# Find a controller for a control system with a cubic state that minimizes an integral function

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.

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

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

• 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 $\lambda_0=-1$. – Dmitry Apr 28 at 18:27

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 $$u$$.