Mathematical Physics Vol 1

7.8 Examples

397

Given that η is an arbitrary function, we chose it to be as simple as possible for further use, but also that the Jacobian is not equal to zero. The partial derivatives are

∂ξ ∂ x

∂ξ ∂ y

∂η ∂ x

∂η ∂ y

= 1 ,

= 1 ,

= 0 ,

= 1 ,

∂ 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 ,

= 0 ,

= 0 ,

= 0 .

The Jacobian is

1 1 0 1

J =

= 1̸ = 0 .

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

+ 2 a 12

∂ξ ∂ x

∂ξ ∂ y

+ a 22

∂ξ ∂ y

2

=

= 1 ( 1 ) 2 + 2 ( − 1 )( 1 )+ 1 2 = 0 ,

¯ a 22 = a 11

∂η ∂ x

+ 2 a 12

∂η ∂ x

∂η ∂ y

+ a 22

∂η ∂ y

2

2

=

=( 0 ) 2 − 2 ( 0 )( 1 )+ 1 2 = 1 ,

¯ a 12 = a 11

∂ξ ∂ x

∂η ∂ x

+ a 12

∂η ∂ x

∂ξ ∂ x

∂η ∂ y

∂ξ ∂ y

∂ξ ∂ y

∂η ∂ y

+ a 22

+

=

= 1 ( 0 )+( − 1 )( 1 )+ 1 = 0 ,

∂ 2 ξ ∂ x 2

∂ 2 ξ ∂ x ∂ y

∂ 2 ξ ∂ y 2

∂ξ ∂ x

∂ξ ∂ y

¯ a 1 = a 1

+ a 2

+ a 11

+ 2 a 12

+ a 22

=

= 3 · 1 + 1 · 1 = 4 ,

∂ 2 η ∂ x 2

∂ 2 η ∂ x ∂ y

∂ 2 η ∂ y 2

∂η ∂ x

∂η ∂ y

¯ a 2 = a 1

+ a 2

+ a 11

+ 2 a 12

+ a 22

=

= 0 + 1 + 0 + 0 + 0 = 1 . The transformed equation u ( x , y ) → v ( ξ , η ) ¯ a 11 + 2¯ a 12 + ¯ a 22 + ¯ a 1

∂ 2 v ∂ξ 2

∂ 2 v ∂ξ∂η

∂ 2 v ∂η 2

∂ v ∂ξ

∂ v ∂η

+ ¯ bv + ¯ c = 0 ⇒

+ ¯ a 2

gains the form

∂ 2 v ∂η 2

∂ v ∂ξ

∂ v ∂η

+ 4

+ 2 v = 0 .

+