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2nd-Order Non-Linear ODE: Where to Start?

  1. Sep 8, 2008 #1

    I have the 2nd-order nonlinear ODE below:

    k(v)=\frac{\phi ''(v)}{\phi (v) (\phi ' ^2 (v) +1)^2}

    Where k(v) is some function. I would like to investigate for what functions k there can exist solutions on a given interval [Latex][a,b][/Latex]. For example, if [Latex]k(v)=0[/Latex], then [Latex]\phi '' (v)=0[/Latex] which implies that [Latex]\phi (v)=C_1 v + C_2[/Latex]. The DE gets very complicated very quickly, though, and I'm not sure how to approach the problem.

    I do not know very much about differential equations, so I need help in figuring out what to learn. What are the methods for analyzing such a DE? Are there any existence theorems or uniqueness theorems already out there? Can anyone recommend some good literature?

    Thanks for the help.
  2. jcsd
  3. Sep 9, 2008 #2


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    Hi timo1023! :smile:

    For LaTeX on this forum, you have to type [noparse][tex] and [/tex] (or [itex] and [/itex] for "inline tex" that fits into an ordinary line).[/noparse] :wink:

    I'll let someone else answer the actual question … :rolleyes:
  4. Sep 10, 2008 #3
    Bump (sorry).
  5. Sep 11, 2008 #4
    The equation can be written like:

    [tex] -2k(v)\phi' \phi=(1/(1+\phi '^2))' [/tex]

    If k(v) is such that the left hand side is exact derivative wrt v (in this context it would be more appropriate to look for k as function of phi, not v), then you can integrate the equation once and obtain a first order equation. Any k that multiplied by phi gives an integrable function of phi (integrable with respect to phi) would do since the left hand side contains the derivative of phi as multiplicative factor.
    Last edited: Sep 11, 2008
  6. Sep 13, 2008 #5
    smallphi, I have a strong calculus background, so I am very familiar with integration, however I don't know very much about differential equations. How would integrating the equation give you a 1st order DE? I need to see a simple example of how that would work.

    Thanks for the response.
  7. Sep 13, 2008 #6
    The trivial example would be k=const, so let's take a little more complicated example, k(phi) = phi. The left hand side looks like the chain rule (and it will always look like that as long as k is function of phi) i.e. d( )/dphi * dphi/dv.= d( )/dv. You find what's inside ( ) by integrating wrt phi of what's in front of phi_prime:

    [tex](-2 \phi^2 )\frac{d\phi}{dv}= \frac{d (-2\phi^3 /3)}{d\phi} \frac{d\phi}{dv}=\frac{d (-2\phi^3 /3)}{dv}[/tex]

    Now you can easily integrate left and right hand sides wrt v since they are derivatives wrt v:

    [tex]-2\phi^3 /3= 1/(1+\phi ' ^2) + const[/tex]

    That is your equation of first order, solve for phi', it will be non-linear in general, although certain choices of k(phi) can render it linear.

    A nice book to study would be
    "An introduction to the theory of differential equations", by Walter Leighton, 1952, about 150 pages with short clear chapters including examples and exercises. It covers the basics like linear ordinary diff. equations of first and second order.
    Last edited: Sep 13, 2008
  8. Sep 14, 2008 #7
    The only k that I can find which renders it linear is [itex]k(\phi )=0[/itex]; are there others?

    Are there any existence theorems that would be applicable for this equation?

    Thanks for the help.
  9. Sep 14, 2008 #8
    The first order equation would be linear if after solving for phi'^2, the other side is an exact square. That depends both on the choice of k(phi) and the integration constant. This is algebra problem and I believe the answer to that is within your reach so I leave it to you.

    Theorems for existence and uniqueness of solutions for first order diff. eq. are those of Pickard and Peano:

  10. Sep 14, 2008 #9
    I appreciate the reply.

    I have a question that is related to the context of the problem. The original equation that I posted is one that I calculated for the Gaussian curvature of a surface of revolution obtained by revolving the curve [itex]\phi (v)[/tex] about the z axis. As previously stated, I obtained:
    [tex] k(v)=\frac{\phi ''(v)}{\phi (v) (\phi ' ^2 (v) +1)^2} [/tex]
    You said that the DE would be easier to solve if I let k be a function of [itex]\phi[/itex]. The question I have is whether or not this makes sense in terms of a function for the curvature of a surface. I suppose it does, but it seems to only be a very special case of the original problem.
  11. Sep 14, 2008 #10
    According to wikipedia the power in the denominator should be 3/2 not 2:

    If you want to design a surface with pre-specified curvature as function of v, then k(v) makes more sense but the equation becomes unsolvable analitically except the simplest case k=const. You can still solve it numerically for any k(v).
  12. Sep 14, 2008 #11
    The wikipedia article does not reference my specific scenario. I first parametrized all surfaces of revolution, then calculated the first and second fundamental forms, and then used that to find the Gaussian curvature. I simplified it heavily, and got the answer I showed to you. I am nearly 100% certain it is correct. My results also match up with the following:

    I guess the case where k is a function of phi isn't really useful in terms of the original problem: prescribing curvature. The case where k(phi)=const is useful, though, in prescribing constant Gaussian curvature, however if k(phi) is not a constant map then this isn't truly prescribing curvature.

    What methods are there for solving a DE "numerically"? I know of elementary methods, like Euler's, however it doesn't give me an equation as a solution, it just lets me approximate a specific value.
  13. Sep 14, 2008 #12
    The reason I though it was power 3/2 is that I thought one of the principal curvatures is 1/phi which is not true so probably your formula is OK.

    k(phi) prescribes the curvature as function of the radial distance = phi to the axis of rotation. If you want prescription k(v) you have to either solve the problem analytically, and I don't see how that's going to happen, or use numeric solution. Numerical solutions are just bunch of values phi(v), for a numerically specified k(v), not a formula. For those best is to use already made routines in Maple, Matlab or Mathematica. I don't like Mathematica cause it has a painful syntax, the first two are user - friendlier.
    Last edited: Sep 14, 2008
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