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Math Help - find the extremals of the integrand..

  1. #1
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    find the extremals of the integrand..

    1/y * sqrt(1+ (y')^2 )

    I tried this with Euler equation but I end up with a very hard formula to integrate...I did several similar problem but keep on getting stuck at the final integration. Please help.
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  2. #2
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    Hey Tennis, I believe we can obtain an implicit solution. Some background for those not familiar with the problem: We wish to minimize the following integral over some domain D:

    I=\mathop\int\limits_D F(x,y,y')dx;\quad F(x,y,y')=\frac{\sqrt{1+(y')^2}}{y}. Euler's necessary requirement to do this is the following:

    \frac{\partial F}{\partial y}-\frac{d}{dx}\frac{\partial F}{\partial y'}=0

    Plugging all that in and remembering the partials treat y and y' as just regular variables but the ordinary derivative \frac{d}{dx} treats them as functions of x, I get (if I didn't make any errors) the following ODE:

    y^2y'y^{''}-(y')^4-2(y')^2-1=0

    Ok, that's a little intimidating but we can do the first integration by making the change of variable y'=p and integrating. I get:

    H(p)=\left[\frac{\arctan(p)}{2}-\frac{p}{2(1+p^2)}\right]=c_1-\frac{1}{2y}

    Now integrate the inverse of H at the point c_1-\frac{1}{2y} to obtain an implicit albeit not very satisfying expression for the solution:

    \int_{c_2}^{y(x)} H^{-1}(c_1-\frac{1}{2y})dy=\int_{x_0}^{x} dx
    Last edited by shawsend; December 5th 2008 at 05:58 AM.
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  3. #3
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    I've made some additional progress with this which I find really interesting and hope some of you will too. Ignoring for the moment extremizing the integral, I'll focus on just the IVP:

    y^2y'y''-(y')^4-2(y')^2-1=0; \quad y(0)=1,\; y'(0)=0.1

    Letting p=y' I obtain:

    \int \frac{p^2}{p^4+2p^2+1}dp=\int \frac{dy}{y^2}

    or:

    \frac{\arctan(p)}{2}-\frac{p}{2(1+p^2)}=c_1-\frac{1}{y}

    This gives:

    y(p)=\left(\frac{p}{2(1+p^2)}-\frac{\arctan(p)}{2}+c_1\right)^{-1}

    Now note that dx=\frac{dy}{p}

    Therefore if I differentiate y with respect to p, divide by p, I'll have dx or:

    \frac{dy}{dp}=\frac{d}{dp}\left\{\left(\frac{p}{2(  1+p^2)}-\frac{\arctan(p)}{2}+c_1\right)^{-1}\right\} =\frac{4p^2}{\left(p+2c_1(1+p^2)-(1+p^2)\arctan(p)\right)^2}

    and therefore:

    dx=\frac{dy}{p}=\frac{4p}{\left(p+2c_1(1+p^2)-(1+p^2)\arctan(p)\right)^2}dp

    Integrating:

    x(p)-x_0=\int_{p_0}^{p}\frac{4p}{\left(p+2c_1(1+p^2)-(1+p^2)\arctan(p)\right)^2}dp

    I now have x(p) and y(p) as the parametric solution to the DE in terms of the variable p which is still the derivative of the function. So I've expressed the solution parametrically as a function of it's derivative! That's pretty cool I think. However, being the skeptical person I am about all this, I checked it numerically. Solving for c_1 above, I solved the DE numerically and then superimposed the results against the parametric solution (x(p) calculated numerically). The numerical solution is red, parametric solution is blue. Pretty close.
    Attached Thumbnails Attached Thumbnails find the extremals of the integrand..-extermal-solution.jpg  
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