Mathematical Physics Vol 1

7.8 Examples

399

The Jacobian is

1 2 y cos −

2 x

x 2

1 2

x 2̸

2 tg

y cos − 2

J =

= 0 .

=

0

1

Calculation of new coefficients ¯ a 11 = a 11 ∂ξ ∂ x 2 + 2 a 12 ∂ξ

∂ x

∂ξ ∂ y

+ a 22

∂ξ ∂ y

2

=

x 2

= sin 2 x

2

1 2

1 2

x 2

x 2

x 2

y cos − 2

y cos − 2

+ y 2 tg 2

+ 2 ( − y sin x )

tg

= 0 ,

¯ a 22 = a 11

∂η ∂ x

+ 2 a 12

∂η ∂ x

∂η ∂ y

+ a 22

∂η ∂ y

2

2

=

= 0 + 0 + y 2 = = y 2 = η 2 ,

¯ a 12 = a 11

∂ξ ∂ x

∂η ∂ x

+ a 12 + y 2 tg

∂η ∂ x

∂ξ ∂ x

∂η ∂ y

∂ξ ∂ y

∂ξ ∂ y

∂η ∂ y

+ a 22

+

=

1 2

1 2

1 2

y cos − 2

= − y sin x

= 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

=

= sin 2 x = − 2 y tg x 2

x 2

1 2

x 2

1 2

x 2

y cos − 3

cos − 2

sin

+ 2 ( − y sin x )

=

x 2

x 2

1 1 + tg 2 x 2

cos 2

= − 2 y tg

.

Given that (from the transformations)

ξ η

x 2

tg

=

,

for ¯ a 1 we obtain:

η 2 ξ 2 + η 2

¯ a 1 = − 2 ξ

(7.302)

.

∂ 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 + 0 + 0 + 0 + 0 = 0 . 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 −

2 ξ ξ 2 + η 2

∂ v ∂ξ

= 0 .