Finding extremal of the functional

Question

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.

Answer 1

Your general solution is correct:

$$y(x)=c_1\dfrac{1}{x^2}+c_2x.$$

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 $y(x)=2x$.