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Thread: 2nd Order ODE "Contradiction"?

  1. #1 2nd Order ODE "Contradiction"? 
    Cool Dude ostkef's Avatar
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    To solve a 2nd order ODE, we can follow the steps as shown below. (Image 2 is a continuation from Image 1, apologies for the size difference.)








    The method to obtain the solution is straightforward.








    Let's say





    If k = -1, a possible solution is y = sin x. If k = 1, a possible solution is y = e^x.




    How do we obtain these two different solutions from one straightforward method?


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    Quote Originally Posted by ostkef View Post
    Let's say





    If k = -1, a possible solution is y = sin x. If k = 1, a possible solution is y = e^x.




    How do we obtain these two different solutions from one straightforward method?
    I'm not sure I understand your question. Are you wondering why different signs of k lead to different solutions? If so, stop wondering. It would be far more surprising if the solution were not sensitive to the sign of a coefficient! Signs matter, in differential equations as in others.

    In the specific example above, you have a linear second-order ODE. You could imagine it arising from, say, a description of an undamped spring-mass system. In the first case, there is no net energy supplied to the undamped system, and the solution is a constant-amplitude sinusoid. In the second case, there is net energy supplied, leading to exponential growth (the other eigenmode decays). Surely you must understand that "adding no energy" and "adding energy" must result in different solutions? There is no "contradiction" here. I'm frankly confused as to why you're confused...

    ETA: Since exponentials are eigenfunctions of linear systems, you can also gain more insight into the nature of solutions by letting y = exp(sx), and solving for the eignenvalues s that satisfy the resulting equation. You will note that the exponential factors cancel out, leaving you with a trivially solved algebraic equation. What's happened is that we've surreptitiously used a Laplace transform to get there. The eigenvalues, perhaps calculated with the aid of a little bit of Euler, allow you to find the general solutions with ease; these subsume the special cases you mention in your example. The dependence on k (both sign and magnitude) then become much more obvious, perhaps.


    Last edited by tk421; August 6th, 2013 at 09:57 PM.
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  4. #3  
    Cool Dude ostkef's Avatar
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    Quote Originally Posted by tk421 View Post
    I'm not sure I understand your question. Are you wondering why different signs of k lead to different solutions? If so, stop wondering. It would be far more surprising if the solution were not sensitive to the sign of a coefficient! Signs matter, in differential equations as in others.
    No, that does not bother me at all.


    What I'm puzzled about is that there's one supposedly straightforward method to solve the ODE; how do we get these two different solutions from that one method?

    I know it's correct, but what's the underlying "mechanism"?
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  5. #4  
    KJW
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    Instead of picking two particular solutions, why don't you find the general solution? Also, don't forget that:

    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by ostkef View Post
    Quote Originally Posted by tk421 View Post
    I'm not sure I understand your question. Are you wondering why different signs of k lead to different solutions? If so, stop wondering. It would be far more surprising if the solution were not sensitive to the sign of a coefficient! Signs matter, in differential equations as in others.
    No, that does not bother me at all.


    What I'm puzzled about is that there's one supposedly straightforward method to solve the ODE; how do we get these two different solutions from that one method?

    I know it's correct, but what's the underlying "mechanism"?
    Your puzzlement seems to derive from an expectation that there should only ever be one particular solution to a general problem. I don't know whence this expectation comes. As I've suggested, and KJW reiterates, perhaps what would help is for you simply to derive the general solution, with k as a free parameter. To do that, follow the steps I outlined earlier. If you get stuck, and the hints about Euler don't help, post back.

    Just understand that the eigenvalues -- and hence the solution -- will depend very much on the signs of the coefficients in the differential equation, just as the roots of a polynomial depend on the coefficients. If the latter dependency does not also puzzle you, then the present situation ought not, either, for they are exactly the same!
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    Quote Originally Posted by ostkef View Post
    What I'm puzzled about is that there's one supposedly straightforward method to solve the ODE; how do we get these two different solutions from that one method?

    I know it's correct, but what's the underlying "mechanism"?
    Because you didn't comment on my previous explanation regarding energy, I'll assume that you didn't understand its significance. Allow me to try again, in more detail.

    Your 2nd-order ODE applies to a dissipation-free mass-on-a-spring. For one sign of k, your spring is an ideal Hooke's-law spring ("restoring force is proportional to, and opposes, the displacement"), and you get a familiar sinusoidal solution.

    For the other sign of k, the spring also exerts a force proportional to displacement, but instead of being a restoring force, it's directed along the displacement -- the greater the displacement, the greater the force exacerbating that displacement. That leads to a runaway solution -- the exponential that you found.

    These are different behaviors, you would agree, and the difference is fundamentally due to the sign change.

    Now, if you follow up on my and KJW's suggestion, you'll solve the ODE generally, making use of the idea that exponentials are linear-system eigenfunctions. Since exponentials subsume periodic solutions (by allowing for complex exponents -- which is why I mentioned Euler, and why KJW explicitly presented Euler's identity) you'll see that you may write the general solution in a single form, if you wish.

    But again, the reason you get different answers is simply that you have two different equations! Sure, the form is the same, but so what? If I give you the equation x + k = 0, doesn't the specific solution depend on the value of k? That shouldn't surprise anyone, so why be surprised that the value of k similarly affects the solution of a differential equation?
    Last edited by tk421; August 7th, 2013 at 03:38 PM. Reason: punctuation
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