Extremising a functional with conditions Extremise the functional:
$$ J[y]=\int_0^1 (yy')^2 dx$$
subject to the constraint 
$$ \int_0^1 y^2 dx=3, $$
And the boundary conditions $y(0)=1$ and $y(1)=2$.
 A: This is an isoperimetric problem with 
$$L(y,y')=\int F(y,y')dx=\int (y\,y')^2 dx$$
subject to
$$\int G(y)dx=\int y^2\,dx=3$$
The first order condition is
$$\frac{\partial}{\partial y}\big(F-\lambda G)-\frac{d}{dx}\bigg(\frac{\partial}{\partial y'}\big(F-\lambda G)\bigg)=0$$
$$\frac{\partial}{\partial y}\big((y\,y')^2-\lambda y^2)-\frac{d}{dx}\bigg(\frac{\partial}{\partial y'}\big((y\,y')^2-\lambda y^2)\bigg)=0$$
$$2\,y\,y'^2-2\lambda y-\frac{d}{dx}\bigg(2\,y^2\,y'\bigg)=0$$
$$2\,y\,y'^2-2\lambda y-2y^2y''-4y\,y'^2=0$$
$$-2y\big(\lambda+y'^2+y\,y''\big)=0$$
which can be solved for $y=0$ and
$$y=\sqrt{\frac{e^{2A}}{\lambda}-\lambda(x+B)^2}=\sqrt{C-\lambda(x+B)^2}$$
To satisfy constraint
$$\int_0^1 y^2\,dx=\int_0^1 \big(C-\lambda(x+B)^2\big)\,dx=C-\frac{\lambda}{3}-B\lambda(1-B)=3$$
and boundary conditions
$$y(0)=\sqrt{C-\lambda(B)^2}=1$$
$$y(1)=\sqrt{C-\lambda(1+B)^2}=2$$
By solving these three equations it follows that $C=4$, $B=-1$ and $\lambda=3$. And the solution
$$y=\sqrt{3-3(x-1)^2}$$
