304 Chapter 5 Higher Dimensions and Other Coordinates
variables by seeking solutions in the form
u(x,y,t)=φ(x,y)T(t).On substitutinguin product form into Eq. (1), we find that it becomes
(
∂^2 φ
∂x^2
+∂
(^2) φ
∂y^2
)
T=^1
kφT′.Separation can be achieved by dividing through byφT,whichleaves
(∂ (^2) φ
∂x^2 +
∂^2 φ
∂y^2
) 1
φ=T′
kT.We may argue, as usual, that the common value of the members of this equa-
tion must be a constant, which we expect to be negative(−λ^2 ).Theequations
that result are
T′+λ^2 kT= 0 , 0 <t, (5)
∂^2 φ
∂x^2 +∂^2 φ
∂y^2 =−λ(^2) φ, 0 <x<a, 0 <y<b. (6)
In terms of the product solutions, the boundary conditions become
φ(x, 0 )T(t)= 0 ,φ(x,b)T(t)= 0 ,
φ( 0 ,y)T(t)= 0 ,φ(a,y)T(t)= 0.
In order to satisfy all four equations, eitherT(t)≡0 for alltorφ=0onthe
boundary. We have seen many times that the choice ofT(t)≡0wipesoutour
solution completely. Therefore, we require thatφsatisfy the conditions
φ(x, 0 )= 0 ,φ(x,b)= 0 , 0 <x<a, (7)
φ( 0 ,y)= 0 ,φ(a,y)= 0 , 0 <y<b. (8)
We are not yet out of difficulty, because Eqs. (6)–(8) constitute a new prob-
lem, a two-dimensional eigenvalue problem. It is evident, however, that the
partial differential equation and the boundary conditions are linear and ho-
mogeneous; thus separation of variables may work again. Supposing thatφ
has the form
φ(x,y)=X(x)Y(y),
we find that the partial differential equation (6) becomes
X′′(x)
X(x) +
Y′′(y)
Y(y)=−λ
(^2) , 0 <x<a, 0 <y<b.