Mathematical Physics Vol 1

7.8 Examples

429

Problem 278 Laplace equation, in 2D space, is of the form u xx + u yy = 0 .

(7.441) We will show that it can also be solved by the method of separating variables. Let us find the solution of this equation (7.441), if the boundary conditions are u ( x , 0 )= 0 , u ( x , 1 )= x − x 2 , (7.442) u ( 0 , y )= 0 , u ( 1 , 0 )= 0 . (7.443)

Solution If we assume that the solution has the form

u ( x , t )= X ( x ) Y ( y ) ,

(7.444)

then substituting into (7.441) and dividing by XY we obtain X ′′ X + Y ′′ Y = 0 .

(7.445)

Given that the left hand side is a function of the independent variable x only, and the right hand side of variable y only, it can be concluded that both sides are constant, i.e. X ′′ X = λ = − k 2 , (7.446) Y ′′ Y = − λ = − k 2 . (7.447) From (7.442) and (7.443), for boundary conditions we obtain

u ( x , 0 )= X ( x ) · Y ( 0 )= 0 ⇒ Y ( 0 )= 0 , u ( 0 , y )= X ( 0 ) · Y ( y )= 0 ⇒ X ( 0 )= 0 , u ( 1 , 0 )= X ( 1 ) · Y ( 0 )= 0 ⇒ X ( 1 )= 0 , u ( x , 1 )= X ( x ) · Y ( 1 )= x − x 2 ⇒ Y ( 1 )=

(7.448)

x − x 2 X ( x )

.

From (7.446) we obtain

X = c 1 sin kx + c 2 cos kx .

(7.449)

Using the initial conditions (7.448), we obtain

X ( 0 )= c 2 = 0 ∧ X ( 1 )= c 1 sin k = 0 ⇒ k = n π , k ∈ Z + ,

and we further obtain

X n = c 1 sin ( n π x ) .

From (7.447) we obtain

ky + c

ky .

4 · e −

Y ( y )= c 3 · e

(7.450)