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.

  • $\begingroup$ 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 $(c_1,c_2)$? $\endgroup$ – Kwin van der Veen Oct 18 '17 at 4:50

Your general solution is correct:


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 $y(x)=2x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.