Finding the minimizer of the variational problem $J[y]= \int_{0}^{2} \left( 1-y'^2 \right)^2dx $ with $y(0)=0$ and $y(2)=0$.

Question

This is a question from a math contest.

Find the minimizer of $J[y]= \int_{0}^{2} \left( 1-y'^2 \right)^2dx$ with the conditions: $y(0)=0$ and $y(2)=0$.

Now as $F$ is $\geq 0$, so to minimize we let $y'^2=1$, $i.e. y'= \pm 1 $ $\implies y= \pm x +c$

but then the initial conditions are not satisfied. So is the only minimizer is $y \equiv 0$?

Also for finding the extremal, using the Euler's condition we have:

$$ \frac{\partial F}{\partial y}- \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] =0 $$ $$ i.e. y''(3y'^2 -1)=0 $$ But I am unable to solve the differential equation.

Any help is highly appreciated.

Answer 1

$\delta{J}[y]= \delta\int_{0}^{2} \left( 1-y'^2 \right)^2dx=-2\int_{0}^{2} \left( 1-y'^2 \right)y'(\delta{y})'dx=2\int_{0}^{2} \frac{d}{dx}\Bigl(\left( 1-y'^2 \right)y'\Bigr)\delta{y}\,dx$

As $\delta{y}$ is arbitrary $\Rightarrow$ $\frac{d}{dx}\Bigl(\left( 1-y'^2 \right)y'\Bigr)=0$, or $\left( 1-y'^2 \right)y'=C=const$

We got a cubic equation for $y'$ which can be resolved. All three roots are constants, so we get $y'=C_{1,2,3} = const$ and $y=C_{1,2,3}x+b\,$ ($b\,$ is another constant).

Due to the requirement $y(0)=y(2)=0$ $\,\Rightarrow$ $\,y\equiv0$ at $x\in[0,2]$

Answer 2

Perhaps they allow for continuous and piece-wise differentiable maps? Then $y(x)=x$, $x\in[0,1]$ and $y(x)=2-x$, $x\in [1,2]$, seems a good candidate. Since you may approximate this arbitrarily well by a $C^1$ function the inf of the functional is zero but is not achieved on $C^1$ functions.