Mathematical Physics Vol 1

Chapter 7. Partial differential equations

430

The initial condition is

Y ( 0 )= c 3 + c 4 = 0 ⇒ c 3 = − c 4 ,

so that

Y n = c 3 e

ky − e − ky = c

e ky − e − ky 2

= 2 c 3 sh ( n π y ) .

3 2

Nowwe have

u n = X ( x ) n · Y ( y ) n ⇒ u = ∞ ∑ n = 1

∞ ∑ n = 1

sin ( n π x ) sh ( n π y ) .

u n =

2 c 1 c 3 |{z} a n

(7.451)

From the remaining boundary condition it follows

∞ ∑ n = 1

u ( x , 1 )= x − x 2 =

a n sh ( n π ) | {z } A n

sin ( n π x ) ,

i.e.

∞ ∑ n = 1

A n sin ( n π x )= x − x 2 .

(7.452)

The left hand side represents a Fourier sine series, and thus, by expanding the function at the right hand side, for coefficients A n we obtain

2 1 Z

2

16 n 3 π 3

0 ( x − x 2 ) sin ( n π x ) d x =

( 1 − cos ( n π )) .

A n =

(7.453)

Given that A n = a n sh ( n π ) , for these coefficients we obtain

4 ( 1 − ( − 1 ) n ) n 3 π 3 sh ( n π ) ,

a n =

(7.454)

and the final solution is

∞ ∑ n = 1

1 − 1 ( − 1 ) n n 3

sh ( n π y ) sh ( n π )

4 π 3

sin ( n π x )

u ( x , y )=

(7.455)

.

7.8.1 Appendix

When solving second order partial differential equation of special importance is the form of function f ( r ) , where r 2 = x i x i , i = 1 , 2 ,..., n , n > 1, that satisfies the Laplace equation 16 :

∂ 2 f ∂ x i ∂ x j

∆ f ( r )= 0 , or δ i j

= 0 .

Given that

∂ 2 r ∂ x i ∂ x j

δ i j

∂ r ∂ x i

x i x j r 2

x i r

r −

=

,

=

,

16 The proof of this generalization was proposed and done by prof. J. Jaric´.