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Math Help - hyperbolic pde

  1. #1
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    hyperbolic pde

    Hello. This is my problem:
    Let A,B and C be positive constants and the hyperbolic partial differential equation: Auxx+Buxy+Cuyy=0. Get a general solution of the form:
    u(x,y) = f(ax+by)+g(ax+by) where a,b,c,d are real constants and f and g are C2 functions.

    Well, I have started computing the derivatives of the given solution and putting them into the equation:
    uxx=a2fxx+a2gxx
    uxy=abfxy+abgxy
    uyy=b2fyy+b2gyy

    A
    a2fxx+Aa2gxx+Babfxy+Babgxy+Cb2fyy+Cb2gyy = 0

    But I am not sure how to continuate. Any idea?? Thanks
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  2. #2
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    Re: hyperbolic pde

    Hyperbolic partial differential equation - Wikipedia, the free encyclopedia

    I think the basic idea is to find a new basis where there are no cross derivatives. Not too unlike rotating a non-standard conic section into standard form.
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  3. #3
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    Re: hyperbolic pde

    Quote Originally Posted by Lolyta View Post
    Hello. This is my problem:
    Let A,B and C be positive constants and the hyperbolic partial differential equation: Auxx+Buxy+Cuyy=0. Get a general solution of the form:
    u(x,y) = f(ax+by)+g(ax+by) where a,b,c,d are real constants and f and g are C2 functions.
    We can assume that $C\not = 0$ otherwise the PDE is easy to solve.

    Let $a,b,c,d$ be real numbers to be determined. Define a new function $U(ax+bt,cx+dy) = u(x,y)$. [Note, for this function to be well-defined we require that $ad - bc\not = 0$, confirm this for yourself.]

    Thus,
    $$ u_x = aU_x + cU_y \text{ and }u_y = bU_x + dU_y$$
    Therefore, we have,
    $$ u_{xx} = a(aU_{xx}+cU_{xy}) + c(aU_{xy}+cU_{yy}) = a^2U_{xx} + 2acU_{xy} + c^2U_{yy}$$
    And,
    $$ u_{yy} = b(bU_{xx} + dU_{yx}) + d(bU_{yx} + dU_{yy}) = b^2U_{xx} + 2bdU_{xy} + d^2U_{yy} $$
    In the same way,
    $$ u_{xy} = ab U_{xx} + (ad + bc) U_{xy} + cd U_{yy} $$

    It follows from here that,
    $$ Au_{xx} + Bu_{xy} + Cu_{yy} = (Aa^2 + Bab + Cb^2) U_{xx} + (2Aac + 2Cbd + B(ad+bc))U_{xy} + (Ac^2 + Bcd + Cd^2)U_{yy}$$

    As this is hyperbolic PDE it means $B^2 - 4AC > 0$, so we can let $\delta = \pm\sqrt{B^2-4AC}$.

    Now let $a=1,c=1$ with $b = \frac{-B + \delta }{2C}$ and $d = \frac{-B - \delta}{2C}$.

    Try doing that and see where that leads you to.
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