# Thread: find the extremals of the integrand..

1. ## 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.

2. 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:

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

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

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

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

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

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

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

$\displaystyle \int_{c_2}^{y(x)} H^{-1}(c_1-\frac{1}{2y})dy=\int_{x_0}^{x} dx$

3. 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:

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

Letting $\displaystyle p=y'$ I obtain:

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

or:

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

This gives:

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

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

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

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

and therefore:

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

Integrating:

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

I now have $\displaystyle x(p)$ and $\displaystyle y(p)$ as the parametric solution to the DE in terms of the variable $\displaystyle p$ 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 $\displaystyle c_1$ 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.