- #1

- Thread starter eahaidar
- Start date

- #1

- #2

Svein

Science Advisor

- 2,109

- 676

- #3

- 71

- 1

Thanks for the response. However I get hyperbolic sinh which just makes it much worse. :(

- #4

Samy_A

Science Advisor

Homework Helper

- 1,241

- 510

Maybe first set $$arcosh(I(z))=\frac {1}{2} \int_z^L I(x)dx $$ and then take the derivate?

- #5

pasmith

Homework Helper

- 1,857

- 513

If you introduce [tex]J(z) = \sinh\left(\frac12 \int I(z)\,dz\right)[/tex] then you can get the two-dimensional system [tex]Thanks for the response. However I get hyperbolic sinh which just makes it much worse. :(

I' = \frac12 IJ \\

J' = \frac12 I^2

[/tex] which can be solved numerically subject to given initial conditions (which, like cosh and sinh, must satisfy [itex]I(0)^2 - J(0)^2 = 1[/itex]). Alternatively, you can construct power series iteratively by starting with [itex]I_0(z) = I(0)[/itex], [itex]J_0(z) = J(0)[/itex] and using the recurrence relation [tex]

I_{n+1}(z) = I(0) + \frac12 \int_0^z I_n(t)J_n(t)\,dt, \\

J_{n+1}(z) = J(0) + \frac12 \int_0^z I_n^2(t)\,dt.[/tex]

- Replies
- 7

- Views
- 967

- Replies
- 17

- Views
- 1K

- Last Post

- Replies
- 10

- Views
- 760

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 21K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 14

- Views
- 7K

- Last Post

- Replies
- 3

- Views
- 635

- Last Post

- Replies
- 7

- Views
- 2K