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3.2 Solution of the Vibrating String Problem 221

and

bn

nπ

ac=

2

a

∫a

0

g(x)sin

(nπx

a

)

dx

or

bn=nπ^2 c

∫a

0

g(x)sin

(

nπx

a

)

dx. (11)

If the functionsf(x)andg(x)are sectionally smooth on the interval 0<x<
a, then we know that the initial conditions really are satisfied, except possibly
at points of discontinuity offorg. By the nature of the problem, however, one
would expect thatf, at least, would be continuous and would satisfyf( 0 )=
f(a)=0. Thus we expect the series forfto converge uniformly.

Example.
If the string is lifted in the middle and then released, appropriate initial condi-
tions are

u(x, 0 )=f(x)=







h·^2 ax, 0 <x<a 2 ,

h

(

2 −

2 x

a

)

,

a

2 <x<a,

∂u

∂t(x,^0 )=g(x)≡^0 ,^0 <x<a.

Thenbn=0 forn= 1 , 2 , 3 ,...,and

an=^2

a

[∫a/ 2

0

h·^2 x

a

sin

(nπx

a

)

dx+

∫a

a/ 2

h

(

2 −^2 x

a

)

sin

(nπx

a

)

dx

]

=^8 h

π^2

sin(nπ/ 2 )

n^2

.

Therefore the complete solution is

u(x,t)=

8 h

π^2

∑∞

n= 1

sin(nπ/ 2 )

n^2 sin

(nπx

a

)

cos

(nπct

a

)

. (12)

The CD shows an animated version of this solution.

Although the solution in the example can be considered valid, it is difficult
to see, in the present form, what shape the string will take at various times.
However, because of the simplicity of the sines and cosines, it is possible to
rewrite the solution in such a way thatu(x,t)may be determined without
summing a series.