234 Chapter 3 The Wave Equation
The functionw(x,t), being the difference betweenu(x,t)andv(x),satisfies
the initial value–boundary value problem
∂
∂x(
s(x)∂w∂x)
=p(cx 2 )∂(^2) w
∂t^2 , l<x<r,^0 <t, (6)
α 1 w(l,t)−α 2 ∂w
∂x
(l,t)= 0 , 0 <t, (7)
β 1 w(r,t)+β 2 ∂w
∂x
(r,t)= 0 , 0 <t, (8)
w(x, 0 )=f(x)−v(x), l<x<r, (9)
∂w
∂t(x,^0 )=g(x), l<x<r. (10)
Since the equation and the boundary conditions are homogeneous and lin-
ear, we attempt a solution by separation of variables. Ifw(x,t)=φ(x)T(t),we
find in the usual way that the factor functionsφandTmust satisfy
T′′+c^2 λ^2 T= 0 , 0 <t, (11)
(
s(x)φ′
)′
+λ^2 p(x)φ= 0 , l<x<r, (12)
α 1 φ(l)−α 2 φ′(l)= 0 , (13)
β 1 φ(r)+β 2 φ′(r)= 0. (14)
The eigenvalue problem represented in the last three lines is a regular
Sturm–Liouville problem, because of the assumptions we have made about
s,p, and the coefficients. We know that there are an infinite number of non-
negative eigenvaluesλ^21 ,λ^22 ,...and corresponding eigenfunctionsφ 1 ,φ 2 ,...
that have the orthogonality property
∫r
lφn(x)φm(x)p(x)dx= 0 , n=m.The solution of the equation forTis
Tn(t)=ancos(λnct)+bnsin(λnct).From here it is clear that the frequencies of vibration that occur in the solution
of Eqs. (1)–(3) areλnc/ 2 π(cycles per time unit). Thus, it is the eigenvalues
coming from Eqs. (12)–(14) that determine these frequencies.
Having solved the subsidiary problems that arose after separation of vari-
ables, we can begin to assemble the solution. The functionwwill have the
form
w(x,t)=∑∞
n= 1φn(x)(
ancos(λnct)+bnsin(λnct)