Mathematical Physics Vol 1

Chapter 7. Partial differential equations

384

Finally we obtain the general solution f = f y x + 1 , y z − 1

or g = g

y

x + 1 y

z − 1

,

,

where f and g are arbitrary differentiable functions by corresponding variables.

R Note that the functions f and g are solutions of our problem and any of these two functions is called the general solution.

Problem 249 Find the solution of the differential equation √ x + √ y

∂ u ∂ x

∂ u ∂ y

∂ u ∂ z

+ √ z

= 0 ,

which for x = 1 becomes u = y − z .

Solution This is the so-called Cauchy problem. We will first find the general solution. To the initial equation corresponds the system d x √ x = d y √ y = d z √ z , which has two first integrals d x √ x = d y √ y ⇒ √ x − √ y = c 1 = ψ 1 and, similarly, the second first integral √ x − √ z = c 2 = ψ 2 . Let is now find the Cauchy solution (Cauchy integral). According to the theory ψ i ( x 1 ,..., x n )= c i , i = 1 ,..., n − 1 ψ 1 ( x 0 , y , z )= ¯ ψ 1 ψ 2 ( x 0 , y , z )= ¯ ψ 2 ⇒ y = λ 2 ( ¯ ψ 1 , ¯ ψ 2 ) , z = λ 3 ( ¯ ψ 1 , ¯ ψ 2 ) is a solution of equation F [ z ]= 0 that satisfies the condition z ( x 0 , y )= ϕ ( y ) , u = ϕ [ λ 2 ( ψ 1 , ψ 2 ) , λ 3 ( ψ 1 , ψ 2 )] . In our case − √ x + √ y = ψ 1 ( x , y , z ) ⇒ ψ 1 ( 1 , y , z )= − 1 + √ y = ¯ ψ 1 ⇒

y =( 1 + ¯ ψ 1 ) 2

= λ 2 ,

− √ x + √ z = ψ 2 ( x , y , z )

√ z = ¯ ψ 2

⇒ ψ 2 ( 1 , y , z )= − 1 +

⇒

z =( 1 + ¯ ψ 2 ) 2

= λ 3 ,