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636 CHAPTER 17 The Heat Equation

17.5 Heat Conduction in an Infinite Cylinder

We will determine the temperature distribution in a solid, infinitely long, homogeneous cylinder
of radius R with its axis along the z-axis in 3-space.
In cylindrical coordinates, the heat equation for the temperature distribution U(r,θ, z,t) is
2
2
2
∂U ∂ U 1 ∂U 1 ∂ U ∂ U
= k + + + .
2
∂t ∂r 2 r ∂r r ∂θ 2 ∂z 2
This is a formidable equation to solve at this stage, and we will restrict it to the special case
that the temperature at any point in the cylinder depends only on the time t and the horizontal
distance r from the z-axis. This symmetry means that ∂U/∂θ =∂U/∂z =0, and the heat equation
becomes
2
∂U ∂ U 1 ∂U
= k +
∂t ∂r 2 r ∂r
for 0 ≤ r < R,t > 0. We will write U(r,t), with dependence only on r and t and assume the
boundary condition
U(R,t) = 0for t > 0.

The initial condition is
U(r,0) = f (r) for 0 ≤r < R.
Put U(r,t) = F(r)T(t) and separate the variables, obtaining
T F + (1/r)F (r)

= =−λ.
kT F(r)
Then
1

T + λT = 0 and F + F + λF = 0.

r
Since U(R,t) = F(R)T (t) = 0, then F(R) = 0. The problem for F is a singular Sturm-Liouville
problem on [0, R]. To determine the eigenvalues and eigenfunctions take cases on λ.
Case 1: λ = 0
Then
1

F + F = 0
r
with general solution of the form F(r) = c ln(r) + d. Since ln(r) →−∞ as r → 0 (the center

of the cylinder), we must choose c = 0. Then F(r) = d. Since T (t) = 0if λ = 0, then T (t) =
constant also. In this case, U(r,t)= constant, and this constant must be zero because U(R,0)=0.
U(r,t) = 0 is indeed the solution if f (r) = 0. If f (r) is not identically zero, then λ = 0 does not
contribute to the solution.

Case 2: λ< 0
Write λ =−ω with ω> 0. Now T − kω T = 0 has a general solution
2
2

2
T (t) = ce ω kt ,
and this is unbounded as t increases. To have a bounded solution, we reject this case.

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October 14, 2010 15:25 THM/NEIL Page-636 27410_17_ch17_p611-640