Page 652 - Advanced_Engineering_Mathematics o'neil

632 CHAPTER 17 The Heat Equation

be in terms of x with the transformed t variable s carried along as a parameter. The second
derivative term with respect to x passes through the transform, which is with respect to t because
x and t are independent. From the partial differential equation, we obtain
∂ 2
sU(x,s) − u(x,0) = k U(x,s).
∂x 2
If we write
∂U(x,s)

= U (x,s),
∂x
the diffusion equation transforms to

sU(x,s) − T 0 = kU (x,s).
This is the second-order ordinary differential equation

s 1

U (x,s) − U(x,s) =− T 0 .
k k
By inspection, a particular solution of this equation is U p = T 0 /s. For the general solution, we
need the general solution of the associated homogeneous equation
s
U (x,s) − U(x,s) = 0.

k
This has the characteristic equation
s
2
λ − = 0
k
√
with roots ± s/k. Therefore, the general solution for U(x,s) is
√
√ s/kx − s/kx 1
U(x,s) = c 1 e + c 2 e + T 0 .
s
Now use the boundary conditions. Take the transform of u(L,t) = 0 to obtain U(L,s) = 0.
Similarly, (∂/∂x)u(0,t) = 0 gives us (∂U/∂x)(0,s) = 0. From the second of these conditions,
we have

∂U s s
(0,s) = c 1 − c 2 = 0,
∂x k k
so c 1 = c 2 and U has the form

s
T 0
U(x,s) = + c cosh x
s k
in which c can be any constant. From the first boundary condition, we have

T 0 s
U(L,s) = 0 = + c cosh L ,
s k
implying that

T 0
c =− √
.
s cosh s/kL
We now have the transform of the solution:
√
cosh s/kx
T 0 T 0
U(x,s) = − √
.
s s cosh s/kL

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October 14, 2010 15:25 THM/NEIL Page-632 27410_17_ch17_p611-640