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The question is a reformulation of an incomplete version.

Consider the following dynamic dividing a dollar game where agent 1 claims $x(t)$ of the dollar and agent 2 $y(t)$ (paper).

\begin{align} &\max_{\{u(t)\}^T_{t=0}} e^{-rT}x(T)\\[2mm] &\max_{\{v(t)\}^T_{t=0}} e^{-rT}y(T)\\[2mm] \text{s.t. }& \dot x(t) = -u(t)\in[0,1-x(t)]\\ & \dot y(t) = -v(t)\in[0,1-y(t)]\\ & x(0) + y(0) > 1 \end{align}

where $y(T) = 1 - x(T)$ with $T:=\min\{t \in \mathbb{R}_{++} : x(t) + y(t) = 1\}$ and $r > 0$ is the time preference rate. Optimal feedback control is given by $u(t) = r(1-x(t))$ and $v(t) = r(1-y(t))$. This solution is somehow motivated by guess and verify.

  • Can we obtain the solution by some first order conditions (e.g. maximum principle)?

Transformation of objective function

I tried to somehow transform the objective function into a common functional. Can someone confirm the following: \begin{align} &1 + \int^T_0{-re^{-rt}dt} = 1 + \left[e^{-rt}\right]^T_0 = 1 + e^{-rT} - e^0 = e^{-rT}\\ & x(0) - \int^T_0{u(t)dt} = x(0) + \int^T_0{\dot x(t) dt} = x(T)\\ \Longrightarrow \qquad& e^{-rT} x(0) + \int^T_0{e^{-rt}u(t)(r - e^{rt})dt} = e^{-rT}x(T) \end{align}

  • What next?
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  • $\begingroup$ It seems something is missing: for any given r,T, x(0), you are maximizing x(T), which is done by setting u=0. Right? $\endgroup$ – Fix.B. May 11 '16 at 17:51
  • $\begingroup$ Are you looking for a single $u$ or for a sequence of varying $u$'s along the $t$-index? $\endgroup$ – Alecos Papadopoulos May 11 '16 at 20:04
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    $\begingroup$ Ok, I see your points. Actually it's a dividing a dollar game, and $x(T)$ is the final claim of player 1 and $y(T) = 1-x(T)$ the final claim of player 2. He (player 2) maximizes $\max_v y(T)$ s.t. $\dot y = -v$ where $x(0) + y(0) > 1$ given. The markovian solution is given in the paper, however it is more or less guess and verify. I wanted to construct the solution by some first order conditions. Should I restate the entire question and delete this thread? $\endgroup$ – clueless May 23 '16 at 9:00

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