Mathematical Physics Vol 1

Chapter 7. Partial differential equations

352

1 ◦ either

∂ a ∂ x

∂ b ∂ x

∂ a ∂ y

∂ b ∂ y

= 0 ,

=

=

=

and we obtain that a and b are constant, that is, that the function g ( x , y , z , a , b ) is the complete solution , 2 ◦ or b = ϕ ( a ) , where ϕ is an arbitrary function. In that case

=

∂ b ∂ a

∂ g ∂ a

∂ a ∂ x ∂ a ∂ x̸

∂ g ∂ b

∂ b ∂ a

∂ a ∂ x

∂ g ∂ a

∂ g ∂ b

∂ a ∂ x

= 0 ,

+

+

that is, provided that

= 0,

∂ g ∂ a

∂ g ∂ b

ϕ ′ ( a )= 0 ,

(7.62)

+

to which both equations (7.37) are reduced. From (7.62) we can express a = ψ ( x , y ) , so we obtain b = ϕ ( ψ ( x , y ))= µ ( x , y ) . which is the general solution . As can be concluded from the previous presentation, both singular and general solutions (if they exist) can be obtained from the complete solution of the first order partial differential equation. Therefore, the basic task is to find a complete solution. It can be determined by applying the Lagrange-Charpit 6 method. Lagrange-Charpit method Observe the general first order partial differential equation (of two independent variables x and y ) f ( x , y , z , p , q )= 0 . (7.63) The main idea of the Lagrange-Charpit method in finding the complete solution is to determine a function g ( x , y , z , p , q )= c 1 (7.64) that is functionally independent of f , where c 1 is an arbitrary constant. The function g should be such that from the system of partial equations

f = 0 ,

g = c 1

(7.65)

p and q can be calculated, i.e.

p = ϕ ( x , y , z , c 1 ) ,

q = ψ ( x , y , z , c 1 ) .

(7.66)

From the assumption that p and q can be determined from (7.65), it follows that

∂ f ∂ p ∂ g ∂ p

∂ f ∂ q

D pq =

= 0 .

(7.67)

∂ g ∂ q ̸

6 Charpit