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Math Help - Nonlinear System of two ODEs

  1. #16
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    I really have no idea. I've never seen an equation like this.
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  2. #17
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    Ok. This is a bit of a subtle point, and it is the essential idea behind separation of variables for pde's (we do not have a pde here, I know, but the essential idea is the same). The idea here is that each side of the equation has only one coordinate in it. Therefore, both sides must be constant. Otherwise, we could change just one side without changing the other, thus violating the equation. Does that make sense?
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  3. #18
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    Yes it does. I'm familiar with the idea. But doesn't that work only if u is independent of v. But here that is not the case. If we set u^{-1}(u)=t
    to be the inverse.

    v(t)=v(u^{-1}(u))=\alpha (u)

    Just like reparametrisation in differential geometry. You can't change the left hand size without changing the right hand size.
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  4. #19
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    Hmm. Do you have any initial/boundary conditions for the system?
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  5. #20
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    yep.

    x(0)=y(0)=0
    x'(0)=const
    y'(0)=const
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  6. #21
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    Hmm. Two questions:

    1. Can you tell me the physics of the situation? What's the problem setup?
    2. Is it the case that x'(0) = y'(0) = const, or are x'(0) and y'(0) equal to different constants?
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  7. #22
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    x'(0) is different than y'(0). The physics is a projectile with friction. Initially at the origin projected with constant velocity.
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  8. #23
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    By "friction" do you mean air resistance? If so, how are you modeling air resistance? I'm wondering because the equations here don't match up with yours. Are you sure your original DE's are correct?
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  9. #24
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    I'm modelling it as F=-kv^2

    Here is a numerical solution for k=1, m=1, g=10, x'(0)=y'(0)=1

    Nonlinear System of two ODEs-solution.jpg
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  10. #25
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    Ok. In that case, I'd agree that your original system of DE's is correct. I think you're also correct in thinking that the "separation of variables" trick I recommended won't work. Based on some preliminary research, I'd say that a complete analytic solution does not exist. You can do the first integral with the substitution, as outlined in your post #9. That reduced DE might be more stable with respect to numerical integration than the original DE.

    I'm afraid that's as much as I know.

    Cheers.
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  11. #26
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    Yeah, I thought that would be the case. Thanks for your help.
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