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Solving second order PDE by separation of variables (getting 2 ODE's)

  1. Jun 3, 2009 #1
    Hi all,

    For my thesis I would like to solve the following second order nonlinear PDE for [tex]V(x,\sigma,t)[/tex]:

    [tex]\frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial x^2}+\frac{1}{2}B^2\frac{\partial^2 V}{\partial \sigma^2}+a\frac{\partial V}{\partial \sigma}=0,[/tex]

    subject to the following boundary conditions:
    [tex]V(x,0,t)=\epsilon^{3/2}K \max(x,0)e^{-r(T-t)}[/tex] and
    [tex]V(x,\infty,t)=K (1+\epsilon^{3/2}x)[/tex]

    and with terminal condition
    V(x,\sigma,T)=\epsilon^{3/2} K \max(x,0).

    I've tried separation of variables, by writing [tex]V=X(x)Y(\sigma)[/tex]
    which gives 2 ODEs
    [tex]X''(x)=kX(x)[/tex] (which is nicely solvable) and
    [tex]B^2Y''(\sigma)+2aY'(\sigma)+k\sigma^2Y(\sigma)=0[/tex], which is quite difficult, because it's still nonlinear.

    Can anyone help me solve this? I don't know if this is the right way to do it, so other suggestions are also welcome!


    PS: Help, I don't get the tex-code in my post in the right way... how does this work?
    Last edited: Jun 3, 2009
  2. jcsd
  3. Jun 3, 2009 #2


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    End with [/tex], not [\tex].
  4. Jun 3, 2009 #3


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    No, that's a linear equation. It just has a variable coefficient. You might be able to do it with a series solution.

  5. Jun 3, 2009 #4
    Thanks for your help with the tex-parts.

    Unforunately, I already tried to construct a power series solution, and I didn't succeed so far.
    This is what I tried:

    We have to solve

    Expand [tex]Y(\sigma)=\sum_{n=0}^\infty c_n \sigma^n [/tex], and substitute this into (1) to get

    [tex]B^2\sum_{n=2}^\infty n(n-1)c_n \sigma^{n-2}+2a \sum_{n=1}^\infty n c_n \sigma^{n-1}+k \sigma^2 \sum_{n=0}^\infty c_n \sigma^n = 0. [/tex]

    If we let all sums start at n=0, we get
    [tex]\sum_{n=0}^\infty \left( B^2(n+2)(n+1)c_{n+2} + 2a (n+1) c_n + k \sigma^2 c_n \right) \sigma^n = 0. [/tex]

    Now how do I solve this? I need to find a recurrency relation for the coefficients [tex]c_n[/tex], but I don't know what to do with the [tex]\sigma^2[/tex] in the last term on the LHS.
    Last edited: Jun 3, 2009
  6. Jun 3, 2009 #5
    The last term is not well stated, the equation for the powers should read

    [tex]B^2\sum_{n=2}^\infty n(n-1)c_n \sigma^{n-2}+2a \sum_{n=1}^\infty n c_n \sigma^{n-1}+k \sum_{n=0}^\infty c_n \sigma^{n+2} = 0. [/tex]

    Now, changing coefficients so all powers are the same, we have

    [tex]\sum_{n=0}^\infty \bigl( (n+2) B^2 c_{n+2} + 2a c_n\bigr)(n+1)\sigma^n + \sum_{n=2}^\infty k c_{n-2} \sigma^n= 0.[/tex]

    Can you take it from here?
  7. Jun 3, 2009 #6
    If a is zero, then your equation for Y is closely related to the quantum-mechanical simple harmonic oscillator in one dimension, and its polynomial solutions should be something like Gaussians multiplied by Hermite polynomials. Perhaps this can get you further along.
  8. Jun 4, 2009 #7
    When I take out the first two terms (for n=0 and n=1), we have
    [tex]2B^2c_2 + 2a c_0 =0 [/tex] and
    [tex](6B^62 c_3 + 4a c_1) \sigma =0, [/tex]
    such that
    [tex]c_2 = -\frac{a}{B^2}c_0[/tex] and
    [tex]c_3 = -\frac{2a}{3B^2}c_1.[/tex]
    Is that right?

    So now both summation terms start at n=2, and can be combined to one summation as follows:

    [tex]\sum_{n=2}^\infty ( ((n+2)B^2 c_{n+2} + 2a c_n)(n+1) + kc_{n-2} ) \sigma^n = 0[/tex]

    Next, I have to find this recurrency relation for all terms with n>1.. but how?
    Is it given by

    [tex]c_{n+2}=\frac{-2a(n+1) c_n - k c_{n-2}}{(n+1)(n+2)B^2}[/tex]?

    And how do I get my boundary conditions for [tex]X(x)[/tex] and [tex]Y(\sigma)[/tex]? Because we need them to determine the constants, right?
    Last edited: Jun 4, 2009
  9. Jun 4, 2009 #8
    [tex]c_2 = -\frac{a}{B^2}a_0,[/tex]

    [tex]c_4 = \frac{2 \cdot 3 a^2-k}{3 \cdot 4 B^4}a_0,[/tex]

    and so on.

    For the constants, I believe the best way is to express the boundary conditions in power series, but I'm not sure.
  10. Jun 4, 2009 #9
    Do you mean [tex]c_0[/tex] instead of [tex]a_0[/tex]?

    I understand how you got the first expression, for [tex]c_2[/tex], but how did you get the one for [tex]c_4[/tex]??
  11. Jun 30, 2009 #10
    Anyone who can help me out with this? I'm still stuck at the same point...
  12. Jul 1, 2009 #11
    Sorry for the late reply, didn't saw the post.

    First of all, you are right, I meant [itex]c_0[/itex]. Second, there is a little mistake in my algebra.


    \sum_{n=0}^\infty \bigl( (n+2) B^2 c_{n+2} + 2a c_n\bigr)(n+1)\sigma^n + \sum_{n=2}^\infty k c_{n-2} \sigma^n= 0,


    c_2 = -\frac{a}{B^2} c_0,

    c_3 = -\frac{2a}{3B^2}c_1,


    c_{n+2}=- \frac{2 (n+1) a c_n + k c_{n-2}}{(n + 1) (n + 2) B^2}.

    So far so good. Now, For [itex]c_4[/itex],

    c_4 = c_{2+2} = - \frac{2 (2 + 1) a c_2 + k c_0}{(2 + 1) (2 + 2) B^2} = - \frac{ 2 \cdot 3 a c_2 + k c_0}{ 3 \cdot 4 B^2}.

    Substituting [itex]c_2[/itex], then

    c_4 = \frac{2 \cdot 3 a^2 - k B^2}{3 \cdot 4 B^4} c_0.

    In the same way you can calculate [itex]c_5,\,c_6,...[/itex]. The reason why I let the constants expressed as multiplications is because it might be easier to spot factorials in the recurrence relation.

    What is clear is that the general expression for [itex]c_{2n}[/itex] is going to be complicated, as it will be a sum of n terms, while [itex]c_{2n+1}[/itex] will be a sum of n-1 terms (i think).

    Now that is clear how to calculate the coefficients, have you figured out how to deal with the boundary conditions?

    P. S. Please redo all the math, just to make sure there's no errors this time.
  13. Jul 1, 2009 #12
    Thanks for your reply :) In the mean time, I indeed figured out how to calculate the coefficients, using the recursion, so on that I agree with you (or at least we do it the same way, tomorrow I'll check the outcomes with my own paperwork).

    About the boundary conditions: still have no idea how to do this!!

    And about the recurrency: I don't expect to see a nice (well known) series or something in that. Do you think it's a logical step to just take the first x (some reasonable number) terms and use that as an approximation of the solution?
  14. Jul 1, 2009 #13
    Well, how good the approximation is depends on how big the error of taking the truncated series. Remember that the solution is only valid whitin the radius of convergence of the series.

    Let me ask you a question, is [itex]\epsilon[/itex] a small parameter?
  15. Jul 3, 2009 #14
    Yes, it is: [itex]\epsilon[/itex] is a small scaling parameter.
  16. Jul 3, 2009 #15
    Last edited by a moderator: May 4, 2017
  17. Jul 3, 2009 #16

    I know, that's exactly what I've been doing. By applying this asymptotic theory (using the method of asymptotic expansions to solve a singular perturbation problem) I got to the problem I stated here... So applying it twice would be a bit strange, wouldn't it?

    About the books: Holmes indeed is a good one, I've been using it. The other ones I don't really know, but thanks for the advice: maybe I'll take a look at them too...
    Last edited by a moderator: May 4, 2017
  18. Jul 4, 2009 #17
    The whole point of perturbation techniques is to simplify a problem by using the small parameter. How come you ended up with a simpler problem that is still dependent of the small parameter?
  19. Jul 5, 2009 #18
    Let me show you the complete problem then. Maybe that will make things a bit more clear. In the attachment you can find a part (scrap version of a chapter) of my report so far. Here I describe the problem completely. It's an application of perturbation methods on a financial model.

    Section 1.3 is what this problem is about. Here we are doing some boundary layer analysis.

    Hope you can help me out with this. If you know any other method to solve the inner problem: that is also fine!

    Attached Files:

  20. Jul 13, 2009 #19
    I'll try a fourier transform on the original problem.. maybe that will help
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