Mathematical Physics Vol 1

7.8 Examples

385

u = ϕ ( λ 2 , λ 3 )= ϕ h 1 + √ y −

√ x 2 i

√ x 2

− 1 +

√ z

−

.

Given that

u ( 1 , y , z )= ϕ ( y − x )= y − x ,

we finally obtain

√ x 2

− 1 +

√ x 2

u = 1 + √ y −

√ z

−

.

Problem 250 Find the solution z = z ( x , y ) of equation

∂ z ∂ x −

∂ z ∂ y

y

x

= 0 ,

that satisfies the condition

z ( x , 0 )= x 2 .

Solution This task is the Cauchy problem.

Let us find first the complete solution. The observed equation is a homogeneous linear first order partial equation.We assign to it the system of ordinary equations d x y = d y − x = d z 0 . From this system we obtain two first integrals d x y = d y − x ⇒ − Z x d x = Z y d y ⇒ x 2 + y 2 = c 1 and, as d z / 0, the second first integral z = c 2 , where c 1 and c 2 arbitrary constants. As we have already mentioned, if the first integrals are ψ 1 = c 1 and ψ 2 = c 2 , than any differentiable function F ( ψ 1 , ψ 2 )= 0 is also a first integral. In this task, this means that the solution (general solution) is z = F x 2 + y 2 , where F ∈ C 1 . The Cauchy integral (solution of the Cauchy problem) is obtained from the condi tion z ( x , 0 )= F ( x 2 )= x 2 ⇒ F ( x )= x , so we finally obtain the Cauchy integral z ( x , y )= x 2 + y 2 .