Higher Engineering Mathematics

518 DIFFERENTIAL EQUATIONS

andBn

(cnπ

L

)

is twice the mean value of

g(x)sin

nπx

L

betweenx=0 andx=L

i.e. Bn=

L

cnπ

(

2

L

)∫L

0

g(x)sin

nπx

L

dx

or Bn=

2

cnπ

∫L

0

g(x)sin

nπx

L

dx (9)

Summary of solution of the wave equation

The above may seem complicated; however a prac-
tical problem may be solved using the following
8-point procedure:

1. Identify clearly the initial and boundary
   conditions.


2. Assume a solution of the formu=XTand express
   the equations in terms ofX andT and their
   derivatives.


3. Separate the variables by transposing the equation
   and equate each side to a constant, say,μ;two
   separate equations are obtained, one inxand the
   other int.


4. Letμ=−p^2 to give an oscillatory solution.


5. The two solutions are of the form:

X=Acospx+Bsinpx

and T=Ccoscpt+Dsincpt.

Thenu(x,t)={Acospx+Bsinpx}{Ccoscpt+

Dsincpt}.

6. Apply the boundary conditions to determine con-
   stantsAandB.


7. Determine the general solution as an infinite sum.


8. Apply the remaining initial and boundary condi-
   tions and determine the coefficientsAnandBn
   from equations (8) and (9), using Fourier series
   techniques.

Problem 5. Figure 53.2 shows a stretched

string of length 50 cm which is set oscillating by

displacing its mid-point a distance of 2 cm from

its rest position and releasing it with zero veloc-

ity. Solve the wave equation:

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

wherec^2 =1, to determine the resulting motion

u(x,t).

u = f(x)

0 25

2

4

u

(x

, 0)

50 x(cm)

Figure 53.2

Following the above procedure,

1. The boundary and initial conditions given are:

u(0,t)= 0

u(50,t)= 0

}

i.e.fixed end points

u(x,0)=f(x)=

2

25

x 0 ≤x≤ 25

=−

2

25

x+ 4 =

100 −2x

25

25 ≤x≤ 50

(Note:y=mx+cis a straight line graph, so the

gradient,m, between 0 and 25 is 2/25 and the

y-axis intercept is zero, thusy=f(x)=

2

25

x+0;

between 25 and 50, the gradient=− 2 /25 and the

y-axis intercept is at 4, thusf(x)=−

2

25

x+4).

[

∂u

∂t

]

t= 0

= 0 i.e. zero initial velocity.

2. Assuming a solutionu = XT, where X is a
   function ofxonly, andTis a function oftonly,

then

∂u

∂x

=X′Tand

∂^2 u

∂x^2

=X′′Tand

∂u

∂y

=XT′and

∂^2 u

∂y^2

=XT′′. Substituting into the partial differen-

tial equation,

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

gives:

X′′T=

1

c^2

XT′′ i.e.X′′T=XT′′sincec^2 =1.