Higher Engineering Mathematics

520 DIFFERENTIAL EQUATIONS

Now try the following exercise.

Exercise 202 Further problems on the wave

equation

1. An elastic string is stretched between two
   points 40 cm apart. Its centre point is dis-
   placed 1.5 cm from its position of rest at
   right angles to the original direction of the
   string and then released with zero velocity.
   Determine the subsequent 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, 80 cm apart, is deflected
   a distance of 1 cm from its position of rest
   perpendicular toPQand released initially
   with zero velocity. Apply the wave equation
   ∂^2 u
   ∂x^2

=

1

c^2

∂^2 u

∂t^2

wherec=8, to determine the

motion of a point 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 equa-

tion; 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

physical properties of the bar, i.e. the thermal con-

ductivity,h, the specific heat of the material,σ, and

the mass per unit length,ρ, of the bar. In the above

equation,c^2 =

h

σρ

With a uniform bar insulated, except at its ends,

any heat flow is along the bar and, at any instant, the

temperatureuat a pointPis a function of its distance

xfrom one end, and of the timet. Consider such a

u

=f

(x

, t

)

0

P

u(x, t)

L x

x

Figure 53.4

bar, shown in Fig. 53.4, where the bar extends from

x=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 function

ofxonly 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

thenX′′=−p^2 XorX′′+p^2 X=0,

givingX=Acospx+Bsinpx

and if−p^2 =

1

c^2

T′

T

then

T′

T

=−p^2 c^2 and integrat-

ing 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).