Mathematical Physics Vol 1

7.8 Examples

393

The transformed partial equation u ( x , y ) → v ( ξ , η ) is now of the form ¯ a 11 + 2¯ a 12 + ¯ a 22 + ¯ a 1 + ¯ a 2 + ¯ bv + ¯ c =

∂ 2 v ∂ξ 2

∂ 2 v ∂ξ∂η

∂ 2 v ∂η 2

∂ v ∂ξ

∂ v ∂η

( ξ + η )

∂ v ∂η

∂ 2 v ∂ξ∂η −

∂ v ∂ξ

1 2

− 2 · 8

= 0 ⇒

+

( ξ + η )

∂ v ∂η

∂ 2 v ∂ξ∂η

∂ v ∂ξ

1 32

= 0 .

+

+

Problem 256

∂ 2 u ∂ x ∂ y

∂ 2 u ∂ y 2

∂ 2 u ∂ x 2

∂ u ∂ x

∂ u ∂ y

+ 6

+ 10

+ 3

= 0 .

+

Solution

a 11 = 1 , a 12 = 3 , a 22 = 10 , a 1 = 1 , a 2 = 3 D = a 2 12 − a 11 a 22 = 9 − 10 = − 1 < 0 . Given that D < 0 in the entire plane xy , the equation is of elliptic type . Characteristic equation is

√ 36

6 ±

− 40

2 − 2 a 12 ( y ′ )+ a 22 = 0 ⇒ ( y ′ ) 1 , 2 =

a 11 ( y ′ )

= 3 ± i .

2

We have two solution – complex conjugated. The characteristics are

= 3 ± and ⇒

d y d x

d y =( 3 + i ) d x ⇒ y =( 3 + i ) x + c 1 d y =( 3 − i ) d x ⇒ y =( 3 − i ) x + c 2 .

Transformations (real and imaginary parts of the characteristics) are ξ = y − 3 x , η = − x . Partial derivatives are

∂ξ ∂ x

∂ξ ∂ y

∂η ∂ x

∂η ∂ y

= − 3 ,

= 1 ,

= − 1 ,

= 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 ,

= 0 ,

= 0 ,

= 0 .