Mathematical Physics Vol 1

Chapter 7. Partial differential equations

396

¯ a 12 = a 11

∂ξ ∂ x

∂η ∂ x

+ a 12

∂η ∂ x

∂ξ ∂ x

∂η ∂ y

∂ξ ∂ y

∂ξ ∂ y

∂η ∂ y

+ a 22

= 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

= 0 ,

∂ 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

=

= 1 ( 1 )= 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

1 2 η

= 0 ,

+

+

where we have used that x 2 = 2 η .

Problem 258

∂ 2 u ∂ x 2 −

∂ 2 u ∂ x ∂ y

∂ 2 u ∂ y 2

∂ u ∂ x

∂ u ∂ y

2

+ 3

+ 2 u = 0 .

+

+

Solution

a 11 = 1 , a 12 = − 1 , a 22 = 1 , a 1 = 3 , a 2 = 1 , D = a 2 12 − a 11 a 22 = 1 − 1 = 0 . Given that D = 0 in the entire plane xy , the equation is of parabolic type . The characteristic equation is a 11 ( y ′ ) 2 − 2 a 12 ( y ′ )+ a 22 = 0 ⇒ ( y ′ ) 2 + 2 ( y ′ )+ 1 = 0 ⇒ ( y ′ )

1 , 2 = − 1 .

It has only one solution – one characteristic. The characteristic is

d y d x = − 1 ⇒ d y = − d x ⇒ y = − x + c 1 .

The transformations are

ξ = x + y , η = η ( x , y )= y .