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Unstable solution of differential equation

  1. Dec 7, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider the following differential equation

    $$\frac{\partial^{2}\phi}{\partial t^{2}}-\nabla^{2}\phi = \phi(a-b\phi^{2}), \qquad a>0.$$

    I would like to prove that $\phi=0$ is an unstable solution of this equation.

    2. Relevant equations

    3. The attempt at a solution

    Do you suggest a numerical approach?

    Can this not be done analytically?
     
  2. jcsd
  3. Dec 7, 2016 #2

    hilbert2

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    Here we need to do something that's called linear stability analysis... Suppose that at an initial moment ##t=t_0##, the function ##\phi## is almost ##\phi (x,t_0 ) = 0##, but has a very small-amplitude oscillation in it:

    ##\phi (x, t_0 ) = \epsilon \sin (k x) ##, ##\epsilon << 1##.

    Now, let's also assume that the initial behavior of this solution (in 1st order with respect to ##t##), is such that the amplitude starts growing or decreasing exponentially:

    ##\phi (x, t) = \epsilon e^{\beta (t-t_0 )} \sin (k x)##

    Substitute this trial function ##\phi## in the diff. equation, expand both sides of the resultant equation as Taylor series about ##t_0 ##, discard all terms that are higher than first order in ##\epsilon## and then prove that this gives the parameter ##\beta## a positive value (implying exponential growth).
     
  4. Dec 7, 2016 #3
    Why do we assume, to begin with, that

    ##\phi (x, t_0 ) = \epsilon \sin (k x) ##, ##\epsilon << 1##?

    In other words, why does ##\phi(x)## have a small-amplitude oscillation? Could the perturbation not be just a constant number in space?
     
  5. Dec 7, 2016 #4

    hilbert2

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    The constant function is just the special case where ##k=0##. Using the sine function trial allows finding out whether the system is unstable for some range of wavenumbers ##k## and stable for others.

    EDIT: Sorry, it has to be ##\cos (kx)## for it to contain the constant function case. However, this does not matter if the DE is symmetric in translation ##x \rightarrow x + a##
     
  6. Dec 7, 2016 #5
    I was wondering why you use the words 'in first order in ##t##' when describing the initial behaviour of the system?
     
  7. Dec 7, 2016 #6

    hilbert2

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    Well, for example, the function ##\sqrt{x+1}## is equal to ##1 + \frac{1}{2}x## in 1st order of the variable ##x##. It means that you consider the function only in so small intervals that it's graph is practically linear.
     
  8. Dec 7, 2016 #7
    But, what if the system follows some other solution of the differential equation.

    I mean,

    ##\phi (x, t) = \epsilon e^{\beta (t-t_0 )} \sin (k x)##

    is just an ansatz and it could be that it does not encapsulate the full set of solutions of the differential equation, right?
     
  9. Dec 7, 2016 #8

    hilbert2

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    Note that any function ##\phi (x, t_0 )## can be written as a combination of different sine and cosine functions with different wavelengths. In this linear approximation we only consider a time interval or values of ##\epsilon## that are so small that we can assume the different Fourier components evolve independently, like they would in the case of a linear partial differential equation (note that the equation here is nonlinear).

    EDIT: also note that a single counterexample like this is enough to prove that a solution is not stable, while proving that a solution is stable requires a lot more effort.
     
    Last edited: Dec 7, 2016
  10. Dec 7, 2016 #9
    Why do the different Fourier components evolve independently for a small time interval or small value of ##\epsilon##?
     
  11. Dec 7, 2016 #10

    hilbert2

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    Because if ##\epsilon## is small, then the term ##-b\phi^3## which causes the nonlinearity of the equation is even smaller by several orders of magnitude.
     
  12. Dec 7, 2016 #11
    Ok, so following your prescription,

    Substituting the trial function ##\phi(x,t)## in the differential equation, I get

    ##\partial^{\mu}\partial_{\mu}\phi = a\phi - b\phi^{3}##

    ##(\partial_{t}^{2}-\partial_{i}^{2})\phi = a\phi - b\phi^{3}##

    ##(\beta^{2}+\vec{k}^{2})\phi = a\phi - b\phi^{3}##

    Then,

    ##(\beta^{2}+\vec{k}^{2})\phi = a\phi - b\phi^{3}##

    ##(\beta^{2}+\vec{k}^{2})\left(\epsilon e^{\beta (t-t_0 )} \sin (\vec{k}\cdot{\vec{x}})\right) = a\left(\epsilon e^{\beta (t-t_0 )} \sin (\vec{k}\cdot{\vec{x}})\right) - \underbrace{b\left(\epsilon e^{\beta (t-t_0 )} \sin (\vec{k}\cdot{\vec{x}})\right)^{3}}_{\text{ignore}}##

    I ignored the second term as it is cubic in ##\epsilon##. Also, I have decided to use wavevector ##\vec{k}## instead of wavenumber ##k##.

    I don't why I need to expand in taylor series about ##t_0##. After all, we already have an equation for ##\beta## in terms of ##a## and ##\vec{k}##, right?
     
  13. Dec 7, 2016 #12

    hilbert2

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    Yeah, you don't really need a Taylor series when the nonlinearity is a power of ##\phi## (##\phi^3## in this case) and not some more complicated function. Just divide the equation with ##\epsilon e^{\beta (t-t_0)} \sin kx## and find the conditions for ##\beta## to be a positive real number.

    EDIT: oh, looks like the equation only gives information about ##\beta^2##... Maybe we'd need an ansatz that's proportional to ##e^{\sqrt{\beta}(t- t_0)}##. I've only done this for equations that have a first order time derivative, before...
     
    Last edited: Dec 7, 2016
  14. Dec 7, 2016 #13
    Why can we divide by ##\sin (kx)##? After all, the sine function can be ##0## and then we'd be dividing by ##0##.

    Why can't we just divide by the exponential and ##\epsilon## on both sides of the equation?
     
  15. Dec 7, 2016 #14

    hilbert2

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    I corrected the last post. I'll try to find some article where the linear stability analysis is done for an equation that is second order in time derivative.

    EDIT: You probably need to use a perturbation term that is a plane wave (proportional to ##e^{ikx}##) instead of a sine or cosine function, to get rid of the division by zero problem. I've seen that kind of an approach being used in some articles.
     
    Last edited: Dec 7, 2016
  16. Dec 7, 2016 #15

    hilbert2

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    If you use the ansatz ##\phi (x,t) = \epsilon e^{\sqrt{\beta}t}e^{ikx}##, you get the equation ##\beta = a - k^2##, which means that there's some maximum value the wavenumber ##k## can have for the solution to be unstable. This kind of relations between the "Lyapunov exponent" ##\beta## and the perturbation wavelength are called dispersion relations.
     
  17. Dec 7, 2016 #16
    So, solution is unstable for ##a> \sqrt{k}##?
     
  18. Dec 7, 2016 #17

    hilbert2

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    It appears to be if ##a>k^2## , and particularly the constant solution (##k=0##) is always unstable if ##a## is a positive real number.
     
  19. Dec 7, 2016 #18
    If ##\beta < 0##, then the solution decreases exponentially. Why is this not an unstable solution also?
     
  20. Dec 7, 2016 #19

    hilbert2

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    Because then any deviation from the constant solution disappears really quickly. The situation is similar to comparing these images:

    qT21I.jpg

    and assuming that there's friction between the sphere and the convex/concave surface. The constant solution is similar to the equilibrium situation where the sphere is positioned exactly on the top/bottom of the hill.
     
  21. Dec 7, 2016 #20
    I don't quite understand. How does the deviation disappear? It's still there, right?

    After all, an exponential solution with negative exponent takes the initial value of ##\phi## exponentially (i.e. takes forever) to zero, right?
     
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