I'm trying to solve the following problem:
Determine smooth extremums in
$$\min \int_0^1y^2y'^2\;dx,\;y(0)=0,\;y(1)=1$$
by
(a) using the fact that the functional does not contain explicitly variable $x$
(b) applying straightforward the Euler-Lagrange equation
where $y=y(x), \;y'(x)=\frac{dy}{dx}, \;F(y,y')=y^2y'^2$.
Euler-Lagrange equation:
$$\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)=0$$
If the integrand does not depend on $x$ then E-L equation gets the form:
$$F-y'F_{y'}=C,$$
where $F_{y'}=\frac{\partial F}{\partial y'}$.
I have tried solving this for a while, but I keep getting into complex numbers or into very hairy differential equations. Could someone show me few of the most important steps what to do, where it is easy for me to take on? Thank you for your help!
The (a)-part was more confusing for me. My attempted solution from part (b) was $y=\sqrt[3]{x}$, but I'm not sure of my answer.
P.S. if asked for, I can post my attempts of trying to solve this problem.
a) Since $F$ does not explicitly depend on $x$ there is a first integral
$$\begin{align} \\ y'\frac{\partial F}{\partial y'} - F &= 2y^{2}y'^{2} - y^{2}y'^{2} \\ &= y^{2}y'^{2} \\ &= C \end{align} $$
Rearranging for $y'$ we find
$$y' = \pm \sqrt\frac{C}{y^{2}} $$
Then separating and integrating we get
$$\int y dy = \pm \int C dx \implies \frac{y^{2}}{2} = Cx + D$$
Where $C$ and $D$ are constants. You'll need to use your conditions to solve for the constants.
b) If
$$F(y,y') = y^{2}y'^{2}$$
you should get a Lagrange equation of the form
$$\begin{align} \\ \frac{d}{dx}\bigg(\frac{\partial F}{\partial y'}\bigg) - \frac{\partial F}{\partial y} &= \frac{d}{dx}\bigg(2y^{2}y'\bigg) - 2yy'^{2} \\ &= (4yy'^{2} + 2y^{2}y'') - 2yy'^{2} \\ &= 0 \end{align}$$
Does that help?