c).


Collapse dynamics. In parametrizing solutions as above, we split the dynamics of (11.4) intoa finite-dimensional part describing motion over the manifold, M, and an infinite-dimensionalfluctuation (the error between the solution and the manifold approximation) which is supposedto stay small.

Assuming the initial condition is radially symmetric and substituting the decomposition(11.14) into the equation (11.10) and using the orthogonality condition, 〈∂bcRbc, φ〉 = 0, wearrive at the equations for the parameters b and c and the fluctuation φ, which, in the leadingorder, produce the differential equation

aτ = − 2a2

ln( 1a)

(11.15)

for a, whose solutions, in the leading order, are 1a

(ln 1

a +O(1))

= 2τ which results in ln 1a(τ) =

ln 2τ − ln ln 2τ + ln ln 2τln 2τ +O

(1

ln 2τ

). Recalling that λ(t)λ(t) = a(τ) and using that λ(t)−1λ(t) =

λ(τ(t))−1∂τλ(τ(t)), we obtain, after some derivations, the law

λ(t) = (T − t)−12 e|

12

ln(T−t)|12 (c1 + o(1)). (11.16)

Instead of conclusion. For equivariant wave maps in the dimension 2 + 1, it was shownby Struwe [?] that the existence of a static solution (a nontrivial harmonic map) is in factthe necessary condition for a blowup. To our knowledge, there is no a similar result for theKeller-Segel equations.

Extensions. 1) RKS with dressed static solution Rabc2) RKS for non-radial case:a) Find the spaces (weights) in which the linearized operator is self-adjoint;b) Compute the lowest eigenvalue of La1 by perturbation theory by writing Lam = La0 +

m2/r2, with m=1, and considering m2/r2 as a perturbation. Is the lowest eigenvalue of La1

positive or not? If not , then we have the radial symmetry breaking!c) If the lowest eigenvalue of La1 is positive, then check whether the RKS could be modified

so that it becomes negative, e.g. check whether f(ρ) = ρ + χρ2 etc, would make the lowesteigenvalue of La1 negatived.

d) Look for different βc terms in the second KS eq such that to find two (or whole familyup to rescaling) stationary solutions such that one is collapsing and another is noncollapsing(perhaps stable).

2) RKS (radial or not): Behaviour at the critical mass (infinite time blowup vs dissipationto 0) and the phase transition in M (the universality of the divergence T → ∞ as M → M∗,say T ∼ (M −M∗)−γ , with a universal γ > 0).

3) The full KS (ct case):a) linearized adiabatic theory with the small parameter multiplying ct being the adiabatic

parameter.b) Determine what happened with eigenvalues of the linearized operator in the ct case - will

they change sign?

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗

The static solutions of (11.4) are critical points of the energy functional F , given in (11.6),under the constraint that

∫ρ = M = const.. Thus, they satisfy E ′(ρ) = C, where C is a


constant. Explicitly E ′(ρ) = C reads

log(ρ) +1

∆ρ = C ⇔ ∆ log(ρ) + ρ = 0 ⇔ ∆u+ eu = 0, (11.17)

where u = log(ρ). Solutions to (11.17) can be written in the form of ’Gibbs states’ ρ = M ec∫ec

,

with the concentration c considered as a negative potential (remember that ∆c = −ρ). In twodimensions, this equation has the solution for M = 8π, given by (11.7). This solution is aminimizer of E under the constraint that

∫ρ = 8π.

The equations (11.4) have the family of homogeneous static solutions, (ρ∗, c∗) : ρ∗ =0, c∗ =constant. These solutions have no organism present (zero total mass) and constantconcentration of the chemoattractant. A simple calculation shows that the linearized operatorhas the spectrum filling in the positive semi-axis. This indicates that these static solutions mightbe stable. However, any initial condition ρ0 ≥ 0, ρ0 is not identical 0 will have positive massand therefore, since the total mass is conserved, could converge to (ρ∗, c∗) only in a weak sensewith the loss of the mass.

11.2 Blowup criterion

(??). We note that∫R2 Rdx = 8π. This mass turns out to be the threshold for the blowup of

solutions of (11.4). Indeed, we have the following result:

Theorem 8 (See [3]). Take ρ0 ≥ 0. If the dimension d = 2 and the total mass satisfies

M :=

∫R2

ρ0 dx > 8π, (11.18)

then the solution to (11.4) breaks down in finite time.

Proof. The time derivative of the second moment of mass

W :=

∫R2

x2ρ(x) dx

is, using that ρ is a solution to (11.4),

∂tW =

∫R2

x2(∆ρ+∇

(ρ∇∆−1ρ

))dx.

Integrating by parts and using that∫R2 x

2∆ρdx = 2d∫R2 ρdx = 2d

∫R2 ρ0dx = 2dM , by the

conservation of total mass, we obtain

∂tW = 2(dM − J). (11.19)

where J :=∫ρ(∇∆−1ρ

)· x dx. Using the integral representation of ∆−1ρ (in two dimensions

∆−1ρ = 12π

∫R2 ln |x− y| ρ(y) dy), and symmetrizing the integral we compute

J =1

2π

∫R2d

x · x− y|x− y|d

ρ(y)ρ(x) dy dx

=1

4π

∫R2d

x · x− y|x− y|d

ρ(y)ρ(x) dy dx+1

4π

∫R2d

y · y − x|x− y|d

ρ(y)ρ(x) dy dx


This gives

J =1

4π

∫R2d

1

|x− y|(d−2)ρ(y)ρ(x) dy dx.

For d = 2, this expression gives that J = 12πM

2. Substituting this into (11.19), we obtain that

∂tW = 4M(1− 1

8πM).

If M > 8π, then the right hand side is constant and negative, and hence, W becomes negativein finite time. When this happens we have a contradiction since W is by definition alwayspositive (recall that if ρ0 ≥ 0 then ρ(t) ≥ 0 by the maximum principle). Thus, if M > 8π,then the solution ρ exists only for a finite time (t < t∗, where t∗ is the point of time when Wvanishes).

This result tells us nothing about how the solution break down in finite time. The latterprocess is investigated in the subsequent sections.

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗

11.3 Appendix: Gradient formulation

The Keller-Segel models (11.1) and (11.4) are gradient systems. We begin by formulating anormalized version of (11.1),

∂tρ = ∆ρ−∇ · (f(ρ)∇c)ε∂tc = ∆c+ ρ− γc,

(11.20)

as a gradient system. This system is obtained from (11.1) by setting unimportant constants to1.

Define the energy (or Lyapunov) functional

Ef (ρ, c) :=

∫Ω

1

2|∇c|2 − ρc+

γ

2c2 +G(ρ) dx, (11.21)

where G(ρ) :=∫ ρg(s) ds and g(ρ) :=

∫ ρ 1f(s) ds. The L2-gradient of Ef (ρ, c) is

gradL2

Ef (ρ, c) =

(−c+ g(ρ)−∆c− ρ+ γc

),

and hence, if we define U = (ρ, c), then (11.20) can be written in the form ∂tU = IE ′f (U), where

I =

(∇ · f(ρ)∇ 0

0 −1ε

).

The operator I is non-positive and may be degenerate, however, assuming it is invertible, theoperator I defines the metric 〈v, w〉I := −

⟨v, I−1w

⟩L2⊕L2 . In this metric, grad E(U) = −IE ′(U)

and hence∂tU = −grad Ef (U).

This shows that (11.20) has the structure of a gradient system. A consequence of this is thatthe energy decreases on solutions of the KS system. Indeed, if f > 0, then

∂tEf (ρ, c) = −∥∥∥f(ρ)

12∇ (c− g(ρ) )

∥∥∥2

L2− 1

ε‖∆c+ ρ− c‖2L2 .


The gradient formulation for (11.4) is similar to the one for (11.20). Instead of (11.21),one uses the energy (??). The latter is obtained from (11.21) by dropping the quadratic term12c

2, replacing c with −∆−1ρ in the remaining terms and using that f(ρ) = ρ. The formalGateaux derivative of E is ∂ρE(ρ)φ =

∫(∆−1 ρ + ln ρ)φ, and therefore the gradient in the metric

〈v, w〉J := −⟨v, J−1w

⟩L2 , where J := ∇ · ρ∇ < 0, is

grad E(ρ) = −∇ · ρ∇(∆−1 ρ + ln ρ) = −∇ · ρ∇∆−1 ρ −∆ρ,

which is the negative of the r.h.s. of the first equation in (11.4) with c = −∆−1ρ. Hencethe equation (11.4) can be written as ∂tρ = −grad E(ρ) in the space with metric 〈v, w〉J :=−⟨v, J−1w

⟩L2 . Again, the energy E decreases on solutions of (11.4):

∂tE =⟨E ′, IE ′

⟩= −

∥∥∥ρ 12∇E ′

∥∥∥2

L2. (11.22)

This can be thought of as an entropy monotonicity formula.

11.4 Appendix: Criterion of break-down in the dimension d ≥ 3

For d ≥ 3, we have the following result

Theorem 9. Take ρ0 ≥ 0. If d ≥ 3 and∫Rd x

2ρ0dx∫Rd ρ0dx

is sufficiently small (this means that ρ0 is

concentrated at x = 0), then the solution to (11.4) blows up in finite time.

Proof. We now prove the case when d ≥ 3. We proceed as above and derive an upper bound on∂tW which is negative for certain initial conditions. To this end we derive a lower bound on J .Let γ = d− 2. We estimate M2 from above using J by first rewriting M2:

M2 =

∫R2d

u(x)u(y) dx dy =

∫R2d

(u(x)u(y)

|x− y|γ

)α (u(x)u(y)|x− y|

αγ1−α)1−α

dx dy.

Holder’s inequality then gives

M2 ≤(∫

R2d

u(x)u(y)

|x− y|γdx dy

)α(∫R2d

u(x)u(y)|x− y|αγ1−α dx dy

)1−α.

The first integral is Jα. We choose αγ1−α = 2 or α = 2

γ+2 so that

M2 ≤(∫

R4

u(x)u(y)

|x− y|γdx dy

) 2γ+2(∫

R2d

u(x)u(y)|x− y|2 dx dy) γγ+2

= J2

γ+2

(∫R2d

u(x)u(y)|x− y|2 dx dy) γγ+2

.

Expanding the square we obtain that∫R2d

u(x)u(y)|x− y|2 dx dy =

∫R2d

u(x)u(y)(x2 + y2 − 2x · y

)dx dy

=

∫R2d

u(x)u(y)(x2 + y2 + 2|x||y|

)= 2WM + 2

(∫Rd|x|u(x) dx

)2

.


By Holder’s inequality again(∫Rd|x|u

12u

12 dx

)2

≤∫Rdx2u dx

∫Rdu dx = WM.

Combining the estimates, one obtains that

M2 ≤ J2

γ+2 (4WM)γγ+2

orJ ≥ CM

γ+42 W−

γ2 .

Substituting into (11.19) gives that

∂tW ≤ 2(dM − CMγ+42 W

−γ2 ).

Thus, if initially M/W (0) 1, then ∂tW is negative and W will decrease. Since it decreases,the time derivative ∂tW becomes more negative and hence W continues to decrease and willbecomes negative in a finite time (since the time derivative will always be less that the initialvalue of ∂tW ). This is again a contradiction since W is by definition always positive.

12 Mean curvature flow

(to be edited)

12.1 Definition and general properties of the mean curvature flow

The mean curvature flow, starting with a hypersurface S0 in Rn+1, is the family of hypersurfacesS(t) given by immersions x(·, t) which satisfy the evolution equation

∂x∂t = −H(x)ν(x)x|t=0 = x0

(12.1)

where x0 is an immersion of S0, H(x) and ν(x) are mean curvature and the outward unit normalvector at x ∈ S(t), respectively. The terms used above are explained in Appendix ??.

In this lecture we describe some general properties of the mean curvature flow, (12.1). Webegin with writing out (12.1) for various explicit representations for surfaces St.

Mean curvature flow for level sets and graphs. We rewrite out (12.1) for the level setand graph representation of S. Below, all differential operations, e.g. ∇,∆, are defined in thecorresponding Euclidian space (either Rn+1 or Rn).

1) Level set representation S = ϕ(x, t) = 0. Then, by (??) of Appendix ??, we have

ν(x) =∇ϕ|∇ϕ|

, H(x) = div

(∇ϕ|∇ϕ|

). (12.2)

We compute 0 = dϕdt = ∇xϕ · ∂x∂t + ∂ϕ

∂t and therefore ∂ϕ∂t = ∇ϕ · ∇ϕ|∇ϕ| div

(∇ϕ|∇ϕ|

), which gives

∂ϕ

∂t= |∇ϕ| div

(∇ϕ|∇ϕ|

). (12.3)


2) Graph representation: S = graph of f . In this case S is the zero level set of the functionϕ(x) = xn+1 − f(u), where u = (x1, . . . , xn) and x = (u, xn+1), and using (12.3) with thisfunction, we obtain

∂f

∂t=√|∇f |2 + 1 div

(∇f√|∇f |2 + 1

). (12.4)

Denote by Hess f the standard euclidean hessian, Hess f :=( ∂2f∂ui∂uj

). Then we can rewrite

(12.4) as∂f

∂t= ∆f − ∇f Hess f∇f

|∇f |2 + 1. (12.5)

Different form of the mean curvature flow. Multiplying the equation (12.1) in Rn+1 byν(x), we obtain the equation

ν(x) · ∂x∂t

= −H(x). (12.6)

In opposite direction we have

Proposition 2. If x satisfies (12.6), then there is a (time-dependent) reparametrization ϕ ofS, s.t. x ϕ satisfies (12.1).

Proof. Denote (∂x∂t )T := ∂x

∂t − (ν · ∂x∂t )ν (the projection of ∂x∂t onto TxS) and let ϕ satisfy the ODE

ϕ = −(dx)−1(∂x∂t ϕ)T (parametrized by u ∈ U). Then ∂∂t(xϕ) = ( ∂∂tx)ϕ+dx ϕ. Substituting

ϕ = −(dx)−1(∂x∂t ϕ)T into this, we obtain ∂∂t(x ϕ) = (ν · ( ∂∂tx) ϕ)ν = −H.

Thus the MCF in the form (12.6) is invariant under reparametrization, while the form (12.1)is obtained by fixing a specific parametrization (fixing the gauge).

Now we describe some general properties of MCF.

Symmetries. Mean curvature is invariant under translations, rotations and scaling. Thelatter is defined as

S(t)→ λS(λ−2t) ⇔ x→ λx, t→ λ−2t.

The invariance under translations and rotations is obvious. To prove the invariance underscaling, we first show that

H(λx) = λ−1H(x). (12.7)

Now, let τ := λ−2t, and xλ(t) := λx(τ). Then

∂txλ = λλ−2(∂τx)(τ) = −λ−1H(x(τ))ν(x(τ)).

On the other hand, νλ ≡ ν(xλ(τ)) = ν ≡ ν(x(τ)) and Hλ ≡ H(xλ(τ)) = λ−1H(x(τ)) ≡ λ−1H(by (12.7)), which gives

∂txλ = −Hλνλ. (12.8)

We can also use the scaling x(u, t) → λx(λ−1u, λ−2t), together with the fact that H → H,under u → λ−1u. Indeed, B → λ−2B and G → λ−2G if ψ(u) → ψ(λ−1u), and thereforeG−1B → G−1B.

For S = graph f , f(u)→ λf(λ−1u)⇔ ψ(u)→ λψ(λ−1u

), where ψ(u) := (u, f(u)). Indeed,

ψ(u) := (u, f(u))→(u, λf(λ−1u)

)= λ

(λ−1u, f(λ−1u)

)=: λψ

(λ−1u

).


Static solutions. Static solutions satisfy the equation H(x) = 0 on S, which, as shown below,is the equation for critical points of the volume functional V (ψ) (the Euler - Lagrange equationfor V (ψ)). Thus, static solutions of the MCF (12.1) are minimal surfaces.

Volume functional. Recall that locally the surface volume functional can be written as

V (S ∩ U) =

∫U

√gdnu,

where g := det(gij). We want to compute the Gateaux derivative of the area functional V .Because of reparametrization (see Proposition 2), it suffices to look only at normal variations,ψs, of the immersion ψ, i.e. generated by vector fields η, directed along the normal ν: η = fν.A standard computation (see Appendix ?? gives For a surface S given locally by an immersionψ and normal variations, η = fν, we have

dV (ψ)η =

∫UHν · η √gdnu. (12.9)

This shows that

• If a closed surface S evolves according to (12.1), then its area, V (S), decreases. In fact,∂tV (S) = −

∫S H

2(x) < 0.

• Critical points of V (S) are static solution to the MCF (12.1).

The assertion on ψ(U) follows from (12.9) as

∂tV (ψ) =

∫UHν · ∂tψ = −

∫ψH2(x),

which after using a partition of unity proves the statement.

12.2 Solitons: self-similar surfaces

Consider solitons for MCF, i.e. equivariant solutions as defined and discussed in Section 3.2,see, in particular, Theorem 3.1. There we also considered spherically and radially symmetric(equivariant) solutions:spheres and cylinders.

Recalling (3.41) - (3.44), we consider solutions of the MCF of the form S(t) ≡ Sλ(t) := λ(t)S(standing waves), or x(u, t) = λ(t)y(u), where λ(t) > 0. Plugging this into (12.1) and usingH(λy) = λ−1H(y), gives λy = −λ−1H(y)ν(y), or λλy = −H(y)ν(y). Multiplying this by ν(y),we obtain (cf. (3.44))

H(y) = a〈ν, y〉, and λλ = −a. (12.10)

Since H(y) is independent of t, then so should be λλ = −a. Solving the last equation, we findλ =

√λ2

0 − 2at.

i) a > 0⇒ λ→ 0 as t→ T :=λ202a ⇒ Sλ is a shrinker.

ii) a < 0⇒ λ→∞ as t→∞⇒ Sλ is an expander.


Recall, that static solutions of the MCF (12.1) satisfy the equation H(x) = 0 on S andtherefore are minimal surfaces. Hence the minimal surfaces are special cases of self-similar onescorresponding to a = 0. In fact, a = 0 separates two types of evolution: contracting a > 0 ( λdecreasing) and expanding a < 0 ( λ increasing). (Remember that a = −λ∂tλ is the negativeof the speed of scaling λ.) (We see that scaling solitons generalize the notion of the minimalsurface.)

The equation (12.10) has the solutions: a is time-independent and x is one of the following

a) Sphere x = Rx, where R =√

na .

b) Cylinder x = (Rx′, x′′), where x = (x′, x′′) ∈ Rk+1 × Rm, where R =√

ka .

As stated in the following theorem, these solutions are robust.

Theorem 10 (Huisken). Let S satisfy H = ax · ν and H ≥ 0. We have(i) If n ≥ 2, and S is compact, then S is a sphere of radius

√na .

(ii) If n = 2 and S is a surface of revolution, then S is the cylinder of radius√

n−1a .

Now we consider self-similar surfaces in the graph representation. Let

f(u, t) = λχ(λ−1u), λ depends on t. (12.11)

Substituting this into (12.4) and setting y = λ−1u and a = −λλ, we find√1 + |∇yχ|2H(χ) = a(y∂y − 1)χ. (12.12)

Translation solitons. These are solutions of the MCF of the form S(t) ≡ S + h(t) (travelingwaves), or x(u, t) = y(u) + h(t), where h(t) ∈ Rn+1. Plugging this into (12.1) and usingH(y + h) = H(y), gives h = −H(y)ν(y), or

H(y) = v · ν(y), and h = v. (12.13)

Since H(y) is independent of t, then so should be h = v. (more to come)

Breathers. MCF periodic in t.

Self-similar surfaces and rescaled MCF. Minimal surfaces are static solutions of the MCF(12.1). Are more general the self-similar surfaces static solutions of some equation? The answeris yes, the self-similar surfaces are static solutions of the rescaled MCF,

∂τϕ = −(H(ϕ)− aϕ · ν(ϕ))ν(ϕ), (12.14)

where a = −λλ, which is obtained by rescaling the surface ψ and time t as

ϕ(u, τ) := λ−1(t)ψ(u, t), τ =

∫ t

0λ−2(s)ds, (12.15)

and then reparametrizing the obtained surface Sresc(τ) := λ(t)S(t) as in the proof of Proposition2. Indeed, since λH = Hλ (or λH(ψ) = H(λ−1ψ)), we find

∂τϕ = λ2∂tϕ = λ2(−λλ−2ψ + λ−1∂tψ)


= −λλϕ− λH(ψ)ν(ψ) = aϕ−H(ϕ)ν(ϕ).

Then the mean curvature flow equation (12.1) implies

∂τϕ = −H(ϕ)ν(ϕ) + aϕ, (12.16)

which after the reparametrization gives (12.14). This is another analogy with minimal surfaces.For static solutions, a =const. Then solving the equation λλ = −a, we obtain the parabolic

scaling:

λ =√

2a(T − t) and τ(t) = − 1

2aln(T − t), (12.17)

where T := λ20/2T , which was already discussed in connection with the scaling solitons.

Now, we know that minimal surfaces are critical points of the volume functional V (ψ) (by(12.9), the equation H(x) = 0 on S is the Euler - Lagrange equation for V (ψ)). Are self-similarsurfaces critical points of some modification of the volume functional? (Recall that, because ofreparametrization (see Proposition 2), it suffices to look only at normal variations, ψs, of theimmersion ψ, i.e. generated by vector fields η, directed along the normal ν: η = fν.) Theanswer to this question is yes and is given in the following

Proposition 3. Let ρ(x) = e−a2|x|2 and Va(ϕ) :=

∫Sλ ρ. For a surface S given locally by an

immersion ψ and normal variations, η = fν, we have

dVa(ϕ)η =

∫U

(H − aϕ · ν)ν · η ρdnu, (12.18)

Proof. The definition of Va(ϕ) gives Va(ϕ) =∫Sλ ρ(ϕ)

√g(ϕ), where ρ(ϕ) = e−

a2|ϕ|2 . We have

dρ(ϕ)η = −aρ(ϕ)η. Using this formula and the equation d√gη = H

√gν · η proven above (see

(??)) and the fact that we are dealing with normal variations, η = fν, we obtain (12.18).

The equation (12.14) and Proposition 3 imply

Corollary 11. Assume a in (12.14) is constant (i.e. the rescaling (12.15) is parabolic, (12.17)).Then the modified area functional Va(ϕ) is momotonically decaying under the rescaled flow(12.14), more precisely,

∂τVa(ϕ) = −∫Sλρ|H − aν · ϕ|2. (12.19)

The relation (12.19) is the Huisken monotonicity formula (earlier results of this type wereobtained by Giga and Kohn and by Struwe).

Most interesting minimal surfaces are not just critical points of the volume functional V (ψ)but are minimizers for it. What about self-similar surfaces? We see that

(i) inf Va(ϕ) = 0 and, for compact minimal surfaces, Va(ϕ) is minimized by any sequenceshrinking to a point.

(ii) Va(ϕ) is unbounded from above. This is clear for a < 0. To see this for a < 0, we constructa sequence of surfaces lying inside a fixed ball in Rn+1 and folding tighter and tighter.

Thus self-similar surfaces are neither minimizers nor maximizers of Va(ϕ). We conjecture thatthey are saddle points satisfying min-max principle: supV infϕ:V (ϕ)=V Va(ϕ). One can try usethis principle (say in the form of the mountain pass lemma) to find solutions of (12.10).


Remark 3. For minimal surfaces (strict) stability implies the the surface is a (strict) localminimizer of the volume functional V (ψ). As is already suggested by the discussion above,this is not so for self-similar surfaces, they are saddle points possibly satisfying some min-maxprinciple.

12.3 Linearized stability of self-similar surfaces

Normal hessians. Recall that the hessian of a functional E(ϕ) is defined as HessE(ϕ) :=d gradVa(ϕ). Note that unlike the Gateaux derivative, d, the gradient grad and therefore thehessian, Hess, depends on the Riemann metric on the space on which E(ϕ) is defined.

Since the tangential variations lead to reparametrization of the surface, in what follows weare dealing with normal variations, η = fν. (In physics terms, specifying normal variations iscalled fixing the gauge.) We use the following notation for a linear operator, A, on normal vectorfields on S: ANf = A(fν). For instance, HessN E(ϕ)f = HessE(ϕ)(fν) and

dNF (ϕ)f = dF (ϕ)(fν). (12.20)

We consider the hessian of the modified volume functional Va(ϕ), at a self-similar ϕ (i.e.H(ϕ) = aϕ · ν) and in the normal direction (i.e. for normal variations, η = fν) in the Riemannmetric h(ξ, η) :=

∫Sλ ξηρ. In what follows, we call this hessian the normal hessian and denote it

by HessN Va(ϕ).Before we proceed, we mention the following important property of Va(ϕ): the equation

H(ϕ)−aϕ · ν(ϕ) = 0 breaks the scaling and translational symmetry. Indeed, using the relations

H(λϕ) = λ−1H(ϕ), ν(λϕ) = ν(ϕ), ∀λ ∈ R+, (12.21)

H(ϕ+ h) = H(ϕ), ν(ϕ+ h) = ν(ϕ), ∀h ∈ Rn+1, (12.22)

H(gϕ) = H(ϕ), ν(gϕ) = gν(ϕ), ∀g ∈ O(n+ 1), (12.23)

using that g ∈ O(n+1) are isometries in Rn+1and using the notation Ha(ϕ) := H(ϕ)−aϕ ·ν(ϕ),we obtain

Hλ−2a(λϕ) = λ−1Ha(ϕ), ∀λ ∈ R+, (12.24)

Ha(ϕ+ h) + ah · ν(ϕ) = Ha(ϕ), ∀h ∈ Rn+1, (12.25)

Ha(gϕ) = Ha(ϕ), ∀g ∈ O(n+ 1). (12.26)

We want to address the spectrum of the normal hessian, Hess⊥ Va(ϕ). First, we note thatthe tangential variations lead to zero modes of the full hessian, HessVa(ϕ). Indeed, we have

Proposition 4. The full hessian, HessVa(ϕ), of the modified volume functional Va(ϕ), has theeigenvalue 0 with the eigenfunctions which are tangential vector fields on S

Proof. We consider a family αs of diffeormorphisms of U , with α0 = 1 and ∂sϕ αs|s=0 = ξ, atangential vector field, reparametrizing the immersion ϕ, and define the family ϕαs of variationsof ϕ. Then ϕ αs satisfies again the soliton equation, Ha(ϕ αs) = 0. Differentiating the latterequation w.r.to s at s = 0 and using that ∂sϕαs|s=0 = ξ, is a tangential vector field, we obtain

dHa(ϕ)ξ = 0, (12.27)

which proves the proposition.


Theorem 12. The hessian, HessN Va(ϕ), of the modified volume functional Va(ϕ), at a self-similar ϕ (i.e. H(ϕ) = aϕ · ν) and in the normal direction (i.e. for normal variations, η = fν),has

• the eigenvalue −2a with the eigenfunction ϕ · ν(ϕ),

• the eigenvalue −a with the eigenfunctions νj(ϕ), j = 1, . . . , n+ 1, and,

• the eigenvalue 0 with the eigenfunctions σjϕ · ν(ϕ), j = 1, . . . , 12n(n − 1), where σj are

generators of the Lie algebra of SO(n+ 1), unless ϕ is a sphere.

Proof. If an immersion ϕ satisfies the soliton equation H(ϕ) = aϕ · ν(ϕ), then by (12.24), wehave Hλ−2a(λϕ) = 0 for any λ > 0. Differentiating this equation w.r.to λ at λ = 1, we obtaindHa(ϕ)ϕ = −2aϕ · ν(ϕ).

Now, choosing ξ to be equal to the tangential projection, ϕT , of ϕ, and subtracting theequation (12.27) from the last equation, we find dHa(ϕ)(ϕ · ν(ϕ))ν(ϕ) = −aϕ · ν(ϕ). Since bythe definition (12.20), dHa(ϕ)fν(ϕ) = dNHa(ϕ)f , this proves the first statement.

To prove the second statement, we observe that the soliton equation implies, by (12.25),that Ha(ϕ+ sh) + ash · ν(ϕ) = 0 and any constant vector field h. Differentiating this equationw.r.to s at s = 0, we obtain dHa(ϕ)h = −ah · ν(ϕ). Now, choosing ξ to be equal to thetangential projection, hT , of h, and subtracting the equation (12.27) from the last equation, wefind dHa(ϕ)(h ·ν(ϕ))ν(ϕ) = −ah ·ν(ϕ), which together with (12.20) gives the second statement.

Finally, to prove the third statement, we differentiate the equation Ha(g(s)ϕ) = 0, whereg(s) is a one-parameter subgroup of O(n + 1), w.r.to s at s = 0, to obtain dHa(ϕ)σϕ = 0,where σ denotes the generator of g(s). Now, choosing ξ in (12.27) to be equal to the tangentialprojection, (σϕ)T , of σϕ, and subtracting the equation (12.27) from the last equation, we finddHa(ϕ)(σϕ · ν(ϕ))ν(ϕ) = 0, which together with (12.20) gives the third statement.

Remark 4. a) For a 6= 0, the soliton equation, ϕ ·ν(ϕ) = a−1H(ϕ), and Proposition 12.2 implythat the mean curvature H is an eigenfunction of HessN Va(ϕ) with the eigenvalue −2a.

b) Strictly speaking, if the self-similar surface is not compact, then ϕ · ν(ϕ) and νj(ϕ), j =1, . . . , n + 1, generalized eigenfunctions of HessN Va(ϕ). In the second case, the Schnol-Simontheorem (see Appendix ?? or [30]) implies that the points −2a and −a belong to the essentialspectrum of HessN Va(ϕ).

c) We show below that the normal hessian, HessNsph Va(ϕ), on the sphere of the radius√

an ,

given in (12.28), has no other eigenvalues below 2an , besides −2a and −a. A similar statement,

but with n replaced by n− 1, we have for the cylinder.

We call the eigenfunction ϕ · ν(ϕ), νj(ϕ) and σjϕ · ν(ϕ), j = 1, . . . , n + 1, the scaling,translational and rotational modes. They originate from the normal projections, (λϕ)N and(sh)N , of scaling, translation and rotation variations.

Linearized stability. Given a self-similar surface ϕ, we consider for example the manifold ofsurfaces obtained from ϕ by symmetry transformations,

Mϕ := λgϕ+ z : (λ, z, g) ∈ R+ × Rn+1 × SO(n+ 1).

By the spectral theorem above, it has unstable and central manifolds corresponding to theeigenvalues −2a, −a and 0. Hence, we can expect only the dynamical stability in the transversedirection.


Definition 12.1 (Linearized stability of self-similar surfaces). We say that a self-similar surfaceφ, with a > 0, is linearly stable (for the lack of a better term) iff the normal hessian satisfiesHessN Va(ϕ) > 0 on the subspace

(span ϕ · ν(ϕ), νi(ϕ), i = 1, . . . , n + 1, σjϕ · ν(ϕ), j =

1, . . . , 12d(d− 1)

)⊥(i.e. on

(span scaling, translational, rotational modes

)⊥).

(I.e. the only unstable motions allowed are scaling, translations and rotations.)In what follows a > 0. Theorem 12 implies that if φ is not spherically symmetric, then 0 is

an eigenvalue of HessN Va(ϕ) of multiplicity at least n+ 1. This gives the first statement in thefollowing corollary, while

Corollary 13. If a self-similar surface with a > 0 satisfies HessN Va(ϕ) > 0 on the subspacespan ϕ · ν(ϕ), νj(ϕ), j = 1, . . . , n+ 1⊥, then it a sphere or a cylinder.

Theorem 14. For a self-similar surface with a > 0, HessN Va(ϕ) ≥ −2a iff H(ϕ) > 0.

To prove this theorem we will use the Perron-Frobenius theory (see Appendix D.3) and itsextension as given in [?]. We begin with

Definition 12.2. We say that a linear operator on L2(S) has a positivity improving propertyiff either A, or e−A, or (A + µ)−1, for some µ ∈ R, takes non-negative functions into positiveones.

Proposition 5. The normal hessian, HessN Va(ϕ), has a positivity improving property.

Proof. By standard elliptic/parabolic theory, e∆ and (−∆ + µ)−1, for any µ > 0, has strictlypositive integra kernel and therefore is positivity improving. To lift this result to HessN Va(ϕ) :=−∆− |W |2 − a1 + ϕ · ∇ we proceed exactly as in [?].

Since, the operator HessN Va(ϕ) is bounded from below and has a positivity improvingproperty, it satisfies the assumptions of the Perron-Frobenius theory (see Appendix D.3) and itsextension in [?] to the case when the positive solution in question is not an eigenfunction. Thelatter theory, Proposition 12.2 and Remark 4 imply the statement of Theorem 14.

Corollary 15. Let ϕ be a self-similar surface. We have(a) For a < 0 (ϕ is an expander), H(ϕ) changes the sign.(b) For a = 0, inf HessN Va(ϕ) < 0.(c) For a > 0, if ϕ is an entire graph over Rn and is weakly stable, then ϕ is a hyperplane.

Indeed, (a) follows directly from Theorems 12 and 14, while (b) follows from the fact thatfor a = 0, 0 is an eigenvalue of the multiplicity n + 2 and therefore, by the Perron-Frobeniustheory, it is not the lowest eigenvalue of HessN Va(ϕ). (c) is based on the fact that entire graphsover Rn cannot have strictly postive mean curvature. (check, references)

Hessians for spheres and cylinders. We finish this section with the discussion of the normalhessians on the n−sphere and (n, k)−cylinder.

Explicit expressions. 1) For the n−sphere SnR of radius R =√

na in Rn+1, we have

HessNsph Va(ϕ) = −an

∆Sn − 2a, (12.28)

on L2(Sn), where ∆Sk is the Laplace-Beltrami operator on the standard n−sphere Sk.


2) For the n−cylinder CnR = Sn−kR × Rk of radius R =√

n−ka in Rn+1, we have

HessNcyl Va(ϕ) = −∆y − ay · ∇y −a

n− k∆Sn−k − 2a, (12.29)

acting on L2(Cn).

Spectra of Lspha := HessNsph Va(ϕ) and Lcyl

a := HessNcyl Va(ϕ). We describe the spectra of the

normal hessians on the n−sphere and (n, k)−cylinder, of the radii√

an and

√a

n−1 , respectively.

It is a standard fact that the operator −∆ = −∆Sn is a self-adjoint operator on L2(Sn). Itsspectrum is well known (see [?]): it consists of the eigenvalues l(l + n− 1), l = 0, 1, . . . , of the

multiplicities m` =

(n+ ln

)−(n+ l − 2

n

). Moreover, the eigenfunctions corresponding to

the eigenvalue l(l+n−1) are the restrictions to the sphere Sn of harmonic polynomials on Rn+1

of degree l and denoted by Ylm (the spherical harmonics),

−∆Ylm = l(l + n− 1)Ylm, l = 0, 1, 2, 3, . . . , m = 1, 2, . . . ,ml. (12.30)

In particular, the first eigenvalue 0 has the only eigenfunction 1 and the second eigenvalue n hasthe eigenfunctions ω1, · · · , ωn+1.

Consequently, the operator Lspha := HessNsph Va(ϕ) = − a

n(∆Sn + 2n) is self-adjoint and itsspectrum consists of the eigenvalues a

n(l(l + n − 1) − 2n) = a(l − 2) + an l(l − 1), l = 0, 1, . . . ,

of the multiplicities m`. In particular, the first n + 2 eigenvectors of Lspha (those with l = 0, 1)

correspond to the non-positive eigenvalues,

Lspha ω0 = −2aω0, Lsph

a ωj = −aωj , j = 1, . . . , n+ 1, (12.31)

and are due to the scaling (l = 0) and translation (l = 1) symmetries.

We proceed to the cylindrical hessian Lcyla := HessNcyl Va(ϕ), given in (12.29). The variables

in this operator separate and we can analyze the operators −∆y − ay · ∇ and − an−k (∆Sn−k +

2(n−k)) separately. The operator − an−k (∆Sn−k +2n) was already analyzed above. The operator

−∆y−ay ·∇ is the Ornstein - Uhlenbeck generator, which can be unitarily mapped by the gaugetransformation

v(y, w)→ v(y, w)e−a4|y|2

into the the harmonic oscillator Hamiltonian Hharm := −∆y + 14a

2|y|2 − ka. Hence the linear

operator −∆y − ay · ∇ is self-adjoint on the Hilbert space L2(R, e−a2|y|2dy). Since, as was

already mentioned, the operator − an−k (∆Sn−k + 2(n − k)) is self-adjoint on the Hilbert space

L2(Cn), we conclude that the linear operator Lcyla is self-adjoint on the Hilbert space L2(Rk ×

Sn−k, e−a2|y|2dydw). Moreover, the spectrum of −∆y − ay · ∇ is

a∑k

1 si : si = 0, 1, 2, 3, . . .,

with the normalized eigenvectors denoted by φs,a(y), s = (s1, . . . , sk),

(−∆y − ay · ∇)φs,a = ak∑1

siφs,a, si = 0, 1, 2, 3, . . . . (12.32)

Using that we have shown that the spectrum of − an−k (∆Sn−k + 2(n−k)) is a

n−k (l(l+n−k−1)−2(n−k)) = a

n−k l(l+n−k−1)−2a, l = 0, 1, . . . , and denoting r =∑k

1 si, si = 0, 1, 2, 3, . . .,

we conclude the spectrum of the linear operator Lcyla , for k = 1, is

spec(Lcyla ) =

(r − 2)a+

a

n− 1`(`+ n− 2) : r = 0, 1, 2, 3, . . . ; ` = 0, 1, 2, . . .

, (12.33)


with the normalized eigenvectors given by φr,l,m,a(y, w) := φr,a(y)Ylm(w). This equation shows

that the non-positive eigenvalues of the operator Lcyla , for k = 1, are

• the eigenvalue −2a of the multiplicity 1 with the eigenfunction φ0,0,0,a(y) = ( a2π )

14 ((r, l) =

(0, 0)), due to scaling of the transverse sphere;

• the eigenvalue−a of the multiplicity n with the eigenfunctions φ0,1,m,a(y) = ( a2π )

14wm, m =

1, . . . , n ((r, l) = (1, 1)), due to transverse translations;

• the eigenvalue 0 of the multiplicity n with the eigenfunctions φ1,1,m,a(y) = ( a2π )

14√aywm, m =

1, . . . , n ((r, l) = (0, 1)), due to rotation of the cylinder;

• the eigenvalue −a of the multiplicity 1 with the eigenfunction φ1,0,0,a(y) = ( a2π )

14√ay

((r, l) = (1, 0));

• the eigenvalue 0 of the multiplicity 1 with the eigenfunction φ2,0,0,a(y) = ( a2π )

14 (1 − ay2)

((k, l) = (2, 0)).

The last two eigenvalues are not of the broken symmetry origin and are not covered by Theorem12. They indicate instability of the cylindrical collapse

By the description of the spectra of the normal hessians of Lspha := HessNsph Va(ϕ) and Lcyl

a :=

HessNcyl Va(ϕ), we conclude that the spherical collapse is linearly stable while the cylindrical oneis not.

It is shown in [?] that indeed the spherical collapse is (nonlinearly) stable while the cylindrical

one is not. We will also show that the last two eigenvalues of Lcyla in the list above are due to

translations of the point of the neckpinch on the axis of the cylinder and due to shape instability,respectively.

12.4 F−stability of self-similar surfaces

Another notion of stability was introduced by analogy with minimal surfaces in [?]:

Definition 12.3 (F−stability of self-similar surfaces, [?]). We say that a self-similar surfaceφ, with a > 0, is F−stable iff the normal hessian satisfies HessN Va(ϕ) ≥ 0 on the subspacespan ϕ · ν(ϕ), νj(ϕ), j = 1, . . . , n+ 1⊥.

Remark 5. a) The F−stability, at least in the compact case, says that the HessN Va(ϕ) has thesmallest possible negative subspace, i.e ϕ has the smallest possible Morse index.

b) The reason the F−stability works in the non-compact case is that, due to the separation ofvariables for the cylinder = (compact surface) ×Rk, the orthogonality to the negative eigenfunc-tions of the compact factor removes the entire branches of the essential spectrum. This mightnot work for warped cylinders.

c) Remark 4(c) shows that the spheres and cylinders are F−stable. However, it is shown in[28, ?] that cylinders are dynamically unstable.

The following statement follows from Remark 4(c) and the definition of the F−stability (seealso the first part of Remark 5(5)):

There are no smooth, embedded self-similar (a > 0), F−stable surfaces in Rn+1 close to

Sk × Rn−k, where Sk is the round k−sphere of radius√

ka .

A slight but a key improvement of this result is a hard theorem:


Theorem 16 (Colding- Minicozzi). The only smooth, complete, embedded self-similar (a > 0),F−stable surfaces in Rn+1 of polynomial growth are Sk × Rn−k, where Sk is the round

k−sphere of radius√

ka .

Remark 6. The difference between this theorem and (trivial) Corollary 13 is that the latterrequires a slightly stronger condition HessN Va(ϕ) > 0 on the subspace span ϕ ·ν(ϕ), νj(ϕ), j =1, . . . , n+ 1⊥, then the F−stability.

Theorem 14 implies that if it is F−stable, then H(ϕ) > 0.

Theorem 17 (Colding- Minicozzi, Huisken). The only smooth, complete, embedded self-similarsurfaces in Rn+1, with a > 0, polynomial growth and H(ϕ) > 0, are Sk ×Rn−k, where Sk is the

round k−sphere of radius√

ka .

(Compare with surfaces of constant mean curvature)Theorems 14 and 17 imply Theorem 16.There is a considerable literature on stable minimal surfaces. Much of it related to the

Bernstein conjecture:The only entire minimal graphs are linear functions.It was shown it is true for n ≤ 7:

(a) Bernstein for n = 2;

(b) De Georgi for n = 3;

(c) Almgren for n = 4; .

(d) Simons for n = 5, 6, 7.

(b) and (d) used in part results of Fleming on minimal cones. These results were extended bySchoen, Simon and Yau.

Bombieri, De Georgi and Giusti constructed a contra example for n > 7.

13 Asymptotic stability of kinks in the Allen-Cahn equation

We are interested in the asymptotic stability of the families of kink solutions of the non-homogeneous Allen-Cahn equation on R

∂u∂t = ∆u− g(u),u|t=0 = u0(x),

(13.1)

where u : R → R and g : R → R is the derivative of a double-well potential (i.e. G(u) ≥ 0and has two non-degenerate global minima with the minimum value 0, see Figure 1, specifically,g(u) = u3 − u), and we set (say by rescaling) the parameter ε to 1.

Recall that the equation (13.1) has a family (manifold) of stationary solutions, χ(x−a), a ∈R, where χ(z) is a smooth function satisfying χ(z)→ ±1, as z → ±∞. Such solutions are calledthe kinks. (χ(−x+ a) are also solutions and they are called anti-kinks.)

As stated above, we are interested in the asymptotic stability of χ(x). Since χ(x) are not inL2, we consider (13.1) (in a weak sense) on the space X := χa +H1, where χa is the translateof χ, with the notation fa(x) := f(x− a), a ∈ R, and H1 is the standard Sobolev space of order1.


Our result below concerns asymptotic stability of the family (manifold) of the kink solutions

Mkink := χ(x− a) | a ∈ R. (13.2)

We assume that g is such that the problem (13.1) is locally well-posed in X in H2(Rn) for anyinitial condition u0 ∈ X, sufficiently close toMkink. Moreover, we assume that the nonlinearityg(u) satisfies the following conditions:

|N (ξ)| . |ξ|2 + |ξ|k, k ≥ 2, (13.3)

where N (ξ) is defined by the following equation:

N (ξ) := g(χ+ ξ)− g(χ)− g′(χ)ξ. (13.4)

Then we have

Theorem 18. (Asymptotic stability of kinks) Under the assumptions above, the kink manifoldMkink is asymptotically stable. More precisely, there are δ > 0 and a(t) s.t. if dist(u0,Mkink) ≤δ, then the solution u to (13.1) with u|t=0 = u0 satisfies

‖u− χa(t)‖H1 . e−ρ4t

for all times t ≥ 0. Moreover, there exists a real number a∞ such that the function a(t)→ a∞,as t→∞.


A Banach and Hilbert Spaces

In this section, we introduce the simplest and most commonly used spaces, Banach and Hilbertspaces, and describe their most important examples.

A.1 Vector Spaces

We begin with the key definitions. A vector space, X, is a collection of elements, denoted, u, v, ...,for which the operations of addition, (u, v) → u + v and multiplication by a (real or complex)number, (α, u)→ αu, are defined in such a way that

(α+ β)u = αu+ βu

andα(u+ v) = αu+ αv.

Recall that the operations of addition and multiplication have the following properties

u+ v = v + u (commutativity)

andα(βu) = (αβ)u (associativity).

Elements of a vector space are called vectors. As will be clear from the context most of thevector spaces we consider in these lectures are defined for multiplication by complex number(they are said to be vector spaces over complex numbers).

Let Ω ⊂ Rn be an open set of Rn. Main examples of vector spaces are:


(a) Rn = x = (x1, ..., xn)| −∞ < xj <∞ ∀ j– the Euclidean space of dimension n;

(b) Ck(Ω) – the space of k times continuously differentiable complex functions on Ω;

(c) Lp(Ω) – the space of measurable complex functions on Ω whose modulus taken to the p–thpower degree is integrable;

(e) S(Rn) – space of C∞ functions vanishing at ∞ together with all their derivatives fasterthan |x|−n for all n.

The spaces Lp(Ω), 1 ≤ p ≤ ∞, s ∈ R are introduced and studied in Sections A.3.The sets above are vector spaces if we define addition and multiplication by real/complex

numbers in the pointwise way. Namely, for Rn we define

(x+ y)j = xj + yj and (αx)j = αxj ∀ jand, for the remaining spaces in (b) – (e), we define

(f + g)(x) := f(x) + g(x) and (αf)(x) := αf(x) ∀x ∈ Ω.

A.2 Banach Spaces

To measure the size of vectors, one uses the notion of norm. A norm is defined to be a map

X 3 u→ ||u|| ∈ [0,∞)

which has the following properties:

(a) ‖u‖ = 0 ⇐⇒ u = 0;

(b) ‖αu‖ = |α|‖u‖;

(c) ‖u+ v‖ ≤ ‖u‖+ ‖v‖.

The last inequality is called the triangle inequality. (Observe an unusual notation for the norm‖ · ‖, not f(·) or n(·) as one would denote other maps). A vector space equipped with a norm iscalled a normed vector space.

Having defined a norm, we can define the notion of (norm–) convergence as follows. Letfn ⊂ X be a sequence. We say that fn converges to f (∈ X), if and only if ‖fn− f‖ → 0. Wewrite fn → f .

A sequence fn ⊂ X is called a Cauchy sequence iff ‖fn − fm‖ → 0, as m,n→∞.A normed vector space X is called complete if and only if every Cauchy sequence converges,

i.e. if fn ⊂ X is a Cauchy sequence then fn converges (i.e. there is a f ∈ X such that||fn − f || → 0, as n → ∞). Remark that the converse is always true: any convergent sequenceis necessarily a Cauchy sequence. A complete normed vector space is called a Banach space.

Completeness is a very important property of a normed vector space, e.g. often one solves anequation by successive approximations, and one wants to know that such approximate solutionsconverge to an actual solution.

Examples: The vector spaces Rn and Lp(Ω) defined above are Banach spaces under thenorms

|x| := (

n∑i=1

x2i )

1/2, ‖f‖p :=

(∫|f |p

)1/p

if 1 ≤ p <∞.


Denote by Ckb (Ω) the subspace of Ck(Ω) consisting of all functions in Ck(Ω) which arebounded together with all their derivatives up to the order k. We equip the space Ckb (Ω) withthe norm

‖f‖Ck = supx∈Ω

∑|α|=k

|∂αf(x)|,

where α = (α1, ..., αn) with αj non-negative integers, ∂α :=∏nj=1 ∂

αjxj and |α| =

∑ni=1 αj .

Exercise 9. Show that

(a) Ckb (Ω) is a vector space;

(b) ‖f‖Ck is a norm on Ckb (Ω).

It is shown in [27], Proposition 4.13 and Exercise 5.7 that the spaces Ckb (Ω) are complete,i.e., that they are Banach spaces.

A.3 Lp–spaces

Consider an open subset Ω of the Euclidean space Rn. In particular Ω can be a bounded subsetor the entire Rn. Let dx denote the Lebesgue measure on Rn. We define the Lp–space for1 ≤ p <∞:

Lp(Ω) := f : Ω→ C | f is measurable, and

∫|f |pdx <∞. (A.1)

In other words, f ∈ Lp(Ω)⇔ |f |p ∈ L1(Ω). We define the L∞-space:

L∞(Ω) := f : Ω→ C | f is measurable, and ess sup |f | <∞, (A.2)

where, recall that ess sup |f | := infsup |g| : g = f a.e.. We often use the abbreviation Lp forLp(Ω).

Strictly speaking, elements of Lp(Ω) are equivalence classes of measurable functions: twofunctions define the same elements of Lp(Ω) if they differ only on a set of measure 0.

Lp, 1 ≤ p ≤ ∞ is a vector space. For p = ∞, this is obvious, and for 1 ≤ p < ∞, it easilyfollows from the inequality

|f + g|p ≤ 2p (|f |p + |g|p) . (A.3)

The latter inequality is obtained as follows: |f + g|p ≤ (2 max(|f |, |g|))p ≤ 2p(|f |p + |g|p).We define for every f ∈ Lp:

‖f‖p :=

(∫|f |p

)1/pif 1 ≤ p <∞,

ess sup|f | if p =∞.(A.4)

Clearly, ‖f‖p = 0 ⇔ f = 0 a.e., and ‖cf‖p = |c| ‖f‖p, ∀c ∈ C. We have also the triangleinequality ‖f + g‖p ≤ ‖f‖p + ‖g‖p, which we prove later.

From these properties, it follows that the map f 7→ ‖f‖p is a norm. Hence Lp is a normedvector space for every 1 ≤ p ≤ ∞. In fact, it is a Banach space. The proof is given in [27],Theorem 6.6, Proposition 6.7 and Theorem 6.8.

There are three basic inequalities for Lp spaces:

1. Holder’s inequality: Let 1 ≤ p, q, r ≤ ∞, p−1 +q−1 = r−1, and f ∈ Lp, g ∈ Lq then fg ∈ Lrand ‖fg‖r ≤ ‖f‖p‖g‖q.


2. Minkowski’s inequality: Let 1 ≤ p <∞ and f, g ∈ Lp, then ‖f + g‖p ≤ ‖f‖p + ‖g‖p.

3. The Hausdorff-Young inequality : Let 1 ≤ p ≤ 2 and p−1 + q−1 = 1. Then ‖f‖q ≤ ‖f‖p.

We prove the first two inequalities at the end of this appendix. Here we prove Holder’s inequalityin easy, but instructive, special cases, (p−1, q−1) = (0, 1) and (p−1, q−1) = (1/2, 1/2), and sketchthe proof of the Hausdorff-Young inequality.

For (p−1, q−1) = (0, 1), we have∫|fg| ≤ sup |f |

∫|g| = ‖f‖∞‖g‖1.

In the case (p−1, q−1) = (1/2, 1/2), we define first the unit vectors f = f/‖f‖2 and g =g/‖g‖2 (here we can assume f 6= 0 and g 6= 0 otherwise the result is trivial), so that ‖f‖2 = 1and ‖g‖2 = 1 (check this!). Integrating the inequality

|f |2 + |g|2 − 2|f g| ≥ 0,

and using that∫|f |2 = 1 =

∫|g|2, we obtain

1 ≥∫|f g|. (A.5)

Since the integral on the right hand side is∫|fg|/‖f‖2‖g‖2,

we find after multiplying (A.5) by ‖f‖2‖g‖2,

‖f‖2‖g‖2 ≥∫|fg|

as claimed. Note that this inequality, which is a special case of the Holder inequality, has its ownname - the Schwarz inequality (with p = q = 2 and r = 1). One can prove the general Holderinequality by interpolating between the (0, 1) and (1/2, 1/2) cases. But it is more elementary toprove it directly as we do in the next paragraph.

We sketch the proof of the Hausdorff-Young inequality. Clearly, ‖f‖∞ ≤ ‖f‖1. Moreover, wehave shown that ‖f‖2 = ‖f‖2. For 1 < p < 2, the result follows from an interpolation theorem(see e.g. [27]).

Note that the Hausdorff-Young inequality implies that F extends to a bounded operatorfrom Lp to Lq. We omit the proof of this statement.

Theorem 19 (Riemann-Lebesgue Lemma). Suppose that g is s.t. g ∈ L1. Then

i) g is bounded and continuous,

ii) g decays at infinity: lim|x|→∞ g(x) = 0.

Proof. i) The boundedness is easily seen: ∀x,

|g(x)| =∣∣∣∣∫ eikxg(k)

∣∣∣∣ ≤ ∫ |g(k)| = ‖g‖L1 .


Next, we show continuity. Since g ∈ L1, then

limh→0

(g(x+ h)− g(x)) = limh→0

∫eik·x

(eik·h − 1

)g(k)dk = 0,

by the dominated convergence theorem (|eik·h − 1||g| ≤ 2|g|). This shows that g is continuous.Next, let us show ii). Since the Schwartz space S is dense in L1, there is a sequence ϕj ∈ S suchthat ||ϕj − g||1 → 0 as j → 0. Thus

||ϕj − g||∞ ≤∫|ϕj(k)− g(k)| = ||ϕj − g||1 → 0,

which shows that ϕj → g uniformly on Rn. But ϕj ∈ S, so ϕj → 0 as |x| → ∞, and thereforeg → 0 as |x| → ∞.

We can further generalize the C and Lp spaces to spaces of continuous and Lp integrablefunctions from an open set Ω ⊂ Rn to a Banach space X. We denote such spaces as C(Ω, X)and L(Ω, X), respectively.

A.4 Proofs of Holder’s and Minkowski’s inequalities

First, we prove Holder’s inequality. Observe first that it suffices to prove the Holder inequalityfor the case r = 1 (this follows easily from ‖fg‖r = (

∫|fg|r)1/r = (

∫|f |r|g|r)1/r, and e.g. calling

f r = f1 and gr = g1).We notice that the result is trivial if ||f ||p = 0 or ||g||p = 0 (for then f = 0 a.e. or g = 0

a.e.), or if ||f ||p =∞ or ||g||p =∞. If neither of these cases hold, then we can define

a =

∣∣∣∣ f(x)

‖f‖p

∣∣∣∣p , b =

∣∣∣∣ g(x)

‖g‖q

∣∣∣∣q and λ =1

p.

Below, we will show that for any a, b > 0, and 0 < λ < 1:

aλb1−λ ≤ λa+ (1− λ)b. (A.6)

Applying this to our case, we get

|f(x)g(x)|‖f‖p‖g‖q

≤ |f(x)|p

p∫|f |p

+|g(x)|q

q∫|g|q

.

Integrating this inequality, and using p−1 + q−1 = 1, we arrive at Holder’s inequality.We finish the proof by showing (A.6). We can assume b 6= 0, so we can divide (A.6) by b on

both sides. (A.6) is then equivalent to the inequality (where t = a/b)

tλ − λt− 1 + λ ≤ 0.

Since 0 < λ < 1, the maximum of the function on the l.h.s. is reached at t = 1, and is equal to0.

Now we prove Minkowski’s inequality. If either p = ∞ or p = 1, then the result is obvious.Now assume p > 1 and f + g 6= 0. Then we estimate

|f + g|p ≤ (|f |+ |g|)|f + g|p−1


Integrating this inequality and applying Holder inequality with exponents p and p′ (1/p+1/p′ =1) we obtain

∫|f + g|p ≤

∫|f ||f + g|p−1 +

∫|g||f + g|p−1

≤ ‖f‖p‖ |f + g|p−1‖p′ + ‖g‖p‖ |f + g|p−1‖p′

Taking into account that (p− 1)p′ = p, we find furthermore that∫|f + g|p ≤ (‖f‖p + ‖g‖p)‖f + g‖

pp′p

Now remember that∫|f + g|p = ‖f + g‖pp and observe that p/p′ = p− 1. Hence dividing both

sides of the latter inequality by ‖f + g‖p−1p gives Minkowski’s inequality.

Remark 1. Inequality (A.6) is a special case of the very useful Jensen’s inequality : let ϕ be aconvex function on [a, b] (i.e., ϕ satisfies ϕ(tx+ (1− t)y) ≤ tϕ(x) + (1− t)ϕ(y) for all t ∈ [0, 1]and x, y ∈ [a, b]), and pk positive numbers satisfying

∑n1 pk = 1. We can think about pk as

probabilities. Then

ϕ(n∑1

pktk) ≤n∑1

pkϕ(tk), (A.7)

for all tk ∈ [a, b]. This inequality indeed implies (A.6) for ϕ(t) = et, n = 2, p1 = λ, t1 = ln aand t2 = ln b.

Exercise A.1. Prove Jensen’s inequality (A.7).

A.5 Hilbert Spaces

To measure angles between vectors one introduces the inner product (sometimes called the scalarproduct ).

The map (f, g)→ 〈f, g〉 ∈ C is called an inner product if and only if

• 〈f, g〉 is linear in the second argument, i.e. 〈f, αg + βh〉 = α 〈f, g〉 + β 〈f, h〉, for anyα, β ∈ C,

• 〈f, g〉 = 〈g, f〉 (this together with the above implies that 〈f, g〉 is anti-linear in the firstargument: 〈αf + βh, g〉 = α 〈f, g〉+ β 〈h, g〉),

• 〈f, f〉 ≥ 0 with equality if and only if f = 0.

We remark that sometimes, the scalar product is taken to be linear in the first argument andanti-linear in the second one.

The space L2(Ω) is a Hilbert space if we define the inner product as

〈f, g〉 :=

∫Ωfg. (A.8)

Exercise 10. Show that (A.8) is an inner product.

An inner product defines a norm according to

‖f‖ =√〈f, f〉. (A.9)


The Schwarz inequality

| 〈f, g〉 | ≤ ‖f‖ ‖g‖.

It follows from the obvious relation

0 ≤ ‖u± v‖2 = 〈u± v, u± v〉 = ‖u‖2 + ‖v‖2 ± 〈u, v〉 ± 〈v, u〉 (A.10)

applied to u = ±f/‖f‖ and v = g/‖g‖ and to u = ±if/‖f‖ and v = g/‖g‖.

Exercise 11. Check that (A.9) defines a norm. (Hint: to prove the triangle inequality, use theSchwarz inequality).

Thus a space with an inner product, or an inner product space, is also a normed space.A complete inner product space is called a Hilbert space. Clearly a Hilbert space is also aBanach space. An inner product, in addition to measuring sizes and distances, measures relativedirections, e.g.,

f ⊥ g ⇔ 〈f, g〉 = 0.

Given a norm ‖f‖, how can we tell whether this norm comes from an inner product? Theanswer to this question is that ‖f‖ comes from an inner product if and only if it satisfies theparallelogram law

‖f + g‖2 + ‖f − g‖ = 2‖f‖2 + 2‖g‖2.

In this case the inner product is given by (in the real case)

〈f, g〉 =1

2

(‖f + g‖2 − ‖f‖2 − ‖g‖2

)(A.11)

(see (A.10)).

Exercise 12. Check that (A.11) defines a (real) inner product.

The space Rn has an inner product defined by x · y =∑n

i=1 xiyy.

A.6 Sobolev spaces

Now we want to introduce an additional structure on Lp–spaces which measures smoothness,similar to the smoothness properties of functions in Ck. We do so only for p = 2, i.e. for thespace L2(Rn). This is the most useful space among the Lp–spaces as it has an inner product:

〈f, g〉 :=

∫fg

and therefore it is a Hilbert space (i.e., recall, an inner product space which is complete withrespect to the norm ||f || :=

√〈f, f〉 induced by the inner product). Another advantage of the

L2–space is that the Fourier transform leaves it invariant (i.e. f ∈ L2 ⇒ f ∈ L2).

We now define for s integer, s ≥ 0, the new spaces

Hs(Rn) = f ∈ L2(Rn) : ∂αf ∈ L2(Rn) ∀α s.t. |α| ≤ s. (A.12)

This definition is very similar to the definition of the Cs(Rn)–spaces: in fact, by replacingL2(Rn) in (A.12) by C(Rn), one obtains the definition of Cs(Rn). But there is one crucial


difference: in the Cs(Rn)–case, the functions f are assumed to be s times continuously differ-entiable, but in the Hs–case, they are not. Namely, the derivatives ∂αf in the above definitionare understood in the distributional sense:

∂αf(ϕ) := (−1)|α|f(∂αϕ),

where f(ϕ) =∫fϕ, and ϕ ∈ S(Rn). In other words ∂αf is a linear functional defined by the

r.h.s. of the above equation and the relation in (A.12) says that ∀α with |α| ≤ s, there exists agα ∈ L2(Rn) s.t. ∂αf(ϕ) =

∫gαϕ ∀ϕ.

The space Hs(Rn) equipped with the inner product

〈f, h〉(s) =∑|α|≤s

〈∂αf, ∂αg〉 (A.13)

is a Hilbert space.There is another way of defining the spaces Hs(Rn) using the Fourier transform defined in

the next appendix:Hs(Rn) = f ∈ L2(Rn) : 〈k〉sf(k) ∈ L2(Rn), (A.14)

where 〈k〉 = (1 + |k|2)1/2. The definition (A.14) has the advantage that it makes sense for anarbitrary s ∈ R. Besides, it does not require extra explanations. Of course we have to show thatdefinitions (A.12) and (A.14) are equivalent for positive integers s. One can show easily thatthe definitions (A.12) and (A.14) are equivalent for positive integers s.

B Fourier transform

In this section, we describe one of the most powerful tools in analysis – the Fourier transform.This transform allows us to analyze a fine structure of functions and to solve differential equa-tions. The Fourier transform takes functions of time to functions of frequencies, functions ofcoordinates to functions of momenta, and vice versa.

B.1 Definitions and properties

Initially, we define the Fourier transform on the Schwartz space S(Rn) = S:

S = f ∈ C∞(Rn) : 〈x〉N |∂αf(x)| is bounded ∀N and ∀α, (B.1)

where 〈x〉 = (1 + |x|2)1/2 and α = (α1, ..., αn), with αj non-negative integers, ∂α :=∏nj=1 ∂

αjxj

and |α| =∑n

i=1 αj . On S, we define the Fourier transform F : f 7→ f by

f(k) := (2π)−n/2∫f(x)e−ik·xdx. (B.2)

Define also the inverse Fourier transform of f(k) as

f(x) := (2π)−n/2∫f(k)eix·kdk. (B.3)

Some key properties of the Fourier transform are collected in the following

Theorem 20. Assume f, g ∈ S(Rn). Then we have:


(a) (−i∂)αf 7→ kαf , and xαf 7→ (−i∂)αf ,

(b) fg 7→ (2π)−n/2f ∗ g, and f ∗ g 7→ (2π)n/2f g,

(c) (f ) = f = (f ) ,

(d) f = f ,

(e)∫fg =

∫fg,

(f)∫f g =

∫fg.

Properties (a) - (f) hold (possibly, with signs changed in (a)) also whenˆ is replaced by .

We give a formal proof. Integrating by parts, we compute

−i(∂xjf ) (k) = (2π)−n/2∫

(−i)∂xjf(x)e−ik·xdx

= (2π)−n/2∫f(x)i∂xje

−ik·xdx

= kj f(k).

Exercise 13. Prove the remaining relations in (a)

Now we prove the second relation in (b). Using e−ik·x = e−ik·(x−y)e−ik·y and changing thevariable of integration as x′ = x− y, we obtain

f ∗ g (k) := (2π)−n/2∫e−ik·x(

∫f(x− y)g(y)dy)dx

= (2π)−n/2∫

(

∫e−ik·(x−y)f(x− y)dx) e−ik·yg(y)dy

= (2π)n/2f(k′)f(k)g(k).

Exercise 14. Prove the first relation in (b) from the second one and (c).

The proof of (c) is more subtle. We use an approximation of unity ϕt(x) = t−nϕ(x/t) andcompute ϕt ∗ (f ) . Let us define ϕx(y) := ϕ(x− y). Using property (b), we find

ϕt ∗ (f ) =

∫ϕxt · (f ) dy =

∫(ϕxt ) fdy =

∫((ϕxt ) ) fdy.

Exercise 15. Show (formally, without justification of the interchange of the order of integrationetc.) that

((ϕxt ) ) = ((ϕt) )x = t−n(ϕ)

(x− yt

).

Thus we haveϕt ∗ (f ) = ((ϕ) )t ∗ f (B.4)

We can choose ϕ such that (ϕ) ∈ L1, and∫

(ϕ) (x)dx = 1. Indeed, take e.g. ϕ(x) =

(4π)−n/2e−|x|2

and use the fact that ((e−|x|2) ) = e−|x|

2. With this in mind, we take the limit

t→ 0 in (B.4) and use the properties of the approximation of identity to get

ϕt ∗ (f ) → (f ) and ((ϕ) )t ∗ f → f as t→ 0

to obtain (f ) = f . Similarly one shows that (f ) = f .


Exercise 16. Prove the relations in (d) – (f).

By definition of the Dirac δ–function, we obtain

F : δ(x− x0)→ (2π)−n/2e−ik·x0 .

Hence the property (c) implies that F−1 : (2π)−n/2e−ik·x0 → δ(x − x0), and, by taking thecomplex conjugate (remember that F(f) = F∗(f) = F−1(f)), we arrive at

F : (2π)−n/2e−ik0·x 7→ δ(k − k0). (B.5)

Exercise 17. Using (B.5), prove formally that (f ) = f = (f ) , and that (fg) = (2π)−n/2f ∗ g.

Statement (f) is called the Plancherel Theorem. The adjoint F∗ of the Fourier transform isdefined by 〈F∗u, v〉 = 〈u,Fv〉 for all u, v ∈ S(Rn), where 〈·, ·〉 is the standard inner product inL2(Rn). Then (d) and (e) show that F∗ = F−1. This together with (e) implies that FF∗ =id = F∗F on S, which is a restatement of the Plancherel theorem.

Corollary B.1. F extends to a unitary operator on L2, i.e. to a bounded operator satisfyingF∗ = F−1.

The next theorem gives the important example of the Fourier transform - the Fourier trans-form of a Gaussian :

Theorem 21. Let A be a n × n matrix s.t. ReA := (A + A∗)/2 is positive definite (i.e.x · ReAx > 0 if x 6= 0). Then we have

F : e−x·Ax/2 7→ (detA)−1/2e−k·A−1k/2 (B.6)

Proof. We prove the theorem only for positive definite matrices. If A is positive definite (i.e. ifx ·Ax > 0 for x 6= 0), then there is an orthogonal matrix U (i.e. U is real and UUT = UTU = id)s.t. A := UTAU is diagonal, say A = diag(λ1, . . . , λn). Letting x = Uy and noticing thatx ·Ax = y · UTAUy, and that detU = 1, we get∫

e−x·Ax/2e−ik·xdx =

∫e−y·Ay/2eik

′·ydy =

n∏1

∫e−λjy

2j /2eik

′jyjdyj ,

where k′ = UTk, and we have used k · Uy = UTk · x. It is left as an exercise to show that forn = 1,

F : e−λx2/2 7→ λ−1/2e−k

2/2λ. (B.7)

The last two relations imply the desired statement.

Exercise 18. Show (B.7).

The function e−x·Ax is called a Gaussian. It is one of the most common functions in applica-tions. There is another important function whose Fourier transform can be explicitly computed:

F : |x|−α 7→Cn,α|k|−n+α if α 6= n,Cn,n ln |k| if α = n.

(B.8)

The coefficients are given for α = 2 by

Cn,2 =

((2− n)σn−1)−1 for n 6= 2,−(σn−1)−1 = −(2π)−1 for n = 2,

(B.9)


where σn is the volume of the n–dimensional unit sphere Sn = x ∈ Rn+1 : |x| = 1. Onecan easily deduce formula (B.8) modulo the constants (B.9). Indeed, since |x|−α is rotationallyinvariant, then so is its Fourier transform (see Exercise 20 below). Also, since |x|−α is homoge-neous of degree −α, then its Fourier transform is homogeneous of degree −n+ α (see Exercise20 below). Hence (B.8) follows. Though it is easy to compute the Fourier transform of |x|−α,it is not easy to justify it. Indeed, the function |x|−α is rather singular and definitely does notbelong to S(Rn).

Exercise 19. For n = 1, compute the Fourier transform of the characteristic function χ(−a,a)(x),using definition (B.2).

As an example we show the following relation

((|k|2 + µ2)−1) =e−µ|x|

4π|x|, for n = 3, (B.10)

which appears often in applications. Indeed, let f(k) = (|k|2 + µ2)−1 and n = 3. We have

f(x) = (2π)−3/2

∫R3

eik·x

|k|2 + µ2dk

= limR→∞

(2π)−1/2

∫ R

0

∫ 1

−1

eir|x|vr2

r2 + µ2drdv

= limR→∞

(2π)−1/2

i|x|

∫ R

−R

eir|x|r

r2 + µ2dr.

In the second equality, we change to spherical coordinates v = cosφ = k·x|k||x| with r = |k|. Now,

the last integral can be computed by changing to a contour integral over a rectangle in the upperhalf complex plane with top vertices at −R+ i

√R and R+ i

√R and taking the limit as R→∞.

By the Residue theory we have (B.10) (show this).

Exercise 20. Let fh(x) := f(x−h) for h ∈ Rn, f (λ)(x) := λn/2f(λx) for λ ∈ R+ and f (g)(x) :=f(gx) for ga ∈ SO(n) be the translation, dilation and rotation of f(x). Show that

F : fh(x) 7→ e−ik·hf(k), (B.11)

F : f (λ)(x) 7→ f (1/λ)(k). (B.12)

F : f (g)(x) 7→ f (g−1)(k). (B.13)

Define the unitary operators Th : f(x) 7→ f(x − h) (the operator of translation by h) andSλ : f(x) 7→ λn/2f(λx) (the operator of dilation by λ). Then (B.11) and (B.12) imply

F Th = Me−ik·h F ,

where we recall that Me−ik·h is the operator of multiplication by e−ik·h, and

F Sλ = S1/λ F .

Note the following important property of the Fourier transform, called the uncertainty principle:let f ∈ S(Rn) be a Schwartz function on Rn, s.t. ‖f‖L2(Rn) = 1. Then(∫

|x|2|f(x)|2dnx)1/2(∫

|k|2|f(k)|2dnk)1/2

≥ 1/2. (B.14)


Exercise 21. Prove inequality (B.14) (Hint: in the case n = 1, notice that

〈f, (−i∂x)x− x(−i∂x)f〉 = −i‖f‖2,

where the scalar product and the norm are in L2(R). On the other hand,

〈f, (−i∂x)x− x(−i∂x)f〉 = 2iIm 〈(−i∂x)f, xf〉 .

Use these two observations to show that ‖f‖2 ≤ 2‖(−i∂x)f‖ ‖xf‖, and finish the proof by invok-ing Plancherel’s theorem).

Let w ∈ L2(Rn) be a given function. Then the integral

(2π)−n/2∫w(x− y)f(x)e−ik·xdnx (B.15)

is called the windowed Fourier transform of f (with the window function w). In signal analysis,i.e. for n = 1, one often uses the Gaussian (2π)−1/2e−x

2/(2α) for w. In this case, the transform(B.15) is called the Gabor transform.

For more details see [38], Chapter 5.

B.2 Applications of Fourier transform to partial differential equations

Our goal in this section is to apply the Fourier transform in order to solve elementary but verybasic partial differential equations (PDE’s).

The Poisson equation on Rn:−∆u = f, (B.16)

where u : Rn → R is an unknown function, f : Rn → R is a given function, and ∆ is the Laplaceoperator (the Laplacian):

∆u :=n∑j=1

∂2u

∂x2j

.

The Poisson equation first appeared in the problem of determining the electric potential u(x),created by a given charge distribution ρ(x) = f(x)/(4π). Since then, it came up in various fieldsof mathematics, physics, engineering, chemistry, biology and economics.

In order to solve the Poisson equation, we apply the Fourier transform to both sides of (B.16)to obtain:

|k|2u(k) = f(k).

This equation can be easily solved: u = f/|k|2. We can now apply the inverse Fourier transformto the last equality to get

u = g ∗ f, where g(k) = |k|−2. (B.17)

But the inverse Fourier transform of g(k) = |k|−2 is known:

g(x) =

[(2− n)σn−1]−1|x|−n+2 if n 6= 2[2π]−1 ln |x| if n = 2,

where σn is the volume of the unit–sphere Sn = x ∈ Rn+1 : |x| = 1 in dimension n.


Explicitly, (B.17) can be written as

u(x) = [(2− n)σn−1]−1

∫f(y)

|x− y|n−2dy,

for n 6= 2, and similarly for n = 2. In particular, for n = 3, we have the celebrated Newtonformula

u(x) = − 1

4π

∫f(y)

|x− y|dy.

Of course, the functions appearing in the above derivation are not necessarily from the Schwartzspace S and therefore these manipulations must be justified. We leave this as an exercise, whileproceeding in a similar fashion with other equations.

The heat equation on Rn:

∂u

∂t= ∆u and u|t=0 = u0, (B.18)

where u : Rnx × R+t → R is an unknown function, and u0 : Rn → R is a given initial condition.

Problem (B.18) is called an initial value problem. It first appeared in the theory of heat diffusion.In that case, u0(x) is a given distribution of temperature in a body at time t = 0, and u(x, t)is the unknown temperature–distribution at time t. Presently, this equation appears in variousfields of science, including mathematical modeling of stock markets.

As before, we apply the Fourier transform to (B.18) and solve the resulting equation

∂u

∂t= −|k|2u and u|t=0 = u0

to get u = e−|k|2tu0. Applying the inverse Fourier transform, and using that (e−|k|

2t) =(4πt)−n/2e−|x|

2/(4t), we obtain

u = g√2t∗ u0, (B.19)

where gs(x) = s−ng(x/s) with g(x) = (2π)−n/2e−|x|2/2. In particular, u → u0 as t → 0, as it

should be.

The Schrodinger equation on Rn:

i∂ψ

∂t= −∆ψ and ψ|t=0 = ψ0. (B.20)

This is an initial value problem for the unknown function ψ : Rnx × R+t 7→ C. Equation (B.20)

describes the motion of a free quantum particle. Proceeding as with the heat equation, we obtain

u = φ√2t∗ ψ0, (B.21)

where φs(x) = s−nϕ(x/s) with φ(x) = (2πit)−n/2ei|x|2/(2t). Observe that this formula can be

obtained from (B.19) by performing the substitution t→ t/i.

Exercise 22. Derive equation (B.21) using the Fourier transform.


The wave equation on Rn:

∂2u

∂t2= ∆u with u|t=0 = u0 and ∂tu|t=0 = u1. (B.22)

This is a second order equation in time and consequently, it has two initial conditions u0 and u1.The wave equation (B.22) describes various wave phenomena: propagation of light and sound,oscillations of strings, etc. Proceeding as with the heat equation, we find

u = ∂tWt ∗ u0 +Wt ∗ u1, (B.23)

where Wt(x) is the inverse Fourier transform of the function sin(|k|t)/|k|. The latter can becomputed explicitly for n = 1, 2, 3:

Wt(x) =

12χρ2≥0 for n = 1,(2π)−1ρ−1χρ2≥0 for n = 2,(2π)−1δ(ρ2) for n = 3,

where ρ2 := t2 − |x|2, and χρ2≥0 stands for the characteristic function of the set (x, t) ∈ R3+1 :ρ2 ≥ 0, i.e.

χρ2≥0 =

1 if ρ2 ≥ 0,0 otherwise,

and δ(x) is the Dirac δ–function, a generalized function, or distribution.Thus the dependence of W on x and t comes through the combination ρ2 = t2 − |x|2,

which is the Minkowski–distance in space–time, playing a crucial role in relativity. Observe thatχρ≥0 = χ|x|≤t and δ(ρ2) = (2t)−1δ(t− |x|).

Exercise 23. Prove (B.23), and find Wt(x) for n = 1.

We examine closer the special case when n = 3 and u0 = 0. Then we get

u = Wt ∗ u1 ≡ 1

4πt

∫δ(|x− y| − t)u1(y)dy

=1

4πt

∫S(x,t)

u1(y)dS(y)

=t

4π

∫S(0,1)

u1(x+ tz)dS(z),

where S(x, t) = y ∈ R3 : |y − x| = t is a sphere of radius t centered at x. We see that onlythe initial condition evaluated on the sphere S(x, t) matters in order to determine the solutionat time t and at position x. This is called the Huygens’ principle.

B.3 Estimates on propagators

In this subsection we derive estimates on solution of the Schrodinger equation (B.20) which playan important role in applications.

Theorem 22. Let ψ := ei∆tψ0 be the solution to the Schrodinger initial value problem (B.20).Then we have the estimate

‖ei∆tψ0‖p ≤ (4πt)−n

2+np ‖ψ0‖p′ (B.24)

where p and p′ are indices satisfying 1p + 1

p′ = 1 and 2 ≤ p ≤ ∞.


Proof. We use representation (B.21) of the solution ψ:

ψ(x, t) = (4πit)−n/2∫ei|x−y|

2/4tψ0(y) dy.

A simple estimate gives |ψ(x, t)| ≤ (4πt)−n/2∫|ψ0(y)|dy, i.e.,

‖ψ‖∞ ≤ (4πt)−n/2‖ψ0(y)‖L1 .

This is estimate (B.24) for p = ∞ and p′ = 1. In the general case expanding |x − y|2 =|x|2 − 2x · y + |y|2 we write

ψ(x, t) = (4πit)−n/2ei|x|2/4t

∫e−ix·y/2tei|y|

2/4tdy.

We rewrite this expression as

ψ = (2i)−n/2MfStFMfψ0, (B.25)

where f = ei|x|2/4t, Mf is the multiplication operator by the function f , F is the Fourier

transform and St is the rescaling operator

(Stu)(x) = t−n/2u(xt).

Since ‖Mfu‖p ≤ ‖f‖∞‖u‖p for any 1 ≤ p ≤ ∞, we have by the Hausdorff–Young inequality (seeSection B)

‖Fu‖p ≤ ‖u‖p′

for any p and p′ as in the theorem. Finally, we use that, for any 1 ≤ p ≤ ∞,

‖Stu‖p ≤ t−n2

+np ‖u‖p.

(Prove the above inequality.) Recalling equation (B.25) and applying the three inequalities weobtain the desired estimate (B.24).

C Linear operators

Linear operators or simply operators are linear maps from one vector space Y into anothervector space X. We denote linear operators usually by capital roman letters, A,B, . . . and usethe notation

A : Y → X (C.1)

and Au to denote an operator A mapping Y into X and application of A to a vector u ∈ Y ,respectively. To define an operator A, we have to give a rule that prescribes to each elementof Y an element of X (the image of u). We require this rule to be linear, i.e. ∀u, v ∈ Y , andα, β ∈ C:

A(αu+ βv) = αAu+ βAv. (C.2)

To fix ideas here and in what follows, we consider vector spaces over the complex numbers C,i.e. complex vector spaces. All the material of this section, except for spectral theory, remainsunchanged if we substitute R for C.


If the space Y in (C.1) is a subset of the space X then sometimes one calls Y the domainof A (in X) and denotes it D(A) ≡ Y . In this case we say that A is defined in X with domainD(A)(= Y ). The range (or image) of A is defined as

Ran (A) := Au : u ∈ Y ≡ AY.

Ran (A) is a vector space (show this). We may assume that D(A) is dense in X, i.e. for anyu ∈ X, there is a sequence un ⊂ D(A) s.t. un → u as n → ∞. Indeed, if D(A) is not denseto begin with, we consider instead of the space X simply the space Y := D(A), the closure ofD(A), which is obtained by adding to Y limits of all possible sequences un convergent in X.

Examples.1) The identity operator 1l : Lp → Lp;2) The multiplication operator Mf : L→Lp, u 7→ fu for a fixed f ∈ L∞;3) The differentiation operator ∂

∂xjin L2(Rn) with the domain D( ∂

∂xj) = H1(Rn);

4) The Laplacian ∆ :=∑n

1∂2

∂x2jin L2(Ω) with the domain D(∆) = H2(Ω);

5) Integral operators, i.e., operators of the form

(Ku)(x) =

∫K(x, y)u(y)dy,

for some function K(x, y) (called the kernel or integral kernel). The domain and range of theintegral operator K depend on the properties of the kernel K(x, y).

6) The Fourier transform F : L1(Rn)→ L∞(Rn),

F : u(x)→ (2π)−n/2∫

e−ik·xu(x)dx;

7) Convolution operator Cf : Lp → Lp, Cf : u→ f ∗ u for fixed f ∈ L1, where

(f ∗ u)(x) =

∫f(x− y)u(y)dy;

8) The wavelet transform

Wψ : f →∫ψabfdx,

where ψab = 1√|a|ψ(x−ba

)for a fixed function ψ satisfying

∫ |ψ(k)|2k dk <∞.

Note that the Fourier transform and convolution are integral operators with the integralkernels (2π)−n/2eik·x and f(x− y), respectively.

In fact also in examples 1)-4), the operators can be represented as integral operators, butwith distributional kernels, e.g. K(x, y) = f(x)δ(x − y) for Mf , and K(x, y) = −δ′(xj −yj)∏i 6=j δ(xi − yi) for ∂

∂xj.

For an operator A : Y → X we define the norm

‖A‖ ≡ ‖A‖Y→X = sup‖u‖Y =1

‖Au‖X . (C.3)

If Y = X and ‖A‖ < ∞, then A is said to be bounded (in X). Observe that our definitionimplies that


‖Au‖ ≤ ‖A‖‖u‖ (C.4)

for all u ∈ Y . Now, if there is a constant C (independent of u) such that

||Au|| ≤ C||u||, (C.5)

for all u ∈ D(A), and if the space X is complete (i.e. a Banach space), then the operator A canbe extended by continuity to the whole space X as a bounded operator. The smallest constantC satisfying (C.5) is the norm ‖A‖ of A, and so ||Au|| ≤ ||A|| ||u|| (so bounded operators forma Banach algebra).

Examples of bounded operators are the identity operator, 1, of example 1) above (in fact,‖1‖ = 1), the multiplication operator, Mf , of example 2), the integral operator, K, of example6) with a kernel K(·, ·) ∈ L2(Rn × Rn), as an operator from L2(Rn) to L2(Rn).

Exercise C.1. Show that

(1) ‖AB‖ ≤ ‖A‖‖B‖;

(2) ‖w‖ = sup‖v‖=1 | 〈w, v〉 |, and therefore

‖A‖ = sup‖u‖,‖v‖=1

| 〈Au, v〉 |;

(3) ‖Mf‖ = ‖f‖∞ (Example 2 above);

(4) for integral operators (Example 5 above)

||K||L2→L2 ≤(∫|K(x, y)|2dxdy

)1/2

provided the r.h.s. is finite.

It is easy to see that F : L1(Rn)→ L∞(Rn),∣∣∣∣(2π)−n/2∫eix·kf(x)dx

∣∣∣∣ ≤ (2π)−n/2∫|f(x)|dx.

It is considerably more difficult to show that F extends from L1(Rn) ∩ L2(Rn) to a boundedoperator on L2(Rn). In fact F is an isometry in the sense that ‖Ff‖L2 = ‖f‖L2 (the Planchereltheorem, see Appendix B).

Exercise C.2. Show that the differentiation operator in example 3) is not bounded in L2(Rn)by finding a sequence fn of functions from D( ∂

∂xj) such that ||fn|| ≤ 1, ∀n, and || ∂∂xj fn||2 →∞,

as n→∞.

We say an operator A : Y → X is invertible if and only if there is an operator A−1 : X → Ysuch that A−1A = 1lY and AA−1 = 1lX , where 1lX and 1lY are the identity operators in X andY respectively.


Exercise 24. Show that:

1) A is invertible if and only if for every f ∈ X, the equation Au = f has a unique solutionu(= A−1f) ∈ Y , i.e. if and only if A is one–to–one (Au = 0⇒ u = 0) and onto (Ran A = X).

2) if A is just one-to-one (i.e. not necessarily onto), then A is invertible as an operator fromY to 〉A ⊂ X, i.e. the equation Au = f has a unique solution u = A−1f for any f ∈〉A.

3) if A and B are invertible then so is AB and (AB)−1 = B−1A−1 (generalize to an arbitrarynumber of factors).

Example C.1. We show invertibility of several important operators we encountered above.

1. 1 is clearly invertible.

2. Mf is invertible for essinf |f | > 0 and M−1f = M1/f .

3. ddx (see discussion below).

4. F is invertible with F−1 given in (B.3) (see Theorem 20(f)).

5. −∆ + 1 is invertible with (−∆ + 1)−1f = G ∗ f where G := F−1( 1|k|2+1

) and ∗ denotes

convolution.

6. curl is invertible with curl−1w =∫ (x−y)2

|x−y|2 w(y)dy in 2d (Biot-Savart formula).

Consider the operator A = ddx : H1(R) → L2(R). Then A is one-to-one: Au = 0 and

u ∈ H1(R) imply u ≡ 0. Thus A has the right inverse B: f(x) →∫ xx0f(y)dy for some fixed

x0 ∈ R: AB = 1l. However, B does not map L2(R) into H1(R). In fact it is not defined onthe entire L2(R) but only on L2(R) ∩ L1(R) and maps this space into L∞. Put differently,the operator A is not onto: Ran A = f ∈ L2(R)|

∫∞−∞ fdx = 0 6= L2(R). Hence A in not

invertible.

The above illustrates the importance of finding inverses of operators: existence of an inversefor A : Y → X is equivalent to the equation Au = f having a unique solution for every f ∈ X.The following simple statement gives a powerful criterion for existence of inverses.

Theorem 23. Assume an operator A : X → X is invertible and an operator B : X → X isbounded with the norm satisfying the inequality

‖B‖ < ‖A−1‖−1

Then the operator A + B (defined on the domain of A) is invertible. Moreover, its inverse isgiven by the absolutely convergent series

(A+B)−1 =∞∑n=0

A−1(−BA−1)n, (C.6)

called the Neumann series (for A+B).

Exercise 25. Prove this theorem. Hint: Show that the series (C.6) is absolutely convergent andgives the inverse to A + B and use that if Tn is a Cauchy sequence of operators from X to X,then there is an operator T : X → X s.t. Tn → T (one has to use the fact that the space ofbounded operators equipped with the operator norm is complete, i.e. is a Banach space).


The adjoint. Consider an operator A on a Hilbert space X. With it we associate its adjointA∗ defined by the relation 〈A∗u, v〉 = 〈u,Av〉, for all v ∈ D(A), and for all u’s such thatsupv∈D(A),‖v‖=1 |〈u,Av〉| <∞ (those u’s form the domain of the operator A∗, D(A∗)).

If A is a bounded operator then sup‖v‖=1 |〈u,Av〉| ≤ ||u||||A|| < ∞ and it suffices to checkonly the relation 〈A∗u, v〉 = 〈u,Av〉.

Exercise C.3. Show that if operators A and B are bounded, then

(a) A∗ is a bounded operator and ‖A‖ = ‖A∗‖ (Hint: use that ‖A‖ = sup||u||,||v||=1 |〈u,Av〉|,see Exercise C.1(2)).

(b) (A+B)∗ = A∗ +B∗,

(c) (αA)∗ = αA∗,

(d) (AB)∗ = B∗A∗,

(e) (A−1)∗ = (A∗)−1.

An important class of operators on a Hilbert space is the class of self–adjoint operators.By definition, an operator A is called self–adjoint if and only if A∗ = A. By definition, everyself–adjoint operator is symmetric, i.e. 〈Au, v〉 = 〈u,Av〉, for all u, v ∈ D(A). Notice that theconverse is not true. If an operator A is symmetric then all we know is that D(A) ⊂ D(A∗)(show this!). However, a symmetric operator A obeying D(A) = D(A∗) is also self-adjoint. Thusevery symmetric bounded operator is self–adjoint.

Consider the examples of the operators 1)–5) above. We have the following: Mf is symmetricif and only if f is a real function; ∂

∂xjis anti–symmetric, so the differentiation operator is not

symmetric, but −i ∂∂xj

is symmetric; the identity operator is obviously symmetric; ∆ is symmet-

ric and so is −∆ + V (x) for V (x) real; the integral operator is symmetric if K(x, y) = K(y, x)(cf with matrices!). In addition, if we know that K(x, y) ∈ L2(Rn×Rn), then the operator K isself-adjoint. A point is that while the symmetry property is easy to verify, the self-adjointnessproperty is hard. See [47] or [31] for a proof of self-adjointness of −i ∂

∂xj, ∆ and −∆ + V (x).

We observe that for any operator A on a Hilbert space X, we can write

X = nullA⊕ Ran A∗ (C.7)

Here, the null space is defined as nullA := u ∈ X : Au = 0.

Exercise 26. Show that for a bounded operator A, nullA is a closed set, and show (C.7).

Projections. A bounded operator P on X is called a projection operator (or simply a pro-jection) if and only if it satisfies

P 2 = P.

This relation implies ‖P‖ ≤ ‖P‖2, i.e. ‖P‖ ≥ 1. We have

v ∈ Ran P ⇐⇒ Pv = v and v ∈ (Ran P )⊥ ⇐⇒ P ∗v = 0. (C.8)

Indeed, if v ∈ Ran P , then there is a u ∈ X s.t. v = Pu, so Pv = P 2u = Pu = v; the secondstatement is left as an

Exercise 27. Prove that (a) P ∗v = 0 if and only if v ⊥ Ran P , (b) Ran P is closed and (c)P ∗ is also a projection.


Examples. The following are projection operators:

1) on a space of functions u(x), χx∈E : u(x) 7→ χE(x)u(x), where χE(x) = 1 if x ∈ E andχE(x) = 0 if x /∈ E,

2) on a space of functions u(x), χp∈E = F−1χx∈E F acting as u(x) 7→ (χE(k)u(k)) (x),

3) on any Hilbert space, where ϕ,ψ are two fixed elements s.t. 〈ϕ,ψ〉 = 1: u 7→ 〈ϕ, u〉ψ4) as in 3), but where now ψi is an orthonormal set (i.e. 〈ψi, ψj〉 = δi,j): u 7→

∑N1 〈ψi, u〉ψi.

A projection P is called an orthogonal projection if and only if it is self-adjoint, i.e. if andonly if P = P ∗. Let P be an orthogonal projection, then (C.8) implies that

v ⊥ Ran P ⇐⇒ Pv = 0, i.e., nullP = (Ran P )⊥. (C.9)

The projections in Examples 1), 2) and 4) above are orthogonal. The projection in Example3) is orthogonal if and only if ϕ = ψ.

Exercise 28. Let P be an orthogonal projection. Show that

(a) ||P || ≤ 1, and therefore ||P || = 1 (Hint: Use (C.8)),

(b) 1l− P is also an orthogonal projection and Ran (1l− P )⊥Ran P and null 1l− P = Ran P ,

(c) X = Ran P ⊕ nullP .

Remark. Orthogonal projections on X are in one-to-one correspondence with closed subspacesof a Hilbert space X. This correspondence is obtained as follows. Let V = Ran P . Then V isa closed subspace of X. To show that V is closed, let vn ⊂ V , and vn → v ∈ X, and showthat v ∈ V . Since P is a projection, we have vn = Pvn, so ||v − Pv|| = ||v − vn − P (v − vn)|| ≤||v − vn||+ ||P || ||v − vn|| → 0, as n→∞. Therefore v = Pv, so v ∈ V , and V is closed.

Conversely, given a closed subspace V , define a projection operator P by

Pu = v, where u = v + v⊥ ∈ V ⊕ V ⊥. (C.10)

Exercise 29. Show that P defined in (C.10) is an orthogonal projection with Ran P = V . Forany given V , show that there is only one orthogonal projection (the one given in (C.10)) suchthat Ran P = V .

(*Galerkin approximation (A→ PAP )*)

The space of bounded linear operators L(X,Y ) We assume that X and Y are normedvector spaces over C, and consider the set of all bounded linear operators from X into Y , i.e.each such operator is defined on the entire space X, and its range lies in Y . This set of operatorsis denoted by L(X,Y ).

For A,B ∈ L(X,Y ), we define a new operator, called A+B, by setting (A+B)u := Au+Bu,for all u ∈ X. Also, for λ ∈ C and A ∈ L(X,Y ), we define a new operator λA as (λA)u = λAu,for all u ∈ X. If in addition to these two operations on operators, we equip the set L(X,Y )with the norm introduced in (C.3), then L(X,Y ) is a normed vector space.

Exercise 30. Show that L(X,Y ) is a vector space.

An important question is: when is L(X,Y ) a Banach space? The answer is given in thefollowing theorem, which is not difficult to prove (see e.g. [27], Proposition 5.3):

Theorem 24. If Y is a Banach space, then L(X,Y ) is a Banach space.


The dual space. In the special case when Y = C, the space L(X,Y ) is called the dual spaceof X (or simply the dual, or adjoint space or conjugate space of X), and it is denoted as X ′.Hence the elements of X ′ := L(X,C) are linear maps from X to C, and they are called linearfunctionals. Remark also that since C is complete, then the last theorem shows that X ′ is alwaysa Banach space, whether X is complete or not.

The operator norm induces a norm on X ′: if l ∈ X ′, then

||l|| = sup||x||=1

|l(x)|.

If X is a space of functions, then X ′ can be identified with either a space of functions or a spaceof distributions or a space of measures. Here are some examples of dual spaces:

1) (Lp)′ = Lq, where 1/p+ 1/q = 1, if 1 ≤ p <∞ (space of functions),

2) (L∞)′ is a space of measures which is much larger than L1,

3) (Hs)′ = H−s (space of distributions if s > 0).

Note that (Lp)′ ⊃ Lq, for 1 ≤ p < ∞ follows from the Holder inequality. In fact, givenf ∈ Lq, define lf (u) :=

∫fu. Since |lf (u)| ≤ ||f ||q||u||p, we see that lf is a bounded linear

functional on Lp. It can be shown that in fact any bounded linear functional on Lp can berepresented by lf for some f ∈ Lq.

Spectrum. Consider an operator A acting on a Banach space X with a domain D(A) (i.e.,A : D(A)→ X). The spectrum, σ(A), of an operator A is the set in C defined by

σ(A) := z ∈ C : A− z1l is not invertible . (C.11)

For notational convenience, the operator “multiplication by z ∈ C” will be simply written asz instead of z1l. Clearly, eigenvalues of A belong to σ(A) (in fact, if λ is an eigenvalue, thenAuλ = λuλ for some nonzero uλ ∈ X (uλ is called an eigenvector), so (A− λ)uλ = 0, and A− λis not invertible). In general, the spectrum can also contain continuous pieces and it can takevery peculiar forms.

Exercise 31. The spectrum of the multiplication operator introduced in example 1) above isσ(Mf ) = Ran f , the differentiation operator 2) has spectrum σ( ∂

∂xj) = iR, and the identity

operator 3) has the spectrum consisting of one point σ(1l) = 1.

D Elements of spectral theory

D.1 Definitions

Consider an operator A acting on a Banach space X with a domain D(A) (i.e., A : D(A)→ X).The spectrum, σ(A), of an operator A is the set in C defined by

σ(A) := z ∈ C : A− z1l is not invertible . (D.1)

For notational convenience, the operator “multiplication by z ∈ C” will be simply written asz instead of z1l. Clearly, eigenvalues of A belong to σ(A) (in fact, if λ is an eigenvalue, thenAuλ = λuλ for some nonzero uλ ∈ X (uλ is called an eigenvector), so (A− λ)uλ = 0, and A− λis not invertible). In general, the spectrum can also contain continuous pieces and it can takevery peculiar forms.


Exercise 32. The spectrum of the multiplication operator introduced in example 1) above isσ(Mf ) = Ran f , the differentiation operator 2) has spectrum σ( ∂

∂xj) = iR, and the identity

operator 3) has the spectrum consisting of one point σ(1l) = 1.

The study of the spectra of operators is called spectral analysis.The complement of the spectrum is called the resolvent set ρ(A):

ρ(A) := C\σ(A).

One can show (see [30]) that the set σ(A) is closed (and, consequently, the set ρ(A) is open).The study of the spectra of operators is called spectral analysis.The complement of the spectrum is called the resolvent set ρ(A):

ρ(A) := C\σ(A).

We begin with

Theorem 25. The set σ(A) is closed (and, consequently, the set ρ(A) is open).

Proof. We prove the equivalent statement that the set ρ(A) is open. Let z0 ∈ ρ(A). Then A−z0

is invertible. We write

A− z = A− z0 + z0 − z = (A− z0)[1l + (z0 − z)(A− z0)−1].

If |z0 − z| < ‖(A− z0)−1‖−1, then the operator in the square brackets on the right is invertibleby the Neumann theorem (see Theorem 23). Therefore the operator A − z is invertible for|z − z0| < ‖(A− z0)−1‖−1 as a product of two invertible operators.

For z ∈ ρ(A) the operator A− z has a bounded in X inverse. Denote this inverse as

RA(z) := (A− z)−1.

It is called the resolvent of A at z ∈ ρ(A). It plays an important role in analysis of operators.The proof of the theorem above shows that the resolvent is an analytic operator valued functionin z ∈ ρ(A) in the sense that for any z0 ∈ ρ(A) and for any z such that |z − z0| < ‖RA(z0)‖−1,the resolvent RA(z) can be expanded in the series

RA(z) = RA(z0)

∞∑n=0

((z − z0)RA(z0))n (D.2)

which is absolutely convergent is the sense that

∞∑n=0

‖(z − z0)RA(z0)‖n =∞∑n=0

|z − z0|n‖RA(z0)‖n <∞.

Indeed the series above is just the Neumann series for the inverse of the operator A − z =(A− z0)[1l + (z0 − z)(A− z0)−1] (see Theorem 23).

The resolvent satisfies two equations:

RA(z)−RA(w) = (z − w)RA(z)RA(w) (D.3)

andRA(z)−RB(z) = RA(z)(B −A)RB(z), (D.4)


called the first and second resolvent equations. The first equation follows from the second onewith B = (z − w)1l and the second equation is equivalent to the identity 1

A −1B = 1

A(B − A) 1B

which can be easily verified.

Proposition 6. If A is a bounded operator then σ(A) ⊂ z ∈ C | |z| ≤ ‖A‖.

Proof. Expand formally by the Neumann series

1

A− z=−1

z

1

1−A/z=−1

z

∞∑n=0

(A

z

)n.

This shows that z ∈ ρ(A) (i.e., (A − z)−1 is bounded) if and only if ‖A/z‖ < 1. Equivalently,z ∈ ρ(A) if and only if |z| > ‖A‖. Therefore, z ∈ σ(A) if and only if |z| ≤ ‖A‖.

D.2 Location of the essential spectrum


D.3 Perron-Frobenius Theory

Consider a bounded operator T on the Hilbert space X = L2(Ω).

Definition D.1. An operator T is called positivity preserving/improving if and only if u ≥0, u 6= 0 =⇒ Tu ≥ 0/Tu > 0.

Note if T is positivity preserving, then T maps real functions into real functions.

Theorem 26. Let T be a bounded positive and positivity improving operator and let λ be aneigenvalue of T with an eigenvector ϕ. Thena) λ = ‖T‖ ⇒ λ is simple and ϕ > 0 (modulo a constant factor).b) ϕ > 0 and ‖T‖ is an eigenvalue of T ⇒ λ is simple and λ = ||T ||.

Proof. a) Let λ = ||T ||, Tψ = λψ and ψ be real. Then |ψ|±ψ ≥ 0 and therefore T (|ψ|±ψ) > 0.The latter inequality implies that |Tψ| ≤ T |ψ| and therefore

〈|ψ|, T |ψ|〉 ≥ 〈|ψ|, |Tψ|〉 ≥ 〈ψ, Tψ〉 = λ||ψ||2.

Since λ = ||T || = sup||ψ||=1〈ψ, Tψ〉, we conclude using variational calculus (see e.g. [30] or [?])that

T |ψ| = λ|ψ| (D.5)

i.e., |ψ| is an eigenfunction of T with the eigenvalue λ. Indeed, since λ = ‖T‖ = sup‖ψ‖=1〈ψ, Tψ〉,|ψ| is the maximizer for this problem. Hence |ψ| satisfies the Euler-Lagrange equation T |ψ| =µ|ψ| for some µ. This implies that µ||ψ||2 = 〈|ψ|, T |ψ|〉 = λ||ψ||2 and hence µ = λ. Equation(D.5) and the positivity improving property of T imply that |ψ| > 0.

Now either ψ = ±|ψ| or |ψ| + ψ and |ψ| − ψ are nonzero. In the latter case they areeigenfunctions of T corresponding to the eigenvalue λ : T (|ψ| ± ψ) = λ(|ψ| ± ψ). By thepositivity improving property of T this implies that |ψ| ± ψ > 0 which is impossible. Thusψ = ±|ψ|.

If ψ1 and ψ2 are two real eigenfunctions of T with the eigenvalue λ then so is aψ1 + bψ2 forany a, b ∈ R. By the above, either aψ1 + bψ2 > 0 or aψ1 + bψ2 < 0 ∀a, b ∈ R \ 0, which isimpossible. Thus T has a single real eigenfunction associated with λ.


Let now ψ be a complex eigenfunction of T with the eigenvalue λ and let ψ = ψ1 +iψ1 whereψ1 and ψ2 are real. Then the equation Tψ = λψ becomes

Tψ1 + iTψ2 = λψ1 + iλψ2.

Since Tψ2 and Tψ2 and λ are real (see above) we conclude that Tψi = λψ2, i = 1, 2, and there-fore by the above ψ2 = cψ1 for some constant c. Hence ψ = (1 + ic)ψ1 is positive and uniquemodulo a constant complex factor.

b) By a) and eigenfunction, ψ, corresponding to ν := ||T || can be chosen to be positive, ψ > 0.But then

λ〈ψ,ϕ〉 = 〈ψ, Tϕ〉 = 〈Tψ, ϕ〉 = ν〈ψ,ϕ〉

and therefore λ = ν and ψ = cϕ.

*Question: Can the condition that ||T || is an eigenvalue of T (see b) be removed?*

Now we consider the Schrodinger operator H = −∆ + V (x) with a real, bounded potentialV (x). The above result allows us to obtain the following important

Theorem 27. Let H = −∆ + V (x) have an eigenvalue E0 with an eigenfunction ϕ0(x) and letinf σ(H) be an eigenvalue. Then

ϕ0 > 0 ⇒ E0 = infλ|λ ∈ σ(H) and E0 is non-degenerate

and, conversely,

E0 = infλ|λ ∈ σ(H) ⇒ E0 is non-degenerate and ϕ0 > 0

(modulo multiplication by a constant factor).

Proof. To simplify the exposition we assume V (x) ≤ 0 and let W (x) = −V (x) ≥ 0. Forµ > supW we have

(−∆−W + µ)−1 = (−∆ + µ)−1∞∑n=0

[W (−∆ + µ)−1]n (D.6)

where the series converges in norm as

‖W (−∆ + µ)−1‖ ≤ ‖W‖‖(−∆ + µ)−1‖ ≤ ‖W‖L∞µ−1 < 1

by our assumption that µ > supW = ‖W‖L∞ . To be explicit we assume that d = 3. Then theoperator (−∆ + µ)−1 has the integral kernel

e−√µ|x−y|

4π|x− y|> 0

while the operator W (−∆ + µ)−1 has the integral kernel

W (x)e−√µ|x−y|

4π|x− y|≥ 0.

Consequently, the operator


(−∆ + µ)−1f(x) =1

4π

∫e−√µ|x−y|

|x− y|f(y) dy

is positivity improving (f ≥ 0, f 6= 0⇒ (−∆ + µ)−1f > 0) while the operator

W (−∆ + µ)−1f(x) =1

4π

∫W (x)

e−√µ|x−y|

|x− y|f(y) dy

is positivity preserving (f ≥ 0⇒W (−∆+µ)−1f ≥ 0). The latter fact implies that the operators[W (−∆ + µ)−1]n, n ≥ 1, are positivity preserving (prove this!) and consequently the operator

(−∆ + µ)−1 +∞∑n=1

[W (−∆ + µ)−1]n

is positivity improving (prove this!).

Thus we have shown that the operator (H+µ)−1 is positivity improving. Series (D.6) showsalso that (H + µ)−1 is bounded. Since

〈u, (H + µ)u〉 ≥ (− supW + µ)‖u‖2 > 0,

we conclude that the operator (H+µ)−1 is positive (as an inverse of a positive operator). Finally,‖(H + µ)−1‖ = supσ((H + µ)−1) = (inf σ(H) + µ)−1 is an eigenvalue by the condition of thetheorem. Hence the previous theorem applies to it. Since Hϕ0 = E0ϕ0 ⇔ (H + µ)−1ϕ0 =(E0 +µ)−1ϕ0, the theorem under verification follows. *This paragraph needs details!*.

E Linear evolution and semigroups

Let X be a Banach space and A be a linear (closed) operator on X with dense domain Y = D(A).Our goal is to solve the initial value problem

∂u

∂t= Au, u|t=0 = u0 ∈ Y

for u ∈ C(R, Y ) ∩ C1(R, X). We use the following terminology.

• The family, U(t), t ≥ 0, of operators on X is called a (strongly continuous) semigroup ifand only if

(a) U(t) are bounded ∀t ≥ 0.

(b) U(0) = 1 and U(t+ s) = U(t)U(s).

(c) t→ U(t)ϕ is continuous ∀ϕ ∈ X.

• The family U(t) is called a contraction semigroup if and only if U(t) is a semigroup and||U(t)|| ≤ 1.

• The (closed) operator

Au := lims→0

1

s(U(s)− 1)u (E.1)


on X with the domain

D(A) := u ∈ X | the limit on the r.h.s of (E.1) exists

is called the generator of U(t). If A is the generator of the semigroup U(t), then we write

U(t) = eAt.

Theorem 28. If A is the generator of U(t), then U(t)D(A) ⊂ D(A) and ∀u0 ∈ D(A), u :=U(t)u0 solves the equation

∂u

∂t= Au (E.2)

with the initial condition u|t=0 = u0.

Proof. If u ∈ D(A), then

AU(t)u = lims→0

1

s(U(s)− 1)U(t)u = U(t) lim

s→0

1

s(U(s)− 1)u exists.

Hence U(t)u ∈ D(A). Furthermore,

AU(t)u = lims→0

1

s(U(t+ s)u− U(t)u) ≡ ∂

∂tU(t)u.

Corollary 29. If an operator A is the generator of a semigroup U(t), then the initial valueproblem

∂u

∂t= Au, u|t=0 = u0

has a solution for any u0 ∈ D(A) and this solution is given by the formula u = U(t)u0.

Thus the main question here is: when does an operator A generate a semigroup? First of allbounded operators generate semigroups. Indeed, for a bounded operator B we define

U(t) ≡ etB :=

∞∑n=0

(tB)n

n!. (E.3)

The series on the r.h.s. converges absolutely since

∞∑n=0

∥∥∥∥(tB)n

n!

∥∥∥∥ ≤ ∞∑n=0

tn

n!||B||n = et||B|| <∞.

Exercise 33. Show that equation (E.3) defines a semigroup and that this semigroup is generatedby B.

For unbounded operators the answer to the question above is given by the following.

Theorem 30 (Hille-Yosida). Let A be a closed operator such that (a) (0,∞) ⊂ ρ(A) and (b)||(A− λ)−1|| ≤ 1/λ for any λ > 0. Then A generates a unique semigroup and this semigroup iscontractive.


Proof. The idea is very simple: we approximate the operator A by bounded operators Aλ sothat Aλu→ Au ∀u ∈ D(A) as λ→∞; construct the semigroup, Uλ(t), for Aλ by the formula

Uλ(t) =∞∑n=0

1

n!(tAλ)n (E.4)

(see equation (E.3)); define the semigroup, U(t), for A as the limit

U(t)u = limλ→∞

Uλ(t)u (E.5)

for any u ∈ D(A) and then, by continuity, extend U(t) to the entire space X.

We define Aλ as

Aλ := Aλ(λ−A)−1.

(Note A− λ for λ > 0 is invertible by the condition that (0,∞) ⊂ ρ(A).) Then ∀u ∈ D(A), by(b)

λ(λ−A)−1u− u = (λ−A)−1Au→ 0,

as λ→∞, and

‖λ(λ−A)−1‖ ≤ 1.

Hence, by an ε/3 argument,

λ(λ−A)−1u→ u ∀u ∈ X.

This implies ∀u ∈ D(A),

Aλu = λ(λ−A)−1Au→ Au.

Now we consider the semigroup Uλ(t) defined in (E.4). Due to (b) and the relation Aλ =λ2(λ−A)−1 − λ, we have

‖eAλt‖ ≤ e−λt∞∑n=0

1

n!‖(tλ2(λ−A)−1)n‖

≤ e−λt∞∑n=0

tn

n!λ2n‖(λ−A)−1‖n

≤ e−λt∞∑n=0

tn

n!λn. (E.6)

Hence

||eAλt|| ≤ 1, (E.7)

i.e., eAλt is the contractive semigroup.

Now we show that eAλt, λ > 0 is a Cauchy family in the sense that

||(eAλ′ t − eAλt)u|| → 0 (E.8)


as λ, λ′ → ∞ for any u ∈ X. To prove this we represent the operator acting on u inside thenorm as an integral of a derivative

eAλ′ t − eAλt =

∫ t

0

∂

∂seAλ′seAλ(t−s)ds.

Using the equation

∂

∂seAλs = Aλe

Aλs = eAλsAλ

gives

eAλ′ t − eAλt =

∫ t

0eAλ′seAλ(t−s)(Aλ′ −Aλ)ds.

The last equation together with (E.7) yields

||(eAλ′ t − eAλt)u|| ≤∫ t

0||eAλ′seAλ(t−s)(Aλ′ −Aλ)u||ds

≤∫ t

0||(Aλ′ −Aλ)u||ds = t||(Aλ′ −Aλ)u||

and therefore (E.8) follows for any t ≥ 0 first ∀u ∈ D(A) and then by continuity ∀u ∈ X.

Now equation (E.8) implies that the limit on the r.h.s. of (E.5) exists ∀t ∈ [0,∞) ∀u ∈ Xand satisfies

||U(t)|| ≤ 1.

Equations (E.5) and (E.7) imply also that U(t) is the semigroup: U(t + s) = U(t)U(s) andU(0) = 1. It remains to prove that U(t) is strongly continuous and is generated by the operatorA. Using the relation Uλ(s)− 1 =

∫ s0 dtUλ(t)Aλ, we find for u ∈ D(A)

lims→0||(U(s)− 1)u|| = lim

s→0limλ→∞

||(Uλ(s)− 1)u||

= lims→0

limλ→∞

||∫ s

0dtUλ(t)Aλu||

≤ lims→0

limλ→∞

∫ s

0dt||Aλu||

= lims→0

s||Au|| = 0.

Hence U(s)u→ u as s→ 0 ∀u ∈ D(A). Since U(t) is bounded uniformly in t we conclude thatit is strongly continuous. Similarly, using the relation Uλ(s)− 1−Aλ =

∫ s0 dt(Uλ(t)− 1)Aλ and

the strong continuity of Uλ(t) shown above, we conclude that U(t) is generated by A.

How do we check the conditions of the Hille-Yosida theorem? Usually this is a hard business.However, there are several cases where this can be easily done. Below, X is a Hilbert Space.

A) A = −A∗ (A is anti-self-adjoint). Then σ(A) ⊂ iR and ‖(A− λ)−1‖ ≤ λ−1 for λ > 0.

B) A = A∗ ≤ 0 (A is non-positive). Then σ(A) ⊂ (−∞, 0] and ‖(A− λ)−1‖ ≤ λ−1 for λ > 0.


C) Perturbations of generators. For example, A = A0 +B where A0 generates a semigroup and‖B(A0 − λ)−1‖ ≤ β with β < 1 ∀λ ≥ λ0, for some λ0 > 0. Indeed, in this case we have forλ > λ0

A− λ = [1 + Tλ](A0 − λ) (E.9)

where Tλ = B(A0 − λ)−1. By the condition above ||Tλ|| ≤ β ∀λ ≥ λ0. Since β < 1, 1 + Tλ isinvertible and therefore so is the r.h.s. of (E.9) ∀λ ≥ λ0. Therefore (λ0,∞) ⊂ ρ(A). Moreover,(E.9) implies that

(A− λ)−1 = (A0 − λ)−1(1 + Tλ)−1

and therefore

||(A− λ)−1|| ≤ ||(A0 − λ)−1|| ||(1 + Tλ)−1||.

Since A0 generates a semigroup, we have that ||(A0−λ)−1|| ≤ 1/λ. Since ||Tλ|| ≤ β < 1 we havethat ||(1 + Tλ)−1|| ≤ (1− β)−1. Collecting the last three estimates we conclude that ∀λ ≥ λ0

||(A− λ)−1|| ≤ (1− β)−1λ−1.

This is not quite what we need (remember (b)). However, for the operator Aµ = A − µ withµ = min(λ0, (1− β)−1), we obtain

||(Aµ − λ)−1|| = ||(A− (µ+ λ))−1|| ≤ µ(µ+ λ)−1 < λ−1.

Thus the operator Aµ generates a contraction semigroup eAµt, ||eAµt|| ≤ 1. Now, eAt := eAµteµt

gives a semigroup for the operator A and this semigroup satisfies the estimate

||eAt|| ≤ eµt.

Exercise 34. Check the last statement.

Examples.

1) The Schodinger equation. In this case A = −iH where H is a self-adjoint operator (e.g., aSchrodinger operator H := −∆ + V (x) on L2(Rd)).

2) The heat equation. In this case A = A∗ ≤ 0 or more generally A = −A0−A1 with A0 = A∗0 > 0and

‖A−1/20 A1A

−1/20 ‖ < 1.

For exampleA = −∑∂xiaij(x)∂xj+

∑bi(x)∂xi+c(x) with the matrix (aij(x)) ≥ δ1, (

∑|bi(x)|2)1/2 ≤

γ and c(x) ≥ γ2/δ.3) The wave equation. In this case

A =

(0 1−H 0

)with H = H∗ ≥ 0. (E.10)

Indeed, if v satisfies the wave equation

∂2v

∂t2= −Hv with H ≥ 0,


then the element u = (v, ∂v/∂t) satisfies the equation

∂u

∂t= Au

with the operator A given in (E.10). On the Hilbert space X = D(H) ⊕ L2 with the innerproduct

〈u,w〉X = 〈u1, Hw1〉+ 〈u2, w2〉

where u = (u1, u2) and w = (w1, w2), the operator (E.10) is anti-self-adjoint, A = −A∗, andtherefore it generates a unique contraction semigroup.

Example. The acoustical wave equation.

∂2v

∂t2= c2ρ∇ · 1

ρ∇v.

The operatorH := −c2ρ∇·1ρ∇ is self-adjoint and, in fact, non-negative on the space L2(R3, (c2ρ)−1dx).

Exercise 35. Show that H is symmetric, i.e.,

〈Hu, v〉 = 〈u,Hv〉 ∀u, v ∈ D(H).

4) Maxwell equations. In a vacuum, Maxwell’s equations for the electric and magnetic fields,E(x, t) and H(x, t), read

curlE = −µ∂H∂t

, curlH = ε∂E∂t

div(εE) = 0, div(µH) = 0

where ε and µ are dielectric constant and magnetic permeability, respectively. These equationscan be written as

∂tu = JAu,

where u = (E,H), J =

(0 1−1 0

)and

A =

(µ−1curl 0

0 ε−1curl

)on the Hilbert space (we use that ε and µ are independent of x)

H1 = (E,H) |E,H ∈ H1(R3,R3), divE = 0, divH = 0.

One can show that the operator A with the domain H1 ⊂ L2(R3,R3) ⊕ L2(R3,R3) is self-adjoint (show that it is symmetric) and σess(H) = [0,∞). Hence the operator JA generates acontraction semigroup according to criterion D) above.


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