524 Higher Engineering Mathematics

If lnT=−p^2 c^2 t+c 1 then

T=e−p

(^2) c (^2) t+c 1
=e−p
(^2) c (^2) t
ec^1 i.e. T=ke−p
(^2) c (^2) t
(where
constantk=ec^1 ).
Hence, u(x,t)=XT={Acospx+Bsinpx}ke−p
(^2) c (^2) t
i.e.u(x,t)={Pcospx+Qsinpx}e−p
(^2) c (^2) t
where
P=AkandQ=Bk.
Applying the boundary conditionsu( 0 ,t)=0gives:
0 ={Pcos 0+Qsin0}e−p
(^2) c (^2) t
=Pe−p
(^2) c (^2) t
from which,
P=0andu(x,t)=Qsinpxe−p
(^2) c (^2) t
.
Also,u(L,t)=0 thus, 0=QsinpLe−p
(^2) c (^2) t
and since
Q=0thensinpL=0 from which,pL=nπorp=
nπ
L
wheren= 1 , 2 , 3 ,...
There are therefore many values ofu(x,t).
Thus, in general,
u(x,t)=
∑∞
n= 1
{
Qne−p
(^2) c (^2) t
sin
nπx
L
}
Applyingthe remaining boundary condition, that when
t= 0 ,u(x,t)=f(x)for 0≤x≤L,gives:
f(x)=
∑∞
n= 1
{
Qnsin
nπx
L
}
From Fourier series, Qn= 2 ×mean value of
f(x)sin
nπx
L
fromxtoL.
Hence, Qn=
2
L
∫ L
0
f(x)sin
nπx
L
dx
Thus, u(x,t)=
2
L
∑∞
n= 1
{(∫ L
0
f(x)sin
nπx
L
dx
)
e−p
(^2) c (^2) t
sin
nπx
L
}
Thismethodofsolutionisdemonstratedinthefollowing
worked problem.
Problem 6. A metal bar, insulated along its sides,
is 1m long. It is initially at room temperature of
15 ◦C and at timet=0, the ends are placed into ice
at 0◦C. Find an expression for the temperature at a
pointPat a distancexm from one end at any time
tseconds aftert=0.
The temperatureualong the length of bar is shown in
Fig. 53.5.
The heat conduction equation is
∂^2 u
∂x^2

1
c^2
∂u
∂t
and the
given boundary conditions are:
u( 0 ,t)= 0 ,u( 1 ,t)=0andu(x, 0 )= 15
(^01)
1
15
x
u^ (
x,
0 )
x (m )
(^0) x (m )
P
u(x, t )
u(
x, t
)
Figure 53.5
Assuming a solution of the formu=XT, then, from
above,
X=Acospx+Bsinpx
and T=ke−p
(^2) c (^2) t
.
Thus, the general solution is given by:
u(x,t)={Pcospx+Qsinpx}e−p
(^2) c (^2) t
u( 0 ,t)=0 thus 0=Pe−p
(^2) c (^2) t
from which,P=0andu(x,t)={Qsinpx}e−p
(^2) c (^2) t
.
Also,u( 1 ,t)=0 thus 0={Qsinp}e−p
(^2) c (^2) t
.
Since Q= 0 ,sinp=0 from which, p=nπ where
n= 1 , 2 , 3 ,...
Hence,u(x,t)=
∑∞
n= 1
{
Qne−p
(^2) c (^2) t
sinnπx
}
The final initial condition given was that at t= 0 ,
u=15, i.e.u(x, 0 )=f(x)=15.
Hence, 15=
∑∞
n= 1
{Qnsinnπx} where, from Fourier
coefficients,Qn= 2 ×mean value of 15sinnπxfrom
x=0tox= 1 ,
i.e. Qn=
2
1
∫ 1
0
15sinnπxdx= 30
[
−
cosnπx
nπ
] 1
0
=−
30
nπ
[cosnπ−cos0]