Di usion and Advection: some PDE models in Spatial Ecology ...fli/Lecture_Notes/2013/02 Lou Lecture...

Center for Partial Differential Equations, ECNU No. 2013-02

Diffusion and Advection: some PDE models inSpatial Ecology

Lecture notes from Program on Nonlinear Equations inPopulation Biology,

Center for PDE, ECNU, mid April-June, 2013

Yuan LouThe Ohio State University, [email protected]

The Center receives funding from Cente for PDE 500 Dongchuan Road Administration Building 12th floor Minhang Campus, East China Normal University Shanghai, 200241, China Email: [email protected]

DIFFUSION AND ADVECTION: SOME PDE MODELS IN SPATIALECOLOGY

Yuan Lou

Abstract. This series of lectures will focus on the dynamics of some reaction-diffusionadvection models from spatial ecology. Mathematically we are interested in the effect ofdiffusion and advection on population dynamics in spatially heterogeneous environment.Biologically we are interested in understanding the evolution of dispersal; i.e., looselyspeaking, to investigate what kind of dispersal strategies are optimal.

LECTURE 1

1. Logistic model

(1.1)dN

dt= rN(1− N

K)

r : intrinsic growth rate (no unite);

K : carrying capacity (same unit as population size).

Discrete-time model

Nt = N(t) : populationsizeattimet, t = 0, 1, 2, · · ·Geometric growth Nt+1 = RNt

R =Nt+1

Nt

=offspring ♯

parents ♯, ♯ : numbers

r = lnR

General model Nt+1 = f(Nt)Beverton-Holt model

(1.2) Nt+1 =RNt

1 + R−1KNt

K: population size where the parent vs offspring ratio

1991 Mathematics Subject Classification. Primary: 35B44, 35K57; Secondary: 35B30, 35K51.Key words and phrases. Diffusion, Advection, Logistic, Lotka-Volterra, Spatial Ecology.

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For geometric model,Nt

Nt+1

=1

RWhat is the next level of models in terms of complexity?

Nt

Nt+1

= linear function of Nt

= line passing through (K, 1) and (0,1

R).

Then we get (1.2).Change (1.2),

Nt+h =RhNt

1 + Rh−1K

Nt

, h > 0.

Nt+h −Nt

h=

1

h

[RhNt

1 + Rh−1K

Nt

−Nt

]

=1

h

[(Rh − 1)Nt − Rh−1

KN2t

1 + Rh−1K

Nt

]

=Rh − 1

h

[(1− Nt

K)Nt

1 + Rh−1K

Nt

]→ lnRNt(1−

N

K), as h→ 0.

Remark 1.1. Two species modelN1(t+ 1) =

R1N1(t)

1 + α1N1(t) + β1N2(t)

N2(t+ 1) =R2N2(t)

1 + α2N1(t) + β1N2(t).

which is equivalent to N1(t)

N1(t+ 1)=

1 + α1N1(t) + β1N2(t)

R1

N2(t)

N2(t+ 1)=

1 + α2N1(t) + β1N2(t)

R2

.

Corresponding continuous-time modeldN1

dt= lnR1(1− C1N1 −D1N2)

dN2

dt= lnR2(1− C2N1 −D2N2),

where Ci, Di, i = 1, 2 depend upon αi, βi i = 1, 2.

DIFFUSION AND ADVECTION: SOME PDE MODELS IN SPATIAL ECOLOGY 3

2. Diffusion models of single species

(2.1) ut = ∇ · [d(x)∇u] +−→b · ∇u+ uf(x, u), x ∈ Ω ⊂ Rn, t > 0.

u(x, t) : density at location x and time t.

d(x) > 0, smooth,−→b = (b1, b2, · · · , bn) Holder continuous.

Boundary condition:

(2.2) ∇u · −→n = 0 on ∂Ω

where −→n is the outward unit normal vector.Obvious, u ≡ 0 is a steady state of (2.1)Stability of u ≡ 0: It is determined by the smallest eigenvalue (denoted by σ1).

(2.3)

∇ · [d(x)∇φ] +−→b · ∇φ+ f(x, 0)φ+ σφ = 0, in Ω

∇φ · −→n = 0 on ∂Ω.

Proposition 2.1.If σ1 > 0, then u ≡ 0 is locally stable;If σ1 < 0, then u ≡ 0 is unstable.

Proof. Sub-solution: Consider σ1 < 0. Set u(x) = εφ1(x) where ε > 0, φ1 > 0 is aneigenfunction of σ1.

Recallut = ∇ · [d(x)∇u] +

−→b · ∇u+ uf(x, u).

It suffices to show:

ut ≤ ∇ · [d(x)∇u] +−→b · ∇u+ uf(x, u)

⇔0 ≤ ε∇ · [d∇φ1] + ε−→b · ∇φ1 + εφ1f(x, εφ1)

⇔0 ≤ −f(x, 0)φ1 − σ1φ1 + φ1f(x, εφ1)

⇔0 ≤ [f(x, εφ1)− f(x, 0)]− σ1.

The last inequality holds for 0 < ε << 1 since σ1 < 0.

Remark 2.1.ut = ∇ · [d(x)∇u] +

−→b · ∇u+ uf(x, u), x ∈ Ω ⊂ Rn, t > 0,

u(x, 0) = εφ1 (sub− solution).

By maximum principles, u(x, t) is increasing in t for every x ⇒ u ≡ 0 is unstable.

Exercise: To check, if σ1 < 0, u ≡ 0 is locally stable (construct super-solution).

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3. Logistic model (PDE)

(3.1)

ut = d∆u+ u(m(x)− u), in Ω× (0,∞),

∇u · −→n = 0, on ∂Ω× (0,∞).

Examples 1, ru(1− uK)

r(x)=K(x)======== u(r(x)− u).

2, ru(1− uK)− h(x)u = u(r − h(x)− r

Ku) (h(x)u harvesting).

Assumption:

• m ∈ Cα(Ω) for some α ∈ (0, 1).• Ω+ = x ∈ Ω : m(x) > 0 has positive Lebesgue measure (⇔ m is positivesomewhere), Ω− = x ∈ Ω : m(x) < 0.

Consider following ODE

(3.2)

dUdt

= U(m(x)− U), t > 0,

U(0) > 0.

⇒ limt→∞ U(x, t) = m+(x) := maxm(x), 0. (≈ PDE (3.1) with d << 1.)Consider ODE

(3.3)

dVdt

= V (m(x)− V ), t > 0,

V (0) > 0.

⇒ limt→∞ V (x, t) =

1|Ω|

∫Ωm,

∫Ωm ≥ 0,

0,∫Ωm < 0.

(≈ PDE (3.1) with d >> 1.)

Theorem 1. If∫Ωm ≥ 0, then (3.1) has a unique positive steady state, denoted by θd

such that it is GAS (globally asymptotically stable).

limd→0 θd = m+(x), in L

∞,

limd→+∞ θd =1|Ω|

∫Ωm, in L∞.

Open problems: Considerd∆θd + θd(m(x)− θd) = 0, in Ω,

∇θd · −→n = 0, on ∂Ω,

θd > 0, in Ω.

1, ∃C > 0, independent of d such that ∥θd∥Cα(Ω) ≤ C?2,

Remark 3.1. (Averill, Lam, Lou) m ∈ C2(Ω),m ≥ 0 in Ω, then θd → m in W 1,2(Ω).


3,(He, Ni) Example: minΩ θd >1|Ω|

∫Ωm for some range of d.

4,(Liang, Lou, 2012, DCDS-B)∫Ω|∇θd|2 decreasing in d. Is

∫Ω(θd − θd)

2 decreasing ind?

Theorem 2. If∫Ωm < 0, then ∃d∗ > 0 such that (3.1) has a positive steady state if and

only if d < d∗. Moreover, there exists at most one positive steady state.

Eigenvalue problem: (how d∗ is determined.)d∆φ+m(x)φ+ σ1φ = 0, in Ω,

∇φ · −→n = 0, on ∂Ω.

Thmalmost⇔ σ1 =

+, d > d∗,

0, d = d∗,

−, d < d∗.

Lemma 3.1.1, σ1 is strictly increasing in d.2, lim

d→0σ1 = −maxΩm < 0.

3, limd→∞

σ1 = − 1|Ω|

∫Ωm > 0.

Remark 3.2. This result of Altenberg illustrats connection of convexity of σ1 in d and λ.

Altenberg Lemma: Considerd∆φ+ λm(x)φ+ σ1φ = 0, in Ω,

∇φ · −→n = 0, on ∂Ω.

Lemma 3.2. The first eigenvalue σ1 is concave in λ. (and σ1 is concave in d.)

Kato’s Theoremd(Σi,jaij

∂2φ

∂xi∂xj+ Σibi

∂φ

∂xi

)+ λm(x)φ+ σ1φ = 0, in Ω,

Σi,jaij∂φ∂xi

−→n j = 0, on ∂Ω.

Then σ1 is concave in λ (and also concave in d by Altenberg’s Lemma which will be shownlater.)

Lemma 3.3 (Altenberg’s result). If the function f(x, y) : (0,∞) × [0,∞) → (−∞,∞)satisfies:1, For every fixed y, f is convex in x,2, f(αx, αy) = αf(x, y), α > 0.Then ∀x, f is convex in y.

Proof. (Altenberg, 2012, PNAS) Fix any y > 0. Take y1, y2, y1 = y2 and δ ∈ (0, 1). Bycondition 1, we have

f(x, (1− δ)y1 + δy2) ≤ (1− δ)f(x, y1) + δf(x, y2).

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Using condition 2 on both sides,

(1− δ)y1 + δy2y

f(xy

(1− δ)y1 + δy2, y) ≤ (1− δ)y1

yf(xy

y1, y) +

δy2yf(xy

y2, y).

Define x1 =xyy1, x2 =

xyy2.

Claim: ∃h ∈ (0, 1) such that

xy

(1− δ)y1 + δy2= (1− h)x1 + hx2 ⇔ h =

my2(1− δ)y1 + δy2

∈ (0, 1).

Define ϕ = (1−δ)y1+δy2y

, then y1y(1− δ) = (1− h)ϕ, y2

yδ = hϕ. Hence,

ϕf((1− h)x1 + hx2, y) ≤ (1− h)ϕf(x1, y) + hϕf(x2, y).

Dived by ϕ,

f((1− h)x1 + hx2, y) ≤ (1− h)f(x1, y) + hf(x2, y).

Corollary 1. σ1 is concave in d.

Proof. Because σ1(αd, αλ) = ασ1(d, λ) and σ1 is concave in λ, we get the conclusiondirectly by Altenberg’s result.

Population size d∆θd + θd(m(x)− θd) = 0, in Ω,

∇θd · −→n = 0, on ∂Ω,

θd > 0, in Ω.

• m > 0, m ≡ constant,m ∈ Cα(Ω).

Then limd→0

θd = m, limd→∞

θd =1|Ω|

∫Ωm in L∞

Corollary 2. limd→0

∫Ωθd =

∫Ωm, lim

d→∞

∫Ωθd =

∫Ωm.

It suffices to prove∫Ωθd >

∫Ωm.

∆θdθd

+m− θd = 0∫Ω

|∇θd|2

θ2d+

∫Ω

m−∫Ω

θd = 0 ⇒∫Ω

θd >

∫Ω

m, ∀d > 0.

Remark 3.3. (He, Ni) There exists m, such that minΩ θd >1|Ω|

∫Ωm for large d.

V.Hutson: He constructed an example of∫Ωθd with exactly two local maximum.

Theorem 3. There exists smooth function g(x) with∫Ωg = 0 such that for m = 1+εg, 0 <

|ε| << 1, has a strict local minimum for d ∈ (0,∞).


Proof. Since m = 1 + εg, then θ = 1 + εθ1 + ε2θ2 + O(ε3). Let λ0 = 0 < λ1 ≤ λ2 · · · beeigenvalues of −∆ with zero Neumann boundary condition, and φi∞i=0 be correspondingeigenfunctions, normalized

∫Ωφ2i dx = 1. Expand g = Σ∞

i=1giφi, then θ = |Ω| + ε2∫θ2 +

O(ε3), and

θ = |Ω|+ ε2Σ∞i=1

g2i dλi(1 + dλi)2

+O(ε3).

Choose g(x) = gpφp + grφr, where p, r are chosen such that λr ≥ 16λp. Denote

G(d) :=g2pdλp

(1 + dλp)2+

g2rdλr(1 + dλr)2

,

then

G′(d) =g2p(1− dλp)λp

(1 + dλp)3+g2r(1− dλr)λr(1 + dλr)3

.

Since λr ≥ 16λp, we can choose λ ∈ (λp, λr) such that 4λp ≤ λ ≤ 14λr. To construct a

local minimum of G, we want

G′(1

lambda) = 0,

and

G′′(1

lambda) > 0.

G′(1

lambda) = 0 ⇔

g2p(1− dλp)λp

(1 + dλp)3+g2r(1− dλr)λr(1 + dλr)3

= 0.

Now, choose gp, gr such that G′( 1λ) = 0, we find that

G′′(1

λ) =

2g2rλr(λp − λr)(2 +2λpλr

λ2 − λp

λ− λr

λ)

(1 + λrλ)4(1− λp

λ)(1 + λp

λ)

< 0.

In fact, just note that 4λp ≤ λ ≤ 14λr implies

λp − λr < 0

and

2 +2λpλr

λ2 − λp

λ− λr

λ< 2 +

2λpλr

λ2 − λp

λ< (2− λr

2λ) +

λr

λ(2λr

λ− 1

2) < 0.

References

[1] L. Altenberg, Resolvent Positive Linear operators exhibit the reduction phenomennon. Proceedingof the National Academy of Science, USA, 109(10) (2012):3705-3710.

[2] R. S. Cantrell, C. Cosner, Spacial Ecology via reaction-diffusion equations, Series in Mathematicaland Computational Biology, John Wiley and Sons, CHichester, UK, 2003.

[3] V. Hutson, Personal communications.[4] S. Liang, Y. Lou, On the dependence of the population size onthe dispersal rate, DCDS-A,

17(2012):2771-2788.

8 YUAN LOU

[5] W.-M. Ni, The mathematics of Diffusion, CBMS Reg. Conf. Ser. Appl. Math. 82, SIAM, Philadelphia,2011.


Yuan Lou

LECTURE 2

1. Homogeneous space

ut = d∆u+ u(1− u), in Ω× (0,∞),

∇u · −→n = 0, on ∂Ω× (0,∞),

u(x, 0) ≥ 0, ≡ 0.

⇒ u(x, t) → 1 in L∞(Ω), as t→ ∞.

System: Lotka-Volterra modelut = d1∆u+ u(a1 − b1u− c1v), in Ω× (0,∞),

vt = d2∆v + v(a2 − b2u− c2v), in Ω× (0,∞),∂u∂−→n = ∂v

∂−→n = 0, on ∂Ω× (0,∞),

u(x, 0) = u0(x) ≥ 0, ≡ 0, v(x, 0) = v0(x) ≥ 0, ≡ 0.

Special case: b2 = c1 = 0 (reduced to single equation) Weak competition: ai, bi, ci > 0

(1.1)a1a2

>b1b2>c1c2(⇔ 0 < b2 < b1 ·

a2a1, 0 < c1 < c2 ·

a1a2

)

Theorem 1. (Hsu) Suppose (1.1) holds, limt→∞

(u(x, t), v(x, t)) = (u∗, v∗)

u∗ =a2c1 − a1c2b2c1 − b1c2

, v∗ =a2b1 − a1b2b2c1 − b1c2

.

Goal:

(1.2)

ut = d1∆u+ u(a1(x)− b1(x)u− c1(x)v), in Ω× (0,∞),

vt = d2∆v + v(a2(x)− b2(x)u− c2(x)v), in Ω× (0,∞).

Consider two ODE1

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dU

dt= U(a1 − b1U − c1V ),

dV

dt= V (a2 − b2U − c2V ),

U(0) = maxΩ

u(x, 0), V (0) = minΩv(x, 0).

(Without loss of generality, assume u(x, 0) > 0, v(x, 0) > 0)dU

dt= U(a1 − b1U − c1V ),

dV

dt= V (a2 − b2U − c2V ),

U(0) = minΩu(x, 0), V (0) = min

Ωv(x, 0).

By comparison principle for 2-species competition model

U(t) ≤ u(x, t) ≤ U(t),

with

limt→∞

U(t) = limt→∞

U(t) = u∗,

and

V (t) ≤ v(x, t) ≤ V (t),

with

limt→∞

V (t) = limt→∞

V (t) = v∗.

Thus

limt→∞

(u(x, t), v(x, t)) = (u∗, v∗).

Consider ai(x), bi(x), ci(x).

2. Large diffusion

Intuitively, PDE system (1.2) with d1, d2 >> 1 behaves likedU

dt= U(a1 − b1U − c1V ),

dV

dt= V (a2 − b2U − c2V ),

From now on, we will use the notation f := 1|Ω| infΩ fdx.


Theorem 2. If b1b2> a1

a2> c1

c2,∀x ∈ Ω, then if both d1 and d2 are sufficiently large, then

(1.2) has a positive steady state, denoted by (U , V ), which is globally asymptotic stable,and

limd1,d2→∞

(U(x), V (x)) = (a2c1 − a1c2b2c1 − b1c2

,a2b1 − a1b2b2c1 − b1c2

).

3. Small diffusion(d1, d2 << 1)

Intuitively, (1.2) behaves asdU

dt= U(a1(x)− b1(x)U − c1(x)V ),

dV

dt= V (a2(x)− b2(x)U − c2(x)V ),

limt→∞

(U , V ) = (u∗, v∗).

Theorem 3. Assume weak competition. If d1 and d2 are sufficiently small, then (1.2)

has a unique positive steady state (U(x), V (x)) which is globally asymptotic stable. Fur-thermore,

limd1,d2→∞

(U(x), V (x)) = (u∗, v∗)inC(Ω)× C(Ω),

ai(x) = r(x)ai, bi(x) = r(x)bi, ci(x) = r(x)ci where ai, bi, ci are positive constants.( Forthis special case, u∗, v∗ are constant.)

Interesting eigenvalue problem:−d∆φ+ q(x)φ = λ1φ, in Ω

∇φ · −→n = 0, on ∂Ω,

φ > 0 in Ω.

Theorem 4. limd→0 λ1 = minΩq(assume that q ∈ C(Ω)).

−d1∆φ+ q11(x)φ+ q12(x)ψ = λ1φ, in Ω

−d2∆ψ + q21(x)φ+ q22(x)ψ = λ1ψ, in Ω

∇φ · −→n = ∇ψ · −→n = 0, on ∂Ω,

φ > 0, ψ > 0 in Ω.

Question: If q12(x) > 0, q21(x) > 0.Claim:

limd1,d2→∞

λ1 = minx∈Ω

λ∗(q(x)),

where q(x) := (qij)2×2 and λ∗(q(x)) denote the smaller eigenvalue of matrix q(x). (K.-Y.Lam, Y. Lou)

4 YUAN LOU

proof of Theorem 3. Step 1,Let (u(x), v(x)) be any positive steady state of (1.2). Thenas d1 → 0, d2 → 0. (u(x), v(x)) → (u∗(x), v∗(x)) in L∞(Ω)× L∞(Ω).

Idea: Hutson, Lpez-Gomez, Mischaikow, Vickers(1995).Step 2, Show that any positive steady state of (1.2) is locally stable, provided d1, d2 <<

1. The linearized problem at the steady state (u, v):d1∆φ+ (a1 − 2b1u− c1v)φ+ (−c1u)ψ + λ1φ = 0, in Ω

d2∆ψ + (−b2v)φ+ (a2 − b2u− 2c2v)ψ + λ1ψ = 0, in Ω

∇φ · −→n = ∇ψ · −→n = 0, on ∂Ω,

Formally,

≈

d1∆φ+ (a1 − 2b1u

∗ − c1v∗)φ+ (−c1u∗)ψ + λ1φ = 0, in Ω

d2∆ψ + (−b2v∗)φ+ (a2 − b2u∗ − 2c2v

∗)ψ + λ1ψ = 0, in Ω

∇φ · −→n = ∇ψ · −→n = 0, on ∂Ω,

⇔

d1∆φ+ (−b1u∗)φ+ (−c1u∗)ψ + λ1φ = 0, in Ω

d2∆ψ + (−b2v∗)φ+ (−c2v∗)ψ + λ1ψ = 0, in Ω

∇φ · −→n = ∇ψ · −→n = 0, on ∂Ω,

Then calculate eigenvalues of

(b1u

∗ c1u∗

b2v∗ c2v

∗

)and apply the claim.

Step 3,(Monotone dynamical system theory, Smith, 1995)⇒ ∃! positive steady state which is globally stable.

4. Intermediate diffusion

4.a Evolution of slow dispersal.a1(x) = a2(x) := m(x) > 0, b1 = c2 = 1,

ut = d1∆u+ u(m(x)− u− bv), in Ω× (0,∞),

vt = d2∆v + v(m(x)− cu− v), in Ω× (0,∞),∂u∂−→n = ∂v

∂−→n = 0, on ∂Ω× (0,∞),

u(x, 0) = u0(x) ≥ 0, ≡ 0, v(x, 0) = v0(x) ≥ 0, ≡ 0.

Weak competition, 0 < b < 1, 0 < c < 1.An interesting case: b = c = 1. (Dckory et. al. , 1998, J. Math. Biol.)

Theorem 5. Suppose that m > 0 in Ω, Holder continuous, non-constant. If d1 < d2,then (u∗, 0) is global asymptotic stable, where u∗ is the unique positive solution of

d1∆u+ u(m(x)− u) = 0, in Ω× (0,∞),

∇u · −→n = 0, on ∂Ω× (0,∞).


Proof. Step 1. Instability of (0, v∗)d2∆v

∗ + v∗(m− v∗) = 0, in Ω× (0,∞),

v∗ > 0, in Ω,

∇v∗ · −→n = 0, on ∂Ω× (0,∞).

The linearized problem is

⇔

d1∆φ+ (m− v∗)φ+ λ1φ = 0, in Ω

d2∆ψ + (−v∗)φ+ (m− 2v∗)ψ + λ1ψ = 0, in Ω

∇φ · −→n = ∇ψ · −→n = 0, on ∂Ω,

The stability of (0, v∗) is determined by the sign of the smallest eigenvalue of

⇔

d1∆φ+ (m− v∗)φ+ λ1φ = 0, in Ω

∇φ · −→n = 0, on ∂Ω,

The sign of λ1(d1, d2) Denote λ′1 =

∂λ1∂d1,∫Ωφ2 = 1

φ′ = ∂φ∂d1

(A. Lazer’s idea: φ is differentiable w.r.t. to d1.)∫Ω

φ[∆φ+ d1∆φ′ + (m− v∗)φ′ + λ′1φ+ λ1φ

′] = 0

−∫Ω

|∇φ|2 +∫Ω

φ′[d1∆φ+ (m− v∗)φ+ λ1φ] +

∫Ω

λ′1φ2 = 0,

λ′1 =∫Ω|∇φ|2 > 0. (A. Hastings, Theo. Pop. Biol. 1983)

Remark 4.1. λ1(d1, d2)-Invasion exponentlimd1→0 λ1 = min

Ω(v∗ −m) < 0,

limd1→∞ λ1 =1|Ω|

∫Ω(v∗ −m) > 0,

limd2→0 λ1 = 0 (d2 → 0 ⇒ v∗ → m)∀d1 > 0, λ1(d1, d2) has a local maximum, 0 < d2 < d1.

Question: What’s are biological meaning of such maximum?Step 2 lim supt→∞ V (x, t) ≤ v∗(x).

vt = d2∆v + v(m(x)− u− v) ≤ d2∆v + v(m(x)− v).

Consider ∂V∂t

= d2∆V + V (m− V ), in Ω× (0,∞),

V (x, 0) = V (x, 0), in Ω,

∇V · −→n = 0, on ∂Ω× (0,∞).

By comparison principle, V (x, t) ≤ V (x, t), and hence

lim supt→∞

V (x, t) ≤ lim supt→∞

V (x, t) = v∗(x).

6 YUAN LOU

Step 3 By Step 2, ∀ε > 0, ∃T1(ε) such that

V (x, t) ≤ (1 + ε)v∗(x).

Consider Ut = d1∆U + U(m(x)− U − V ), in Ω× [T1,∞),

Vt = d2∆V + V (m(x)− U − V ), in Ω× [T1,∞),

U(x, T1) = δφ, V (x, T1) = (1 + ε)v∗, ≡ 0.

Claim: (δφ, (1 + ε)v∗) is a sub-super solution.

d2∆[(1 + ε)v∗] + (1 + ε)v∗(m− δφ− (1 + ε)v∗) = (1 + ε)[d2∆v∗ + v∗(m− v∗)− (δφ+ εv∗)v∗]

≤ 0

d1∆(δφ) + δφ(m− δφ− (1 + ε)v∗) ≥ 0

⇔d1∆φ+ φ(m− v∗) + φ(−δφ− εv∗) ≥ 0

⇔− λ1φ+ φ(−δφ− εv∗) ≥ 0 ( since λ1 < 0, choose δ, ε small)

⇒U(x, t) is increasing in t, V (x, t) is decreasing in t

the last result is a consequence of comparison principle for competitive system.Step 4 limt→∞(U, V ) = (u∗, 0). It suffices to show that there is no co-existence state.

This follows from comparison principle for eigenvalues.Step 5 U(x, T1) = δφ ≤ u(x, T1), V (x, T1) ≤ (1 + ε)v∗ = V (x, T1)

⇒ U(x, t) ≤ u(x, t), v(x, t) ≤ V (x, t)

⇒ limt→∞

v(x, t) = 0, limt→∞

u(x, t) = u∗.

Open problems:1 a), Three or more species∂ui∂t

= di∆ui + ui(m(x)− Σkj=1uj), in Ω× (0,∞),

∂ui∂−→n = 0, on ∂Ω× (0,∞),

k ≥ 3, i = 1, 2, · · · , k.

Conjecture: limt→∞ ui(x, t) =

u∗, i = 1,

0, i ≥ 2.

1 b), 2-species, d1 < d2. No Lyapunov functional has been found.02, Add mutations2 a), Discrete traits

∂ui∂t

= di∆ui + ui(m(x)− Σkj=1uj) + εΣk

j=1mijuj.

Dockory et. al. (1998, JMB)k = 2, 0 < ε << 1, d1 < d22 b), continuous traits (B. Perthame)


∂u

∂t= µ∆u+ u(m(x)−

∫ D

0

u(x, t;µ)dµ) + ε∂2u

∂µ2.

µ: trait,∫ D0u(x, t;µ)dµ: competition among all phenotypes, ε∂

2u∂µ2

: mutation.

Question: limt→∞ u(x, t;µ) =? (slowest diffuser wins)∇xu · −→n = 0 on ∂Ω ∂u

∂µ(0) = ∂u

∂µ(D) = 0.

5. Heterogeity vs. Homogeneity

(5.1)

ut = d1∆u+ u(m(x)− u− v), in Ω× (0,∞),

vt = d2∆v + v(m− u− v), in Ω× (0,∞),∂u∂−→n = ∂v

∂−→n = 0, on ∂Ω× (0,∞),

u(x, 0) = u0(x) ≥ 0, ≡ 0, v(x, 0) = v0(x) ≥ 0, ≡ 0.ut = d1∆u+ u(r − u− v)− h(x)u, in Ω× (0,∞),

vt = d2∆v + v(r − u− v)− hv, in Ω× (0,∞),∂u∂−→n = ∂v

∂−→n = 0, on ∂Ω× (0,∞),

u(x, 0) = u0(x) ≥ 0, ≡ 0, v(x, 0) = v0(x) ≥ 0, ≡ 0.

h(x)u: inhomogeneous harvesting, hu: homogeneous harvesting.Dynamics of (5.1)

Theorem 6. (He, Ni) Suppose that m ≥ 0, m non-constant and m is Holder continuous.(a) (0, v∗) is unstable for any d1, d2 > 0.(b) ∃ a continuous function d∗ := d∗(d1) : (0,∞) → [0,+∞), such that

(u∗, 0) =

stable, d2 > d∗2,

unstable, d2 < d∗2.

Moreover,

limd1→0+

d∗ = +∞, limd1→∞

d∗ = 0.

u∗ is defined as the solution of the following problemd1∆u

∗ + v∗(m− u∗) = 0, in Ω× (0,∞),

∇u∗ · −→n = 0, on ∂Ω× (0,∞).

Open problem: If d1 = d2 := d, ∃d > 0, such that if 0 < d < d, there exists a uniquepositive steady state which is globally stable, for d > d, then (u∗, 0) is globally stable?(True for Two-path model, 2-species model.)

8 YUAN LOU

6. Time-periodic environment

(2001, Hutson et. al. , JMB)ut = d1∆u+ u(m(x)− u− v), in Ω× (0,∞),

vt = d2∆v + v(m(x)− u− v), in Ω× (0,∞).

Results: The following three possibilities can occur:(i) Slower diffuser wins;(ii) Faster diffuser wins;(iii) Co-existence of 2-species.

7. Evolution of faster dispersal

ut = d1uxx − qux + u(1− u− v), 0 < x < L, t > 0,

vt = d2vxx − qvx + v(1− u− v), 0 < x < L, t > 0,

d1ux(0)− qu(0) = d2vx(0)− qv(0) = 0,

ux(L) = vx(L) = 0.

q: water speed drift x = 0: upper stream end, x = L: down stream end.Single species

(7.1)

duxx − qux + u(1− u) = 0, 0 < x < L,

dux(0)− qu(0) = 0, ux(L) = 0.

Theorem 7. (Vasilyeva, Lutscher, 2011, Can. Appl. Math. Q.) ∃ ! q∗ > 0 such that(7.1) has a positive solution ⇔ q < q∗.

Theorem 8. ( Lutscher, 2013) Suppose that d1 > d2. Then (u∗, 0), if exists, is globallyasymptotic stable.

Faster diffuser wins!!1 > u∗ > 0, limd1→∞ u∗ = 1.Conjecture Replace boundary condition at x = L by u(L) = v(L) = 0. ∃d∗ > 0 such

that if d1 = d∗, d2 = d∗, then (u∗, 0) is globally stable.

References

[1] S.B. Hsu, Limiting behavior for competing species, SIAM J. Appl. Math. 34 (1978) 760-763. TheLyapunov functional (page 2-2) can be found in this paper.

[2] W.-M. Ni, The Mathematics of Diffusion, CBMS Reg. Conf. Ser. Appl. Math. 82, SIAM, Philadel-phia, 2011. The proof of Theorem 1 (page 2-4) is given in this book.

[3] R.S. Cantrell and C. Cosner, Spatial Ecology via Reaction-Diffusion Equations, Series in Math-ematical and Computational Biology, John Wiley and Sons, Chichester, UK, 2003. Reference forcomparison principles for reaction-diffusion equations and systems.


[4] V. Hutson, Y. Lou, and K. Mischaikow, Convergence in competition models with small diffusioncoefficients, J. Diff. Eqs. 211 (2005) 135-161. The proof of Theorem 2 (page 2-5) can be found in thispaper.

[5] K.Y. Lam and Y. Lou, Asymptotic behavior of the principal eigenvalue of cooperative system withapplications, preprint. The proof of Claim (page 2-6) can be found in this paper.

[6] H. Smith, Monotone Dynamical Systems. Mathematical Surveys and Monographs 41. AmericanMathematical Society, Providence, Rhode Island, U.S.A., 1995. Reference for Monotone dynamicalsystem theory.

[7] J. Dockery, V. Hutson, K. Mischaikow, and M. Pernarowski, The evolution of slow dispersal rates:a reaction-diffusion model, J. Math. Biol. 37 (1998) 61-83. Proof of Theorem (page 2-8) and discretemutation model (page 2-12) can be found in this paper.

[8] Xiaoiqng He and Wei-Ming Ni, The effects of diffusion and spatial variation in Lotka-Volterracompetition-diffusion system, I: Heterogeneity vs. homogeneity, J. Diff. Eqs. 254 (2013) 528-546.Proof of Theorem (page 2-13) can be found in this paper.

[9] Xiaoiqng He and Wei-Ming Ni, The effects of diffusion and spatial variation in Lotka-Volterracompetition-diffusion system, II: The general case, J. Diff. Eqs. 254 (2013) 4088-4108. This paperdiscusses competition for general resources.

[10] V. Hutson, K. Mischaikow, and P. Polacik, The Evolution of Dispersal Rates in a HeterogeneousTime-Periodic Environment (The final version can be found in J Math Biol 43 (2001) 6, 501-533.This paper discusses evolution of dispersal in time-periodic environment.

[11] O. Vasilyeva and F. Lutscher. Population dynamics in rivers: analysis of steady states. Can. Appl.Math. Quart., 18(4):439-469, 2011. Theorem (page 2-15) for single species is from this paper.

[12] Y. Lou and Frithjof Lutscher, Evolution of dispersal in advective environments, preprint. Theorem(page 2-16) for 2-species is from this paper.


Yuan Lou

LECTURE 3

1. Single species

(1.1)

ut = duxx − qux + ru(1− u), 0 < x < L, t > 0,

dux(0)− qu(0) = 0, ux(L) = 0.

Theorem 1. ∃ q∗ > 0 such that if 0 ≤ q < q∗ (1.1) has a unique positive steady statewhich is global stable.

Intuition: the speed should not too large.

sketch of proof. The key is to determine the stability of u = 0.dφxx − qφx + rφ+ λ1φ = 0, 0 < x < L,

dφx(0)− qφ(0) = 0, φx(L) = 0.

Set φ = elxψφx = elx(ψx + lψ)

φxx = elx(ψxx + 2lφx + l2ψ2)

delx(ψxx + 2lφx + l2ψ2)− qelx(ψx + lψ) + relxψ + λ1elxψ = 0.

dψxx + (2dl − q)ψx + ψ(dl2 − ql + r + λ1) = 0.

Let l = q2d

dψxx + ψ(− q2

4d+ r + λ1) = 0, 0 < x < L,

ψx(0)− q2dψ(0) = ψx(L) +

q2dψ(L) = 0.

ψ(x) = A cos(

√4d(r + λ1)− q2

2dx) +B sin(

√4d(r + λ1)− q2

2dx).

ψx(0)−q

2dψ(0) = 0 ⇒ A = B

√4d(r + λ1)− q2

q

ψx(L) +q

2dψ(L) = 0 ⇒ 0 = A

−√4d(r + λ1)− q2

2dsin(

√4d(r + λ1)− q2

2dL)

1

2 YUAN LOU

+B

√4d(r + λ1)− q2

2dcos(

√4d(r + λ1)− q2

2dL)

+q

2d[A cos(

√4d(r + λ1)− q2

2dL) +B sin(

√4d(r + λ1)− q2

2dL)]

Combining these two boundary conditions,

tan(

√4d(r + λ1)− q2

2dL)[

q

d− 2(r + λ1)

q] +

√4d(r + λ1)− q2

d= 0.

Let λ1 = 0,

tan(

√4dr − q2

2dL∗) =

q√4dr − q2

2rd− q2, L∗ : critical length.

Claim:

λ1 =

+, L > L∗;

0, L = L∗;

−, L < L∗.

Biological meaning: The species can persist if 0 ≤ q < 2√rt (slow flow) and L > L∗(

longer river).

4dr − q2 > 0 ⇔ q < 2√rd.0

L∗(q) =

arctan( q

√4rd−q2

2rd−q2)

√4rd−q2

2d

, 0 ≤ q ≤√2rd,

π+arctan( q√

4rd−q2

2rd−q2)

√4rd−q2

2d

,√2rd < q < 2

√rd.

Clearly, this method provides an estimate on q∗, i.e., q∗ < 2

√rd.

Second method for determining q∗.dφxx − qφx + r(x)φ+ λ1φ = 0,

dφx(0)− qφ(0) = 0, φx(L) = 0.

Claim: ∂λ∂q> 0.

Intuition: q increase ⇔ harder for species to persist ⇔ u = 0 more likely stable ⇔ λ1increase.

By direct calculations, set ϕ = eqdxψ. Then

∂λ

∂q=e

qdL ψ

2(L)2

+ ψ2(0)2

+ qd

∫ L0e

qdx ψ2

2∫ L0e

qdxψ1

> 0

Furthermore, limq→∞ λ1 = +∞. If λ1(0) ≥ 0 ⇒ λ1(q) > 0 for all q > 0.


If λ1(0) < 0, then ∃!q∗ such that

λ1(q) =

−, 0 ≤ q < q∗,

0, q = q∗,

+, q > q∗.

Remark 1.1. When r is constant, q = 0, λ1(0) = −r < 0.

Claim: For any r(x) satisfying λ1(0) < 0, then ∃q∗ > 0 such that

λ1(q) =

− , 0 ≤ q < q∗, (⇔ u = 0 unstable.)

0 , q = q∗,

+ , q > q∗.ut = d1uxx − qux + u(1− u− v), 0 < x < L, t > 0,

vt = d2vxx − qvx + v(1− u− v), 0 < x < L, t > 0,

d1ux(0)− qu(0) = d2vx(0)− qv(0) = 0,

ux(L) = vx(L) = 0.

Theorem 2. If d1 > d2, then (u∗, 0), whenever it exists, is globally stable. ( The fasterdiffuser wins!)

Idea: (u∗, 0) is locally stable for d1 ≈ d2, d1 > d2. The stability of (u∗, 0) is determinedby

d2φxx − qφx + (1− u∗)φ+ λ1(d1, d2)φ = 0, 0 < x < L,

d2φx(0)− qφ(0) = 0, φx(L) = 0,

φ > 0 in (0, L).

where u∗ satisfies

(1.2)

d1u

∗xx − qu∗x + u∗(1− u∗) = 0, 0 < x < L,

d1u∗x(0)− qu∗(0) = u∗x(L) = 0,

u∗ > 0 in (0, L).

Calculate∂λ1∂d2

|d2=d1 =∫ L0(e

− qd1xu∗)xu

∗dx∫ L0e− q

d1x(u∗)2dx

< 0.

(Adaptive Dynamics)Claim: 0 < u∗(x) < 1, u∗x > 0, 0 ≤ x ≤ L.

d1u∗xx − qu∗x < 0, 0 < x < L⇒ d1(u

∗x −

q

d1u∗)x < 0

⇒ u∗x −q

d1u∗ is decreasing.

4 YUAN LOU

Since u∗x(0)−qd1u∗(0) = 0, then u∗x −

qd1u∗ < 0, 0 < x ≤ L.

(e− q

d1xu∗)x = e

− qd1x(u∗x −

q

d1u∗) < 0.

Remark 1.2. 1, q = 0

(1.3)

ut = d1uxx + u(1− u− v), 0 < x < L, t > 0,

vt = d2vxx + v(1− u− v), 0 < x < L, t > 0,

ux = vx = 0, at 0, L.

(τ, 1 − τ) : 0 < τ < 1 are positive steady states. The asymptotic behavior of solutionsof (1.3) are unknown. It depends upon initial data.

2, r=r(x): Lots of open problems.

2. Diffusion driven extinction

(2.1)

ut = d1∆u+ u(m(x)− u− cv), 0 < x < L, t > 0,

vt = d2∆v + v(m(x)− bu− v), 0 < x < L, t > 0,∂u∂−→n = ∂v

∂−→n = 0 on ∂Ω.

Theorem 3. (Hsu) 0 ≤ b, c < 1. If m(x) = positive constant, (2.1) has a positive steadystate for any d1, d2 which is global stable.

Assumption: m > 0, m Holder continuous and m is non-constant.Two cases(discussed earlier) (1) d1, d2 >> 1,(2) d1, d2 << 1.

Theorem 4. For every c ∈ (c∗, 1), where c∗ = inf0<d<∞

∫Ωm∫Ω θd

, there exists b∗ positive small

such that for 0 < b < b∗, (θd, 0) is globally stable for some ranges of d1, d2.

Idea: b = 0 ut = d1∆u+ u(m(x)− u− cv), 0 < x < L, t > 0,

vt = d2∆v + v(m(x)− v), 0 < x < L, t > 0.

⇒ v = θd2 ⇒

d1∆u+ u(m(x)− cθd2 − u) = 0 in Ω∂u∂−→n = 0 on ∂Ω.

has a positive solution iff u = 0

is unstable.Consider the eigenvalue problem

d1∆φ+ φ(m(x)− cθd2) + σ1φ = 0, in Ω∂φ∂−→n = 0 on ∂Ω.

σ1 = σ1(c, d1, d2) smallest eigenvalue.Claim: m− cθd2 is positive somewhere.


Proof. m− cθd2 ≥ m− θd2 ,∫Ω∆θd2 +

∫Ωθd2(m− θd2) = 0

⇒∫Ω

θd2(m− θd2)0 ⇒ m− θd2 change sign ⇒ m− θd2 is positive somewhere.

Claim:

∫Ω(m− cθd2) < 0 for some range of d2, provided that C > C∗ = inf

d2>0

∫Ωm∫Ω θd2

.

For d2 ∈ (d2, d2), C > infd2>0

∫Ωm∫Ω θd2

⇔∫Ω(m− cθd2) < 0, ∀d2 < d2 < d2.

By the two claims, we have

Lemma 2.1. If c ∈ (c∗, 1), d2 ∈ (d2, d2), then the following problem∆ψ + µ(m(x)− cθd2)ψ = 0, in Ω∂ψ∂−→n = 0 on ∂Ω,

ψ > 0 in Ω.

has a positive eigenvalue (µ = 0 is always an eigenvalue.)

Denote previous eigenvalue by µ1 > 0 (µ1 = µ1(c, d2)) Go back to original eigenvalueproblem

d1∆φ+ (m− cθd2) + σφ = 0 in Ω

⇒ σ1 = σ1(d1, d2) satisfies σ1(1µ1, d2) ≡ 0, d2 < d2 < d2.

Lemma 2.2. ∂σ1∂d1

> 0 ⇒ σ1(d1, d2) =

+, d1 >

1µ1;

0, d1 =1µ1;

−, d1 < 1µ1.

when d2 < d2 < d2.

⇒ (0, θd2) is stable for d1 >1

µ1

, d2 ∈ (d2, d2)

Conjecture: For every c ∈ (c∗, 1), 0 < b ≤ 1, (0, θd2), if locally stable, then it isglobally stable.

1, Lam, Ni (SIAM,2012) The conjecture holds for small b. Roughly speaking, there aretwo alternatives:

(i) (0, θd2) locally stable ⇒ global stable.(ii) (0, θd2) locally unstable ⇒ ∃! positive steady state which is globally stable.2, Lam, NiFor any 0 ≤ b, c < 1,∃δ > 0 such that if |d1 − d2| < δ, then there is a unique positive

steady state that is globally stable. (This result connect d1, d2 << 1 case and d1, d2 >> 1case.)

Assumption 0 ≤ b, c < 1,m > 0 and m non-constant.Conjecture: ∃δ > 0 such that if d1 < delta or d2 < δ, then ∃! positive steady state

that is globally stable.

Remark 2.1. If d1, d2 < δ, then above holds.

6 YUAN LOU

Shadow System: d2 → ∞.

(2.2)

d1∆u+ u(m− u− v) = 0, in Ω∫Ω(m− cu) = |Ω|v,

v > 0.

(Work on (2.2): F. Li, L. Wang, Y. Wang, DCDS-B.)

3. Directed movement of single species

(1995, Belgacem, Cosner, Canadian Appl. Math. Quarterly)

ut = ∇ · [D∇u− αu∇m] + u(m(x)− u), α > 0.

No-flux boundary condition

[D∇u− αu∇m] · −→n = 0, on ∂Ω.

In their paper, Belgacem, Cosner also considered Dirichlet boundary condition.

Theorem 5. m ∈ C2(Ω) non-constant, positive somewhere. If u ≡ 0 is locally stable,then it is globally stable; If u ≡ 0 is unique positive steady state that is globally stable.

I. stability of u ≡ 0.The stability of u ≡ 0 is determined by the smallest eigenvalue of

∇ · [D∇φ− αφ∇m] +mφ+ σ1φ = 0 in Ω

[D∇φ− αφ∇m] · −→n = 0 on ∂Ω.

Question: limα→∞ σ1 =?(Asymptotic behavior of eigenvalues with large advection rate)Divide both sides by D ( α

D→ α, σ1

D→ σ1, D = 1

λ)

∇ · [D∇φ− αφ∇m] + λmφ+ σ1φ = 0 in Ω

[D∇φ− αφ∇m] · −→n = 0 on ∂Ω.

Theorem 6. ∃λ∗ ≥ 0 such that(i) u = 0 is unstable for λ > λ∗;(ii) u = 0 is stable for 0 < λ ≤ λ∗.

λ = λ∗ ⇔ σ1 = 0

⇔

∇ · [D∇φ− αφ∇m] + λ∗mφ = 0 in Ω

[D∇φ− αφ∇m] · −→n = 0 on ∂Ω.

⇔ λ∗ = infψ∈W 1,2,

∫Ωme

αmψ2>0

∫Ωeαm|∇ψ|2∫

Ωmeαmψ2

.


Cosner: Is λ∗ decreasing in α?Species can persist iff λ ∈ (λ∗,∞)Cosner’s intuition:

α ⇒ λ∗ ⇒ (λ∗,∞) species more likely to persist.

Conjecture of Cosner: dλ∗dα

≤ 0.

Remark 3.1. 1, If∫Ωm ≥ 0, then λ∗ ≡ 0,∀α ≥ 0.∫

Ω

m ≥ 0 ⇒∫Ω

meαm > 0,∀α ≥ 0.⇒ λ∗,∀α > 0.

2,∫Ωm < 0.

d

dα

∫meαm =

∫m2eαm > 0.⇒ ∃!α > 0 such that

∫Ω

meαm =

+, α > α,

0, α = α,

−, α < α.

⇒ λ∗ =

0, α ≥ α,

+, 0 < α < α.

Theorem 7. (Cosner, Lou)1, If Ω is convex, dλ∗

dα|α=0 < 0,∀m,

∫m < 0.

2, For some non-convex Ω and some m, dλ∗dα

|α=0 > 0.

Sketch of proof. Step 1, Calculate dλ∗dα

|α=0.

dλ∗dα

|α=0 < 0 = −∫ψ∇m · ∇ψ∫

mψ2

where ψ satisfies

(3.1)

−∆ψ = µm(x)ψ in Ω∂ψ∂−→n = 0 on ∂Ω,

ψ > 0 in Ω.

Since∫Ωm < 0, m change sign, we know that (3.1) has a positive eigenvalue µ > 0.

Claim: If Ω convex, for any m,∫Ωψ∇m · ∇ψ > 0.

One dimension

−ψxx = µmψ

8 YUAN LOU

Differentiate w.r.t. x,−ψxxx = µmxψ + µmψx∫ 1

0

ψx(−ψxxx) =∫ 1

0

µmxψψx +

∫ 1

0

µmψ2x

µ

∫ 1

0

ψψxmx =

∫ 1

0

ψ2xx − µ

∫ 1

0

mψ2x

µ = infφ∈W 1,2,

∫ 10 mφ

2>0

∫ 1

0φ2x∫ 1

0mφ2 > 0

⇒∫ 1

0

φ2x ≥ µ

∫ 1

0

mφ2

Choose φ = ψx, we get ∫ 1

0

ψ2xx > µ

∫ 1

0

mψ2x.

non-convex domain

Ω = (x1, x2) : 0 < x1 < 1, 0 < x2 < εa(x1).

∆x1,x2 ≈1

a(x1)

d

dx1(

1

a(x1)

d

dx1)

(J. Hale · · · lots of work for such thin domains)

m(x1, x2) = m(x1)

∇m = (m′(x1), 0)

⇒ directed movement may hit geographical barrier and may not be beneficial. For moredetails, see Cosner, Lou, JMAA, 2003.

References

[1] O. Vasilyeva and F. Lutscher. Population dynamics in rivers: analysis of steady states. Can. Appl.Math. Quart., 18(4):439-469, 2011.

[2] Y. Lou and Frithjof Lutscher, Evolution of dispersal in advective environments, preprint.[3] Y. Lou, On the effects of migration and spatial heterogeneity on single and multiple species, J. Diff.

Eqs. 223 (2006) 400-426.[4] K.Y. Lam and W-M Ni, Uniqueness and complete dynamics of the Lotka-Volterra competition

diffusion system, SIAM J. Appl. Math. 72 (2012), no.6, 1695-1712[5] F. Li, L.P. Wang and Y. Wang, On the effects of migration and inter-specific competions in steady

state of some Lotka-Volterra model, Disc. Cont. Dynam. Sys. Series-B 15 (2011) 669-686.[6] F. Belgacem. Elliptic boundary value problems with indefinite weights: variational formulations of

the principal eigenvalue and applications. Pitman Research Notes in Mathematics, vol. 368 (Long-man, 1997).

[7] F. Belgacem and C. Cosner. The effects of dispersal along environmental gradients on the dynamicsof populations in heterogeneous environment. Can. Appl. Math. Q. 3 (1995), 379-397.

[8] C. Cosner and Y. Lou. When does movement toward better environment benefit a population? J.Math. Analysis Applic. 277 (2003), 489-503.

[9] J. K. Hale and G. Raugel. Reaction-Diffusion equation on thin domains. J. Math. Pures Appl. 71(1992), 32-95.


Yuan Lou, Ohio State University

Lecture 7 Idea free distribution

1. Discrete models

Two key assumptions in Ideal Free Distribution theory (abbreviated as IFD: Fretwelland Lucas [8]):

• Organisms have complete knowledge about habitat (food, competitor, predators)• No cost associated with movement

Prediction: Organisms will aggregate proportionally to resource available.To motivate the theory of IFD and evolution of dispersal, we first consider a 2-patch

model [10].Let ui denote the density of population in patch i. Without dispersal, the dynamics of

ui is described byduidt

= ri(1−uiKi

), t > 0

where the unique positive equilibrium is given by (u∗1, u∗2) = (K1, K2). Note that

K1

u∗1=K2

u∗2(= 1),

i.e., without dispersal organisms will aggregate proportionally to resource available.With dispersal, the dynamics of ui is determined by

(1.1)

du1dt

= m21u2 −m12u1 + r1u1(1− u1K1

)du2dt

= m12u1 −m21u2 + r2u2(1− u2K2

),

where m12 > 0 is the rate of movement from patch 1 to 2, andm21 is the rate of movementfrom patch 2 to 1.

Question: What kind of (m12,m21) can ensure that equilibrium of (1.1) is ideal freedistribution? That is,

(1.2)K1

u1=K2

u2,

1

2 YUAN LOU

where (u1, u2) is the unique positive equilibrium of (1.1), i.e.,m21u2 −m12u1 + r1u1(1− u1

K1) = 0,

m12u1 −m21u2 + r2u2(1− u2K2

) = 0.

Claim: (1.2) holds if and only if m21K2 = m12K1. If (1.2) holds, then

K1

u1=K2

u2= c

for some c. Hence,

r1u1(1− c) + r2u2(1− c) = 0 ⇒ c = 1

⇒ u1 = K1, u2 = K2

⇒ m21K2 = m12K1.

It is easy to see that the converse also holds.

Definition 1.1. We say (m12,m21) is an ideal free strategy if m21K2 = m12K1.

Question: Can ideal free strategies be invaded by other strategies?Consider

dv1dt

= m21v2 − m12v1 + r1v1(1− u∗1+v1K1

),

dv2dt

= m12v1 − m21v2 + r2v2(1− u∗2+v2K2

),

where v1, v2 are the density of a mutant in patches 1 and 2. Recall that u∗1 = K1, u∗2 = K2

is the unique positive equilibrium when (m12,m21) is an ideal free strategy.Question: If (v1, v2) = (0, 0) stable or unstable?Let v1 = εeλtη1, v2 = εeλtη2, we find

λη1 = m21η2 − m12η1,

λη2 = m12η1 − m21η2.

We find that λ1 = 0, λ2 = −(m21 + m12). Hence, the Lyapunov method does not yield anaffirmative answer. We will shown later on that ideal free strategies can not be invadedby any non-ideal free strategies.

2. IFD for multiple species

The materials of this section can be found in Cantrell et al. [2].Let uki denote the density of species k in patch i. The dynamics of the organisms is

described by

dukidt

= Fki(u)uki +n∑j=1

[dkijukj − dkjiuki]

u = (u11, · · · , u1n, · · · , umn), 1 ≤ i ≤ n, 1 ≤ k ≤ m.n: total number of patches; m: total number of species.


dkij: rate of movement from patch j to i for species k. Note that the notation ofmovement rate is different from two patch case.

Remark 2.1. Note that the disperse operator can be rewritten as

•∑n

j=1,j =i[dkijukj − dkdiuki];

•∑

j akijuj, aij =

dkij, j = i,

−∑n

l=1,k =i dkli, j = i.

Let u∗ be an asymptotically stable equilibrium of

dukidt

= Fki(u)uki.

The following general definition for multiple species is given by Cantrell et al [2]:

Definition 2.1. We say (dkij)1≤i,j≤n, 1 ≤ k ≤ m is ideal free if

n∑j=1

[dkiju∗kj − dkdiu

∗ki] = 0, ∀1 ≤ i ≤ n, 1 ≤ k ≤ m.

3. Two-species competition model

The following model includes the 2-patch model in Section 1.duidt

= riui(1− ui+viKi

) +∑n

j=1[dijuj − djiui],

dvidt

= rivi(1− ui+viKi

) +∑n

j=1[Dijvj −Djivi].

ui: density of species 1 in patch i; vi: density of species 2 in patch i.Assumptions: (dij) is ideal free but (Dij) is not ideal free, i.e.1.∑n

j=1 dijKj =∑n

j=1 djiKi,∀1 ≤ i ≤ n.

2.∑n

j=1DijKj =∑n

j=1DjiKi, for at least one i.Example: n = 2, i = 1, d11K1 + d12K2 = d11K1 + d21K1 ⇔ d12K2 = d21K1.i = 2, d21K1 + d22 = d12K2 + d22K2 ⇔ d12K2 = d21K1. This shows that for 2-patch

case, our assumptions are equivalent to

d12K2 = d21K1, D12K2 = D21K1.

Theorem 1. ([4]) Suppose that (dij), (Dij) are non-negative and irreducible. Then (u∗, 0)is globally stable, where u∗ = (K1, · · · , Kn).

Lemma 3.1. ([7]) Let A = (aij)n×n be n × n non-negative matrix. Then the followingstatements are equivalent:

(i)∑n

j=1 aij =∑n

j=1 aji,∀1 ≤ i ≤ n.

(ii)∑

i,j aijxixj

≥∑

i,j aij,∀xi > 0.

4 YUAN LOU

Proof. (i)⇒(ii) (Z. Shuai)∀y > 0, y ≥ 1 + ln y ⇒ xi

xj≥ 1 + ln xi

xj⇒ aij

xixj

≥ aij + aij lnxixj. Summation,

aijxixj

≥ aij + aij lnxixj

and use the fact∑

ij aij lnxixj

= 0.

Remark 3.1. Suppose that A is irreducible. The equality in (ii) holds if and only ifxi = xj, ∀i, j.

Proof of Theorem 1. The idea is to use the Lyapunov functional (Goh, S-B Hsu.)

E(t) :=∑i

ui +∑i

vi −∑i

Ki lnui.

Lemma 3.2. dE(t)dt

≤ 0,∀t ≥ 0.

By the equation, we have

d

dt(∑i

ui) = r∑i

ui(1−ui + viKi

),

d

dt(∑i

vi) = r∑i

vi(1−ui + viKi

)

andd

dt(∑i

Ki lnui) = r∑i

Ki(1−ui + viKi

) +∑i,j

Ki(dijujui

− dji).

Denote aij = dijKj, we have∑i,j

Ki(dijujui

− dji) =∑i,j

(aij

Ki

uiKj

uj

− aji

)≥ 0,

where the last inequality follows from Lemma (3.1). Thus

dE(t)

dt≤ r

∑i

(ui + vi)(1−ui + viKi

)− r∑i

Ki(1−ui + viKi

)

= − r

Ki

∑i

(ui + vi −Ki)2 ≤ 0.

Set M = (u, v) : dE(t)dt

= 0. Then we find that (u, v) : ui = cKi, vi = (1 − c)Ki. LetM ′ be the largest invariant set of M .

Claim: M ′ = (u∗, 0). This assertion follows from∑

j DijKj =∑

j DjiKi for some i.Apply Lasalle’s invariant principle, we get

limt→∞

(u(t), v(t)) = (u∗, 0).

(For more details, see Cantrell et. al. [4])


Open problems: 1. Consider positive solution of

(3.1)

∑nj=1[dijuj − djiui] + riui(1− ui+vi

Ki) = 0,∑n

j=1[Dijvj −Djivi] + rivi(1− ui+viKi

) = 0.

Question: Does (3.1) have at most one positive solution?For n = 2, if d12K2 > d21K1 and D12K2 < D21K1, it can be proved that there exist at

least one positive solution. Uniqueness is even open for this very special case.2. Consider

duidt

= (1− ε)∑n

j=1 dijuj −∑n

j=1 djiui + riui(1− ui+viKi

),

dvidt

= (1− ε)∑n

j=1Dijvj −∑n

j=1Djivi + rivi(1− ui+viKi

).

Question: 0 < ε < 1, what’s the dynamics of such system with travel loss?3. Consumer-resource models (2 consumers and 1 resource)

duidt

=∑n

j=1[dijuj − djiui] + ui(gi(Ri)− di),

dvidt

=∑n

j=1[Dijvj −Djivi] + vi(gi(Ri)− di),dRi

dt= Ri[ri(1− Ri

τi)− ui − vi] +

∑nj=1[LijRj − LjiRi].

4. Nonlocal models

Consider

ut =

∫Ω

k(x, y)u(y, t)dy − (

∫Ω

k(y, x)dy)u(x) + ru(1− u

K(x))

where K(x, y) is continuous, positive in Ω× Ω, Ω is a bounded domain in Rn. See Cosneret al. [6].

Definition 4.1. We say k is an ideal free dispersal strategy if∫Ω

k(x, y)K(y)dy = [

∫Ω

k(y, x)dy]K(x),∀x ∈ Ω.

Question: Given K(x), does such k exist?Examples: k(x, y) = Kα(x)Kβ(y).∫

Ω

Kα(x)Kβ+1(y)dy = [

∫Ω

Kα(y)dy]Kβ+1,

which holds if and only if α = β + 1. Hence we may choose k(x, y) = Kα(x)Kα−1(y),where α is any real number. The followings special cases are of interest:

• α = 0, k(x, y) = 1K(y)

≈ ∆( uK);

• α = 12, k(x, y) =

√K(x)K(y)

;

• α = 1, k(x, y) = K(x) ≈ div(K2∇( uK)).

6 YUAN LOU

Consider ut =∫Ωk(x, y)u(y, t)dy − (

∫Ωk(y, x)dy)u(x) + ru(1− u+v

K(x)),

vt =∫Ωk(x, y)v(y, t)dy − (

∫Ωk(y, x)dy)v(x) + rv(1− u+v

K(x)).

Assumptions:

(1) k, k positive and continuous in Ω× Ω.

(2) k is ideal free, k is not ideal free.

Remark 4.1. Since k is ideal free, (u, v) = (K, 0) is an equilibrium.

Theorem 2. (Cantrell et. al. [5]) The equilibrium (K, 0) is globally stable.

One technical difficulty: Solution trajectories are not pre-compact in C(Ω×C(Ω)).The idea is to use comparison principles.

5. PDE models

Example 1 (single species). Considerut = ∇ · (∇u− u∇P ) + u(m(x)− u) in Ω× (0,∞),

[∇u− u∇P ] · −→n = 0 on ∂Ω× (0,∞).

Question: What kind of function P (x) can produces IFD?

IFD⇔ equilibrium solution u ≡ m. ⇔

∇ · [∇m−m∇P ] = 0 in Ω,

[∇m−m∇P ] · −→n = 0 on ∂Ω.Hence,∇m−

m∇P = −−→V , where ∇ ·

−→V = 0 in Ω,

−→V · −→n = 0. Therefore we have

∇P = ∇(lnm) +1

m

−→V .

In particular, P = lnm+ constant can always produce ideal free distribution.Example 2: 2-competing species with neutral competition

ut = div(∇u− u∇P ) + u(m(x)− u− v) in Ω× (0,∞),

vt = div(∇v − v∇Q) + v(m(x)− u− v) in Ω× (0,∞)

[∇u− u∇P ] · −→n = 0 on ∂Ω× (0,∞),

[∇v − v∇Q] · −→n = 0 on ∂Ω× (0,∞).

Question: What kinds of P,Q can produce IFD?As we regard m− u− v as the fitness of both species, positive steady state (u, v) is an

IFD if and only if m − u − v = c for some constant c in Ω. It can be shown that c = 0since

∫Ωu(−m− u− v) = 0. Hence, IFD ⇔ m− u− v = 0. Therefore u satisfies

∇ · [∇u− u∇P ] = 0 in Ω,

[∇u− u∇P ] · −→n = 0 on ∂Ω,


which can be rewritten as ∇ · [eP∇(e−Pu)] = 0 in Ω,

∇(e−Pu) · −→n = 0 on ∂Ω.

By the maximum principle, e−Pu = C1 ⇒ u = C1eP . Similarly, v = C2e

Q. Thus,

(5.1) C1eP + C2e

Q = m.

Given any P , there exist C1, C2, Q such that (5.1) holds.Example 3. 2-competing species with weak competition. Consider

∇ · [∇u− u∇P ] + u(m− u− bv) = 0 in Ω,

∇ · [∇v − v∇Q] + v(m− cu− v) = 0 in Ω

where 0 < b, c < 1 (weak competition case), and (u∗, v∗) = ( (1−b)m1−bc ,

(1−c)m1−bc ) is an equilib-

rium for the corresponding ODE.Question: What kind of P and Q can produce IFD?

By similar argument as for Example 2, IFD ⇔

u = (1−b)m

1−bc ,

v = (1−c)m1−bc .

Therefore we have

u = C1eP , v = C2e

Q ⇔

P = ln (1−b)m

1−bc ,

Q = ln (1−c)m1−bc .

This means that both P − lnm and Q− lnm

must be constants, which is the same as the single species case.Example 4: Predator-Prey models.

∂u∂t

= µ∇ · [∇u− u∇P ] + u(a(x)− u− bv),∂v∂t

= ν∇ · [∇v − v∇Q] + v(−d(x) + cu),

where u is the density for the prey species, and v is the density of the predator species.

IFD ⇔

a(x)− u− bv = 0,

−d(x) + cu = 0.⇔

u = d(x)

c,

v =a(x)− d(x)

c

b.

Hence, the ideal free dispersal strategies for predator and prey are

P = ln d(x) + constant,

Q = ln(a(x)− d(x)

c

b) + constant = ln(a(x)− d(x)

c) + constant.

Case 1 d(x) = d > 0.

For this case we have P =constant, Q = ln(a(x) − dc). Hence, the prey species adopts

random dispersal and the predator is tracking prey’s resource (not the density of theprey!) In biology literatures this phenomenon is referred as “leaping frog” (see the worksof A. Shih).

Case 2. d ≡constant, a ≡constant, a.For this case we have P = ln d(x), Q = ln(a − d

c). This case is intuitively clear as it is

beneficial for the predator species to stay away from regions with large mortality, and it

8 YUAN LOU

can also be beneficial for the prey to have tendency to stay in such regions where the thedensity of the predator might be relatively low. Such biased movement of predators andprey might help some sort of weak spatial segregation of both species.

6. Mathematical results for PDE models

In this section we considerut = div(∇u− u∇(lnm)) + u(m(x)− u− v) in Ω× (0,∞),

vt = div(∇v − v∇Q) + v(m(x)− u− v) in Ω× (0,∞)

[∇u− u∇(lnm)] · −→n = [∇v − v∇Q] · −→n = 0 on ∂Ω× (0,∞).

Theorem 3. (Cantrell et. al. [3] and Averill et. al. [1]) Suppose that m ∈ C2(Ω),m > 0in Ω. If Q− lnm ≡constant, then (m, 0) is globally stable.

Proof. E(t) =∫Ωu+

∫Ωv −

∫Ωm lnu.

d

dt

∫Ω

u =

∫Ω

u(m− u− v).

d

dt

∫Ω

v =

∫Ω

v(m− u− v).

d

dt

∫Ω

m lnu =

∫Ω

u

m∇ · [m∇(

u

m)] + u(m− u− v)

=

∫Ω

m|∇( um)|2

( um)2

+

∫Ω

m(m− u− v).

⇒ dE

dt= −

∫Ω

m|∇( um)|2

( um)2

−∫Ω

(m− u− v)2 ≤ 0.

For more details of proof see Averill et al [1].

Consider

(6.1)

ut = ∇ · [∇u− u∇P ] + u(m(x)− u− v),

vt = ∇ · [∇v − v∇Q] + v(m(x)− u− v)

Question: If P − lnm ≡constant, Q− lnm ≡ constant, what’s the dynamics of (6.1)?

Theorem 4. ([1]) Suppose that P = α lnm,Q = β lnm. If (α−1)(β−1) < 0, then (6.1)has at least one stable positive steady states.

For more details about the dynamics of (6.1) we refer to Gejji et al. [9].

Remark 6.1. picture.


References

[1] I. Averill, Y. Lou and D. Munther, On several conjectures from evolution of dispersal, J. Biol. Dyn.6 (2012) 117-130.

[2] R.S. Cantrell, C. Cosner, D.L. DeAngelis, and V. Padron, The ideal free distribution as an evolu-tionarily stable strategy, J. Biol. Dyns. 1 (2007) 249-271.

[3] R.S. Cantrell, C. Cosner and Y. Lou, Evolution of dispersal and ideal free distribution, Math Bios.Eng. 7 (2010) 17-36.

[4] R.S. Cantrell, C. Cosner and Y. Lou) Evolutionary stability of ideal dispersal strategies in patchyenvironments, J. Math. Biol. 65 (2012) 943-965.

[5] R.S. Cantrell, C. Cosner, Y. Lou and D. Ryan, Evolutionary stability of ideal dispersal strategies:A nonlocal dispersal model, Canadian Applied Mathematics Quarterly, 20 (2012), 16-38.

[6] C. Cosner, J. Davilla and S. Martinez, Evolutionary stability of ideal free nonlocal dispersal, J. Biol.Dyn. 6 (2012) 395-405.

[7] B. Eaves, A. Hoffman, U. Rothblum and H. Schneider, Mathematical Programming Study 25 (1985)124-141.

[8] S.D. Fretwell, and H.L. Lucas, On territorial behavior and other factors influencing habitat selectionin birds. Theoretical development, Acta Biotheretica 19 (1970), 16-36.

[9] R. Gejji, Y. Lou, D. Munther and J. Peyton, Evolutionary Convergence to Ideal Free DispersalStrategies and Coexistence, Bull. Math Biol., Vol. 74 (2012), 257-299

[10] M.A. McPeek and R.D. Holt, The evolution of dispersal in spatially and temporally varying envi-ronments, Am. Nat. 140 (1992) 1010-1027.


Yuan Lou, Ohio State University

Lecture 8

1. Single equation

Consider model (2005, TPB)

ut = −∇(u∇f(x, u)) + uf(x, u)

Directed movement along the gradient of fitness (effective growth rate)

f(x, u) = m(x)− u

(1.1)

ut = ∇[u∇u− u∇m] + u(m− u) in Ω,

u∇(m− u)−→n = 0 on ∂Ω.

∇m: advection along resource gradient.Steady state (weak formulation)∫

Ω

u∇φ∇(m− u) =

∫Ω

φu(m− u).

Look for u ∈ W 1,2 ∪ L∞, ∀φ ∈ W 1,2 u = m+ is a weak solution.(Cantrell et. al. JDE 2008)

ut = ∇[µ∇u− αu∇(m− u)] + u(m− u) in Ω× (0,∞),

[µ∇u− αu∇(m− u)−→n = 0 on ∂Ω× (0,∞).

⇔ ut = ∇[(µ+ αu)∇u− αu∇m] + u(m− u), µ > 0.

Theorem 1. m ∈ C2,γ(Ω) and ∂Ω ∈ C2,γ for some γ ∈ (0, 1). For µ > 0, α ≥ 0, (1.1)has a unique classical solution u ∈ C2,1(Ω× (0,∞)).

Focus on steady state

Theorem 2. (Existence) If∫Ωm > 0, then for µ > 0, α ≥ 0, (1.1) has at least one stable

positive steady state.1

2 YUAN LOU

Proof. Super/sub solution method.1, If

∫Ωm > 0, then ∀µ > 0, α ≥ 0, u ≡ 0 is unstable.(⇒ existence of sub-solution

u(x) = εφ(x)).2,

∇[u∇(µ ln u+ αu− αm)] + u(m− u) ≤ 0.

Choose u: µ ln u+ αu− αm = constant,

m− u ≤ 0.

Theorem 3. For any positive steady state u, as α

µ→ ∞, u → m+ weakly in W 1,2 and

strongly in L2.

Remark 1.1. ∀η > 0 if α ≥ η and αµ→ ∞, then u→ m+ in Cγ(Ω), γ ∈ (0, 1).

Remark 1.2. Fix α > 0, let µ→ 0+, then u→ m+ in Cγ(Ω).

Lemma 1.1. For any positive steady state u,

minΩm ≤ u ≤ max

Ωm

Remark 1.3.∇ · (µ∇u− αu∇m) + u(m− u) = 0

maxΩ > maxΩm can happen.

Proof.∇ · (µ∇u− αu∇(m− u)) + u(m− u) = 0

Let w = ue−αµ(m−u),

µ∇ · [eαµ∇w] + u(m− u) = 0 in Ω,

∂w∂−→n = 0 on ∂Ω.

Let w(x) = maxΩw. Then, by maximum principle u(m− u) x ≥ 0 i.e. u(x) ≤ m(m).

maxΩ

w = w(x) = u(x)e−αµ[m(x)−u(x)] ≤ u(x) ≤ m(x) ≤ max

Ωm.

minΩm ≤ ue−

αµ(m−u) ≤ max

Ωm.

Claim: maxΩ u ≤ maxΩm.If not, u(x∗) > maxΩm for some x∗.

u(x∗) ≤ maxΩ

m · eαµ (m(x∗)− u(x∗)) ≤ max

Ω,

a contradiction. Corollary 1. Suppose that m > 0 in Ω. Then

lnminΩm

maxΩm≤ α

µ[u−m] ≤ ln

maxΩm

minΩm.


Proof.

lnu− α

µ[m− u] ≤ lnmax

Ωm.

u ≥ minΩm⇒ lnmin

Ωm+

α

µ≤ lnmax

Ωm

⇒ α

µ(u−m) ≤ lnmax

Ωm.

Step 3. For any positive steady state u,∫

Ω

|∇u|2 ≤∫Ω

|∇m|2.

Proof. Multiply∇ · [µ∇u− αu∇(m− u)] + u(m− u) = 0

by m− u and integrate by parts

⇒ −µ∫

∇u · ∇(m− u) + α

∫u|∇(m− u)|2 +

∫u(m− u)2 = 0

⇒∫Ω

|∇u|2 ≤∫Ω

∇m · ∇u ≤ ∥∇m∥L2∥∇u∥L2 .

Now, we know u→ u∗ weakly in W 1,2 and strongly in L2 since ∥∇u∥W 1,2 ≤ C.

Lemma 1.2. (Non-degeneracy Lemma)For any positive steady state u∫Ω

(m− u)+ ≤ µ

α|Ω|.

Remark 1.4. u∗ ≥ 0, ≡ 0.

Proof. Let w = u · e−αµ(m−u), then

µ∇ · [eαµ(m−u)∇w] + u(m− u) = 0.

Multiply it by 1wand integrate by parts

µ

∫Ω

eαµ(m−u)|∇w|2

w2+

∫Ω

eαµ(m−u)(m− u) = 0.

⇒∫Ω

eαµ(m−u)(m− u) ≤ 0.

Let F (y) = y · eαµy, then F (y) ≥ α

µy2,∀y > 0 and F (y) ≥ −µ

α. Hence

α

µ

∫x∈Ω:m−u≥0

(m− u)2 ≤ µ

α|x ∈ Ω : m− u < 0| ≤ µ

α|Ω|.

4 YUAN LOU

∫Ω

(m− u∗) = 0 ⇒ u∗ ≥ m, a.e.∫Ω

u∗(m− u∗) = 0 ⇒ u∗ = m+, a.e.

Theorem 4. Suppose m > 0 in Ω. Then for sufficiently large αµhas a unique positive

steady state.

Proof. Suppose that there are 2 different solutions u1, u2. By super-subsolution discussion,we can suppose u1 ≥ u2, u1 ≡ u2.

(1.2)

∫Ωu1(m− u1) = 0,∫

Ωu2(m− u2) = 0.

⇒∫Ω(u1−u2)[m− (u1+u2)] = 0. Since m > 0, u1, u2 → m in L∞(Ω), we have m− (u1+

u2) → −m < 0 for αµ>> 1.

Remark 1.5. If m > 0 in Ω and maxΩm

minΩm≤ 2 then for any µ, α ≥ 0 has a unique positive

steady state.

2. Fitness-dependent dispersal vs. random dispersal

(2.1)

ut = ∇[µ∇u− α∇(m− u− v)] + u(m− u− v),

vt = ν∆v + v(m− u− v) in Ω× (0,∞),

[µ∇u− α∇(m− u− v)] · −→n = ∇v−→n = 0 on ∂Ω.

(Cantrell et. al. 2013, JDE)1, Global existence.

Theorem 5. (2.1) has a unique solution u ∈ C2,1(Ω × (0,∞)) if one of the followingholds(1) N ≤ 2;(2) N ≥ 3, ν > µ;(3) (Lou, Tao, Winkler) N ≥ 3, Ω convex,

∫Ωup +

∫Ω|∇v|q

2, stability of (u, 0).

Lemma 2.1. If µ < ν, then (u, 0) is stable for 0 ≤ α << 1. If µ > ν, (u, 0) is unstablefor 0 ≤ α << 1.

Theorem 6. Suppose that m changes sign or m > 0 in Ω but non-constant. Then forα >> 1, (u, 0) is stable.


Proof. ut = ∇[µ∇u− α∇(m− u− v)] + u(m− u− v),

ν∆φ+ (m− u)φ+ σφ = 0 in Ω,

∇φ−→n = 0 on ∂Ω.

As α → +∞, u→ m+,Case 1, m changes sign

ν∆ψ + (m−m+)ψ + σψ = 0 in Ω ⇒ σ > 0.

Case 2, m > 0 in Ω.

ν∆ψ + σψ = 0 ⇒ σ1 = 0.

Lemma 2.2. Suppose that m > 0 in Ω, m ∈ C2(Ω).

α

µ(m− u) →

∫Ωm lnm∫Ωm

− lnm.

⇔ u = m+µ

α(lnm− c) + o(

µ

α).

Proof.

lnminΩm

maxΩm≤ α

µ≤ ln

maxΩm

minΩm

∇ · [µ∇u− αu∇(m− u)] + u(m− u) = 0.

µ∇[u∇(ln u+α

µ(u−m))] + u(m− u) = 0.

w := ln u+α

µ(u−m) ⇔ α

µ(ln u− w) = m− u.

µ∇[u∇w] + α

µu(ln u− w) = 0.

α∇[u∇w] + u(ln u− w) = 0.

α→ ∞, w → w =constant. ∫Ω

u(ln u− w) = 0.

∫Ω

m(lnm− w) = 0 ⇒ w =

∫Ωm lnm∫Ωm

⇒ α

µ(m− u) → lnm−

∫Ωm lnm∫Ωm

.

As α → ∞, u→ m, φ→ 1 after normalization,

σ

∫φ = −

∫(m− u)φ

6 YUAN LOU

α

µσ

∫φ = −

∫α

µ(m− u)φ

⇒ limα→∞

α

µσ

∫φ = |Ω|

∫Ωm lnm∫Ωm

−∫

lnm > 0.

Lemma 2.3. If m > 0, m ≡constant,∫Ωm lnm∫Ωm

>

∫lnm

|Ω|.

Proof. Jensen’s inequality.

Remark 2.1. limα→∞ σα = C1 > 0 ⇒ (u, 0) is stable with slow convergence rate.

3, Local stability of (0, v).v is the unique positive solution of

ν∆v + v(m− v),∂φ∂−→n = 0 on ∂Ω.

Lemma 2.4. Define H(τ) =∫Ω(m− v)eτ(m−v), then H(τ) = 0 has a unique positive root.

Proof. H(0) =∫Ω(m− v) < 0, limτ→∞H(τ) = +∞, m− v is positive somewhere, H ′(τ) =∫

Ω(m− v)2eτ(m−v) > 0 ⇒ H(τ ∗) = 0.

Theorem 7. If αµ≥ τ ∗, then (0, v) is unstable.

Proof. Consider

∇[µ∇φ− αφ∇(m− v)] + φ(m− v) + σ1φ = 0.

µ[eαµ(m−v)∇(e−

αµ(m−v))φ] + φ(m− v) + σ1φ = 0.

µ

∫Ω

eαµ(m−v)|∇(e−

αµ(m−v)φ)|2

(e−αµ(m−v)φ)2

+H(α

µ) + σ1

∫e

αµ(m−v) = 0

⇒ σ1

∫e

αµ(m−v) < −H(

α

µ) ≤ −H(τ ∗) = 0.

Theorem 8. Suppose Ω is convex, ( ∂2m∂xi∂xj

) is strictly negative definite in Ω. Then ∃γ0 =γ0(m,Ω) >> 1, such that if µ > ν ≥ ν0, then ∃!α∗ > 0 such that

(0, v) =

stable, 0 ≤ α < α∗,

unstable, α > α∗.


3. Non-existence of positive steady state for large α

Theorem 9. If m change sign, then for α >> 1 (2.1) has no positive steady state.

Remark 3.1. 1, If m ≥ 0, it is unknown whether there is no positive steady state forα >> 1.

2, If m change sign, is (u, 0) globally stable?

4. Global bifurcation of positive steady state

Theorem 10. Suppose that Ω is convex, ( ∂2m∂xi∂xj

) is strictly negative definite, m change

sign µ > ν > ν0. Then there exists a branch of positive steady states (α, u, 0) at α∗∗α = α(s)

u = u(x; s)

v = v(x; s),

with 0 ≤ s ≤ 1, α(0) = α∗, α(1) = α∗∗.

• Rabinowitz.• Shi, Wang (JDE, 2009).• Dancer, Fitzapatricle, Lopez-Gomez.• Pejschowicz, Rabier.

5. Bifurcation points (α∗, α∗∗)

Open problem: (u, 0) changes stability exactly once?

Theorem 11. Fix ν > 0. Passing to a subsequence if neccessary,

(a) limµ→∞α∗(µ,ν)

µ= τ ∗(ν).

(b) limµ→∞α∗∗(µ,ν)

µ= Λ(ν) > 0.

(5.1)

Λ(m− u∗) = lnu∗ −

∫u∗ lnu∗∫u∗

,

ν∆φ+ (m− u∗)φ = 0, φ > 0.

Find (Λ, u∗) such that (5.1) has a solution φ > 0.

6. Fitness-dependent dispersal

ut = ∇ · [µ∇u− αu∇(m− u− v)] + u(m− u− v),

vt = ∇ · [ν∇v − αv∇(m− u− v)] + v(m− u− v).

Case 1 µ = ν, α > β, Conjecture: limt→∞ v(x, t) = 0.Case 2 µ < ν, α = β, Conjecture: limt→∞ v(x, t) = 0.

8 YUAN LOU

References

[1] Xinfu Chen, R. Hambrock and Y. Lou, Evolution of conditional dispersal: a reaction-diffusion-advection model, J. Math. Biol. Vol. 57 (2008) 361-386.

[2] I. Averill, K.Y. Lam and Y. Lou, The effect of intermediate advection on two competing species, inpreparation.

[3] R. Hambrock and Y. Lou, Evolution of mixed dispersal strategy in spatially heterogeneous habitat,Bull. Math. Biol., Vol. 71 (2009) 1793-1817.

[4] Y. Lou and D. Munther, Dynamics of a three species competition model. Discrete and ContinuousDynamical Systems-A, Vol. 32 (2012), 3099-3131.

[5] X.F. Chen, K.Y. Lam and Y. Lou, Dynamics of a reaction-diffusion-advection model for two com-peting species, Discrete Continuous Dynamical Systems, Series A, Vol. 32 (2012), 3841-3859.

[6] K.Y. Lam and Y. Lou, Evolution of conditional dispersal: Evolutionarily stable strategies in spatialmodels, J. Math Biol., in press (DOI 10.1007/s00285-013-0650-1).

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