1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

150 Chapter 2 The Heat Equation

anddeterminedthatwsatisfies the boundary value–initial value problem

∂^2 w

∂x^2 =

1

k

∂w

∂t,^0 <x<a,^0 <t, (8)

w( 0 ,t)= 0 , 0 <t, (9)

w(a,t)= 0 , 0 <t, (10)

w(x, 0 )=f(x)−v(x)≡g(x), 0 <x<a. (11)

Our objective is to determine the transient temperature distribution,w(x,t),
and — sincev(x)is already known — the unknown temperature will be

u(x,t)=v(x)+w(x,t). (12)
The problem inwcan be attacked by a method calledproduct method, sepa-
ration of variables,orFourier’s method. For this method to work, it is essential
to have homogeneous partial differential equation and boundary conditions.
Thus, the method may be applied to the transient distributionwbutnotto
the original functionu. Of course, because both the partial differential equa-
tion and the boundary conditions satisfied byw(x,t)are homogeneous, the
functionw≡0 satisfies them. Because this solution itself is obvious and is of
no help in satisfying the initial condition, it is called thetrivial solution.We
are seeking the unobvious, nontrivial solutions, so we shall avoid the trivial
solution at every turn.
The general idea of the method is to assume that the solution of the par-
tial differential equation has the form of a product:w(x,t)=φ(x)T(t).We
require that neither of the factorsφ(x)andT(t)be identically 0, since that
would lead back to the trivial solution. Now, each of the factors depends on
only one variable, so we have

∂^2 w

∂x^2 =φ

′′(x)T(t), ∂w

∂t =φ(x)T

′(t).

The partial differential equation becomes

φ′′(x)T(t)=^1

k

φ(x)T′(t),

and on dividing through byφTwe find

φ′′(x)

φ(x)

=T

′(t)

kT(t)

, 0 <x<a, 0 <t.

Here is the key argument: The ratio on the left contains functions ofxalone
and cannot vary witht. On the other hand, the ratio on the right contains
functions oftalone and cannot vary withx. Since this equality must hold for