r = (partial^2 z)/(partial x^2)
s = (partial^2 z)/(partial x partial y)
t =(partial^2 z)/(partial y^2)
x( (partial^2 z)/(partial x^2)- (partial^2 z)/(partial x partial y))=y((partial^2 z)/(partial y^2)- (partial^2 z)/(partial x partial y))
This equation can be arranged as
x (partial^2 z)/(partial x^2) + (y-x) (partial^2 z)/(partial x partial y)-y(partial^2 z)/(partial y^2)=0 or
A (partial^2 z)/(partial x^2) + B (partial^2 z)/(partial x partial y)+C(partial^2 z)/(partial y^2)=0
now according with
B^2-4AC {(<0 -> "elliptic"),(=0->"parabolic"),(>0-> "hyperbolic"):}
In our case B^2-4AC = (y-x)^2+4xy = (x+y)^2 ge 0 -> "hyperbolic" uu "parabolic"
Now joining the equations
((A,B,C),(dx,dy,0),(0,dx, dy))(((partial^2 z)/(partial x^2)),( (partial^2 z)/(partial x partial y)),((partial^2 z)/(partial y^2)))=((0),(d((partial z)/(partial x))),(d((partial z)/(partial y))))
we obtain the associated characteristic equations by solving
A dy^2-B dx dy + C dx^2=0 giving
dy/dx = (B pm sqrt(B^2-4AC))/(2A) = {(y/x),(-1):}
so the characteristic lines are generated by
(dy)/(dx) = y/x rArr x-C_1y = 0 and
(dy)/(dx) = -1 rArr x+y = C_2
then C_1 = 1 and C_2 = 0
and the general solution is any phi, psi
z(x,y) = C_3phi(x-y)+C_4 psi(x+y)
Compatible with the boundary conditions.
NOTE
Substituting phi(y+m x) into the PDE we get
(m+1)(m x-y)phi''(y+mx) = 0 rArr m = -1
or with phi(y-mx)
(m-1)(y+mx)phi''(y-mx) = 0 rArr m = 1
