Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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.

share|improve this answer
add comment

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 $y$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.