# Finding optimal way to steer a system to origin with input constraints

For the system $$\ddot{x}+x=u$$ with $$\|u\| \le 1$$, find the optimal way to steer the system from

(a) $$(x(0),\dot{x}(0))$$ to $$(0,0)$$

(b) $$(x(0),\dot{x}(0))$$ to $$x=0$$

and minimize $$\int_0^{t_f} \|u(t)\| \,\mathrm{d} t$$

Attemp (for part a) the system can be written as $$\dot{x}_1=x_2~ ,~ \dot{x}_2=-x_1+u$$ and the hamiltonian can be written as $$H(x,t,u,\lambda)=|u(t)|+x_2\lambda_1-x_1\lambda_2+u(t)\lambda_2$$then i'll have $$\begin{bmatrix}\dot{\lambda}_1 \\ \dot{\lambda}_2 \end{bmatrix}= \begin{bmatrix}0 &1 \\ -1 &0 \end{bmatrix}\begin{bmatrix}\lambda_1 \\ \lambda_2 \end{bmatrix},$$ With initial conditions $$\lambda_1(0)=A\cos \theta, \lambda_2(0)=A\sin \theta$$ i have $$\lambda_1(t)=A\cos(t-\theta), \lambda_2(t)=A\sin(t-\theta)$$ and $$u(t)=-sign(\lambda_2(t))=+1 ~\text{or} -1$$.

Is my approach correct ? i'm confused about the range of optimal $$u(t)$$ for various $$\lambda$$. And i'm also confused about approaching part b, when $$x=0$$ then i've the set $$S =\{(x_1,x_2)|x_1=0, -\infty , then i think i need to find a vector $$q=[K ~0]^{\top}$$ such that $$<\lambda(t_f),q>=0$$ ?

You also have to take the absolute value of $$u(t)$$ into account. Namely say $$\lambda_2=0.5$$ then your solution gives $$u(t)=-1$$ with $$|u(t)|+u(t)\,\lambda_2=0.5$$. However, $$u(t)=0$$ yields $$|u(t)|+u(t)\,\lambda_2=0$$ which is lower. Namely $$|u(t)|$$ is the dominant term when $$|\lambda_2|<1$$, which has the minimum $$u(t)=0$$. So the expression which would minimize the Hamiltonian would be

$$u(t) = \left\{ \begin{array}{ll} 1, & \text{if}\ \lambda_2 < -1 \\ -1, & \text{if}\ \lambda_2 > 1 \\ 0, & \text{otherwise} \end{array} \right. .$$

You still need to find values for $$\lambda_1(0)$$ and $$\lambda_2(0)$$, however I am not sure there would be a closed form solution for it. There isn't even always a solution, for example if $$t_f$$ is too small such that the constrained input can't drive the system from its initial conditions to zero. In general this is solved with the shooting method. In the case that $$(x(t_f),\dot{x}(t_f))=(0,0)$$ then $$(\lambda_1(t_f),\lambda_2(t_f))$$ is allowed to be anything. In the case that only $$x(t_f)=0$$ ($$\dot{x}(t_f)$$ can be anything) then $$\lambda_1(t_f)$$ can still be anything, however the other co-state has to satisfy

$$\lambda_2(t_f) = \left[\frac{\partial g_{t_f}}{\partial \dot{x}}\right]_{x=x(t_t)},$$

with $$g_{t_f}$$ the terminal cost, which in this case is zero.

• yes that's right, but how do i find $\lambda_1(0), \lambda_2(0)$ such that $x(t_f)=0$ ? and what about part (b) how do i approach with the transversality conditions ? – Zeno San Sep 14 at 5:17

It is a tricky problem to solve with traditional methods, because the problem doesn't have a nice optimal control function. See the reasoning.

We rewrite the initial equation as: $$0=\ddot x+x-u=\ddot x\dot x+\dot x x+u\dot x=\frac d{dt}\frac{\dot x^2+x^2}2+u\dot x=0.$$

The value $$E=(\dot x^2+x^2)/2$$ is strictly non-negative, and goes from the positive value to zero by the influence of $$u\dot x$$. It doesn't matter what $$\dot x$$ and $$x$$ are, because if there is optimal trajectory from some $$(\dot x, x)=(a,b)|_{E=E_0}$$ there is optimal trajectory from other $$(\dot x, x)=(c,d)|_{E=E_0}$$, since we can simply wait doing nothing until $$(\dot x, x)=(a,b)$$ and then use the optimal control (we have not restriction or penalty on time).

Thus we can think only about $$E$$. When is there the least penalty to decrease $$E$$ to $$E-dE$$? Of course when $$\dot x$$ is maximal (or $$x=0$$). Thus any non-zero control outside of $$x=0$$ is sub-optimal. (You can show this in a more formal way).

Finally, there is an infinite number of optimal controlling functions: $$u^*(t) = \sum_{k=1}^\infty -a_k\mathop{\mathrm{sign}} \dot x(t_k)\delta(t-t_k),$$ where $$t_k$$ is the time when $$x=0$$ at $$k$$-th time and $$\sum a_k=\sqrt{\dot x(0)^2+x(0)^2}=L$$ with the same penalty $$L$$. In other words you need a total boost of $$L$$ but you can split it over as many passings through zero as you want.