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I Wrote problems and solutions, I need just few explanations.


$$J(x)=\int_{0}^{1}x'^{2}dt,\quad x(0)=0, x(1)=1. $$ 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 $J$?


$$J(x)=\int_{0}^{1}x'^{3}dt ,\quad x(0)=0, x(1)=1. $$

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.

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I have those definitions, but I don't know how to determine if it's local/global, weak/strong in concrete problem. – user23709 May 15 '13 at 20:14

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.

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