What is the solution for the attached equation?

[eahaidar]

Good afternoon,
i was just wondering if this equation is possibly solvable where I(z) is a function of z. The equation is:
I(z)=cosh(1/2 ∫I(z)dz)
I know it looks stupid but is it possible? How would you approach this problem?
Thank you.

 

[Svein]

Try to derive both side with respect to z. This will give you an ordinary differential equation. Then try to solve it in another way.

 

[eahaidar]

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

 

[Samy_A]

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

 

[pasmith]

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

If you introduce $$J(z) = \sinh\left(\frac12 \int I(z)\,dz\right)$$ then you can get the two-dimensional system $$I' = \frac12 IJ \\<br /> J' = \frac12 I^2$$ which can be solved numerically subject to given initial conditions (which, like cosh and sinh, must satisfy $I(0)^2 - J(0)^2 = 1$). Alternatively, you can construct power series iteratively by starting with $I_0(z) = I(0)$, $J_0(z) = J(0)$ and using the recurrence relation $$I_{n+1}(z) = I(0) + \frac12 \int_0^z I_n(t)J_n(t)\,dt, \\<br /> J_{n+1}(z) = J(0) + \frac12 \int_0^z I_n^2(t)\,dt.$$