Mathematical Tools for Physics

10—Partial Differential Equations 303

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

(39)

To put this into a real form that is easier to interpret, change variables, lettingm=−nin the second sum and
m=nin the first, finally changing the sum so that it is over only 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

(40)