# Calculus of Variation of Parameters.

I come across these 2 problems in my self very first study to calculus of variation of parameters and I faced two problems. I solved the fist one, I hope my solution is right, but I do not know how to attack the second one which is somehow similar to the problem in the following link; Why $\frac{d}{dx}f_\xi=f_u$ and $f(b,\bar u(b), \bar u'(b))=0$.

Problem one For the functional

$$J[u] = \int_0^1 (2 x u(x) - (u'(x))^2 + 3 u'(x) (u(x)^2))\,dx,$$

find all the $$C^1 ([0,1])$$ stationary curves (i.e solutions of Euler-Lagrange equation) satisfying the boundary conditions $$u(0)=0, u(1)=\frac{1}{3}.$$

My solution Let $$F(u)= 2 x u - (u')^2 + 3 u' u^2.$$ Finding the followings

$$\frac{\partial F}{\partial u} = 2 x + 6u u'.$$

$$\frac{\partial F}{\partial u'} = -2 u' + 3 u^2$$

Now we plug the above two equations into Euler-Lagrange equation

\begin{align*} \frac{\partial F}{\partial u} - \frac{d}{dx} \left( \frac{\partial F}{\partial u'} \right)&=0,\\ 2x + 6 u u' + 2u''-6uu'&=0. \end{align*}

Thus we have

$$u''(x) =-x \Rightarrow u'(x) = -\frac{1}{2}x^2 + c_1 \Rightarrow u(x) = -\frac{1}{6} x^3 + c_1 x + c_2.$$

From $$u(0)=0$$ we have $$c_2 =0$$ and from $$u(1)= \frac{1}{ 3}$$ we have

$$\frac{1}{3} = u(1) = - \frac{1}{6} + c_1 \Rightarrow c_1 =\frac{1}{3}.$$

Then the curve is;

$$u(x) = \frac{x}{3}-\frac{x^3}{6}.$$

Problem two For $$f \in C^2 ([a,b] \times \mathbb{R} \times \mathbb{R})$$ we consider the minimization problem

$$\inf_{u \in X}J[u],\,\, J[u] = \int_a^b f(x,u(x),u'(x)) dx,\,\,\, X=\{ u \in C^1([a,b]); u(a)=\alpha \}.$$

Assume that a minimizer $$\overline{u} \in C^2([a,b]) \cap X$$ attaining the infimum of the problem exists, find the condition that the minimizer $$\overline{u}$$ fulfills at the right endpoint $$x=b$$ of the interval $$[a,b]$$ (i.e. find the relation satisfied by $$\overline{u}(b)$$, $$\overline{u}'(b)$$).

1. Essential/Dirichlet BC: $$~~u(b)~=~\beta,$$
1. Natural BC: $$~~\left.\frac{\partial f}{\partial u^{\prime}}\right|_{x=b}~=~0,$$