Mathematical Physics Vol 1

Chapter 7. Partial differential equations

364

R Note that this sum should be taken symbolically, because the convergence of the series that appear here has not been tested.

Depending on the value under the square root, there are two possible cases d 2 = 4 cf ;

(7.156) (7.157)

d 2̸ = 4 cf .

In the first case, the value under the square root is a full square ck 2 + dk + f =( mk + n ) 2 ,

(7.158)

which yields

∞ ∑ k = 0

∞ ∑ k = 0

[ kx +( a + m ) ky +( b + n ) y ] +

[ kx +( a − m ) ky +( b − n ) y ] ,

u =

C k · e

D k · e

(7.159)

that is

∞ ∑ k = 0

∞ ∑ k = 0

u = e ( b + n ) y ·

[ kx +( a + m ) ky ] + e ( b − n ) y ·

[ kx +( a − m ) ky ] .

C k · e

D k · e

(7.160)

7.6 The variable separation method

The variable separation method (Fourier method) is one of the most commonly used methods for solving partial differential equation that satisfy given initial and/or boundary (contour) conditions. The Fourier method can be applied to equations of the form a 11 u xx + a 22 u yy + b 1 u x + b 2 u y +[ F ( x )+ G ( y )] u = 0 , (7.161) with initial conditions u ( x , 0 )= f ( x ) , u y ( x , 0 )= g ( x ) (7.162) and boundary conditions au ( 0 , y )+ bu x ( 0 , y )= 0 , cu ( l , y )+ du x ( l , y )= 0 , (7.163) where f and g are given functions, and a , b , c and d known constants. Assume that the solution has the form u ( x , y )= X ( x ) · Y ( y ) . (7.164) Let us now find the corresponding derivatives u x = X ′ · Y , u y = Y ′ · X , (7.165) u xx = d 2 X d x 2 · Y ( y ) ≡ X ′′ · Y , u yy = d 2 Y d y 2 · X ( x ) ≡ Y ′′ · X (7.166) and substitute then into the initial equation (7.161) a 11 · X ′′ · Y + a 22 · X · Y ′′ + b 1 · X ′ · Y + b 2 · X · Y ′ +[ F ( x )+ G ( y )] XY = 0 . From here, dividing by XY , we obtain

X ′′ X

Y ′′ Y

X ′ X

Y ′ Y

a 11

+ a 22

+ b 1

+ b 2

+ F ( x )+ G ( y )= 0 ,

(7.167)