Jost J. Partial Differential Equations

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4.1 The Heat Equation: Deﬁnition and Maximum Principles 87

with

g(t):=



−1

t>0, for some k>1,

0 t =0,

v(x, t):=0 forall(x, t) ∈ R × (0, ∞).

Then u and v are solutions of (4.1.1) with f(x) = 0. For further details we

refer to the book of F. John [10].

Proof of Theorem 4.1.2: Since we can divide the interval (0,T) into subinter-

vals of length τ<

4λ

, it suﬃces to prove the claim for T<

4λ

, because we

shall then get

sup

×[0,kτ ]

u ≤ sup

×[0,(k−1)τ ]

u ≤···≤sup

f(x).

Thus let T<

4λ

. We may then ﬁnd ε>0with

T + ε<

4λ

. (4.1.23)

For ﬁxed y ∈ R

and δ>0, we consider

(x, t):=u(x, t) − δΛ(x, y, t, T + ε), 0 ≤ t ≤ T. (4.1.24)

It follows that

− Δv

= u

− Δu ≤ 0, (4.1.25)

since Λ is a solution of the heat equation. For Ω

:= B(y, ρ), we thus obtain

from Theorem 4.1.1

(y, t) ≤ max

∂

∗

. (4.1.26)

Moreover,

(x, 0) ≤ u(x, 0) ≤ sup

f, (4.1.27)

and for |x − y| = ρ,

(x, t) ≤ Me

λ|x|

− δ

(4π(T + ε − t))

exp



4(T + ε − t)



≤ Me

λ(|y|+ρ)

− δ

(4π(T + ε))

exp



4(T + ε)



88 4. Existence Techniques II: Parabolic Methods. The Heat Equation

Because of (4.1.23), for suﬃciently large ρ, the second term has a larger

exponent than the ﬁrst, and so the whole expression can be made arbitrarily

negative; in particular, we can achieve that it is not larger than sup

Consequently,

≤ sup

f on ∂

∗

. (4.1.28)

Thus, (4.1.26) and (4.1.28) yield

(y, t)=u(y,t) − δΛ(y,y, t, T + ε)=u(y,t) −δ

(4π(T + ε − t))

≤ sup

The conclusion follows by letting δ → 0. 

Actually, we can use the representation formula (4.1.12) to obtain a strong

maximum principle for the heat equation, in the same manner as the mean

value formula could be used to obtain Corollary 1.2.3:

Theorem 4.1.3: Let Ω ⊂ R

be open and bounded and

Δu − u

=0 in Ω

with the regularity properties speciﬁed at the beginning of this section. Then

if there exists some (x

) ∈ Ω × (0,T] with

u(x

)=max

u (or with u(x

)=min

u),

then u is constant in

Proof: The proof is the same as that of Lemma 1.2.1, using the representation

formula (4.1.12). (Note that by applying (4.1.12) to the function u ≡ 1, we

obtain



∂M(y,T;μ)

|x − y|

2(T −t)

do(x, t)=1,

and so a general u that solves the heat equation is indeed represented as some

average. Also, M (y,T; μ

) ⊂ M(y, T; μ

)forμ

≤ μ

,andasμ →∞,the

sets M(y, T; μ) shrink to the point (y,T).) 

Of course, the maximum principle also holds for subsolutions, i.e., if

Δu − u

≥ 0inΩ

In that case, we get the inequality “≤” in place of “=” in (4.1.12), which

is what is required for the proof of the maximum principle. Likewise, the

statement with the minimum holds for solutions of

Δu − u

≤ 0.

Slightly more generally, we even have

4.1 The Heat Equation: Deﬁnition and Maximum Principles 89

Corollary 4.1.2: Let Ω ⊂ R

be open and bounded and

Δu(x, t)+c(x, t)u(x, t) − u

(x, t) ≥ 0 in Ω

with some bounded function

c(x, t) ≤ 0 in Ω

. (4.1.29)

Then if there exists some (x

) ∈ Ω × (0,T] with

u(x

)=max

u ≥ 0, (4.1.30)

then u is constant in

Proof: Our scheme of proof still applies because, since c is nonpositive, at

a nonnegative maximum point (x

)ofu, c(x

)u(x

) ≤ 0which

strengthens the inequality used in the proof. 

Again, we obtain a minimum principle when we reverse all signs.

For use in §5.1 below, we now derive a parabolic version of E.Hopf’s boundary

point lemma 2.1.2. Compared with §2.1, we shall reverse here the scheme of

proof, that is, deduce the boundary point lemma from the strong maximum

principle instead of the other way around. This is possible because here we

consider less general diﬀerential operators than the ones in §2.1sothatwe

could deduce our maximum principle from the representation formula. Of

course, one can also deduce general Hopf type maximum principles in the

parabolic case, in a manner analogous to §2.1, but we do not pursue that

here as it will not yield conceptually or technically new insights.

Lemma 4.1.1: Suppose the function c is bounded and satisﬁes c(x, t) ≤ 0 in

.Letu solve the diﬀerential inequality

Δu(x, t)+c(x, t)u(x, t) − u

(x, t) ≥ 0 in Ω

and let (x

) ∈ ∂

∗

. Moreover, assume

(i) u is continuous at (x

);

(ii) u(x

) ≥ 0 if c(x) ≡ 0;

(iii) u(x

) >u(x, t) for all (x, t) ∈ Ω

;

(iv) there exists a ball

B((y, t

),R) ⊂ Ω

with (x

) ∈ ∂B((y, t

),R).

We then have, with r := |(x, t) − (y, t

)|,

∂u

∂r

) > 0, (4.1.31)

provided that this derivative (in the direction of the exterior normal of Ω

)

exists.

90 4. Existence Techniques II: Parabolic Methods. The Heat Equation

Proof: With the auxiliary function

v(x):=e

−γ(|x−y|

+(t−t

)

− e

−γR

the proof proceeds as the one of Lemma 2.1.2, employing this time the max-

imum principle Theorem 4.1.3. 

I do not know of any good recent book that gives a detailed and systematic

presentation of parabolic diﬀerential equations. Some older, but still useful,

references are [6], [16].

4.2 The Fundamental Solution of the Heat Equation 91

4.2 The Fundamental Solution of the Heat Equation.

The Heat Equation and the Laplace Equation

We ﬁrst consider the so-called fundamental solution

K(x, y, t)=Λ(x, y, t, 0) =

(4πt)

−

|x−y|

, (4.2.1)

and we ﬁrst observe that for all x ∈ R

, t>0,



K(x, y, t)dy =

(4πt)

dω



∞

−

d−1

dr =

dω



∞

−s

d−1



−|y|

dy =1. (4.2.2)

For bounded and continuous f : R

→ R, we consider the convolution

u(x, t)=



K(x, y, t)f(y)dy =

(4πt)



−

|x−y|

f(y)dy. (4.2.3)

Lemma 4.2.1: Let f : R

→ R be bounded and continuous. Then

u(x, t)=



K(x, y, t)f(y)dy

is of class C

∞

on R

× (0, ∞), and it solves the heat equation

= Δu. (4.2.4)

Proof: That u is of class C

∞

follows, by diﬀerentiating under the inte-

gral (which is permitted by standard theorems), from the C

∞

property of

K(x, y, t). Consequently, we also obtain

∂

∂t

u(x, t)=



∂

∂t

K(x, y, t)f(y)dy =



K(x, y, t)f(y)dy = Δ

u(x, t).



Lemma 4.2.2: Under the assumptions of Lemma 4.2.1, we have for every

x ∈ R

lim

t→0

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

92 4. Existence Techniques II: Parabolic Methods. The Heat Equation

Proof:

|f(x) − u(x, t)| =



f(x) −



K(x, y, t)f(y)dy





K(x, y, t)(f(x) − f(y))dy



with (4.2.2)



(4πt)



∞

−

d−1



d−1

(f(x) − f(x + rξ)) do(ξ) dr





∞

−s

d−1



d−1



f(x) − f(x +2

√

tsξ)



do(ξ) ds



···



···+ ···



∞

···



≤ sup

y∈B(x,2

√

tM)

|f(x) − f(y)| +2sup

|f|

dω



∞

−s

d−1

ds.

Given ε>0, we ﬁrst choose M so large that the second summand is less

than ε/2, and we then choose t

> 0sosmallthatforallt with 0 <t<t

the ﬁrst summand is less than ε/2 as well. This implies the continuity. 

By (4.2.3), we have thus found a solution of the initial value problem

(x, t) − Δu(x, t)=0 forx ∈ R

,t>0,

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

for the heat equation. By Theorem 4.1.2 this is the only solution that grows

at most exponentially.

According to the physical interpretation, u(x, t) is supposed to describe

the evolution in time of the temperature for initial values f(x). We should

note, however, that in contrast to physically more realistic theories, we here

obtain an inﬁnite propagation speed as for any positive time t>0, the

temperature u(x, t)atthepointx is inﬂuenced by the initial values at all

arbitrarily far away points y, although the strength decays exponentially with

the distance |x − y|.

In the case where f has compact support K, i.e., f(x)=0forx/∈ K,the

function from (4.2.3) satisﬁes

|u(x, t)|≤

(4πt)

−

dist(x,K)



|f(y)|dy, (4.2.5)

whichgoesto0ast →∞.

Remark: (4.2.5) yields an explicit exponential rate of convergence!

More generally, one is interested in the initial boundary value problem for

the inhomogeneous heat equation:

4.2 The Fundamental Solution of the Heat Equation 93

Let Ω ⊂ R

be a domain, and let ϕ ∈ C

(Ω × [0, ∞)), f ∈ C

(Ω), g ∈

(∂Ω × (0, ∞)) be given. We wish to ﬁnd a solution of

∂u(x, t)

∂t

− Δu(x, t)=ϕ(x, t)inΩ ×(0, ∞),

u(x, 0) = f(x)inΩ, (4.2.6)

u(x, t)=g(x, t)forx ∈ ∂Ω, t ∈ (0, ∞).

In order for this problem to make sense, one should require a compatibility

condition between the initial and the boundary values: f ∈ C

(

Ω), g ∈

(∂Ω × [0, ∞)), and

f(x)=g(x, 0) for x ∈ ∂Ω. (4.2.7)

We want to investigate the connection between this problem and the Dirichlet

problem for the Laplace equation, and for that purpose, we consider the

case where ϕ ≡ 0andg(x, t)=g(x) is independent of t. For the following

consideration whose purpose is to serve as motivation, we assume that u(x, t)

is diﬀerentiable suﬃciently many times up to the boundary. (Of course, this

is an issue that will need a more careful study later on.) We then compute



∂

∂t

− Δ



= u

− u

Δu

−



i=1

= u

∂

∂t

− Δu) −



i=1

= −



i=1

≤ 0. (4.2.8)

According to Theorem 4.1.1,

v(t):=sup

x∈Ω



∂u(x, t)

∂t



then is a nonincreasing function of t.

We now consider

E(u(·,t)) =





i=1

and compute

∂

∂t

E(u(·,t)) =





i=1

= −



Δudx, since u

(x, t)=

∂

∂t

g(x)=0 forx ∈ ∂Ω

= −



dx ≤ 0. (4.2.9)

94 4. Existence Techniques II: Parabolic Methods. The Heat Equation

With (4.2.8), we then conclude that

∂

∂t

E(u(·,t)) = −



∂

∂t

dx = −



Δu

dx +2





i=1

= −



∂Ω

∂

∂ν

do(x)+2





i=1

dx.

Since u

≥ 0inΩ, u

=0on∂Ω,wehaveon∂Ω,

∂

∂ν

≤ 0.

It follows that

∂

∂t

E(u(·,t)) ≥ 0. (4.2.10)

Thus E(u(·,t)) is a monotonically nonincreasing and convex function of t.In

particular, we obtain

∂

∂t

E(u(·,t)) ≤ α := lim

t→∞

∂

∂t

E(u(·,t)) ≤ 0. (4.2.11)

Since E(u(·,t)) ≥ 0 for all t,wemusthaveα = 0, because otherwise for

suﬃciently large T ,

E(u(·,T)) = E(u(·, 0)) +



∂

∂t

E(u(·,t))dt ≤ E(u(·, 0)) + αT < 0.

Thus it follows that

lim

t→∞



dx =0. (4.2.12)

In order to get pointwise convergence as well, we have to utilize the maximum

principle once more. We extend u

(x, 0) from Ω to all of R

as a nonnegative,

continuous function l with compact support and put

v(x, t):=



(4πt)

−

|x−y|

l(y)dy. (4.2.13)

We then have

− Δv =0,

and since l ≥ 0, also

v ≥ 0,

4.2 The Fundamental Solution of the Heat Equation 95

and thus in particular

v ≥ u

on ∂Ω.

Thus w := u

− v satisﬁes

∂

∂t

w − Δw ≤ 0inΩ, (4.2.14)

w ≤ 0on∂Ω,

w(x, 0) = 0 for x ∈ Ω,t =0.

Theorem 4.1.1 then implies

w(x, t) ≤ 0,

i.e.,

(x, t) ≤ v(x, t) for all x ∈ Ω,t > 0. (4.2.15)

Since l has compact support, from Lemma 4.2.2

lim

t→∞

v(x, t) = 0 for all x ∈ Ω,

and thus also

lim

t→∞

(x, t) = 0 for all x ∈ Ω. (4.2.16)

We thus conclude that provided that our regularity assumptions are valid

the time derivative of a solution of our initial boundary value theorem with

boundary values that are constant in time goes to 0 as t →∞.Thus,ifwe

can show that u(x, t) converges for t →∞with respect to x in C

, the limit

function u

∞

needs to satisfy

Δu

∞

=0,

i.e., be harmonic. If we can even show convergence up to the boundary, then

∞

satisﬁes the Dirichlet condition

∞

(x)=g(x)forx ∈ ∂Ω.

From the remark about (4.2.5), we even see that u

(x, t) converges to 0 ex-

ponentially in t.

If we know already that the Dirichlet problem

Δu

∞

=0 inΩ,

∞

= g on ∂Ω, (4.2.17)

admits a solution, it is easy to show that any solution u(x, t) of the heat

equation with appropriate boundary values converges to u

∞

. Namely, we

even have the following result:

96 4. Existence Techniques II: Parabolic Methods. The Heat Equation

Theorem 4.2.1: Let Ω be a bounded domain in R

,andletg(x, t) be con-

tinuous on ∂Ω ×(0, ∞), and suppose

lim

t→∞

g(x, t)=g(x) uniformly in x ∈ ∂Ω. (4.2.18)

Let F (x, t) be continuous on Ω × (0, ∞), and suppose

lim

t→∞

F (x, t)=F (x) uniformly in x ∈ Ω. (4.2.19)

Let u(x, t) be a solution of

Δu(x, t) −

∂

∂t

u(x, t)=F (x, t) for x ∈ Ω, 0 <t<∞,

u(x, t)=g(x, t) for x ∈ ∂Ω, 0 <t<∞. (4.2.20)

Let v(x) be a solution of

Δv(x)=F (x) for x ∈ Ω,

v(x)=g(x) for x ∈ ∂Ω. (4.2.21)

We then have

lim

t→∞

u(x, t)=v(x) uniformly in x ∈ Ω. (4.2.22)

Proof: We consider the diﬀerence

w(x, t)=u(x, t) − v(x). (4.2.23)

Then

Δw(x, t) −

∂

∂t

w(x, t)=F (x, t) − F (x)inΩ × (0, ∞),

w(x, t)=g(x, t) − g(x)in∂Ω × (0, ∞), (4.2.24)

and the claim follows from the following lemma:

Lemma 4.2.3: Let Ω be a bounded domain in R

,letφ(x, t) be continuous

on Ω ×(0, ∞), and suppose

lim

t→∞

φ(x, t)=0 uniformly in x ∈ Ω. (4.2.25)

Let γ(x, t) be continuous on ∂Ω × (0, ∞), and suppose

lim

t→∞

γ(x, t)=0 uniformly in x ∈ ∂Ω. (4.2.26)

Let w(x, t) be a solution of