194 Chapter 2 The Heat Equation
Using the same techniques as before, we look for solutions in the form
u(x,t)=φ(x)T(t)so that the heat equation (1) becomes
φ′′(x)
φ(x)=T′(t)
T(t)=constant.As in the previous section, the constant must be nonpositive (say,−λ^2 )in
order for the solutions to be bounded. Thus, we have the singular eigenvalue
problem
φ′′+λ^2 φ= 0 , −∞<x<∞,
φ(x) bounded asx→±∞.It is easy to see thateverysolution ofφ′′/φ=−λ^2 is bounded. Thus, our fac-
torsφ(x)andT(t)are
φ(x;λ)=Acos(λx)+Bsin(λx),
T(t;λ)=exp(
−λ^2 kt)
.
We c o m b i n e t h e s o l u t i o n sφ(x)T(t)intheformofanintegraltoobtain
u(x,t)=∫∞
0(
A(λ)cos(λx)+B(λ)sin(λx))
exp(
−λ^2 kt)
dλ. (4)At timet=0, the exponential factor becomes 1, and the initial condition is
∫∞
0(
A(λ)cos(λx)+B(λ)sin(λx))
dλ=f(x), −∞<x<∞.As this is clearly a Fourier integral problem, we must chooseA(λ)andB(λ)to
be the Fourier integral coefficient functions,
A(λ)=^1
π∫∞
−∞f(x)cos(λx)dx, B(λ)=^1
π∫∞
−∞f(x)sin(λx)dx. (5)Then the functionu(x,t)in Eq. (4) satisfies the partial differential equation (1)
and the initial condition (2), provided thatfis sectionally smooth and|f(x)|
has a finite integral. It can be proved that the boundedness condition (3) is
also satisfied, provided that the initial valuef(x)is bounded asx→±∞.
Example.
Solve the problem posed in Eqs. (1)–(3) with
f(x)={ 0 , x<−a,
T 0 , −a<x<a,
0 , a<x.