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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.

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  • $\begingroup$ 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). $\endgroup$ – Hans Engler Nov 25 '12 at 22:00
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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: $$ \boxed{y = \sqrt{-3x^2+6x+1}}. $$

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  • $\begingroup$ Is there a general order in which one should try to solve for the parameters? $\endgroup$ – user415535 Jan 10 '18 at 20:10
  • $\begingroup$ @user21312 No. But in general we should try firstly to get a general form of the solution, or a general ODE/PDE the solution satisfies. $\endgroup$ – Shuhao Cao Jan 10 '18 at 22:20
  • $\begingroup$ you mean a solution $y=\pm \sqrt{\lambda x^{2}+ax+b}$ when you say that? Or do you men that we solved for $\lambda$ and get an usual ODE ? $\endgroup$ – user415535 Jan 11 '18 at 7:18
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    $\begingroup$ @user21312 I mean you have to solve for an ODE first, the parameters actually comes from the solution procedure of the ODE. $\endgroup$ – Shuhao Cao Jan 12 '18 at 18:00

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