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Help with Greens functions

  1. Jun 21, 2011 #1

    hunt_mat

    User Avatar
    Homework Helper

    I have to solve the following PDE:
    [tex]
    \frac{\partial^{2}u}{\partial t^{2}}+2\frac{\partial^{2}u}{\partial t\partial x}+\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t}+k^{2}u=f
    [/tex]
    I use the Greens function method and examine the equation:
    [tex]
    \frac{\partial^{2}G}{\partial t^{2}}+2\frac{\partial^{2}G}{\partial t\partial x}+\frac{\partial G}{\partial x}+\frac{\partial G}{\partial t}+k^{2}G=4\pi\delta (x-x')\delta (t-t')
    [/tex]
    I then write:
    [tex]
    G=\frac{1}{2\pi}\int_{-\infty}^{\infty}g(X|X')e^{i\omega (T-T')}d\omega\quad \delta (T-T') =\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega (T-T')}d\omega
    [/tex]
    The equation then becomes:
    [tex]
    (1+2\omega i)\frac{\partial g}{\partial X}+(k^{2}-\omega^{2}+i\omega)g=4\pi\delta (X-X')
    [/tex]
    Take the Fourier transform to obtain:
    [tex]
    i\xi (1+2\omega i)\hat{g}+(k^{2}-\omega^{2}-\omega i)\hat{g}=4\pi e^{i\xi X'}
    [/tex]
    Rearrange and take the inverse Fourier transform to obtain:
    [tex]
    g=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4\pi e^{-i\xi (X-X')}}{i\xi (1+2\omega i)+k^{2}-\omega^{2}-\omega i}d\xi
    [/tex]
    Am I on the right track here?
     
  2. jcsd
  3. Jun 21, 2011 #2

    hunt_mat

    User Avatar
    Homework Helper

    I think that I can find g by using contour integration. Write:
    [tex]
    \frac{1}{2\pi}\oint_{\gamma}\frac{4\pi e^{-iz(X-X')}}{iz(1+2\omega i)+k^{2}-\omega^{2}-\omega i}dz
    [/tex]
    Which can then be evaluated via Cauchy's integral formula:
    [tex]
    g=4\pi ie^{i(X-X')h(\omega )},\quad h(\omega )=\frac{k^{2}-\omega^{2}-\omega i}{1+2\omega i}
    [/tex]

    Thoughts?
     
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