Local and global extremes I Wrote problems and solutions, I need just few explanations.  
1.Let
$$J(x)=\int_{0}^{1}x'^{2}dt,\quad x(0)=0, x(1)=1. $$
   Find the extrema value for $J$.   
I'm doing this using Euler equation $\frac{\partial f}{\partial x}-\frac{d}{dt} \frac{\partial f}{\partial x'}=0$, where $f=x'^2$.
Applying Euler equation, I got:  
$$-2x''=0  \Rightarrow x''=0  \Rightarrow x'=C_1 \Rightarrow x=C_1 t+C_2.  $$
Using initial conditions, I got that $x=t$. This is candidate for extreme.  
Let $h\in\ C^1[0,1]$ such that $h(0)=h(1)=0$. Then,   
$$J(x(t)+h(t))-J(x(t))= J(t+h(t))-J(t)=  \int_{0}^{1}(1+h')^{2}dt-\int_{0}^{1}dt$$
$$=  \int_{0}^{1}h'^{2}dt\ge 0.  $$
My question is: Why is $x=t$ weak global minimum of $J$?
2.Let
$$J(x)=\int_{0}^{1}x'^{3}dt ,\quad x(0)=0, x(1)=1.  $$
Doing the same procedure, I got that $x=t$ is a candidate for extreme.  Then I got :
$$J(x(t)+h(t))-J(x(t))=  J(t+h(t))-J(t)=  \int_{0}^{1}(1+h')^{3}dt-\int_{0}^{1}dt=  \int_{0}^{1}(3h'^2+h'^3)dt=  3\int_{0}^{1}h'^2(1+\frac{1}{3} h')^{2}dt.$$
Question for this part is when $J(x(t)+h(t))-J(x(t)) \geq 0$  ?
(I have solution that it $J(x(t)+h(t))-J(x(t)) \geq0$  when $||h||_1\leq3$ and that for such $h$ I have weak local minimum). I don't know when it is local/global and weak/strong.
 A: If OK I give a more proper example. Let
$$L(y)=\int_0^2dx\,y'^2(1-y')^2$$
with b.c. $y(0)=0$ and $y(2)=1$.
For weak variation we can use Euler-Lagrange equation such as
$$\frac{d\,F}{d\,y'}=2y'(1-y')^2-y'^2(1-y')=k$$
where $k$ is any constant and $F=y'^2(1-y')^2$. Solving the nonlinear ordinary differential (you can use WolframAlpha) and using boundary conditions you can find the solution
$$y=\frac x2\Rightarrow L=\frac 18$$
For strong variation we don't need a smooth function and we can break the the integral into integral with $0\le c\le 2$
$$L(y)=\int_0^cdx\,y'^2(1-y')^2+\int_c^2dx\,y'^2(1-y')^2$$
By solving Euler Lagrange equations we can find two curves
$$y_1=m_1\,x+d_1\qquad 0\le x\le c$$
$$y_2=m_2\,x+d_2\qquad c\le x\le 2$$
Applying boundary conditions $y(0)=0$ and $y(2)=1$ we get
$$y_1=m_1\,x\qquad 0\le x\le c$$
$$y_2=m_2\,(x-2)+1\qquad c\le x\le 2$$
We now need corner point conditions and continuity equation
$$\lim_{x\to c_-} F-y'F_{y'}=\lim_{x\to c_+} F-y'F_{y'}\Rightarrow m_1^2(1-m_1)(1-3m_1)=m_2^2(1-m_2)(1-3m_2)$$
$$\lim_{x\to c_-} F_{y'}=\lim_{x\to c_+} F_{y'}\Rightarrow m_1(1-m_1)(1-2m_1)=m_2(1-m_2)(1-2m_2)$$
$$m_1\,c=m_2(c-2)+1$$
By solving three equations we find two solution sets such as
$$m_1=1\quad m_2=0\quad c=1$$
$$m_1=0\quad m_2=1\quad c=1$$
Let's replace the first set into $L(y)$ to see the result
$$L=\int_0^1dx\,1^2(1-1)^2+\int_1^2dx\,0^2(1-0)^2=0$$
which is a global minimum.
