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I have to find the extremal and natural boundary conditions for the following functional: $$J[y]=\int_0^1e^y(y')^2 dx. $$

This is subject to $y(0)=0$ and the right endpoint varying along $x=1$.

I have found the Euler-Lagrange equation to be $$2y''-(y')^2=0.$$

Please could you confirm if this is correct and how to complete the problem from here.

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up vote 1 down vote accepted

This is a somewhat silly problem, since the solution can be found without the calculus of variations: The functional is non-negative and is zero when $y'=0$, so the solution subject to $y(0)=0$ is $y(x)\equiv0$.

To do this with the calculus of variations, you can either use the natural boundary condition

$$ J_{y'}(1,y(1),y'(1))=0\;, $$

which yields $y'(1)=0$, or you can solve the problem with $y(1)$ unspecified and then extremize the functional with respect to the parameter of the solution. In either case, the Euler-Lagrange equation is $2y''+y'^2=0$ (you got the sign wrong), and as Peter pointed out this can be solved by substituting $u=y'$. The result is $y(x)=2\log(x+c)+c'$. The boundary condition $y(0)=0$ yields $c'=-2\log c$ and thus $y(x)=2\log (x/c+1)$.

Now applying the natural boundary condition gives $c=\infty$ and thus $y(x)\equiv0$; or you can calculate the functional, $J(c)=4/c^2$, and again $c=\infty$ gives the extremum.

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I got $2y''+(y')^{2}=0$ for the E-L equation. Then just substitute $u=y'$ to get $2u'=u^{2}$, which is seperable, and then solve for $y$.

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