Mathematical Physics Vol 1

7.8 Examples

389

from where we obtain the first two integrals

p = const . = a i q = const . = b .

Substituting in the initial equation we obtain the complete solution

z = ax + by + ab .

Let us look now for the singular solution. If it exists, it is obtained from the system of equations (7.54). In our case

∂ g ∂ a

∂ g ∂ b

g = ax + by + ab − z = 0 ,

= x + b i

= − y + a

from where we obtain

b = − x i a = y .

Thus, there exists a singular solution

xy + z = 0 .

For the set of examples, which follow without text, determine the type of partial equation and reduce it to the canonical form:

Problem 254

∂ 2 u ∂ x 2 −

∂ 2 u ∂ x ∂ y −

∂ 2 u ∂ y 2 −

∂ u ∂ y

cos 2 x

2sin x

cos x

= 0 .

Solution The discriminant of the characteristic equation, given that a 12 = − sin x , a 11 = 1 and a 22 = − cos 2 x , is D = a 2 12 − a 11 a 22 = sin 2 x + cos 2 x = 1 > 0 , thus positive in the entire plane x , y . The observed equation is hence of the hyperbolic type in the entire x , y plane. It has two real characteristics, which are obtained by solving the characteristic equation a 11 d y 2 − 2 a 12 d x d y + a 22 d x 2 = 0 , which in this case has the form d y 2 + 2sin x d x d y − cos 2 x d x 2 = 0 , from where we obtain ( y ′ ) 2 + 2sin xy ′ − cos 2 x = 0 ⇒ y ′ 1 , 2 = − 2sin x ± p 4sin 2 x + 4cos 2 x 2 = − sin x ± 1 .