2
$\begingroup$

I'm having trouble with the following problem:

The system

$$\dot{x} = -x+4u$$

where $u=u(t)$ is not subject to any constraint, is to be controlled from $x(0)=2$ to $x(1)=1$ in such a way as to minimize

$$\begin{equation*} J = \int_0^1(x+2xu+u^2)\, \mathrm{d}t \end{equation*}$$

Find the optimal control $u^*$ and the optimal path $x^*$.

I've tried using Pontryagin's maximum principle by writing the Hamiltonian and maximizing it with respect to the control $u$, but it doesn't seem to get me anywhere.

Any help would be appreciated. Thanks.

$\endgroup$
3
$\begingroup$

From the system $u = (\dot{x} + x)/4$, you can then replace it $J$ to get

$$ J(x,\dot{x}) = \int_0^1 dt\; \left[ x + \frac{5}{8}x\dot{x} + \frac{1}{16}\dot{x}^2 + \frac{9}{16}x^2\right] = \int_0^1dt\;\mathcal{L}(x,\dot{x}) $$

which can be minimized by solving

\begin{eqnarray*} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}} \right) - \frac{\partial \mathcal{L}}{\partial x} &=& 0 \\ \Rightarrow\quad \ddot{x} - 9x &=& 8 \end{eqnarray*}

which is a second order-linear ODE in $x$.


Alternatively you can define

$$ v = \frac{\partial \mathcal{L}}{\partial \dot{x}} $$

and solve the system

\begin{eqnarray*} \dot{x} &=& \frac{\partial \mathcal{H}}{\partial v}\\ -\dot{v} &=& \frac{\partial \mathcal{H}}{\partial x}\\ \end{eqnarray*}

which are two first-order linear ODE, with

$$ \mathcal{H}(x, v) = \dot{x} v - \mathcal{L}(x, \dot{x}) $$

$\endgroup$
  • $\begingroup$ I got the ODE $\ddot x - 9 x = 8$ from the Euler-Lagrange equation. One of us made an error. $\endgroup$ – Rodrigo de Azevedo Nov 4 '16 at 11:00
  • $\begingroup$ I also got $\ddot{x}-9x=8$ $\endgroup$ – Sam42 Nov 4 '16 at 12:09
  • $\begingroup$ @RodrigodeAzevedo you are right, I just updated my comment. Thanks for spotting the mistake $\endgroup$ – caverac Nov 4 '16 at 12:20
  • $\begingroup$ @Sam42 Thanks Sam, already corrected the mistake $\endgroup$ – caverac Nov 4 '16 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.