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Solved Problems

Consider 1. The initial condition is small, it satisfies |u(x, 0)| < . However the
solution for any positive time can be made arbitrarily large by choosing a sufficiently

large, positive value of λ. We can make the solution exceed the value M at time t
by choosing a value of λ such that

e λκt > M

1 M
λ > ln .
κt
Thus we see that Equation 9.1 is ill posed because the solution does not depend

continuously on the initial data. A small change in the initial condition can produce
an arbitrarily large change in the solution for any fixed time.

Exercise 180. Derive the heat equation for a general 3 dimensional body, with

non-uniform density ρ(x), specific heat c(x), and conductivity k(x). Show that

∂u(x, t) 1
= ∇ · (k∇u(x, t))
∂t cρ

where u is the temperature, and you may assume there are no internal sources or

sinks.
Solution: Consider a Region of material, R. Let u be the temperature and φ be
the heat flux. The amount of heat energy in the region is

Z
cρu dx.
R

We equate the rate of change of heat energy in the region with the heat flux across
the boundary of the region.

d Z Z
cρu dx = − φ · n ds
dt R ∂R

We apply the divergence theorem to change the surface integral to a volume integral.
Z Z
d
cρu dx = − ∇ · φ dx
dt R R
Z
∂u
cρ + ∇ · φ dx = 0
R ∂t

Since the region is arbitrary, the integral must vanish identically.

∂u
cρ = −∇ · φ
∂t

We apply Fourier’s law of heat conduction, φ = −k∇u, to obtain the heat equation.

∂u 1
= ∇ · (k∇u)
∂t cρ

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