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One-Dimensional Heat Equation
.

Solution. Setting L(y(x, t)) = Y (x, s) and taking the Laplace transform of
both sides of (5.16) give

bx 2
+
xY (x, s) + sY (x, s) − y(x, 0 ) + aY (x, s) = ,
x
s
that is,
dY bx 2
x + (s + a)Y =
dx s
or
dY (s + a) bx
+ Y = (s > 0).
dx x s
Solving this first-order ODE using an integrating factor gives

bx 2
Y (x, s) = + cx −(s+a) (x > 0, s > −a).
s(s + a + 2)

Taking the Laplace transform of the boundary condition y(0, t) = 0 gives

Y (0, s) = L(y(0, t)) = 0,

and thus c = 0. Therefore,

bx 2
Y (x, s) = ,
s(s + a + 2)

and inverting,
bx 2
y(x, t) = (1 − e −(a+2)t .
a + 2
. . . .

One-Dimensional Heat Equation

The heat flow in a finite or semi-infinite thin rod is governed by the PDE
2
∂u ∂ u
= c ,
∂t ∂x 2
where c is a constant (called the diffusivity), and u(x, t) is the temperature at

position x and time t. The temperature over a cross-section at x is taken to be
uniform. Many different scenarios can arise in the solution of the heat equation;
.
we will consider several to illustrate the various techniques involved.

Example 5.20, Solve

2
∂u ∂ u
= , x > 0, t > 0, (5.17)
∂t ∂x 2
. . . . .

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