Mathematical Tools for Physics

10—Partial Differential Equations 299

the central axis of the box, aboutx=a/ 2. That it is even about this point implies that the hyperbolic functions
ofxshould be even aboutx=a/ 2. You can do this simply by using acoshabout that point.

Asinky

(

Dcoshk(x−a/ 2 )

)

Put these together and you have a sum

∑∞

n=0

ansin

(

(n+^1 / 2 )πy

b

)

cosh

(

(n+^1 / 2 )π(x−a/ 2 )

b

)

(29)

Each of these terms satisfies Laplace’s equation, satisfies the boundary conditions aty= 0andy=b, and is even
about the centerlinex=a/ 2. It is now a problem in Fourier series to match the conditions atx= 0. They’re
then automatically satisfied atx=a.

∑∞

n=0

ansin

(

(n+^1 / 2 )πy

b

)

cosh

(

(n+^1 / 2 )πa

2 b

)

=−F 0

y

κ

(30)

The sines are orthogonal by the theorem Eq. (5.12), so you can pick out the componentanby the orthogonality
of these basis functions.

un= sin

(

(n+^1 / 2 )πy

b

)

, then

〈

um,left side

〉

=

〈

um,right side

〉

or, am

〈

um,um

〉

cosh

(

(m+^1 / 2 )πa

2 b

)

=−

F 0

κ

〈

um,y

〉

Write this out; do the integrals, add the linear term, and you have

T(x,y) =F 0

y

κ

−

8 F 0 b

κπ^2

∑∞

n=0

(−1)n

(2n+ 1)^2

× (31)

sin

(

(n+^1 / 2 )πy

b

)

cosh

(

(n+^1 / 2 )π(x−a/ 2 )

b

)

sech

(

(n+^1 / 2 )πa

2 b

)