Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PDES: GENERAL AND PARTICULAR SOLUTIONS

discussed in the next chapter. Nevertheless, we now considerD’Alembert’s solution

u(x, t) of the wave equation subject to initial conditions (boundary conditions) in

the following general form:

initial displacement,u(x,0) =φ(x); initial velocity,

∂u(x,0)

∂t

=ψ(x).

The functionsφ(x)andψ(x) are given and describe the displacement and velocity

of each part of the string at the (arbitrary) timet=0.

It is clear that what we need are the particular forms of the functionsfandg

in (20.26) that lead to the required values att= 0. This means that

φ(x)=u(x,0) =f(x−0) +g(x+0), (20.27)

ψ(x)=

∂u(x,0)

∂t

=−cf′(x−0) +cg′(x+0), (20.28)

where it should be noted thatf′(x−0) stands fordf(p)/dpevaluated, after the

differentiation, atp=x−c×0; likewise forg′(x+0).

Looking on the above two left-hand sides as functions ofp=x±ct, but

everywhere evaluated att= 0, we may integrate (20.28) between an arbitrary

(and irrelevant) lower limitp 0 and an indefinite upper limitpto obtain

1

c

∫p

p 0

ψ(q)dq+K=−f(p)+g(p),

the constant of integrationKdepending onp 0. Comparing this equation with

(20.27), withxreplaced byp, we can establish the forms of the functionsfand

gas

f(p)=

φ(p)

2

−

1

2 c

∫p

p 0

ψ(q)dq−

K

2

, (20.29)

g(p)=

φ(p)

2

+

1

2 c

∫p

p 0

ψ(q)dq+

K

2

. (20.30)

Adding (20.29) withp=x−ctto (20.30) withp=x+ctgives as the solution to

the original problem

u(x, t)=

1

2

[φ(x−ct)+φ(x+ct)]+

1

2 c

∫x+ct

x−ct

ψ(q)dq, (20.31)

in which we notice that all dependence onp 0 has disappeared.

Each of the terms in (20.31) has a fairly straightforward physical interpretation.

In each case the factor 1/2 represents the fact that only half a displacement profile

that starts at any particular point on the string travels towards any other position

x, the other half travelling away from it. The first term^12 φ(x−ct) arises from

the initial displacement at a distancectto the left ofx; this travels forward

arriving atxat timet. Similarly, the second contribution is due to the initial

displacement at a distancectto the right ofx. The interpretation of the final