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Nonlinear diff. eq. involving Besselfunction of first kind

  1. Jul 1, 2009 #1
    Hi there, I hope my post is not against forum rules (not sure if this section is only intended for general questions or if it is ok to ask about specific problems .. it is definitely not Homework). Anyway, I was hoping that there were some guru's out there that can help me with the following tough first order nonautonomous nonlinear differential equation.

    [tex]
    \frac{dx}{dt}+xJ_0(x t)=0
    [/tex]

    where [itex]J_0(y)[/itex] is the zeroth order Bessel function of the first kind.
    Any kind of help would be greatly appreciated.
     
    Last edited: Jul 1, 2009
  2. jcsd
  3. Jul 3, 2009 #2
    Yes it doesn't look easy to solve. I presume you are only interested with analytical solution.

    What methods have you tried? May be we get think of other methods. We are not interested in reinventing unsuccessful stories :wink:

    Is there any initial condition that comes with this DE?
     
  4. Jul 6, 2009 #3
    Thanks for the reply matematikawan, sorry for my late reply. I've had some internet issues lately.

    My experiences with nonlinear diff eqns is quite limited so I am not aware of many methods for solving such equations, I know most autonomous nonlinear diff eqns can be solved by simple integration, but other than that my mind is pretty much blank.

    Ideally I would of course like a full analytical solution for arbitrary boundary conditions but I think that is too much to ask for in this case. I would be very happy with an analysis of the long time asymptotics (t being time of course).

    It seems from my numerical calculations that the solution always tends to [itex]x\rightarrow 0[/itex] as [itex]t\rightarrow \infty[/itex]. Is it possible to prove this analytically?

    Any help is appreciated.

    /Jens
    /Jens
     
  5. Jul 7, 2009 #4
    It is quite easy to rearrange the equation into an integral, which cannot be calculated whatsoever.

    So we use the following basic property of Bessel functions:

    [tex]t\cdot x \cdot J_0(xt)=\frac{d}{dt}\left(tJ_1(xt)\right) \Longrightarrow x \cdot J_0(xt)=\frac{1}{t}\frac{d}{dt}\left(tJ_1(xt)\right) [/tex]

    plugging this into our DE, we can integrate at once :

    [tex]x+C = -\int_0^t \frac{1}{t'}\frac{d}{dt'}\left(t'J_1(xt')\right) dt' [/tex]

    using partial integration:

    [tex] x+C = -J_1(xt) -\int_0^t \frac{J_1(xt')}{t'}dt' [/tex]

    Now the above integration can be expressed at most by the hypergeometric function.. but that is not much use to us.. Or you can try a series expansion.

    The asymptotic form is straightforward.., if [tex]t\to \infty [/tex] the first Bessel goes to zero (of course t has to be really big.., as we now the Bessels decay as [tex]1/\sqrt{t}[/tex]) in this case the integral can be evaluated exactly:

    [tex]\int_0^{\infty} \frac{J_1(xt')}{t'}dt' = \pm 1 [/tex]

    (if x > 0 and x<0 respectively)

    So in this case we see that x converges to some constant depending on the initial conditions.

    If you want some deeper theory then take a look in "Watson: treatise on the theory of Bessel functions".
     
  6. Jul 7, 2009 #5
    Thanks for your reply Thaakisfox,

    However, I believe you have made some critical errors in your solution:


    This relation would be correct if x did not depend on time, but since we want to solve for x(t) this is obviously not the case. (Perhaps I am wrong and this is a more sophisticated relation but I cannot see how it could hold for time-dependent x).




    This expression again holds only if x were independent of time. Since x depends on time we can not compute this integral (as far as I can see). In fact we cannot even conclude that the integral converges since there is no way of knowing that the product [itex]xt\rightarrow \infty[/itex] as [itex]t\to \infty[/itex], (consider for example if the asymptotic form of x would go as [itex]x(t)\sim 1/t[/itex], then [itex]xt\rightarrow \text{const.}[/itex])

    I will take a look at that book, title sounds promising. Let me know if I have misunderstood something about your derivation.

    /Jens
     
  7. Jul 7, 2009 #6
    Well werent you looking for an asymptotic solution??,

    I considered the case where t is great, and x is quite smooth.. which means that [tex]x'/x << 1[/tex]

    for the integral i checked the cases where where [tex]xt \to \infty [/tex], of course this doesnt exhaust all the cases for x.., i am glad that at least something could be said in this case..

    I should have clarified my approximation conditions a bit more, then you would have understood it :D


    Of course an analytic solution can be ruled out as we take a first glimpse at the problem.. So we are left with approximating methods..
     
  8. Jul 7, 2009 #7
    Series solution near [tex]t=0[/tex], with initial condition [tex]x(0)=a[/tex] ...

    [tex]x(t) = a-a t+((1/2) a) t^2+(-(1/6) a+(1/12) a^3) t^3+
    (-(5/24) a^3+(1/24)a) t^4+((4/15) a^3-(1/120) a-(1/320) a^5) t^5+O(t^6)[/tex]

    [done with Maple]
     
  9. Jul 7, 2009 #8
    Thaakisfox, don't get me wrong. I very much appreciate your effort to help me.

    Yes definitely, I simply am not convinced by your analysis of the asymptotic solution. Main reason being that your predictions do not agree with numerical calculations. All numerical solutions tend towards x=0. My initial analysis of this equation also seemed to lead to the conclusion that there should exist solutions which tend to a non-zero value. However, I have now revised my analysis and now the predictions agree with the numerical calculations.

    Unfortunately my analysis is mainly graphical so I cannot quite write it down here, but I can summarize the asymptotic forms of the solutions:

    The asymptotic solutions depend strongly on the initial conditions. There exist a countable set of critical values of initial value [itex]x(0)[/itex] which define the boundary between different asymptotic forms. When the initial value is lower than the lowest critical value then the asymptotic solution will have an exponential form. When the initial value is larger than the lowest critical value then the asymptotic solution will go as [itex]x\sim 1/t[/itex] with a proportionality constant given by the the 2n:th zeropoint of the Bessel function when the [itex]x(0)[/itex] is larger than the n:th critical value.

    This analysis fits perfectly with the numerical solution so I am quite happy with what I have now.

    Thanks guys.

    /Jens
     
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