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.

Among all the admissible functions $y = y(x)$, find those that extremise the functional $$J[y] = \int_0^1 (yy')^2dx$$ subject to the constraint $\int_0^1 y^2 dx =3$ and the boundary conditions $y(0)=1, y(1)=2$.

Am I correct in saying denoting $F=(yy')^2 + \lambda y^2$. Thus the extrema corresponding to this problem are the extrema for the functional, $$J[y] =\int_0^1 F dx$$ and therefore they are solutions of the Euler-Lagrange equation $$\lambda y -y^2y''-(yy')^2=0.$$ Could some clarify whether my Euler-Lagrange equation is correct. I am pretty confused as to where to go from here. Any help would be grand.

share|improve this question
    
Before you try to analyze the Euler-Lagrange equations, think about the fact that $yy'$ is essentially the derivative of $u = y^2$. So try to rephrase the problem in terms of this $u$ and then think about it. (This is not an exercise in setting up Euler-Lagrange equations, it's an exercise in non-convex calculus of variations). –  Hans Engler Nov 25 '12 at 22:00

1 Answer 1

Am I correct in saying denoting $F=(yy')^2 + \lambda y^2$. Thus the extrema corresponding to this problem are the extrema for the functional, $$J[y] = \int_0^1 F dx.$$

You are correct. If we wanna minimize a functional $J[y]$ subject to a constraint $H[y]=\text{const}$, then the new functional we want to minimize is $$ J_{\lambda}[y]:= J[y]+\lambda H[y] = \int^1_0 \left((yy')^2+\lambda y^2\right) dx.\tag{1} $$


Therefore they are solutions of the Euler-Lagrange equation $$\lambda y -y^2y''-(yy')^2=0$$ Could some clarify whether my Euler-Lagrange equation is correct.

Upon my checking, I think this is not correct, equation $(\star)$ in blue color is what I obtained.

Instead of plugging the formula, I still prefer the old school way of deriving the E-L equation, aka the first variation of the functional in (1) is zero when the extremum is attained at $y$: $$ \lim_{\epsilon\to 0} \frac{d}{d\epsilon} J_{\lambda}[y+\epsilon v] = 0,\\ \text{for any smooth test function } v \text{ that does not change the boundary value of } y. $$ This is to say $v=0$ on both end points.

Firstly computing $J_{\lambda}[y+\epsilon v]$ yields: $$ \int^1_0 \left((y+\epsilon v)^2(y'+\epsilon v')^2+ \lambda (y+\epsilon v)^2\right) dx. $$ Taking derivative w.r.t. $\epsilon$ yields: $$ \int^1_0 \left(2(y+\epsilon v)v(y'+\epsilon v')^2+2(y+\epsilon v)^2 (y'+\epsilon v') v' + 2\lambda (y+\epsilon v) v\right) dx. $$ Letting $\epsilon\to 0$ and setting the integral to be zero yield: $$ \int^1_0 \left( y(y')^2v + \color{red}{y^2 y' v'} + \lambda y v\right) dx = 0. $$ Integrating by parts on the red term and using $v=0$ on the end points: $$ \int^1_0 \left( y(y')^2v - 2y(y')^2 v - y^2y''v + \lambda y v\right) dx = 0. $$ By fundamental theorem of calculus of variation: $$ \color{blue}{y(y')^2 + y^2y'' =\lambda y}.\tag{$\star$} $$ Suppose $y\neq 0$ for any $x$, we have $$ (y')^2 + y y'' =\lambda y \implies (y^2)'' = 2\lambda. $$
Solving this yields: $$ y =\pm \sqrt{\lambda x^2 + ax + b}. $$ The constraint $\int^1_0 y^2\,dx = 3$, the boundary condition $y(0)=1$, $y(1)=2$ will pin down the parameters, and we reached the final answer: $$ \boxed{y = \sqrt{-3x^2+6x+1}}. $$

share|improve this answer

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.