How to find the extremum in this variational problem? I am finding this difficult because this is quite a unique question. In this one, I have to find the Extremum of the functional,
$$I(y)=\int_0^1(xy+y^2-2y^2y')dx $$
Boundary conditions, $y(0)=1, ~ y(1)=2$
Obviously, I have started with Euler-Lagrangian equation
$$ 
\frac{\partial f}{\partial y}-\frac{d}{dx}\bigg(\frac{\partial f}{\partial y'}\bigg)=0 \\ \\
x+2y-4yy'-\frac{d}{dx}(-2y^2)=0 \\ \\
x+2y-4yy'+4yy'=0 \\ 
x+2y=0
$$
Well, I don't understand what to do with this. And how to proceed from now. And also, explain whether this kind of result has any significance or something.
 A: Note that we can compute a part of the integral:
$$
\int_0^1 2y^2\; y'\; dx
=
\frac 23\int_0^1\left(\ y^3\ \right)'\; dx
=
\frac 23\left[\ y^3\ \right]_0^1
=\frac 23(\ y^3(1)-y^3(0)\ )\ ,
$$
which is a constant.
It remains to minimize / maximize the remained functional:
$$
J(y) = \int_0^1 (xy + y^2)\; dx\ ,
$$
also under the conditions $y(0)=1$, $y(1)=2$. Suppose that $y$ is such a global extremal point. Then a deformation of the shape $y+\epsilon h$, $\epsilon \in \Bbb R$, $h$ (variation) function on $[0,1]$ of class
$\mathcal C^2$
with $h(0)=h(1)=0$,
would lead to a function  $\epsilon\to J(y+\epsilon h)$ having an extremal value in $\epsilon=0$. This function is:
$$
\epsilon\to
\int_0^1 (\ x(y+\epsilon h)+(y+\epsilon h)^2\ )\; dx
$$
and its Taylor expansion around zero is
$$
J(y) +\epsilon\cdot \int_0^1 h\cdot(x+2y)\; dx + O(\epsilon^2)\ ,
$$
so we need $x+2y\equiv 0$ as a necessary condition for an extremal value in $y$. But the corresponding solution does not fulfill the boundary conditions.
A: The Euler-Lagrangian equation you obtained is
$$ x+2y=0$$
which does not satisfy the boundary conditions $y(0)=1, y(1)=2$. This means the functional does not have extremum.
A: Let consider $u(x)=y+\frac{x}2$. Then $y=u-\frac{x}2$, $u(0)=1$, $u(1)=\frac52$ and
$$I=\int_0^1 \left(2u^2-2u^2u'+2xuu'-xu-\frac{x^2}2u'\right)\,dx=\\
\int_0^1 u^2\,dx+\int_0^1\left((u^2+2uu'x)-(2u^2u')-\left(xu+\frac{x^2}2u'\right)\right)\,dx=\\
\int_0^1 u^2\,dx+\left.\left(xu^2-\frac23 u^3-\frac{x^2}2u\right)\right|_0^1=\int_0^1 u^2\,dx-\frac{19}4$$
Minimum value of $I$ is obtained when $u$ is as much near to 0 as possible. We can consider $u=a^x-ax+\frac52\left(b^{1-x}-b(1-x)\right)$ with some small $a$ and $b$. Then minimum value of $I$ is obtained when $a$ and $b$ are infinitely small.
