But if the PDE is any hyperbolic second order linear PDE with constant coefficients (e.g. u_xx- 4 u_yy = 0), then my argument above works, right?
dy/dx=2
y-2x=c
dy/dx=-2
y+2x=k
Then we can immeidately say that the general solution is u=f(y-2x)+g(y+2x), right?
Why?
But for the original PDE, we just have
a=1
b=0
c=-x^2
b^2 - 4ac = x^2 >0 (since we are given x>0) as well
So it falls into your case, but it does not reduce to the canonical form u_rs =0. WHY?
Is there any reason why the canonical form is different for constant (e.g. u_xx- x^2 u_yy = 0) and non-constant coefficients (ue.g. _xx- 4 u_yy = 0) hyperbolic PDEs?
Thanks!
It has to do with the change of variable formulas
If the PDE is where A, B and C are functions of x and y, then with a change of variables is
.
Substituting and re-arranging gives
To transform to your hyperbolic form you want to choose and such that
,
noting that since (it's hyperbolic), you can factor these choosing one factor for and the other for . Both are linear PDEs!
With this choice, the lower order terms above (in blue) are typically not zero.