The Mean Field Type Control Problems
Chapter
First Online:
Abstract
We need to assume that the and we use the notation \(\dfrac{\partial f} {\partial m}(x,m,v)(\xi )\) to represent the derivative, so that Here, x, v are simply parameters. Coming back to the definition ( 2.9)–( 2.11), consider a feedback v(x) and the corresponding trajectory defined by ( 2.9). The probability distribution m v(. )(t) of x v(. )(t) is a solution of the FP equation and the objective functional J(v(. ), m v(. )) can be expressed as follows Consider an optimal feedback \(\hat{v}(x)\) and the corresponding probability density \(m_{\hat{v}(.)}(x,t) = m(x,t)\).
$$\displaystyle\begin{array}{lll}\,\, m \rightarrow f(x,m,v),\;g(x,m,v),\;h(x,m) \\ \text{are} \text{ differentiable in}\;m \in {L}^{2}({\mathbb{R}}^{n}){}\end{array}$$
(4.1)
$$\displaystyle{\frac{d} {d\theta }f(x,m +\theta \tilde{ m},v)_\vert {}_{\theta =0} =\int _{{\mathbb{R}}^{n}} \dfrac{\partial f} {\partial m}(x,m,v)(\xi )\tilde{m}(\xi )\,d\xi.}$$
$$\displaystyle\begin{array}{rcl} \frac{\partial m_{v(.)}} {\partial t} + {A}^{{\ast}}m_{ v(.)} + \text{div }(g(x,m_{v(.)},v(x))m_{v(.)})& =& 0, \\ m_{v(.)}(x,0)& =& m_{0}(x){}\end{array}$$
(4.2)
$$\displaystyle\begin{array}{rcl} J(v(.),m_{v(.)}(.))& =& \int _{0}^{T}\int _{{ \mathbb{R}}^{n}}f(x,m_{v(.)}(x),v(x))m_{v(.)}(x)dxdt\; \\ & & \quad +\int _{{\mathbb{R}}^{n}}h(x,m_{v(.)}(x,T))m_{v(.)}(x,T)dx.{}\end{array}$$
(4.3)
Keywords
Nash Equilibrium Feedback Control Optimal Feedback Field Type Convexity Assumption
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
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© Alain Bensoussan, Jens Frehse, Phillip Yam 2013