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Given a linear time-invariant system:

$$ \dot{x}(t)=Ax(t)+Bu(t) $$

with initial state $ x(0)=x_0 $ and final state $ x(T)=x_T $.

The performance measure to be minimized is:

$$ \int_{0}^{T} ((x_T-x(t))^T(x_T-x(t))+\rho u(t)^Tu(t) dt $$ with $ \rho=100 $

$(x_T-x(t)) $ is the difference between the state of the system at time $ t $ and the final state. To compute an optimal control $ u^* $ that induces a transition from the initial state $x_0 $ to the target state $x_T$, it is necessary to define the Hamiltonian as:

$$ H(p,x,u,t)=(x_T-x)^T(x_T-x)+\rho u^Tu+p^T(Ax+Bu) $$

from there the next steps are unclear to me. I know I have to apply the Pontryagin minimum principle somehow, but I'm lost and seek some help finding the solution (if possible with rather more detailed explanations than less detailed).


It always helps me to understand a concrete example (same performance measure). So let

$$ A=\begin{bmatrix} -1 & 0.5 \\ 0.3 & -1 \end{bmatrix}, B=\begin{bmatrix} 1\\ 1 \end{bmatrix} $$ and let the initial state be $ x_0=\begin{bmatrix} 1 \\ 0 \end{bmatrix} $ and the final state be $ x_T=\begin{bmatrix} 0 \\ 1 \end{bmatrix} $. Since the eigenvalues of $ A $ are both negative, the system is stable and we control both variables so the system is controllable.

What would be the optimal control trajectories in this specific case?

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  • $\begingroup$ Except for the control constraint $|u(t)|\leq1$ this is a linear-quadratic regulator problem, for which there exists a Wikipedia entry. Your Hamiltonian function should be $(x_T-x)^T(x_T-x)+\rho u^Tu+p^T(Ax+Bu)$. $\endgroup$ – Gerhard S. Oct 19 '17 at 10:59
  • $\begingroup$ Thanks, Gerhard, I changed the Hamiltonian. Still not sure how to solve it though :(. $\endgroup$ – holistic Oct 19 '17 at 13:30
  • $\begingroup$ Also removed the constraint on $ u(t) $ $\endgroup$ – holistic Oct 19 '17 at 13:42
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    $\begingroup$ Maximum principle: $2\rho u(t)^T+p(t)^TB=0$. Adjoint equation: $\dot p(t)^T=2[x_T-x(t)]-p(t)^TA$. Solve the maximum principle for $u(t)$ and substitute this into the state equation. Then you have a two-point boundary value problem in $(x(t),p(t))$. $\endgroup$ – Gerhard S. Oct 19 '17 at 13:47
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    $\begingroup$ If you want to reach the terminal state in minimal time, then you have to treat $T$ as a decision variable and the cost function would have to be simply $T$. That problem would have an optimal solution only if you impose constraints on $u$, for example those that you had in your earliest version of the post. The solution would then probably be of the bang-bang type, that is, it would be a boundary solution in which $u(t)$ takes only values in $\{-1,1\}$. $\endgroup$ – Gerhard S. Oct 21 '17 at 10:24
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The general optimal control problem that Pontryagin minimum principle can solve is of the following form

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

with

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

It is sometimes also called the Pontryagin maximum principle. The naming convention I believe depending on if you prefer to minimize cost or maximize profit. But in both cases the acronym would be PMP. But back to the solution. The PMP states that this problem can be solved with

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

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 some (or all) of the components of the state are constraint at the terminal time, $x_i(T) = c_i$, then the variables $\lambda_i(T)$ can be chosen freely. But if $x_j(T)$ is free, then $\lambda_j(T)$ will be constraint, 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 $g_T(x(T)) = 0$, all components of the terminal states are constraint $x(T) = x_T$ (so both $\lambda(0)$ and $\lambda(T)$ are unknown), and $g(t,x(t),u(t))$ and $f(t,x(t),u(t))$ are defined as

$$ g(t,x(t),u(t)) = (x_T - x(t))^\top (x_T - x(t)) + \rho\,u(t)^\top u(t) \tag{8} $$

$$ f(t,x(t),u(t)) = A\,x(t) + B\,u(t). \tag{9} $$

I will be omitting the time dependency from now on, so $x$ instead of $x(t)$. Substituting $(8)$ and $(9)$ this into $(3)$ indeed gives

$$ H(t, x, u, \lambda) = \lambda^\top \left(A\,x + B\,u\right) + (x_T - x)^\top (x_T - x) + \rho\,u^\top u. \tag{10} $$

Substituting $(10)$ into $(4)$ and $(6)$ gives

$$ \dot{\lambda} = -A^\top \lambda - 2\,(x - x_T) \tag{11} $$

$$ 0 = \lambda^\top B + 2\,\rho\,u^\top. \tag{12} $$

Solving $(12)$ gives the optimal control as a function of the co-state

$$ u = -\frac{1}{2\,\rho}B^\top \lambda. \tag{13} $$

By substituting $(13)$ and $(9)$ into $(2)$ and combining it with $(11)$ then the dynamics of the whole system can be written as

$$ \underbrace{ \begin{bmatrix} \dot{x} \\ \dot{\lambda} \end{bmatrix} }_\dot{z} = \underbrace{ \begin{bmatrix} A & -\frac{1}{2\,\rho}B\,B^\top \\ -2\,I & -A^\top \end{bmatrix} }_{\hat{A}} \underbrace{ \begin{bmatrix} x \\ \lambda \end{bmatrix} }_z + \begin{bmatrix} 0 \\ 2\,I \end{bmatrix} x_T. \tag{14} $$

The "steady state" of this system can be used to write it into a form for which the solution to its differential equation is easy the solve, and can be found by setting the time derivative to zero and then solve for the full state

$$ z_{ss} = -\hat{A}^{-1} \begin{bmatrix} 0 \\ 2\,I \end{bmatrix} x_T. \tag{15} $$

By defining new coordinates as $\hat{z} = z - z_{ss}$ then its dynamics can be written as

$$ \dot{\hat{z}} = \hat{A}\,\hat{z} \tag{16} $$

whose solution is simply

$$ \hat{z}(t) = e^{\hat{A}\,t} \hat{z}(0) \tag{17} $$

and therefore by using the definition of $\hat{z}$ using $(15)$ the solution to $(14)$ can be written as

$$ \begin{bmatrix} x(t) \\ \lambda(t) \end{bmatrix} = e^{\hat{A}\,t} \left( \begin{bmatrix} x(0) \\ \lambda(0) \end{bmatrix} + \hat{A}^{-1} \begin{bmatrix} 0 \\ 2\,I \end{bmatrix} x_T \right) - \hat{A}^{-1} \begin{bmatrix} 0 \\ 2\,I \end{bmatrix} x_T. \tag{18} $$

In this case when you have an analytical solution it might be possible to find $\lambda(0)$ explicitly. For instance if the dynamics would have been non-linear then this might not be the case and you might need to resort the something like the shooting method in order to find the $\lambda(0)$ that would result in $x(T) = x_T$. In this case an explicit solution for $\lambda(0)$ (and $\lambda(T)$) exists and can be found by evaluating $(18)$ at $t = T$ and substituting in the boundary values for $x$

$$ \begin{bmatrix} x_T \\ \lambda(T) \end{bmatrix} = e^{\hat{A}\,T} \begin{bmatrix} x_0 \\ \lambda(0) \end{bmatrix} + \left(e^{\hat{A}\,T} - I\right) \hat{A}^{-1} \begin{bmatrix} 0 \\ 2\,I \end{bmatrix} x_T. \tag{19} $$

Now by defining $e^{\hat{A}\,T}$ as a matrix constructed out of four sub-matrices

$$ \begin{bmatrix} M_{11} & M_{12} \\ M_{21} & M_{22} \end{bmatrix} = e^{\hat{A}\,T} \tag{20} $$

it is possible to rewrite $(19)$ and solve for $\lambda(0)$ and $\lambda(T)$

$$ \begin{bmatrix} \lambda(0) \\ \lambda(T) \end{bmatrix} = \begin{bmatrix} -M_{12} & 0 \\ -M_{22} & I \end{bmatrix}^{-1} \left( \begin{bmatrix} M_{11} & -I \\ M_{21} & 0 \end{bmatrix} + \left(e^{\hat{A}\,T} - I\right) \hat{A}^{-1} \begin{bmatrix} 0 & 0 \\ 0 & 2\,I \end{bmatrix} \right) \begin{bmatrix} x_0 \\ x_T \end{bmatrix}. \tag{21} $$


For instance if we apply this to your example using $T=1$, then equation $(21)$ yields that $\lambda_1(0) = 7.6584\cdot 10^{4}$ and $\lambda_2(0) = -6.5076\cdot 10^4$ and the resulting trajectories as a function of time, which can be obtained by substituting in the initial condition for $x$ and $\lambda$ into $(18)$ and can be seen below.

Resulting trajectories

However due to limited numerical accuracy it might for your example actually be better to write the optimal control as a time varying feedback controller. So at each time instance solve the optimal problem but using $T-t$ and $x(t)$ instead of $T$ and $x_0$ respectively. These can then be used instead to solve $(21)$ for $\lambda(t)$, which enables you find the optimal control input at time $t$ using $(13)$. For instance if I just solve for $\lambda(0)$ and then plug that into $(18)$ then for $T=1$ the final error is $\|x(T) - x_T\| = 0.0153$, but if I use $T=5$ instead I get $\|x(T) - x_T\| = 1.8754$. But when I use the time varying feedback controller for $T=1$ I get $\|x(T) - x_T\| = 7.4196\cdot 10^{-5}$ and for for $T=5$ I get $\|x(T) - x_T\| = 2.7624\cdot 10^{-5}$. Solving for the time varying feedback controller is more computationally expensive, because you have to basically calculate the optimal trajectory every time instance but also because you no longer can use the solution from $(18)$. However it is much more numerically stable.

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  • $\begingroup$ Thank you so much, that helped me understand a lot! But I'm still having trouble understanding how you applied equation 21 to my example. That is, how did you get from equation 21 to the solutions of $x_1$, $x_2$ and $\lambda_1$ and $\lambda_2$? Specifically, how did you calculate the block matrix in equation 20? It would be nice, if you could elaborate on the single steps to finding the solution based on equation 21 for my example if possible (sorry, no mathematician and never dealt with such complex matrix differential equations before. I understand it better based on an easy example) $\endgroup$ – holistic Oct 23 '17 at 14:04
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    $\begingroup$ @holistic Equation $(20)$ might have been confusing because of the order, which I corrected. Namely you calculate the matrix exponential of $\hat{A}\,T$, which should be in $\mathbb{R}^{2\,n \times 2\,n}$ with $n$ the dimension of $x$. Then for example $M_{11}$ is just the upper left part or this matrix which is of size $\mathbb{R}^{n \times n}$. The other matrices are also of the same size. Using these matrices it is possible to solve for $\lambda(0)$ in $(21)$. Then using this initial condition the trajectory as a function of time can be found using $(18)$. $\endgroup$ – Kwin van der Veen Oct 23 '17 at 15:26
  • $\begingroup$ Thanks again! I would forever thankful if you can add the steps where you calculated $\lambda(0)$ based on $M_{11}$ and $M_{12}$ and how you got $x(t)$ and $\lambda(t)$ from equation 18/21. That is, filling in the values based on the example. Again sorry if those seem like very basic requests, but I really want to be sure I got it right and that I understood all the steps of the solution. Since I'm much more comfortable with reading code you can also post the code for the plots you made, if it is in matlab/mathematica/R/C/python or whatever language you used :) $\endgroup$ – holistic Oct 23 '17 at 18:59
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    $\begingroup$ @holistic Since my post is already not that short I did not wanted to add my code to the answer directly. Instead I will link to a pastebin of of my MATLAB code. This also includes the code for the time varying feedback. I also added some comments, so hopefully you will be able to understand it. $\endgroup$ – Kwin van der Veen Oct 23 '17 at 19:57
  • $\begingroup$ That is perfect, thank you sooo much for making the effort! Tried to understand this for weeks now based on books, but that made it much clearer!! $\endgroup$ – holistic Oct 23 '17 at 20:12

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