228 Chapter 3 The Wave Equation
It is actually possible to find the general solution of this last equation. Put in
another form it says
∂
∂z
(∂v
∂w)
= 0 ,
which means that∂v/∂wis independent ofz,or
∂v
∂w=θ(w).Integrating this equation, we find that
v=∫
θ(w)dw+φ(z).Here,φ(z)plays the role of an integration “constant.” Since the integral of
θ(w)is also a function ofw,wemaywritethegeneralsolution of the partial
differential equation foregoing as
v(w,z)=ψ(w)+φ(z),whereψandφarearbitraryfunctions with continuous derivatives. Trans-
forming back to our original variables, we obtain
u(x,t)=ψ(x+ct)+φ(x−ct) (1)as a form for the general solution of the one-dimensional wave equation. This
is known asd’Alembert’s solutionor thetraveling wave solution. It represents
the solution as the superposition of two waves, one moving to the left and the
other to the right, with propagation speedc.
Now let us look at the vibrating string problem:
∂^2 u
∂x^2
=^1
c^2∂^2 u
∂t^2, 0 <x<a, 0 <t, (2)
u( 0 ,t)= 0 , u(a,t)= 0 , 0 <t, (3)
u(x, 0 )=f(x), 0 <x<a, (4)
∂u
∂t(x, 0 )=g(x), 0 <x<a. (5)We already know a form foru. The problem is to chooseψandφin such a way
that the initial and boundary conditions are satisfied. We assume then that
u(x,t)=ψ(x+ct)+φ(x−ct).The initial conditions are
ψ(x)+φ(x)=f(x), 0 <x<a,
cψ′(x)−cφ′(x)=g(x), 0 <x<a.