166 Chapter 2 The Heat Equation
Figure 5 Solution of the example, Eq. (20):u(x,t)is shown as a function ofx
for various times, which are chosen so that the dimensionless timekt/a^2 takes the
values 0.001, 0.01, 0.1, 1.0. For the illustration,T 0 has been chosen equal to 0 and
T 1 =100.
Nowthatwehavebeenthroughthreemajorexamples,wecanoutlinethe
method we have been using to solve linear boundary value–initial value prob-
lems. Up to this moment we have seen only homogeneous partial differential
equations, but a nonhomogeneity that is independent oftcan be treated by
the same technique.
Summary of Separation of Variables
Prepare
If the partial differential equation or a boundary condition or both are not
homogeneous, first find a functionv(x), independent oft, that satisfies the
partial differential equation and the boundary conditions. Sincev(x)does not
depend ont, the partial differential equation applied tov(x)becomes an ordi-
nary differential equation. Findingv(x)is just a matter of solving a two-point
boundary value problem.
Determine the initial value–boundary value problem satisfied by the “tran-
sient solution”w(x,t)=u(x,t)−v(x).Thismustbeahomogeneous problem.
That is, the partial differential equation and the boundary conditions (but not
usually the initial condition) are satisfied by the constant function 0.
Separate
Assuming thatw(x,t)=φ(x)T(t), with neither factor 0, separate the partial
differential equation into two ordinary differential equations, one forφ(x)and
one forT(t), linked by the separation constant,−λ^2. Reduce the boundary
conditions to conditions onφalone.