Higher Engineering Mathematics, Sixth Edition

An introduction to partial differential equations 523

Now try the following exercise

Exercise 201 Further problemson the

waveequation

1. An elastic string is stretched between two
   points40cm apart. Its centre pointis displaced
   1.5cm from its position of rest at right angles
   to the original direction of the string and then
   released with zero velocity.Determinethesub-
   sequent motionu(x,t)by applying the wave

equation

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

withc^2 = 9.

[

u(x,t)=

12

π^2

∑∞

n= 1

1

n^2

sin

nπ

2

sin

nπx

40

cos

3 nπt

40

]

2. The centre point of an elastic string between
   two pointsPandQ, 80cm apart, is deflected
   a distance of 1cm from its position of
   rest perpendicular to PQand released ini-
   tially with zero velocity. Apply the wave

equation

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

wherec=8, to deter-

minethemotionofapoint distancexfromPat

timet.

[

u(x,t)=

8

π^2

∑∞

n= 1

1

n^2

sin

nπ

2

sin

nπx

80

cos

nπt

10

]

53.7 The heat conduction equation

The heat conduction equation

∂^2 u

∂x^2

=

1

c^2

∂u

∂t

is solved

in a similar manner to that for the wave equation; the
equation differs only in that the right hand side contains
a first partial derivative instead of the second.
The conduction of heat in a uniform bar depends on
the initial distribution of temperature and on the phys-
ical properties of the bar, i.e. the thermal conductivity,
h, the specific heat of the material,σ,andthemass
per unit length,ρ, of the bar. In the above equation,

c^2 =

h
σρ
With a uniformbar insulated,except at itsends, any heat
flow is along the bar and, at any instant, the temperature
uat a pointPis a function of its distancexfrom one
end, and of the timet. Consider such a bar, shown in

0

x

x

y

P

L

u u(x, t )

5

f^ (

x, t

)

Figure 53.4

Fig. 53.4, where the bar extends fromx=0tox=L,the

temperature of the ends of the bar is maintained at zero,

and the initial temperature distribution along the bar is

defined byf(x).

Thus, the boundary conditions can be expressed as:

u( 0 ,t)= 0

u(L,t)= 0

}

for allt≥ 0

and u(x, 0 )=f(x)for 0≤x≤L

As with the wave equation, a solution of the form

u(x,t)=X(x)T(t)is assumed, whereXis a functionof

xonly andTis a function oftonly. If the trial solution

is simplified tou=XT,then

∂u

∂x

=X′T

∂^2 u

∂x^2

=X′′Tand

∂u

∂t

=XT′

Substituting into the partial differential equation,

∂^2 u

∂x^2

=

1

c^2

∂u

∂t

gives:

X′′T=

1

c^2

XT′

Separating the variables gives:

X′′

X

=

1

c^2

T′

T

Let −p^2 =

X′′

X

=

1

c^2

T′

T

where−p^2 is a constant.

If−p^2 =

X′′

X

then X′′=−p^2 X or X′′+p^2 X= 0 ,

givingX=Acospx+Bsinpx

and if−p^2 =

1

c^2

T′

T

then

T′

T

=−p^2 c^2 and integrating

with respect totgives:

∫

T′

T

dt=

∫

−p^2 c^2 dt

from which, lnT=−p^2 c^2 t+c 1

The left hand integral is obtained by an algebraic

substitution (see Chapter 39).