
#1
Mar812, 08:08 PM

P: 2

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
Mar812, 08:18 PM

Mentor
P: 20,933

This is a very simple differential equation to solve, and the solution is not necessarily the trivial solution, although that is one solution. 



#3
Mar812, 08:47 PM

P: 2

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. 



#4
Mar912, 09:32 AM

Mentor
P: 20,933

Does this differential equation have a solution?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) = R_{max}, then C = R_{max} If you also know that R(L/2) = R(L/2) = 0, then C = R_{max} = 0. Am I missing something? 


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