Jost J. Partial Differential Equations

Подождите немного. Документ загружается.

5.3 The Turing Mechanism 137

b>a.

+ g

< 0thenimples

0 <b−a<(a + b)

, (5.3.18)

while df

+ g

> 0 implies

d(b − a) > (a + b)

. (5.3.19)

Finally, (df

+ g

)

− 4d(f

− f

) > 0requires



d(b − a) − (a + b)



> 4d(a + b)

. (5.3.20)

The parameters a, b, d satisfying (5.3.18), (5.3.19), (5.3.20) constitute the

so-called Turing space for the reaction-diﬀusion system investigated here.

For many case studies of the Turing mechanism in biological pattern for-

mation, we recommend [19].

Summary

In this chapter, we have studied reaction-diﬀusion equations

(x, t) − Δu(x, t)=F (x, t, u)forx ∈ Ω,t > 0

as well as systems of this structure. They are nonlinear because of the u-

dependence of F . Solutions of such equations combine aspects of the linear

diﬀusion equation

(x, t) − Δu(x, t)=0

and of the nonlinear reaction equation

(t)=F (t, u),

but can also exhibit genuinely new phenomena like travelling waves.

The Turing mechanism arises in systems of the form

= Δu + γf(u, v),

= dΔv + γg(u, v).

under appropriate conditions, in particular when an inhibitor v diﬀuses at a

faster rate than an enhancer u, that is, when d>1 and certain conditions

on the derivatives f

are satisﬁed. A Turing instability means that

for such a system, a spatially homogeneous state becomes unstable. Thus,

spatially nonconstant patterns will develop. This is obviously a genuinely

nonlinear phenomenon.

138 5. Reaction-Diﬀusion Equations and Systems

Exercises

5.1 Consider the nonlinear elliptic equation

Δu(x)+σu(x) − u

(x)=0inadomainΩ ⊂ R

u(y)=0fory ∈ ∂Ω.

(5.3.21)

Let λ

be the smallest Dirichlet eigenvalue of Ω (cf. Theorem 9.5.1 below).

Show that for σ<λ

, u ≡ 0 is the only solution (hint: multiply the

equation by u and integrate by parts and use Corollary 9.5.1 below).

5.2 Consider the nonlinear elliptic system

Δu

(x)+F

(x, u)=0 forx ∈ Ω, α =1,...,n, (5.3.22)

with homogeneous Neumann boundary conditions

∂u

(x)

∂ν

=0forx ∈ ∂Ω, α =1,...,n. (5.3.23)

Assume that

δ = λ

min

α=1,...,n

− L>0 (5.3.24)

as in Theorem 5.2.1. Show that u ≡ const.

5.3 Determine the Turing spaces for the Gierer-Meinhardt and Thomas sys-

tems.

5.4 Carry out the analysis of the Turing mechanism for periodic boundary

conditions.

6. The Wave Equation and its Connections

with the Laplace and Heat Equations

6.1 The One-Dimensional Wave Equation

The wave equation is the PDE

∂

∂t

u(x, t) − Δu(x, t)=0 forx ∈ Ω ⊂ R

,t∈ (0, ∞)ort ∈ R. (6.1.1)

As with the heat equation, we consider t as time and x as a spatial variable.

For illustration, we ﬁrst consider the case where the spatial variable x is

one-dimensional. We then write the wave equation as

(x, t) − u

(x, t)=0. (6.1.2)

Let ϕ, ψ ∈ C

(R). Then

u(x, t)=ϕ(x + t)+ψ(x −t) (6.1.3)

obviously solves (6.1.2).

This simple fact already leads to the important observation that in con-

trast to the heat equation, solutions of the wave equation need not be more

regular for t>0 than they are at t = 0. In particular, they are not necessarily

of class C

∞

. We shall have more to say about that issue, but right now we

ﬁrst wish to motivate (6.1.3):

ϕ(x + t) solves

− ϕ

=0, (6.1.4)

ψ(x − t) solves

+ ψ

=0, (6.1.5)

and the wave operator

L :=

∂

∂t

−

∂

∂x

(6.1.6)

canbewrittenas

140 6. The Wave Equation

L =



∂

∂t

−

∂

∂x



∂

∂t

∂

∂x



, (6.1.7)

i.e., as the product of the two operators occurring in (6.1.4), (6.1.5). This

suggests the transformation of variables

ξ = x + t, η = x −t. (6.1.8)

The wave equation (6.1.2) then becomes

ξη

(ξ,η)=0, (6.1.9)

and for a solution, u

has to be independent of η, i.e.,

= ϕ



(ξ)(where“



” denotes a derivative as usual),

and consequently,

u =





(ξ)+ψ(η)=ϕ(ξ)+ψ(η). (6.1.10)

Thus, (6.1.3) actually is the most general solution of the wave equation

(6.1.2).

Since this solution contains two arbitrary functions, we may prescribe two

data at t = 0, namely, initial values and initial derivatives, again in contrast

to the heat equation, where only initial values could be prescribed. From the

initial conditions

u(x, 0) = f(x),

(x, 0) = g(x),

(6.1.11)

we obtain

ϕ(x)+ψ(x)=f(x),



(x) − ψ



(x)=g(x),

(6.1.12)

and thus

ϕ(x)=

f(x)



g(y)dy + c,

ψ(x)=

f(x)

−



g(y)dy −c

(6.1.13)

with some constant c. Hence we have the following theorem:

Theorem 6.1.1: The solution of the initial value problem

(x, t) − u

(x, t)=0 for x ∈ R,t>0,

u(x, 0) = f(x),

(x, 0) = g(x),

6.1 The One-Dimensional Wave Equation 141

is given by

u(x, t)=ϕ(x + t)+ψ(x −t)

{f(x + t)+f(x − t)} +



x+t

x−t

g(y)dy.

(6.1.14)

(For u to be of class C

, we need to require f ∈ C

, g ∈ C

.) 

The representation formula (6.1.14) emphasizes another diﬀerence be-

tween the wave and the heat equations. For the latter, we had found an

inﬁnite propagation speed, in the sense that changing the initial values in

some local region aﬀected the solution for arbitrary small t>0 in its en-

tire domain of deﬁnition. The solution u of the wave equation from formula

(6.1.14), however, is determined at (x, t) already by the values of f and g in

the interval [x −t, x + t]. The value u(x, t) thus is not aﬀected by the choice

of f and g outside that interval. Conversely, the initial values at the point

(y, 0) on the x-axis inﬂuence the value of u(x, t) only in the cone

y − t ≤ x ≤ y + t.

Since the rays bounding that region have slope 1, the propagation speed for

perturbations of the initial values for the wave equation thus is 1.

In order to compare the wave equation with the Laplace and the heat

equations, as in Section 4.1, we now consider some open Ω ⊂ R

and try to

solve the wave equation on

= Ω × (0,T)(T>0)

by separating variables, i.e., writing the solution u of

(x, t)=Δ

u(x, t)onΩ

u(x, t)=0 forx ∈ ∂Ω,

(6.1.15)

u(x, t)=v(x)w(t) (6.1.16)

as in (4.1.2). This yields, as in Section 4.1,

(t)

w(t)

Δv(x)

v(x)

, (6.1.17)

and since the left-hand side is a function of t, and the right-hand side one of

x, each of them is constant, and we obtain

Δv(x)=−λv(x), (6.1.18)

(t)=−λw(t), (6.1.19)

142 6. The Wave Equation

for some constant λ ≥ 0.

As in Section 4.1, v is thus an eigenfunction of the Laplace operator on Ω

with Dirichlet boundary conditions, to be studied in more detail in Section 9.5

below. From (6.1.19), since λ ≥ 0, w is then of the form

w(t)=α cos

√

λt+ β sin

√

λt. (6.1.20)

As in Section 4.1, referring to the expansions demonstrated in Section 9.5,

we let 0 <λ

≤ λ

... denote the sequence of Dirichlet eigenvalues of

Δ on Ω,andv

,... the corresponding orthonormal eigenfunctions, and

we represent a solution of our wave equation (6.1.15) as

u(x, t)=



n∈N



cos



t + β

sin





(x). (6.1.21)

In particular, for t =0,wehave

u(x, 0) =



n∈N

(x), (6.1.22)

and so the coeﬃcients α

are determined by the initial values u(x, 0). Like-

wise,

(x, 0) =



n∈N



(x), (6.1.23)

and so the coeﬃcients β

are determined by the initial derivatives u

(x, 0) (the

convergence of the series in (6.1.23) is addressed in Theorem 9.5.1 below). So,

in contrast to the heat equation, for the wave equation we may supplement

the Dirichlet data on ∂Ω by two additional data at t = 0, namely, initial

values and initial time derivatives.

From the representation formula (6.1.21), we also see, again in contrast to

the heat equation, that solutions of the wave equation do not decay exponen-

tially in time, but rather that the modes oscillate like trigonometric functions.

In fact, there is a conservation principle here; namely, the so-called energy

E(t):=





(x, t)



i=1

(x, t)



dx (6.1.24)

is given by

E(t)=



⎧

⎨

⎩







−α



sin



t + β



cos





(x)





i=1







cos



t + β

sin





∂

∂x

(x)



⎫

⎬

⎭



(α

+ β

), (6.1.25)

6.2 The Mean Value Method 143

since



(x)v

(x)dx =



1forn = m,

0 otherwise,

and



i=1



∂

∂x

(x)

∂

∂x

(x)=



for n = m,

0 otherwise

(see Theorem 9.5.1). Equation (6.1.25) implies that E does not depend on t,

and we conclude that the energy for a solution u of (6.1.15), represented by

(6.1.21), is conserved in time. This issue will be taken up from a somewhat

diﬀerent perspective in Section 6.3.

6.2 The Mean Value Method: Solving the Wave

Equation Through the Darboux Equation

Let v ∈ C

), x ∈ R

, r>0. As in Section 1.2, we consider the spatial

mean

S(v,x, r)=

dω

d−1



∂B(x,r)

v(y)do(y). (6.2.1)

For r>0, we put S(v, x,−r):=S(v, x,r), and S(v,x, r)thusisaneven

function of r ∈ R.Since

∂

∂r

S(v,x, r)|

r=0

= 0, the extended function remains

suﬃciently many times diﬀerentiable.

Theorem 6.2.1 (Darboux equation): For v ∈ C



∂

∂r

d − 1

∂

∂r



S(v,x, r)=Δ

S(v,x, r). (6.2.2)

Proof: We have

S(v,x, r)=

dω



|ξ|=1

v(x + rξ) do(ξ),

and hence

∂

∂r

S(v,x, r)=

dω



|ξ|=1



i=1

∂v

∂x

(x + rξ)ξ

do(ξ)

dω

d−1



∂B(x,r)

∂

∂ν

v(y) do(y),

where ν is the exterior normal of B(x, r)

dω

d−1



B(x,r)

Δv(z) dz

by the Gauss integral theorem.

(6.2.3)

144 6. The Wave Equation

This implies

∂

∂r

S(v,x, r)=−

d − 1

dω



B(x,r)

Δv(z)dz +

dω

d−1



∂B(x,r)

Δv(y) do(y)

= −

d − 1

∂

∂r

S(v,x, r)+

dω

d−1



∂B(x,r)

v(y) do(y),

(6.2.4)

because



∂B(x,r)

v(y) do(y)=Δ



∂B(x

,r)

v(x − x

+ y) do(y)



∂B(x

,r)

v(x − x

+ y) do(y)



∂B(x,r)

Δv(y) do(y).

Equation (6.2.4) is equivalent to (6.2.2). 

Corollary 6.2.1: Let u(x, t) be a solution of the initial value problem for the

wave equation

(x, t) − Δ(x, t)=0 for x ∈ R

,t>0,

u(x, 0) = f(x),

(x, 0) = g(x).

(6.2.5)

We deﬁne the spatial mean

M(u, x, r, t):=

dω

d−1



∂B(x,r)

u(y, t) do(y). (6.2.6)

We then have

∂

∂t

M(u, x, r, t)=



∂

∂r

d − 1

∂

∂r



M(u, x, r, t). (6.2.7)

Proof: By the ﬁrst line of (6.2.4),



∂

∂r

d − 1

∂

∂r



M(u, x, r, t)=

dω

d−1



∂B(x,r)

u(y, t) do(y)

dω

d−1



∂B(x,r)

∂

∂t

u(y, t) do(y),

since u solves the wave equation, and this in turn equals

∂

∂t

M(u, x, r, t).



6.2 The Mean Value Method 145

For abbreviation, we put

w(r, t):=M (u, x, r, t). (6.2.8)

Thus w solves the diﬀerential equation

= w

d − 1

(6.2.9)

with initial data

w(r, 0) = S(f,x, r),

(r, 0) = S(g,x, r).

(6.2.10)

If the space dimension d equals 3, for a solution u of (6.2.9), v := rw then

solves the one-dimensional wave equation

= v

(6.2.11)

with initial data

v(r, 0) = rS(f,x,r),

(r, 0) = rS(g, x,r).

(6.2.12)

By Theorem 6.1.1, this implies

rM(u, x, r, t)=

{(r + t)S(f,x,r + t)+(r − t)S(f,x,r −t)}



r+t

r−t

ρS(g,x, ρ)dρ. (6.2.13)

Since S(f,x, r)andS(g,x, r) are even functions of r, we obtain

M(u, x, r, t)=

{(t + r)S(f,x,r + t) − (t − r)S(f, x, t − r)}



t+r

t−r

ρS(g,x, ρ)dρ. (6.2.14)

We want to let r tend to 0 in this formula. By continuity of u,

M(u, x, 0,t)=u(x, t), (6.2.15)

and we obtain

u(x, t)=tS(g, x, t)+

∂

∂t

(tS(f,x,t)). (6.2.16)

By our preceding considerations, every solution of class C

of the initial value

problem (6.2.5) for the wave equation must be represented in this way, and

we thus obtain the following result:

146 6. The Wave Equation

Theorem 6.2.2: The unique solution of the initial value problem for the

wave equation in 3 space dimensions,

(x, t) − Δu(x, t)=0 for x ∈ R

,t>0,

u(x, 0) = f(x),

(x, 0) = g(x),

(6.2.17)

for given f ∈ C

), g ∈ C

), can be represented as

u(x, t)=

4πt



∂B(x,t)



tg(y)+f (y)+



i=1

(y)(y

− x

)



do(y). (6.2.18)

Proof: First of all, (6.2.16) yields

u(x, t)=

4πt



∂B(x,t)

g(y)do(y)+

∂

∂t



4πt



∂B(x,t)

f(y)do(y)



. (6.2.19)

In order to carry out the diﬀerentiation in the integral, we need to transform

the mean value of f back to the unit sphere, i.e.,

4πt



∂B(x,t)

f(y)do(y)=

4π



|z|=1

f(x + tz)do(z).

The Darboux equation implies that u from (6.2.19) solves the wave equation,

and the correct initial data result from the relations

S(w,x, 0) = w(x),

∂

∂r

S(w,x, r)|

r=0

satisﬁed by every continuous w. 

An important observation resulting from (6.2.18) is that for space dimen-

sions 3 (and higher), a solution of the wave equation can be less regular

than its initial values. Namely, if u(x, 0) ∈ C

, u

(x, 0) ∈ C

k−1

, this implies

u(x, t) ∈ C

k−1

, u

(x, t) ∈ C

k−2

for positive t.

Moreover, as in the case d = 1, we may determine the regions of inﬂuence

of the initial data. It is quite remarkable that the value of u at (x, t) depends

on the initial data only on the sphere ∂B(x, t), but not on the data in the

interior of the ball B(x, t). This is the so-called Huygens principle. This prin-

ciple, however, holds only in odd dimensions greater than 1, but not in even

dimensions. We want to explain this for the case d = 2. Obviously, a solution

of the wave equation for d = 2 can be considered as a solution for d = 3 that

happens to be independent of the third spatial coordinate x

We thus put x

= 0 in (6.2.19) and integrate on the sphere ∂B(x, t)=

{y ∈ R

:(y

− x

)

+(y

− x

)

+(y

)

= t

} with surface element

do(y)=