AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS 521

I

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 conditions u(0,t)= 0
gives: 0={Pcos 0+Qsin 0}e−p
(^2) c (^2) t
=Pe−p
(^2) c (^2) t
from
which,P=0 andu(x,t)=Qsinpxe−p
(^2) c (^2) t
.
Also, u(L,t)=0 thus, 0=QsinpLe−p
(^2) c (^2) t
and
sinceQ=0 then sinpL=0 from which,pL=nπ
orp=
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
}
Applying the remaining boundary condition, that
whent=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
}
This method of solution is demonstrated in the
following worked problem.
Problem 6. A metal bar, insulated along its
sides, is 1 m long. It is initially at room tem-
perature 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 timetseconds 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)= 0 and u(x,0)= 15
u
(x
, t
)
u(
x,0)
0
0
1
1
x(m)
x(m)
15
u(x, t)
P
x
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=0 andu(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
wheren=1, 2, 3,...
Hence,u(x,t)=
∑∞
n= 1
{
Qne−p
(^2) c (^2) t
sinnπx
}
The final initial condition given was that att=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 15 sinnπxfrom
x=0tox=1,
i.e. Qn=
2
1
∫ 1
0
15 sinnπxdx= 30
[
−
cosnπx
nπ
] 1
0
=−
30
nπ
[cosnπ−cos 0]