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Finding Dispersion relation

  1. Sep 4, 2011 #1
    [itex]\imath\frac{\partial u}{\partial t} + \frac{\partial^2 u}{\partial x^2}=0[/itex]

    [itex]\left(x,t\right) = \int^{\infty}_{-\infty}A\left(k\right)e^{\imath\left(kx-wt\right)}dk[/itex]

    [itex]u\left(x,0\right)=\delta\left(x\right) [/itex]

    This is what I am working with. I am supposed to find the dispersion relation. So far I have gotten

    [itex]A\left(k\right) = \frac{1}{2\pi}\int^{\infty}_{-\infty}\delta\left(x\right)e^{-\imath\left(kx\right)}dx = \frac{1}{2\pi}[/itex]

    plugging this in to u(x,t) do I work with

    [itex] u\left(x,t\right) =\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{\imath\left(kx-wt\right)}dk[/itex]

    This is where I am stuck. I know w(k) is the dispersion relation. If I put in the pde do I just deal with

    [itex] \imath \left(-\imath w\right) + \frac{d^{2}u}{dt^{2}} = w +\frac{d^{2}u}{dt^{2}}=0 [/itex]

    not sure how to pull out the dispersion equation or if I am even going the right route. Any clues on how to proceed would be most appreciated. Solving this equation does not seem to get me to where I want to be. Thanks!
    Last edited: Sep 4, 2011
  2. jcsd
  3. Sep 4, 2011 #2
    Am I making this too hard on myself. I just saw a similar problem and according to this book I should just get

    [itex] w-k^2=0\Rightarrow w=k^2 [/itex]

    I figured this out I believe.
    Last edited: Sep 4, 2011
  4. Sep 5, 2011 #3


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    Homework Helper

    You have:
    i\frac{\partial u}{\partial t}+\frac{\partial^{2}u}{\partial x^{2}}dk
    u=\int_{-\infty}^{\infty}A(k)e^{i(kx-\omega (k)t)}
    So we just differentiate!
    \frac{\partial u}{\partial t}=\int_{-\infty}^{\infty}(-i\omega (k))A(k)e^{i(kx-\omega (k)t)}dk,\quad\frac{\partial^{2}u}{\partial x^{2}}=\int_{-\infty}^{\infty}A(k)(-k^{2})e^{i(kx-\omega (k)t)}dk
    Insert the above into your equation to obtain the dispersion relation.
  5. Sep 7, 2011 #4
    Is that dk a typo? Not sure what it is there for.

    My professor said to just use

    [itex] \phi\left(x,t\right)=A\left(k\right)e^\left(i\left(kx-wt\right)\right) [/itex]

    to find the dispersion relation, therefore leaving me with

    [itex] w=k^2 [/itex]

    and that the integral is just the "sum" of all the solutions.

    Thanks for the help!
  6. Sep 7, 2011 #5


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    Homework Helper

    You have to integrate over all possible wavenumbers k, so when do the substitution as I suggested you can get everything to one side as:
    \int_{-\infty}^{\infty}(\omega (k)-k^{2})A(k)e^{i(kx-\omega (k)t)}dk=0
    The only way for this integral to be zero of if the integrand is zero and hence you have your dispersion relation.
  7. Sep 7, 2011 #6
    Ahhhh...makes sense now. Thank you for the clarification.
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