3.3 d’Alembert’s Solution 227
Solve the eigenvalue problem, sketch the first two eigenfunctions, and
compare them to the figure.
In Exercises 14–16, find a solution by separation of variables.
14.∂^2 u
∂x^2 =1
c^2(∂ (^2) u
∂t^2 +^2 k
∂u
∂t
)
,0<x<a,0<t,u( 0 ,t)=0, u(a,t)=0, 0 <t,u(x, 0 )=f(x), ∂∂ut(x, 0 )=0, 0 <x<a,
wheref(x)is as in Eq. (11). (Assume thatkis small and positive.)- ∂
(^2) u
∂x^2
=^1
c^2∂^2 u
∂t^2+γ^2 u,0<x<a,0<t,
u( 0 ,t)=0, u(a,t)=0, 0 <t,u(x, 0 )=h, ∂u
∂t(x, 0 )=0, 0 <x<a,wherehandγ^2 are constants.- ∂
(^2) u
∂x^2 =
1
c^2∂^2 u
∂t^2 ,0<x<a,0<t,
u( 0 ,t)=0,∂u
∂x(a,t)=0,^0 <t,
u(x, 0 )=0,∂u
∂t(x,^0 )=1,^0 <x<a.
17.Does the series in Eq. (12) converge uniformly?
18.In the text, we assumed that the ratioφ′′/φhad to be a negative con-
stant. Show that, ifφ′′/φ=p^2 >0 (or equivalently, ifφ′′−p^2 φ=0), then
the only function that also satisfies the boundary conditions, Eq. (7), is
φ(x)≡0.3.3 d’Alembert’s Solution
In Section 2 we saw that, in some cases, we could express the solution of the
wave equation directly in terms of the initial data. From this evidence we might
suspect that there is something special aboutx+ctandx−ct.Totestthisidea,
we change variables and see what the wave equation looks like. Letw=x+ct,
z=x−ct,andu(x,t)=v(w,z). Then a calculation using the chain rule shows
that the wave equation becomes (see Exercise 11)
∂^2 v
∂z∂w=^0.