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CHAPTER 6

The Heat Equation

We introduce several PDE techniques in the context of the heat equation:

The Fundamental Solution is the heart of the theory of infinite domain prob-lems. The fundamental solution also has to do with bounded domains, whenwe introduce Greens functions later.

The Maximum Principle applies to the heat equation in domains boundedin space and time. It is an important property of parabolic equations usedto deduce a variety of results such as uniqueness of solutions, comparisonprinciples. (E.g., if boundary conditions are changed in a way that suggestsintuitively the resulting temperature should be smaller, this can be provedusing the maximum principle.)

The Energy Method works analogously to the wave equation, except thatthe physical (heat) energy is less interesting than a mathematical energy,which typically decays. As for the wave equation, this leads to straightfor-ward uniqueness results. It is also useful for obtaining estimates on solutions

that are part of the existence and regularity theory for parabolic equations.

Initial Boundary Value Problems. We will spend some time describing ex-plicit solutions, expressed as infinite series of functions, of the heat equationplus initial and boundary conditions. There is a general technique frequentlyreferred to as separation of variables, or as eigenfunction expansions. Thedevelopment of this technique leads us to an analysis of eigenvalue problemsfor ordinary and partial differential equations, and to the analysis of Fourierseries.

6.1. The Fundamental Solution

To start with, we consider the heat equation in one space variable, plus time.In this section, we derive the fundamental solution and show how it is usedto solve the Cauchy problem:

ut = kuxx, |x| < , t >0u(x, 0) = g(x), |x| <

59

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60 6. THE HEAT EQUATION

The heat equation has a scale invariance property that is analogous to scaleinvariance of the wave equation or scalar conservation laws, but the scaling

is different.

Let a > 0 be a constant. Under the scaling x ax, t a2t theheat equation is unchanged. More precisely, if we introduce the change ofvariables: t= a2t, x= ax, then the heat equation becomes

ut= kuxx

This scale invariance suggests that we seek solutions v depending on the

similarity variable x2

t , or on x

t. However, there is a property of the heat

equation we would like to preserve in our similarity solution, that of con-servation of energy. Suppose u is a solution of the heat equation with theproperty that

|

u(x, 0) dx

| 0 for all x R, t >0

2. is C in (x, t), t >0

3.Rn

(x, t)dx= 1 for all t >0.

Properties 1 and 2 are obvious; here is the proof of property 3 for n= 1.

(x, t)dx= 1

4kt

ex2

4kt dx

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= 1

ey

2dy (let y =

x

4kt, dy=

dx

4kt)

= 1.

x

y

1

4 kt

0

y= (x,t)

Figure 6.1. Graph of the fundamental solution for the heatequation, witht >0. .

Therefore, (x, t) is a probability distribution for each t >0,with interestingdependence on t in the limits t and t 0 :The area under the graph is 1 for all t >0,yet ast ,maxx (x, t) 0;the tail spreads out to maintain

= 1

Ast 0 the maximum (at x = 0) blows up like 1t

,but the integral remains

constant. We also observe (x, t) 0 for x = 0, as t 0 +.

6.2. The Cauchy Problem for the Heat Equation

As for the wave equation, the Cauchy problem is the pure initial valueproblem, here stated in the one-dimensional case, n= 1:

(2.5)ut = kuxx, |x| < , t >0 (a)

u(x, 0) = g(x), |x| < (b)

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6.2. THE CAUCHY PROBLEM FOR THE HEAT EQUATION 63

Recall that the fundamental solution

(x, t) = 1

4kte

x2

4kt

satisfies (2.5(a)) for t >0.

Now (x y, t) is a solution of (2.5(a)) for all y, by translation invariance:x x y does not change the heat equation. Thus,

(x y, t)g(y)is also a solution of (2.5(a)). For later reference, we note that the heatequation is invariant under time translation also.

By linearity and homogeneity of the PDE, we can also take linear combina-tions of solutions. This suggests that

(2.6) u(x, t) =

(x y, t)g(y)dy

should also be a solution. Moreover, properties of suggest that as t 0+,u(x, t) g(x), since (x y, t) collapses to zero away from y = x, andblows up at y = x in such a way (i.e., preserving

= 1) that the initial

condition is satisfied in the sense u(x, t) g(x) as t 0 +.It is straightforward to check that the integrals for u, ut, uxx all convergeprovided g C(R) is bounded. Then

ut=

t (x y, t)g(y)dy; uxx =

2

x2(x y, t)g(y)dy,so that u satisfies the PDE for t >0.

It is more complicated to check that the initial condition is satisfied. Weneed to show u(x, 0) =g(x). But t= 0 is a singular point for : ( x, t) isnot defined at t = 0. To get an idea of why limt0+ u(x, t) =g(x),lets fixx.

Then, for >0,

(x y, t)g(y)dy

= (xy) (x y, t)g(y)dy |xy|

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Theorem: Let g C(R) be bounded, and let u(x, t) be given be the formula(2.6). Then

1. u is C in (x, t) for t >0;

2. u satisfies the heat equation

ut= kuxx, x R, t >0;3. lim

(x, t) (x0, 0)t >0

u(x, t) = g(x0) for all x0 R.

Proof:

Property 1 follows because is C fort >0, and derivatives of all decayexponentially as|x| , so the integrals converge.Property 2 follows from t= kxx, t >0.

To prove property 3, we look at the difference |u(x, t) g(x0)|, estimateand show that the pieces we get behave as we expect, i.e., as in the roughargument preceeding the proof.

Let >0.(This measures |u(x, t)g(x0)|.) Then, since

dx = 1,we have

(2.7) |u(x, t) g(x0)| =

(x y, t)(g(y) g(x0)) dy,

Let >0 (we choose below), and break up the integrals in (2.7):(2.8)

|u(x, t)g(x0)| |x0y|0 so that|g(y) g(x0)| < for|y x0| < (by continuity ofgat x0).

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6.2. THE CAUCHY PROBLEM FOR THE HEAT EQUATION 65

Then

I

|x0y|

(x y, t)|g(y) g(x0)|dy

2K|x0y|>

14kt

e(xy)2

4kt dy

Here is the second use of : Consider x satisfying|

x

x0|

<

2. Then

|x y| > 2 in the range of integration|x0 y| > .But this is not a goodenough estimate of the exponential, because we would still be left with anintegral over an infinite interval of a small but positive quantity. So, weneed to observe that in the region of integration,|x y| 12 |y x0|. Then

J Kk|x0y|>

1t

e(x0y)

2

16kt dy

C|z| 4kt

ez2 dz < ,

for t >0 sufficiently small.

Thus,

|u(x, t)

g(x0)

|0 sufficiently small.

This proves property 3.

Remarks

1. limt0+(x, t) is not a function in the usual sense, but is a distributionor generalized functioncalled the Dirac delta function(x) :

R

(x y)g(y) dy= g(x),

where the integral is a notational convenience. One way to interpret the deltafunction and the integral is to recognize the delta function as a measure that

places unit mass at x= 0 and zero mass at each x = 0.2. (x y, t) g(y) dy is the convolution of the function (, t) with g

More generally, for integrable functions , on R the convolution productof and is also a function defined by

(x) =

(x y)(y) dy.

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6.2.1. Using the Fundamental Solution to Solve Quarter-plane

Problems. :

Consider the so-called quarter-planeproblem:

1) ut = kuxx, x >0, t >0

2) u(0, t) = 0, t >0. (Homogeneous Dirichlet boundary condition.)

3) u(x, 0) =g(x), x >0

As for the wave equation, we reflect the initial data so the solution satisfiesthe boundary conditions:

Let g(x) be the odd extension ofg(x).

g(x) = g(x), x >0

g(x), x 0

u(0, t) =

(y, t)g(y)dy = 0, t >0,

since (y, t) is an even function ofy ,and g is an odd function.

It is straightforward to check directly that u(x, t) is an odd function ofx R. That is, the symmetry in the initial data is carried through to thesame symmetry in the solution.

Now replace g(y) with g(y) using g(y) = g(y), y

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6.3. THE ENERGY METHOD. 67

With a homogeneous Neumann boundary condition instead, we extend theinitial data to be even:

ut = kuxx, x >0, t >0ux(0, t) = 0 (Homogeneous Neumann Condition)

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

Solution: Extend g using the even extension, so that the first derivative iszero at x= 0.

Then u(x, t) =

0

((x y, t) + (x +y, t)) g(y) dy.

To check the boundary condition:

ux(0, t) = 0 (x(y, t) + x(y, t) g(y) dy.(y, t) is even in y so x(y, t) is odd. Thus, ux(0, t) = 0.

6.3. The Energy Method.

Consider

ut = k uxx, a x b, t >0.multiplying by u and integrating over [a, b], b

auut dx= k

ba

uuxxdx.

Therefore, d

dt

ba

1

2u2dx= kuux|ba k

ba

u2xdx.

Thus, the energy integral (not the heat energy) E(t) =ba

12 u

2(x, t)dx is

decreasing in time ifuux|ba 0.For example, if either u or ux is zero at each end pointx = a, x= b, thenthe energy integral decreases in t.

In this case, we have the important comparison to the initial data:E(t) E(0), t >0.

I.e.,

(3.10)

ba

u2(x, t)dx ba

u2(x, 0)dx, t >0.

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6.3.1. Using the Energy Method to prove Uniqueness. Considerthe initial boundary value problem

(IBVP)

ut = kuxx+f(x, t), 0 < x < l, t >0

u(0, t) = g(t), t >0u(L, t) = h(t), t >0

u(x, 0) = (x), 0< x < L

where f, g, h, are given functions.

The energy inequality works in any number of dimensions (will involve di-vergence theorem and Greens identities instead of integration by parts).

Theorem. If u1, u2 solve (IBVP) and are C2 functions, then u1 = u2everywhere.

Proof: Let u = u1 u2Then u satisfies (IBVP) with zero data: f gh0. But we knowthat E(t) is decreasing, from the energy inequality

0 L

0(u(x, t))2dx

L0

u(x, 0)2 dx= 0

Therefore,u 0, so that u1 u2.

Remark. In general, the existence of solutions is harder to establish thanuniqueness.

6.3.2. The Energy Principle in Higher Dimensions: Consider abounded open subset U ofRn,and the initial boundary value problem

ut = ku, x U t >0

u(x, t) = 0, x U, t >0

u(x, 0) = (x), x UWe begin by defining the energy integral as in one dimension:

(3.11) E(t) =1

2

U

u(x, t)2 dx.

Then

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6.4. THE MAXIMUM PRINCIPLE 69

E(t) =U

uut dx= k

U

uu dx

= k u

uu n dS k U

u u dx

= k U |u|2dx 0.Exercise: Prove the energy inequality for ut =.(k(x, u)u) wherek(x, u) Rn is a given positive function. The only difference is that k getsmoved under the integral sign. Does this enable us to prove uniqueness ofsolutions for the quasilinear heat equation?

6.4. The Maximum Principle

For T > 0, and U Rn, we use the notation UT = U (0, T]. Note thatUT includes the top U {t = T}. The following theorem, the MaximumPrinciple, states that the maximum of any (smooth) solution of the heatequation occurs either initially (at t = 0), or on the boundary of the domain.These parts of the boundary ofUTare known as theparabolic boundaryT :

T =UT UT = (U [0, T]) (U {t= 0}).Solutions of the heat equation should have two spatial derivatives and onetime derivative, so we define the appropriate space of functions on UT:

C21 (UT) = {u= u(x, t) :u, ut, Dxu, D2xu C(UT)}.In order to compare values of u in UT with values on the boundary, werequire in the theorem that u should be continuous on UT.

Theorem. The Maximum Principle. Letu C(UT) C21 (UT) satisfy

ut = ku, (x, t) UT.Then

maxUT

u(x, t) = maxT

u(x, t).

Remarks:

1. Supposeu has a local maximum at (x, t)UT. Then ut = 0 = ux = 0,u 0. If u

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2. UT is closed, so u achieves its maximum somewhere in UT: there is an(x0, t0) such that u(x0, t0) = max

UT

u(x, t).

Proof: Let M= maxT

u(x, t)

UT

Lx

T

t

0

Goal: Prove that U(x, t) Mfor all (x, t) UTTo deal with the possibility u= 0 at a maximum, we perturb u a bit:

Let v(x, t) =u(x, t) +|x|2

, >0. Then

vt kv = ut ku 2 k n

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6.5. DUHAMELS PRINCIPLE FOR THE INHOMOGENEOUS HEAT EQUATION 71

Therefore maximum ofv on UToccurs on T. I.e., v(x, t) max(y,t)UT

v(y, t)

for all (x, t)

UT. We have proved

u(x, t) +|x|2 max(y,t)T

(u(y, t) +|y|2) maxT

u+C for all (x, t) UT,

where C= maxyU

|y|2.Thus,u(x, t) M+ (C |x|2)

M+ C

Since >0 is arbitrary, we have u(x, t) Mfor all (x, t) UT.Remarks:

1. The weak maximum principle is easy to prove. The related strong maxi-mum principle is somewhat harder to prove. The strong maximum principlestates that, provided U is connected, then the maximum of u is achievedonlyon the parabolic boundary, unless u is constant throughout UT.

2. By applying the maximum principle tou, which also satisfies the con-ditions of the Theorem, we see that there is a corresponding minimum prin-ciple:

minUT

u(x, t) = minT

u(x, t).

6.5. Duhamels Principle for the Inhomogeneous Heat Equation

Consider the example of the heat equation on the whole real line, with aheat source or sink represented by the density function f(x, t) :

(P)

ut = kuxx+f(x, t), |x| < , t >0

u(x, 0) = 0 |x| <

The fundamental solution (x, t) of the homogeneous heat equation satisfiesut =kuxx, for t > 0. But then, for any y,s > 0 the shifted function (x y, t

s) satisfies the equation fort > s, and we can multiply by an amplitude

f(y, s),so that we have a collection of solutions (x y, t s)f(y, s) of theheat equation, with y R and t > s . Summing (i.e., integrating) thesesolutions over y R, with s fixed, we have that

u(x, t; s) =

(x y, t s)f(y, s) dy is a solution for all t > s,

satisfying u(x, t= s; s) =f(x, s)

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Now the idea is to integrate u(x, t; s) with respect to s from 0 to t:

Define u(x, t) = t0

u(x,t,s)ds. Then

ut kuxx = u(x, t; t) + t

0(ut(x, t; s) kuxx(x, t; s))ds

= f(x, t)

Note that this calculation is slightly misleading, because we have differen-tiated under the integral sign, when (x y , ts) has a singularity atx = y, t = s, on the boundary of the domain of integration. However, thissingularity can be handled with the appropriate limit, and the result is thesame. (See Evans [PDEs], page 50 for the details.)

Finally, we observe

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

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