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Boltzmann equation with collision term

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the Boltzmann equation for a homogeneous plasma with not external forces present when the collision term is
    [itex]\frac{\partial f(v,t)}{\partial t} = -\nu (f(v,t) - f_0(v))[/itex],
    where [itex]\nu[/itex] and [itex]f_0[/itex] are constants.

    2. Relevant equations
    Boltzmann equation
    [itex]\frac{\partial f}{\partial t} + v\cdot \nabla f + a \cdot \nabla_v f = -\nu (f(v,t) - f_0(v))[/itex]
    where [itex]\nabla_v f[/itex] is the gradient of f in respect to vx, vy and vz.

    3. The attempt at a solution
    Right, first off, i'm assuming that [itex]v \cdot \nabla f = 0[/itex] since in the collision term it claims that the function f(v,t) doesn't depend on x. This follows that
    [itex]\frac{\partial f}{\partial t} + a \cdot \nabla_v f = -\nu (f(v,t) - f_0(v))[/itex]

    After this i'd assume that the function can be represented as [itex]f(v,t)=g(v)h(t)[/itex]
    Now with a little algebra we can get the equation in the form of
    [itex]\frac{1}{h(t)}\frac{\partial h(t)}{\partial t} + \nu = \nu \frac{f_0(v)}{g(v)} - \frac{1}{g(v)}a \cdot \nabla_v g(v) = constant = \lambda[/itex]

    Now i have a pair of equations and calculating h(t) is fairly simple
    [itex]\frac{1}{h(t)}\frac{\partial h(t)}{\partial t} + \nu = \lambda[/itex]
    which by integrating leads to
    [itex]h(t) = Ce^{(\lambda - \nu)t}[/itex] where C is the integration constant.

    However, solving the other equation (below) is a bit harder
    [itex]\nu \frac{f_0(v)}{g(v)} - \frac{1}{g(v)}a \cdot \nabla_v g(v) = \lambda[/itex]

    I can guess the solution to this would be
    [itex]g(v) = e^{(\lambda/a)v}+\nu \frac{f_0(v)}{a}+D[/itex] where D is again some constant.
    And as stated before, connecting the two solutions like [itex]f(v,t)=g(v)h(t)[/itex], would give the answer... maybe.

    Now the problem is, i have no clue if this correct or even close to the solution. Could anyone check if i'm correct and if not (which is very likely), where have i messed up and how could i do the calculation right?
  2. jcsd
  3. Feb 16, 2013 #2


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    Homework Helper

    I would definitely use a different symbol than ##\nu## for that constant. It took me seeing the LaTeX to notice that there was a ##\nu## and a ##v##. I could definitely see myself mixing the two up by accident during a hand calculation.

    It looks like here is where you made your mistake. When you separate variables and divide by ##h(t)g(v)##, you'll end up with a term ##f_0(v)/(h(t)g(v))##, not ##f_0(v)/g(v)## as you have written. Because of this inhomogeneous term, separation of variables is probably not the best approach to take.

    There are a few potential ways to go about solving this problem. One would be to Laplace transform the equation over the time variable. This would leave you with a PDE in v only for the Laplace-transform ##\tilde{f}(v,s)##. Alternatively, assuming the components of v can run from ##-\infty## to ##+\infty##, you could do a 3d fourier transform over v, leaving you with an ODE in time for the fourier transform ##\hat{f}(k,t)##. These methods also work well if ##f_0(v)## is an unspecified function, as you could do a Green's function calculation.

    Do you know any of these methods?

    Unless you just simplified the problem to one dimension without mentioning it, this cannot be correct. The parameter "a" in your original equation is a vector, as is "v", so we should expect to see some sort of dot product in the exponentials if this were the solution. If we plug this back into the PDE, we'll also see that this solution would generate a term ##\nabla_v f_0(v)##, which doesn't appear in the original equation, so something is amiss. If you're not sure of a solution, plugging it back into the PDE is always a good thing to check!

    As a general comment on the separation of variables technique, ##f(v,t) = h(t)g(v)## itself would not likely be the solution. Typically, in solving the problem (if there were no inhomogeneous term), one would find a set of allowed values for ##\lambda## and sets of functions ##h_\lambda(t)## and ##g_\lambda(v)## to go with them, and the full solution would be

    $$f(v,t) = \sum_\lambda c_\lambda h_\lambda (t)g_\lambda (v),##
    where the ##c_\lambda## are ##\lambda##-dependent constants that help the series fit the overall boundary conditions you want to impose on the function f(v,t). Note also that the sum in this equation could also be an integral if the set of ##\lambda## are continuous, or the sum could be a sum plus an integral if ##\lambda## has a discrete set of allowed values and a continuous set.

    Last edited: Feb 16, 2013
  4. Feb 16, 2013 #3
    I'm somewhat familiar with the laplace transform and thus had it a go.
    I'll use "k" to replace that nu mark for the sake of convenience.

    Now [itex]\frac{\partial f(v,t)}{\partial t} + a \cdot \nabla_v f(v,t) = -k (f(v,t) - f_0(v))[/itex].
    What was probably clear before but what i didn't state, was that "v" and "a" are vectors standing for velocity and acceleration, "k" is just a constant and "f_0" is equilibrium distribution (as well a constant).
    Using the laplace transform in respect to "t" yields (L stands for the laplace function symbol)
    [itex]sL(f(v,t))-f(v,0) + a \cdot \nabla_v L(f(v,t)) = -kL(f(v,t)) - kL(f_0(v))[/itex]
    What's the biggest concern here for me is the [itex]a \cdot \nabla_v[/itex] term. Could it be just as simple as [itex]\frac{\partial v_x}{\partial t}\hat{i} \cdot \frac{\partial}{\partial v_x}\hat{i}+... = 3\frac{\partial}{\partial t}[/itex] when done over 3 dimensions? The PDE would be very simple - at a quick glance - to solve even without laplace transform if that's so.

    Actually i'll stop here now because the rest of the problem relies on that term and how it behaves. So could i get an answer to the question underlined above? If it behaves in other ways, any help on that would be appreciated.
  5. Feb 16, 2013 #4


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    Homework Helper

    Ah, right, it didn't occur to me that ##a## was acceleration. It's been too long since I solved a Boltzmann equation, apparently. This will actually make the problem a lot simpler! First, a mathematical comment:

    $$a \cdot \nabla_v \neq 3 \frac{\partial}{\partial t},$$
    but there's a similar rule that is true: for a function f(x(t),v(t),t), the total derivative with respect to time can be written

    $$\frac{d}{dt} f(x,v,t) = \frac{\partial}{\partial t} f(x,v,t) + \frac{dx}{dt} \cdot \nabla_x f(x,v,t) + \frac{dv}{dt} \cdot \nabla_v f(x,v,t),$$
    where as usual the gradients of f(x,v,t) are done with t held fixed.

    You'll notice that this is basically just the Boltzmann equation! The only additional information we're using is that the total time derivative is equation to the 'collision' term and that there is some force we're imposing on the system, which is what sets ##a = dv/dt##.

    ...But your initial problem tells us there's no external force on the plasma, which means ##F = ma = 0##. So, basically that troublesome term just goes away! This should make the equation a lot easier to solve. You probably don't even have to Laplace transform.

    Note, however, that if there were an external force on the plasma, the acceleration in the ##a \cdot \nabla_v## is set by the external force, and you can't write this term in terms of a time derivative. For example, if you had an external force due to an electric field, ##F = qE = qE_0 \hat{z}##, then ##a \cdot \nabla_v f(v,t) = qE_0 \partial_{v_z} f(v,t)##, and you'd have to solve the equation using the Laplace transform method, for example. Note, however, that if the acceleration from the external force were time-dependent, the Laplace transform of ##a(t) \cdot \nabla_v f(v,t)## is not ##a(t) \cdot \nabla_v L(f(v,t))## because the Laplace integral has to act on a(t) as well, so a Laplace transform wouldn't be a good approach in that case.

    But anyways, you have a much simpler case to deal with right now. Give it a shot and let us know if you get stuck again.
  6. Feb 16, 2013 #5
    Oh nice! Everything seems so much clearer now :). Thanks a lot Mute!
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