Mathematical Physics Vol 1

Chapter 7. Partial differential equations

398

Problem 259

∂ 2 u ∂ x ∂ y

∂ 2 u ∂ y 2

∂ 2 u ∂ x 2 −

+ y 2

sin 2 x

2 y sin x

= 0 .

Solution

2 x , a

2 , a

a 11 = sin

12 = − y sin x , a 22 = y

1 = 0 , a 2 = 0

D = a 2 2 sin 2 x − ( sin 2 x ) y 2 = 0 . Given that D = 0 in the entire plane xy , the equation is of parabolic type . The characteristic equation a 11 ( y ′ ) 2 − 2 a 12 ( y ′ )+ a 22 = 0 ⇒ ( y ′ ) 2 − 2 ( − y sin x )( y ′ )+ y 2 = 0 ⇒ ( y ′ ) 1 , 2 = − y sin x . It has only one solution – one characteristic. The characteristic is 12 − a 11 a 22 = y

d y d x

y sin x ⇒

d y y

d x sin x ⇒

x 2

y tg

= c 1 .

= −

= −

We have used here

Z d x

= Z

d x = Z

sin 2 x

2 x 2

2 + cos

1 2sin x

d x =

x 2

2sin x

x 2

sin x

2 cos

2 cos

x 2

= lntg

.

The transformations are

x 2

ξ = y tg , η = η ( x , y )= y .

R Note that , 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 simplest choice would be η = const . , but then J = 0. The next choices would be η = x or η = y . We have chosen the latter.

The partial derivatives are ∂ξ ∂ x = 1 2 y cos − 2

∂ξ ∂ y

∂η ∂ x

∂η ∂ y

x 2

x 2

= tg

= 0 ,

= 1 ,

,

,

∂ 2 ξ ∂ x 2

∂ 2 ξ ∂ y 2

∂ 2 ξ ∂ x ∂ y

1 2

x 2

x 2

1 2

x 2

y cos − 3

cos − 2

sin

= 0 ,

=

,

=

,

∂ 2 η ∂ x 2

∂ 2 η ∂ y 2

∂ 2 η ∂ x ∂ y

= 0 ,

= 0 ,

= 0 .