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Solved Problems

Exercise 181. Verify Duhamel’s Principal: If u(x, t, τ) is the solution of the initial
value problem:
u = κu , u(x, 0, τ) = f(x, τ),
t
xx
then the solution of

w = κw xx + f(x, t), w(x, 0) = 0
t

is
t
Z
w(x, t) = u(x, t − τ, τ) dτ.
0
Solution: We verify Duhamel’s principal by showing that the integral expression for

w(x, t) satisfies the partial differential equation and the initial condition. Clearly
the initial condition is satisfied.

0
Z
w(x, 0) = u(x, 0 − τ, τ) dτ = 0
0

Now we substitute the expression for w(x, t) into the partial differential equation.

∂ Z t ∂ 2 Z t
u(x, t − τ, τ) dτ = κ u(x, t − τ, τ) dτ + f(x, t)
∂t 0 ∂x 2 0
t t
Z Z
u(x, t − t, t) + u (x, t − τ, τ) dτ = κ u (x, t − τ, τ) dτ + f(x, t)
xx
t
0 0
t t
Z Z
f(x, t) + u (x, t − τ, τ) dτ = κ u (x, t − τ, τ) dτ + f(x, t)
xx
t
0 0
t
Z
(u (x, t − τ, τ) dτ − κu (x, t − τ, τ)) dτ
t
xx
0
Since u (x, t − τ, τ) dτ − κu (x, t − τ, τ) = 0, this equation is an identity.
t
xx
Exercise 182. Modify the derivation of the diffusion equation
k
2
2
φ = a φ , a = , (9.2)
xx
t
cρ
so that it is valid for diffusion in a non-homogeneous medium for which c and k are
functions of x and φ and so that it is valid for a geometry in which A is a function
of x. Show that Equation (9.2) above is in this case replaced by

cρAφ = (kAφ ) .
t
x x
Recall that c is the specific heat, k is the thermal conductivity, ρ is the density, φ
is the temperature and A is the cross-sectional area.

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