# Finding extremal of the functional

Find the stationary point of the functional $$J[y]=\int \left( x^2y'^2+2y^2 \right) dx$$ where $y(0)=0, y(1)=2.$

My Solution:

E-L equation: $x^2y''+2xy'-2y=0.$

This is also Cauchy-Euler equation.

Let $y(x)=x^m$. Substituting to eqn. , we get $m_1=-2, m_2=1$ and I found the general solution $y(x)=c_1x^{-2}+c_2x$.

Now, we will find $c_1,c_2.$

Since $y(1)=2$, we have $c_1+c_2=2$. But when I write $x=0$ to general solution, $0=y(x)=c_1.0^{-2}+c_20$, there is a uncertainty. Please help me.

• I used Pontryagin's maximum principle and arrived at a slightly different solution (but I might also have made a mistake) but that solution also has the same problem near $t = 0$. Have you tried evaluating the integral for different valid values for $(c_1,c_2)$? – Kwin van der Veen Oct 18 '17 at 4:50

$$y(x)=c_1\dfrac{1}{x^2}+c_2x.$$
From $y(x=0)=0$ we see that $c_1=0$ must be given or your solution will explode. One way to show this would be to multiply the equation with $x^2$ to obtain
$$x^2y(x)=c_1+c_2x^3 \implies 0 = c_1 + c_2 \cdot 0^3 \implies c_1=0.$$
Using $y(x=1)=2=c_2\cdot 1$, will give you $c_2=2$.
The final solution is $y(x)=2x$.