10—Partial Differential Equations 256
without bound. Terms such aseπz/Lhowever increase withz. This means that the coefficients of the
terms that increase exponentially inzcannot be there.
An= 0forn > 0 , and Bn= 0forn < 0
V(x,z) =A 0 +B 0 +
∑∞
n=1
enπix/LBne−nπz/L+
∑−^1
n=−∞
enπix/LAnenπz/L (10.40)
The combined constantA 0 +B 0 is really one constant; you can call itC 0 if you like. Now use the
usual Fourier techniques given that you know the potential atz= 0.
V(x,0) =C 0 +
∑∞
n=1
Bnenπix/L+
∑−^1
n=−∞
Anenπix/L
The scalar product ofemπix/Lwith this equation is
〈
emπix/L,V(x,0)
〉
=
2 LC 0 (m= 0)
2 LBm (m > 0 )
2 LAm (m < 0 )
(10.41)
Now evaluate the integral on the left side. First,m 6 = 0:
〈
emπix/L,V(x,0)
〉
=
∫L
−L
dxe−mπix/L
{
0 (−L < x < 0 )
V 0 ( 0 < x < L)
=V 0
∫L
0
dxe−mπix/L=V 0
L
−mπi
e−mπix/L
∣∣
∣
L
0
=V 0
L
−mπi
[
(−1)m− 1
]
Then evaluate it separately form= 0, and you have
〈
1 ,V(x,0)
〉
=V 0 L.
Now assemble the result. Before plunging in, look at what will happen.
Them= 0term sits by itself.
For the other terms, only oddmhave non-zero values.
V(x,z) =
1
2
V 0 +V 0
∑∞
m=1
1
− 2 mπi
[
(−1)m− 1
]
emπix/Le−mπz/L
+V 0
∑−^1
m=−∞
1
− 2 mπi
[
(−1)m− 1
]
emπix/Le+mπz/L
(10.42)
To put this into a real form that is easier to interpret, change variables, lettingm=−nin the second
sum andm=nin the first, finally changing the sum so that it is over just the odd terms.
V(x,z) =
1
2
V 0 +V 0
∑∞
n=1
1
− 2 nπi
[
(−1)n− 1
]
enπix/Le−nπz/L
+V 0
∑∞
1
1
+2nπi
[
(−1)n− 1
]
e−nπix/Le−nπz/L
=
1
2
V 0 +V 0
∑∞
n=1
[
(−1)n− 1
] 1
−nπ
sin(nπx/L)e−nπz/L
=
1
2
V 0 +
2
π
V 0
∑∞
`=0
1
2 `+ 1
sin
(
(2`+ 1)πx/L
)
e−(2`+1)πz/L
(10.43)
