Mathematical Physics Vol 1

7.8 Examples

391

∂ 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

=

=( − cos x )+ cos x = = 0 .

Given that b = 0 it follows that also ¯ b = 0, and the transformed equation, when u ( x , y ) → v ( ξ , η ) , finally obtains the form ¯ a 12 ∂ 2 v ∂ξ∂η = 0 ⇒ ∂ 2 v ∂ξ∂η = 0 .

Problem 255

∂ 2 u ∂ x 2 −

∂ 2 u ∂ x ∂ y −

∂ 2 u ∂ y 2 −

∂ u ∂ y

( 3 + sin 2 x )

2cos x

y

= 0 .

Solution The old coefficients

2 x ) , a

a 11 = 1 , a 12 = − cos x , a 22 = − ( 3 + sin

1 = 0 , a 2 = − y .

The discriminant

D = cos 2 x + 3 + sin 2 x = 4 > 0 . Given that the discriminant is greater than zero in the entire plane x , y , this equation is of hyperbolic type in the entire plane x , y .

Given that D > 0 we have two real characteristics. The characteristic equation, in this case, has the form

d y 2 + 2cos x d x d y − ( 3 + sin 2 x ) d x 2 = 0 ⇒ − 2cos x ± q 4cos 2 x + 4 ( 3 + sin 2 x ) 2

y ′

= − cos x ± 2 .

1 , 2 =

The characteristics are

y = − sin x + 2 x + c 1 , y = − sin x − 2 x + c 2 .

The substitutions are

c 1 = ξ = y − 2 x + sin x , c 2 = η = y + 2 x + sin x .