What is the solution for the attached equation?

  • #1
71
1
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.
 

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Answers and Replies

  • #2
Svein
Science Advisor
Insights Author
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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.
 
  • #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
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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
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Thanks for the response. However I get hyperbolic sinh which just makes it much worse. :(
If you introduce [tex]J(z) = \sinh\left(\frac12 \int I(z)\,dz\right)[/tex] then you can get the two-dimensional system [tex]
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]
 

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