3.2 Solution of the Vibrating String Problem 219
∂^2 u
∂x^2
=^1
c^2
∂^2 u
∂t^2
, 0 <x<a, 0 <t, (1)
u( 0 ,t)= 0 , u(a,t)= 0 , 0 <t, (2)
u(x, 0 )=f(x), 0 <x<a, (3)
∂u
∂t(x,^0 )=g(x),^0 <x<a, (4)
contains a linear, homogeneous partial differential equation and linear, homo-
geneous boundary conditions. Thus we may apply the method of separation of
variables with hope of success. If we assume that^1 u(x,t)=φ(x)T(t), Eq. (1)
becomes
φ′′(x)T(t)=
1
c^2 φ(x)T
′′(t).
Dividing through byφT,weobtain
φ′′(x)
φ(x)=
T′′(t)
c^2 T(t),^0 <x<a,^0 <t.
For the equality to hold, both members of this equation must be constant.
We write the constant as−λ^2 and separate the preceding equation into two
ordinary differential equations linked by the common parameterλ^2 :
T′′+λ^2 c^2 T= 0 , 0 <t, (5)
φ′′+λ^2 φ= 0 , 0 <x<a. (6)
The boundary conditions become
φ( 0 )T(t)= 0 ,φ(a)T(t)= 0 , 0 <t
and, sinceT(t)≡0 gives a trivial solution foru(x,t),wemusthave
φ( 0 )= 0 ,φ(a)= 0. (7)
TheeigenvalueproblemEqs.(6)and(7)isexactlythesameastheonewe
have seen and solved before. (See Chapter 2, Section 3.) We know that the
eigenvalues and eigenfunctions are
λ^2 n=
(
nπ
a
) 2
,φn(x)=sin(λnx), n= 1 , 2 , 3 ,....
Equation (5) is also of a familiar type, and its solution is known to be
Tn(t)=ancos(λnct)+bnsin(λnct),
(^1) Tno longer symbolizes tension.