# Math Help - 2nd order linear hyperbolic PDE?!

1. ## 2nd order linear hyperbolic PDE?!

[note: also under discussion in sos math cyberboard]

2. Originally Posted by kingwinner

[note: also under discussion in sos math cyberboard]
If you let $r = 2y + x^2$ and $s = 2y - x^2$, then your PDE becomes

$
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

$
u_{rs} = 0.
$

3. 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?

4. Originally Posted by kingwinner
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?
In general, if the PDE is

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

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

then the original PDE transforms to

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

$u = F\left(\frac{-b - \sqrt{b^2-4ac}}{2a} x + y\right) + G\left(\frac{-b + \sqrt{b^2-4ac}}{2a} x + y\right)$

5. Originally Posted by Danny
In general, if the PDE is

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

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

then the original PDE transforms to

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

$u = F\left(\frac{-b - \sqrt{b^2-4ac}}{2a} x + y\right) + G\left(\frac{-b + \sqrt{b^2-4ac}}{2a} x + y\right)$
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!

6. Originally Posted by kingwinner
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 $Au_{xx} + B u_{xy} + C u_{yy} = 0$ where A, B and C are functions of x and y, then with a change of variables is

$
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
$

$
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
$

$
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

$
\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} +$

$\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 $r$ and $s$ such that

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

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