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Math Help - 2nd order linear hyperbolic PDE?!

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    2nd order linear hyperbolic PDE?!



    [note: also under discussion in sos math cyberboard]
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    Quote Originally Posted by kingwinner View Post


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

     <br />
u_{rs} - \frac{u_r - u_s}{4(r-s)} = 0,<br />
a Possion - Euler - Darboux equation.

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

     <br />
u_{rs} = 0.<br />
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    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?
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    Quote Originally Posted by kingwinner View Post
    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)
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    Quote Originally Posted by Danny View Post
    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!
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    Quote Originally Posted by kingwinner View Post
    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

     <br />
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<br />

     <br />
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<br />

     <br />
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<br />
.

    Substituting and re-arranging gives

     <br />
\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<br />

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

     <br />
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<br />
,

    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.
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