#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#
