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How would I solve a DiffEq of the form

  1. Jul 24, 2008 #1
    \frac{dn(t)}{dt} = A sin(B*n(t)*t) n(t)

    Or a more general
    \frac{dn(t)}{dt} = F(n(t)) n(t)

    I'm not even sure what method I could use, or what it would be called.

    A first order, non-linear equation?

    Maybe it looks neater as:
    \frac{dn}{dt}=A n Sin(n t)

    EDIT : This isn't homework. I'm just looking for insight.
  2. jcsd
  3. Jul 24, 2008 #2


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    That is extremely non-linear because the dependent function, u(t), occurs inside the non-linear function cosine. The fact that you then have that function multiplied by u(t) just makes it worse.

    There are no general methods to solve non-linear equations or even any special classes of non-linear equations. (You can sometimes use "quadrature" for equations where the independent variable does not appear explicitly but they typically result in an integration that cannot be done in closed form.)
  4. Jul 24, 2008 #3
    Hi, K.J.Healey!

    Where did you get this equation from? I mean, does it describe a certain physical system, or you 'invented' it by yourself?
  5. Jul 24, 2008 #4
    It comes from solving a large particle system decay rate due to (B=2) transition (oscillation) and subsequent annhilation. Its actually already a first term of an expansion on a much much more difficult equation. The potential splitting the energies of the particle-antiparticle is a function of the density of the system, which itself is a function of time.

    That's where the amplitude's, as well as the frequency's, dependence one the density of states "n" comes into play. Usually the method is to take for small times "t" and just do an approx, or for t>>0 and do a sin^2 -> (1/2). But unfortunately I cannot. I have no explicit time-dependent extremes with which to expand about, so I need a complete solution.
  6. Jul 24, 2008 #5
    Perhaps I can do this numerically...
  7. Jul 24, 2008 #6


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    Does the * denote convolution? I was thinking either Fourier or Laplace transform, but I can't come up with the transform for the RHS.
  8. Jul 24, 2008 #7
    No, it was merely for a multiplication, because it looked really messy in tex (squshed everything together:
    \frac{dn(t)}{dt} = A Sin(B n(t) t) n(t)

    hard to read.
  9. Jul 24, 2008 #8
    This DE is pretty intense
  10. Jul 25, 2008 #9
    It's not possible to solve, the reason is that inside the sine function there is the function n itself and the independent variable t. There is no way to seperate them and the only way to solve this is using taylor polynomials with an approximation. The first degree approximation is only valid if the product nt is very small and leads to,
    [tex]\frac{dn}{dt}\approx ABn^2t[/tex].
    Divide by [itex]n^2[/tex] and multiply by [itex]dt[/itex] and then integrate and use algebra to get [itex]n=n(t)[/itex].

    Hope that helps, a little.
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