Conjecture Steady State from limit properties

Question

The question is related to this thread. I'd like to derive a unique steady state for an optimal control problem.

Consider the following programm \begin{align} &V(x_0) := \max_u \int^\infty_0 e^{-\rho t}F(x(t),u(t))dt\\ s.t.~&\dot x(t)=f(x(t),u(t))\\ &x(0) = x_0 \end{align} where $\rho > 0$ denotes time preference, $V(\cdot)$ is the value and $F(\cdot)$ an objective function. $x\in X$ is the state variable and $u\in U=[0,1]$ the control. The state is governed by $f(\cdot)$. The Hamilton-Jacobi-Bellman equation reads \begin{align} \rho V(x)=\max_u [F(x,u) + V'(x)f(x,u)],\quad \forall t\in[0,\infty) \end{align}

Now presume that the feedback control is given by \begin{align} u(x) = \frac{1}{1+V'(x)} = \arg\max_u [F(x,u) + V'(x)f(x,u)],\quad \forall x\in X. \end{align}

Suppose there is a fixed point at $x = \tilde x$ and we can derive an alternative representation for the optimal control in the fixed point with \begin{align} u(\tilde x) = \frac{\rho + u'(\tilde x)}{\rho + u'(\tilde x) + 1}. \end{align}

Suppose additionaly the HJB in the fixed point is given by \begin{align} \rho V(x)=\ln\left(\frac{1}{1+V'(x)}\right) + 1 - \frac{1}{1+V'(x)}. \end{align}

If $\rho\to 0$ approches zero we must have $V'(\tilde x) = 0 \Rightarrow u(\tilde x) = 1 \Rightarrow u'(\tilde x) = \infty$. On the other hand if $\rho\to\infty$ approches infinity we must have $V(\tilde x) = 0$ by the definition of the value function and thus again $V'(\tilde x) = 0 \Rightarrow u(\tilde x) = 1$. Summarizing we have the following properties in equilibrium \begin{align} \lim_{\rho\to 0} u(\tilde x)=\lim_{\rho\to\infty} u(\tilde x) = 1. \end{align}

Well, that's at odds with \begin{align} \frac{\partial u(\tilde x)}{\partial \rho}=\frac{1}{(\rho + u'(\tilde x) + 1)^2} > 0 \end{align}

being a strictly monoton increasing function, contradicting our previous result. Note, however, that we may resolve the issue by observing \begin{align} \lim_{u'(\tilde x)\to\infty}\frac{\partial u(\tilde x)}{\partial \rho}= 0 \end{align}

So, can we finally conjecture that we must have in the fixed point $u'(\tilde x) = \infty \Rightarrow u(\tilde x) = 1$ such that \begin{align} \rho V(\tilde x)=\ln(1) + 1 - 1 \Leftrightarrow V(\tilde x) = 0. \end{align}

Answer 1

Upgrading an exchange of comments, a critical point in the question is the expression \begin{align} \frac{\partial u(\tilde x)}{\partial \rho}=\frac{1}{(\rho + u'(\tilde x) + 1)^2} \end{align}

which is mistaken , because from

\begin{align} u(\tilde x) = \frac{\rho + u'(\tilde x)}{\rho + u'(\tilde x) + 1}. \end{align}

we obtain

\begin{align} \frac{\partial u(\tilde x)}{\partial \rho}=\frac{1+\partial u'(\tilde x)/\partial \rho}{(\rho + u'(\tilde x) + 1)^2} \end{align}

The OP notes that the sign of $\partial u'(\tilde x)/\partial \rho$ is indeterinate, and so $\partial u(\tilde x)/\partial \rho$ cannot be said to be monotonically increasing, which was what drove the pinning-down of the steady state.