Lecture Notes on Calculus of Variations I - Uni Kassel

Lecture Noteson

Calculus of Variations I

Karoline Disser

Winter term 2020/2021

Address: Universität KasselFB 10 Mathematik und NaturwissenschaftenAG Analysis und Angewandte MathematikHeinrich-Plett-Str. 4034131 KasselGermany

Office: 3318

Phone: +49 (0)561 804 4613

E-mail: [email protected]

Acknowledgements: The contents of this course and these lecture notes are based on thecourses taught and notes prepared on this topic by Prof. Dr. Alex Mielke andProf. Dr. Dorothee Knees. Corrections by Philipp Käse, Steffen Polzer, JanSchröder, Konstantin Sitnikov. Thank you!

1

CONTENTS 2

Contents1 Introduction 4

1.1 Mission statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Examples: sketches and pictures . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Brachistochrone curve – the fastest marble run . . . . . . . . . . . 41.2.2 Catenoid – the laziest chain . . . . . . . . . . . . . . . . . . . . . . 41.2.3 Soap bubble – the smallest surface, the largest volume . . . . . . 51.2.4 Elasticity – the most relaxed sponge . . . . . . . . . . . . . . . . . 5

1.3 Examples: famous functionals and some modelling . . . . . . . . . . . . . 51.3.1 Brachistochrone curve . . . . . . . . . . . . . . . . . . . . . . . . . 51.3.2 Fermat’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3.3 Minimal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Reminder: the case X = Rn, I = F ∈ C2(Rn,R). . . . . . . . . . . . . . 71.5 Reminder: Weierstraß’ principle. . . . . . . . . . . . . . . . . . . . . . . . 71.6 Strategy for Chapter 3 – how can this be fixed? . . . . . . . . . . . . . . . 91.7 Outlook and Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Classical Methods in the Calculus of Variations 102.1 Problem set-up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 (Multidimensional) notation . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 The Euler-Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3.1 The Dirichlet integral. . . . . . . . . . . . . . . . . . . . . . . . . . 152.3.2 Minimal surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 Different types of extrema . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4.1 The double-well functional . . . . . . . . . . . . . . . . . . . . . . . 17

2.5 Some necessary and sufficient conditions . . . . . . . . . . . . . . . . . . 212.6 Types of Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.7.1 Minimal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.7.2 Brachistochrone curve . . . . . . . . . . . . . . . . . . . . . . . . . 402.7.3 Linear elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3 The Direct Method in the Calculus of Variations 543.1 Abstract existence theorems from functional analysis . . . . . . . . . . . . 543.2 Reminder: Lebesgue and Sobolev spaces . . . . . . . . . . . . . . . . . . . 58

3.2.1 Lebesgue spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.2.2 Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . 593.2.3 Weak derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.2.4 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.3 Properties of I(u) =∫

Ωf(x, u(x),∇u(x)) dx on W 1,p(Ω) . . . . . . . . . . 71

3.3.1 Quasiconvexity and weak lower semicontinuity of I . . . . . . . . . 743.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

3.4.1 p-Laplace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.4.2 Optimal Poincaré constant . . . . . . . . . . . . . . . . . . . . . . 823.4.3 Phase separation (Cahn-Hilliard energy) . . . . . . . . . . . . . . . 833.4.4 Nonlinear Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 85

CONTENTS 3

Literature

(Introductory) Books on the Calculus of Variations[Rin18] F. Rindler. Calculus of Variations. Springer, 2018.

[Dac08] B. Dacorogna. Direct Methods in the Calculus of Variations. Springer, 2nded., 2008.

[Dac04] B. Dacorogna. Introduction to the Calculus of Variations. Imperial CollegePress, 2004.

[EkT] I. Ekeland and R. Temam. Convex Analysis and Variational Problems.North Holland, 1976.

Further References[Adams] R.A. Adams. Sobolev Spaces, Academic Press, 1975.

[Alt] W. Alt. Lineare Funktionalanalysis, Springer, 4th ed., 2002.

[Ciarlet] P. Ciarlet. Mathematical Elasticity, North Holland, 2004.

[Evans] L.C. Evans. Partial Differential Equations, AMS Graduate Studies in Math-ematics, 2nd ed., 2010.

[Klenke] A. Klenke. Wahrscheinlichkeitstheorie, Springer, 3rd ed., 2013.

1 INTRODUCTION 4

1 Introduction

1.1 Mission statementConsider (non)linear functionals

I : M → R ∪ +∞,on a set M ⊂ X, where X is an infinite-dimensional Banach space.

Find (local or global) minimizers of I.

1.2 Examples: sketches and pictures1.2.1 Brachistochrone curve – the fastest marble run

1.2.2 Catenoid – the laziest chain

1 INTRODUCTION 5

1.2.3 Soap bubble – the smallest surface, the largest volume

1.2.4 Elasticity – the most relaxed sponge

Note: (almost) every solid material is somehow elastic!

1.3 Examples: famous functionals and some modelling1.3.1 Brachistochrone curve

Find a curve S that starts at A = (0, h) and ends at B = (l, 0) such that the travelingtime of a marble running along S is minimal (no friction!). The curve S = (x, u(x)) isgiven as the graph of a function u : [0, l]→ R with u(0) = h the initial height and u(l) = 0.The total energy of the marble at position x is given by the sum of its potential energy

Epot(x) = mgu(x),

where m is its mass and g the gravitation constant, and its kinetic energy,

Ekin(x) =1

2m|v(x)|2,

where v(x) is the velocity of the marble’s center of gravity. Here, as is customary for thisproblem in mathematics, we disregard the rotational energy

Erot(x) =1

5m|v(x)|2,

1 INTRODUCTION 6

of the marble. It would only modidy the factor “2′′ that appears in the following to be“ 10

7

′′. By conservation of energy, for all x, we obtain the equation

E(0) = mgh+1

2m|v(0)|2 = mgu(x) +

1

2m|v(x)|2 = E(x).

Since the marble starts at zero velocity, v(0) = 0, this gives

|v(x)|2 = 2g(h− u(x)). (1.1)

At the same time, the velocity of the curve is

v(t) =

(x(t)

ddtu(x(t))

)chain rule

= x(t)

(1

u′(x(t))

),

so|v(t)|2 = x(t)2(1 + u′(x(t))2).

Combining this with (1.1), we get

|x(t)| =√

2g(h− u(x(t)))

1 + u′(x(t))2.

Hence, the Brachistochrone curve is given by the function u with u(0) = h and u(l) = 0that minimizes the functional

T (u) =

∫ T

0

dtt 7→x(t)

=

∫ l

0

t′(x) dxt′(x)= 1

x(t(x))=

∫ l

0

1

x(t(x))dx

=

∫ l

0

√1 + u′(x)2

2g(h− u(x))dx.

The problem of finding this curve was first solved by Jakob Bernoulli in 1696. The solutionwill be discussed in Subsection 2.7.2.

1.3.2 Fermat’s principle

“Light takes the path of shortest time.” Pierre de Fermat (1607 – 1665) proved thisprinciple using the calculus of variations. A possible formalization of this prinicple is thefollowing: Again, look for a parameterized curve (x(t), u(x(t))) that is the graph of afunction u and connects a point A = (x0, u0) to a point B = (x1, u1). Light should travelalong this curve in a medium with index of refraction n(x, u) > 0. As before, we have

|v(t)| = |x(t)|√

1 + u′(x(t))2.

In addition, a constitutive equation as in (1.1) is needed. In this case, it is given by thelaw of optics,

v(x, u) =c

n(x, u),

where c is the (constant) speed of light in vacuum. Thus, we need to find u such thatu(x0) = u0 , u(x1) = u1, and such that the total time

T (u) =

∫ T

0

dtx=x(t)

=

∫ x1

x0

1

x(t(x))dx =

∫ x1

x0

√1 + u′(x)2

n(x, u(x))

cdx

is minimal.

1 INTRODUCTION 7

1.3.3 Minimal surface

Consider a bounded domain Ω ⊂ R2 with boundary ∂Ω. Given a function u0 : Ω → R,find a function u : Ω→ R, u ∈ C1(Ω) such that u = u0 on ∂Ω and such that the surfacearea

I(u) =

∫

Ω

√1 + |∇u(x)|2 dx

is minimal. (The surface area of the surface given by the graph of u over Ω is calculatedby integrating the cross product of its tangential vectors in this parameterization:

Remark.

‖

10

∂∂x1

u(x)

×

01

∂∂x2

u(x)

‖ = ‖

− ∂∂x1

u(x)

− ∂∂x2

u(x)

1

‖ =

√1 + |∇u(x)|2.)

In this course, we study these and other examples with two different strategies.In the first part of the lecture, in Chapter 2, we use indirect or classical methods. Theyare based on the idea of generalizing necessary and sufficient conditions for the existenceof minimizers from the finite-dimensional to the infinite-dimensional situation. As a briefreminder of these conditions, let us look at the finite-dimensional case:

1.4 Reminder: the case X = Rn, I = F ∈ C2(Rn,R).(For proofs see “Analysis II” and Exercise 1.IV.)

Necessary Conditions: If x0 ∈ X is a local minimizer of F , then

1. x0 is a critical point, i.e. ∂∂xi

F (x0) = 0 for all i = 1, . . . , n, and,

2. the Hessian HF (x0) is positive semi-definite.

Sufficient Condition: If x0 is a critical point of F and HF (x) is positive definite, thenx0 is a local minimizer of F .

Aims of Chapter 2: Generalize and adapt these ideas to infinite-dimensional X andspecific types of I.

In the second part of the course, Chapter 3 in these notes, we get to know the directmethod in the calculus of variations. Roughly speaking, it is an adaptation of Weierstraß’Principle to specific infinite-dimensional settings. In the following three short sections,we look at some aspects and a rough guideline for this adaptation.

1.5 Reminder: Weierstraß’ principle.(Karl Weierstraß (1815-1897), for proofs recall “Analysis I and/or II”:) A real-valuedcontinuous function on a compact set attains its maximum and minimum.

1 INTRODUCTION 8

Aims of Chapter 3: For the problems we would like to solve, the assumptions ofWeierstraß’ Principle (compactness and continuity) do not fully apply. Thus, we try tomodify them by using more specific information on M ⊂ X and I.

Simple examples: Non-compact X ⊂ Y = R and non-continuous, bounded F –what can got wrong?

Problem 1. M = X = R (non-bounded) and F (x) = e−x, then there is no minimizer ofF :

Problem 2. M = [0, 1)∪(1, 2] (non-closed) and F (x) = 0.5+(x−1)2, then infx∈X F (x) =0.5, but there is no minimizer in M :

1 2

0.5

1.5

Problem 3. M = [0, 2] and

F (x) =

0.5 + (x− 1)2, x ∈ [0, 1),

2, x ∈ [1, 2],

(non-continuous), then again infx∈X F (x) = 1, but there is no minimizer:

1 INTRODUCTION 9

1 2

0.5

1.5

1.6 Strategy for Chapter 3 – how can this be fixed?Let X be a Banach space, M ⊂ X, and let I : M → R be a functional that is boundedfrom below. Then there is an infimizing sequence (un)n ⊂M for I, i.e.

limnI(un) = inf

u∈XI(u) = I .

Now adopt the following strategy:

Step 1. Assume that I is coercive, i.e. for every sequence (un)n ⊂ M , if ‖un‖X →∞,then I(un) → ∞. This implies that (un)n is bounded without assuming that M isbounded. In particular, this assumption deals with Problem 1.

Step 2. Assume that M is (weakly sequentially) closed in X or that, for some otherreason, there is a (weakly) convergent subsequence of (un)n with (weak) limit M 3 u =limk unk . This assumption addresses Problem 2 – except that the assumption of weakclosedness is stronger than the assumption of closedness of M .

Step 3. Now if I is (weakly sequentially) continuous on X, then limk I(unk) = I(u), sou is a minimizer and we are done. However, for the applications we have in mind, thisassumption is far too strong. If, instead, we assume that I is (weakly sequentially) lowersemicontinuous, then still

I = limkI(unk) ≥ lim inf

kI(unk) ≥ I(u) ≥ I,

so u is a minimizer. In particular, this assumption is sufficient for avoiding Problem 3and can be shown to hold in relevant cases (think of the norm on a Hilbert space as afirst example of a weakly sequentially lower semicontinuous functional).

1.7 Outlook and MotivationThis Introduction shows a little bit of everything we will encounter during this lecture.In particular,

2 CLASSICAL METHODS IN THE CALCULUS OF VARIATIONS 10

• we will need

– (multidimensional) calculus, and

– some functional analysis, and,

• we will get to know

– some convex analysis,

– some PDE (Partial Differential Equations) theory, and

– some modelling, and,

• the problems, theorems and examples in this lecture are strongly inspired by andconnected to many other interesting topics in

– analysis,

– geometry,

– (nonlinear) optimization,

– mechanics, and, more generally, physics,

– . . .

2 Classical Methods in the Calculus of Variations

2.1 Problem set-upLet n,m ∈ N be (spatial) dimensions for our problem (e.g. “n = 1,m = 1” for theBrachistochrone curve, “n = 2,m = 1” for the minimal surface, “n = 3,m = 3” for thesponge) and let Ω ⊂ Rn be a bounded domain with smooth boundary ∂Ω (see Remarkbelow) and outer normal vector field ν. Now consider functionals of the form

I(u) =

∫

Ω

f(x, u(x),∇u(x)) dx+

∫

∂Ω

g(x, u(x)) dx′ (2.1)

and find (differentiable) functions u : Ω→ Rm that minimize I.In this Chapter,

• the function f : Ω × Rm × Rm×n → R, (x, u,A) 7→ f(x, u,A) is called a Lagrangefunction or (energy) density and satisfies f ∈ C2(Ω× Rm × Rm×n;R), and

• the function g : ∂Ω×Rm → R is called a boundary density and satisfies g ∈ C2(∂Ω×Rm;R).


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It often happens that the sought-for function u ∈ C1(Ω;Rm) =: X should be restrictedto fixed values u0 on some part Γ ⊂ ∂Ω of the boundary (e.g. u(0) = h, u(l) = 0 for theBrachistochrone curve). Hence, for given u0 ∈ X, we define the set

M := v ∈ X : v|Γ = u0|Γ.

Then we denote byX0 := v ∈ X : v|Γ = 0

the corresponding linear space of test functions with M = u0 + X0 and consider thefunctional I as a map

I : X ⊃M → R. (2.2)

Remark. The assumption on the smoothness of the boundary ∂Ω is “technical” – it can berelaxed and it doesn’t really have an impact on what we would like to do. We will discussthis point a bit more in Chapter 3. Here, we just need to know that we can integrate over∂Ω in a reasonable way. The notation dx′ is used for this integral.

The examples in Chapter 1 fit into the form of (2.2) with suitable X,M and I as in(2.1). You can check this as an exercise even before we discuss them again.

2.2 (Multidimensional) notationIn this Section, we collect specific multidimensional notation used in this course.

In the previous Section and in the following, ∇u(x) ∈ Rm×n denotes the (transposed)gradient and the Jacobian matrix of u,

(∇u(x))ij =∂ui∂xj

(x), i ∈ 1, . . . ,m, j ∈ 1, . . . , n.

We adopt this (transposed) notation because it is used frequently in the literature and inapplications.

On the set Rm×n of real m× n-matrices we use the scalar product

B : C = trace(BTC) = Σmi=1Σnj=1BijCij

for two matrices B,C ∈ Rm×n and the Frobenius norm |B| =√B : B.


For f as before and x ∈ Ω, u ∈ Rm, A ∈ Rm×n, we write:

∂xf(x, u,A) =

(∂

∂xjf(x, u,A)

)

1≤j≤n∈ Rn,

∂uf(x, u,A) =

(∂

∂uif(x, u,A)

)

1≤i≤m∈ Rm,

∂Af(x, u,A) =

(∂

∂Aijf(x, u,A)

)

1≤i≤m,1≤j≤n∈ Rm×n.

For second derivatives, and v, w ∈ Rm, A,B,C ∈ Rm×n, we write

∂2uf(x, u,A)v · w =

m∑

i,k=1

∂2

∂ui∂ukf(x, u,A)viwk,

∂2Af(x, u,A)[B,C] =

∑

1≤i,k≤m,1≤j,l≤n

∂2

∂Aij∂Aklf(x, u,A)BijCkl,

∂A∂uf(x, u,A)[w,C] =∑

1≤i,k≤m,1≤l≤n

∂2

∂Akl∂uif(x, u,A)wiCkl.

For a differentiable function A : Ω→ Rm×n, the divergence operator “acts on rows”:

(divA(x))i = Σnj=1

∂

∂xjAij(x).

For a k-times continuously differentiable function v ∈ Ck(Ω;Rm), the correspondingnorm ‖v‖Ck of v is given by

‖v‖Ck =

k∑

l=0

max|α|=k

supx∈Ω

|Dαv(x)|. (2.3)

2.3 The Euler-Lagrange EquationsIn the following, let Ω,Γ, u0, X,M,X0 and f, g and I be as in the previous sections.

Definition 4. (First Variation) Let u ∈M and v ∈ X0. Consider the map

R ⊇ Bδu(0) 3 t 7→ I(u+ tv) ∈ R,

with Bδu(0) an open interval such that u+ tv ∈ M for all t ∈ Bδu(0) and v ∈ X0. Thenthe expression

DI(u)[v] :=d

dtI(u+ tv)|t=0

is called the first variation of I at u in the direction of v. Moreover, we have

DI(u)[v] =

∫

Ω

∂uf(x, u(x),∇u(x)) · v(x) + ∂Af(x, u(x),∇u(x)) : ∇v(x) dx

+

∫

∂Ω

∂ug(x, u(x)) · v(x) dx′


The first variation can be considered as a directional derivative of I and it is equal tothe Gâteaux differential of I. The expression for DI(u)[v] contains the definition (first:=) which works for general functionals, and it contains a small Proposition (second =)that follows from the chain rule if we differentiate the integral form (2.1) of I. We caninterchange differentiation and integration as f, g are continuously differentiable. Notethat the directions v need to be in the linear space of test functions X0.

With the first variation at hand, it is clear how to define critical points of I. As infinite dimensions, critical points are candidates for local minimizers.

Definition 5. (Critical points of I) A point u ∈ M is called a critical point of I, iffor all v ∈ X0, DI(u)[v] = 0.

From this definition, it is usually not clear how to find critical points of I. In particular,we cannot reduce the problem to finitely many equations. The aim of this Section is toshow a necessary criterion for u being a critical point of I, namely being a solution of theEuler-Lagrange Equations. We start with two important preliminary results.

Theorem 6. (Version of Gauss’ Theorem (Gauss)) Let u ∈ X. Then for all 1 ≤ i ≤ m,1 ≤ j ≤ n, ∫

Ω

∂ui∂xj

(x) dx =

∫

∂Ω

ui(x)νj(x) dx′.

Proof. cf. Analysis II/IV!

Theorem 7. (Fundamental Lemma of the Calculus of Variations (FL)) If a : Ω → Rmand b : ∂Ω→ Rm are continuous and for all v ∈ C∞(Ω;Rm) such that v|Γ = 0 we have

∫

Ω

a(x) · v(x) dx+

∫

∂Ω

b(x) · v(x) dx′ = 0,

then a(x) = 0 and b(x′) = 0 for all x ∈ Ω and x′ ∈ ∂Ω \ Γ.

Proof. We use an indirect proof. Assume that x0 ∈ Ω is such that a(x0) 6= 0. Since a iscontinuous, then there exists δ > 0 such that Bδ(x0) ⊂ Ω and a(x) · a(x0) ≥ 1

2 |a(x0)|2 forall x ∈ Bδ(x0). Here and in the following, Bδ(x0) = x ∈ Rn : |x − x0| < δ is the openball of radius δ with center x0. Define χ ∈ C∞(Rn;R) by

χ(x) =

e− 1

1−|x|2 , |x| < 1,

0, |x| ≥ 1,

so that suppχ = x ∈ Ω : χ(x) 6= 0 = B1(0). Let v(x) = a(x0)χ(x−x0

δ ). Thenv ∈ C∞(Ω;Rm), v|∂Ω = 0 and

0 =

∫

Ω

a(x) · v(x) dx =

∫

Bδ(x0)

a(x) · a(x0)χ(x− x0

δ) dx

≥ 1

2|a(x0)|2

∫

Bδ(x0)

χ(x− x0

δ) dx > 0,

a contradiction. It follows that a(x) = 0 for all x ∈ Ω. Using a = 0 and a similarargument, we can also show that b = 0 on ∂Ω \ Γ.


With these two tools at hand, we can prove the main result of this Section.

Theorem 8. (Euler-Lagrange Equations (ELE)) If u ∈ M is a critical point of I andu ∈ C2(Ω;Rm), then it satisfies the Euler-Lagrange Equations

−div (∂Af(x, u(x),∇u(x))) + ∂uf(x, u(x),∇u(x)) = 0, x ∈ Ω,

∂Af(x, u(x),∇u(x))ν(x) + ∂ug(x, u(x)) = 0, x ∈ ∂Ω \ Γ,

u(x) = u0(x), x ∈ Γ.

In this case, u is called a classical solution of the variational problem DI(u)[v] = 0.

The proof will also show the converse: if u ∈ C2(Ω;Rm)∩M satisfies the ELEs, thenit is a critical point of I.

Proof. By Definition 5 of critical points,

0 = DI(u)[v] =

∫

Ω

∂uf(x, u(x),∇u(x)) · v(x) + ∂Af(x, u(x),∇u(x)) : ∇v(x) dx

+

∫

∂Ω

∂ug(x, u(x)) · v(x) dx′,

short=

∫

Ω

∂uf · v + ∂Af : ∇v dx+

∫

∂Ω

∂ug · v dx′,

for every v ∈ X0 (recall that X0 = v ∈ C1(Ω;Rm) : v|Γ = 0). Here, the last line is justfor introducing a short way of writing the integrands. Using Gauss’ Theorem 6 and thedefinition of the divergence operator in Subsection 2.2, we have

∫

∂Ω

∑

1≤i≤m,

∑

1≤j≤n(∂Af)ijviνj dx′

Gauss=

∫

Ω

∑

1≤i≤m

∑

1≤j≤n

∂

∂xj

((∂Af)ij vi

)dx

Chain Rule=

∫

Ω

∑

1≤i≤mdiv (∂Af)i vi dx+

∫

Ω

∑

1≤i≤m

∑

1≤j≤n(∂Af) : ∇v dx.

Understanding this index notation and how Gauss’ Theorem needs to be used is the mostdifficult part of the proof. We then have

0 =

∫

Ω

(∂uf − div(∂Af)) · v dx+

∫

∂Ω

((∂Af)ν + ∂ug) · v dx′, (2.4)

for every v ∈ X0. Since u ∈ C2(Ω;Rm), f ∈ C2(Ω × Rm × Rm×n;R), and g ∈ C2(∂Ω ×Rm;R), and the normal vector field ν is (at least) continuous on ∂Ω \ Γ, the functions

Ω 3 x 7→ (∂uf − div(∂Af))(x, u(x),∇u(x))

and∂Ω 3 x 7→ ((∂Af)ν + ∂ug)(x, u(x))

are continuous as well. The theorem now follows directly from (2.4) and the FundamentalLemma (FL).


Of course, the question is: how do the ELEs help us find critical points? Solving theELEs may also be a difficult task. For varying dimensions m,n ∈ N, the Euler-LagrangeEquations are:

For n = 1,m = 1: an ordinary differential equation,

for n = 1,m ≥ 2: a system of ordinary differential equations,

for n ≥ 2,m = 1: a partial differential equation, and

for n ≥ 2,m ≥ 2: a system of partial differential equations.

Next, we look at two typical examples of Euler-Lagrange Equations that are a PDE. Thefollowing subsection contains a lot of terms from PDE theory. They appear here so thatif you have seen them before, you can make the connection – but it is not necessary thatyou know them already.

2.3.1 The Dirichlet integral.

Here, n ≥ 2, m = 1, the functional I is quadratic and the corresponding ELEs are a linearpartial differential equation. For a matrix-valued (coefficient) function

K ∈ C2(Ω;Rn×n)

and functions f0 ∈ C2(Ω;R), g0 ∈ C0(∂Ω;R), define

f(x, u,A) =1

2ATK(x)A− f0(x)u,

g(x′, u) = −g0(x′)u,

where x ∈ Ω, x′ ∈ N := ∂Ω \ Γ, u ∈ R and A ∈ R1×n = Rn so that

I(u) =

∫

Ω

1

2(∇u)T (x)K(x)∇u(x)− f0(x)u(x) dx+

∫

∂Ω

g0(x)u(x) dx′.

Then the ELEs for the critical points of I are

−div(

12 (K(x) +K(x)T )∇u(x)

)= f0(x), in Ω,

12 (K(x) +K(x)T )∇u(x) · ν(x) = g0(x), on N,u(x) = u0(x), on Γ.

(Exercise!) This is a linear elliptic PDE in u if K(x) is positive definite, with Dirichletboundary conditions on Γ and Neumann or ’natural’ boundary conditions on N . If

K(x) =

1 0 0

0. . . 0

0 0 1

,

then−div

(1

2(K(x) +K(x)T )∇·

)= −∆·,

the Laplace operator. In this case, the ELEs are called the Poisson equation (−∆u = f0)or the Laplace equation (∆u = 0) on Ω.


2.3.2 Minimal surfaces

As a second important example, we check the ELEs for the minimal surface problem fromSubsection 1.3.3. Recall that we have a bounded domain Ω ⊂ R2 with boundary ∂Ω anda function u0 ∈ C0(Ω;R) is given. The graph of this function on ∂Ω is a frame for oursoap film. The soap film itself is given by the graph F = (x, u(x)) ∈ R3 : x ∈ Ω of afunction u : Ω → R, such that u = u0 on ∂Ω and such that F has minimal surface area,i.e.

I(u) =

∫

Ω

√1 + |∇u(x)|2 dx

should be minimal. In this setting, the energy density is f(x, u,A) =√

1 + |A|2 withA ∈ R2 and M = u ∈ C1(Ω;R) : u|∂Ω = u0|∂Ω. Since

∂Af(x, u,A) =1√

1 + |A|2

(A1

A2

),

the ELEs are −div

(∇u(x)√

1+∂1u(x)2+∂2u(x)2

)= 0, in Ω,

u(x) = u0(x), on ∂Ω.

Again, we have a PDE because n = 2,m = 1, but now it is nonlinear in u. Thisdistinction is important as, like in the finite-dimensional case, the theory for solvingnonlinear equations is much less complete than the theory for solving linear problems(solving nonlinear PDE is a key application of the Calculus of Variations). Here, the

expression div

(∇u(x)√

1+∂1u(x)2+∂2u(x)2

)is the mean curvature of F . So the ELEs show that

a minimal surface has zero mean curvature.More examples will appear soon: an ODE in the next Section, systems of PDEs at a

later point in time...

2.4 Different types of extremaWe have seen that the solutions of the ELEs are critical points of the functional I. Butdo critical points provide extrema? This need not be true, even in finite dimensions.Critical points may be saddle points. The aim of this Section is to derive necessary andsufficient conditions for critical points to be minimizers. As a first step, we need to definedifferent types of extrema and will give an extensive, but simple example that motivatesthe distinction (and that will be used again).Remark. We only talk about minima and minimizers, but everything could be adoptedto the problem of finding maxima and maximizers, for example, by changing the sign ofI and its boundary conditions.

Definition 9. (Global, strong, and weak minimizers) Let M ⊂ C1(Ω;Rm) be the set ofadmissible functions and let I : M → R ∪ +∞ be a functional (slightly more generalthan the setting from Section 2.1). Then a point u ∈M is called

1. a weak local minimizer, if

(∃δ > 0)(∀v ∈M)‖u− v‖C1 < δ ⇒ I(u) ≤ I(v),


2. a strong local minimizer, if

(∃δ > 0)(∀v ∈M)‖u− v‖C0 < δ ⇒ I(u) ≤ I(v),

3. a global minimizer, if(∀v ∈M)I(u) ≤ I(v).

Here, the norms ‖ · ‖C1 and ‖ · ‖C0 are defined as in (2.3).

Remark. In finite dimensions, there would not be any distinction between weak andstrong local minimizers, because all norms on the space of admissible vectors would beequivalent. The relevance of this distinction in the infinite-dimensional setting derivesfrom the examples that we look at and from the applications that we have in mind (inother words: it need not/should not be clear now, but will become clear as we move on).

As a first context for the definition, consider the following implications:

u global min.1)⇒ u strong loc. min.

2)⇒ u weak loc. min.3)∗⇒ u crit. pt.,

and their converse implications:

u global min.4): u strong loc. min.

5)∗: u weak loc. min.6): u crit. pt.

Then 1) and 2) follow directly from the definition, and we will show 3)∗ in Theorem 13. 4)is known already from finite-dimensional counterexamples (not every minimum is a globalminimum), 5)∗ will be shown in the Example below and 6) is again obvious, because acritical point may also be a maximum. To summarize, the “unstarred” implications aretrivial, but we will have a second look at the “starred” ones.

To illustrate the definition, we use the following example:

2.4.1 The double-well functional

Let Ω = (0, 1) and define the energy density f(a) = (1− a2)2 (double-well potential).

−1 1


Then consider the functional

Iα(u) =

∫ 1

0

f(u′(x)) dx =

∫ 1

0

(1− u′(x)2)2 dx

on the setMα = u ∈ C1([0, 1];R) : u(0) = 0 and u(1) = α

of functions with fixed/clamped/Dirichlet boundary conditions u(0) = 0 and u(1) = αthat vary with some parameter α ≥ 0.

Of course, the question is: What are minimizers of Iα in Mα?

Claim 10. We will show the following:

1. The only critical points of Iα in Mα are uα defined by uα(x) = αx.

2. If α = 1, then u1 is a global minimizer.

3. If α < 1, then infv∈MαIα(v) = 0, but there is no global minimizer.

4. If α ≥ 1, then uα is a global minimizer.

5. If 1√3< α < 1, then uα is weak local minimizer, but not a strong local minimizer.

6. If 0 ≤ α < 1√3, then there is no weak local minimizer.

7. If α = 1√3, then .... Exercise!

Remark. This example also clearly illustrates the importance of boundary conditions!

Proof. First, we observe that regardless of the choice of α and u, Iα(u) ≥ 0. In the proofof 1., we use the following result related to the Fundamental Lemma and proved in thesecond Exercise.

Lemma 11. (Lemma of du Bois-Reymond) Consider

X0 = v ∈ C1([a, b];Rm) : v(a) = v(b) = 0

and h,H ∈ C0([a, b];Rm). Then if

∫ b

a

h(x) · v(x) +H(x) · v′(x) dx = 0

for all v ∈ X0 , then H ∈ C1([a, b];Rm) and H ′(x) = h(x) for all x ∈ (α, β).

1. For all a ∈ R, we have f ′(a) = −4a(1− a2), so by definition, a critical point u mustsatisfy

DIα(u)[v] = −∫ 1

0

4u′(x)(1− u′(x)2)v′(x) dx

for all v ∈ X0. By the du Bois-Reymond Lemma,

2u′(x)3 − 2u′(x) + c = 0 (2.5)


for all x ∈ (0, 1) and a constant c ∈ R (plug in h = 0, H(x) = f ′(u′(x))). Since(2.5) is a third-order polynomial in u′(x), it has at most three distinct real roots.If u′(x) takes on the values of two different roots at two different places, then sinceu′ is continuous it must take on all values that lie inbetween, a contradiction. Itfollows that u′(x) = const. From the boundary conditions for u ∈ Mα, it followsthat uα(x) = αx is the only possible choice as a critical point. Note that by solvingthe corresponding ELEs

(4u′(1− (u′)2))′(x) = 0, for all x ∈ (0, 1),

with boundary conditions u(0) = 0, u(1) = α, we also turn up uα, but we have notproved that it is the only critical point in Mα.

2. In this case, I1(u1) = 0, so u1 is a global minimizer because I1(u) ≥ 0 for all u ∈M1.

3. In order to show the first statement, we construct a sequence of functions un ∈Mα

such that limn→∞ Iα(un) = 0. A first observation is that if u can be constructedsuch that u′(x) = ±1 for all x ∈ (0, 1), then Iα(u) = 0. For example, the functionu with

u(x) =

x, 0 ≤ x ≤ c,−x+ 2c c ≤ x ≤ 1,

and c = α+12 satisfies Iα(u) = 0 and u(0) = 0, u(1) = α. However, u is not

continuously differentiable, its derivative has a jump, and thus u /∈ Mα (note thatwe have u ∈ PC1([0, 1];R), the space of piecewise C1-functions). So, the idea isto approximate u by a sequence un ∈ Mα for which the “tip” is smoothened by aparabola on smaller and smaller intervals around x = c.

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c 1

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In particular, we set

u′n(x) =

1, 0 ≤ x ≤ c− 1n ,

−1, c+ 1n ≤ x ≤ 1,

−n(x− c), c− 1n ≤ x ≤ c+ 1

n ,

with

un(x) =

x, 0 ≤ x ≤ c− 1n ,

−x+ 2c, c+ 1n ≤ x ≤ 1,

−n2 (x− c)2 − 12n + c, c− 1

n ≤ x ≤ c+ 1n .


Then, un ∈ Mα, there is pointwise convergence of un(x) to u(x) for all x ∈ [0, 1],and

Iα(un) =

∫ c+1/n

c−1/n

(1− n2(x− c)2)2 dx ≤∫ c+1/n

c−1/n

1 dx =2

n

n→∞→ 0.

Thus, we have shown that infu∈Mα Iα(u) = 0. The fact that there is no globalminimizer follows from 5. and 6., so we do not give a separate proof here.

4. This follows from the fact that if α ≥ 1, then for all b ∈ R,

f(b) ≥ f(α) + f ′(α)(b− α).

Hence, for all u ∈Mα,

Iα(u) =

∫ 1

0

f(u′(x)) dx ≥∫ 1

0

f(α) + f ′(α)(u′(x)− α) dx

= Iα(uα) + f ′(α)

(∫ 1

0

u′(x) dx− α)

= Iα(uα).

5. To show that uα is a weak local minimizer, note that if 1√3< α < 1, then f ′′(α) > 0,

so there exists a δα > 0 such that for all |b− α| < δα,

f(b) ≥ f(α) + f ′(α)(b− α),

as in 4. above. So for all u ∈Mα such that

supx∈[0,1]

|u′(x)− α| = ‖u′ − α‖C0 ≤ ‖u− uα‖C1 < δα,

we haveIα(u) ≥ Iα(uα)

as in 4.To show that uα is not a strong minimizer, the idea is the following: construct asequence of zig-zag (highly oscillating) functions vn ∈Mα, such that

(a) ‖vn − uα‖C0 → 0 as n→∞, and(b) Iα(vn) = 0.

Then vn approximates uα in C0 but Iα(vn) < Iα(uα), so uα cannot be a strong localminimizer. To get (b), we must have v′n(x) = ±1 for all x ∈ (0, 1). We set v1 = uand then make vn “more zig-zag” (more changes in the sign of v′n) as n increases sothat the difference to uα becomes smaller:






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More precisely, let

w1(x) =

(1− α)x, 0 ≤ x < c,

2c(1− x), c ≤ x ≤ 1,

so that u = uα + w1. Extend w1 periodically to be a continuous function on all ofR. Then set

wn(x) =1

nw1(nx), so that w′n(x) = w′1(nx),

and set vn ' uα+wn, where'means that to get vn ∈Mα, we need it to be smoothedat the tips, as in the parabola-construction in 3.! We ignore this “technical” detailin the following. We get that

‖vn − uα‖C0 = supx∈[0,1]

‖wn(x)‖ ≤ 1

n(1− α)c

n→∞→ 0

and that

Iα(vn) =

∫ 1

0

f(v′n(x)) dx

=

∫ 1

0

f(α+ w′1(nx)) dx

y=nx=

1

n

∫ n

0

f(α+ w′1(y)) dy

w1 is 1−periodic=

∫ 1

0

f(α+ w′1(y)) dy = Iα(v1) = 0,

which proves the claim.

6. If 0 ≤ α < 1√3, then f ′′(α) < 0, i.e. f is locally at α strictly concave. As in the proof

of 5., where we showed that uα is a weak local minimizer, we can now show thatuα is a weak local maximizer. Thus, it is not a weak local minimizer. By Theorem13 (next Section!), it is necessary for a weak local minimizer to be a critical point,so by 1., there can be no other weak minimizer.

2.5 Some necessary and sufficient conditionsIn the finite-dimensional case, the definiteness of the second derivative or of the Hessiangive us necessary and sufficient conditions for critical points to be extrema, cf. Section1.4 and Exercise 1.IV. Here, the idea is to modify these conditions to apply to weak localminima of the functional I.

Definition 12. (Second variation) Let u ∈M and v, w ∈ X0. Then the map

D2I(u)[v, w] :=d

dtDI(u+ tw)[v]|t=0


is called the second variation of I at u along v, w. If I is as before with integral densitiesf, g, we have

D2I(u)[v, w]

=

∫

Ω

∂2uf(x, u(x),∇u(x))v(x) · w(x)

+ ∂2Af(x, u(x),∇u(x))[∇v(x),∇w(x)]

+ ∂u∂Af(x, u(x),∇u(x))[v(x),∇w(x)]

+ ∂u∂Af(x, u(x),∇u(x))[w(x),∇v(x)] dx

+

∫

∂Ω

∂2ug(x, u(x))v(x) · w(x) dx′

(notation from Section 2.2!).

Proof. As for the first variation, we use the chain rule and the fact that differentiationand integration can be interchanged. This follows from the fact that f and g are twicecontinuously differentiable.

Theorem 13. Let Ω,Γ, u0,M,X0, f, g as before. Then the following holds:

1. If u ∈M is a weak local minimizer, then

(a) for all v ∈ X0 , DI(u)[v] = 0, i.e. u is a critical point of I,

(b) for all v ∈ X0, D2I(u)[v, v] ≥ 0, i.e. the second variation is positive semi-definite.

2. If u ∈ M = u0 + X0 is a critical point and there exists γ > 0 such that for allv ∈ X0,

D2I(u)[v, v] ≥ γ(

∫

Ω

|v(x)|2 + |∇v(x)|2 dx+

∫

∂Ω

|v(x)|2 dx′),

then u is a weak local minimizer.

Proof. To prove the two necessary conditions, we observe that a minimizer of a func-tional should be a minimizer along every direction and thus reduce the problem to aone-dimensional situation:

1. For u ∈ M , v ∈ X0 define ϕv : R → R, t 7→ I(u + tv). Since u is a weak localminimizer, there is a δ > 0 such that for all w ∈M ,

‖w − u‖C1 ≤ δ ⇒ I(u) ≤ I(w).

In particular, for all |t| ≤ δ‖v‖C1

, ‖u− (u+ tv)‖C1 ≤ δ, so

ϕv(t) = I(u+ tv) ≥ I(u) = ϕv(0).

Hence, 0 is a local minimizer of ϕv. By Exercise 1.IV (or Analysis II), necessarily,

ϕ′v(0) = 0 and ϕ′′v(0) ≥ 0.


Hence, necessarily,

DI(u)[v] = ϕ′v(0) = 0 and D2I(u)[v, v] = ϕ′′v(0) ≥ 0.

Note, however, that being a minimizer for every ϕv does not imply being a localminimizer for I, not even in the finite-dimensional situation, cf. Exercise 1.II!

2. Let u ∈ M be a critical point of I and w ∈ X0. Then by Taylor’s Theorem in onedimension, there is a θ ∈ [0, 1] such that

I(u+ w) = ϕw(1) = ϕw(0) + ϕ′w(0) +1

2ϕ′′w(θ) (2.6)

= I(u) +DI(u)[w] +1

2D2I(u+ θw)[w,w]

= I(u) +1

2D2I(u)[w,w]

+1

2(D2I(u+ θw)[w,w]−D2I(u)[w,w]), (2.7)

where for the last equality, we added a suitable zero and used that u is a criticalpoint. To prove the claim, it is now sufficient to show that the sum of all termsadding to I(u) in the last equality is ≥ 0 for sufficiently small ‖w‖C1 . We useDefinition 12 to obtain

D2I(u+ θw)[w,w]−D2I(u)[w,w] (2.8)

=

∫

Ω

(∂2uf(x, u+ θw,∇u+ θ∇w)− ∂2

uf(x, u,∇u))w · w (2.9)

+ 2(∂u∂Af(x, u+ θw,∇u+ θ∇w)− ∂u∂Af(x, u,∇u))[w,∇w]

+ (∂2Af(x, u+ θw,∇u+ θ∇w)− ∂2

Af(x, u,∇u))[∇w,∇w] dx

+

∫

∂Ω

(∂2ug(x, u+ θw)− ∂2

ug(x, u))w · w dx′.

Now we use that the function

∂2uf

u : Ω× Rm × Rm×n → Rm×m

(x, v,A) 7→ ∂2uf(x, u(x) + v,∇u(x) +A)

is uniformly continuous on compact subsets of Ω×Rm×Rm×n, so that there existsa δγ1 > 0 such that for all w ∈ X0 and x ∈ Ω, if ‖w‖C1 ≤ δγ1 , then

|∂2uf(x, u(x) + θw(x),∇u(x) + θ∇w(x))− ∂2

uf(x, u(x),∇u(x)))| ≤ γ

4.

In the same way, we choose δγ2 , δγ3 , δ

γ4 > 0 for the other terms that appear in (2.8).

Adding up, there exists a δ = min(δγ1 , δγ2 , δ

γ3 , δ

γ4 ) > 0 such that for all w ∈ X0, if

‖w‖C1 < δ, then

D2I(u+ θw)[w,w]−D2I(u)[w,w]

≤ γ

4

∫

Ω

|w(x)|2 + 2|w(x)||∇w(x)|+ |∇w(x)|2 dx+γ

4

∫

∂Ω

|w(x)|2 dx′

≤ γ

2

∫

Ω

|w(x)|2 + |∇w(x)|2 dx+γ

4

∫

∂Ω

|w(x)|2 dx′.


We insert this estimate in (2.6). Together with the assumption on D2I(u)[v, v], weget that

I(u+ w) ≥ I(u)

for all ‖w‖C1 < δ, i.e. u is a weak local minimizer for I.

As a first application of Theorem 13, we look at the double-well functional Iα from thelast section. We have shown that uα are the only critical points, so they are necessarilythe only candidates for weak local (and thus strong local or even global) minimizers.We have f ′′(a) = 4(3a2 − 1) for the double-well density f , so the second variation of Iαat uα is

D2Iα(uα)[v, v] =

∫ 1

0

4(3α2 − 1)(v′(x))2 dx.

Clearly, D2Iα(uα)[v, v] ≥ 0 if and only if |α| ≥ 1√3, so there are no weak local minimizers

for 0 < α < 1√3. This is an alternative proof of 6. in the last Section. Moreover, if

α > 1√3, then 4(3α2 − 1) = γ′ > 0 and so

D2Iα(uα)[v, v] = γ′∫

Ω

v′(x)2 dx ≥ γ∫

Ω

v(x)2 + v′(x)2 dx

for some γ > 0 by the Poincaré Inequality (the Poincaré Inequality may be known fromfunctional analysis or some PDE or geometry course. If not, don’t worry, it will bediscussed in more detail later!) In particular, by Theorem 13, if α > 1√

3, then uα is a

weak local minimizer (compare 3. and 4. in the last Section).We have seen that the ELEs (Theorem 8) provide a necessary condition for the nec-

essary condition of u being a critical point for being a local minimizer (Theorem 13).Similarly, with the next result, we replace the second necessary condition in Theorem13 on the second variation by a weaker, but pointwise necessary condition on f , theLegendre-Hadamard (LH) condition. The idea is the following:

1. In the second variation, the first term

D2I(u)[v, v] =

∫

Ω

∂2Af(x, u(x),∇u(x))[∇v(x),∇v(x)] dx+ · · · ≥ 0

is the principal part, the most relevant part. The LH-condition is a necessarycondition that asks for the non-negativity of this part only.

2. The condition∂2Af(x, u(x),∇u(x))[B,B] ≥ 0 (2.10)

for all matrices B ∈ Rm×n would be sufficient for the non-negativity of this part, butit would also be too strong to be necessary, as the integral term has more structure.Instead, the LH-condition replaces B ∈ Rm×n with rank-1-matrices ξ ⊗ η ∈ Rm×nfor ξ ∈ Rm, η ∈ Rn.


Theorem 14. (Legendre-Hadamard condition (LH)) If u ∈ X and

D2I(u)[v, v] ≥ 0

for all v ∈ X0, then for all x ∈ Ω, for all ξ ∈ Rm and for all η ∈ Rn,m∑

i,k=1

n∑

j,l=1

∂2

∂Aij∂Aklf(x, u(x),∇u(x))ξiξkηjηl ≥ 0,

in short:∂2Af(x, u(x),∇u(x))[ξ ⊗ η, ξ ⊗ η] ≥ 0.

In particular, if u is a weak local minimizer of I, then the Legendre-Hadamard conditionholds.

Proof. Fix ξ ∈ Rm and η ∈ Rn. The idea is to consider the limit

limδ%0

D2I(u)[vδ, vδ],

where

• ∇vδ ' ξ ⊗ η – not exactly equal, but highly oscillating,

• vδ ∈ X0 and supp vδ ⊂ Bδ(x0) for given x0 ∈ Ω,

• vδ = O(δ), so that lower-order terms in D2I(u)[vδ, vδ] disappear in the limit.

Step 1: “Freeze coefficients in the integral and get rid of lower-order terms”: weshow that if D2I(u)[v, v] ≥ 0 for all v ∈ X0, then for all x0 ∈ Ω andw ∈ C1

0 (B1(0)) := w ∈ C1(B1(0)) : w|∂B1(0) = 0,∫

B1(0)

∂2Af(x0, u(x0),∇u(x0))[∇w(y),∇w(y)] dy ≥ 0. (2.11)

Fix w ∈ C10 (B1(0)), x0 ∈ Ω and δ0 > 0 such that Bδ0(x0) ⊂ Ω. For all

0 < δ < δ0 definewδ(x) = δw(

x− x0

δ).

Then ∇wδ(x) = ∇w(x−x0

δ ) and wδ ∈ X0. By assumption, for all 0 < δ ≤ δ0,we have

0 ≤ 1

δn

∫

Bδ(x0)

∂2Af(x, u(x),∇u(x))[∇w(

x− x0

δ),∇w(

x− x0

δ)]

+ 2δ∂A∂uf(x, u(x),∇u(x))[w(x− x0

δ),∇w(

x− x0

δ)]

+ δ2∂2uf(x, u(x),∇u(x))[w(

x− x0

δ), w(

x− x0

δ)] dx

=

∫

B1(0)

∂2Af(x0 + δy, u(x0 + δy),∇u(x0 + δy))[∇w(y),∇w(y)] dy

+O(δ),


with the substitution y = x−x0

δ in the last equality. In this estimate, we nowapply the limit δ % 0 and use the boundedness and continuity of

y 7→ ∂2Af(x0 + δy, u(x0 + δy),∇u(x0 + δy))

to obtain (2.11).

Step 2: “Construction of vδ”: Fix ξ ∈ Rm, η ∈ Rn. Let χ ∈ C∞0 (B1(0);R) with χ > 0.For example, choose

χ(x) =

e− 1

1−|x|2 , |x| < 1,

0, |x| ≥ 1,

as in the proof of (FL). Then define

vδ(y) := δχ(y) cos(η · y

δ

)ξ ∈ C1

0 (B1(0)),

with y 7→ cos(η·yδ ) a suitable “Wellblech”-function. It follows that

∇vδ(y) = δ cos(η · yδ

)ξ ⊗∇χ(y)− χ(y) sin(η · yδ

)ξ ⊗ η.

We can consider the first term in the derivative as of lower order for δ % 0.Thus, plugging vδ into (2.11), we get

0 ≤∫

B1(0)

∂2Af(x0, u(x0),∇u(x0))[ξ ⊗ η, ξ ⊗ η]χ2(y) sin2(

η · yδ

) dy

+O(δ) +O(δ2).

Taking the limit δ % 0 in this equality, we obtain

0 ≤ ∂2Af(x0, u(x0),∇u(x0))[ξ ⊗ η, ξ ⊗ η] lim

δ%0

∫

B1(0)

χ2(y) sin2(η · yδ

) dy.

Now note that if η = 0, then (LH) holds automatically. If η 6= 0, then (LH)now follows from the following Lemma.

Lemma 15. Let χ ∈ C10 (Ω;R) and η ∈ Rn \ 0. Then

limδ%0

∫

B1(0)

χ(y) sin2(η · yδ

) dy =1

2

∫

B1(0)

χ(y) dy.

(Later, we will show a general result that provides: sin2(η·yδ )L2(B1(0)) 1

2)

Proof. Note that for all α ∈ R, sin2(α) = 12 (1− cos(2α)), so

∫

B1(0)

χ(y) sin2(η · yδ

) dy =1

2

∫

B1(0)

χ(y) dy − 1

2

∫

B1(0)

χ(y) cos(2η · yδ

) dy.


We use Gauss’ Theorem to deal with the second term on the right-hand side. By thechain rule,

divy(sin(2η · yδ

)η)divy(η)=0

= η · ∇y sin(2η · yδ

) =2

δcos(

2η · yδ

)|η|2.

It follows that

limδ%0

∫

B1(0)

χ(y) cos(2η · yδ

) dy

= limδ%0

δ

2|η|2∫

B1(0)

χ(y)divy

(sin(

2η · yδ

)

)dy

Gauss= − lim

δ%0

δ

2|η|2∫

B1(0)

∇χ(y) · η sin(2η · yδ

) dy = 0,

where in the last line, we have used that χ(y) = 0 on ∂B1(0) and that the remainingintegral is bounded independently of δ > 0.

Remark 16. Some special cases and applications of the (LH)-condition:

• If m = n = 1, then clearly, the (LH)-condition is equivalent to

∂2af(x, u(x), u′(x)) ≥ 0,

the second derivative of f with respect to the last component being non-negative.As an example, we look at the double-well-potential:

∂2af(α) = 12α2 − 4 ≥ 0⇔ α ≥ 1√

3.

This is a quick proof of the necessity of α ≥ 1√3for uα to be a weak local minimizer.

• If n = 1,m ≥ 1, or n ≥ 1,m = 1, then the (LH)-condition asks that the A-component of the Hessian of f ,

∂2Af(x, u(x),∇u(x)) ∈ Rk×k, k = m or k = n,

is positive semi-definite at every x ∈ Ω. As an example, for the Dirichlet integral,this means that the symmetric part of K(x),

1

2(K(x) +K(x)T ),

needs to be positive semi-definite at every x ∈ Ω.

• More generally, for m,n ≥ 2, and quadratic functionals I, where the ELEs are alinear system of PDEs, the (LH)-condition is a particular type of ellipticity condi-tion.

• The (LH)-condition states that, roughly speaking, f should be rank-one-convex inthe gradient component, cf. the next Section.


As a next step, we introduce more explicit necessary conditions on f for u to be a stronglocal minimizer of I. This leads to the concept of quasiconvexity of f . Recall that, roughlyspeaking, being a strong local minimizer means being a minimizer in spite of oscillations,as highly oscillating functions drop out of the C1-neighbourhood of a potential minimizer,but may still be in a C0-neighbourhood. In the following, we “quantify” this idea bylooking at C0-perturbations u+ wε of u that satisfy ‖wε‖C0 ∼ ε and ‖∇wε‖C0 = O(1).

Let u ∈ M be a strong local minimizer of I. In particular, there exists a δ0 > 0 suchthat for every w ∈ C1

0 (Ω;Rm),

‖w‖C0 ≤ δ0 ⇒ u+ w ∈M and I(u+ w) ≥ I(u).

We derive a localized necessary condition. Fix x0 ∈ Ω and δ ≤ δ0 such that Bδ(x0) ⊂ Ω.Now for any w ∈ C1

0 (Bn1 (0);Rm) and ε ≤ minδ, δ‖w‖C0

= δ, define

wε(x) := εw(x− x0

ε).

Then ∇wε(x) = ∇w(x−x0

ε ), u+ wε ∈M and ‖u− (u+ wε)‖C0 = ε‖w‖C0 ≤ δ so

I(u+ wε) ≥ I(u).

It follows that

0 ≤ limε→0

1

εn(I(u+ wε)− I(u))

= limε→0

1

εn

∫

Bε(x0)

f(x, u+ wε(x),∇u(x) +∇w(x− x0

ε))− f(x, u(x),∇u(x)) dx

y=x−x0ε= lim

ε→0

∫

B1(0)

f(x0 + εy, u(x0 + εy) + εw(y),∇u(x0 + εy) +∇w(y))

− f(x0 + εy, u(x0 + εy),∇u(x0 + εy)) dy

=

∫

B1(0)

f(x0, u(x0),∇u(x0) +∇w(y))− f(x0, u(x0),∇u(x0)) dy

=

∫

B1(0)

f(x0, u(x0),∇u(x0) +∇w(y)) dy − vol(Bn1 (0))f(x0, u(x0),∇u(x0)).

This motivates the following definition.

Definition 17. (quasiconvexity, Morrey 1952) A function F : Rm×n → R is called qua-siconvex in A0 ∈ Rm×n, if for all w ∈ PC1

0 (Bn1 (0);Rm) (“piecewise C10 ”),

∫

B1(0)

F (A0 +∇w(y)) dy ≥∫

B1(0)

F (A0) dy = vol(Bn1 (0))F (A0).

We say that F is quasiconvex, if F is quasiconvex in every A0 ∈ Rm×n.Remark 18. There are several (equivalent) versions of this definition and we will use themas needed, without proof:

• Bn1 (0) could be replaced by other suitable types of bounded domainsD, for example,cubes.


• w ∈ PC10 (D) can be replaced by w ∈ C1

0 (D) or even w ∈ C∞c (D).

A direct consequence of our considerations is the following necessary condition forstrong local minimizers:

Theorem 19. (Morrey 1952) Let u∗ ∈ M be a strong local minimizer of I. Then themap

f(x0, u∗(x0), ·) : Rm×n → R

is quasiconvex in ∇u∗(x0) for all x0 ∈ Ω.

Example 20. We look at the (standard) double-well example and claim that f is qua-siconvex in α if and only if |α| ≥ 1. In particular, this reproves that uα is no strong localminimizer of Iα if α < 1.

Proof. To check quasiconvexity of f : R 3 a 7→ (1 − a2)2 ∈ R in α ∈ R, let w ∈PC1

0 ([−1, 1]). By definition, we need to verify or disprove∫ 1

−1

f(α+ w′(y)) dy!≥ 2f(α).

If |α| ≥ 1, then, cf. the proof of Claim 10.4., for all β ∈ R,

f(α+ β) ≥ f(α) + f ′(α)β.

It follows that∫ 1

−1

f(α+ w′(y)) dy ≥∫ 1

−1

f(α) dy + f ′(α)

∫ 1

−1

w′(y) dy = 2f(α),

using∫ 1

−1w′(y) dy = 0.

If |α| < 1, construct wα ∈ PC1([−1, 1]) such that∫ 1

−1

f(α+ w′α(y)) dy = 0 :

choose wα as a piecewise affine “hat”, so that for all y ∈ [−1, 1], either α + w′α(y) = 1 orα+ w′α(y) = −1. More precisely, set

wα(x) =

(1− α)(x+ 1), −1 ≤ x ≤ α,(−1− α)(x− 1), α ≤ x ≤ 1.

Then ∫ 1

−1

f(α+ w′α(y)) dy =

∫ α

−1

f(1) dy +

∫ 1

α

f(−1) dy = 0,

and, at the same time, 2f(α) > 0, so f is not quasiconvex in any |α| < 1. By Morrey’sTheorem, in this case, uα is no strong local minimizer of Iα. This quickly reproves someof the results in Claim 10.5.


2.6 Types of ConvexityThe conditions derived in the last Section suggest that we look more closely at differentnotions of convexity for the “A-”component of the density function f for given x, u:

F : Rm×n → R, F (A) = f(x, u,A).

Since it will be helpful later, we allow the function F to take on the value “+∞”, if itdoesn’t interfere with the definition.

Definition 21. (Types of Convexity)

1. F : Rm×n → R ∪ +∞ is called convex, if for all A,B ∈ Rm×n and θ ∈ [0, 1],

F (θA+ (1− θ)B) ≤ θF (A) + (1− θ)F (B).

2. A locally bounded, measurable function F : Rm×n → R is called quasiconvex (qc),if for all A ∈ Rm×n and for all w ∈ PC1

0 (Bn1 (0);Rm),∫

B1(0)

F (A+∇w(y)) dy ≥ vol(Bn1 (0))F (A).

3. F : Rm×n → R ∪ +∞ is called rank-1-convex (r1c), if for all θ ∈ [0, 1] and for allA,B ∈ Rm×n such that rank(B −A) = 1,

F (θA+ (1− θ)B) ≤ θF (A) + (1− θ)F (B).

Even now, the relations between these different notions of convexity are not fullyunderstood. We know the following.

Theorem 22. (Types of Convexity) Let F : Rm×n → R, then

1. F convex ⇒F quasiconvex ⇒F rank-1-convex.

2. If min(m,n) = 1, then

F convex ⇔ F quasiconvex ⇔ F rank-1-convex.

3. If F ∈ C2(Rm×n), thenF rank-1-convex ⇔ F satisfies (LH) in every A ∈ Rm×n, i.e. (

(∀A ∈ Rm×n)(∀ξ ∈ Rm)(∀η ∈ Rn) ∂2AF (A)[ξ ⊗ η, ξ ⊗ η] ≥ 0.

Before we proceed to the proof, we discuss Statement 1. in the Theorem in some moredetail: In general, we can say that the converse implications fail,

F convex : F quasiconvex : F rank-1-convex.

More precisely,


• If m = n = 2, then the determinant det : R2×2 → R with det

(a bc d

)= ad − bc

is quasiconvex, but not convex.

Proof. We show first that det is quasiconvex: For A =

(a bc d

)∈ R2×2, define

adj(A) =

(d −b−c a

)the adjugate of A. Then for all A, A ∈ R2×2,

det(A+ A) = ad− bc+ ad− bc+ ad+ ad− bc− bc= detA+ det A+ adj(A)T : A.

Let now w ∈ PC10 (B1(0);R2), then, using this equality,

Proof.∫

B1(0)

det(A+∇w(y)) dy

= vol(B1(0)) detA

+

∫

B1(0)

det(∇w(y)) dy + adj(A) :

∫

B1(0)

∇w(y) dy. (2.12)

For the last term, we get

adj(A) :

∫

B1(0)

∇w(y) dyGauss

= adj(A) :

∫

∂B1(0)

w(y)⊗ ν dy′w|∂B1(0)=0

= 0.

For the middle term on the right-hand-side of (2.12), we get∫

B1(0)

det(∇w(y)) dy =

∫

B1(0)

∂1w1(y)∂2w2(y)− ∂1w2(y)∂2w1(y) dy

Gauss= −

∫

B1(0)

∂2∂1w1(y)w2(y) dy

+

∫

∂B1(0)

∂1w1(y)w2(y)ν2(y) dy′

+

∫

B1(0)

∂1∂2w1(y)w2(y) dy

−∫

∂B1(0)

∂2w1(y)w2(y)ν1(y) dy′

w|∂B1(0)=0= 0.

Here, we have actually used w ∈ C20 (B1(0);R2) to apply Gauss’ Theorem. It remains

to use Remark 18 to argue that this is sufficient.


Now we show by counterexample that det is not convex: Let A =

(−1 00 1

)and

B =

(1 00 −1

). Then for θ = 1

2 ,

θA+ (1− θ)B =

(0 00 0

),

sodetA = detB = θ detA+ (1− θ) detB = −1 < det(0) = 0.

• If m ≥ 3, n ≥ 2, then there is an example of a function F : Rm×n → R that isrank-1-convex, but not quasiconvex [Dac08, Sect 5.3.7]

• If m = 2, n ≥ 3, then it is not known whether

F quasiconvex : F rank-1-convex.

Proof. (of Theorem 22)To prove 2., assume 1.: Then it is sufficient to show that

F rank-1-convex ⇒ F convex,

but this is also trivially satisfied since all A,B ∈ R1×n or A,B ∈ Rm×1 have rank one.To prove 3., first note that for all C ∈ Rm×n,

rank(C) = 1⇔ (∃ξ ∈ Rm)(∃η ∈ Rn) C = ξ ⊗ η.

To see this, let C be given such that Im(C) = span(ξ). Then for all canonical basis vectorsej ∈ Rn, eji := δij , there exists ηj ∈ R such that Cej = ηjξ. In particular, Cej = (ξ⊗η)ej .Now proceed through the following equivalences:

F rank-1-convex I⇔ (∀ξ ∈ Rm)(∀η ∈ Rn)(∀A ∈ Rm×n) the map ϕξ,ηA : R→ R,

given by ϕξ,ηA (t) = F (A+ tξ ⊗ η), is convex,II⇔ (∀ξ ∈ Rm)(∀η ∈ Rn)(∀A ∈ Rm×n)∀t ∈ R,

∂2AF (A+ tξ ⊗ η)[ξ ⊗ η, ξ ⊗ η] = (ϕξ,ηA )′′(t) ≥ 0,

III⇔ (∀ξ ∈ Rm)(∀η ∈ Rn)(∀A ∈ Rm×n),

∂2AF (A)[ξ ⊗ η, ξ ⊗ η] ≥ 0


I⇒: Given t1, t2 ∈ R and θ ∈ [0, 1], we have

ϕξ,ηA (θt1 + (1− θ)t2) = F (A+ θt1ξ ⊗ η + (1− θ)t2ξ ⊗ η)

= F (θ(A+ t1ξ ⊗ η) + (1− θ)(A+ t2ξ ⊗ η))

F r1c≤ θF (A+ t1ξ ⊗ η) + (1− θ)F (A+ t2ξ ⊗ η)

= θϕξ,ηA (t1) + (1− θ)ϕξ,ηA (t2).

I⇐: Given θ ∈ [0, 1] and A,B ∈ Rm×n such that rank(B − A) = 1, there existξ ∈ Rm, η ∈ Rn such that B −A = ξ ⊗ η, so

F (θB + (1− θ)A) = F (A+ θ(B −A)) = ϕξ,ηA (θ)

ϕξ,ηA convex≤ θϕξ,ηA (1) + (1− θ)ϕξ,ηA (0)

= θF (B) + (1− θ)F (A).

II⇔: This follows from the fact that for all ϕ ∈ C2(R;R),

ϕ convex ⇔ ∀a ∈ Rϕ′′(a) ≥ 0.

III⇒ : This is clear, just set t = 0.

III⇐ : This is clear since A+ tξ⊗η ∈ Rm×n holds for all A ∈ Rm×n, ξ ∈ Rm, η ∈ Rnand t ∈ R.

It remains to prove 1.: To show that convex F are quasiconvex, we use Jensen’s inequality(for a proof, see Exercise 3):

Lemma 23. (Jensen’s Inequality) Let Ω ⊂ Rn be a bounded domain in Rn, let N ∈ Nand let F : RN → R be convex, then for all u ∈ C0(Ω;RN ) and for all u ∈ L1(Ω;RN ) ,

F (1

vol(Ω)

∫

Ω

u(y) dy) ≤ 1

vol(Ω)

∫

Ω

F (u(y)) dy.

Let now w ∈ PC10 (B1(0);Rm) and let F : Rm×n → R be convex. In particular, F

is locally bounded and measurable. Then by Gauss,∫B1(0)

∇w(y) dy = 0, so for allA ∈ Rm×n,

1

vol(B1(0))

∫

B1(0)

F (A+∇w(y)) dyJensen≥ F (

1

vol(B1(0))

∫

B1(0)

A+∇w(y) dy) = F (A),

and thus F is quasiconvex.We “extensively sketch” the proof of the fact that quasiconvex functions F are rank-1-convex. A full proof is in [Dac08, Chapter 3.]

Let A,B ∈ Rm×n such that rank(B−A) ≤ 1. Then for all θ ∈ [0, 1], we need to show

F (A+ θ(B −A)) ≤ (1− θ)F (A) + θF (B).


Rough idea: Using quasiconvexity of F , with B1(0) replaced by cubes

D1(0) = x ∈ Rn : 0 < xi < 1, vol(D1(0)) = 1,

we would get

F (A+ θ(B −A)) ≤∫

D1(0)

F (A+ θ(B −A) +∇w(y)) dy (2.13)

=

∫

ΩA

F (A) dy +

∫

Ω

F (B) dy

= (1− θ)F (A) + θF (B),

if w ∈ PC10 (D1(0);Rm) can be constructed suitably, i.e. such that there exist ΩA,ΩB ⊂

D1(0) with vol(ΩA) = 1− θ, vol(ΩB) = θ and

∇w(y) =

−θ(B −A) =: RA, y ∈ ΩA,

(1− θ)(B −A) =: RB , y ∈ ΩB .

Clearly, these properties of w can only be achieved approximately. The following lemmashows how this can be done.

Lemma 24. (Dacorogna: Lemma 3.11) Let RA, RB ∈ Rm×n such that

rank(RA −RB) ≤ 1

and such that 0 ∈ co(RA, RB), where

co(RA, RB) = C ∈ Rm×n : ∃s ∈ [0, 1]C = sRA + (1− s)RB

is the line segment joining RA and RB. Then for all ε > 0, there exists a

wε ∈ PC10 (D1(0);Rm)

and ΩεA,ΩεB ⊂ D1(0) such that

1. 1− θ − ε < |vol ΩεA| < 1− θ and θ − ε < |vol ΩεB | < θ,

2. ‖wε‖C0 < ε,

3. ∇w(y) =

RA, y ∈ ΩεA,

RB , y ∈ ΩεB .

4. for all y ∈ D1(0), dist(∇wε(y), co(RA, RB)) < ε.

Using this Lemma, we can prove rank-1-convexity of F in the following way: we applythe Lemma to RA, RB defined as above. This is possible, as we have 0 ∈ co(RA, RB)


since (1− θ)RA + θRB = 0. Following (2.13), for all ε > 0, we have

F (A+ θ(B −A))

≤∫

D1(0)

F (A+ θ(B −A) +∇wε(y)) dy

=

∫

D1(0)\(ΩεA∪ΩεB)

F (A+ θ(B −A) +∇wε(y)) dy

+ vol(ΩεA)F (A) + vol(ΩεB)F (B)

≤ εC + (1− θ)F (A) + θF (B),

where in the last inequality, we have used 4. from Lemma 24 and the local boundednessof F . Taking the infimum over all ε > 0 then finishes the proof of Theorem 22.

Before we sketch the proof of Lemma 24, here is a brief discussion:

• There is a heuristics, why rank(RB − RA) ≤ 1 is (somewhat) necessary for a con-

struction of this kind: Let ∇w(y) =

LA, y ∈ ΩA,

LB , y ∈ ΩB .and Γ = ∂ΩA ∩ ∂ΩB , where,

in general Γ may contain n linearly independent vectors xj ∈ Ω. Then for all x ∈ Γ,as w is continuous,

w(x) = LAx+ cA = LBx+ cB ,

so for all xj , (LB − LA)xj = cA − cB and thus rank(LB − LA) ≤ 1.

• In fact, the functions wε given in Lemma 24 are piecewise affine (see sketch ofproof).

• The Lemma holds not only for D1(0), but for bounded open sets (see Step 2 in thesketch of proof)

Proof. (Sketch of the proof of Lemma 24)

Step 1: First, we assume that RB − RA =(ξ 0 . . . 0

)for some ξ ∈ Rm, i.e.

η = e1. Then all components wεi of wε should be constant except in e1-direction with slopes−θξi and (1−θ)ξi. In dependence of y1, we thus constructwεi as a Sägezahn-function:




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For all i = 1, . . . ,m, we set

∂1wεi (y) '

−θξi, y1 ∈ I := ∪N :=b 1εc

k=1 Ik,

(1− θ)ξi, y1 ∈ J := ∪Nk=1Jk,

0, otherwise,

where Ik, Jk are defined as in the picture. Here, ' is used as we must decreasethe height of the “Säge” suitably near the boundaryD1(0) to get wε|∂D1(0) = 0,in accordance with 4. For this, the assumption 0 ∈ co(RA, RB) is used. Set

ΩεA := y ∈ D1(0) : y1 ∈ I and ΩεB := y ∈ D1(0) : y1 ∈ J.

It follows that 1− θ − ε < |vol ΩεA| < 1− θ and θ − ε < |vol ΩεB | < θ (1.),

∇wε(y) '

−θ(B −A), y ∈ ΩεA,

(1− θ)(B −A), y ∈ ΩεB ,

0, otherwise,

(3.) with wε piecewise affine, and ‖wε‖C0 < εmaxi |ξi| (see picture)(2.).

Step 2: It remains for us to see that the situation with general η ∈ Rn in the Lemmacan be reduced to the case η = e1 treated in Step 1:

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D"(yk")

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Set ξ := |η|ξ and η := η|η| , so that ξ ⊗ η = ξ ⊗ η and thus w.l.o.g. |η| = 1.

Then we can choose O ∈ SO(n) orthogonal such that η = e1O and e1 = ηOT .Set Ω = OD1(0) and RA = RAOT , RB = RBOT , then

RB − RA = (RA −RB)OT = ξ ⊗ e1.

Now we split Ω into many cubes Dδε(ykδε

) with suitable length and midpointsykδε , as in the picture, to get that the remaining grey part of Ω is small. Thenwe apply Step 1 to RB , RA on each of the Dδε(y

kδε

) — after rescaling andtranslating to D1(0) – and glue together the resulting functions wε,k – afterre-rescaling and translating back – to get a suitable function wε on Ω (notethat we used wε,k(∂Dδε(y

kδε

)) = 0 for the glueing). The function

wε : D1(0)→ Rm, wε(y) = wε(Oy),

should then be the right one.


After this extensive discussion and sketch of the proof of the general Theorem 22, we nowlook at the much more simple case of quadratic F .

Lemma 25. Let M ∈ R(m×n)×(m×n) be symmetric and F : Rm×n → R given by F (A) =MA : A, i.e.

F (A) =∑

i,k∈1,...,m,j,l∈1,...,nM(ij)(kl)AijAkl.

ThenF quasiconvex ⇔ F rank-1-convex.

Proof. “⇒” follows from Theorem 22. To show “⇐”, we first observe that if F is quadratic,then for all A ∈ Rm×n and w ∈ PC1

0 (Bn1 (0);Rm),∫

Bn1 (0)

F (A+∇w(y)) dyM symmetric

= |Bn1 (0)|MA : A+ 2MA :

∫

Bn1 (0)

∇w(y) dy

+

∫

Bn1 (0)

F (∇w(y)) dy

= |Bn1 (0)|F (A) +

∫

Bn1 (0)

F (∇w(y)) dy,

where we have used that by Gauss’ Theorem,∫Bn1 (0)

∇w(y) dy = 0. It follows that if F isquadratic, then

F quasiconvex

⇔ ∀w ∈ PC10 (Bn1 (0);Rm)

∫

Bn1 (0)

F (∇w(y)) dy ≥ 0. (2.14)

On the other hand, if F is quadratic, in particular, F ∈ C2(Rm×n) and by Theorem 22.3.,if F is rank-1-convex, then for all ξ ∈ Rm, η ∈ Rn,

F (ξ ⊗ η) = M(ξ ⊗ η) : (ξ ⊗ η)A∈Rm×n

= ∂2AF (A)[ξ ⊗ η, ξ ⊗ η]

(LH)

≥ 0. (2.15)

It is interesting that (2.14) can now be shown via Fourier transform: for all w ∈ PC10 (Bn1 (0);Rm),

define the extension w0 ∈ PC10 (Rn;Rm) by

w0(y) =

w(y), y ∈ Bn1 (0),

0, y ∈ Rn \Bn1 (0),

and letw0(η) =

∫

Rnw0(y)e−2πi〈y,η〉 dy


be the Fourier transform of w0. Then for k = 1, . . . ,m, l = 1, . . . , n,

(∇w0)kl(η) =

∫

Rn(∂lw0,k(y))e−2πi〈y,η〉 dy

Gauss= −

∫

Rnw0,k(y)(∂le

−2πi〈y,η〉) dy

= 2πi

∫

Rnw0,k(y)ηle

−2πi〈y,η〉 dy

= 2πiw0,k(y)ηl.

Summing up, we get∫

B1(0)

F (∇w(y)) dy =

∫

RnM∇w0(y) : ∇w0(y) dy

M symm. + Plancherel=

∫

RnM∇w0(η) : ∇w0(η) dη

= 4π2

∫

RnF (w0(η)⊗ η) dη

(2.15)≥ 0.

By (2.14), this finishes the proof.

Note that the function det : R2×2 → R is quadratic, quasiconvex, but not convex, sothe Lemma doesn’t extend to convexity of F . Recall that if f : Rm×n → R is quadratic,then the ELEs corresponding to I(u) =

∫Ωf(∇u(x)) dx are linear.

We now go back to studying the functional I. Combining the results of this Sectionand of Section 2.5, we have seen how quasiconvexity of the “A-”component of the densityfunction f , F : Rm×n → R, F (A) = f(x, u,A), is necessary for the existence of stronglocal minimizers, and how rank-1-convexity of F is necessary for the existence of weaklocal minimizers. Now we look at the situation when I or f are convex. “As usual”, wesay that

I : M 3 u 7→ I(u) ∈ R ∪ ∞is convex, if for all u, v ∈M , for all θ ∈ [0, 1],

I(θu+ (1− θ)v) ≤ θI(u) + (1− θ)I(v).

If we can put < instead of ≤ for u 6= v, then I is strictly convex.

Theorem 26. (Convexity of I)

1. If I is convex, then every critical point of I is a global minimizer.

2. If I is strictly convex, then there is only one minimizer and it is global.

3. If for all x ∈ Ω, x′ ∈ ∂Ω, f(x, ·, ·) : Rm × Rm×n → R and g(x′, ·) : Rm → R areconvex, then I is convex.

Proof. The proof is straightforward:


1. If I is convex, then for all θ ∈ [0, 1] and u, v ∈M ,

θ(I(v)− I(u)) ≥ −I(u) + I(u+ θ(v − u)).

For θ > 0, we get

I(v) ≥ I(u) +1

θ[I(u+ θ(v − u))− I(u)]

θ%0→ I(u) +DI(u)[v − u]

u crit. pt.= I(u).

2. Since all weak local minimizers are critical points, by 1., they are all global mini-mizers. If there are two distinct global minimizers, u, u, then for all θ ∈ (0, 1), bystrict convexity,

I(θu+ (1− θ)u) < θI(u) + (1− θ)I(u) = minv∈M

I(v),

a contradiction.

3. This follows directly by applying the definition(s).

The following figure wraps up some of the results of this Chapter:

sufficient

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u strong loc. min.<latexit sha1_base64="QPBGTM4xv2vWyanN8wIuuj7HESo=">AAACBHicbVA9SwNBEN3zM8avU8s0i0GwkONOA1oGbCwjmA9IjrC32SRL9naP3TkxHCls/Cs2ForY+iPs/Ddukis08cHA470ZZuZFieAGfP/bWVldW9/YLGwVt3d29/bdg8OGUammrE6VULoVEcMEl6wOHARrJZqROBKsGY2up37znmnDlbyDccLCmAwk73NKwEpdt5TiDrAHyLABreQAC0W9zlnMpTfpumXf82fAyyTISRnlqHXdr05P0TRmEqggxrQDP4EwIxo4FWxS7KSGJYSOyIC1LZUkZibMZk9M8IlVerivtC0JeKb+nshIbMw4jmxnTGBoFr2p+J/XTqF/FWZcJikwSeeL+qnAoPA0EdzjmlEQY0sI1dzeiumQaELB5la0IQSLLy+TxrkXXHjBbaVcreRxFFAJHaNTFKBLVEU3qIbqiKJH9Ixe0Zvz5Lw4787HvHXFyWeO0B84nz/+Q5ef</latexit>

u weak loc. min.<latexit sha1_base64="HSCTZ8mW5nScZdzprgfaPIBwVUQ=">AAACAnicbVDLSgNBEJyNrxhfUU/iZTAIHmTZ1YAeA148RjAPSEKYnXSSIbOzy0yvGpbgxV/x4kERr36FN//GyeOgiQUNRVU33V1BLIVBz/t2MkvLK6tr2fXcxubW9k5+d69qokRzqPBIRroeMANSKKigQAn1WAMLAwm1YHA19mt3oI2I1C0OY2iFrKdEV3CGVmrnDxLaRHjAlN4DG1AZcbd5Ggrljtr5gud6E9BF4s9IgcxQbue/mp2IJyEo5JIZ0/C9GFsp0yi4hFGumRiIGR+wHjQsVSwE00onL4zosVU6tBtpWwrpRP09kbLQmGEY2M6QYd/Me2PxP6+RYPeylQoVJwiKTxd1E0kxouM8aEdo4CiHljCuhb2V8j7TjKNNLWdD8OdfXiTVM9c/d/2bYqFUnMWRJYfkiJwQn1yQErkmZVIhnDySZ/JK3pwn58V5dz6mrRlnNrNP/sD5/AEqOZaW</latexit>

u crit. point<latexit sha1_base64="eWIUzgyTzlpDl9NbPjbgAErmpjw=">AAAB/3icbVBNS8NAEN34WetXVPDiZbEIHiQkWtBjwYvHCvYD2lA22027dLMJuxOxxB78K148KOLVv+HNf+OmzUFbHww83pthZl6QCK7Bdb+tpeWV1bX10kZ5c2t7Z9fe22/qOFWUNWgsYtUOiGaCS9YADoK1E8VIFAjWCkbXud+6Z0rzWN7BOGF+RAaSh5wSMFLPPkxxF9gDZJgqDk73LIm5hEnPrriOOwVeJF5BKqhAvWd/dfsxTSMmgQqidcdzE/AzooBTwSblbqpZQuiIDFjHUEkipv1sev8Enxilj8NYmZKAp+rviYxEWo+jwHRGBIZ63svF/7xOCuGVn3GZpMAknS0KU4EhxnkYuM8VoyDGhpD8f04xHRJFKJjIyiYEb/7lRdI8d7wLx7utVmrVIo4SOkLH6BR56BLV0A2qowai6BE9o1f0Zj1ZL9a79TFrXbKKmQP0B9bnD7YQldw=</latexit>

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Thm 14.2

<latexit sha1_base64="oFIh9bD+nzE70UjUbi6RSg/xE7M=">AAAB+nicbVBNS8NAEN34WetXqkcvi0XwICGpBT0WvHis0C9oQ9lsN+3S3STsTtQS+1O8eFDEq7/Em//GbZuDtj4YeLw3w8y8IBFcg+t+W2vrG5tb24Wd4u7e/sGhXTpq6ThVlDVpLGLVCYhmgkesCRwE6ySKERkI1g7GNzO/fc+U5nHUgEnCfEmGEQ85JWCkvl3qAXuErDGSvYsp9qpOpW+XXcedA68SLydllKPet796g5imkkVABdG667kJ+BlRwKlg02Iv1SwhdEyGrGtoRCTTfjY/fYrPjDLAYaxMRYDn6u+JjEitJzIwnZLASC97M/E/r5tCeO1nPEpSYBFdLApTgSHGsxzwgCtGQUwMIVRxcyumI6IIBZNW0YTgLb+8SloVx7t0vLtquVbN4yigE3SKzpGHrlAN3aI6aiKKHtAzekVv1pP1Yr1bH4vWNSufOUZ/YH3+AMcEkvk=</latexit>

I convex<latexit sha1_base64="aHe5Pqqi7sO95g8N00jQteooOwk=">AAAB+XicbVDLSgNBEJyNrxhfqx69DAbBU9jVgB4DXvQWwTwgCWF20kmGzM4uM70hYcmfePGgiFf/xJt/4yTZgyYWNBRV3XR3BbEUBj3v28ltbG5t7+R3C3v7B4dH7vFJ3USJ5lDjkYx0M2AGpFBQQ4ESmrEGFgYSGsHobu43xqCNiNQTTmPohGygRF9whlbquu4DbSNMMKU8UmOYzLpu0St5C9B14mekSDJUu+5XuxfxJASFXDJjWr4XYydlGgWXMCu0EwMx4yM2gJalioVgOuni8hm9sEqP9iNtSyFdqL8nUhYaMw0D2xkyHJpVby7+57US7N92UqHiBEHx5aJ+IilGdB4D7QkNHOXUEsa1sLdSPmSacbRhFWwI/urL66R+VfKvS/5juVgpZ3HkyRk5J5fEJzekQu5JldQIJ2PyTF7Jm5M6L86787FszTnZzCn5A+fzB2L4k3U=</latexit>

Thm 28.1<latexit sha1_base64="I0/HiQFgcdm3XG9jMuhRgLtQQGQ=">AAAB+nicbVDLSsNAFJ3UV62vVJduBovgQkJSC3ZZcOOyQl/QhDKZTtqhkwczN2qJ/RQ3LhRx65e482+ctllo64ELh3Pu5d57/ERwBbb9bRQ2Nre2d4q7pb39g8Mjs3zcUXEqKWvTWMSy5xPFBI9YGzgI1kskI6EvWNef3Mz97j2TisdRC6YJ80IyinjAKQEtDcyyC+wRstY4dC9nuFq3nIFZsS17AbxOnJxUUI7mwPxyhzFNQxYBFUSpvmMn4GVEAqeCzUpuqlhC6ISMWF/TiIRMedni9Bk+18oQB7HUFQFeqL8nMhIqNQ193RkSGKtVby7+5/VTCOpexqMkBRbR5aIgFRhiPM8BD7lkFMRUE0Il17diOiaSUNBplXQIzurL66RTtZwry7mrVRq1PI4iOkVn6AI56Bo10C1qojai6AE9o1f0ZjwZL8a78bFsLRj5zAn6A+PzB80fkv0=</latexit>

Thm 14.1a)<latexit sha1_base64="4vnrmXOLdW+4Qji925tJamGUj8U=">AAACBXicbVBNS8NAEN3Ur1q/oh71sFgEBQmJFvRY8OKxQr+gCWWz3bZLd5OwOxFL6MWLf8WLB0W8+h+8+W/ctD1o9cHA470ZZuaFieAaXPfLKiwtr6yuFddLG5tb2zv27l5Tx6mirEFjEat2SDQTPGIN4CBYO1GMyFCwVji6zv3WHVOax1EdxgkLJBlEvM8pASN17UMf2D1k9aH0zybYqzieLwkMlczI6aRrl13HnQL/Jd6clNEcta796fdimkoWARVE647nJhBkRAGngk1KfqpZQuiIDFjH0IhIpoNs+sUEHxulh/uxMhUBnqo/JzIitR7L0HTmJ+pFLxf/8zop9K+CjEdJCiyis0X9VGCIcR4J7nHFKIixIYQqbm7FdEgUoWCCK5kQvMWX/5LmueNdON5tpVytzOMoogN0hE6Qhy5RFd2gGmogih7QE3pBr9aj9Wy9We+z1oI1n9lHv2B9fANf55fN</latexit>

Thm 14.1b)<latexit sha1_base64="vhXoB85IzVQmwN80FK65LfLzdO0=">AAACBXicbVBNS8NAEN3Ur1q/oh71sFgEBQmJFvRY8OKxQr+gCWWz3bZLd5OwOxFL6MWLf8WLB0W8+h+8+W/ctD1o9cHA470ZZuaFieAaXPfLKiwtr6yuFddLG5tb2zv27l5Tx6mirEFjEat2SDQTPGIN4CBYO1GMyFCwVji6zv3WHVOax1EdxgkLJBlEvM8pASN17UMf2D1k9aH0zybYqzieLwkMlczC00nXLruOOwX+S7w5KaM5al370+/FNJUsAiqI1h3PTSDIiAJOBZuU/FSzhNARGbCOoRGRTAfZ9IsJPjZKD/djZSoCPFV/TmREaj2WoenMT9SLXi7+53VS6F8FGY+SFFhEZ4v6qcAQ4zwS3OOKURBjQwhV3NyK6ZAoQsEEVzIheIsv/yXNc8e7cLzbSrlamcdRRAfoCJ0gD12iKrpBNdRAFD2gJ/SCXq1H69l6s95nrQVrPrOPfsH6+AZhbZfO</latexit>

Thm 15<latexit sha1_base64="eIlTLmZQTu/mQ/BUzI6LChNZMsY=">AAAB+HicbVBNS8NAEN3Ur1o/GvXoZbEIHqQkWtFjwYvHCv2CJpTNdtMu3U3C7kSsob/EiwdFvPpTvPlv3LY5aOuDgcd7M8zMCxLBNTjOt1VYW9/Y3Cpul3Z29/bL9sFhW8epoqxFYxGrbkA0EzxiLeAgWDdRjMhAsE4wvp35nQemNI+jJkwS5ksyjHjIKQEj9e2yB+wRsuZIeudT7F717YpTdebAq8TNSQXlaPTtL28Q01SyCKggWvdcJwE/Iwo4FWxa8lLNEkLHZMh6hkZEMu1n88On+NQoAxzGylQEeK7+nsiI1HoiA9MpCYz0sjcT//N6KYQ3fsajJAUW0cWiMBUYYjxLAQ+4YhTExBBCFTe3YjoiilAwWZVMCO7yy6ukfVF1L6vufa1Sr+VxFNExOkFnyEXXqI7uUAO1EEUpekav6M16sl6sd+tj0Vqw8pkj9AfW5w/h7ZKG</latexit>

Thm 20<latexit sha1_base64="F9tcjCG4VuaAh/UsuVKW6Tu3los=">AAAB+HicbVBNS8NAEN3Ur1o/GvXoZbEIHqQktaDHghePFfoFTSib7aZdupuE3YlYQ3+JFw+KePWnePPfuG1z0NYHA4/3ZpiZFySCa3Ccb6uwsbm1vVPcLe3tHxyW7aPjjo5TRVmbxiJWvYBoJnjE2sBBsF6iGJGBYN1gcjv3uw9MaR5HLZgmzJdkFPGQUwJGGthlD9gjZK2x9C5nuOYM7IpTdRbA68TNSQXlaA7sL28Y01SyCKggWvddJwE/Iwo4FWxW8lLNEkInZMT6hkZEMu1ni8Nn+NwoQxzGylQEeKH+nsiI1HoqA9MpCYz1qjcX//P6KYQ3fsajJAUW0eWiMBUYYjxPAQ+5YhTE1BBCFTe3YjomilAwWZVMCO7qy+ukU6u6V1X3vl5p1PM4iugUnaEL5KJr1EB3qInaiKIUPaNX9GY9WS/Wu/WxbC1Y+cwJ+gPr8wfb3pKC</latexit>

Thm 23.3<latexit sha1_base64="rHsPI4cllZRko9V37PbYWMo2HMk=">AAAB+nicbVDLSsNAFJ3UV62vVJduBovgQkLSFnRZcOOyQl/QhDKZTtqhkwczN2qJ/RQ3LhRx65e482+ctllo64ELh3Pu5d57/ERwBbb9bRQ2Nre2d4q7pb39g8Mjs3zcUXEqKWvTWMSy5xPFBI9YGzgI1kskI6EvWNef3Mz97j2TisdRC6YJ80IyinjAKQEtDcyyC+wRstY4dC9nuFqzagOzYlv2AnidODmpoBzNgfnlDmOahiwCKohSfcdOwMuIBE4Fm5XcVLGE0AkZsb6mEQmZ8rLF6TN8rpUhDmKpKwK8UH9PZCRUahr6ujMkMFar3lz8z+unEFx7GY+SFFhEl4uCVGCI8TwHPOSSURBTTQiVXN+K6ZhIQkGnVdIhOKsvr5NO1XJqlnNXrzTqeRxFdIrO0AVy0BVqoFvURG1E0QN6Rq/ozXgyXox342PZWjDymRP0B8bnD8iJkvo=</latexit>

D2I(u)<latexit sha1_base64="+9MoCnoVtgGG93iDkZZwYMFy/n8=">AAAB7nicbVBNSwMxEJ2tX7V+VT16CRahXspuLeixoAe9VbAf0K4lm2bb0Gw2JFmhLP0RXjwo4tXf481/Y9ruQVsfDDzem2FmXiA508Z1v53c2vrG5lZ+u7Czu7d/UDw8auk4UYQ2Scxj1QmwppwJ2jTMcNqRiuIo4LQdjK9nfvuJKs1i8WAmkvoRHgoWMoKNldo3j9W7cnLeL5bcijsHWiVeRkqQodEvfvUGMUkiKgzhWOuu50rjp1gZRjidFnqJphKTMR7SrqUCR1T76fzcKTqzygCFsbIlDJqrvydSHGk9iQLbGWEz0sveTPzP6yYmvPJTJmRiqCCLRWHCkYnR7Hc0YIoSwyeWYKKYvRWREVaYGJtQwYbgLb+8SlrVindR8e5rpXotiyMPJ3AKZfDgEupwCw1oAoExPMMrvDnSeXHenY9Fa87JZo7hD5zPH+ykjpg=</latexit>

pos. def.<latexit sha1_base64="25Udbew9IPw/7QaPYyWYMu/wSeY=">AAAB+nicbVBNS8NAEN34WetXqkcvwSJ4kJBoQY8FLx4r2A9oQ9lsJu3SzQe7E7XE/hQvHhTx6i/x5r9x2+agrQ8GHu/NMDPPTwVX6Djfxsrq2vrGZmmrvL2zu7dvVg5aKskkgyZLRCI7PlUgeAxN5Cigk0qgkS+g7Y+up377HqTiSXyH4xS8iA5iHnJGUUt9s9JDeMQ8TZTdOwsgtCd9s+rYzgzWMnELUiUFGn3zqxckLIsgRiaoUl3XSdHLqUTOBEzKvUxBStmIDqCraUwjUF4+O31inWglsMJE6orRmqm/J3IaKTWOfN0ZURyqRW8q/ud1MwyvvJzHaYYQs/miMBMWJtY0ByvgEhiKsSaUSa5vtdiQSspQp1XWIbiLLy+T1rntXtjuba1arxVxlMgROSanxCWXpE5uSIM0CSMP5Jm8kjfjyXgx3o2PeeuKUcwckj8wPn8AAr6TyA==</latexit>




(8x) f(x, u(x), ·) : Rmn ! R<latexit sha1_base64="aNYyN4BZY/z7ezFWBSevKF7JhDs=">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</latexit>

quasiconvex<latexit sha1_base64="43Y7sYkK4066ErSMEwYYRXNbwIE=">AAAB+3icbVBNT8JAEN3iF+JXxaOXRmLiibRKokcSLx4xETCBhmyXKWzYbuvulEAa/ooXDxrj1T/izX/jAj0o+JJJXt6bycy8IBFco+t+W4WNza3tneJuaW//4PDIPi63dJwqBk0Wi1g9BlSD4BKayFHAY6KARoGAdjC6nfvtMSjNY/mA0wT8iA4kDzmjaKSeXe4iTDB7SqnmLJZjmMx6dsWtugs468TLSYXkaPTsr24/ZmkEEpmgWnc8N0E/owo5EzArdVMNCWUjOoCOoZJGoP1scfvMOTdK3wljZUqis1B/T2Q00noaBaYzojjUq95c/M/rpBje+BmXSYog2XJRmAoHY2cehNPnChiKqSGUKW5uddiQKsrQxFUyIXirL6+T1mXVu6p697VKvZbHUSSn5IxcEI9ckzq5Iw3SJIxMyDN5JW/WzHqx3q2PZWvBymdOyB9Ynz8xp5Uj</latexit>

D2I(u)[v, v] 0<latexit sha1_base64="tTqeEBCLu0X+XNqB1q1w9GSfSrs=">AAAB/HicbVDLSgNBEOz1GeNrNUcvg0GIIGE3BvQY0IPeIpgHJGuYncwmQ2YfzswGliX+ihcPinj1Q7z5N06SPWhiQUNR1U13lxtxJpVlfRsrq2vrG5u5rfz2zu7evnlw2JRhLAhtkJCHou1iSTkLaEMxxWk7EhT7Lqctd3Q19VtjKiQLg3uVRNTx8SBgHiNYaalnFq4fKrel+LQzPhs73QF9RBbqmUWrbM2AlomdkSJkqPfMr24/JLFPA0U4lrJjW5FyUiwUI5xO8t1Y0giTER7QjqYB9ql00tnxE3SilT7yQqErUGim/p5IsS9l4ru608dqKBe9qfif14mVd+mkLIhiRQMyX+TFHKkQTZNAfSYoUTzRBBPB9K2IDLHAROm88joEe/HlZdKslO3zsn1XLdaqWRw5OIJjKIENF1CDG6hDAwgk8Ayv8GY8GS/Gu/Exb10xspkC/IHx+QMkdpMa</latexit>

f<latexit sha1_base64="cA4z5t5YXPjXNuy9lAFYBHjhTGc=">AAAB6HicbVBNS8NAEJ3Ur1q/qh69LBbBU0m0oMeCF48t2A9oQ9lsJ+3azSbsboQS+gu8eFDEqz/Jm//GbZuDtj4YeLw3w8y8IBFcG9f9dgobm1vbO8Xd0t7+weFR+fikreNUMWyxWMSqG1CNgktsGW4EdhOFNAoEdoLJ3dzvPKHSPJYPZpqgH9GR5CFn1FipGQ7KFbfqLkDWiZeTCuRoDMpf/WHM0gilYYJq3fPcxPgZVYYzgbNSP9WYUDahI+xZKmmE2s8Wh87IhVWGJIyVLWnIQv09kdFI62kU2M6ImrFe9ebif14vNeGtn3GZpAYlWy4KU0FMTOZfkyFXyIyYWkKZ4vZWwsZUUWZsNiUbgrf68jppX1W966rXrFXqtTyOIpzBOVyCBzdQh3toQAsYIDzDK7w5j86L8+58LFsLTj5zCn/gfP4Ax7eM3w==</latexit>

satisfies (LH) in every<latexit sha1_base64="EjEYzY543SfrR7X+qlDviMlAV/M=">AAACCHicbVA9T8MwFHT4LOUrwMiARYVUliqBSjBWYunAABKFSm1VOe4LWDhOZL8goqgjC3+FhQGEWPkJbPwb3JIBWk6ydL57J/tdkEhh0PO+nJnZufmFxdJSeXlldW3d3di8NHGqObR4LGPdDpgBKRS0UKCEdqKBRYGEq+D2ZORf3YE2IlYXmCXQi9i1EqHgDK3Ud3e6CPeYG3s1oQBDq6fNfSoUBZvK6LDvVryaNwadJn5BKqTAWd/97A5inkagkEtmTMf3EuzlTKPgEoblbmogYfyWXUPHUsUiML18vMiQ7lllQMNY26OQjtXfiZxFxmRRYCcjhjdm0huJ/3mdFMPjXi5UkiIo/vNQmEqKMR21QgdCA0eZWcK4FvavlN8wzTja7sq2BH9y5WlyeVDzD2v+eb3SqBd1lMg22SVV4pMj0iBNckZahJMH8kReyKvz6Dw7b877z+iMU2S2yB84H9/OJJkh</latexit>

(x, u(x),ru(x))<latexit sha1_base64="VuecWsF6XBWiCJlEV5xci1PhFgM=">AAAB/XicbZDLSgMxFIbP1Futt3rZuQkWoYVSZrSgy4IblxXsBdqhZNJMG5rJDElGWofiq7hxoYhb38Odb2Om7UJbfwh8/OcczsnvRZwpbdvfVmZtfWNzK7ud29nd2z/IHx41VRhLQhsk5KFse1hRzgRtaKY5bUeS4sDjtOWNbtJ664FKxUJxrycRdQM8EMxnBGtj9fInxXE5Lo5L5a7AHsco5VIvX7Ar9kxoFZwFFGChei//1e2HJA6o0IRjpTqOHWk3wVIzwuk0140VjTAZ4QHtGBQ4oMpNZtdP0blx+sgPpXlCo5n7eyLBgVKTwDOdAdZDtVxLzf9qnVj7127CRBRrKsh8kR9zpEOURoH6TFKi+cQAJpKZWxEZYomJNoHlTAjO8pdXoXlRcS4rzl21UKsu4sjCKZxBERy4ghrcQh0aQOARnuEV3qwn68V6tz7mrRlrMXMMf2R9/gDZ6pN/</latexit>


f = F 2 C2(Rmn, R)<latexit sha1_base64="6WdZf7sY4HNtK3chSuuzymFMfGo=">AAACF3icbVDLSgMxFM34rPVVdekmWIQKMszUgm6EQkFcVrEP6LQlk2ba0CQzJBmhDPMXbvwVNy4Ucas7/8b0AWrrgQuHc+7l3nv8iFGlHefLWlpeWV1bz2xkN7e2d3Zze/t1FcYSkxoOWSibPlKEUUFqmmpGmpEkiPuMNPxhZew37olUNBR3ehSRNkd9QQOKkTZSN2cHl/DKowJWOsWCx5Ee+H5ym3YS7mnKiYIiPf2RT7q5vGM7E8BF4s5IHsxQ7eY+vV6IY06Exgwp1XKdSLcTJDXFjKRZL1YkQniI+qRlqEBmZzuZ/JXCY6P0YBBKU0LDifp7IkFcqRH3Tef4RDXvjcX/vFasg4t2QkUUayLwdFEQM6hDOA4J9qgkWLORIQhLam6FeIAkwtpEmTUhuPMvL5J60XbPbPemlC+XZnFkwCE4AgXggnNQBtegCmoAgwfwBF7Aq/VoPVtv1vu0dcmazRyAP7A+vgHK0p8P</latexit>

rank-one convex<latexit sha1_base64="JSYNFIpYVTyiH3et71ZQy1ZfodQ=">AAAB/3icbVDLSgNBEJz1GeMrKnjxMhgEL4ZdFfQY8OIxgnlAsoTZSUeHzM4sM72SsObgr3jxoIhXf8Obf+Mk2YOvgoaiqpvuriiRwqLvf3pz8wuLS8uFleLq2vrGZmlru2F1ajjUuZbatCJmQQoFdRQooZUYYHEkoRkNLiZ+8w6MFVpd4yiBMGY3SvQFZ+ikbmm3gzDEzDA1ONIKKNfqDobjbqnsV/wp6F8S5KRMctS6pY9OT/M0BoVcMmvbgZ9gmDGDgksYFzuphYTxAbuBtqOKxWDDbHr/mB44pUf72rhSSKfq94mMxdaO4sh1xgxv7W9vIv7ntVPsn4eZUEmKoPhsUT+VFDWdhEF7wgBHOXKEcSPcrZTfMsM4usiKLoTg98t/SeO4EpxUgqvTcvU0j6NA9sg+OSQBOSNVcklqpE44uSeP5Jm8eA/ek/fqvc1a57x8Zof8gPf+BYqplmM=</latexit>

necessary

2.7 ExamplesHow do the abstract results of this chapter apply? Here, a brief discussion of the minimalsurface problem and a longer discussion of the Brachistochrone:


2.7.1 Minimal surface

As before, given a bounded domain Ω ⊂ R2 and u0 ∈ C0(Ω;R), set

M = u ∈ C1(Ω;R) : u|∂Ω = u0|∂Ω

and X0 = C10 (Ω;R), so n = 2, m = 1. Consider the functional

I : M → R, I(u) =

∫

Ω

√1 + |∇u(x)|2 dx.

In Section 2.3.2 we derived the ELEs for this problem – but what are the extremalproperties of its solutions? Are critical points minimizers? Here, Theorem 26.3 applies.Since g = 0, it suffices to show that f : R2 3 A → f(A) =

√1 + |A|2 is convex. This

follows if ∂2Af(A) is positive semi-definite. For i, j = 1, 2, we calculate

∂Af(A)i =1√

1 + |A|2Ai,

∂2Af(A)ij =

(1 + |A|2

)− 12 δij −

(1 + |A|2

)− 32 AjAi,

with the Kronecker symbol δij = 1 if i = j and δij = 0 otherwise. To see that the secondderivative is positive semi-definitie, for all η ∈ R2, just apply Cauchy-Schwartz to get

∂2Af(A)[η, η] =

(1 + |A|2

)− 12 |η|2 −

(1 + |A|2

)− 32 |A · η|2

≥(1 + |A|2

)− 32((

1 + |A|2)|η|2 − |A|2|η|2

)≥ 0.

It follows that f is convex, so all critical points of I are global minimizers.

2.7.2 Brachistochrone curve

Recall the functional

I(u) =

∫ l

0

√1 + u′(x)2

2g(h− u(x))dx

associated to the Brachistochrone problem for

u ∈M = w ∈ C1([0, l],R) : w(0) = h,w(l) = 0.

It turns out that the abstract results we obtained so far don’t apply well in this case.But it is still interesting to see how they fail and to do some explicit calculations thatquantify the fastest marble run:

1. What are the ELEs?

The function

f : R2 → R

(u, a) 7→√

1 + a2

2g(h− u)


is the energy density for the Brachistochrone problem. We have

∂uf(u, a) =1

2√

2g(1 + a2)1/2(h− u)−3/2,

∂af(u, a) =a√2g

(1 + a2)−1/2(h− u)−1/2.

After taking out the factor 1√2g

and omitting (x) in u(x), u′(x), the ELEs are:

− d

dx

(u′(1 + u′2))−1/2(h− u)−1/2

)+

1

2(1 + u′2)1/2(h− u)−3/2 = 0,

u(0) = h,

u(l) = 0,

where the first equality should hold for all x ∈ (0, l).

2. Noether’s Theorem and a simplification of the ELEs:

The ELEs look ugly and it is in fact not straightforward to solve them. The followingresult is a special case of Noether’s Theorem that is very helpful in this and othercases.

Proposition 27. If n = 1 and f does not depend on x, ∂xf = 0, then the function

E(u, u′) = u′ · ∂Af(u, u′)− f(u, u′)

is constant along all solutions u of the ELEs.

Proof. This follows by a direct calculation. In Exercise 4 on “Eshelby’s tensor”,a higher-dimensional result is shown, which includes Proposition 27 as a specialcase.

We apply this result:

E(u, u′) =1√2gu′2(1 + u′2)−1/2(h− u)−1/2 − 1√

2g(1 + u′2)1/2(h− u)−1/2

= − 1√2g

(1 + u′2)−1/2(h− u)−1/2,

so to find a solution of the ELEs, we can look for solutions of the simpler equation

− ∂

∂x

((1 + u′2)−1/2(h− u)−1/2

)= 0. (2.16)

3. Verification of the parameterized cycloid solution:

There are several strategies for solving the ELEs or (2.16). All of them involveseveral tricks and they get quite messy1. The solution is a cycloid, usually given in

1For a reference, see the manuscript “Variationsrechnung und Sobolevräume” by Professor Hans-DieterAlber, p. 49,https://www.mathematik.tu-darmstadt.de/media/analysis/lehrmaterial_anapde/alber/vari.pdf.


parameterized form through the angle ϕ ∈ (ϕ0, ϕl):

u(x(ϕ)) = y(ϕ) = h− 1

2k′(cos(2ϕ) + 1),

x(ϕ) =1

2k′ sin(2ϕ) + k′ϕ+ k,

where k, k′, ϕ0, ϕl are constants determined by the boundary conditions

x(ϕ0) = 0, (2.17)x(ϕl) = l,

y(ϕ0) = h,

y(ϕl) = 0.

We verify that this gives a solution of the ELEs (this is messy already....): In orderto calculate u′, we use that by the chain rule,

u′(x) = ϕ′(x)y′(ϕ(x)),

where ϕ = x−1 is the inverse function of x(ϕ) with

ϕ′(x) =1

x′(ϕ(x))=

1

k′(cos(2ϕ) + 1)=

1

2(h− u(x)).

Since y′(ϕ) = k′ sin(2ϕ), we get

u′(x) =k′ sin(2ϕ(x))

2(h− u(x)).

Next, calculate and simplify

1 + u′2 =4(h− u)2 + k′2 sin2(2ϕ)

4(h− u)2

=k′2(cos2(2ϕ) + 2 cos(2ϕ) + 1) + k′2 sin2(2ϕ)

4(h− u)2

=2k′2(cos2(2ϕ) + 1)

4(h− u)2=

k′

h− u.

It follows that

u′(1 + u′2))−1/2(h− u)−1/2 =k′ sin(2ϕ)

2(h− u)

(h− u)1/2

√k′

1

(h− u)1/2.

=

√k′ sin(2ϕ)

2(h− u)=

u′√k′,

hence, the first term in the ELE is

− 1√k′

d

dxu′(x) = −ϕ

′(x)√k′

d

dϕ

k′ sin(2ϕ)

2(h− y(ϕ))

= −√k′

2

k′(cos(2ϕ)(cos(2ϕ) + 1) + sin2(2ϕ)

)

2(h− u(x))3

= −√k′

2

1

(h− u(x))2.


On the other hand, the second term in the ELE also simplifies to

1

2(1 + u′2)1/2(h− u)−3/2 =

√k′

2(h− u)−2.

The boundary conditions are verified in the next paragraph.Just for fun, we show that it is much simpler to see that u solves (2.16): In fact, bythe above calculations,

E(u, u′) = − 1√2g

(1 + u′2)−1/2(h− u)−1/2 =1√2gk′

= const.

4. Determine parameters from boundary conditions:We use (2.17) to determine k, k′, ϕ0, ϕl. We get

I) k′

2 sin(2ϕ0) + k′ϕ0 + k = 0,

II) k′

2 sin(2ϕl) + k′ϕl + k = l,

III) k′

2 (cos(2ϕ0) + 1) = 0,

IV) k′

2 (cos(2ϕl) + 1) = h.

From III), we get ϕ0 = π2 . Plugging this into I), we have k = −π2 k′. With this, II)

is equivalent to

sin(2ϕl) + 2ϕl − π =2l

k′,

and from IV), using cos(2ϕl) + 1 = 2 cos2(ϕl), we get

k′ =2h

cos(2ϕl) + 1=

h

cos2(ϕl). (2.18)

Combining these two, ϕl satisfies

g(ϕl) :=sin(2ϕl) + 2ϕl − π

cos(2ϕl) + 1=

cos(ϕl) sin(ϕl) + ϕl − π/2cos2(ϕl)

=l

h. (2.19)

In the range π2 < ϕl <

3π2 , g is bijective. To see this, we show that g is monotone

which follows from

g′(ϕl) =2 cos3(ϕl) + 2 sin(ϕl)(cos(ϕl) sin(ϕl) + ϕl − π/2)

cos3(ϕl)≥ 0.

For ϕl ∈ (π/2, 3π/2), this is equivalent to the numerator being non-positive,

n(ϕl) = 2 cos3(ϕl) + 2 sin(ϕl)(cos(ϕl) sin(ϕl) + ϕl − π/2)

= 2 cos(ϕl) + 2 sin(ϕl)(ϕl − π/2) ≤ 0.

To see this, note that n(π/2) = 0 and

n′(ϕl) = 2 cos(ϕl)(ϕl − π/2) ≤ 0

in this range. Thus,

(π

2,

3π

2) 3 ϕl = g−1(

l

h).

In particular, g(π2 ) = 0 and limϕl→ 3π2g(ϕl) = +∞, so any positive ratio l

h is covered.From there, we can also solve for k′ and k.


5. Before we proceed, we should relate the solution x, y of the ELEs to the standardcycloid parameterization

x(t) = r(t− sin t),

y(t) = r(1− cos t).

The curve (x(t), y(t)) is traced out by a fixed point on a turning wheel’s rim. Thewheel has radius r and t = 0 with x(0) = 0 = y(0) is where the wheel starts turning.At t = 2π, the wheel has turned exactly once. The easiest way to see the cycloid isto write (

xy

)(t) = r

(t1

)− r

(sin tcos t

),

where m(t) = r

(t1

)is the position of the wheel’s axis.

Now from solving the ELEs we have

x(ϕ) =1

2k′(sin(2ϕ) + 2ϕ− π),

y(ϕ) = h− 1

2k′(cos(2ϕ) + 1).

This is a cycloid upside down, translated on the y-axis by h. So we set

x(t) = x(ϕ) =1

2k′(sin(2ϕ) + 2ϕ− π) =

k′

2(t− sin t),

y(t) = −(y(ϕ)− h) =1

2k′(cos(2ϕ) + 1) =

k′

2(1− cos t),

for t = 2ϕ− π. We obtain 0 ≤ t ≤ tl = 2ϕl − π and r = k′

2 .The calculations in 4. can thus be rephrased in the following way: Given h and l,find the radius r of the wheel and the time tl such that

h = y(tl) = r(1− cos tl),

l = x(tl) = r(tl − sin tl).

(the calculations do not appear to become simpler though).

6. Examples:

(a) Simple explicit values for ϕl, k′, k are ϕl = 54π with l

h = 1 + 1.5π ' 5.71,k′ = 2h, k = −πh, so h = r. Then

u(x(ϕ)) = −h cos(2ϕ).

(b) Another example is the case h = 0: then with ϕl = 32π, k

′ = lπ (from II)) and

k = − l2 , we get a solution

u(x(ϕ)) = − l

2π(cos(2ϕ) + 1),

a symmetric curve with minimum y(π) = − lπ (see 7.).


7. When does u have a negative minimum?

It is the first striking feature of the Brachistochrone curve that its minimum maybe below the target height u(l) = 0. When does this happen (we have some ideasabout this as we know what the cycloid looks like)?

To get local extrema of u, we check that u′(x) = 0 if

y′(ϕ) = k′ sin(2ϕ) = 0.

For π/2 < ϕ < 3π/2, this happens at ϕ = π. In fact,

y′′(π) = 2k′ cos(2π) > 0,

so π gives a local minimum. The value is

y(π) = h− k′,

as k′ = 2r = umax − umin. By (2.18), y(π) < 0. In Example (a), −h = −r is theminimum! The minimum is attained at x(π) = π

2 k′ = πr, after a half turn of the

wheel. However, the negative local minimum is never reached if ϕl ≤ π. Since g ismonotone, this happens if g(π) = π

2 ≥ lh , or, more intuitively, if the average slope

hl ≥ π

2 is sufficiently large.

8. What is the marble’s speed?

Using the relations we determined when we modelled I, the marble’s speed is

|v(x)| = |x|(1 + u′(x)2) =√

2g(h− u(x)).

In particular, by assumption, |v(0)| = 0, and we see that |v| is maximal where uis minimal (cf. 7.), and that |v(l)| =

√2gh. This shows that in Example (b), as

expected, the end point (l, 0) is reached with zero velocity.Regarding the slope of u, we note that

u′(0) = limϕ→π/2

tan(ϕ) = +∞,

supporting the intuition that the marble should gain speed as quickly as possible.

9. So, how long does it take?

With u the solution of the ELEs from above, and using the calculations from above,

I(u) =

∫ l

0

√1 + u′(x)2

2g(h− u(x))dx =

√k′

2g

∫ ϕl

π2

(h− u(x))−12(h− u(x)) dϕ

=

√2k′

g(ϕl −

π

2).

10. The Brachistochrone is also the Tautochrone.

The Tautochrone curve describes the marble run (more commonly, it is associatedto a pendulum) with the property that no matter where the marble starts (with


zero initial velocity), the time it takes to get to the end is always the same. We showthat the cycloid Brachistochrone curve u is also the Tautochrone curve if the endpoint is the lowest point at (x(π), y(π) = uπ), (cf. 7): The time it takes a marblethat starts at any point (ϕ∗, u(x∗) = h∗ = y(ϕ∗)) and runs along u to get to (π, uπ)is, using the calculations from above,

I∗(u) =

∫ x(π)

l∗

√1 + u′(x)2

2g(h∗ − u(x))dx

=

√2k′

g

∫ π

ϕ∗

(h∗ − y(ϕ))−1/2(h− y(ϕ))1/2 dϕ

=

√2k′

g

∫ π

ϕ∗

√cos(2ϕ) + 1√

cos(2ϕ) + 1 + κdϕ,

with κ = 2h∗−hk′ . We use the substitution w(ϕ) =√

cos(2ϕ) + 1 + κ. Then√

cos(2ϕ) + 1 =√y2 − κ

and

w(ϕ∗) =

√2

k′√h∗ − y(ϕ∗) = 0,

w′(ϕ) = − sin(2ϕ)

w(ϕ), dϕ = − w

sin(2ϕ)dw,

2ϕ = arccos(w2 − κ− 1),

sin(2ϕ) = sin(arccos(w2 − κ− 1))

=√

1− (w2 − κ− 1)2 =√

2 + κ− w2√w2 − κ,

so

I∗(u) = −√

2k′

g

∫ w(π)

0

1√2 + κ− w2

dw

=

√2k′

g

[arctan

(w√

2 + κ− w2

w2 − κ− 2

)]w(π)

0

.

Since arctan(0) = 0, it remains to insert w(π). We get

w(π) =

√2

k′√h∗ − y(π)

7=

√2

k′√h∗ − h+ k′,

so w(π)2 − κ− 2 = 0, so

arctan

(w(π)

√2 + κ− w(π)2

w(π)2 − κ− 2

)= arctan(+∞) =

π

2

independently of (ϕ∗, h∗). This shows that the time I∗(u) =√

k′

2gπ it takes themarble to get “to the bottom” is the same, no matter where the marble starts onthe Brachistochrone curve u.


11. Are these cycloids minimizers?In the context of this course, the actual question we should ask is whether thecycloid solutions u of the ELEs are minimizers of I. We have seen two sufficientconditions for this:

• In Theorem 13: if D2I(u)[·, ·] is positive definite on H10 ((0, l);R), then u is a

weak local minimizer.• In Theorem 26: if I were convex, then u would be a global minimizer.

Unfortunately, neither of these criteria seem to apply:

• To check the first condition, we calculate

D2I(u)[v, v] =

∫ l

0

∂2af(u, u′)v′2 + 2∂u∂af(u, u′)vv′ + ∂2

uf(u, u′)v2 dx.

for v ∈ C10 ((0, l);R), where we have

∂a∂uf(u, a) =a

2√

2g(1 + a2)−1/2(h− u)−3/2,

∂2uf(u, a) =

3

4√

2g(1 + a2)1/2(h− u)−5/2,

∂2af(u, a) =

1√2g

((1 + a2)−1/2 − a2(1 + a2)−3/2

)(h− u)−1/2.

To have a chance of D2I(u)[v, v] ≥ γ‖v‖H1 for some γ > 0, we essentially needto look at ∂2

af(u, u′). From the calculations in 2.,

∂2af(u(x), u′(x)) =

1√2g

(1 + u′2)−3/2(h− u)−1/2

=1√

2gk′3(h− u)

=1

2√

2gk′(cos(2ϕ(x)) + 1)

x→0→ 0.

This shows that the estimate∫ l

0∂2af(u, u′)v′2 dx ≥ γ′‖v′‖2L2 will fail as v′ con-

centrates near 0.• If we wanted to use Theorem 26 to show that I is convex, we would need to

show that f is convex on R2. By Theorem 22.1, it would be necessary thatf is rank-1-convex at every point, meaning that, also by Theorem 22.3, theHessian

Hf =

(∂2uf ∂a∂uf

∂a∂uf ∂2af

)

is everywhere positive semi-definite. Using the calculations from above,

Hf (u, a)

=1√2g

(34 (1 + a2)1/2(h− u)−5/2 a

2 (1 + a2)−1/2(h− u)−3/2

a2 (1 + a2)−1/2(h− u)−3/2 (1 + a2)−3/2(h− u)−1/2

).

= c(u, a)

(34 (1 + a2)2 a

2 (1 + a2)(h− u)a2 (1 + a2)(h− u) (h− u)2

),


with

c(u, a) =(1 + a2)−3/2(h− u)−5/2

√2g

≥ 0

for the most relevant u ≤ h. So for all w ∈ R2,

wT ·Hf (u, a)w

= c(u, a)

(9

16w2

1(1 + a2)2 + w1w2a(1 + a2)(h− u) + w22(h− u)2

)

= c(u, a)

([3

4w1(1 + a2) +

2

3w2a(h− u)

]2

+

[1− 4

9a2

]w2

2(h− u)2

).

Clearly, Hf (u, a) is not positive semi-definite if |a| > 32 (for a numerically

simple example, set a = 3, h − u = h/2, w2 = 1/h and w1 = −2/15 to get avalue < 0).

12. What about necessary conditions?It is straightforward to show that the LH-condition is satisfied at u: for all x ∈ (0, l),

∂2af(u(x), u′(x)) = (1 + u′(x)2)−3/2(h− u(x))−1/2 ≥ 0.

In fact, we have > 0, so the “negative” LH-condition for u is not satisfied. At least,we have thus shown that u is not a (weak local) maximizer. It seems considerablymore difficult to check that, in fact, D2I(u)[v, v] ≥ 0, which is also necessary for ubeing a minimizer. So, in this sense, the LH-condition is helpful here.

13. Conclusion:Many elementary, but not necessarily “straightforward” calculations reveal interest-ing facts about the Brachistochrone curve. Classical methods in the calculus ofvariations are about solving these types of problems. Even in this completely one-dimensional situation, they are hard to summarize in an abstract theory. We haveseen a glimpse of further things that can be done: for example, Noether’s Theoremand other ideas for gaining understanding of the ELEs.At the same time, we have seen the limitations of the abstract criteria we provedin this chapter. They will be more relevant, in a slightly different sense, in thefollowing chapter on the direct method, in particular, for multidimensional settings.

2.7.3 Linear elasticity

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Basic strategies, thoughts and notation on/for modelling elastic behaviour of an n-dimensional body:

• aim: model the deformation of an elastic body subject to external forces

• The configuration ϕ(Ω) of the body is specified by: the open set Ω ⊂ Rn as areference configuration and ϕ : Rn ⊂ Ω → Rn as its deformation. The functionF : Ω→ Rn×n, x 7→ ∇ϕ(x) is called the deformation gradient.

• Depending on its deformation, the elastic body stores elastic energy

I(ϕ) =

∫

Ω

W (x, ϕ(x),∇ϕ(x)) dx−∫

∂Ω\Γϕ(x) · g0(x) dx′ −

∫

Ω

ϕ(x) · h(x) dx,

whereW : Ω× Rn × Rn×n → R∞

is a suitable energy density, g0 is a boundary density for external loading, and hcan be added as a volume density for volume forcings. At Γ ⊂ ∂Ω, the body isfixed/clamped, so ϕ(x) = x for x ∈ Γ.

• Idea: the elastic body is deformed in such a way that it stores the least amount ofelastic energy → we need to model and then (locally) minimize I!

• in this Subsection, we focus on the modelling of W . This is different from what wedid before: in previous examples from geometry and classical mechanics, the modelwas given, the mathematics was to be done.

• we should think about: how is elastic behaviour of a body different from rigid,plastic, fluidic, gaseous, ... behaviour?

• we use continuum mechanics: there are no distinguishable particles – it must bereasonable to associate a new position ϕ(x) ∈ Rn to every point x ∈ Ω ⊂ Rn.

Typcial (reasonable?) properties of W :

1. translation invariance: W (x, ϕ, F ) = W (x, F ),i.e. F has all the relevant information on ϕ, translating the body doesn’t take orproduce elastic energy

2. rotation invariance: (∀F ) (∀Q ∈ SO(n)) W (x,QF ) = W (x, F ),i.e. I(ϕ) = I(Qϕ), I is independent of an orthogonal change of coordinates. This isalso called objectivity.

From these two assumptions, it follows that (see [Ciarlet]),

W (x, F ) = W (x,C), where C = FTF is the “right Cauchy-Green deformation tensor′′.

3. no self-interprenetation: detF ≤ 0 ⇒ W (x, F ) =∞.This is to avoid that the material folds in on itself. This is additionally supportedby the condition

detFn > 0 with detFnn→∞→ 0 ⇒ W (x, Fn)

n→∞→ ∞.


Note that this condition implies non-convexity of W in F : For example, if

F1 =

1 0 00 1 00 0 1

and F2 =

−1 0 00 −1 00 0 1

,

then detF1 = detF2 = 1, but det( 12F1 + 1

2F2) = 0, so W (x, 12F1 + 1

2F2) = ∞. Onthe other hand, FT1 F1 = FT2 F2 = Idn×n, so W (x, F1) = W (x, F2) = W (x, Idn×n)and we should have W (x, Idn×n) 6=∞.As a consequence, non-convex energies are needed to model “large” deformationsthat may get close to detF = 0. (What are typical situations/materials that al-low/need detF = 0?)

From here: we assume “small” deformation, and show how to linearize the model in thissetting. The corresponding energy and material behaviour is called Linear Elasticity:

• we replace the deformation ϕ by the displacement u : Ω→ Rn, given by

ϕ(x) = x+ u(x).

ThenC = ∇ϕT∇ϕ = Idn×n + (∇u+∇uT ) +∇uT∇u,

withe(u) :=

1

2(∇u+∇uT )

the linearized deformation tensor (with respect to the undeformed state ϕ0 = IdΩ.)

• A Taylor expansion of W in Idn×n gives

W (∇ϕT∇ϕ) = W (Idn×n) +DW (Idn×n)[2e(u) +∇uT∇u]

+1

2D2W (Idn×n)[2e(u) +∇uT∇u, 2e(u) +∇uT∇u]

+ terms of higher order.

• now assume that the undeformed state ϕ0 = IdΩ is a local minimizer of W and thus,in particular, a critical point of I. Then with ∇ϕ0 = Idn×n, we have DW (Idn×n) =0. In addition, assume that ∇uT∇u is sufficiently small to be ignored. Then wemay approximate W by the quadratic functional

fel(x,A) := 2D2W (x, Idn×n)[A,A],

defined on the symmetric tensors A ∈ Rn×nsym . Recall that the fact that fel isquadratic means that the ELEs are linear.

• we have modelled the Problem of Linear Elasticity : Minimize

I(u) =

∫

Ω

2D2W (Idn×n)e(u) : e(u) dx−∫

∂Ω\Γg · udx′ −

∫

Ω

h · udx

onM0 = u ∈ C1(Ω;Rn) : u|Γ = 0.


As an important example, we consider linear elasticity models for Isotropic material : inthis case, the material response to loading should not depend on the direction of theloading (see Picture):

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Then (see [Ciarlet]):

fel(x,A) =λ(x)

2|trA|2 +

µ(x)

4|A+AT |2

=λ(x)

2|tr e(u)|2 + µ(x)|e(u)|2,

where λ, µ ∈ R are called Lamé-constants. They depend on the elastic material properties– see the discussion below! As fel is quadratic in A, we have

∂Afel(A)[B] = λtrA trB +1

2µ(A+AT ) : (B +BT ),

∂2Afel(A)[B,B] = 2f(B).

The ELEs are called Lamé-Navier Equations:

div(σ(u)) + h = 0, in Ω,

u = 0, on Γ,

σν + g = 0, on ∂Ω \ Γ,

whereσ(u) = ∂Af(x,∇u) = λtr∇u Idn×n + 2µe(u)

is the corresponding stress tensor.Now: isfel convex in A?

Lemma 28. (Convexity for linear isotropic elasticity) We fix µ, λ ∈ R. Then

1. fel(A) = λ2 |trA|2 + µ

4 |A+AT |2 satisfies (LH) if and only if µ ≥ 0 and λ+ 2µ ≥ 0,and

2. fel is convex if and only if µ ≥ 0 and nλ+ 2µ ≥ 0.

Since fel is smooth and quadratic, the (LH)-condition is equivalent to rank-one con-vexity and rank-one convexity and quasiconvexity are the same (Lemma 25)! But clearly,also in this case, convexity is a stronger property than rank-one convexity (as long asn ≥ 2).


Proof. (compare to Exercise 5 on Assignment 4.)

1. For all B = ξ ⊗ η, ξ, η ∈ Rn, we have

|ξ ⊗ η + η ⊗ ξ|2 = 2|ξ|2|η|2 + 2|ξ · η|2,

hence

D2f(A)[B,B] = 2f(B) =λ

2(ξ · η)2 +

µ

4|ξ ⊗ η + η ⊗ ξ|2

=λ+ µ

2(ξ · η)2 +

µ

2|ξ|2|η|2.

Moreover,

∀ξ, η ∈ Rn 1

2

((λ+ µ)(ξ · η)2 + µ|ξ|2|η|2

)≥ 0

⇔ ∀ξ, η ∈ Rn \ 0 (λ+ µ)(ξ · η)2

|ξ|2|η|2 + µ ≥ 0

⇔ ∀α ∈ (0, 1] (λ+ µ)α+ µ ≥ 0

⇔ µ ≥ 0 and λ+ 2µ ≥ 0.

2. First note that convexity of fel is equivalent to fel(B) ≥ 0 for all B ∈ Rn×n. As atool in the proof, we decompose

Rn×n = RIdn×n + Rn×ndev ,

in the sense that

Rn×n 3 B =1

ntrB Idn×n + (B − trB

nIdn×n).

Here, the scalar expression trBn can be seen as a linearized measure of the change

of volume associated to B (see example below!), and the matrix

dev(B) = (B − trB

nIdn×n) ∈ Rn×ndev

is called the deviatoric part of B. Note that

Idn×n : dev(B) = Idn×n : B − trB = 0,

so RIdn×n⊥Rn×ndev . Hence

fel(B) =λ

2|trB|2 +

µ

4| 2n

trB Idn×n + (B +BT − 2

ntrB Idn×n)|2

=nλ+ 2µ

n|trB|2 +

µ

4|dev(B) + dev(B)T |2,

so fel(B) ≥ 0 for all B ∈ Rn×n holds if and only if µ ≥ 0 and nλ+ 2µ ≥ 0.


We close this discussion of linear elasticity with two prototypes of deformations, theirgeometric/physical interpretation in dimension 2, and their representation in fel.

Let n = 2 and u(x) =

(α β0 α

)x, so that the gradient has the decomposition

A =

(α β0 α

)= αId2×2 +

(0 β0 0

)= αId2×2 + dev(A).

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If α > 0, β = 0, this corresponds to a stretching of the body in every direction,ϕ(x) = x + αx, with a corresponding change in volume (Figure on the left). If α = 0,β > 0, this corresponds to shearing of the body in e1-direction, ϕ(x) = x + dev(A)x,without any change in volume (Figure on the right). We have

fel(A) = (λ+ µ)α2 +µ

2β2,

where the first part is the gain of energy due to volume change, and the second part isthe energy absorbed due to shear stress.

3 THE DIRECT METHOD IN THE CALCULUS OF VARIATIONS 54

3 The Direct Method in the Calculus of VariationsIn this part of the course, we look at

(DP) (Direct Problem in the calculus of variations):Let (X, ‖ · ‖) be a real Banach space and I : X → R∞ a functional.Can we find a global minimizer u ∈ X such that I(u) = infv∈X I(v)?

Here, R∞ = R∪+∞ indicates that I can also take on the value +∞. This can be usedto model constraints: if actually we are looking for minimizers u ∈M ⊂ X of a functionalI : X ⊃M → R, then we can define

I(v) =

I(v), v ∈M,

+∞, otherwise.

Clearly, the minimizers and infimum/minimum values of I and I in M coincide. If theset where I(v) 6= +∞ is nonempty, we say that I is proper. We always assume (tacitly)that I is proper.

3.1 Abstract existence theorems from functional analysisTo solve (DP), we get a first existence theorem for reflexive Banach spaces X that followsfrom the strategy outlined in Section 1.6. In the Theorem, the functional I is assumed tobe coercive and weakly sequentially lower semicontinuous. First, we define these notions.

Definition 29. A functional I : X → R∞ on a Banach space X is called coercive, if forall sequences (vn)n ⊂ X, if ‖vn‖ → +∞, then I(vn)→ +∞.

Definition 30. A functional I : X → R∞ on a Banach space X is called

1. sequentially weakly continuous, if for all weakly converging sequences (vn)n ⊂ X,vn v, we have

limn→∞

I(vn) = I(v),

2. sequentially weakly lower semicontinuous ((s)wlsc), if for all weakly converging se-quences (vn)n ⊂ X, vn v, we have

lim infn→∞

I(vn) ≥ I(v).

Remark 31. A short discussion and examples for Definition 30:

1. For brevity, we will omit “sequentially” in “sequentially weakly lower semicontinu-ous” and the first “s” in (swlsc). However, we always use this (sequential) definition.

2. Strongly converging sequences are also weakly converging, so clearly, the propertyof I being weakly continuous (or weakly lower semicontinuous) is stronger than theproperty of I being continuous (or lower semicontinuous).


3. Sequential weak continuity of a functional is a ‘rare’ property. We will come backto this later in the course.

4. The norm ‖ · ‖ : X → R is (wlsc).

Proof. For all x′ ∈ X ′, the space of bounded linear functionals onX, and convergentsequences, xn → x in X,

‖x′‖X′‖xn‖X ≥ x′(xn)n→∞→ x′(x),

so lim infn ‖x′‖X′‖xn‖X ≥ x′(x). Now choose (Hahn-Banach!) x′ such that ‖x′‖X′ =1 and x′(x) = ‖x‖X .

5. In many infinite-dimensional Banach spaces, the norm ‖ · ‖ : X → R is not sequen-tially weakly continuous, i.e. weak convergence does not imply norm convergence(Exercise: example + proof?)

Theorem 32. (Tonelli, “Adapted Weierstraß Principle”) Let (X, ‖ · ‖) be a reflexive realBanach space. Assume that the functional I : X → R∞ is bounded from below, coerciveand weakly lower semicontinuous. Then (DP) has at least one solution.

Proof. We follow the strategy from Section 1.6. Let infv∈X I(v) = I ∈ R (I is boundedfrom below). Then there is an infimizing sequence (vn)n ⊂ X for I, i.e.

limnI(vn) = inf

v∈XI(v) = I .

Since I is coercive, (vn)n is bounded. Since X is reflexive, (vn)n has a weakly convergingsubsequence vnk u ∈ X. Since I is weakly lower semicontinuous,

I = limkI(vnk) ≥ lim inf

kI(vnk) ≥ I(u) ≥ I,

so u is a global minimizer.

Now, there are two natural questions:

1. How do we see if I is coercive?

2. How do we characterize I that are wlsc?

We will find answers to the first question in the more specific setting

I(u) =

∫

Ω


∫

∂Ω\Γg(x, u(x)) dx′ (3.1)

that we already used in the previous chapter. This is the content of Section 3.3 (aftera brief reminder of Lebesgue and Sobolev spaces in Section 3.2). Regarding the secondquestion, there are a few things that we can say now, in a general/abstract setting, beforewe reconsider this problem for I of type (3.1) in Section 3.4.

Lemma 33. (Mazur) Let X be a real Banach space and let M ⊂ X be closed and convex.Then M is weakly sequentially closed.


Proof. This is a direct consequence of Separation Theorems/the Hahn-Banach Theoremfrom functional analysis. We recall one version that is suitable:

Theorem. (Separation Theorem) Let X be a real Banach space, let M ⊂ X be closedand convex and let x0 ∈ X \M . Then there exist x′ ∈ X ′ and α ∈ R such that for allx ∈M ,

x′(x) ≤ α and x′(x0) > α.

Proof. See, for example, [Alt, p. 223, Section 6.11].

To prove Mazur’s Lemma, let (un)n ⊂M, un u ∈ X be weakly convergent. Assumeu /∈M . ThenM is closed and convex and by the Separation Theorem, there exist x′ ∈ X ′and α ∈ R such that

x′(un) ≤ α and x′(u) > α.

Clearly, this is in contradiction to the weak convergence of un to u.

Mazur’s Lemma allows us to prove the following criterion for I being wlsc. Geomet-rically, we use that convexity of I implies convexity of its sublevels

Lα(I) = u ∈ X : I(u) ≤ α.

Theorem 34. Let X be a real Banach space and let I : X → R∞ be strongly lowersemicontinuous and convex. Then I is wlsc.

Proof. Let (un)n ⊂ X, un u, be any weakly convergent sequence. We need to showthat

I(u) ≤ lim infn

I(un) =: α.

We distinguish three cases.

1. If α = +∞, then, clearly, I(u) ≤ α.

2. If α ∈ R, choose a subsequence (unk)k such that limk I(unk) = α. For ε > 0,consider the sublevels

Lα+ε(I) = v ∈ X : I(v) ≤ α+ ε.

By convexity of I, the Lα+ε(I) are convex. By strong lower semicontinuity of I, theLα+ε(I) are strongly closed. Thus, by Mazur’s Lemma, the sublevels are weaklyclosed. Moreover, for all ε > 0, there is a kε such that unk′ ∈ Lα+ε(I) for all k′ ≥ kε.Thus, u ∈ Lα+ε(I) for all ε > 0 and hence I(u) ≤ α.

3. If α = −∞, consider L−R(I) for R → +∞. For all R > 0, we find u ∈ L−R as in2.. Hence I(u) = −∞, a contradiction.

Now we briefly look at a related but different existence theorem for (DP) with “moreconvexity but different continuity”. We start with a definition.


Definition 35. The functional I : X → R∞ is called uniformly convex, if there exists amonotone function g : (0,∞)→ (0,∞] such that for all x, y ∈ X, x 6= y, θ ∈ (0, 1),

I(θx+ (1− θ)y) ≤ θI(x) + (1− θ)I(y)− θ(1− θ)g(‖x− y‖).

Theorem 36. Let I : X → R∞ be uniformly convex and continuous (w.r.t. the norm inX). Then (DP) has a unique solution, i.e. a global minimizer.

Proof. Let (un)n be an infimizing sequence for I and I = infv∈X I(v) = limn I(un). Weshow that (un)n is a Cauchy sequence. We have

g(‖um − un‖) ≤ 2I(um) + 2I(un)− 4I(1

2um +

1

2un)

and I( 12um + 1

2un) ≥ I, so

lim supm,n

g(‖um − un‖) ≤ 4I − 4I = 0. (3.2)

Since g is positive and monotone, this implies that (un)n is a Cauchy sequence in X: If(un)n wasn’t Cauchy, there would be ε > 0 with

lim supm,n

‖um − un‖ ≥ ε.

By monotonicity and positivity of g, this would imply

lim supm,n

g(‖um − un‖) ≥ g(ε) > 0,

in contradiction to (3.2). We set u = limn un. Since I is continuous,

I = limnI(un) = I(u),

so u is a global minimizer. Since I is uniformly convex, it is also strictly convex. Therefore,u is unique.

To close this abstract Section, we summarize different notions of differentiability in(in)finite-dimensional Banach spaces.

Definition 37. (Gâteaux differential and Fréchet derivative) Let X,Y be Banach spacesand F : X → Y .

1. The Gâteaux differential of F at x0 ∈ X in the direction of v ∈ X is the limit

DF (x0)[v] = limt0

F (x0 + tv)− F (x0)

t∈ Y

if it exists.

2. If for x0 ∈ X, there exists A0 ∈ L(X,Y ) (the space of bounded linear operatorsfrom X to Y ) such that for all v ∈ X,

DF (x0)[v] = A0v,

then F is Gâteaux differentiable at x0 with Gâteaux derivative A0.


3. F is called Fréchet differentiable at x0 ∈ X if there exists A ∈ L(X,Y ) such thatfor all (vn)n ⊂ X with limn→∞ vn = 0,

limn→∞

‖F (x0 + vn)− F (x0)−Avn‖Y‖v‖X

= 0.

A is then called the Fréchet derivative of F at x0.

4. As usual, F is Gâteaux/Fréchet differentiable if it is Gâteaux/Fréchet differentiableat every x0 ∈ X.

Remark 38. 1. If F is Fréchet differentiable at x0, then it is Gâteaux differentiable atx0. The converse need not be true.

2. Even if all directional derivatives DF (x0)[v] exist, F need not be Gâteaux differen-tiable (counterexample for X = R2, Y = R).

3. If Y = R, then DF (x0)[v] is the first variation of F at x0 in the direction of v. TheGâteaux derivative would be DF (x0) ∈ L(X,R) = X ′.

4. Recall that if u0 ∈ X is a solution of (DP) and DI(u0)[v] exists for all v ∈ X, thenDI(u0)[v] = 0 for all v ∈ X, i.e. I is Gâteaux differentiable at u0 with derivative 0and u0 is a critical point.

5. Note that if I : X → R is convex and Gâteaux differentiable, then for all u, v ∈ X,

I(u+ v) ≥ I(u) +DI(u)[v].

Proof.

DI(u)[v] = limt0

I(u+ tv)− I(u)

t= limt0

I(t(u+ v) + (1− t)u)− I(u)

tI convex≤ lim

t0

tI(u+ v) + (1− t)I(u)− I(u)

t= I(u+ v)− I(u).

6. Note that even uniformly convex functionals need not be (Gâteaux) differentiable.

3.2 Reminder: Lebesgue and Sobolev spacesWe would like to apply the results of the last Section to functionals I : X → R∞ of type(3.1). Thus, we need that X is reflexive and contains functions that are differentiable (insome sense). Moreover, it may be useful if the norm of X also “quantifies” the derivatives.Sobolev spaces appear as a natural choice for X. In this Section, we recall properties ofLebesgue and Sobolev spaces that will be important for the remainder of the course thatdeals with the coercivity and weak lower semicontinuity of I. Compactness results willalso be useful.


3.2.1 Lebesgue spaces

Let Ω ⊂ Rn be a domain. Recall the Lebesgue spaces Lp(Ω) = Lp(Ω;R) of (equivalenceclasses of) functions f : Ω→ R for p ∈ [1,∞] with norms

‖f‖p =

(∫

Ω

|f(x)|p dx

)1/p

, p 6=∞,

‖f‖∞ = ess supx∈Ω|f(x)| = inf

r>0|x ∈ Ω : |f(x)| > r| = 0.

Some facts we will use throughout this Section and the remainder of the course, see, e.g.[Alt]:

• For all p ∈ [1,∞], Lp(Ω) are Banach spaces.

• For p ∈ (1,∞), Lp(Ω) is reflexive.

• For p = 2, L2(Ω) is a Hilbert space.

• For p ∈ [1,∞), 1 = 1p + 1

p′ , p′ =∞ if p = 1, the dual (Lp(Ω))

′ of Lp(Ω) is isomorphicto Lp

′(Ω), i.e. for all ξ ∈ (Lp(Ω))

′ there exists a unique vξ ∈ Lp′(Ω) such that for

all u ∈ Lp(Ω),

ξ(u) =

∫

Ω

vξ(x)u(x) dx,

and ‖ξ‖(Lp(Ω))′ = ‖vξ‖Lp′ (Ω). For this, recall Hölder’s inequality: for all f ∈ Lp(Ω),g ∈ Lp′(Ω), ∫

Ω

|f(x)g(x)|dx ≤ ‖f‖p‖g‖p′ .

• For p ∈ [1,∞), C∞c (Ω) is dense in Lp(Ω).

• For p ∈ [1,∞], Ω bounded, the span of step functions is dense in Lp(Ω).

3.2.2 Convergence Theorems

• Monotone convergence: Let u ∈ L1(Ω), (un)n ⊂ L1(Ω) be such that

0 ≤ u1(x) ≤ · · · ≤ un(x) ≤ · · · u(x) ∈ L1(Ω) a.e.,

thenun

L1

→ u and limn

∫

Ω

un(x) dx =

∫

Ω

u(x) dx.

(Sketch of proof:∫Ω un is monotone and bounded, so convergent. Hence for k ≥ l,∫

Ω|uk − ul| =

∫Ωuk −

∫Ωul → 0,

so (un)n is Cauchy sequence in L1(Ω). )


• Fatous’ Lemma: Let (un)n be a sequence of measurable functions with 0 ≤ un a.e.Then

lim infn→∞

∫

Ω

un(x) dx ≥∫

Ω

lim infn→∞

un(x) dx.

(Sketch of proof: Apply monotone convergence to fn = infk≥n uk lim infn un yielding

lim infn→∞

∫Ωun ≥ lim

n

∫Ω

infk≥n

uk =

∫Ω

lim infn

un.)

• Lebesgue dominated convergence: Let (un)n be a sequence of measurable functionswith |un| ≤ h ∈ L1(Ω) and pointwise convergence un(x)→ u(x) almost everywhere.

Then u ∈ L1(Ω) and unL1

→ u.(Sketch of proof: Apply Fatou’s Lemma to gn := h− 1

2|un − u| ≥ 0 to get

lim supn→∞

∫Ω|un − u| = 0.)

Lemma 39. (Weak convergence) Let p ∈ [1,∞), u ∈ Lp(Ω). Then the following areequivalent:

• un u in Lp(Ω), and,

• Both

1. ∃c > 0 ∀n ∈ N ‖un‖p < c, and,

2. for all sets S ⊂ (Lp(Ω))′ = Lp′(Ω) such that span(S) is dense in Lp

′(Ω),

∀v ∈ S,∫

Ω

vun →∫

Ω

vu.

In particular, we can choose S = C∞c (Ω) ⊂ Lp′(Ω) (asking p 6= 1), or span(S) the set of

step functions (asking p 6= 1 if Ω is unbounded).

Proof. “⇒”:

1. by Banach-Steinhaus

2. by definition of weak convergence.

“⇐”: For v ∈ Lp′(Ω) and ε > 0, choose vε ∈ span(S) such that ‖v − vε‖p′ < ε. Set

c0 := ‖u‖p. Then, for sufficiently large n,

|∫

Ω

unv −∫

Ω

uv| ≤ |∫

Ω

un(v − vε)|+ |∫

Ω

unvε −∫

Ω

uvε|+ |∫

Ω

u(vε − v)|Hölder,1.≤ c‖v − vε‖p′ + |

∫

Ω

(un − u)vε|+ c0‖v − vε‖p′2.≤ Cε.


Example 40. (“Concentrations”) Let Ω = (−1, 1), α ∈ R and

un(x) =

nα, x ∈ (0, 1

n ),

0, otherwise.

Then:

• for all x ∈ Ω, un(x)→ 0, and

• ‖un‖pp =∫ 1/n

0nαp dx = nαp−1 <∞ for p ∈ [1,∞)

Now distinguish four cases:

1. αp > 1, then ‖un‖p →∞, so we have neither strong nor weak (Lemma 39) conver-gence in Lp(Ω),

2. αp < 1, then ‖un‖p → 0, so we have norm convergence and strong convergence inLp(Ω),

3. α = 1p , then ‖un‖p = 1, so we have boundedness. For p ∈ (1,∞), use Lemma 39:

let ϕ ∈ C∞c (Ω), then

|∫ 1

−1

unϕ| = |∫ 1/n

0

n1/pϕ| = |n1/p

∫ 1/n

0

ϕ| ≤ n1/p−1‖ϕ‖∞ p>1→ 0,

so un 0.

4. α = p = 1, then ‖un‖1 = 1. Check what happens for ϕ ∈ C∞c (Ω):

∫ 1

−1

unϕ = n

∫ 1/n

0

ϕ→ ϕ(0),

(Lebesgue differentiation), but there is no u ∈ L1(Ω) such that ϕ(0) =∫ 1

−1uϕ for

all ϕ ∈ C∞c (Ω). We show that (un)n doesn’t converge: Let v(x) =∑∞k=1 χ[ 1

4k, 2

4k]

with χM the characteristic function for the set M . Then v ∈ L∞(Ω) = (L1(Ω))′

and∫ 1

−1

unvn= 4k

2=

∫ 2/4k

0

4k

2v(x) dx ≥

∫ 2/4k

1/4k

4k

2=

1

2,

∫ 1

−1

unvn=4k

=

∫ 1/4k

0

4kv(x) dx =

∫ 2/4k+1

0

4kv(x) dx ≤ 4k2

3

2

4k+1=

1

3,

so (un)n does not converge weakly.

Example 41. (Oscillations) Let Ω = (0, 1) and u ∈ Lp(Ω), p ∈ [1,∞). Extend uperiodically to u : R→ R,

u(x+ k) = u(x), x ∈ (0, 1), k ∈ Z.


Consider the sequence (un)n ⊂ Lp(Ω) with

un(x) = u(nx), x ∈ (0, 1), n ∈ N.

Then (un)n converges weakly to its average, unLp u : x 7→

∫ 1

0u(x) dx.

Proof. Note that the set of step functions on (0, 1) is given by

span(χ[0,α] : 0 < α < 1),

and it is dense in Lp′(0, 1) for p ∈ [1,∞). Thus, by Lemma 39, to show weak convergence,

it is sufficient to show:

1. the sequence (un)n is bounded, as by a direct calculation,

‖un‖pp =

∫ 1

0

|u(nx)|p dxy=nx

=1

n

∫ n

0

|u(y)|p dy = ‖u‖pp,

and,

2. for all α ∈ (0, 1), we have∫ 1

0

χ[0,α]un →∫ 1

0

χ[0,α]u = αu.

This follows from∫ 1

0

χ[0,α](x)un(x) dxy=nx

=1

n

∫ nα

0

u(y) dy

=1

n

∫ bnαc

0

u(y) dy +1

n

∫ nα

bnαcu(y) dy → αu,

where1

n

∫ bnαc

0

u(y) dy =bnαcn

u→ αu

as| bnαcn− α| = | bnαc − nα

n| ≤ 1

n→ 0,

and1

n

∫ nα

bnαcu(y) dy ≤ 1

n

∫ bnαc+1

bnαc|u(y)|dy =

1

n‖u‖1 → 0.

Corollary 42. (“weak continuity is too strong”) Let g : R→ R be measurable with |g(u)| ≤C(1 + |u|p) for some p ∈ [1,∞). Consider the functional

I : Lp(0, 1)→ R, I(u) =

∫ 1

0

g(u(x)) dx.

Then the following are equivalent:


1. I is sequentially weakly continuous on Lp(0, 1), and,

2. I is affine, and there exists b ∈ R such that I(u) = I(0) + bu.

Proof. 2.⇒ 1. follows directly from the definition of weak convergence.1.⇒ 2. in two steps:

Step 1: Consider arbitrary u ∈ Lp(0, 1) with periodic extension u : R → R andun(x) := u(nx) for x ∈ (0, 1). By Example 41, un

Lp u. Thus, by weak

continuity,

I(un) =

∫ 1

0

g(un(x)) dx→ I(u) = g(u).

On the other hand, I(un)y=nx

= 1n

∫ n0g(u(y)) dy = I(u), so for all u ∈ Lp(0, 1),

I(u) = g(u).

Step 2: g is convex and concave and hence, g is affine: Let λ ∈ (0, 1), α, β ∈ R, and

v(x) :=

α, x ∈ (0, λ),

β, x ∈ (λ, 1).

By Step 1,

g(λα+ (1− λ)β) = g(v) = I(v) = λg(α) + (1− λ)g(β),

so g is concave and convex. It follows that there exist a, b ∈ R such that forall u ∈ R, g(u) = a+ bu. (Exercise!) This implies that for all u ∈ Lp(0, 1),

I(u) = a+ b

∫ 1

0

u(x) dx = I(0) + bu.

In particular, I is affine linear.

3.2.3 Weak derivatives

We say that a function u ∈ Lp(Ω), p ∈ [1,∞] has weak derivative Dαu ∈ Lp(Ω), α amultiindex, if there exists w ∈ Lp(Ω) such that

∫

Ω

w(x)ϕ(x) dx = (−1)|α|∫

Ω

u(x)Dαϕ(x) dx

for all ϕ ∈ C∞c (Ω) with (classical) derivative Dαϕ. Then we set w = Dαu and note thatw is unique in Lp(Ω).Note that if u is classically differentiable, then its classical and weak derivatives coincide.Note that if u is classically differentiable almost everywhere, this does not imply its weakdifferentiability (Exercise).


Example 43. Let Ω = (−1, 1) and

u(x) =

0, −1 < x ≤ 0,

xδ, 0 < x < 1,

so that u ∈ Lp(Ω) if and only if δ > − 1p . Does u′ ∈ Lp(Ω) exist?

• Consider ϕ ∈ C∞c ((−1, 0)) ⊂ C∞c (Ω), then∫ 1

−1

wϕ =

∫ 0

−1

wϕ!= −

∫ 0

−1

0ϕ′ = 0,

so w|(−1,0) = 0 in Lp((−1, 0)).

• Let ϕ ∈ C∞c ((0, 1)) ⊂ C∞c (Ω) such that suppϕ ⊂ [ε, 1). Then∫ 1

0

wϕ =

∫ 1

ε

wϕ!= −

∫ 1

ε

xδϕ′(x) dx =

∫ 1

ε

δxδ−1ϕ(x) dx,

so w|(0,1) = δxδ−1 and w ∈ Lp(0, 1)⇔ δ > 1− 1/p.

• Assume δ > 1− 1p and set

w(x) =

0, −1 < x < 0,

δxδ−1, 0 < x < −1,π2

17 , x = 0.

Then we show that w = u′ ∈ Lp(−1, 1): For all ϕ ∈ C∞c (−1, 1),∫ 1

−1

wϕ = δ

∫ 1

0

xδ−1ϕ(x) dx.

At the same time,

−∫ 1

−1

uϕ′ =

∫ 1

0

δxδ−1ϕ+ xδϕ(x)|x=0 + xδϕ(x)|x=1

=

∫ 1

−1

wϕ, if δ > 0.

• Note that u is classically differentiable if and only if δ > 1.

3.2.4 Sobolev spaces

For p ∈ [1,∞] and a domain Ω ⊂ Rn, the Sobolev space of order k ∈ N is

W k,p(Ω) := u ∈ Lp(Ω) : ∀α ∈ Nn with |α| ≤ k,Dαu ∈ Lp(Ω),with norm

‖u‖k,p = ‖u‖Wk,p =

∑

|α|≤k‖Dαu‖pp

1/p

.

Some facts:


• W k,p(Ω) are Banach spaces.

• If p ∈ (1,∞), then W k,p(Ω) are reflexive.

• If p = 2, then W k,2(Ω) =: Hk(Ω) are Hilbert spaces.

• If p ∈ [1,∞), then u ∈ C∞(Ω) : ‖u‖k,p <∞ is dense in W k,p(Ω). In particular,

W k,p(Ω) = Hk,p(Ω) = completion of Ck(Ω) w.r.t. W k,p-norm.

Note that in general, C∞(Ω) is dense in W k,p(Ω) only if ∂Ω is sufficiently regular(for example, in the case discussed in the paragraph on boundary regularity below).

For proofs of these facts and Theorems 46, 47 and 48, we refer to [Adams]. It is acomprehensive and well-written monograph on the topic of Sobolev Spaces that alsocontains some history, sharp results and counterexamples, and details on the requirementsof the geometry of Ω that extensively generalize what we are using here. You can findcompact introductions to the topic in [Alt] and in [Evans]. The first one is close to (but notidentical to) what is in these notes, but, for example, defines Sobolev spaces via density.The second one provides an additional point-of-view, also inspired by applications toPDEs as well as variational problems.

Lemma 44. (Weak convergence in W k,p) Let p ∈ [1,∞), un, u ∈ W k,p(Ω). Then thefollowing are equivalent:

• un u in W k,p(Ω), and,

• ∀α ∈ Nn with |α| ≤ k, Dαun Dαu in Lp(Ω).

Proof. (only k = 1) Embed W 1,p(Ω) → Lp(Ω)n+1 via

J :

W 1,p(Ω) → Lp(Ω)n+1

u 7→ (u, ∂1u, . . . , ∂nu).

Then J is injective and norm-preserving. Moreover, J(W 1,p(Ω)) is a closed subspaceof Lp(Ω)n+1 (Exercise!). Hence, J is an isomorphism of Banach spaces (Exercise!).This shows that weak convergence in W 1,p(Ω) must be the same as weak convergencein Lp(Ω)n+1.

Example 45. Let Ω = (0, 2π), un(x) = 1n sin(nx), u′n(x) = cos(nx). Then

‖un‖22 =

∫ 2π

0

1

n2sin2(nx) dx =

π

n2→ 0,

‖u′n‖22 =

∫ 2π

0

cos(nx)2 dx = π,

so ‖un‖21,2 = π(1 + 1n2 ) ≤ 2π. Abstractly, since H1(0, 2π) is reflexive, the bounded

sequence (un)n has a weakly convergent subsequence. However, we know more: We have

unL2

→ 0 (by norm convergence) and by Example 41, u′nL2

cos = 0, hence, by Lemma 44,

unH1

0.


Boundary regularity: In the following, unless stated otherwise, we fix Ω to be abounded domain with Lipschitz (graph) boundary, in short, bounded Lipschitz domain, seee.g. [Alt, A 6.2]:A bit more precisely, this means that there is a finite number of open sets U1, . . . , U l ⊂ Rnthat cover ∂Ω in such a way that the sets ∂Ω ∩ U j , j = 1, . . . , l, are each given as a partof the graph of a Lipschitz continuous function gj : Rn−1 → R, and Ω ∩ U j is located onone side of this graph.

This assumption is stronger than what is needed for most of the following Theorems.They may also hold for less regular and/or unbounded domains/unbounded boundaries.However, the assumptions needed for sharper statements quickly become technically in-volved. In particular, there may be an interaction between the types of unboundednessand the types of non-regularity that are admitted. For a comprehensive overview andtreatment, including important counterexamples, we refer to [Adams]. For the func-tionals we study in this course, the assumption of bounded Lipschitz domains seems anacceptable compromise between achieving maximal generality and avoiding too manytechnicalities.In particular, due to the local Lipschitz graph property, for bounded Lipschitz domainsΩ, it is straightforward to define the integrals, Lp-norms and spaces Lp(∂Ω) for functionsf : ∂Ω → R via coordinate transform to Rn−1. For example, if supp(f) ⊂ U j , then wecan set ∫

∂Ω

f(x) dx′ =

∫

Rn−1

f(

n−1∑

i=1

yiei + gj(y)en)

√1 + |∇gj(y)|2 dy,

which is well-defined due to Lipschitzianity of gj and, e.g. Rademacher’s Theorem. Fordetails, see e.g. [Alt, A 6.5.1 and A 6.5.2].Moreover, if Ω is a bounded Lipschitz domain, the unit outer normal vector field

ν : ∂Ω→ Rn, ν ∈ L∞(∂Ω;Rn), (3.3)

is well-defined, i.e. it is independent of the choice of a local representation of ∂Ω, cf. [Alt,A 6.5.3].

Theorem 46. (Sobolev and Hölder embeddings) Let Ω ⊂ Rn be a bounded Lipschitzdomain. Let k > k ∈ N0. Then

1. If p, p ∈ [1,∞) satisfyk − n

p≥ k − n

p,

then W k,p(Ω) → W k,p(Ω), i.e. there is a constant C(Ω, k, k, p, p) > 0 such that forall u ∈W k,p(Ω), also u ∈W k,p(Ω) with

‖u‖k,p ≤ C‖u‖k,p.

2. Ifk − n

p≥ k + γ,

with p ∈ [1,∞) and γ ∈ (0, 1), then

W k,p(Ω) → C k,γ(Ω),


with C k,γ(Ω) the Hölder space given by

C0,γ(Ω) := u ∈ C(Ω) : ‖u‖C0,γ <∞,

‖u‖C0,γ := ‖u‖∞ + supx,y∈Ω,x 6=y

|u(x)− u(y)||x− y|γ ,

or, for k 6= 0,

C k,γ(Ω) := u ∈ C k(Ω) : ‖u‖Ck,γ <∞,

‖u‖Ck,γ := ‖u‖Ck +∑

|α|=k

supx,y∈Ω,x 6=y

|Dαu(x)−Dαu(y)||x− y|γ .

The number k − np is called Sobolev number of W k,p(Ω).

The proof of this Theorem is elaborate. Here, we just sketch two heuristics:

1. How can we trade weak differentiability for larger integrability exponents (and howdoes the spatial dimension come in)?We look at one specific example: Let n = 2, k = 1, p = 1 and assume that u ∈W 1,1(Ω) with u = 0 on ∂Ω. Then we can extend u by zero to u ∈ W 1,1(R2). Wesee that u ∈ L2(Ω) by the following estimates:We have

u(x1, x2) =

∫ x1

−∞∂1u(y1, x2) dy1 =

∫ x2

−∞∂2u(x1, y2) dy2,

hence|u(x)| ≤

∫

R|∇u(xj , yi)|dyi, für i = 1, 2, j 6= i. (3.4)

Consequently,

|u(x)|2 ≤(∫

R|∇u(x1, y2)|dy2

)(∫

R|∇u(y1, x2)|dy1

).

Integrating with respect to x1 and x2 gives the estimate we were looking for,

‖u‖22 ≤(∫

R

∫

R|∇u(x1, y2)|dy2dx1

)(∫

R

∫

R|∇u(y1, x2)|dy1dx2

)≤ ‖u‖21,1.

2. How can we trade weak differentiability for Hölder regularity?We just look at the case n = 1, k = 1. Then

|u(x)− u(y)| = |∫ y

x

u′(s) ds|Hölder≤ ‖1‖Lp′ ((x,y))‖u′‖p = |x− y|1−1/p‖u′‖p,

with 0 ≤ γ = 1− 1/p ≤ 1 the Hölder exponent and ‖u′‖p the Hölder norm.

Also worth thinking about: there is clearly no ‘partial’ vs ‘total’ weak differentiability –the definition is ‘partial’, but ‘global’ (what does this mean?). The Embedding Theoremshows that sufficient (for example, second-order, large p) Sobolev regularity implies (total)classical differentiability.

The following compactness theorem will also be crucial for our analysis of (DP).


Theorem 47. (Rellich Compactness Theorem) With the same assumptions as in Theo-rem 46, if holds for the conditions on the Sobolev number in 1. or 2., then the corre-sponding embeddings are compact.

In particular, note that this means that if a sequence, e.g. an infimizing sequence, is (apriori) bounded in W k,p(Ω), it has a strongly convergent subsequence in W k,p(Ω), withk, p, k, p subject to the assumptions of Rellich’s Theorem.Since we are interested in functionals of type (3.1), defined via volume and surface integraldensities for functions u ∈W 1,p(Ω), we need to understand and quantify the properties ofu on ∂Ω. If u ∈W 1,p(Ω) is insufficient for u ∈ C(Ω), we need to think carefully about thedefinition of the integral

∫∂Ωg(u(x)) dx′. The essential tool is the following trace operator

that uniquely determines u|∂Ω.

Theorem 48. (Trace operator) Let Ω ⊂ Rn be a bounded Lipschitz domain. For p ∈[1,∞], define

p∗ =

∞, if p > n,

arbitrary, but <∞, if p = n,p(n−1)n−p ≥ p, if p < n.

Then there is a bounded linear operator

γ : W 1,p(Ω)→ Lp∗(∂Ω)

u 7→ γ(u) = u|∂Ω

that is the extension of the restriction operator

γ : C1(Ω)→ C(∂Ω)

u 7→ u|∂Ω.

For u ∈W 1,p(Ω), we call u|∂Ω ∈ Lp∗(Ω) the trace of u on ∂Ω.

We see immediately that if p > n, then W 1,p(Ω) → C(Ω) by the embedding theoremand the trace of u is just its continuous extension. Otherwise, more work is needed toobtain the right estimate, see e.g. [Adams, Thm 5.22] that can then be combined with adensity argument.

Heuristics for the right estimate in the case n = 2, p = p∗ = 1, locally, assuming thatU j ⊂ span(e1) is flat. As in (3.4), we have

|u(x1, 0)| ≤∫

R|∂2u(x1, y2)|dy2,

for every (x1, 0) ∈ U j , so‖u(·, 0)‖L1(R) ≤ ‖∇u‖L1(R2) ≤ ‖u‖1,1.

Now for general bounded Ω ⊂ Rn and k ≥ 1, p ∈ [1,∞), recall the spaces

W k,p0 (Ω) = C∞c (Ω)

‖·‖k,p ⊂W k,p(Ω),

given by closure of the set of C∞c (Ω)-functions in W k,p(Ω). Without proof, we note thatif Ω is a bounded Lipschitz domain, then they relate to the usual Sobolev space in asensible way, i.e.

W 1,p0 (Ω) = u ∈W 1,p(Ω): u|∂Ω = 0.


Theorem 49. (Poincaré inequality) Let Ω ⊂ Rn be a bounded domain and p ∈ (1,∞).Then

1. For δ = diam(Ω) = sup|x − y| : x, y ∈ Ω < ∞, for all u ∈ W 1,p0 (Ω) (defined

above), we have‖u‖p ≤ δ‖∇u‖p. (3.5)

2. Let Ω be a bounded Lipschitz domain. Assume that V ⊂W 1,p(Ω) is a closed subspacewith the property

u ∈ V and ‖∇u‖Lp(Ω) = 0⇒ u = 0.

Then there exists a constant Cp > 0, called Poincaré constant, such that

‖u‖1,p ≤ Cp‖∇u‖p (3.6)

for all u ∈ V .

Proof. For these two versions of “Poincaré”, it is convenient to use two very differentstrategies:

1. By density, it suffices to prove (3.5) for u ∈ C∞c (Ω). Extend u to Rn by 0 (and stillcall this function u). For x ∈ Ω and e1 the first unit vector, we have x+ δe1 /∈ Ω, so

u(x) = u(x+ δe1)−∫ δ

0

∂1u(x+ se1) ds = −∫ δ

0

∂1u(x+ se1) ds,

and thus,

|u(x)|p =

∣∣∣∣∣

∫ δ

0

∂1u(x+ se1) ds

∣∣∣∣∣

p

Holder≤

∣∣∣∣∣

∫ δ

0

1 ds

∣∣∣∣∣

p/p′ ∣∣∣∣∣

∫ δ

0

|∂1u(x+ se1)|p ds

∣∣∣∣∣

1

≤ δp/p′∫ δ

0

|∇u(x+ se1)|p ds.

Hence, by Fubini,∫

x∈Ω

|u(x)|p dx ≤ δp/p′∫

x∈Ω

∫ δ

0

|∇u(x+ se1)|p dsdx

y=x+se1

= δp/p′∫ δ

0

∫

y∈Ω+se1|∇u(y)|p dy ds

≤ δp/p′+1

∫

Rn|∇u(y)|p dy

= δp‖u‖pp.

2. By contradiction: Assume that for all n ∈ N, there exists un ∈ V such that

‖un‖1,p ≥ n‖∇un‖p.


Now consider vn = un‖un‖1,p with ‖vn‖1,p = 1 and

‖∇vn‖p =‖∇un‖p‖un‖1,p

≤ 1

n. (3.7)

Since W 1,p(Ω) is reflexive and (vn)n is bounded, there exists a weakly convergingsubsequence (vnk)k of (vn)n with limit

vnkW 1,p

v∗.

Since V ⊂ W 1,p(Ω) is a weakly closed subspace (is is linear! – apply Mazur), weobtain v∗ ∈ V . By (3.7), we conclude ∇vnk → 0 in Lp(Ω). Using ∇vnk ∇v∗(Lemma 44!), this means that ∇v∗ = 0, so v∗ = 0 by assumption. At the sametime, by Rellich’s Compactness Theorem, there is a subsequence of (vnk)k withvnk′ → v∗ = 0 strongly in Lp(Ω) (here, some boundary regularity is used). Thiscontradicts ‖vn‖1,p = 1.

Remark 50. What are choices for V in Poincaré’s Theorem?

• Let Γ 6= ∅ be a part of the boundary Γ ⊂ ∂Ω of a bounded Lipschitz domain Ω.Choose V = X0 = u ∈W 1,p(Ω) : u|Γ = 0. Then if u ∈ V and ∇u = 0, u = const.Depending on Γ, e.g. if it is open in ∂Ω, this implies u = 0. Weaker assumptionson Ω and Γ may suffice.

• Another typical choice: V = u ∈ W 1,p(Ω) : u = 1|Ω|∫

Ωu = 0. In particular,

Theorem 49 implies that for all u ∈W 1,p(Ω),

‖u− u‖1,p ≤ Cp‖∇u‖p.

More useful facts that close out this section:

• Gauss’ Theorem holds, i.e. in Theorem 6, it is sufficient to ask that Ω is boundedLipschitz and u ∈W 1,1(Ω), to get

∫

Ω

∂ju(x) dx =

∫

∂Ω

u|∂Ω(x)νj(x) dx′

with ν ∈ L∞(∂Ω;Rn) the outer normal vector field as in (3.3) and u|∂Ω ∈ L1(∂Ω)by the Trace Theorem.

• Product rule: if u ∈ W 1,p(Ω), v ∈ W 1,p′(Ω), then uv ∈ W 1,1(Ω) and (uv)′ =u′v + uv′ (just use “Hölder”).

• Chain rule: (take care!) If f : R → R is uniformly Lipschitz and u ∈ W 1,p(Ω),then f u ∈ W 1,p(Ω) and (f u)′ = (f ′ u)u′. For p ∈ [1,∞), the mappingTf : W 1,p(Ω) 3 u 7→ f u ∈W 1,p(Ω) is bounded.


3.3 Properties of I(u) =∫

Ωf(x, u(x),∇u(x)) dx on W 1,p(Ω)

The aim of this section is to provide necessary and sufficient conditions on p,Ω, f and gfor the functionals

I : W 1,p(Ω) ⊇ X → R∞

u 7→∫

Ω


∫

∂Ω

g(x, u(x)) dx

of the form we looked at in Chapter 2, to allow us to solve (DP). The strategy is to applyTonelli’s Theorem 32, so, in particular, we want

• X to be reflexive (choose p ∈ (1,∞) and X = W 1,p(Ω) or a suitable subspace, seeExamples),

• I to be well-defined (Proposition 52),

• I to be coercive (Theorem 54), and

• I to be weakly lower semi-continuous (see Subsection 3.3.1).

For simplicity (and clarity), for the moment, assume that g = 0. Criteria for general gare discussed in the Exercises.

We start from the following definition.

Definition 51. (Carathéodory function) Let Ω ⊂ Rn be open and N ∈ N. Then f : Ω×RN → R∞ is called a Carathéodory function if

1. for a.e. x ∈ Ω, f(x, ·) : RN → R∞ is continuous, and,

2. for all a ∈ RN , f(·, a) : Ω→ R∞ is measurable.

The point of this definition is the following result on I being well-defined if f isCarathéodory and satisfies a suitable growth condition.

Proposition 52. Let f : Ω× RN → R∞ be a Carathéodory function. Then

1. If u : Ω→ RN is measurable, then

h : Ω→ R∞x 7→ f(x, u(x)),

is measurable.

2. For p ∈ [1,∞), if there are a ∈ L1(Ω), b ∈ R such that for all u ∈ RN , a.e. x ∈ Ω,

f(x, u) ≥ a(x) + b|u|p, (3.8)

then the functional

I : Lp(Ω;RN )→ R∞, I(u) =

∫

Ω

f(x, u(x)) dx,

is well-defined.


Proof.

1. Recall that measurable functions can be approximated by step functions [Klenke,Satz 1.96] and that pointwise limits of measurable functions are measurable [Klenke,Satz 1.92]. We first show 1. in the case that u is a step function,

u(x) =

k∑

i=1

αiχAi(x)

where αi ∈ R∞ and Ai ⊂ Ω are disjoint measurable sets with ∪ki=1Ai = Ω. Thenfor every a ∈ R, the set

x ∈ Ω : h(x) ≤ a = ∪ki=1x ∈ Ai : f(x, u(x)) ≤ a= ∪ki=1(Ai ∩ f−1((−∞, a);αi)

is measurable, as it is the union of measurable sets, as the functions f(·, αi) aremeasurable by the Carathéodory assumption. Now we approximate general mea-surable u by step functions un and use the continuity of f(x, ·) to get that for a.e.x ∈ Ω,

h(x) = f(x, u(x)) = limn→∞

f(x, un(x)).

Hence h is measurable as the limit of measurable functions.

2. What is needed for I to be well-defined? By 1., it remains to see that the in-tegral

∫Ωf(x, u(x)) dx does not return the value −∞, i.e. (f(·, u(·)))− should be

integrable. But this follows directly from (3.8).

Example 53. Let f(x, a) = κ(x)|a|p with κ ∈ L∞(Ω) and p ∈ [1,∞). Then

I(u) =

∫

Rf(x,∇u(x)) dx

is well-defined on W 1,p(Ω).

Theorem 54. (Coercivity) Let Ω ⊂ Rn be bounded Lipschitz. Assume that

f : Ω× Rm × Rm×n → R∞

is a Carathéodory function (N = m+ n ·m) that satisfies

f(x, u,A) ≥ c|A|p − δ(x)|u|r − h(x),

for a.e. x ∈ Ω, all u ∈ Rm and A ∈ Rm×n. Here, p ∈ (1,∞), c > 0 is a constant,r ∈ [1, p), δ ∈ L p

p−r (Ω) and h ∈ L1(Ω).Then there exist constants C > 0 and β ∈ R such that

I(u) =

∫

Ω

f(x, u(x),∇u(x)) dx ≥ C‖u‖p1,p − β

for all u ∈W 1,p0 (Ω). Hence, I is coercive on X = u0 +W 1,p

0 (Ω) for any u0 ∈W 1,p(Ω).


Proof. Let u ∈W 1,p0 (Ω). By assumption,

I(u) ≥ c‖∇u‖pp −∫

Ω

δ(x)|u(x)|r dr − ‖h‖1.

By Poincaré’s Inequality, c‖∇u‖pp ≥ cCP‖u‖p1,p. Together with Hölder’s Inequality with

(pr

)> 1,

(pr

)′= p

p−r , this gives

I(u) ≥ c

Cp‖u‖p1,p − ‖δ‖ p

p−r‖u‖rp − ‖h‖1.

Note that this can be shown for any u ∈ V where V is a subspace of W 1,p(Ω) such thatthe Poincaré Inequality holds. Note that since p > r ≥ 1, there exists an s0 > 0 suchthat c

2CPsp ≥ ‖δ‖ p

p−rsr for all s ≥ s0. Set β = ‖δ‖ p

p−rsr0, then

c

2CPsp + β − ‖δ‖ p

p−rsr ≥ 0

for all s ≥ 0. Then we have shown

I(u) ≥ c

2CP‖u‖p1,p + β − ‖h‖1 =: C‖u‖p1,p − β.

Example 55. (Dirichlet problem) Let Ω ⊂ Rn be bounded Lipschitz and

I(u) =

∫

Ω

|∇u(x)|2 dx.

Then I is coercive on W 1,p0 (Ω) if and only if p ≤ 2.

Proof. There are two cases to be considered:

p ≤ 2 Then there exists β ∈ R such that for all s ≥ 0,

s2 ≥ sp − β.

It follows that

I(u) ≥∫

Ω

|∇u(x)|p dx− β|Ω|Poincaré≥ 1

Cp‖u‖p1,p − β|Ω|,

so I is coercive on W 1,p0 (Ω).

p > 2 Let u0 ∈ W 1,20 (Ω) \W 1,p

0 (Ω) and choose (un)n ⊂ C∞c (Ω) such that un → u0

strongly in W 1,2(Ω). Then

I(un) = ‖∇un‖22 → I(u0) <∞,

but‖un‖1,p →∞,

so I is not coercive on W 1,p0 (Ω).


Example 56. (Is r < p necessary?) Let p ∈ (1,∞), λ ∈ R and

Iλ(u) =

∫

Ω

|∇u(x)|p dx+ λ

∫

Ω

|u(x)|p dx.

Let CP be the optimal constant in the Poincaré inequality (3.6) on W 1,p0 (Ω). Then Iλ is

coercive on W 1,p0 (Ω) if and only if λ − 1

CpP−1=: −c. This shows that we cannot just

omit the condition r < p in Theorem 54.

Proof. Note that from (3.6), we get that c is the optimal constant in the inequality‖∇u‖pp ≥ c‖u‖pp for all u ∈W 1,p

0 (Ω).

λ > −c : Then Iλ is coercive (Exercise!)

λ ≤ −c: We show that Iλ is not coercive on W 1,p0 (Ω). Let u ∈ W 1,p

0 (Ω) \ 0 with‖∇u‖pp = c‖u‖pp. (We prove later that the optimal constant c > 0 and theminimizer of constants u exist.) Let α ∈ R. Clearly,

‖αu‖1,p →∞, as |α| → ∞.

On the other hand,

Iλ(αu) = |α|p(∫

Ω

|∇u(x)|p dx+ λ

∫

Ω

|u(x)|p dx

)

= |α|p(c+ λ)‖u‖pp.

Since c+λ ≤ 0, this shows that Iλ(αu) ≤ 0 for all α ∈ R, so Iλ is not coercive.

3.3.1 Quasiconvexity and weak lower semicontinuity of I

In this Subsection, we show that, roughly,

1. if I is wlsc in W 1,p, then f is qc in the gradient component,

2. if f is qc in the gradient component and satisfies a growth condition, then I is wlscin W 1,p.

We start out from the necessity of quasiconvexity and show only the simpler case of nox- and u-dependence in f .

Theorem 57. Let Ω ⊂ Rn be open and bounded and let f : Rm×n → R be continuousand

I(u) :=

∫

Ω

f(∇u(x)) dx.

Then for p ∈ [1,∞), if I is wlsc on W 1,p(Ω), then f is quasiconvex.


Proof. We show that for [0, 1]n = D ⊂ Rn the unit cube, for every ξ ∈ Rm×n and forevery w ∈W 1,p

0 (D;Rm),

1

|D|

∫

D

f(ξ +∇w(y)) dy ≥ f(ξ). (3.9)

(Recall Definition 17 of quasiconvexity of f). Let w ∈W 1,p0 (D;Rm) be extended period-

ically to Rn, i.e.w(x+ z) = w(x), for all x ∈ D and z ∈ Zn.

Consider the sequence

wk(x) :=1

kw(kx) ∈W 1,p

0 (D;Rm).

Then wk 0 in W 1,p0 (D;Rm) by Lemma 44 and Example 41 and an exercise: the result

of Example 41 modified to hold on cubes D (instead of intervals (0, 1)). Now defineuξ(x) := ξx ∈W 1,p(Ω;Rm) (Ω is bounded!) and

uk(x) :=

uξ(x), x ∈ Ω \D,uξ(x) + wk(x), x ∈ D.

Then uk uξ in W 1,p(D : Rm) and

I(uk) =

∫

Ω

f(∇uk(x)) dx =

∫

Ω\Df(ξ) dx+

∫

D

f(ξ +∇wk(kx)) dx

= |Ω \D|f(ξ) +1

kn

∫

kD

f(ξ +∇w(y)) dy

= |Ω \D|f(ξ) +

∫

D

f(ξ +∇w(y)) dy.

Taking the limit and using wlsc’ity of f gives (3.9) via

I(uξ) = |Ω|f(ξ) ≤ lim infk

I(uk) = |Ω \D|f(ξ) +

∫

D

f(ξ +∇w(y)) dy.

Corollary 58. If n = 1 or m = 1 in Theorem 57, then f is necessarily convex.

Remark 59. Theorem 57 holds more generally for f that also depend on x and u [Dac08,Lemma 3.18] and it can be shown also for unbounded Ω [Dac08, Theorem 3.13].

We now want to prove something like a converse statement to Theorem 57. We firstprove a simple preliminary results on continuity of separately convex functions that satisfya growth condition:

Definition 60. (Separate Convexity – should have been included in Chapter 2!) Afunction f : RN → R∞ is separately convex or convex in each variable, if for every x =(x1, . . . , xN ), for every i = 1, . . . , N , the functions

fyi : R→ R∞yi 7→ f(x1, . . . , xi−1, yi, xi+1, . . . , xN )

are convex.


Remark 61. Clearly, a rank-1-convex function is separately convex. The converse is nottrue, consider, for example, the function f : R2 → R given by f(x1, x2) = x1x2. It isseparately convex but not convex. This example also shows how separate convexity is notobjective in its dependence of the choice of coordinates.

Lemma 62. Let f : RN → R be separately convex and such that for some α ≥ 0, p ≥ 1,for every x ∈ RN ,

|f(x)| ≤ α(1 + |x|p). (3.10)

Then there exists β ≥ 0 such that for every x, y ∈ RN ,

|f(x)− f(y)| ≤ β(1 + |x|+ |y|)p−1|x− y|.

Proof. In three steps:

Step 1: If a function g : R→ R is convex, then for every λ > µ > 0 and every t ∈ R,g(t± µ)− g(t)

µ≤ g(t± λ)− g(t)

λ. (3.11)

(Picture!) This is proved by

g(t± µ) = g(µ

λ(t± λ) + (1− µ

λ)t)

≤ µ

λg(t± λ) + (1− µ

λ)g(t),

using convexity of g.

Step 2: Now fix x1 = (x2, . . . , xN ) ∈ RN−1 and define

g(t) := f(t, x1)

for t ∈ R. Then we show that there is β1 ≥ 0 such that

|g(x1)− g(y1)| ≤ β1(1 + |x|+ |y|)p−1|x1 − y1|). (3.12)

Wlog, let x1 < y1. In (3.11), set

λ := 1 + |x|+ |y|, µ := y1 − x1,

then

g(y1)− g(x1) = g(x1 + (y1 − x1))− g(x1)

≤ (y1 − x1)g(x1 + 1 + |x|+ |y|)− g(x1)

1 + |x|+ |y|and

g(x1)− g(y1) = g(y1 − (y1 − x1))− g(y1)

≤ (y1 − x1)g(y1 − (1 + |x|+ |y|))− g(y1)

1 + |x|+ |y| .

Using (3.10), this proves (3.12).


Step 3: With

f(x)− f(y) = (f(x)− f(y1, x2, . . . , xN )) +

. . . (f(y1, . . . , yi−1, xi, . . . , xN )− f(y1, . . . , yi, xi+1, . . . , xN )) . . .

+ (f(y1, . . . , yN−1, xN )− f(y)) ,

the result follows from Step 2.

Theorem 63. (Quasiconvexity and growth condition imply wlsc’ity) Let p ∈ (1,∞) andf : Rm×n → R be quasiconvex, additionally satisfying the growth condition

0 ≤ f(ξ) ≤ α(1 + |ξ|p) (3.13)

for every ξ ∈ Rm×n and some α ≥ 0. Let Ω ⊂ Rn be open and bounded and

I(u) :=

∫

Ω

f(∇u(x)) dx.

Then I is wlsc in W 1,p(Ω;Rm).

Proof. In two steps:

1. show wlsc’ity only for un u with u(x) = Ax+ b affine.

2. show wlsc’ity for general u using approximation by piecewise affine functions andStep 1.

Note that a consequence of condition (3.13) is that f is quasiconvex in A if and only if∫Df(A + ∇w(y)) dy ≥ |D|f(A) holds for all w ∈ W 1,p

0 (D;Rm) (instead of: only for allw ∈ PC1

0 (D;Rm)).

Step 1: Let (un)n ⊂W 1,p(Ω) with un u and u(x) = Ax+b for some A ∈ Rm×n andb ∈ Rm. If we knew un|∂Ω = u|∂Ω, then un = u + wn where wn ∈ W 1,p

0 (Ω).Then from quasiconvexity, we would get

I(un) =

∫

Ω

f(A+∇wn(x)) dx ≥ |Ω|f(A) = I(u) (3.14)

and we would be done. However, in general, un|∂Ω 6= u|∂Ω so we need addi-tional arguments:Consider

• a set Ω0 ⊂ Ω such that R := 12dist(Ω,Ω0) > 0,

• arbitrary L ∈ N \ 0,• for integers 1 ≤ i ≤ L, the sets

Ωi := x ∈ Ω : dist(x,Ω0) i

LR,


• cut-off functions ϕi ∈ C10 (Ωi), with 0 ≤ ϕi ≤ 1, ϕi|Ωi−1

= 1, and |∇ϕi| ≤2(L+1)R ,

as in the Figure below:

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.Now let

vi,n := u+ ϕi(un − u),

then I(vi,n) ≥ I(u) as in (3.14). Moreover, (omitting the argument “(x)” inthe following),

I(u) ≤ I(vi,n)

=

∫

Ω\Ωif(∇u) dx+

∫

Ωi\Ωi−1

f(∇vi,n) dx+

∫

Ωi−1

f(∇un) dx

f≥0

≤∫

Ω\Ω0

f(∇u) dx+

∫

Ωi\Ωi−1

f(∇vi,n) dx+

∫

Ω

f(∇un) dx.

Since∇vi,n = (1− ϕi)∇u+ ϕi∇un +∇ϕi(un − u),

by (3.13),∫

Ωi\Ωi−1

f(∇vi,n) dx

≤∫

Ωi\Ωi−1

C(1 + |∇vi,n|p) dx

Young≤

∫

Ωi\Ωi−1

C(1 + |∇u|p + |∇un|p +

(2(L+ 1)

R

)p|un − u|p) dx.

Putting the estimates above together, summing over i and dividing by L, weget

∫

Ω0

f(∇u) dx

≤ C

L

∫

Ω

(1 + |∇u|p + |∇un|p +

(2(L+ 1)

R

)p|un − u|p) dx

+

∫

Ω

f(∇un) dx.


Now recall that un u weakly in W 1,p(Ω), so∫

Ω|∇un(x)|p dx is bounded

and by Rellich, un → u strongly in Lp(Ω), so taking n large,(

2(L+ 1)

R

)p ∫

Ω

|un − u|p(x) dx

is bounded independently of L. Taking the liminf and using wlsc’ity of thenorm, we obtain

∫

Ω0

f(∇u) dx ≤ C

L+ lim inf

n→∞

∫

Ω

f(∇un) dx.

As Ω0 ⊂ Ω and L ∈ N were arbitrary, conclude∫

Ω

f(∇u) dx ≤ lim infn→∞

∫

Ω

f(∇un) dx.

Step 2: Now we look at general (un)n ⊂ W 1,p(Ω) with un u ∈ W 1,p(Ω) not nec-essarily affine. We approximate Ω by S(l) cubes Dl

s of size 1l , l ∈ N. In

particular, for given δ > 0, l can be made so large and the cubes can beplaced so that

H l = ∪S(l)i=1D

li ⊂ Ω with |Ω \H l| < δ, (3.15)

see Figure below:

Dli




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Set ξli := 1|Dli|

∫Dli∇u(x) dx. Then for ε > 0 we can choose l ∈ N such that

l∑

i=1

∫

Dli

|∇u− ξli|p dx < ε. (3.16)

(To see this, approximate ∇u ∈ Lp(Ω;Rm×n) by wε ⊂ C(Ω;Rm×n) such that‖∇u−wε‖pp < ε. Since wε is continuous, it follows directly that for sufficientlysmall l,

l∑

i=1

∫

Dli

|w(x)− 1

|Dli|

∫

Dli

w(y) dy|p dx < ε.

It is then straightforward to show (3.16).)


Now look at

I(un)− I(u) =

∫

Ω

f(∇un)− f(∇u) dx

=

∫

Ω\Hlf(∇un)− f(∇u) dx

+

∫

Hlf(∇u+∇(un − u))− f(∇u) dx

= Jn,l1 + Jn,l2 .

From (3.13), using (3.15), we get

Jn,l1 ≥ −C∫

Ω\Hl(1 + |∇u(x)|p) dx

l→∞→ 0.

To estimate Jn,l2 , let ξl be the piecewise constant function with ξl(x) = ξli ifx ∈ Dl

i and set

Jn,l2 =

∫

Hlf(ξl +∇(un − u) dx

+

∫

Hlf(∇u+∇(un − u))− f(ξl +∇(un − u)) dx

+

∫

Hl−f(∇u) dx.

=: Jn,l21 + Jn,l22 + J l23.

From Step 1, for all Dli, we obtain

lim infn→∞

∫

Dli

f(ξli +∇(un − u)) dx ≥∫

DlI

f(ξli) dx,

so we have

lim infn→∞

Jn,l21 + J l23 ≥∫

Hlf(ξl(x))− f(∇u(x)) dx =: J l3.

Now by Lemma 62, there exists β,C ≥ 0 such that

|J l3| ≤∫

Hl|f(ξl(x))− f(∇u(x)) dx

≤ β∫

Hl(1 + |ξl(x)|+ |∇u(x)|)p−1|ξl(x)−∇u(x)|dx

Hölder≤ β‖(1 + |ξl|+ |∇u|)p−1‖Lp′ (Hl)‖ξl −∇u‖Lp(Hl)

= β‖1 + |ξl|+ |∇u|‖p−1Lp(Hl)

‖ξl −∇u‖Lp(Hl)

≤ C‖ξl −∇u‖Lp(Hl),


where we used that ‖ξl‖p is bounded by ‖∇u‖p considering (3.16). In thesame way, for large n,

|Jn,l22 | ≤ β′∫

Hl(1 + |ξl|+ |∇u|+ |∇un|)p−1|ξl −∇u|dx

≤ C‖ξl −∇u‖Lp(Hl).

Putting everything together, we obtain

lim infn→∞

I(un)− I(u) ≥ lim infn→∞

Jn,l1 + Jn,l2

≥ lim infn→∞

Jn,l1 + Jn,l22 + J l3

≥ −C∫

Ω\Hl(1 + |∇u(x)|p) dx− C‖ξl −∇u‖Lp(Hl)

l→∞→ 0,

and have proved the claim.

Remark 64. The growth condition (3.13) can be weakened, but it cannot be omitted, asthe example f(∇u) = det(∇u) shows. If m = n = 2, then f is quasiconvex, but clearly,(3.13) is not satisfied, as the determinant could be negative. Moreover, I is not wlsc inW 1,2(Ω) [Dac08, Example 8.6].

Remark 65. Theorem 63 holds more generally for f that also depend on x and u [Dac08,Theorem 8.8].

Remark 66. In spite of the results of this Subsection, for a given non-convex function itremains difficult to decide whether it is quasiconvex and/or weakly lower semicontinuous.Moreover, the assumption of quasiconvexity and the growth condition (3.13) may be toorestrictive. The need for functionals

f : Ω× Rm×n → [0,∞]

that are Carathéodory, non-convex, but with more straightforward structure than qua-siconvex functions, and such that I : W 1,p(Ω) → R∞ is wlsc for suitable p leads to theconcept of polyconvexity, see e.g. [Dac08, Chapter 5].

3.4 Examples3.4.1 p-Laplace

Let Ω ⊂ Rn be bounded Lipschitz, p ∈ (1,∞). Consider the functional

I(u) =

∫

Ω

1

p|∇u(x)|p dx−

∫

Ω

h(x)u(x) dx

in W 1,p0 (Ω). Then for every h ∈ Lp′(Ω), there exists a unique minimizer of I in W 1,p

0 (Ω).Proof and discusssion in four steps:


1. I is well-defined and coercive in W 1,p(Ω)

Consider f : Ω × R × Rn → R, f(x, u,A) := 1p |A|p − h(x)u. Clearly, f is a

Carathéodory function. We can estimate f from below,

f(x, u,A) ≥ 1

p|A|p − |h(x)||u| ≥ 1

p(|A|p − |u|p)− 1

p′|h(x)|p′ .

In the second estimate, if h ∈ Lp′(Ω), then |h|p′ ∈ L1(Ω), so the condition in

Proposition 52 is satisfied and thus I is well-defined. The first estimate shows thatthe growth condition in Theorem 54 is satisfied with r = 1 if h ∈ Lp

′=p/p−1(Ω).Hence, I is coercive on W 1,p

0 (Ω).

2. I is wlsc

The function f1 : Rn → [0,∞), A 7→ 1p |A|p is convex and satisfies the growth condi-

tion in Theorem 63, so for a weakly convergent sequence un u in W 1,p0 (Ω), using

h ∈ Lp′(Ω) ⊂ (W 1,p0 (Ω))′, we have

lim infn

I(un) ≥∫

Ω

f1(∇u(x)) dx− limn

∫

Ω

h(x)un(x) dx

=

∫

Ω

f1(∇u(x)) dx−∫

Ω

h(x)u(x) dx = I(u),

so I is wlsc in W 1,p0 (Ω).

By Tonelli’s Theorem, 1. and 2. give the existence of a minimizer of I in W 1,p0 (Ω).

3. Uniqueness of the minimizer follows from strict convexity: Exercise!

4. Why p-Laplace?

The Euler-Lagrange equations for I are

div(|∇u|p−2∇u(x)) + h(x) = 0, x ∈ Ω,

u(x) = 0, x ∈ ∂Ω.

They generalize Laplace’s Equation (the case p = 2). By the direct method, wehave shown the existence of a unique (weak) solution.

3.4.2 Optimal Poincaré constant

Let Ω ⊂ Rn be bounded Lipschitz, p ∈ (1,∞). Then there exist c > 0 and u ∈W 1,p0 (Ω) \

0 such that for all v ∈W 1,p0 (Ω),

‖∇v‖pp ≥ c‖v‖pp, (3.17)

and

‖∇u‖pp = c‖u‖pp. (3.18)

We prove this in three steps:


1. Reformulation as a minimization problem with constraintIf (3.17) is true for all v ∈W 1,p

0 (Ω) with ‖v‖p = 1, then it is true for all u ∈W 1,p0 (Ω)

(if (3.17) is true for some u ∈ W 1,p0 (Ω), it must hold for v := u

‖u‖p .) Thus, weminimize

I(u) :=

∫

Ω

|∇u(x)|p dx = ‖∇u‖pp

in the set M := v ∈W 1,p0 (Ω) : ‖v‖p = 1.

2. Application of the Direct MethodLet (un)n ⊂M be an infimizing sequence for I.

• Since I is coercive (Poincaré’s Inequality), (un)n is bounded.• Since W 1,p

0 (Ω) is reflexive, (un)n is weakly convergent.• If M is weakly sequentially closed, then the weak limit un u is again in M .• Since I is wlsc on M (wlsc on W 1,p

0 (Ω) by convexity and Theorem 63), thenlim infn I(un) ≥ I(u), so u ∈M is a minimizer.

It remains to show that M is weakly sequentially closed. Since M is not convex,Mazur’s Lemma does not apply. Instead we use Rellich’s Theorem: Let (un)n ⊂Msuch that un u0 ∈ W 1,p(Ω). Then u0 ∈ W 1,p

0 (Ω) since W 1,p0 (Ω) is a linear and

thus weakly sequentially closed subspace of W 1,p(Ω). By Rellich’s Theorem, thereis a subsequence unk

k→ u0 that converges strongly in Lp(Ω), so

‖u0‖p = limk‖unk‖p = 1

and hence u0 ∈M .

3. ConclusionBy 2., there is an u ∈M such that ‖∇u‖pp ≤ ‖∇v‖pp for all v ∈M . Define c := I(u),then

I(v) = ‖∇v‖pp ≥ c‖v‖ppfor all v ∈M and hence for all v ∈W 1,p

0 (Ω) by 1. In addition,

‖∇u‖pp = c · 1 = c‖u‖pp,so (3.17) and (3.18) are proved. It remains to check that c 6= 0. Assume c = 0, then∇u = 0 in Ω, so u = const. on each connected component of Ω. Since u ∈W 1,p

0 (Ω),this implies u = 0. But this is in contradiction to ‖u‖p = 1.

3.4.3 Phase separation (Cahn-Hilliard energy)

Let Ω ⊂ R3 be bounded Lipschitz and κ, ε > 0. Define

I(u) :=

∫

Ω

ε

2|∇u(x)|2 +

κ

ε(1− u(x)2)2 dx (3.19)

on the setV := u ∈W 1,2(Ω) :

∫

Ω

u(x) dx = 0

– the double-well potential has returned! Then, I has a minimizer in the space V .


1. ProofSince the Poincaré Inequality holds in the reflexive Banach space V and for allu ∈ R, (1− u2)2 ≥ 0, I is well-defined and coercive. If we show that I is wlsc in V ,then by Tonelli’s Theorem, we have a minimzer u ∈ V . To show that I is wlsc, itremains to show that

I2 : W 1,2(Ω) → Ru 7→

∫Ω

(1− u(x)2)2 dx

is wlsc in V . This follows from the following proposition using that

g : R ⊃ u 7→ (1− u2)2 ∈ R

is continuous with |g(u)| ≤ C(1 + |u|4) for some C > 0 and that the embeddingW 1,2(Ω)

c→ Lr(Ω) is compact for r = 4.

Proposition 67. (Non-convex functionals of lower order) Let Ω ⊂ Rn be boundedLipschitz, p ∈ (1,∞) and r ∈ [1,∞) such that the embedding W 1,p(Ω)

c→ Lr(Ω) is

compact, i.e. 1 − np > −nr . Moreover, let g : Rm → R be continuous with |g(u)| ≤

C(1 + |u|r) for all u ∈ Rm. Then the functional

Ig : W 1,p(Ω) → Ru 7→

∫Ωg(u(x)) dx

is weakly sequentially continuous.

Proof. Consider unW 1,p

u. Let (unk′ )k′ be an infimizing subsequence of (unk)k suchthat

lim infn

Ig(un) = limkIg(unk).

By Rellich’s Theorem, there is a strongly converging subsequence unk′Lr(Ω)→ u, so,

in particular,∫

Ω

(1 + |unk′ (x)|r) dx = |Ω|+ ‖unk′‖rrk′→∞→ |Ω|+ ‖u‖rr

and we can choose (unk′ )k′ such that it converges pointwise a.e. to u. We obtain

lim infn

Ig(un) = limk′Ig(unk′ ) + C(|Ω|+ ‖unk′‖rr)− C(|Ω|+ ‖unk′ ‖rr)

Fatou≥

∫

Ω

limk′

(g(unk′ (x)) +

C

|Ω| (|Ω|+ ‖unk′‖rr)

)dx− C(|Ω|+ ‖unk′‖rr)

=

∫

Ω

g(u(x)) dx = Ig(u).

Here, the condition |g(u)| ≤ C(1 + |u|r) ensures that Fatou is applicable as theintegrand is positive. In the same way, we prove that

− lim supn

Ig(un) = lim infn−Ig(un) ≥ −Ig(u),

hence, limn Ig(un) = Ig(u).


Remark 68. Note the importance of g acting in the lower order: Otherwise, g andIg would be affine (Corollary 42) or g would have to be convex (Theorem 57).

2. Discussion – why “phase separation”?

Consider u as the volume density of a given compound material, re-scaled, so that

u(x) = 1 ' material A at x (e.g. zinc),u(x) = −1 ' material B at x (e.g. lead),

u(x) ∈ (−1, 1) ' mixture of materials A and B at x.

The assumption u ∈ M just means that the total ratio of 50% zinc and 50% leadis fixed. Then I in (3.19) is an energy associated to the phases’ wish to separate:the states u(x) = 1 and u(x) = −1 are preferable, due to the double-well potential.Mixed states are more costly, as well as changes of material composition (∇u 6=0), due to the first integrand. The minimizer will thus be mostly constant witha minimal size of the separating layer Sε (see Picture). S can be considered toapproximate (ε→ 0) the free surface that separates two natural phases.

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3.4.4 Nonlinear Elasticity

Recall the modelling and notation for elasticity from Section 2.7.3. Recall the condition

detFn > 0 with detFnn→∞→ 0 ⇒ W (x, Fn)

n→∞→ ∞ (3.20)

on the deformation gradient F = ∇ϕ that was introduced to avoid self-interprenetation.This condition implied that elastic energies would need to be non-convex.

For n = 2, set

W (A) =

1p‖A‖p + 1

detA , detA > 0,

+∞, detA ≤ 0,

as a model energy density for nonlinear elasticity that satisfies (3.20). Set

I(ϕ) =

∫

Ω

W (∇ϕ(x)) dx

as the corresponding nonlinear elastic energy. What are properties of I? Do minimizersexist (in what space)?

As usual, we want to apply Tonelli’s Theorem.


• For p > 1, I is well-defined inW 1,p0 (Ω) by Proposition 52 and coercive by Poincaré’s

Inequality,

I(ϕ) ≥ 1

p‖∇ϕ‖pp ≥ C‖ϕ‖pW 1,p(Ω).

• To show wlsc’ity for p > 2, it remains to show wlsc’ity of

I2(ϕ) =

∫

Ω

f2(∇ϕ(x)) dx,

where

f2(A) =

1

detA , detA > 0,

+∞, detA ≤ 0.

We decompose I2 on W 1,p0 (Ω) into the operator

B : W 1,p(Ω) 3 ϕ 7→ det(∇ϕ) ∈ Lp/2(Ω),

(this uses p > 2), and the functional

J2 : Lp/2(Ω;R) 3 θ 7→ J2(θ) =

∫

Ω

g2(θ(x)) dx ∈ [0,∞],

where

g2(s) =

1s , s > 0,

+∞, s ≤ 0.

Note that g2 is positive, continuous and convex. Thus, wlsc’ity of J2 holds by thefollowing result.

Proposition 69. (Convexity and “+∞”) Let Ω ⊂ Rn be open and bounded. Assumethat

f : Ω× RN → [0,∞]

is a Carathéodory function that satisfies

f(x, ·) : RN → [0,∞] is convex for a.e.x ∈ Ω.

Then for all p ∈ (1,∞), the functional

I(u) =

∫

Ω

f(x, u(x)) dx

is wlsc in Lp(Ω).

Proof. Note that by assumption, I is well-defined. Since f(x, ·) is convex, I isconvex. Thus, by Theorem 34, it suffices to show that I is strongly lower semicon-tinuous. Hence, let (un)n ⊂ Lp(Ω) such that un → u ∈ Lp(Ω). Let (unk)k be asubsequence such that

limkI(unk) = lim inf

nI(un).


Let (unk′ )k′ be a further subsequence that converges pointwise a.e. to u. Since thef(x, ·) are continuous and non-negative, by Fatou, we obtain

lim infn

∫

Ω

f(x, un(x)) = lim infk′

∫

Ω

f(x, unk′ (x)) dx

≥∫

Ω

limk′f(x, unk′ (x)) dx

=

∫

Ω

f(x, u(x)) dx.

To prove wlsc’ity of I2, it now suffices to show that the operator B is weaklysequentially continuous. So assume ϕn ϕ weakly in W 1,p

0 (Ω). We need to showthat det(∇ϕn) det(∇ϕ) weakly in Lp/2(Ω). By Lemma 39, it suffices to show

1. ‖ det(∇ϕn)‖p/2 ≤ C for C > 0 independent of n ∈ N (this also shows that Bis well-defined), and

2.∫

Ωdet(∇ϕn(x))ψ(x) dx→

∫Ω

det(∇ϕ(x))ψ(x) dx for all ψ ∈ C∞c (Ω).

To show 1., note that for any v ∈W 1,p(Ω),

‖det(∇v)‖p/2p/2 ≤ C∫

Ω

|∂1v1∂2v2|p/2 + |∂1v2∂2v1|p/2 dx

≤ C∫

Ω

|∂1v1|p + |∂2v2|p + |∂1v2|p + |∂2v1|p dx

= C‖∇v‖pp < +∞.

Since (ϕn)n is weakly convergent, it is bounded, and this proves the claim.To show 2., apply Gauss’ Theorem:

∫

Ω

det(∇ϕn)ψ =

∫

Ω

(∂1ϕn1∂2ϕn2 − ∂2un1∂1un2)ψ

=

∫

Ω

div(ϕn1

(∂2ϕn2

−∂1ϕn2

))ψ

Gauss= −

∫

Ω

ϕn1

(∂2ϕn2

−∂1ϕn2

)·(∂1ψ∂2ψ

)

→ −∫

Ω

ϕ1

(∂2ϕ2

−∂1ϕ2

)·(∂1ψ∂2ψ

)

=

∫

Ω

det(∇ϕ)ψ.

In conclusion, we have shown that the elastic energy defined above has a minimizer inW 1,p

0 (Ω). Its density of the form f(A) = g(A,detA) with convex g is a special polyconvexfunction. The notion of polyconvexity is important in nonlinear elasticity theory. Roughlyspeaking,

f convex ⇒ f polyconvex ⇒ f quasiconvex ⇒ . . . ,


and, at least in general,

f convex : f polyconvex : f quasiconvex . . .

Lecture Notes on Calculus of Variations I - Uni Kassel

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