next up previous contents index
Next: 3.3.3 Target set Up: 3.3 Optimal control problem Previous: 3.3.1 Control system   Contents   Index


3.3.2 Cost functional

We will consider cost functionals of the form

$\displaystyle J(u):=\int_{t_0}^{t_f} L(t,x(t),u(t))dt+K(t_f,x_f)$ (3.20)

which is an exact copy of (1.2). Here $ t_f$ and $ x_f:=x(t_f)$ are the final (or terminal) time and state, $ L:\mathbb{R}\times\mathbb{R}^n\times U\to\mathbb{R}$ is the running cost (or Lagrangian), and $ K:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}$ is the terminal cost. We will explain how the final time $ t_f$ is defined in Section 3.3.3 below. Since the cost depends on the initial data and the final time as well as on the control, it would be more accurate to write $ J(t_0,x_0,t_f,u)$ , but we write $ J(u)$ for simplicity and to reflect the fact that the cost is being minimized over the space of control functions. Note that even if $ L$ does not depend on $ u$ , the cost $ J$ depends on the control $ u(\cdot)$ through $ x(\cdot)$ which is the trajectory that this control generates. The reader might have remarked that our present choice of arguments for $ L$ deviates from the one we made in calculus of variations; it seems that $ L(t,x,\dot x)$ would have been more consistent. However, we can always pass from $ (t,x,\dot x)$ to $ (t,x,u)$ by substituting $ f(t,x,u)$ for $ \dot x$ , while it may not be possible to go in the opposite direction. Thus it is more general, as well as more natural, to let $ L$ depend explicitly on $ u$ . In contrast with Section 3.3.1, where the regularity conditions on $ f$ and $ u$ were dictated by the goal of having a well-posed control system, there are no such a priori requirements on the functions $ L$ and $ K$ . All derivatives that will appear in our subsequent derivations will be assumed to exist, and depending on the analysis method we will eventually see what differentiability assumptions on $ L$ and $ K$ are needed.

Optimal control problems in which the cost is given by (3.21) are known as problems in the Bolza form, or collectively as the Bolza problem. There are two important special cases of the Bolza problem. The first one is the Lagrange problem, in which there is no terminal cost: $ K\equiv0$ . This problem--and its name--of course come from calculus of variations. The second special case is the Mayer problem, in which there is no running cost: $ L\equiv
0$ . We can pass back and forth between these different forms by means of simple transformations. Indeed, given a problem with a terminal cost $ K$ , we can write

$\displaystyle K(t_f,x_f)$ $\displaystyle =K(t_0,x_0)+ \int_{t_0}^{t_f}\frac d{dt}K(t,x(t))dt$    
  $\displaystyle =K(t_0,x_0)+\int_{t_0}^{t_f}\big( {K}_{t}(t,x(t))+{K}_{x}(t,x(t))\cdot f(t,x(t),u(t))\big)dt.$    

Since $ K(t_0,x_0)$ is a constant independent of $ u$ , we arrive at an equivalent problem in the Lagrange form with $ {K}_{t}(t,x)+{K}_{x}(t,x)\cdot f(t,x,u)$ added to the original running cost. On the other hand, given a problem with a running cost $ L$ satisfying the same regularity conditions as $ f$ , we can introduce an extra state variable $ x^0$ via

$\displaystyle \dot x^0=L(t,x,u),\qquad x^0(t_0)=0
$

(we use a superscript instead of a subscript to avoid confusion with the initial state). This yields

$\displaystyle \int_{t_0}^{t_f} L(t,x(t),u(t))dt=x^0(t_f)
$

thus converting the problem to the Mayer form. (The value of $ x^0(t_0)$ can actually be arbitrary, since it only changes the cost by an additive constant.) Note that the similar trick of introducing the additional state variable $ x_{n+1}:=t$ , which we already mentioned in Section 3.3.1, eliminates the dependence of $ L$ and/or $ K$ on time; for the Bolza problem this gives $ J(u)=\int_{t_0}^{t_f} L(x(t),u(t))dt+K(x_f)$ with $ x\in\mathbb{R}^{n+1}$ .


next up previous contents index
Next: 3.3.3 Target set Up: 3.3 Optimal control problem Previous: 3.3.1 Control system   Contents   Index
Daniel 2010-12-20