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Homework Help: Does this differential equation have a solution?

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm examining the equation Ohm_Max + dOhm/dr = Ohm_Max - dOhm/dr and can't find any solutions other than the trivial one, Ohm(r) = 0 for all r.

    It's meant to determine if it is possible to build a length of conductor such that, upon dividing it at any arbitrary point, you'll find that the resistance behind is the same as the resistance ahead.

    3. The attempt at a solution

    Ohm(r) = 0. You can't have one because if Ohm(r) is even then dOhm/dr is odd.

    Ohm(r) = Ohm_Max at r = 0 for bounds -L/2 to L/2, captured in the use of Ohm_Max the constant.

    So the integral from -L/2 to 0 must equal the integral from 0 to L/2 meaning that dOhm/dr has to be even. This can't be as if Ohm(r) is odd then Ohm(0) must be 0 and not Ohm_Max. Odd functions cannot be valued at 0.

    Now, mathematically why can't this work? I apologize for the absence of TeX. I'm still getting used to the forum interface.
  2. jcsd
  3. Mar 8, 2012 #2


    Staff: Mentor

    Instead of using Ohm as a variable, let's switch to R, for resistance. Your equation simplifies to 2 dR/dr = 0.

    This is a very simple differential equation to solve, and the solution is not necessarily the trivial solution, although that is one solution.
  4. Mar 8, 2012 #3
    Ohm is resistance. I apologize for the confusion.

    So R(r) = C1 * exp(0 * r) = C1

    Particular solution for R_Max is 0 as all of its derivatives are zero, putting in our constraint at R(0) we solve for our constant C1.

    R(0) = R_Max => C1 = R_Max except...

    R(-L/2) = 0

    R(L/2) = 0

    So C1 must be a function of r except C1 cannot be a function of r.

    This means that R does not depend on r. Meaning that dR/dr has to be zero (easily verified) and I just realized where my last post went off the rails.

    I meant to add that for bounds -L/2 and L/2 that R(r) must be zero.

    Let me add that.

    So the only way you can do it is if you stick a slider on some rails next to a circuit containing two perfectly matched resistors and call the resistors the system because I forgot the above constraint and I apologize.

    So, with this added constraint, what is the grand mathgalactic reason why it isn't a solvable problem? It's something fundamental. It has to do with the problem type and I can't put my finger on it.
  5. Mar 9, 2012 #4


    Staff: Mentor

    This is really the long way around.
    R'(r) = 0 ==> R(r) = C
    The idea is that if the derivative of something is zero, the something must be a constant.

    Now, using the initial condition R(0) = Rmax, then C = Rmax

    If you also know that R(L/2) = R(-L/2) = 0, then C = Rmax = 0.

    Am I missing something?
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