An introduction to partial differential equations 525

=

30

nπ

( 1 −cosnπ)

=0(whennis even) and

60

nπ

(whennis odd)

Hence, the required solution is:

u(x,t)=

∑∞

n= 1

{

Qne−p

(^2) c (^2) t
sinnπx
}

60
π
∑∞
n(odd)= 1
1
n
(sinnπx)e−n
(^2) π (^2) c (^2) t
Now try the following exercise
Exercise 202 Further problemson the heat
conduction equation

1. A metal bar, insulated along its sides, is 4m
   long. It is initially at a temperature of 10◦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.
   ⎡
   ⎣u(x,t)=^40
   π

∑∞

n(odd)= 1

1

n

e−

n^2 π^2 c^2 t

(^16) sin
nπx
4
⎤
⎦

2. An insulated uniform metal bar, 8m long,
   has the temperature of its ends maintained at
   0 ◦C, and at timet=0 the temperature dis-
   tribution f(x)along the bar is defined by
   f(x)=x( 8 −x).Ifc^2 =1, solve the heat con-

duction equation

∂^2 u

∂x^2

=

1

c^2

∂u

∂t

to determine

the temperatureuat any point in the bar at

timet.

⎡

⎣u(x,t)=

(

8

π

) (^3) ∑∞
n(odd)= 1
1
n^3
e−
n^2 π^2 t
(^64) sin
nπx
8
⎤
⎦

3. The ends ofan insulatedrodPQ, 20unitslong,
   are maintained at 0◦C. At timet=0, the tem-
   perature within the rod rises uniformly from
   each end reaching 4◦C at the mid-point of
   PQ.Find an expression for the temperature
   u(x,t)at any point in the rod, distantxfrom
   Pat any timetaftert=0. Assume the heat

conduction equation to be

∂^2 u

∂x^2

=

1

c^2

∂u

∂t

and

takec^2 = 1.

⎡

⎣u(x,t)=^320

π^2

∑∞

n(odd)= 1

1

n^2

sin

nπ

2

sin

nπx

20

e

−

(

n^2 π^2 t

400

)⎤

⎦

53.8 Laplace’s equation

The distribution of electrical potential, or temperature,

over a plane area subject to certain boundaryconditions,

can be described by Laplace’s equation. The potential

at a pointPin a plane (see Fig. 53.6) can be indicated

by an ordinate axis and is a function of its position, i.e.

z=u(x,y),whereu(x,y)is the solution of the Laplace

two-dimensional equation

∂^2 u

∂x^2

+

∂^2 u

∂y^2

=0.

The method of solution of Laplace’s equation is similar

to the previous examples, as shown below.

Figure 53.7 shows a rectangle OPQR bounded by

the lines x= 0 ,y= 0 ,x=a, and y=b, for which

we are required to find a solution of the equation

∂^2 u

∂x^2

+

∂^2 u

∂y^2

=0. The solutionz=(x,y)will give, say,

0 x

z y

P

Figure 53.6

0 x

z

y

u

(x, y

)

P

Q

R

y 5 b

x 5 a

Figure 53.7