Mathematical Physics Vol 1

7.8 Examples

395

Problem 257

∂ 2 u ∂ x 2

∂ 2 u ∂ y 2

+ x 2

= 0 , x̸ = 0 .

Solution

2 , a

a 11 = 1 , a 12 = 0 , a 22 = x

1 = 0 , a 2 = 0

D = a 2 2 < 0 . Given that D < 0 in the entire plane xy , except at point x = 0, which is excluded, the equation is of elliptic type . The characteristic equation is a 11 ( y ′ ) 2 − 2 a 12 ( y ′ )+ a 22 = 0 ⇒ ( y ′ ) 1 , 2 = ± xi . We have two solutions – complex conjugated. The characteristics are 12 − a 11 a 22 = − x

= ± xi ⇒

d y = xi d x ⇒ y = 1 d y = − xi d x ⇒ y = − 1

d y d x

2 xi + c 1 ,

2 xi + c 2 .

Transformations (real and imaginary part of the characteristics) ξ = y , η = x .

1 2

The partial derivatives are ∂ξ ∂ x = 0 ,

∂ξ ∂ y

∂η ∂ x

∂η ∂ y

= 1 ,

= x ,

= 0 ,

∂ 2 ξ ∂ x 2 ∂ 2 η ∂ x 2

∂ 2 ξ ∂ y 2 ∂ 2 η ∂ y 2

∂ 2 ξ ∂ x ∂ y ∂ 2 η ∂ x ∂ y

= 0 ,

= 0 ,

= 0 ,

= 1 ,

= 0 ,

= 0 .

The Jacobian is

x 1 x 0

J = −

= − x̸ = 0 ,

because x = 0 is excludeed.

Calculation of new coefficients: ¯ a 11 = a 11 ∂ξ ∂ x 2

+ 2 a 12

∂ξ ∂ x

∂ξ ∂ y

+ a 22

∂ξ ∂ y

2

=

=( 0 ) 2 + 2 ( 0 )( − x )+ x 2 = x 2 ,

¯ a 22 = a 11

∂η ∂ x

+ 2 a 12

∂η ∂ x

∂η ∂ y

+ a 22

∂η ∂ y

2

2

=

=( x ) 2 + 2 ( 0 )( x ) 0 + x 2 ( 0 ) 2 = x 2 ,