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Hyperbolic partial differential equation

  1. Jul 8, 2015 #1
    What is the general solution of the following hyperbolic partial differential equation:
    upload_2015-7-8_15-21-26.png
    The head (h) at a specified distance (x) is a sort of a damping function in the form:
    upload_2015-7-8_15-28-5.png
    Where, a, b, c and d are constants. And the derivatives are with respect to t (time) and x (distance).

    Thanks in advance.
     
  2. jcsd
  3. Jul 8, 2015 #2

    BiGyElLoWhAt

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    Well, just looking at it, you know it has to be exponential in t. If we guess ##h_1=e^{-a*t}##, that works as one homogeneus solution, although the second 2 terms become arbitrary. So h_1 is part of the general solution, we just need other solutions to make the second 2 terms non arbitrary. So the second derivative with respect to x must differ from partial x partial t only by a constant. x works, as it's derivative is zero for all partials above, so far we have ##h_{1+2}=k_1e^{-a*t}+k_2x## where ##k_{n}## is an arbitrary constant. Now you try to throw a few out there, I have a feeling there are a lot, or at least one more function of x and t. I honestly don't know though.
     
  4. Jul 8, 2015 #3
    I know that the governing differential equation of a free vibration of a damped system has this form
    upload_2015-7-8_18-59-34.png
    and its solution (subject to initial conditions) is
    upload_2015-7-8_19-0-5.png
    where, upload_2015-7-8_19-0-22.png , ω_n is the natural frequency and ζ is a damping factor <1 and if drawn it gives
    upload_2015-7-8_19-2-27.png
    But, the problem here is the third and the fourth terms.
    Is it right to just add this term k_2 x to the previous solution?
     

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  5. Jul 8, 2015 #4

    BiGyElLoWhAt

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    The all cancel out. The general solution is the sum of all homogenous solutions (they all equal zero). This isnt an enveloped oscillator, at least as far as I can see.
     
  6. Jul 9, 2015 #5

    BiGyElLoWhAt

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    Basically, the general solution to this differential equation is any terms that you can add to gether that, when you plug the whole thing into this PDE, equals zero. I will almost guarantee that there are more terms. One way you can check is by taking 2 initial conditions and plugging them into the function, see if this function satisfies both conditions, if not, then we're missing stuff.
     
  7. Jul 10, 2015 #6

    pasmith

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    There are wave-like solutions [itex]h(x,t) = f(x - vt)[/itex] where [itex]f[/itex] must satisfy [tex]
    (v^2 - bv + c)f'' - av f' = 0.[/tex]

    There are also separable solutions of the form [itex]h(x,t) = X(x)e^{-kt}e^{i\omega t}[/itex] where [itex]X[/itex] must satisfy [tex]
    cX'' + b(-k + i\omega)X' + ((-k + i\omega)^2 + a(-k + i\omega))X = 0.[/tex]
     
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