Higher Engineering Mathematics

AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS 515

I

7. Solve

∂^2 u

∂x∂t

=sin(x+t) given that

∂u

∂x

= 1

whent=0, and whenu= 2 twhenx= 0.

[u=−sin(x+t)+x+sinx+ 2 t+sint]

8. Show thatu(x,y)=xy+

x

y

is a solution of

2 x

∂^2 u

∂x∂y

+y

∂^2 u

∂y^2

= 2 x.

9. Find the particular solution of the differ-

ential equation

∂^2 u

∂x∂y

=cosxcosygiven the

initial conditions that when y =π,

∂u

∂x

=x,

and whenx=π,u=2 cosy.

[

u=sinxsiny+

x^2

2

+2 cosy−

π^2

2

]

10. Verify thatφ(x,y)=xcosy+exsinysatis-
    fies the differential equation

∂^2 φ

∂x^2

+

∂^2 φ

∂y^2

+xcosy=0.

53.4 Some important engineering

partial differential equations

There are many types of partial differential equa-

tions. Some typically found in engineering and

science include:

(a) The wave equation, where the equation of
motion is given by:

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

wherec^2 =

T

ρ

, withTbeing the tension in a string

andρbeing the mass/unit length of the string.

(b) Theheat conduction equationis of the form:

∂^2 u

∂x^2

=

1

c^2

∂u

∂t

wherec^2 =

h

σρ

, withhbeing the thermal conduc-

tivity of the material,σthe specific heat of the

material, andρthe mass/unit length of material.

(c)Laplace’s equation, used extensively with elec-

trostatic fields is of the form:

∂^2 u

∂x^2

+

∂^2 u

∂y^2

+

∂^2 u

∂z^2

= 0.

(d) Thetransmission equation, where the poten-

tialuin a transmission cable is of the form:

∂^2 u

∂x^2

=A

∂^2 u

∂t^2

+B

∂u

∂t

+CuwhereA,BandCare

constants.

Some of these equations are used in the next sections.

53.5 Separating the variables

Letu(x,t)=X(x)T(t), whereX(x) is a function of

xonly andT(t) is a function oftonly, be a trial

solution to the wave equation

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

. If the

trial solution is simplified tou=XT, then

∂u

∂x

=X′T

and

∂^2 u

∂x^2

=X′′T. Also

∂u

∂t

=XT′and

∂^2 u

∂t^2

=XT′′.

Substituting into the partial differential equation

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

gives:

X′′T=

1

c^2

XT′′

Separating the variables gives:

X′′

X

=

1

c^2

T′′

T

Letμ=

X′′

X

=

1

c^2

T′′

T

whereμis a constant.

Thus, sinceμ=

X′′

X

(a function ofxonly), it must be

independent oft; and, sinceμ=

1

c^2

T′′

T

(a function

oftonly), it must be independent ofx.

Ifμis independent ofx and t, it can only be a con-

stant. Ifμ=

X′′

X

thenX′′=μXorX′′−μX=0 and

ifμ=

1

c^2

T′′

T

thenT′′=c^2 μT orT′′−c^2 μT=0.

Such ordinary differential equations are of the form

found in Chapter 50, and their solutions will depend

on whetherμ>0,μ=0 orμ<0.