and relabeling x ” ξ and t ” „ , we have

∞ ∞ t

u(x, t) = G(x, t; ξ, 0)u0(ξ) dξ + G(x, t; ξ, „ )q(ξ, „ )dξd„. (6.105)

’∞ ’∞ 0

Note how the e¬ects of any heat source q(x, t) active prior to the initial-data

epoch at t = 0 have been subsumed into the evolution of the initial data.

6.3.3 Duhamel™s Principle

Often, the temperature of the spatial boundary of a region is speci¬ed in

addition to the initial data. Dealing with this type of problem leads us to a

new strategy.

Suppose we are required to solve

‚2u

‚u

=κ 2 (6.106)

‚t ‚x

for a semi-in¬nite rod 0 ¤ x < ∞. We are given a speci¬ed temperature,

u(0, t) = h(t), at the end x = 0, and for all other points x > 0 we are given

an initial condition u(x, 0) = 0.

u

u(x,t)

h(t)

x

Semi-in¬nite rod heated at one end.

168 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

We begin by ¬nding a solution w(x, t) that satis¬es the heat equation with

w(0, t) = 1 and initial data w(x, 0) = 0, x > 0. This solution is constructed

in the problems, and is

x

√

w = θ(t) 1 ’ erf . (6.107)

2t

Here erf(x) is the error function

2 x 2

e’z dz.

erf(x) = √ (6.108)

π 0

which obeys erf(0) = 0 and erf(x) ’ 1 as x ’ ∞.

erf(x)

1

x

Error function.

If we were given

h(t) = h0 θ(t ’ t0 ), (6.109)

then the desired solution would be

u(x, t) = h0 w(x, t ’ t0 ). (6.110)

For a sum

hn θ(t ’ tn ),

h(t) = (6.111)

n

the principle of superposition (i.e. the linearity of the problem) tell us that

the solution is the corresponding sum

hn w(x, t ’ tn ).

u(x, t) = (6.112)

n

We therefore decompose h(t) into a sum of step functions

t

™

h(t) = h(0) + h(„ ) d„

0

∞

™

θ(t ’ „ )h(„ ) d„.

= h(0) + (6.113)

0

6.4. LAPLACE™S EQUATION 169

It is should now be clear that

t

™

w(x, t ’ „ )h(„ ) d„ + h(0)w(x, t)

u(x, t) =

0

‚

t

=’ w(x, t ’ „ ) h(„ ) d„

‚„

0

‚

t

w(x, t ’ „ ) h(„ ) d„.

= (6.114)

‚t

0

This is called Duhamel™s solution, and the trick of expressing the data as a

sum of Heaviside functions is called Duhamel™s principle.

We do not need to be as clever as Duhamel. We could have obtained

this result by using the method of images to ¬nd a suitable causal Green

function for the half line, and then using the same Lagrange-identity method

as before.

6.4 Laplace™s Equation

The topic of potential theory, as problems involving the Laplacian are known,

is quite extensive. Here we will only explore the foothills.

Poisson™s equation, ’ 2 χ = f (r), r ∈ „¦, and the Laplace equation

to which it reduces when f (r) ≡ 0, come with various kinds of boundary

conditions, of which the commonest are

χ = ρ(x) on ‚„¦, (Dirichlet)

(n · )χ = q(x) on ‚„¦. (Neumann) (6.115)

2

A function for which χ = 0 in some region „¦ is said to be harmonic there.

6.4.1 Separation of Variables

Cartesian Coordinates

Let

‚2χ ‚2χ

+ 2 = 0, (6.116)

‚x2 ‚y

and write

χ = X(x)Y (y), (6.117)

170 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

so that

1 ‚2X 1 ‚2Y

+ = 0. (6.118)

X ‚x2 Y ‚y 2

Since the ¬rst term is a function of x only, and the second of y only, both

must be constants and the sum of these constants must be zero. Therefore

‚2X

+ k 2 X = 0,

2

‚x

‚2X

’ k2 X = 0. (6.119)

‚x2

The solutions are X = e±ikx and Y = e±ky . Thus