Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear wave equation with moving point source

  1. Aug 10, 2010 #1
    I'll need some help and clarification about solving this equation.

    After some non-dimensionalization, I can arrive at the following wave equation with a moving point source. The initial conditions are zero.

    [itex]\Delta P - \frac{\partial^2 P}{\partial \tau^2} = - A \cos(\tau) \delta^3(\vec{r} - \vec{r}_s(\tau))[/itex]

    Whether the source is a cosine, a sine, or something else that is periodic is unimportant.

    [itex]\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}[/itex] or something to that effect. [itex]\vec{r}_s[/itex] is the location of the moving source.

    I wanted to solve this equation, so I tried a change of variables to make the source stationary. I try [itex]\vec{r}' = \vec{r} - \vec{r}_s(t)[/itex]. In Cartesian coordinates this means [itex]x' = x - x_s(t)[/itex], [itex]y' = y - y_s(t)[/itex], [itex]z' = z - z_s(t)[/itex].


    [itex]\Delta P = \frac{\partial^2 P}{\partial x^2} + \frac{\partial^2 P}{\partial y^2} + \frac{\partial^2 P}{\partial z^2} = \frac{\partial^2 P}{\partial x'^2} \left(\frac{\partial x'}{\partial x}\right)^2 + \frac{\partial^2 P}{\partial y'^2} \left(\frac{\partial y'}{\partial y}\right)^2 + \frac{\partial^2 P}{\partial z'^2} \left(\frac{\partial z'}{\partial z}\right)^2[/itex]

    But [itex]\frac{\partial x'}{\partial x} = \frac{\partial y'}{\partial y} = \frac{\partial z'}{\partial z} = 1[/itex]. So [itex]\Delta P = \frac{\partial^2 P}{\partial x'^2} + \frac{\partial^2 P}{\partial y'^2} + \frac{\partial^2 P}{\partial z'^2}[/itex]

    [itex]\frac{\partial^2 P}{\partial x'^2} + \frac{\partial^2 P}{\partial y'^2} + \frac{\partial^2 P}{\partial z'^2} - \frac{\partial^2 P}{\partial \tau^2} = - A \cos(t) \delta^3(\vec{r}')[/itex]

    This seems simpler than I thought it would be. I would expect additional source terms because the source is moving. I'm convinced there's something fundamentally wrong with my change of variables. If this is wrong, what am I missing?

    Also, I'll need some help solving the equation once I get it into a good form. I have Strauss's PDE book, which details the inhomogeneous wave equation with zero initial conditions in section 9.3, however, I've always found Strauss to be too terse to be useful.
    Last edited: Aug 10, 2010
  2. jcsd
  3. Aug 10, 2010 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    What's wrong is that your change of variables involves four coordinates; not three. So

    [tex]\frac{\partial}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} + \frac{\partial \tau'}{\partial x} \frac{\partial}{\partial \tau'}[/tex]


    [tex]\frac{\partial}{\partial \tau} = \frac{\partial x'}{\partial \tau} \frac{\partial}{\partial x'} + \frac{\partial \tau'}{\partial \tau} \frac{\partial}{\partial \tau'}[/tex]

    So, using

    [tex]x'(x, \tau) = x - x_s (\tau)[/tex]
    [tex]\tau'(x, \tau) = \tau[/tex]

    some of the above terms cancel. Your space derivatives transform just as you computed; but your time derivative transforms in a nontrivial way.

    By the way, I think the easiest way to solve this equation is to use the coordinates you started with. Then you simply integrate the source term against the retarded Green function.

    Alternatively, if your source is moving in a straight line at constant velocity, then you can use a Lorentz transformation, which leaves the wave operator invariant. If the source velocity is less than the wave velocity, you can transform to a stationary source; if the source velocity is greater, you can transform to an instantaneous line source.
  4. Aug 10, 2010 #3
    Ah, that makes sense. Thank you.

    Could you recommend a website or book which details the integration against the retarded Green's function approach? My PDE class several semesters ago did not detail that and the book I have is not particularly helpful.
  5. Aug 10, 2010 #4


    User Avatar
    Homework Helper

    As long as your spatial boundary conditions are at infinity, an alternative approach you might consider is to Fourier transform in the spatial variables. This gives you

    [tex]-|\mathbf{k}|^2 \tilde{P} - \frac{d^2 \tilde{P}}{d \tau^2} = -A\cos(\tau)\exp({i\mathbf{k} \cdot \mathbf{r}_s(\tau)})[/tex]
    where [itex]\mathbf{k} = (k_x,k_y,k_z)[/itex].

    Solve this, then inverse Fourier transform the solution.
  6. Aug 12, 2010 #5
    Mute, this problem is unbounded, so that's a good suggestion. I've attempted a few things with the Fourier transform now, but I don't think any of them will lead to an acceptable solution. To solve what you've derived practically requires me to know the position of the source, or at least something about its movement. I'm trying to make as few assumptions about the movement of the source as possible.

    I even tried my original change of variables (corrected, of course), and I dropped a few terms (which are okay if the Mach number of the point source is low), but that doesn't seem to lead anywhere. I attempted to develop a perturbed solution, but the solution got messy. Perhaps I'm too picky.

    My ultimate goal is to determine the movement of the source given some of the pressure field. I'm going to start looking at books about inverse methods.

    Could someone suggest a book that explains the retarded Green's function approach for the wave equation in more detail than Strauss's PDE book?
  7. Aug 18, 2010 #6
    Could someone check this change of variables? I'm suspicious that I've made a mistake.

    For simplicity I'll focus on the case with one spatial variable.

    [itex]\frac{\partial^2 P}{\partial x^2} - \frac{\partial^2 P}{\partial \tau^2} = - A \cos(\tau) \delta(x - x_s(\tau))[/itex]

    With the change of variables [itex]x' = x - x_s(\tau)[/itex] and [itex]\tau' = \tau[/itex] I get the following equation in the new variables.

    [itex]\left(1 - \left(\frac{d x_s}{d \tau'}\right)^2 \right) \frac{\partial^2 P}{\partial x'^2} - \frac{\partial^2 P}{\partial \tau'^2} + \frac{d^2 x_s}{d \tau'^2} \frac{\partial P}{\partial x'}= - A \cos(\tau') \delta(x')[/itex]

    Does this seem right?

    If this is right, the [itex]\left(\frac{d x_s}{d \tau}\right)^2[/itex] term can be dropped. My non-dimensionalization made all velocities Mach numbers, and as I'll only be looking at the low Mach number (of the moving source) case (10 m/s at most!), M^2 seems even less significant. This could lead to an accurate solution for low Mach numbers.
    Last edited: Aug 18, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook