Higher Engineering Mathematics, Sixth Edition

520 Higher Engineering Mathematics

have solutions:X=Aepx+Be−pxand

T=Cecpt+De−cptwhereA,B,CandDare constants.

ButX=0atx= 0 ,hence 0=A+Bi.e.B=−Aand

X=0atx=L,hence

0 =AepL+Be−pL=A(epL−e−pL).

Assuming (epL–e−pL)is not zero, thenA=0 and since

B=−A,thenB=0also.

This corresponds to the string being stationary; since it

is non-oscillatory, this solution will be disregarded.

Case 2:μ= 0

In this case, sinceμ=p^2 =0,T′′=0andX′′=0. We

will assume thatT(t)=0. SinceX′′=0,X′=aand

X=ax+bwhereaandbare constants. ButX=0at

x=0, henceb=0andX=axandX=0atx=L, hence

a=0. Thus, again, the solutionis non-oscillatoryand is

also disregarded.

Case 3:μ< 0

For convenience,

letμ=−p^2 thenX′′+p^2 X=0 from which,

X=Acospx+Bsinpx (1)

andT′′+c^2 p^2 T=0 from which,

T=Ccoscpt+Dsincpt (2)

(see worked Problem 4 above).

Thus, the suggested solutionu=XTnow becomes:

u={Acospx+Bsinpx}{Ccoscpt+Dsincpt}

(3)

Applying the boundary conditions:

(i) u=0whenx=0 for all values oft,

thus 0={Acos 0+Bsin0}{Ccoscpt

+Dsincpt}

i.e. 0 =A{Ccoscpt+Dsincpt}

from which,A= 0 ,(since{Ccoscpt

+Dsincpt} = 0 )

Hence, u={Bsinpx}{ Ccoscpt

+Dsincpt} (4)

(ii) u=0whenx=Lfor all values oft

Hence, 0={BsinpL}{Ccoscpt+Dsincpt}

NowB=0oru(x,t)would be identically zero.

Thus sinpL=0i.e.pL=nπorp=

nπ

L

for inte-

ger values ofn.

Substituting in equation (4) gives:

u=

{

Bsin

nπx

L

}{

Ccos

cnπt

L

+Dsin

cnπt

L

}

i.e. u=sin

nπx

L

{

Ancos

cnπt

L

+Bnsin

cnπt

L

}

(where constantAn=BCandBn=BD).There

will be many solutions, depending on the value of

n. Thus, more generally,

un(x,t)=

∑∞

n= 1

{

sin

nπx

L

(

Ancos

cnπt

L

+Bnsin

cnπt

L

)}

(5)

To findAnandBnwe put in the initial conditions

not yet taken into account.

(i) Att= 0 ,u(x, 0 )=f(x)for 0≤x≤L

Hence, from equation (5),

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

∑∞

n= 1

{

Ansin

nπx

L

}

(6)

(ii) Also att= 0 ,

[

∂u

∂t

]

t= 0

=g(x)for 0≤x≤L

Differentiatingequation(5)withrespecttotgives:

∂u

∂t

=

∑∞

n= 1

{

sin

nπx

L

(

An

(

−

cnπ

L

sin

cnπt

L

)

+Bn

(

cnπ

L

cos

cnπt

L

))}

and whent= 0 ,

g(x)=

∑∞

n= 1

{

sin

nπx

L

Bn

cnπ

L

}

i.e. g(x)=

cπ

L

∑∞

n= 1

{

Bnnsin

nπx

L

}

(7)

From Fourier series (see page 638) it may be shown that:

Anis twice the mean value off(x)sin

nπx

L

between

x=0andx=L

i.e. An=

2

L

∫L

0

f(x)sin

nπx

L

dx

forn= 1 , 2 , 3 ,...(8)