Mathematical Physics Vol 1

Chapter 7. Partial differential equations

390

The characteristics are y ′

1 = − sin x + 1 ⇒ y = x + cos x + c 1 that is y ′ 2 = − sin x + 1 ⇒ y = − x + cos x + c 2 .

the substitutions are

c 1 = y − x − cos x = ξ i c 2 = y + x − cos x = η . The corresponding partial derivatives, necessary for calculating the new coefficients, are

∂ξ ∂ x

∂ξ ∂ y

∂η ∂ x

∂η ∂ y

= − 1 + sin x ,

= 1 ,

= 1 + sin x ,

= 1 ,

∂ 2 ξ ∂ x ∂ y ∂ 2 η ∂ x ∂ y

∂ 2 ξ ∂ y 2 ∂ 2 η ∂ y 2

∂ 2 ξ ∂ x 2 ∂ 2 η ∂ x 2

= cos x ,

= 0 ,

= 0 ,

= cos x ,

= 0 ,

= 0 .

Check of the Jacobian:

∂ξ ∂ x ∂η ∂ x

∂ξ ∂ y ∂η ∂ y

1 + sin x 1 1 + sin x 1

= −

J =

= − 2̸ = 0 .

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

+ a 22

∂ξ ∂ y

2

∂ξ ∂ x

∂ξ ∂ y

+ 2 a 12

=

=( − 1 + sin x ) 2 + 2 ( − sin x )( − 1 + sin x )+( − cos 2 x )= 0 , ¯ a 22 = a 11 ∂η ∂ x 2 + 2 a 12 ∂η ∂ x ∂η ∂ y + a 22 ∂η ∂ y 2 = =( 1 + sin x ) 2 + 2 ( − sin x )( 1 + sin x )+( − cos 2 x )= 0 ,

∂ξ ∂ x

∂η ∂ x

∂ξ ∂ x

∂η ∂ y

∂ξ ∂ y

∂η ∂ x

∂ξ ∂ y

∂η ∂ y

¯ a 12 = a 11

+ a 12

+ a 21

+ a 22

=

=( − 1 + sin x )( 1 + sin x )+( − sin x )[( − 1 + sin x )+( 1 + sin x )]+( − cos 2 x )= − 2 . Given that a 1 = 0 and a 2 = − cos x , we obtain: ¯ a 1 = a 1 + a 2 + a 11 + 2 a 12 + a 22 =

∂ 2 ξ ∂ x 2

∂ 2 ξ ∂ x ∂ y

∂ 2 ξ ∂ y 2

∂ξ ∂ x

∂ξ ∂ y

=( − cos x )+ cos x = = 0 ,