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Exponential second member

  1. Nov 11, 2007 #1
    Hello,


    I'm trying to solve this equation :

    [tex]\frac{d^2f\left(x\right)}{dx^2}= A\exp\left(f\left(x\right)/B\right)[/tex]

    (which comes from plasma equilibrium)

    I can't find out what's the general solution... is there a method to solve with this kind of second member
     
  2. jcsd
  3. Nov 11, 2007 #2

    arildno

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    Well, you can make some headway as follows:
    [tex]\frac{d^{2}f}{dx^{2}}f'(x)=Ae^{\frac{f}{B}}f'(x)\to\frac{1}{2}((f'(x))^{2}-(f'(x_{0}))^{2})=BA(e^{\frac{f(x)}{B}}-e^{\frac{f(x_{0})}{B}[/tex]
    Whereby it follows that:
    [tex]f'(x)^{2}=Ce^{\frac{f}{B}}+K[/tex]
    where C=2AB, and K depends on the values of the function and its derivative at [itex]x_{0}[/tex]

    There is no particular reason why this should have a simple exact solution.
     
  4. Nov 11, 2007 #3
    lets say that [tex]f\left(x_0=0\right)=f'\left(0\right)=0[/tex]

    so that we have :

    [tex]f'^2 = 2BA\left(\exp\left(f/B\right)-1\right)[/tex]

    or [tex]f'^2 = 4AB\exp\left(f/2B\right)sinsh\left(f/2B\right)[/tex]

    I know there's a trick somewhere because the solution should be something like

    [tex]log\left(cosh\left(\right)\right)[/tex]
     
    Last edited: Nov 11, 2007
  5. Nov 11, 2007 #4

    arildno

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    Okay, let's see:

    Let:
    [tex]u^{2}=e^{\frac{f}{B}}\to{2u}\frac{du}{dx}=u\frac{f'}{B}\to{2Bdu}=f'dx[/tex]
    This substitution seems very promising..
    We get:
    [tex]\frac{f'}{\sqrt{2AB}\sqrt{e^{\frac{f}{B}}-1}}=\pm{1}[/tex]
    which leads to:
    [tex]\sqrt{\frac{2B}{A}}\int\frac{du}{\sqrt{u^{2}-1}}=\pm{x}+C[/tex]
    We set u=Cosh(v), yielding:
    [tex]\sqrt{\frac{2B}{A}}sgn(Sinh(v))v=\pm{x}+C[/tex]
    essentially yielding the result for f you mentioned; just get the signs right.
     
    Last edited: Nov 11, 2007
  6. Nov 11, 2007 #5
    Hum,

    [tex]2u\frac{du}{dx} = \frac{f'}{B}e^{f/B} = \frac{f'}{B}u^2[/tex]

    which means that

    [tex]2B\frac{du}{u} = f'dx[/tex]

    now the integral is :

    [tex]\int\frac{2Bdu}{u\sqrt{u^2-1}}[/tex]

    and with your u = ch(v) you get

    [tex]\int \frac{dv}{ch\left(v\right)}[/tex]
     
  7. Nov 11, 2007 #6

    arildno

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    Okay, made a mistake there.
     
  8. Nov 11, 2007 #7
    I've looked on the web and

    [tex]\int \frac{dv}{ch\left(v\right)} = gd\left(v\right)[/tex]

    where gd(v) is the Gudermannian function :

    [tex]gd\left(v\right) = 2\arctan\left(th\left(\frac{v}{2}\right)\right)[/tex]


    which doesn't look nice if you come back to f
     
  9. Nov 11, 2007 #8

    I haven't seen it the first time but you can't even do this if x=0 (because f(0)=0)
     
  10. Nov 12, 2007 #9

    arildno

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    That need not be destructive, since f' is zero also.
    You have an improper integral.
     
  11. Nov 12, 2007 #10

    arildno

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    So, then we continue!

    We set y=tanh(v/2)[tex]\frac{dy}{dv}=\frac{1}{2}\frac{1}{Cosh^{2}(\frac{v}{2})}=\frac{1}{2}(1-y^{2})\to{dv}=\frac{2dy}{1-y^{2}}[/tex]
    Thereby, we get:
    [tex]\int\frac{1*dv}{Cosh(v)}=\int\frac{Cosh^{2}(\frac{v}{2})-Sinh^{2}(\frac{v}{2})}{Cosh^{2}(\frac{v}{2})+Sinh^{2}(\frac{v}{2})}dv=\int\frac{1-y^{2}}{1+y^{2}}\frac{2dy}{1-y^{2}}=\int\frac{2dy}{1+y^{2}}=2arctan(y)+C[/tex]
    And we are essentially done. :smile:
     
    Last edited: Nov 12, 2007
  12. Nov 12, 2007 #11

    arildno

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    I didn't see this post. Sorry about that.
     
  13. Nov 12, 2007 #12
    Hi :)

    In the mean time I got another solution :

    [tex]I = \int\frac{f'dx}{\sqrt{1-\exp{\left(f/B\right)}}}[/tex]

    let [tex]u^2 = 1-\exp{\left(f/B\right)}[/tex]

    so we have :

    [tex]-2Budu\frac{1}{1-u^2} = f'dx[/tex]

    and we get :

    [tex]I = -2B\int \frac{du}{1-u^2} = -2Bargth\left(u\right)[/tex]

    [tex]I = -2Bargth\left(\sqrt{1-\exp{\left(f/K\right)}}\right) = -\sqrt{2AB}x[/tex]

    [tex]\sqrt{1-\exp{\left(f/B\right)}} = th\left(\sqrt{\frac{A}{2B}x}\right)[/tex]

    [tex]f = -2B\ Ln\left(ch\left(\sqrt{\frac{A}{2B}}x\right)\right)[/tex] :)
     
  14. Nov 12, 2007 #13

    arildno

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    Note that in this case, you have 1-exp beneath the square root sign; not exp-1.

    That will naturally yield a totally different solution, valid whenever AB<0, K>0.
     
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