curve extremizing the functional

Question

Let $y \in C^([0,\pi])$ satisfying $y(0)=y(\pi)=0$ and $\int_0 ^\pi y^2(x)dx=1$ extremizes the functional

$J(y)=\int_0^\pi (y'^2(x))dx$ then

1. $y(x)=\frac{\sqrt{2}}{\pi} \sin x$

2. $y(x)=-\frac{\sqrt{2}}{\pi} \sin x$

3. $y(x)=\frac{\sqrt{2}}{\pi} \cos x$

4. $y(x)=-\frac{\sqrt{2}}{\pi} \cos x$

its answer is (1)(2)

$y(x)=\frac{\sqrt{2}}{\pi}$ and $y(x)=-\frac{\sqrt{2}}{\pi} \cos x$ does not extremizes the functional as both does not satisfy the $y(0)=y(\pi)=0$.

$F(x,y,y')=(y'^2(x))$ applying Euler's equation

$\frac{\partial F}{\partial y'}= C$

solving this i am not getting option 1 and 2 .

Please help, Thanks very much in advance.

Answer 1

Hint

You have forgotten to consider the constraint $\int_0 ^\pi y^2(x)dx=1$. If the our problem is

$$\begin{align} I(y) &= \int_{x_0}^{x_1}f(x,y,y^{'}) \\ y(x_0) &= y_0 \\ y(x_1) &= y_1 \\ \int_{x_1}^{x_2}g(x,y,y') &= C \end{align}$$

then the function satisfying the Euler-Lagrange equation is

$$ \begin{align} f+\lambda g &= h \\ \frac{\partial h}{\partial y}-\frac{d}{dx} \frac{\partial h}{\partial y^{'}} &=0 \end{align}$$

where $\lambda$ is an unknown constant which should be determined.

So in your example we have

$$h = y'^2 + \lambda y^2$$