# Solution to Singular Free Boundary PDE

As part of my research, I have come across the following problem and I am trying to tackle it.
Let $(X_t)_{0 \leq t \leq T}$ be a mean controlled Brownian Motion with the following dynamics $$X_t = B_t + \int_0^t \theta_s ds$$ and we have the following cost function $$C(t,x;\theta) = \mathbb{E}_{t,x} \left[ \int_t^T \left( \frac{1}{2} X_s^2 + |\theta_s|\right) ds \right] %$\frac{1}{2}$$$

I would like to establish that Value function $V(t,x):= \min_{\theta} C(t,x;\theta)$ to be the unique viscosity solution of the following Free Boundary PDE. Although intuitively it's clear, I am not able to prove it formally.

\begin{align} \min \left\lbrace \partial_tV + \frac{1}{2} \partial^2_{xx}V + \frac{1}{2} x^2, 1-|\partial_xV| \right\rbrace &=0 \\ V(T,x) &= 0 \end{align}

Also, I am interested in several aspects of the viscosity solution $V(t,x)$ to the above PDE.

• Are there any sufficient conditions under which we can expect $V$ to be classical solution i.e. $V \in C^{1,2}$?
• From the structure of the PDE, we can see that $V$ must be locally Lipschitz. Moreover, $V(t,\cdot)$ should be linear outside of $(-L(t),L(t))$. Can we prove this rigorously? Also, can we write any ODE for $L(t)$?

Thank you very much!

• Couple of comments. 1. It seems very likely that it is a unique viscosity solution. You can refer to books of Pham and Touzi to prove that (I have both and can share if you don't have them). 2. Given 1, it is hard to expect such regularity. You should be able to argue that $V'_x$ is continuous from the definition of a viscosity solution. But I am almost certain that $V''_{xx}$ is is discontinuous along the free boundary. 3. To show the interval structure of the "no action" region (where $|V'_x|<1$), I would try to prove that $V$ is convex. – zhoraster Jan 6 '17 at 9:55
• The interval structure would lead to some equation on $L$ (however, I would expect it to be an integro-differential equation rather than an ODE). A possible approach is to write the Ito formula for $U(\tau,W(\tau)) - U(t,x)$, where $\tau$ is the time when $W(s)-W(t) + x$ hits the free boundary, and $U(t,x) = \frac12 x^2(T-t) - \frac14 (T-t)^2 - V(t,x)$; then take the expectation. However, it is quite hard to write the distribution of $\tau$. – zhoraster Jan 6 '17 at 10:04
• Overall, this is not doing to be easy, as always with free boundary problems. Some natural things you could try to show is that $L(t)\sim \frac{1}{T-t}$ or $U(t,x)= o((T-t)^3)$ as $t\to T$. By the way, it would be nice if you add a couple of words on how exactly your research leads to this problem. – zhoraster Jan 6 '17 at 10:09