Control $\dot x = x+u, u=u(t)$, from $x(0)=3$ to $x(t_1)=2$, s.t $J = \int_0^{t_1} x^2+ux+u^2 \: dt$ is minimized

Question

Control $\dot x = x+u, u=u(t)$ unconstrained, from $x(0)=3$ to $x(t_1)=2$, s.t $J = \int_0^{t_1} x^2+ux+u^2 \: dt$ is minimized. Find, the optimal control $u^*$, optimal path $x^*$ and $t_1$

Attempt

Using PMP, $H = -(x^2+ux+u^2) + \psi (x+u)$

Costate eqns: $\displaystyle \dot \psi = -\frac{\partial H}{\partial x}= 2x + u -\psi$

Maximize $H$ as a fn of $u$:

\begin{align} 0 &= \frac{\partial H}{\partial u} = -x -2u + \psi \Longrightarrow u = \frac{1}{2}(\psi - x) \end{align} which is a max since $\frac{\partial^2 H}{\partial u^2} =-2$

$\Longrightarrow \dot \psi = 2x + u -\psi = 2x + \frac{1}{2}(\psi - x) - \psi = \frac{3}{2}x -\frac{1}{2}\psi$

and $\dot x = x+u = {x+\psi \over 2}$

Solving the ode system gives $x = Ae^t+Be^{-t}$ and $\psi =Ae^t-3Be^{-t}$

$H(t_1)=0 \Longrightarrow x = 2\pm \sqrt{-12 + 4\psi +\psi^2}$

If $x = 2 - \sqrt{-12+4\psi(0)+\psi(0)^2}$

For $t=0$ \begin{align*} x(0) &= 3 = 2 - \sqrt{-12+4\psi(0)+\psi(0)^2} = A+B \\ \Longrightarrow - 1 &= \sqrt{-12+4(A-3(3-A))+(A-3(3-A))^2} \quad \text{is impossible} \end{align*} so no solution for $x=2 - \sqrt{-12+4\psi(0)+\psi(0)^2}$

If $x= 2 +\sqrt{-12+4\psi(0)+\psi(0)^2}$

For $t=0 \Longrightarrow A= \frac{1}{4}(7\pm\sqrt{17})$

If $A= \frac{1}{4}(7+\sqrt{17}), B = 3-A$, at $x(t_1)=2=Ae^{t_1}+Be^{-t_1} \Longrightarrow t_1 <0$

If $A= \frac{1}{4}(7-\sqrt{17}), B = 3-A$ there is no solution for $t_1$

The system can't be controlled from $x(0)=3$ to $x(t_1)=2$ for $t_1>0$

Answer 1

You have to find a stationary point of $$ L(x,ψ,u,T) = \int_0^T[H(x,ψ,u)-ψ\dot x]dt $$

For the variation of this functional with two final times $T$ and $T+δT>T$ use some smooth continuation of $u$. Then $x(T)=(x+δx)(T+δT)$. The variation can now be computed, ignoring terms quadratic or higher in the increments, as \begin{align} \delta L&=L(x+δx,ψ+δψ,u+δu,T+δT)-L(x,ψ,u,T)\\ &=\int_0^{T+δT}[H_u(x,ψ,u)δu]dt+ \int_0^{T+δT}[H_ψ(x,ψ,u)-\dot x]δψ\,dt \\&~ ~ ~ ~+\int_0^{T+δT}[H_x(x,ψ,u)+\dot ψ]δx\,dt-[δx(T+δT)ψ(T+δT)-δx(0)ψ(0)] \\&~ ~ ~ ~+[H(x(T),ψ(T),u(T))-ψ(T)\dot x(T)]δT \end{align} The last term compensates for the additional integral over $[T,T+δT]$ that is not present in $L(x,ψ,u,T)$ but used in all the other linearizations.

Now with the fixed boundary conditions for $x$ you get $δx(0)=0$. On the other end the situation is a little more involved. Taylor-expanding at $T$ gives $$δx(T+δT)=x_f-x(T+δT)=-\dot x(T)δT+O(δT^2).$$

The other terms but the first give the usual dynamic of Pontryagin's maximum principle, the time variation of the end point gives the additional condition $$ 0=H(x(T),ψ(T),u(T))=−(x(T)^2+u(T)x(T)+u(T)^2)+ψ(T)(x(T)+u(T)) $$ From the local optimality condition for the control $u$ it follows $ψ=x+2u$, so that this last condition is equal $$ 0=−(x(T)^2+u(T)x(T)+u(T)^2)+(x(T)+2u(T))(x(T)+u(T))\\=2u(T)x(T)+u(T)^2=u(T)(2x(T)+u(T)) $$ Following your solution, this means $AB=0$, so that either $x(T)=Ae^t$ or $x(T)=Be^{-t}$. As the boundary values are falling, $A=0$, $B=3$, $e^T=\frac32$.