4
$\begingroup$

I have been learning how to solve optimal control systems this past week and am currently working on solving this one

$$\dot{x}=-x^3+u$$ where $x(0)=1$. Find $u$ with $t \in [0:\infty]$ that minimizes $$\int_0^\infty x^{2}+u^{2}dt$$

but this problem is giving me a lot of trouble because the state is cubed when usually its not (at least in the problems I am used to). Can I solve this using the Hamiltonian with Euler-Lagrange or Ricatti equation or is there another more better approach?

What are the differences I should account for when solving these kinds of unique problems? Would the process be similar if the input was also raised to a power? Also I want to thank the community for all their help. I have been learning a great deal from here.

$\endgroup$
6
$\begingroup$

When the dynamics is nonlinear (or the cost function in the integral is not just quadratic) a good starting point would be Pontryagin's maximum (or minimum) principle (PMP), which is a Hamiltonian approach. The general optimal control problem that PMP can solve is of the following form

$$ \min_{u(t)} \int_0^T g(t, x(t), u(t))\,dt + g_T(x(T)), \tag{1} $$

with

$$ \dot{x} = f(t, x(t), u(t)), \quad x(0) = x_0. \tag{2} $$

PMP states that this problem can be solved with the Hamiltonian

$$ H(t, x(t), u(t), \lambda(t)) = \lambda(t)^\top f(t, x(t), u(t)) + g(t, x(t), u(t)), \tag{3} $$

where $\lambda(t)$ is called the co-state and has the same dimension as $x(t)$. The dynamics of the system to which the optimal control law is applied to can be expressed as using

$$ \dot{\lambda}(t) = -\left[\frac{\partial}{\partial\,x} H(t, x(t), u(t), \lambda(t))\right]^\top, \tag{4} $$

$$ \dot{x}(t) = \left[\frac{\partial}{\partial\,\lambda} H(t, x(t), u(t), \lambda(t))\right]^\top, \tag{5} $$

$$ 0 = \frac{\partial}{\partial\,u} H(t, x(t), u(t), \lambda(t)). \tag{6} $$

It can be noted that after substitution of $(3)$ into $(5)$ you get $(2)$ back. If the input $u(t)$ is constrained $(6)$ can be replaced with solving for $u(t)$ such that the Hamiltonian is minimized, but if $u(t)$ is not constrained and the Hamiltonian is convex in $u(t)$ this is equivalent to $(6)$. If some (or all) of the components of the state are constrained at the terminal time, $x_i(T) = c_i$, the variables $\lambda_i(T)$ can be chosen freely. But if $x_j(T)$ is free $\lambda_j(T)$ will be constrained, which is defined by

$$ \lambda_j(T) = \left[\frac{\partial\,g_T(x)}{\partial\,x}\right]^\top_{x = x(T)}. \tag{7} $$


In your case $f(t,x,u) = -x^3 + u$, $g(t,x,u) = x^2 + u^2$ and $g_T(x(T)) = 0$ which give the following Hamiltonian when using $(3)$

$$ H(x, u, \lambda) = \lambda(t)\,(-x^3 + u) + x^2 + u^2. \tag{8} $$

Plugging this into $(4)$ gives

$$ \dot{\lambda} = x\,(3\,\lambda\,x - 2). \tag{9} $$

Since there are no constraints for $u$, thus $(5)$ can be used

$$ \frac{\partial\,H}{\partial\,u} = 2\,u + \lambda = 0, \tag{10} $$

which yields $u = -\tfrac{1}{2}\lambda$. Plugging this into $f(t,x,u)$ allows use to express the total system dynamics as

\begin{align} \dot{x} &= -x^3 - \frac{1}{2} \lambda, \tag{11a} \\ \dot{\lambda} &= x\,(3\,\lambda\,x - 2), \tag{11b} \end{align}

with $\lambda(T=\infty) = 0$, since $g_T = 0$. Here is where the limitations of PMP come in, namely PMP does not provide a general way through which one can easily solve for $\lambda(0)$ and PMP is normally also not ideal when dealing with $T=\infty$. In order for $\lambda(\infty) = 0$ it also must be true that $\dot{\lambda}(\infty) = 0$, which combined with $(11b)$ also imply $x(\infty) = 0$ (which also makes sense when considering the cost function).

In this case I found it easier to perform a coordinate transformation, namely by differentiation $\dot{x}$ another time and expressing $\lambda$ as a function of $x$ and $\dot{x}$. By doing this and simplifying the expressions it can be shown to yield

$$ \ddot{x} = x + 3\,x^5, \tag{12} $$

$$ \lambda = -2\left(\dot{x} + x^3\right). \tag{13} $$

So one now instead has to find $\dot{x}(0)$ such that $x(\infty)=0$. It can be noted that $(12)$ is a second order differential equation which is only dependent on the position, for which one can define potential energy as minus the "force" integrated over $x$. The total energy of the system can therefore shown to be equal to

$$ E = \frac{1}{2} \dot{x}^2 - \frac{1}{2} x^2 (1 + x^4). \tag{14} $$

If the system would go to the origin ($x(\infty)=0$ and $\dot{x}(\infty)=0$) would require that the energy in the system would be zero $E=0$. Solving $(14)$ set to zero for $\dot{x}$ gives

$$ \dot{x} = \pm x\sqrt{1 + x^4}. \tag{15} $$

When determining the sign of $\dot{x}$ one can reason that the system should move towards the origin, so if $x$ is positive $\dot{x}$ should be negative and vice versa, so the plus minus sign should always be a minus sign. This can now be used to find the initial condition for the co-state by combining $(13)$ with $(15)$

$$ \lambda(0) = -2\left(-x(0)\sqrt{1 + x(0)^4} + x(0)^3\right). \tag{16} $$

However $(12)$ and thus $(11)$ is unstable, so tiny deviations in initial conditions can eventually lead to that the system would blow up. Therefore it is often better give a control policy as a function of the current state instead of only the initial state. This can be done by not only evaluating $(16)$ at $t=0$ but also at all following times, substituting this into the solution for $u$ gives the following optimal control policy

$$ u = -x\sqrt{1 + x^4} + x^3. \tag{17} $$

$\endgroup$
  • $\begingroup$ The right form of (3) would be $H(x,\lambda,u)=\lambda^T f(x,u) + \lambda_0 g(x,u)$. For a (non-degenerate) minimization problem $\lambda_0=-1$. $\endgroup$ – Dmitry Apr 28 at 18:27
2
$\begingroup$

An alternative approach to this problem is achieved by using dynamic programming and the Hamilton-Jacobi-Bellman (HJB) equation.

Our problem is given as an unconstrained ($u\in \mathbb{R}$) optimal control problem with an infinite horizon (the final time $t_\text{f}$ is infinite).

Dynamic programming states the problem unconstrained infinite horizon problem as

$$\text{min: } J = \dfrac{1}{2}\int_{t_0}^{\infty}\mathcal{L}(t,x,u)dt$$ $$\text{s.t.: } \dot{x}=f(t,x,u), x(t_0)=x_0.$$

  1. Step: The solution can be obtained by solving (unconstrained optimization)

$$\dfrac{d}{du}\left[\dfrac{1}{2}\mathcal{L}+\lambda f(t,x,u)\right]=0$$

for $u$ depending on $\lambda$.

  1. Step: Plugging $u(\lambda)$ this into HJB for infinite horizon

$$\dfrac{1}{2}\mathcal{L}+\lambda f(t,x,u)=0.$$

  1. Step: Solve for $\lambda(x)$ Then $u(\lambda(x))=u(x)$ is the optimal control.

Applied to this problem:

  1. Step: Determine $u(\lambda)$ $$\dfrac{d}{du}\left[ \dfrac{1}{2}(x^2+u^2)+\lambda(-x^3+u)\right]=0 \implies u + \lambda = 0 \implies u = -\lambda.$$

  2. Step: Determine $\lambda(x)$ $$\dfrac{1}{2}(x^2+u^2)+\lambda(-x^3+u)=0$$ $$\dfrac{1}{2}(x^2+\lambda^2)+\lambda(-x^3-\lambda)=0$$ $$\lambda^2+2x^3\lambda-x^2=0 \implies \lambda_{1,2}=-x^3\pm \sqrt{x^6+x^2}$$

  3. Step: Determine $u(\lambda(x))=u(x)$ $$u=-\lambda \implies u(x) = x^3\mp \sqrt{x^6+x^2}$$

We have two solutions it turns out that $u(x) = x^3- \sqrt{x^6+x^2}$ is the correct solution, because $u(x) = x^3+\sqrt{x^6+x^2}$ will lead to unstable behavior. This instability can be shown by $V(x) = \dfrac{1}{2}x^2$ as a Lyapunov candidate function. The solution with the minus sign is globally asymptotically stable by the same Lyapunov candidate function.


This method does not force $u$ to be of any type. You can also try the same procedure with other powers of $u$.

$\endgroup$

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.