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Nonlinear System of DEs

  1. Oct 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve the System of DEs:

    [itex]\sqrt{1+y'^{2}+z'^{2}}-\frac{y'^{2}}{\sqrt{1+y'^{2}+z'^{2}}}=C_{1}[/itex]

    [itex]\sqrt{1+y'^{2}+z'^{2}}-\frac{z'^{2}}{\sqrt{1+y'^{2}+z'^{2}}}=C_{2}[/itex]

    2. Relevant equations

    The two equations above are quite relevant.

    3. The attempt at a solution

    I attempted basic substitution to do is this:
    Multiply through by the radical
    Cancel terms
    Solve for y' and z' in terms of each other
    Plug them into each other and then attempt to solve

    I ended up trying to solve for y' first. What I got is a solvable polynomial in terms of y'; a quite tedious looking polynomial at that. I stopped here and began erasing. Maybe I was doing it right, but I don't even want to see what happens when I plug it into the quadratic equation and then attempt to substitute it back into the z' equation.. It sounds like WAYYY to long of a process considering this is physics homework, not math homework.

    Can anybody give me some advice? Any good way to approach problems like these? I'm starting to encounter them a lot and it's always the part of the problem that I spend hours looking at :\
     
  2. jcsd
  3. Oct 18, 2013 #2

    UltrafastPED

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    Create a new variable: [itex]Q^{2}={1+y'^{2}+z'^{2}}[/itex] and carry out the replacement.

    This gives you a new system which looks easier - solve for Q and z; then apply the auxiliary equation above so that you can go back to y and z.
     
  4. Oct 18, 2013 #3
    Ok thanks. That was a pretty straightforward step.

    This is what I did:
    Plugged Q into the original system.
    Solved the system for y'^2 and z'^2
    Plugged my values for y'^2 and z'^2 into the expression for Q
    Solved for Q

    Q again come in the form of a quadratic equation with two solutions. The solutions were both in terms of C1 and C2. At this point I used to following logic to prove that y and z are both linear.

    Q is a constant, thus, y' and z' squared are constants, thus, y' and z' are constants, thus, y and z are linear.

    Considering that there were two values for Q though, is this the correct way of thinking about it? Do the two Q values basically say that there are two solutions of y and z, both solutions being linear? As long as I can prove the solutions are linear then I've successfully answered the question.
     
  5. Oct 18, 2013 #4

    UltrafastPED

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    The substitution generated an additional set of values because Q could be + or -; since your original problem used square root, which implies the + branch, you could just rule out the - by that reason.

    Think of the extra solutions as a bonus - you learned something, but you only have to turn in the + half!

    BTW this is "multiplication" of solutions is common with the method of substitution - though the method really cuts down the work!
     
  6. Oct 19, 2013 #5
    Thanks, I appreciate the help :)

    I'm looking forward to experimenting with these nonlinear systems. It's a shame we never covered them in my ODE class.
     
  7. Oct 19, 2013 #6

    UltrafastPED

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    Nonlinear is an advanced topic; systematic techniques for finding solutions are
    Lacking IIRC.
     
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