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.

  • $\begingroup$ in the solution is $x=0$ impossible $\endgroup$ – Dr. Sonnhard Graubner Oct 12 '17 at 16:22
  • 1
    $\begingroup$ I got simply the solution $y=2x$ $\endgroup$ – Raffaele Oct 12 '17 at 17:31
  • $\begingroup$ dear Raffaele could you write your solution? $\endgroup$ – HD239 Oct 12 '17 at 17:32

With writing $y''+\dfrac2xy'-\dfrac{2}{x^2}y=0$ we see the functions $p(x)=\dfrac2x$ and $q(x)=-\dfrac{2}{x^2}$ aren't continuous in $x=0$. Since solutions of the DE must be exist in every intervals which $p(x)$ and $q(x)$ are continuous, so $0$ isn't in interval of solution. Here, the condition $y(0)$ is irrelevant.

  • $\begingroup$ Original of this question is to find stationary point of a functional. I will edit the question just now. $\endgroup$ – HD239 Oct 12 '17 at 17:16

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