An introduction to partial differential equations 521
andBn
(cnπ
L
)
is twice the mean value of
g(x)sin
nπx
L
betweenx=0andx=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 practi-
cal problem may be solved using the following8-point
procedure:
- Identify clearly the initial and boundary
conditions. - Assume a solution of the formu=XTand express
the equations in terms of X and T and their
derivatives. - 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. - Letμ=−p^2 to give an oscillatory solution.
- The two solutions are of the form:
X=Acospx+Bsinpx
and T=Ccoscpt+Dsincpt.
Thenu(x,t)={Acospx+Bsinpx}{Ccoscpt+
Dsincpt}.
- Apply the boundary conditions to determine con-
stantsAandB. - Determine the general solution as an infinite sum.
- Apply the remaining initial and boundary condi-
tions and determine the coefficientsAn and Bn
from equations (8) and (9), using Fourier series
techniques.
Problem 5. Figure 53.2 shows a stretched string
of length 50cm which is set oscillating by
displacing its mid-point a distance of 2cm from its
rest position and releasing it with zero velocity.
Solve the wave equation:
∂^2 u
∂x^2
=
1
c^2
∂^2 u
∂t^2
where
c^2 =1, to determine the resulting motionu(x,t).
0
2
4
25 50 x (cm)
u(
x,
0 )
u 5 f (x )
y
Figure 53.2
Following the above procedure,
- 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+cisastraightlinegraph,sothegra-
dient,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.
- Assuming a solutionu=XT, where X is a func-
tion ofx only, andT is a function oftonly,
then
∂u
∂x
=X′Tand
∂^2 u
∂x^2
=X′′Tand
∂u
∂y
=XT′and
∂^2 u
∂y^2
=XT′′.Substitutingintothepartialdifferential
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.
- Separating the variables gives:
X′′
X
=
T′′
T
Let constant,
μ=
X′′
X
=
T′′
T
thenμ=
X′′
X
andμ=
T′′
T