Extremal of the variational problem $J[y] = \int_{0}^1 (y'(x))^2 dx$ with the conditions $y(0)=0$, $y(1) = 1$ and $\int_{0}^1 y(x)dx =0$

Question

I want to solve the variational problem $$ J[y] = \int_{0}^1 (y'(x))^2 dx $$ with the conditions $y(0)=0, y(1)=1$ and $\int_{0}^{1}y(x)dx = 0$.

Using Euler's condition (for an extremal): $$ \frac{\partial F}{\partial y}- \frac{\partial^2 F}{ \partial x \partial y'}- \frac{\partial ^2 F}{\partial y \partial y'} -y'' \frac{\partial^2 F}{\partial y'^2} =0. $$

This will give: $ y'' \frac{\partial ^2 F}{\partial y}=0 $

If $y''=0$ then the solution is of the form $Ax+B$, which can not be the case here. After this point I am unable to solve the problem.

Can you please help me understand what am I doing wrong. Any help is highly appreciated.

Answer 1

Consider the lagrangian

$$ \int_0^1(\dot y)^2dt +\lambda\int_0^1 y dt $$

then from Euler-Lagrange

$$ 2\ddot y -\lambda = 0 $$

and solving with boundary conditions

$$ y = \frac 14(4t-\lambda t+\lambda t^2) $$

The solution should satisfy the restriction so

$$ \int_0^1 ydt = \frac 12-\frac{\lambda}{24}=0\Rightarrow \lambda = 12 $$

and the solution is

$$ y = 3t^2-2 t $$