Chapter 9 Parabolic Equations - Texas Tech...

Chapter 9

Parabolic Equations

9.1 Introduction

If u(x, t) represents the temperature in a body Ω ⊂ Rn at a point x ∈ Ω at time t, then thepropagation (or diffusion) of heat is (approximately) described by the heat equation

ut = ∆u.

The heat equation is the usual example of a parabolic equation that one finds in most bookson partial differential equations. In what follows we will attempt to describe some of thebasic ingredients in a classical disscussion of the heat equation. At the same time we try toprovide the student with a brief introduction to some important tools used in PDEs whichhave not been discussed up to this point. One such tool that could have used all along is theFourier transform. One problem is that to use this tool most effectively the student shouldhave had a course in real analysis. Since this is not required for the present class it presentsa situation in which we must strike a delicate balance between what we would like to do andwhat we can really do rigorously. At some point I will present results without proof andsimply give references for details.

Following [3] we first give a few caveats concerning the heat equation in physics. Aspointed out in [3], the heat equation does not say anything about the microscopic physicalprocesses that accompany heat flow. It merely describes a limiting situation in which thesize of the atoms can be considered infinitesimally small. It does not recognize the existenceof an absolute zero temperature – indeed, temperatures for the heat equation can even benegative. Nonetheless, the heat equation is a very important an interesting model to study.

9.2 Some Results from Functional Analysis

At this point we digress to discuss several topics from real and functional analysis.

1

2 CHAPTER 9. PARABOLIC EQUATIONS

Holder’s inequality is used often in analysis and can be found in almost any book on realor functional analysis, see for example [5, 20]. Since a graduate course in real analysis is nota prerequisite for this course we will present several several results, more or less directly fromthe book used for real analysis at Texas Tech, “Real Analysis” by Gerald Folland [5]. For

this presentation when you see an expression like

∫f(x) dµ(x) where we say µ is a measure

just replace dµ by dx and imagine the functions are integrable in the sense you know best.This might be the Riemann integral or even the Lebesgue integral if you had Baby Reals.The strict validity of the results does rely heavily upon the Lebesgue integral but for ourpurposes we are really only interested in using the results and not in really rigorously provingthem – thats what the real analysis is for and you should take it if you have not already. Wewill be interested in the classical Lp function spaces consisting of functions (not really theyare actually equivalence classes of functions that differ on sets of measure zero, but we neednot worry about this) f whose pth power is integrable in Rn, i.e.

Lp(Rn) = f : Rn → C : ‖f‖p <∞

where ‖ · ‖p for 1 ≤ p <∞ is defined by

‖f‖p =

(∫|f(x)|p dx

)1/p

.

The case of p = 2 is of special importance and is given the name Hilbert Space. In this casethe norm ‖ · ‖2 is induced (just as it is in Cn) from an inner product. Namely we define theinner product in L2 by

〈f, g〉 =

∫f(x)g(x) dx

and it follows immediately that

‖f‖22 =

∫|f(x)|2 dx =

∫f(x)f(x) dx = 〈f, f〉.

Theorem 9.2.1 (Holder’s Inequality). Suppose that 1 < p < ∞ and let q satisfy p−1 +q−1 = 1 (in this case we say that p and q are conjugate exponents). If f and g are measureablefunctions on a measure space (X,µ) then

‖fg‖1 ≤ ‖f‖p‖g‖q.

In particular, if f ∈ Lp(X,µ) and g ∈ Lq(X,µ), then fg ∈ L1(X,µ).

The special case p = q = 2 is called the Cauchy-Schwartz Inequality.One result which has far reaching consequences is the following which is taken from [3].

9.2. SOME RESULTS FROM FUNCTIONAL ANALYSIS 3

Theorem 9.2.2 (Generalized Young’ Inequality). Let (S, µ) be a measure space 1 ≤p ≤ ∞ and C > 0. Suppose K is a measureable function on S × S such that∫

S

|K(x, y)|dµ(y) ≤ C, ∀ x ∈ S

and ∫S

|K(x, y)|dµ(x) ≤ C, ∀ y ∈ S

and suppose that f ∈ Lp(S). Then

Tf(x) =

∫S

K(x, y)f(y) dy

is defined almost everywhere and Tf ∈ Lp(S) with

‖Tf‖p ≤ C‖f‖p.

An indispensible tool is the convolution integral. For f and g in L1loc(Rn) we define

formally the convolution of f and g by

f ∗ g(x) =

∫Rnf(x− y)g(y) dy.

Corollary 9.2.1 (Young’s inequality). If f ∈ L1 and g ∈ Lp (1 ≤ p ≤ ∞), then f ∗g(x) ∈ Lp and ‖f ∗ g‖p ≤ ‖f‖1‖g‖p.Proof. Apply Theorem 9.2.2 with S = Rn and K(x, y) = f(x− y).

Remark 9.2.1. If f ∈ Lp and g ∈ Lq where p−1 + q−1 = 1, then (f ∗ g) ∈ L∞. This followsfrom Holder’s inequality:

‖(f ∗ g)‖∞ ≤ ess supx∈Rn

∫Rn|f(x− y)| |g(y)| dy ≤

ess supx∈Rn

(∫Rn|f(x− y)|p dy

)1/p(∫Rn|g(y)|q dy

)1/q

= ‖f‖p‖g‖q.

Here‖ϕ‖∞ = ess supx∈Rn |ϕ(x)| = infα ≥ 0 : µ(x : |ϕ(x)| ≥ α) = 0.

For our course without real analysis, you can assume that ϕ is say piecewise continuous and

‖f‖∞ = supx∈Rn|f(x)|.


Note that from Holder’s inequality, if f ∈ Lq and g ∈ Lp, (1/p+ 1/q = 1), then

f ∗ g ∈ L∞

and ‖f ∗ g‖∞ ≤ ‖f‖q‖g‖p.Definition 9.2.1. If f is a function on Rn and x ∈ Rn we define fx(y) = f(x+ y)

Lemma 9.2.1. For 1 ≤ p <∞ and f ∈ Lp,limx→0‖fx − f‖p = 0.

Proof. Since a continuous function on a compact set is uniformly continuous, if g is a con-tinuous, compactly supported function, then g is also uniformly continuous. Thus for such ag, given ε > 0 there exists an δ so that |g(x+ y)− g(y)| < ε for all y and provided |x| ≤ δ. Ifwe fix a compact set containing the support of gx for all |x| ≤ 1, then we can assume that gand gx are supported on a common compact set for |x| ≤ 1. Thus we have ‖gx− g‖p → 0 asx→ 0. Now for a given f ∈ Lp and ε > 0, choose a continuous g with compact support suchthat ‖f − g‖p ≤ ε/3. Then we also have ‖fx − gx‖p ≤ ε/3 (by a simple change of variablesin the integral), so that for all |x| ≤ 1

‖fx − f‖p ≤ ‖fx − gx‖p + ‖gx − g‖p + ‖g − f‖p ≤ ‖gx − g‖p + 2ε/3.

Now for x sufficiently small we have ‖gx − g‖p ≤ ε/3 and we have

‖fx − f‖p ≤ ε.

Remark 9.2.2. The result is not true for p =∞. In fact the condition that ‖fx− f‖∞ → 0as x→ 0 means that f agrees (a.e.) with a uniformly continuous function.

Remark 9.2.3. Minkowski’s inequality says that the Lp norm of the sum of a finite numberof functions is less than or equal to the sum of the Lp norms of the functions. This result isalso true in an integral form (See J. Folland’s book on real analysis, [5] page 186). Namely,if (X,µ) and (Y, ν) are σ-finite measure spaces, f(x, y) ≥ 0 and 1 ≤ p <∞, then we have[∫ ∣∣∣∣∫ f(x, y) dν(y)

∣∣∣∣p dµ(x)

]1/p

≤∫ [∫

f(x, y)p dµ(x)

]1/p

dν(y). (∗)

Theorem 9.2.3. Let φ ∈ L1(Rn), ε > 0, φε(x) = ε−nφ(xε

)and

∫Rnφ(x) dx = a. If f ∈ Lp,

1 ≤ p <∞, then‖f ∗ φε − af‖p → 0, ε→ 0

If f is bounded and uniformly continuous, then

‖f ∗ φε − af‖∞ → 0, ε→ 0


Proof. By the change of variables x→ εx we see that

∫ϕε(x) dx = a for all ε > 0. Therefore

we have

f ∗ ϕε(x)− af(x) =

∫[f(x− y)− f(x)]ϕε(y) dy =

∫[f(x− εy)− f(x)]ϕ(y) dy.

If f ∈ Lp and p <∞ then we can apply the Minkowski inequality for integrals (∗) given inRemark 9.2.3, to conclude that

‖f ∗ ϕε − af‖p ≤∫‖f−εy − f‖p |ϕ(y)| dy.

But ‖f−εy−f‖p is bounded by 2‖f‖p and tends to zero as ε→ 0 for each y, by Lemma 9.2.1.Thus we can apply the Lebesgue Dominated Convergence Theorem to get

limε→0‖f ∗ ϕε − af‖p ≤ lim

ε→0

∫‖f−εy − f‖p |ϕ(y)| dy =

∫limε→0‖f−εy − f‖p |ϕ(y)| dy = 0.

If p =∞ and f ∈ L∞ is uniformly continuous on a set V . Given δ > 0, choose a compact

set W so that

∫Rn\W

|ϕ| < δ. Then

supx∈V|(f ∗ ϕε)(x)− af(x)| ≤ sup

x∈V,y∈W|f(x− εy)− f(x)|

∫W

|ϕ|+ 2‖f‖∞δ.

The first term on the right tends to zero as ε → 0, and δ is arbitrary, so f ∗ ϕε tendsuniformly to (af) on V .

Definition 9.2.2. If φ ∈ L1 and∫φ = 1, then the functions φε are called an approximation

to the Identity.

Definition 9.2.3. The Schwartz class S of rapidly decreasing functions consists on all C∞

function on Rn which, together with there derivatives, go to zero at infinity faster than anypower of |x|. More precisely, for any nonnegative integer N and any multi-index α we define

‖f‖(N,α) = supx∈Rn

(1 + |x|)N |∂αf(x)|

andS = f ∈ C∞(Rn) : ‖f‖(N,α) <∞, for all N,α.

Theorem 9.2.4. If f ∈ L1 (1 ≤ p ≤ ∞) and φ ∈ S then f ∗ φ ∈ C∞ and ∂α(f ∗ φ) =f ∗ (∂αφ), for all α.


Proof. Let ϕ ∈ S, then for any bounded set V ⊂ Rn we have

supx∈V

∣∣∂αϕ(x− y)∣∣ ≤ Cα,V (1 + |y|)−(n+1) for all y ∈ Rn.

The function (1 + |y|)−(n+1) is in Lq for every q, so the integral

f ∗ ∂αϕ(x) =

∫f(y)∂αϕ(x− y) dy

converges absolutely and uniformly on bounded subsets of Rn. Therefore the derivatives ∂α

can be interchanged with the integration and we conclude that ∂α(f ∗ ϕ) = f ∗ ∂αϕ.

If in the above result we had ϕ ∈ C∞c (Rn) then we would only need to assume thatf ∈ L1

loc(Rn) for the convolution f ∗ϕ to be well defined and the same proof as above showsthat f ∗ ϕ ∈ C∞.

By way of a standard example, we can establish the existence of smooth compactlysupported functions: Namely, let

f(t) =

exp

(1

1− t2), |t| < 1

0 |t| > 1.

(9.2.1)

Then f ∈ C∞c (R) and we can have ψ(x) = f(|x|) for x ∈ Rn. Then

ψ ≥ 0, and 0 < a ≡∫ψ(x) dx <∞

and we define

φ(x) =ψ(x)

a∈ C∞c (Rn) (9.2.2)

Another important result is the following generalization of Urysohn’s Lemma.

Lemma 9.2.2. Let V ⊂ Rn be compact and Ω ⊂ Rn be open such that V ⊂ Ω. Then thereexists f ∈ C∞0 (Ω) such that f = 1 on V and 0 ≤ f ≤ 1 everywhere.

One of the most important concepts in PDEs is the classical partition of unity due toJ.P. Dieudonne. The above result is often used in the construction of a partition of unity.

Once again following [3], we define the Fourier transform and list many of its properties.

Definition 9.2.4. For f ∈ L1(Rn) we define the Fourier transform of f by

F(f)(ξ) = f(ξ) =

∫Rne−2πix·ξf(x) dx (9.2.3)

Clearly f(ξ) is well defined for every ξ.


More precisely, we have

Theorem 9.2.5. For f ∈ L1, f ∈ L∞ and

‖f‖∞ ≤ ‖f‖1.

Proof. This is obvious since

|f(ξ)| =∣∣∣∣∫Rne−2πix·ξf(x) dx

∣∣∣∣ ≤ ∫Rn|f(x)| dx = ‖f‖1.

Theorem 9.2.6. 1. If f, g ∈ L1, then F(f ∗ g) = F(f)F(g).

2. If f ∈ S, then ∂β f = g where g(x) = (−2πix)βf(x).

3. If f ∈ S, then F(∂αf)(ξ) = (2πiξ)αf(ξ).

4. If f ∈ S, then f ∈ S.

Proof. 1. First we note that by the Generalized Young’s inequaltiy, since f, g ∈ L1, then(f ∗ g) ∈ L1. Now the result follows by applying Fubini’s theorem (cf, [3]) in order tojustify an interchange of integrations. Namely we have

F(f ∗ g)(ξ) =

∫∫e−2πix·ξ f(x− y)g(y) dxdy

=

∫∫e−2πi(x−y)·ξ f(x− y) e−2πiy·ξg(y) dxdy

=

∫ (∫e−2πi(x−y)·ξ f(x− y) dx

)e−2πiy·ξg(y) dy, let w = (x− y),

=

∫ (∫e−2πiξ·w f(w) dw

)e−2πiy·ξg(y) dy

= f(ξ)

∫e−2πiy·ξg(y) dy

= f(ξ) g(ξ).

2. Just differentiate under the integral sign.

∂

∂ξjf(ξ) =

∫∂

∂ξje−2πiξ·x f(x) dx =

∫(−2πixj)e

−2πiξ·x f(x) dx = F((−2πixj)f).


3. Write out the integral and integrate by parts.

F(∂xjf)(ξ) =

∫e−2πiξ·x ∂xjf(x) dx

= (2πiξj)

∫e−2πiξ·x f(x) dx

= (2πiξj)F(f)(ξ).

4. By parts 2. and 3.

∂β(ξαf(ξ)) = (−1)|β|(2πi)|β|−|α|F[xβ∂αf

],

so ∂β(ξαf(ξ) is bounded for all α and β. Then using the product rule for derivatives

and mathematical induction β it can be proven that ξα∂β f is bounded for α and βfrom which we see that f ∈ S.

Theorem 9.2.7 (Riemann-Lebesque Lemma). If f ∈ L1(Rn) then f is continuous andtends to zero as |x| → ∞.

Proof. This follows from Theorem 9.2.6 for f ∈ S and the fact that S is dense in L1.Namely, if f ∈ L1 and fj ⊂ S such that fj → f in L1, then fj → f uniformly (since

‖fj − f‖∞ ≤ ‖fj − f‖1) and the result follows.

Theorem 9.2.8. If f(x) = e−aπ|x|2

with a > 0 then f(ξ) = a−n/2e−π|ξ|2/a

Proof. By making the change of variables x→ a−1/2x we may assume that a = 1. Since theexponential function converts sums into products, by Fubini’s Theorem

f(ξ) =

∫e−2πix·ξ−π|x|2 dx =

n∏j=1

∫e−2πixjξj−πx2

j dx,

so it suffices to verify that the j th factor in the product is e−πξ2j . So let us consider the case

n = 1 and we have ∫e−2πixξ−πx2

= e−πξ2

∫e−π(x+iξ)2

dx.

At this point we again appeal to a result that outside the scope of this class. It is known thatthe complex valued function of the complex variable z = (x+ iy) given by ϕ(z) = e−π(z+iξ)2

is an entire function (analytic in the entire complex plane) and has the property that

|ϕ(z)|2 = ϕ(z)ϕ(z) = e−π[(x+i(y+ξ))2+(x−i(y+ξ))2] = e−2πx2

e2π(y+ξ)2


which decreases exponentially to zero as x→ ±∞ as long as y remains bounded (recall thatξ is fixed). Thus we can apply Cauchy’s Theorem to shift the contour of integration fromIm z = 0 to Im z = −ξ. This gives us

F(e−|x|

2)

(ξ) = e−πξ2

∫e−π(x+iξ)2

dx = e−πξ2

∫e−πx

2

dx = e−πξ2

.

Theorem 9.2.9 (Plancherel Formula for S). If f, g ∈ S then∫fg =

∫f g

Proof. By Fubini’s Theorem∫f(x)g(x) dx =

∫∫f(x)g(y)e−2πix·y dy dx =

∫f(y)g(y) dy.

This result can be used to prove the following major result.

Theorem 9.2.10. 1. Let F∗f(x) = Ff(−x), then f ∈ S implies F∗Ff(x) = f(x).

2. The Fourier transform is an isomorphism of S onto S.

3. The Fourier transform extends to a unitary isomorphism of L2 onto itself, i.e., for allf ∈ L2 we have

‖f‖2 = ‖f‖2.

This result is often refered to as The Plancherel Theorem.

Proof. 1. Given ε > 0 and x ∈ Rn, set

ϕ(ξ) = e−2πix·ξ−ε2π|ξ|2 .

Then using Theorem 9.2.8 we have

ϕ(y) =

∫e−2πi(y−x)·ξ e−ε

2π|ξ|2 dξ = ε−ne−π|x−y|2/ε2 .

Thus

ϕ(y) = ε−ng

( |x− y|ε

)= gε(x− y) where g(x) = e− π|x|2 .


Then by Theorem 9.2.9, we have

f ∗ gε(x) =

∫f(x)gε(x− y) dy =

∫f ϕ =

∫f ϕ =

∫e−πε

2|ξ|2 e2πix·ξf(ξ) dξ.

Now by Theorems 9.2.3 and 9.2.4, f ∗ gε → f uniformly as ε → 0 since f is in S andfunctions in S are uniformly continuous. But for each x, by the dominated convergencetheorem, we have∫

e−πε2|ξ|2 e2πix·ξf(ξ) dξ →

∫e2πix·ξf(ξ) dξ = F

(f)

(x).

2. This is just a restatment of part 1 since for f ∈ S define g(ξ) = F(f)(−ξ) = F∗f(ξ) ∈ Sand we have f(x) = F(g)(x) by part 1.

3. Since S is dense in L2, by part 2 we only need to show that ‖f‖2 = ‖f‖2 for all f ∈ S.If f ∈ S, set g(x) = f(−x). We see that

g(ξ) =

∫e−2πix·ξ g(x) dx =

∫e−2πix·ξ f(−x) dx

=

∫e2πix·ξ f(−x) dx

=

∫e−2πix·ξ f(x) dx

= f(ξ).

Thus by Theorem 9.2.6 part 1 and Theorem 9.2.9 we have

‖f‖22 =

∫|f(x)|2 dx =

∫f(x)f(x) dx

= (f ∗ g)(0) = F∗F(f ∗ g)(0)

=

∫e2πi 0·ξF(f ∗ g)(ξ) dξ

=

∫f(ξ) f(ξ) dξ

= ‖f‖22.

9.3. THE HEAT EQUATION IN RN 11

9.3 The Heat Equation in Rn

Now we can begin our discussion of the heat equation on Rn × R+. We first consider theinitial value problem

∂tu = ∆u

u(x, 0) = f(x)(9.3.1)

To solve this problem we first assume that f ∈ S and apply the Fourier transform to constructthe so-called Gaussian or Heat Kernel. We have, by Theorem 9.2.6 part 3,

∂tu(ξ, t) + 4π2|ξ|2u(ξ, t) = 0, u(ξ, 0) = f(ξ).

For each ξ this is an initial value problem in t and we can solve it to get

u(ξ, t) = f(ξ)e−4π2|ξ|2t, (t > 0).

Thus we obtain a product of two functions. Next we apply the inverse Fourier transform toboth sides. If we apply Theorem 9.2.8 we have

Kt(x) ≡ F∗(e−4π2|ξ|2t

)(x) =

1

(4πt)n/2e−|x|

2/4t. (9.3.2)

Theorem 9.2.6 part 1 and Theorem 9.2.10 part 1, imply that the solution to our IVP is

u(x, t) = (Kt ∗ f)(x) =1

(4πt)n/2

∫Rne−|x−y|

2/4t f(y) dy.

Here we have obtained a formal solution. To prove that this is actually a solution requireseither some hard estimates or the use of approximate identities and delta families to showthat the initial condition is actually satisfied. Note that

Kt(x) = t−n/2K1(t−1/2x),

∫RnKt(x) dx = Kt(0) = 1.

Therefore by using ε = t1/2 we see from Theorem 9.2.3, that Ktt>0 is an approximation tothe identity. Thus we have the main result.

Theorem 9.3.1. If f ∈ Lp(Rn), 1 ≤ p ≤ ∞. Then u(x, t) = (Kt ∗ f)(x) satisfies

∂tu−∆u = 0, on Rn × (0,∞)

and


1. If f is bounded and continuous, then u is continuous on Rn×[0,∞) and u(x, 0) = f(x).

2. If f ∈ Lp, p <∞, then u(·, t)→ f in Lp as t→ 0.

Since Kt is rapidly decreasing along with all its derivatives, we see that we can differ-entiate under the integral to show that for all t > 0, u ∈ C∞. So the heat operator isinstantly, infinitely smoothing. This shows that we can not reverse time in the heat equationin general, since if we had a non-smooth initial data and we went back in time to say −t0then used this function as an initial data and propagated it forward in time it would haveto be smooth at time zero. We also note that Kt ∗ f actually makes sense if for t ≤ T wewould only assume that |f(x)| ≤ C exp(|x|2/(4T )).

As for the question of uniqueness for the heat eqaution in Rn, we have the following resultfrom [3].

Theorem 9.3.2. If u ∈ C(Rn × [0,∞)), u ∈ C2(Rn × (0,∞)) and u satisfies

∂tu = ∆u for t > 0, u(x, 0) = 0,

and if for every ε > 0 there exists C > 0 such that

|u(x, t)| ≤ Ceε|x|2

, |∇xu(x, t)| ≤ Ceε|x|2

,

then u(x, t) ≡ 0.

Proof. See Folland [3] Theorem 4.4, page 144.

To see that uniqueness does not hold for the initial value problem (9.3.1) without someconditions at infinity, consider the case n = 1 and let g(t) be any C∞ function on R, thenformally the series

u(x, t) =∞∑k=0

g(k)(t)x2k

(2k)!

9.3. THE HEAT EQUATION IN RN 13

formally satisfies the heat equation:

∂u

∂t=∞∑k=0

g(k+1)(t)x2k

(2k)!

=∞∑k=1

g(k)(t)x2(k−1)

(2(k − 1))!

∂2u

∂x2=∞∑k=1

g(k)(t)x2k−2

(2(k − 1))!

=∞∑k=1

g(k)(t) (2k)(2k − 1)x2k−2

(2k)!

=∞∑k=1

g(k)(t)x2(k−1)

(2(k − 1))!,

and therefore∂u

∂t=∂2u

∂x2.

To produce a nonzero solution to (9.3.1) with u(x, 0) = f(x) = 0, we give a nonzero gsatisfying g(k)(0) = 0 for k = 0, 1, · · · and so that the series converges on Rn × [0,∞. Sucha function is given by

g(t) = e−t−2

.

It is worth commenting that a Theroem due to D. Widder says that if the inital datadata is nonnegative, i.e., f(x) ≥ 0, then the unique nonnegative solution is given by u(x, t) =(Kt ∗ f)(x).

Let us briefly discuss the notion of a Fundamental Solution for a parabolic initial valueproblem.

Remark 9.3.1. Suppose we want to solve a Cauchy problem

ut = Lu, x ∈ Rn, t > 0, (9.3.3)

u(x, 0) = f(x), x ∈ Rn, (9.3.4)

where L is a constant coefficient differential operator.We consider first the “distributional” problem

Ft = LF, x ∈ Rn, t > 0, (9.3.5)

F (x, 0) = δ(x), x ∈ Rn. (9.3.6)


In (9.3.6) δ is the so-called dirac distribution which is defined by way of the requirement∫Rnϕ(x) δ(x) dx = ϕ(0),

for every ϕ ∈ C∞c (Rn). Such an F is called a fundamental solution for the initial valueproblem. We have already constructed such a function above for L = ∆, namely, F = K.

Once we have F , the solution to (9.3.3), (9.3.4) is given by

u(x, t) =

∫RnF (x− y, t) f(y) dy ≡ (F (·, t) ∗ f)(x). (9.3.7)

There is also a version of Duhamel’s Principle for parabolic problems. Consider thenonhomogeneous problem

ut = Lu+ g(x, t), x ∈ Rn, t > 0, (9.3.8)

u(x, 0) = 0, x ∈ Rn.

The solution to (9.3.8) is given by

u(x, t) =

∫ t

0

∫RnF (x− y, t− s) g(y, s) dyds. (9.3.9)

Just to make sure that there is no confusion with the literature, for an operator (∂t−L)the fundamental solution of the initial value problem (9.3.5), (9.3.6) coincides with the “freespace” fundamental solution which satisfies

(∂t − L)F (x, t) = δ(x, t), (9.3.10)

F (x, t) = 0, t < 0. (9.3.11)

For example, for the heat equation the free space fundamental solution is given by

K(x, t) =

Kt(x) t > 0

0 t < 0(9.3.12)

and we have the following result.

Theorem 9.3.3. The kernel K in (9.3.12) is the fundamental solution for the heat operator,i.e.,

(∂t −∆)F (x, t) = δ(x)δ(t).

Proof. See Folland [3] Theorem 4.6, page 146.

9.4. THE HEAT EQUATION IN BOUNDED DOMAINS 15

9.4 The Heat Equation in Bounded Domains

For the heat equation on bounded domains we must once again consider the BoundaryValue Problem (BVP). Generally we have a bounded domain Ω ⊂ Rn and we consider atime interval 0 ≤ t ≤ T < ∞. Thus the data for the problem generally consists of initialdata at time t = 0 given as u(x, 0) and some type of boundary data on ∂Ω × [0, T ]. Thereare three physically motivated BCs often considered:

1. u(x, t) = f(x, t) for (x, t) ∈ ∂Ω× [0, T ]. Thus the temperature on the body is specified.

2.∂u

∂n(x, t) = 0 for (x, t) ∈ ∂Ω× [0, T ]. In this case the body is insulated, i.e., there is no

heat flow in or out of the region.

3.

(∂u

∂n+ c(u− u0)

)(x, t) = 0. This condition cooresponds to Newton’s Law of Cooling

in which outside the region Ω a temperature u0(x, t) is maintained and the rate of heatflow in or out of the region is proportional to (u− u0)(x, t)

For parabolic BVP on bounded domains we once again have a maximum principle givenin [3] by

Theorem 9.4.1. Let Ω be a bounded domain in Rn and 0 < T <∞. Let u be a real-valuedcontinuous function on Ω× [0, T ] satisfying (∂t −∆)u = 0 on Ω× (0, T ). Then

maxx∈Ω×[0,T ]

|u(x, t)| = max

max

x∈∂Ω×[0,T ]|u(x, t)|,max

x∈Ω|u(x, 0)|

.

Proof. For any ε > 0 set v(x, t) = u(x, t) + ε|x|2. Then (∂t − ∆)v = −2nε < 0. Also fixany 0 < T ′ < T . If the maximum of v on Ω × [0, T ′] occurs at an interior point (x0, t0)of Ω × [0, T ′], then from calculus we must have ∂xjv(x0, t0) = 0 and ∂tv(x0, t0) = 0 andthe second derivatives ∂2

xjv(x0, t0) ≤ 0. Thus we must have (∂t − ∆)v(x0, t0) ≥ 0 which

contradicts the result (∂t −∆)v(x0, t0) = −2nε < 0.If the maximum occurs on Ω× T ′, then at the maximum ∂tv ≥ 0 and ∆v ≤ 0 so that,

just as above we have a contradiction. Thus we have

maxx∈Ω×[0,T ′]

|u(x, t)| ≤ maxx∈Ω×[0,T ′]

|v(x, t)|

= maxx∈(Ω×0)∪(∂Ω×[0,T ′])

|v(x, t)|

= maxx∈(Ω×0)∪(∂Ω×[0,T ′])

|u(x, t)|+ εmaxx∈Ω|x|2.


Passing to the limits as ε→ 0 and T ′ → T gives the desired result. We note that T ′ wasneeded because we can only apply the calculus results in the case that u is smooth which wedon’t know at t = T .

Remark 9.4.1. By replacing u by −u we can also see that the minimum is also achievedeither on (∂Ω× [0, T ]) or (Ω× 0).Corollary 9.4.1. There is at most one continuous function u on Ω× [0, T ] solving the BVP

ut −∆u = 0 (x, t) ∈ Ω× (0, T ),

u(x, 0) = f(x) x ∈ Ω,

u(x, t) = g(x, t) (x, t) ∈ ∂Ω× [0, T ].

As you might expect the method of separation of variables once again plays a centralrole. We seek simple solutions of the heat equation in the form

u(x, t) = X(x)T (t)

which givesT (t)

T (t)=

∆X(x)

X(x)= −λ2.

We conclude thatT (t) + λ2T (t) = 0

and∆X(x) + λ2X(x) = 0, X(x)

∣∣x∈∂Ω

= 0.

The first of these problems, an ordinary differential equation has general solution

T (t) = A exp(−λ2t).

At this point recall a general result first expressed in Chapter 7.

Theorem 9.4.2. The Laplacian operator ∆ on a bounded domain Ω with smooth boundary∂Ω admits an orthonormal basis for L2(Ω) consisting of eigenfunctions ψj(x) with associatedeigenvalues (−λ2

j) < 0 satisfying

∆ψj + λ2jψj = 0,

ψ(x) = 0 x ∈ ∂Ω.

Thus⟨ψj, ψk

⟩= δj,k where < ·, · > is the inner product in L2(Ω) given by⟨

f, g⟩

=

∫Ω

f(x)g(x) dx


where z denotes the complex conjugate of z.In particular, for every f ∈ L2(Ω) we have

f =∞∑j=1

⟨f, ψj

⟩ψj,

in the sense of L2(Ω).Similar results hold for the Neumann boundary conditions (∂nu = 0) and the radiation

conditions (∂nu+ cu) = 0.

With this result we can write the solution to

ut −∆u = 0 (x, t) ∈ Ω× (0, T ),

u(x, 0) = f(x) x ∈ Ω,

u(x, t) = 0 (x, t) ∈ ∂Ω× [0, T ].

as

u(x, t) =∞∑j=1

e−λ2j t fjψj(x),

where

f(x) =∞∑j=1

fjψj(x),

andfj =

⟨f, ψj

⟩.

Exercise Set 5: Parabolic Equations

1. Consider the nonhomogeneous heat problems:

(a) ut = uxx + f(x, t), x ∈ R, t > 0, u(x, 0) = 0, x ∈ R. Apply a Duhamel’sprinciple to find the following formula for the solution of this problem:

u(x, t) =

∫ t

0

∫ ∞−∞

1

2√π(t− τ)

e−(x−ξ)2/(4(t−τ))f(ξ, τ) dξdτ.

(b) Solve ut = 4uxx + t+ et, x ∈ R, t > 0, u(x, 0) = 2, x ∈ R.

(c) Solve ut = uxx + e−t cos(x), x ∈ R, t > 0, u(x, 0) = cos(x), x ∈ R.

(d) Solve ut = uxx + et sin(x), x ∈ R, t > 0, u(x, 0) = sin(x), x ∈ R.


2. The heat equation in the semi-infinite rod with its end kept at zero temperature orbeing insulated leads to the initial-boundary value problems

ut = uxx, 0 < x <∞, t > 0

u(x, 0) = φ(x), 0 ≤ x <∞

u(0, t) = 0, t ≥ 0 (1)

or

ux(0, t) = 0 t ≥ 0. (2)

For the boundary condition (1) show that

u(x, t) =1

2√π

∫ ∞0

1√t

[exp

(−(x− ξ)2

4t

)− exp

(−(x+ ξ)2

4t

)]φ(ξ) dξ.

For the boundary condition (2) show that

u(x, t) =1

2√π

∫ ∞0

1√t

[exp

(−(x− ξ)2

4t

)+ exp

(−(x+ ξ)2

4t

)]φ(ξ) dξ.

3. Recall that the solution to the initial value heat problem

ut = uxx, −∞ < x <∞, t > 0

u(x, 0) = f(x)

is given by

u(x, t) =1√4πt

∫ ∞−∞

exp

(−(x− y)2

4t

)f(y)dy.

(a) Prove that the solution depends continuously on the data in the sense that if

|f(x)− f(x)| < ε, x ∈ R,

then the corresponding solutions satisfy

|u(x, t)− u(x, t)| < ε, x ∈ R, t > 0.


(b) Assume that f(x) is continuous and bounded. Show that

limt→0+

u(x, t) = f(x)

by first proving the following result.

Lemma: Let ψ be positive, continuous and integrable on (−∞,∞) and satisfy∫ ∞−∞

ψ(r)dr = 1.

If f is continuous and bounded, then

limδ→0

1

δ

∫ ∞−∞

ψ

(y − xδ

)f(y)dy = f(x).

Hint: Make the obvious change of variable and reduce the problem to showing

limδ→0

∫ ∞−∞

ψ(r) [f(x+ δr)− f(x)] dr = 0.

Prove (b) by applying the Lemma to

ψ(x) =1√πe−x

2

.

4. Consider the initial boundary value problem,

ut = uxx + F (x, t), 0 < x < l, t > 0u(x, 0) = 0, 0 ≤ x ≤ lu(0, t) = 0, u(l, t) = 0, t ≥ 0.

Use Duhamel’s principle and a formal series solution to obtain the following formulafor the solution of the nonhomogeneous problem,

u(x, t) =∞∑k=1

∫ t

0

Fk(s)e− k

2π2(t−s)l2 ds

sin

kπx

l

where

Fk(s) =2

l

∫ l

0

F (x, s) sinkπx

ldx.

If F (x, t) = F (t) sin(πx/l) show that the solution is

u(x, t) =

(∫ t

0

F (s)eπ2sl2 ds

)sin

πx

le−

π2tl2


5. (a) Separate variables to construct a series solution of

ut = uxx, 0 < x < π, t > 0

u(x, 0) = x, 0 ≤ x < π

ux(0, t) = 0 = ux(π, t), t ≥ 0

(b) Carefully justify that the series solution satisfies the boundary conditions and theinitial condition and show that for each t > 0 the function u(x, t) defined by thisseries represents a C∞ function in x that satisfies the heat equation.

6. Show that the solution to the initial value problem

uxx = ut, x ∈ R, t > 0u(x, 0) = ebx, x ∈ R

is given by u(x, t) = ebxeb2t

7. Solve the heat problem

ut = uxx, 0 < x < 1, t > 0

ux(0, t) = 1, ux(1, t) = 0,

u(x, 0) = 0.

8. Do both parts

(a) Let uj(y, t)nj=1 be solutions of the one-dimensional heat equation ut = uyy. Showthat

v(x, t) =n∏j=1

uj(xj, t)

satisfies the n-dimensional heat equation. What is special about the heat equationthat makes this true?

(b) Suppose that uk(x, t)nk=1 are solutions of the nonhomogeneous heat problems

ut = uxx, u(x, t) = fk(xk), x ∈ Rn, t > 0, k = 1, · · · , n.Show that

u(x, t) =n∏k=1

uk(x, t)

is a solution of the problem

ut = ∆u, u(x, 0) =n∏k=1

ff (xk).

Bibliography

[1] R. Dautray and J.L. Lions, Mathematical analysis and numerical methods for scienceand technology,

[2] J. Dieudonne, Foundations of Modern Analysis, Academic press, 1960.

[3] G. Folland, Introduction to partial differential equations,

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21

22 BIBLIOGRAPHY

[18] M. Pinsky, Introduction to partial differential equations with applications,

[19] W. Rudin, Functional analysis, McGraw-Hill, New york, 1973.

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