Does this differential equation have a solution?

In summary, the equation Ohm_Max + dOhm/dr = Ohm_Max - dOhm/dr is examining the possibility of building a length of conductor with the same resistance behind and ahead when divided at any point. The only solution found is the trivial one, Ohm(r) = 0 for all r. The attempt at a solution involves using Ohm as a variable, but switching to R for resistance simplifies the equation to 2 dR/dr = 0. The solution to this differential equation is not necessarily the trivial one, although that is one solution. However, the added constraint that R(-L/2) = R(L/2) = 0 means that R must not depend on r, making d
  • #1
xvudi
2
0

Homework Statement



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.

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.
 
Physics news on Phys.org
  • #2
xvudi said:

Homework Statement



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.

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.

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.
 
  • #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.
 
  • #4
xvudi said:
Ohm is resistance. I apologize for the confusion.

So R(r) = C1 * exp(0 * r) = C1
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?
xvudi said:
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.
 

1. What is a differential equation?

A differential equation is an equation that relates a function to its derivatives. It is used to model many physical phenomena, such as growth, motion, and heat transfer.

2. How do you know if a differential equation has a solution?

There are several ways to determine if a differential equation has a solution. One way is to solve the equation analytically using mathematical techniques such as separation of variables or variation of parameters. Another way is to use numerical methods to approximate the solution.

3. Are there any types of differential equations that do not have solutions?

Yes, there are certain types of differential equations that do not have solutions. For example, some initial value problems may not have a unique solution, or some boundary value problems may not have any solutions at all.

4. Can a differential equation have multiple solutions?

Yes, a differential equation can have multiple solutions. This can occur in cases where the equation has more than one set of initial conditions or when the solution is a periodic function.

5. How can I check if my solution to a differential equation is correct?

One way to check if a solution to a differential equation is correct is to substitute the solution into the original equation and see if it satisfies the equation. Additionally, you can also check if the solution satisfies any initial or boundary conditions given in the problem.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Replies
0
Views
418
  • Calculus and Beyond Homework Help
Replies
2
Views
625
  • Calculus and Beyond Homework Help
Replies
7
Views
133
  • Calculus and Beyond Homework Help
Replies
1
Views
785
  • Calculus and Beyond Homework Help
Replies
1
Views
177
  • Calculus and Beyond Homework Help
Replies
6
Views
274
  • Calculus and Beyond Homework Help
Replies
7
Views
641
  • Calculus and Beyond Homework Help
Replies
7
Views
642
  • Calculus and Beyond Homework Help
Replies
2
Views
171
Back
Top