3.6 Wave Equation in Unbounded Regions 243
u( 0 ,t)=h(t), 0 <t. (10)Asuis to be a solution of the wave equation, it must have the form
u(x,t)=ψ(x+ct)+φ(x−ct). (11)The two initial conditions, Eqs. (8) and (9), can be treated exactly as in the
first problem. Of course,G(x)≡0, and the constantA, being arbitrary, may
be taken as 0. The conclusion is that
ψ(x)= 0 ,φ(x)= 0 , 0 <x.Because bothxandtarepositiveinthisproblem,weseethatψ(x+ct)= 0
always, so Eq. (11) may be simplified to
u(x,t)=φ(x−ct). (12)The boundary condition Eq. (10) will tell us how to evaluateφfor negative
arguments. The equation is
u( 0 ,t)=φ(−ct)=h(t), 0 <t. (13)We now put together what we know of the functionφ:
φ(q)=
0 , q>0,h(
−
q
c)
, q<0.(14)
The argumentqis a dummy, used to avoid association with eitherxort.Equa-
tions (12) and (14) now specify the solutionu(x,t)completely.
Example.
Ta k eh(t)as shown in Fig. 6. The major steps to construct the graph ofφ(q).
Note that the graph ofφfor negative argument is that ofh,reflected.Inother
words, to make the graph ofφ(q), start from the graph ofh(t):(1)Graphthe
even extensionhe(t); (2) replace the right half (from 0 up) with 0; (3) adjust
scales so thatq=−cwheret=−1, etc.
The graphs in Fig. 7 showu(x,t)as a function ofxforvariousvaluesoft.It
is clear from both the graphs and the formula that the disturbance caused by
the variable boundary condition arrives at a fixed pointxat timex/c.Thusthe
disturbance travels with the velocityc, the wave speed. An example is animated
on the CD.
A wave equation accompanied by nonzero initial conditions and time-
varying boundary conditions can be solved by breaking it into two prob-
lems, one like Eqs. (1)–(4) with zero boundary condition, and the other like
Eqs. (7)–(10) with zero initial conditions.