http://sites.google.com/site/asdfasdf23135/pde2.JPG

[note: also under discussion in sos math cyberboard]

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- October 5th 2009, 05:49 PMkingwinner2nd order linear hyperbolic PDE?!
http://sites.google.com/site/asdfasdf23135/pde2.JPG

[note: also under discussion in sos math cyberboard] - October 6th 2009, 07:44 AMJester
- October 8th 2009, 10:55 AMkingwinner
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? - October 8th 2009, 12:40 PMJester
- October 16th 2009, 03:45 AMkingwinner
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! - October 16th 2009, 07:12 AMJester
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.