Optimal control problem proving extremum

Question

Let's consider the integral $F(x) = \int_{1}^{e}(x-t\dot{x})dt$, with initial conditions $x(1) = 1, x(e) = 2$. The problem is to find extremum of the integral above.
My approach: I was able to find function $\hat{x} = \frac{1+e}{2}ln(t) - \frac{t}{2} + \frac{3}{2}$ by standard technique using Eulers equation, but I don't understand how to prove whether this function is a maximum or a minimum of $F(x)$. I will be grateful for any help!

Answer 1

I assume there is a typo in your question, since the Euler-Lagrange equation of the functional $$ F[x] = \int_{1}^{e}(x-t\dot{x})\,dt \tag{1} $$ is $1+1=0$, which has no solution. On the other hand, the function you found, $\hat{x}(t)=\frac{1+e}{2}\ln(t)-\frac{t}{2}-\frac{3}{2}$, is the solution to the Euler-Lagrange equation of $$ \tilde{F}[x]=\int_1^e(x-t\dot{x}^2)\,dt \tag{2} $$ satisfying the boundary conditions $x(1)=1, x(e)=2$, so I'll assume that $(2)$ is the correct functional. Now, let me answer your question.

Let $\eta(\cdot)$ be a sufficiently well-behaved function satisfying the boundary conditions $\eta(1)=\eta(e)=0$, so that $x=\hat{x}+\eta$ satisfy the boundary conditions of the question. Then \begin{align} \tilde{F}[\hat{x}+\eta]-\tilde{F}[\hat{x}]&=\int_1^e[\eta-t(2\dot{\hat{x}}\dot{\eta}+\dot{\eta}^2)]\,dt \\ &=\left.-2t\dot{\hat{x}}\eta\,\right|_1^{e}+\int_1^{e}\left[1+\frac{d}{dt}(2t\dot{\hat{x}})\right]\eta\,dt-\int_1^{e}t\dot{\eta}^2\,dt. \tag{3} \end{align} The first term on the RHS of $(3)$ vanishes because of the boundary conditions satisfied by $\eta$; the second term also vanishes, because $\hat{x}$ satisfies the Euler -Lagrange equation of $\tilde{F}$. Therefore, $$ \tilde{F}[\hat{x}+\eta]-\tilde{F}[\hat{x}]=-\int_1^{e}t\dot{\eta}^2\,dt\leq 0, \tag{4} $$ which implies that $\tilde{F}$ has a maximum at $\hat{x}$.