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

[note: also under discussion in sos math cyberboard]

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- Oct 5th 2009, 04:49 PMkingwinner2nd order linear hyperbolic PDE?!
http://sites.google.com/site/asdfasdf23135/pde2.JPG

[note: also under discussion in sos math cyberboard] - Oct 6th 2009, 06:44 AMJester
If you let $\displaystyle r = 2y + x^2$ and $\displaystyle s = 2y - x^2$, then your PDE becomes

$\displaystyle

u_{rs} - \frac{u_r - u_s}{4(r-s)} = 0,

$ a Possion - Euler - Darboux equation.

The only way to get the solution you mention if the PDE transformed to

$\displaystyle

u_{rs} = 0.

$ - Oct 8th 2009, 09:55 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? - Oct 8th 2009, 11:40 AMJester
In general, if the PDE is

$\displaystyle a u_{xx} + b u_{xy}+ c u_{yy} = 0$ where $\displaystyle b^2 - 4ac > 0$, then it's hyperbolic. If we introduce new coordinates

$\displaystyle r = \frac{-b - \sqrt{b^2-4ac}}{2a} x + y, \;\;\;s = \frac{-b + \sqrt{b^2-4ac}}{2a} x + y$

then the original PDE transforms to

$\displaystyle u_{rs} = 0$ whose general solution is $\displaystyle u = F(r) + G(s)$ and with $\displaystyle r $ and $\displaystyle s$ above gives the general solution

$\displaystyle u = F\left(\frac{-b - \sqrt{b^2-4ac}}{2a} x + y\right) + G\left(\frac{-b + \sqrt{b^2-4ac}}{2a} x + y\right)$ - Oct 16th 2009, 02:45 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! - Oct 16th 2009, 06:12 AMJester
It has to do with the change of variable formulas

If the PDE is $\displaystyle Au_{xx} + B u_{xy} + C u_{yy} = 0$ where A, B and C are functions of x and y, then with a change of variables is

$\displaystyle

u_{xx} = r_x^2 u_{rr} + 2 r_x s_x u_{rs} + s_x^2 u_{ss} + r_{xx} u_r + s_{xx} u_s

$

$\displaystyle

u_{xy} = r_x r_y u_{rr} + (r_x s_y + r_y s_x)u_{rs} + s_x s_y u_{ss} + r_{xy} u_r + s_{xy} u_s

$

$\displaystyle

u_{yy} = r_y^2 u_{rr} + 2 r_y s_y u_{rs} + s_y^2 u_{ss} + r_{yy} u_r + s_{yy} u_s

$.

Substituting and re-arranging gives

$\displaystyle

\left(A r_x^2 + B r_x r_y + C r_y^2 \right)u_{rr} + \left(A r_x r_y + B (r_x s_y + r_y s_x) + C r_y s_y \right) u_{rs} + $

$\displaystyle \left(A s_x^2 + B s_x s_y + C s_y^2 \right) u_{ss} + \left(\color{blue}{ Ar_{xx} + B r_{xy} + C r_{yy}}\; \right) u_r + \left( \color{blue}{ As_{xx} + B s_{xy} + C s_{yy}}\; \right) u_s = 0

$

To transform to your hyperbolic form you want to choose $\displaystyle r$ and $\displaystyle s$ such that

$\displaystyle

A r_x^2 + B r_x r_y + C r_y^2 = 0,\;\;\;A s_x^2 + B s_x s_y + C s_y^2 = 0

$,

noting that since $\displaystyle B^2-4AC > 0$ (it's hyperbolic), you can factor these choosing one factor for $\displaystyle r$ and the other for $\displaystyle s$. Both are linear PDEs!

With this choice, the lower order terms above (in blue) are typically not zero.