AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS 519
I
- Separating the variables gives:
X′′
X
=
T′′
T
Let constant,
μ=
X′′
X
=
T′′
T
thenμ=
X′′
X
andμ=
T′′
T
from which,
X′′−μX= 0 and T′′−μT= 0.
- Lettingμ=−p^2 to give an oscillatory solution
gives:
X′′+p^2 X= 0 and T′′+p^2 T= 0
The auxiliary equation for each is:m^2 +p^2 = 0
from which,m=
√
−p^2 =±jp.
- Solving each equation gives:
X=Acospx+BsinpxandT=Ccospt+Dsinpt.
Thus,
u(x,t)={Acospx+Bsinpx}{Ccospt+Dsinpt}.
- Applying the boundary conditions to determine
constantsAandBgives:
(i)u(0,t)=0, hence 0=A{Ccospt+Dsinpt}
from which we conclude thatA=0.
Therefore,
u(x,t)=Bsinpx{Ccospt+Dsinpt} (a)
(ii)u(50,t)=0, hence
0 =Bsin 50p{Ccospt + Dsinpt}. B=0,
hence sin 50p=0 from which, 50p=nπand
p=
nπ
50
- Substituting in equation (a) gives:
u(x,t)=Bsin
nπx
50
{
Ccos
nπt
50
+Dsin
nπt
50
}
or, more generally,
un(x,t)=
∑∞
n= 1
sin
nπx
50
{
Ancos
nπt
50
+Bnsin
nπt
50
}
(b)
whereAn=BCandBn=BD.
- From equation (8),
An=
2
L
∫L
0
f(x) sin
nπx
L
dx
=
2
50
[∫
25
0
(
2
25
x
)
sin
nπx
50
dx
+
∫ 50
25
(
100 − 2 x
25
)
sin
nπx
50
dx
]
Each integral is determined using integration by
parts (see Chapter 43, page 418) with the result:
An=
16
n^2 π^2
sin
nπ
2
From equation (9),
Bn=
2
cnπ
∫L
0
g(x) sin
nπx
L
dx
[
∂u
∂t
]
t= 0
= 0 =g(x) thus,Bn= 0
Substituting into equation (b) gives:
un(x,t)=
∑∞
n= 1
sin
nπx
50
{
Ancos
nπt
50
+Bnsin
nπt
50
}
=
∑∞
n= 1
sin
nπx
50
{
16
n^2 π^2
sin
nπ
2
cos
nπt
50
+(0) sin
nπt
50
}
Hence,
u(x,t)=
16
π^2
∑∞
n= 1
1
n^2
sin
nπx
50
sin
nπ
2
cos
nπt
50
For stretched string problems as in problem 5 above,
the main parts of the procedure are:
- DetermineAnfrom equation (8).
Note that
2
L
∫L
0
f(x) sin
nπx
L
dxisalwaysequal
to
8 d
n^2 π^2
sin
nπ
2
(see Fig. 53.3)
- DetermineBnfrom equation (9)
- Substitute in equation (5) to determineu(x,t)
y
d
0 Lx
y=f(x)
L
2
Figure 53.3
