Jost J. Partial Differential Equations

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6.3 The Energy Inequality and the Relation with the Heat Equation 147

Since the points (y

)and(y

, −y

) yield the same contributions,

we obtain

u(x

,t)=

2π



B(x,t)

g(y)

−|x − y|

∂

∂t

⎛

⎝

2π



B(x,t)

f(y)

−|x − y|

⎞

⎠

where x =(x

), y =(y

), and the ball B(x, t) now is the two-

dimensional one.

Thevaluesofu at (x, t) now depend on the values on the whole disk

B(x, t) and not only on its boundary ∂B(x, t).

A reference for Sections 6.1 and 6.2 is F. John [10].

6.3 The Energy Inequality and the Relation with the

Heat Equation

Let u be a solution of the wave equation

(x, t) − Δu(x, t)=0 forx ∈ R

,t>0. (6.3.1)

We deﬁne the energy norm of u as follows:

E(t):=





(x, t)



i=1

(x, t)



dx. (6.3.2)

We have







i=1







− Δu)+



i=1

)



(6.3.3)

if u(x, t) = 0 for suﬃciently large |x| (where that may depend on t, so that this

computation may be applied to solutions of (6.3.1) with compactly supported

initial values).

In this manner, it is easy to show the following result about the region of

dependency of a solution of (6.3.1), partially generalizing the corresponding

results of Section 6.2 to arbitrary dimensions:

148 6. The Wave Equation

Theorem 6.3.1: Let u be a solution of (6.3.1) with

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

(x, 0) = 0, (6.3.4)

and let K := supp f



{x ∈ R

: f(x) =0}



be compact. Then

u(x, t)=0 for dist(x, K) >t. (6.3.5)

Proof: We show that f(y) = 0 for all y ∈ B(x, T ) implies u(x, T ) ≥ 0, which

is equivalent to our assertion. We put

E(t):=



B(x,T −t)





i=1



dy (6.3.6)

and obtain as in (6.3.3) (cf. (1.1.1))



B(x,T −t)







dy −



∂B(x,T −t)







do(y)



∂B(x,T −t)



∂u

∂ν

−









do(y).

By the Schwarz inequality, the integrand is nonpositive, and we conclude that

≤ 0fort>0.

Since by assumption

E(0) = 0 and E is nonnegative, necessarily

E(t) = 0 for all t ≤ T,

and hence

u(y, t)=0 for |x − y|≤T − t,

so that

u(x, T )=0

as desired. 

Theorem 6.3.2: As in Theorem 6.3.1, let u be a solution of the wave equa-

tion with initial values

u(x, 0) = f(x) with compact support

and

6.3 The Energy Inequality and the Relation with the Heat Equation 149

(x, 0) = 0.

Then

v(x, t):=



∞

−∞

−

√

4πt

u(x, s)ds

yields a solution of the heat equation

(x, t) − Δv(x, t)=0 for x ∈ R

,t>0

with initial values

v(x, 0) = f (x).

Proof: That u solves the heat equation is seen by diﬀerentiating under the

integral

∂

∂t

v(x, t)=



∞

−∞

∂

∂t



−

√

4πt



u(x, s)ds



∞

−∞

∂

∂s



−

√

4πt



u(x, s)ds

(since the kernel solves the heat equation)



∞

−∞

−

√

4πt

∂

∂s

u(x, s)ds



∞

−∞

−

√

4πt

u(x, s)ds

(since u solves the wave equation)

= Δv(x, t),

where we omit the detailed justiﬁcation of interchanging diﬀerentiation and

integration here. Then v(x, 0) = u(x, 0) = f(x) follows as in Section 4.1. 

Summary

In the present chapter we have studied the wave equation

∂

∂t

u(x, t) − Δu(x, t)=0 forx ∈ R

,t>0

with initial data

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

∂

∂t

u(x, 0) = g(x).

150 6. The Wave Equation

In contrast to the heat equation, there is no gain of regularity compared

to the initial data, and in fact, for d>1, there may even occur a loss of

regularity.

As was the case with the Laplace equation, mean value constructions are

important for the wave equation, and they permit us to reduce the wave

equation for d>1 to the Darboux equation for the mean values, which is

hyperbolic as well but involves only one spatial coordinate.

The propagation speed for the wave equation is ﬁnite, in contrast to the

heat equation. The eﬀect of perturbations sets in sharply, and in odd dimen-

sions greater than 1, it also terminates sharply (Huygens principle).

The energy

E(t)=





(x, t)|

+ |∇

u(x, t)|



is constant in time.

By a certain time averaging, a solution of the wave equation yields a

solution of the heat equation.

Exercises

6.1 We consider the wave equation in one space dimension,

− u

=0 for0<x<π,t>0,

with initial data

u(x, 0) =

∞



n=1

sin nx, u

(x, 0) =

∞



n=1

sin nx

and boundary values

u(0,t)=u(π,t) = 0 for all t>0.

Represent the solution as a Fourier series

u(x, t)=

∞



n=1

(t)sinnx

and compute the coeﬃcients γ

(t).

6.2 Consider the equation

+ cu

for some function u(x, t), x, t ∈ R,wherec is constant. Show that u is

constant along any line

Exercises 151

x − ct = const = ξ,

and thus the general solution of this equation is given as

u(x, t)=f (ξ)=f(x − ct)

where the initial values are u(x, 0) = f(x). Does this diﬀerential equation

satisfy the Huygens principle?

6.3 We consider the general quasilinear PDE for a function u(x, y)oftwo

variables,

+2bu

+ cu

= d,

where a, b, c, d are allowed to depend on x, y, u, u

,andu

. We consider

the curve γ(s)=(ϕ(s),ψ(s)) in the xy-plane, where we wish to prescribe

the function u and its ﬁrst derivatives:

u = f (s),u

= g(s),u

= h(s)forx = ϕ(s),y = ψ(s).

Show that for this to be possible, we need the relation



(s)=g(s)ϕ



(s)+h(s)ψ



(s).

For the values of u

along γ, compute the equations



+ ψ



= g



+ ψ



= h



Conclude that the values of u

,andu

along γ are uniquely de-

termined by the diﬀerential equations and the data f,g,h (satisfying the

above compatibility conditions), unless

aψ



− 2bϕ



+ cϕ



along γ. If this latter equation holds, γ is called a characteristic curve for

the solution u of our PDE au

+2bu

+ cu

= d. (Since a, b, c, d may

depend on u and u

, in general it depends not only on the equation,

but also on the solution, which curves are characteristic.) How is this

existence of characteristic curves related to the classiﬁcation into elliptic,

hyperbolic, and parabolic PDEs discussed in the introduction? What are

the characteristic curves of the wave equation u

− u

=0?

7. The Heat Equation, Semigroups, and

Brownian Motion

7.1 Semigroups

We ﬁrst want to reinterpret some of our results about the heat equation. For

that purpose, we again consider the heat kernel of R

, which we now denote

by p(x, y, t),

p(x, y, t)=

(4πt)

−

|x−y|

. (7.1.1)

For a continuous and bounded function f : R

→ R, by Lemma 4.2.1

u(x, t)=



p(x, y, t)f(y)dy (7.1.2)

then solves the heat equation

Δu(x, t) − u

(x, t)=0. (7.1.3)

For t>0, and letting C

denote the class of bounded continuous functions,

we deﬁne the operator

: C

) → C

)

via

f)(x)=u(x, t), (7.1.4)

with u from (7.1.2). By Lemma 4.2.2

f := lim

t→0

f = f; (7.1.5)

i.e., P

is the identity operator. The crucial point is that we have for any

≥ 0,

= P

◦ P

. (7.1.6)

Written out, this means that for all f ∈ C

154 7. The Heat Equation, Semigroups, and Brownian Motion



(4π (t

+ t

))

−

|x−y|

4(t

)

f(y) dy



(4πt

)

−

|x−z|



(4πt

)

−

|z−y|

f(y) dy dz. (7.1.7)

This follows from the formula

(4π (t

+ t

))

−

|x−y|

4(t

)

(4πt

)

(4πt

)



−

|x−z|

−

|z−y|

dz,

(7.1.8)

which can be veriﬁed by direct computation (cf. also Exercise 4.3).

There exists, however, a deeper and more abstract reason for (7.1.6):

f(x) is the solution at time t

+ t

of the heat equation with initial

values f.Attimet

, this solution has the value P

f(x). On the other hand,

f)(x) is the solution at time t

of the heat equation with initial values

f. Since by Theorem 4.1.2, the solution of the heat equation is unique

within the class of bounded functions, and the heat equation is invariant

under time translations, it must lead to the same result starting at time 0

with initial values P

f and considering the solution at time t

, or starting

at time t

with value P

f and considering the solution at time t

+ t

,since

the time diﬀerence is the same in both cases. This reasoning is also valid

for the initial value problem because solutions here are unique as well, by

Corollary 4.1.1. We have the following result:

Theorem 7.1.1: Let Ω ⊂ R

be bounded and of class C

,andletg : ∂Ω →

R be continuous. For any f ∈ C

(Ω),welet

Ω,g,t

f(x)

be the solution of the initial value problem

Δu − u

=0 in Ω × (0, ∞),

u(x, t)=g(x) for x ∈ ∂Ω, (7.1.9)

u(x, 0) = f(x) for x ∈ Ω.

We then have

Ω,g,0

f = lim

t0

Ω,g,t

f = f for all f ∈ C

(Ω), (7.1.10)

Ω,g,t

= P

Ω,g,t

◦ P

Ω,g,t

. (7.1.11)



Corollary 7.1.1: Under the assumptions of Theorem 7.1.1, we have for all

≥ 0 and for all f ∈ C

(Ω),

Ω,g,t

f = lim

tt

Ω,g,t



7.2 Inﬁnitesimal Generators of Semigroups 155

We wish to cover the phenomenon just exhibited by a general deﬁnition:

Deﬁnition 7.1.1: Let B be a Banach space, and for t>0,letT

: B → B

be continuous linear operators with

(i) T

=Id;

(ii) T

= T

◦ T

for all t

≥ 0;

(iii) lim

t→t

v = T

v for all t

≥ 0 and all v ∈ B.

Then the family {T

}

t≥0

is called a continuous semigroup (of operators).

A diﬀerent and simpler example of a semigroup is the following: Let B be

the Banach space of bounded, uniformly continuous functions on [0, ∞). For

t ≥ 0, we put

f(x):=f(x + t). (7.1.12)

Then all conditions of Deﬁnition 7.1.1 are satisﬁed. Both semigroups (for

the heat semigroup, this follows from the maximum principle) satisfy the

following deﬁnition:

Deﬁnition 7.1.2: A continuous semigroup {T

}

t≥0

of continuous linear op-

erators of a Banach space B with norm ·is called contracting if for all

v ∈ B and all t ≥ 0,

T

v≤v. (7.1.13)

(Here, continuity of the semigroup means continuous dependence of the op-

erators T

on t.)

7.2 Inﬁnitesimal Generators of Semigroups

If the initial values f(x)=u(x, 0) of a solution u of the heat equation

(x, t) − Δu(x, t) = 0 (7.2.1)

are of class C

, we expect that

lim

t0

u(x, t) − u(x, 0)

= u

(x, 0) = Δu(x, 0) = Δf(x), (7.2.2)

or with the notation

u(x, t)=P

f(u)

of the previous section,

lim

t0

− Id)f = Δf. (7.2.3)

We want to discuss this in more abstract terms and verify the following

deﬁnition:

156 7. The Heat Equation, Semigroups, and Brownian Motion

Deﬁnition 7.2.1: Let {T

}

t≥0

be a continuous semigroup on a Banach space

B.Weput

D(A):=



v ∈ B : lim

t0

− Id)v exists



⊂ B (7.2.4)

and call the linear operator

A : D(A) → B,

deﬁned as

Av := lim

t0

− Id)v, (7.2.5)

the inﬁnitesimal generator of the semigroup {T

Then D(A) is nonempty, since it contains 0.

Lemma 7.2.1: For al l v ∈ D(A) and all t ≥ 0, we have

Av = AT

v. (7.2.6)

Thus A commutes with all the T

Proof: For v ∈ D(A), we have

Av = T

lim

τ0

− Id)v

= lim

τ0

− T

)v (since T

is continuous and linear)

= lim

τ0

− T

)v (by the semigroup property)

= lim

τ0

− Id)T

= AT



In particular, if v ∈ D(A), then so is T

v. In that sense, there is no loss of

regularity of T

v when compared with v (= T

v).

In the sequel, we shall employ the notation

v :=



∞

λe

−λs

vds for λ>0 (7.2.7)

for a contracting semigroup {T

}. The integral here is a Riemann integral

for functions with values in some Banach space. The standard deﬁnition of

the Riemann integral as a limit of step functions easily generalizes to the