Freiling G., Yurko V. Lectures on Differential Equations of Mathematical Physics: A First Course

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152 G.

Freiling and V. Yurko

Lemma 2.9.2. The

following relation holds

˙e

= Ae

−4iρ

. (2.9.5)

Proof. By Lemma 2.9.1, the function ˙e

−Ae

is a solution of (2.9.3). Since the

functions {e

−

} form a fundamental system of solutions of (2.9.3), we have

˙e

−Ae

= β

+ β

−

where β

= β

(t, ρ) , k = 1,2 do not depend on x. As x → +∞,

∼ exp(±iρx), ˙e

∼ 0, Ae

∼ 4iρ

exp(iρx),

consequently β

= −4iρ

, β

= 0, and (2.9.5) is proved. 2

Lemma 2.9.3. The following relations hold

˙a = 0,

b = −8iρ

b, ˙s

= 8iρ

, (2.9.6)

= 0,

= 8κ

. (2.9.7)

Proof. According to (2.8.18),

= ag

+ bg

−

. (2.9.8)

Differentiating (2.9.8) with respect to t, we get ˙e

= ( ˙ag

−

) + (a ˙g

+ b ˙g

−

). Using

(2.9.5) and (2.9.8) we calculate

a(Ag

−4iρ

) + b(Ag

−

−4iρ

−

) = ( ˙ag

−

) + (a ˙g

+ b ˙g

−

). (2.9.9)

Since g

∼ exp(±iρx), ˙g

∼ 0, Ag

∼ ±4iρ

exp(±iρx) as x → −∞, then (2.9.9) yields

−8iρ

exp(−iρx) ∼ ˙a exp(iρx)+

bexp(−iρx), i.e. ˙a = 0,

b = −8iρ

b. Consequently, ˙s

8iρ

, and (2.9.6) is proved.

The eigenvalues λ

= ρ

= −κ

, κ

> 0, j =

1,N are

the

zeros of the function a =

a(ρ,t). Hence, according to ˙a = 0, we have

= 0. Denote

= e(x,t, iκ

), g

= g(x,t, iκ

), j =

1,N.

Theorem

2.8.3, g

= d

, where d

= d

(t) do not depend on x. Differentiating the

relation g

= d

with respect to t and using (2.9.5), we infer

˙g

+ d

˙e

+ d

−4κ

˙g

+ Ag

−4κ

As x →

−∞

, g

∼ exp(κ

x), ˙g

∼ 0, Ag

∼ −4κ

exp(κ

x), and consequently

8κ

, or, in view of (2.8.32),

= 8κ

. 2

Hyperbolic P

artial Differential Equations 153

Thus, it

follows from Lemma 2.9.3 that we have proved the following theorem.

Theorem 2.9.2. Let q(x,t) be the solution of the Cauchy problem (2.9.1) −(2.9.2) ,

and let

(

) =

{

(

)

(

)

(

)

}

the

scattering data for

(

)

Then

(t, ρ) = s

(0,ρ)exp(8iρ

t),

(t) = λ

(0), α

(t) = α

(0)exp(8κ

t), j =

1,N (λ

= −κ

)

(2.9.10)

The

formulae

(2.9.10) give us the evolution of the scattering data with respect to t, and

we obtain the following algorithm for the solution of the Cauchy problem (2.9.1)-(2.9.2).

Algorithm 2.9.1. Let the function q(x,0) = q

(x) be given. Then

1) construct the scattering data {s

(0,ρ), λ

(0), α

(0), j =

1,N};

2) calculate {s

(t, ρ), λ

(t), α

(t), j =

1,N} by (2.9.10) ;

3) ﬁnd

the

function q(x,t) by solving the inverse scattering problem ( see Section

2.8) .

We notice once more the main points for the solution of the Cauchy problem (2.9.1)-

(2.9.2) by the inverse problem method:

(1) The presence of the Lax representation (2.9.4).

(2) The evolution of the scattering data with respect to t.

(3) The solution of the inverse problem.

Among the solutions of the KdV equation (2.9.1) there are very important particular

solutions of the form q(x,t) = f (x −ct). Such solutions are called solitons. Substituting

q(x,t) = f (x −ct) into (2.9.1), we get f

000

+ 6 f f

+ c f

= 0, or ( f

+ 3 f

+ c f )

= 0.

Clearly, the function

f (x) = −

cosh

√

satisﬁes

this

equation. Hence, the function

q(x,t) = −

cosh

√

c(x−c

(2.9.11)

soliton (see ﬁg. 2.9.1).

Figure

2.9.1.

154 G.

Freiling and V. Yurko

It is

interesting to note that solitons correspond to reﬂectionless potentials (see Section

2.8). Consider the Cauchy problem (2.9.1)-(2.9.2) in the case when q

(x) is a reﬂectionless

potential, i.e. s

(0,ρ) = 0. Then, by virtue of (2.9.10), s

(t, ρ) = 0 for all t, i.e. the

solution q(x,t) of the Cauchy problem (2.9.1)-(2.9.2) is a reﬂectionless potential for all t.

Using (2.9.10) and (2.8.79), we derive

q(x,t) = −2

∆

(x,t),

∆(x,t) = det

+ α

(0)exp(8κ

exp(−(κ

)x)

k,l=

1,N

particular

, if N = 1, then α

(0) = 2κ

, and

q(x,t) = −

2κ

cosh

(κ

(x −4κ

)

aking c = 4κ

, we get (2.9.11).

Chapter

arabolic Partial Differential

Equations

Parabolic partial differential equations usually describe various diffusion processes. The

most important equation of parabolic type is the heat equation or diffusion equation (see

Section 1.1). The properties of the solutions of parabolic equations do not depend essen-

tially on the dimension of the space, and therefore we conﬁne ourselves to considerations

concerning the case of one spatial variable. In Section 3.1 we study the mixed problem for

the heat equation on a ﬁnite interval. Section 3.2 deals with the Cauchy problem on the line

for the heat equation. We note that in Chapter 6, Section 6.3 one can ﬁnd exercises and

illustrative example for problems related to parabolic partial differential equations.

3.1. The Mixed Problem for the Heat Equation

We consider the following mixed problem

= u

, 0 < x < l, t > 0, (3.1.1)

|x=0

= u

|x=l

= 0, (3.1.2)

|t=0

= ϕ(x). (3.1.3)

This problem describes the one-dimensional propagation of heat in a ﬁnite slender homo-

geneous rod which lies along the x -axis, provided that the ends of the rod x = 0 and x = l

have zero temperature, and the initial temperature ϕ(x) is known.

Denote by D = {(x,t) : 0 < x < l, t > 0}, Γ = ∂D the boundary of the domain D.

Deﬁnition 3.1.1. A function u(x,t) is called a solution of problem (3.1.1)-(3.1.3) if

u(x,t) ∈C(

D) , u(x,t) ∈C

(D), and u(x,t) satisﬁes

(3.1.1)-(3.1.3).

First

we will prove the maximum (and minimum) principle for the heat conduction

equation. Denote D

= {(x,t) : 0 < x < l, 0 < t ≤ T }, Γ

= Γ ∩

(i.e. Γ

the

part

of the boundary of D

without the upper cap (see ﬁg. 3.1.1).

156 G.

Freiling and V. Yurko

0 l

Figure

3.1.1.

Theor

em 3.1.1. Let u(x,t) ∈C(

) , u(x,t) ∈C

), and

let

u(x,t) satisfy equation

(3.1.1) in D

. Then u(x,t) attains its maximum and minimum on Γ

Proof. For deﬁniteness, we give the proof for the maximum. Denote

M = max

u(x,t), m = max

u(x,t).

Clearly

, M ≥ m. W

e have to prove that M = m. Suppose that M > m. Let M = u(x

i.e. the maximum is attained at the point (x

). Consider the function

v(x,t) = u(x,t) +

M −m

Then

v(x

) ≥ M,

|Γ

≤ m +

M −m

< M,

and

consequently

, the maximum of the functions v(x,t) is not attained on Γ

. Let

max

v(x,t)

= v

and (x

) /∈ Γ

. According to the necessary conditions for an maximum we have

) ≥ 0, v

) = 0, v

) ≤ 0.

On the other hand,

−v

= (u

−u

) −

M −m

< 0.

This

contradiction

proves the theorem. 2

As an immediate consequence of the maximum (and minimum) principle we get the

following interesting corollaries.

Corollary 3.1.1 (the comparison principle). Let u

(x,t),u

(x,t) ∈C(

D)∩C

(D) sat-

isfy (3.1.1) .

(x,t) ≤ u

(x,t) on Γ, then u

(x,t) ≤ u

(x,t) in

Parabolic

Partial Differential Equations 157

Corollary

3.1.2 (the uniqueness theorem). If the solution of the problem (3.1.1) −

(3.1.3) exists, then it is unique.

Indeed, let u

(x,t) and u

(x,t) be solutions of (3.1.1)-(3.1.3). Then u

(x,t) = u

(x,t)

on Γ, and consequently, u

(x,t) = u

(x,t) in

Cor

ollary

3.1.3 (the stability theorem). Let u(x,t) and ˜u(x,t) be solutions of

(3.1.1) −(3.1.3) under the initial conditions ϕ and

ϕ, respectively. If |ϕ(x) −

ϕ(x)| ≤ ε ,

0 ≤ x ≤l, then |u(x,t) − ˜u(x,t)| ≤ ε in

Indeed,

denote v

(x,t) = u(x,t)− ˜u(x,t). Then v(x,t) ∈C(

D), satisﬁes equation (3.1.1),

and |v(x,t)| ≤ ε on Γ. Hence |v(x,t)| ≤ ε in

will solve the mixed problem (3.1.1)-(3.1.3) by the method of separation of vari-

ables. First we consider the following auxiliary problem. We will seek non-trivial (i.e. not

identically equal to zero) solutions of equation (3.1.1) satisfying the boundary condition

(3.1.2) and having the form

u(x,t) = Y (x)T (t).

Acting in the same way as in Section 2.2, we obtain

(x)

Y (x)

T (t)

= −λ,

where λ is

complex parameter. Hence

T (t) + λT (t) = 0, (3.1.4)

(x) + λY (x) = 0, Y (0) = Y (l) = 0. (3.1.5)

In Section 2.2 we found the eigenvalues

πn

, n ≥1,

and

the

eigenfunctions

(x) = sin

πn

x, n ≥1,

the

Sturm-Liouville problem (3.1.5). Equation (3.1.4) for λ = λ

has the general solution

(t) = A

exp(−λ

t),

where A

are arbitrary constants. Thus, the solutions of the auxiliary problem have the

form

(x,t) = A

exp

−

πn

sin

πn

x, n ≥1. (3.1.6)

will seek the solution of the mixed problem (3.1.1)-(3.1.3) by superposition of functions

of the form (3.1.6):

u(x,t) =

∞

∑

n=1

exp

−

πn

sin

πn

x. (3.1.7)

158 G.

Freiling and V. Yurko

Formally

, by construction, the function u(x,t) satisﬁes equation (3.1.1) and the boundary

conditions (3.1.2) for any A

. Choose A

such that u(x,t) satisﬁes the initial condition

(3.1.3). Substituting (3.1.7) into (3.1.3) we get

ϕ(x) =

∞

∑

n=1

sin

πn

Using

the

formulas for the Fourier coefﬁcients we calculate (formally)

ϕ(x)sin

πn

x, n ≥1. (3.1.8)

Theorem 3.1.2. Let ϕ(x) ∈C[0,l], ϕ(0) = ϕ(l) = 0. Then the solution of the mixed

problem (3.1.1) −(3.1.3) exists, is unique and is given by the formulae (3.1.7)−(3.1.8) .

Proof. It is sufﬁcient to prove that the function u(x,t), deﬁned by (3.1.7)-(3.1.8), is a

solution of problem (3.1.1)-(3.1.3). Fix δ > 0 and consider the domains Ω

= {(x,t) : 0 ≤

x ≤ l, t ≥ δ} and Ω = {(x,t) : 0 ≤x ≤l, t > 0}. Since |A

| ≤C, we have

∞

∑

n=1

exp

−

πn

< ∞ for

all s ≥ 0

. (3.1.9)

It follows from (3.1.9) that the function u(x,t), deﬁned by (3.1.7)-(3.1.8), is inﬁnitely

differentiable in Ω

, i.e. it has in Ω

continuous partial derivatives of all orders, and these

derivatives can be obtained by termwise differentiation of the series (3.1.7). By virtue of

the arbitrariness of δ we deduce that u(x,t) ∈C

∞

(Ω). Clearly, u(x,t) satisﬁes (3.1.1) and

(3.1.2). It is more complicated to deal with the initial condition (1.3.3), since in the general

case the series (3.1.7) for t = 0 can be divergent.

1) First we consider the particular case when ϕ(x) ∈C

[0,l] , ϕ(0) = ϕ(l) = 0. Then,

using in (3.1.8) integration by parts (as in Section 2.2), we obtain

∞

∑

n=1

| < ∞.

Consequently, the series (3.1.7) converges absolutely and uniformly in

D, and u(x,t) ∈

D). F

or t = 0

we have

u(x,0) =

∞

∑

n=1

sin

πn

hence

u(x,0) sin

πn

Comparing this relation with (3.1.8) we get

(u(x,0) −ϕ(x))sin

πn

x = 0, n ≥1.

On account of the completeness of the system {sin

πn

n≥1

, we

deduce

that

u(x,0) = ϕ(x). Thus, for the case ϕ(x) ∈C

[0,l] Theorem 3.1.2 is proved.

Parabolic

Partial Differential Equations 159

2) Consider

now the general case when ϕ(x) ∈C[0,l] , ϕ(0) = ϕ(l) = 0. Construct a

sequence of the functions ϕ

(x) ∈C

[0,l] , ϕ

(0) = ϕ

(l) = 0 such that

lim

m→∞

max

0≤x≤l

|ϕ

(x) −ϕ(x)|= 0.

Let u

(x,t) be the solution of problem (3.1.1)-(3.1.3) with the initial data ϕ

(x). Using

Corollary 3.1.3 we obtain that in

D the

sequence u

(x,t) con

verges uniformly to some

function ˜u(x,t) :

lim

m→∞

max

(x,t) − ˜u(x,t)| = 0,

and ˜u(x,t) ∈C(

D). Moreo

er,

˜u(x,0) = ϕ(x),

since

ϕ(x) = lim

m→∞

(x) = lim

m→∞

(x,0) = ˜u(x,0).

In the domain Ω

we rewrite (3.1.7) for u(x,t) as follows:

u(x,t) =

∞

∑

n=1

ϕ(ξ)sin

πn

ξdξ

−

sin

πn

and

consequently

u(x,t) =

G(x,ξ,t)ϕ(ξ)dξ,

where

G(x,ξ,t) =

∞

∑

n=1

−

sin

πn

ξsin

πn

The

function G

(x,ξ,t) is called the Green’s function. Furthermore, in Ω

we have

˜u(x,t) = lim

m→∞

(x,t) = lim

m→∞

G(x,ξ,t)ϕ

(ξ)dξ

G(x,ξ,t) lim

m→∞

(ξ)dξ =

G(x,ξ,t)ϕ(ξ)dξ = u(x,t).

By virtue of the arbitrariness of δ > 0, we get ˜u(x,t) ≡ u(x,t) in Ω, hence (after deﬁning

u(x,t) continuously for t = 0 ) u(x,t) ∈C(

D) , u(x,0)

= ϕ

(x). Theorem 3.1.2 is proved.

Deﬁnition 3.1.2. The solution of the mixed problem (3.1.1)-(3.1.3) is called stable

if for each ε > 0 there exists δ > 0 such that if |ϕ(x) −

ϕ(x)| ≤ δ , 0 ≤ x ≤ l, then

|u(x,t) − ˜u(x,t)| ≤ ε in

D. Here ˜u(x,t) is

the

solution of the mixed problem with the

initial condition ˜u(x,0) =

ϕ(x).

It follows from Corollary 3.1.3 that the solution of the mixed problem (3.1.1)-(3.1.3) is

stable (one can take δ = ε ). Thus, the mixed problem (3.1.1)-(3.1.3) is well-posed.

160 G.

Freiling and V. Yurko

3.2. The

Cauchy Problem for the Heat Equation

We consider the following Cauchy problem

= u

, −∞ < x < ∞, t > 0, (3.2.1)

|t=0

= ϕ(x). (3.2.2)

This problem describes the one-dimensional propagation of heat in a inﬁnite slender ho-

mogeneous rod which lies along the x -axis, provided that the initial temperature ϕ(x) is

known.

Denote Q = {(x,t) : −∞ < x < ∞, t > 0}.

Deﬁnition 3.2.1. A function u(x,t) is called a solution of the Cauchy problem (3.2.1)-

(3.2.2), if u(x,t) is continuous and bounded in

Q , u(x,t) ∈ C

(Q), and u(x,t) satisﬁes

(3.2.1)-(3.2.2).

Theor

3.2.1. If a solution of problem (3.2.1) −(3.2.2) exists, then it is unique.

Proof. Let u

(x,t) and u

(x,t) be solutions of (3.2.1)-(3.2.2). Denote v(x,t) =

(x,t) − u

(x,t). Then v(x,t) is continuous and bounded in

Q , v

= v

in Q

, and

v(x,0) = 0. Denote M = sup

v(x,t). Fix L > 0. Let Q

= {(x,t) : |x| < L, t > 0},

= ∂Q

the

boundary of Q

. Consider the function

w(x,t) =

+ 2t).

have

|v(x,0)| = 0 ≤w(x,0),

|v(±L,t)| ≤ M ≤ w(±L,t).

)

(3.2.3)

The function w(x,t) satisﬁes equation (3.2.1), and by virtue of (3.2.3),

−w(x,t) ≤ v(x,t) ≤ w(x,t)

on Γ

. According to Corollary 3.1.1, the last inequality is also valid in

. Thus,

−

+ 2t) ≤ v(x,t) ≤

+ 2t), |x|

≤ L, t ≥0

As L →∞ we obtain v(x,t) ≡ 0, and Theorem 3.2.1 is proved. 2

Derivation of the Poisson formula. We will seek bounded particular solutions of equa-

tion (3.2.1) which have the form u(x,t) = Y (x)T (t). Then

Y (x)

T (t) = Y

(x)T (t)

(x)

Y (x)

T (t)

= −λ

Parabolic

Partial Differential Equations 161

Thus, for

the functions Y (x) and T (t) we get the ordinary differential equations:

T (t) + λ

T (t) = 0,

(x) + λ

Y (x) = 0.

The general solutions of these equations have the form

Y (x) = A

(λ)e

iλx

+ A

(λ)e

−iλx

T (t) = A

(λ)e

−λ

Therefore, the functions

u(x,t,λ) = A(λ)e

−λ

t+iλx

are the desired bounded particular solutions of (3.2.1) admitting separation of variables.

We will seek the solution of the Cauchy problem (3.2.1)-(3.2.2) in the form

u(x,t) =

∞

−∞

A(λ)e

−λ

t+iλx

dλ.

We determine A(λ) from the initial condition (3.2.2). For t = 0 we have

ϕ(x) =

∞

−∞

A(λ)e

iλx

dλ.

Using the formulae for the Fourier transform (see [11]) we obtain (formally)

A(λ) =

2π

∞

−∞

ϕ(ξ)e

−iλξ

dξ,

and

consequently

u(x,t) =

2π

∞

−∞

∞

−∞

ϕ(ξ)e

−iλξ

dξ

−λ

t+iλx

dλ

2π

∞

−∞

ϕ(ξ)

∞

−∞

−λ

t+iλ(x−ξ)

dλ

dξ.

Let

calculate the inner integral. For this purpose we consider for a ﬁxed t the function

g(y) =

∞

−∞

−λ

t+λy

dλ. (3.2.4)

Differentiating (3.2.4) and using integration by parts we deduce

(y) =

∞

−∞

λe

−λ

λy

dλ =

∞

−∞

λy

−λ

d(λ

)

= −

∞

−∞

λy

−λ

∞

−∞

λy

−λ

dλ =

g(y).

Therefore,

(y)

g(y),