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I Solution of Ld2f/dx2=sinh(f)

  1. Jul 15, 2016 #1

    I have a physics article who solve Poisson equation of the form:


    The proposed solution is:

    tanh(f/4)=exp(x/L) tanh(f0/4)

    with f0 a constant

    I suspect an error, something like a forgotten factor.

    How can I verify? (I tried but I failed)

    I forgot the limit conditions:

    x belong to [-∞,0] and f and its first dérivative are null at -∞

    Last edited: Jul 15, 2016
  2. jcsd
  3. Jul 15, 2016 #2


    Staff: Mentor

    Is there some reason this equation involves partial derivatives? From the proposed solution f appears to be a function of x alone.
    If so, the above should be ##L^2\frac{d^2 f}{dx^2} = \sinh(f)##
    Please show us what you tried.
  4. Jul 16, 2016 #3
    Yes you are right, f is a single variable function, so the equation is
    Sorry for the error.

    So I tried to derivate the expression tanh(f/4)=exp(x/L) tanh(f0/4) or f=4*atanh(exp(x/L) tanh(f0/4)) two times (both forms).
    That leads to huge expressions wich have nothing to do with the supposed result 1/L2 sinh(f)

    Thanks for your help
  5. Jul 16, 2016 #4
    Multiple both sides by ##\frac{df}{dx}##. Then you get $$L^2 \frac{d^2f}{dx^2} \frac{df}{dx} = sinh(f) \frac{df}{dx}.$$ Notice that ##\frac{d^2f}{dx^2} \frac{df}{dx} = \frac{1}{2} \frac{d}{dx} (\frac{df}{dx})^2 ##. Integrate both sides with respect to x and you get $$L^2(\frac{df}{dx})^2 = 2\int sinh(f) df = 2 cosh(f) + C.$$

    So, $$L\frac{df}{dx} = \pm \sqrt{2cosh(f) + C}.$$ I think this is as simplified as it gets...
  6. Jul 16, 2016 #5
    Yes OK I see
    If we take the given solution, we have Ldf/dx=1/L (cosh(f/4)) sinh(f/4)
    then by some trigonometric transformation we get
    L df/dx=+/- sqrt(2 cosh(f)-2).
    So the solution is true.

    I will read the article one more time to find where is the bug in my simulations.

    And I will try to learn matjax next time

    Thank you
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