Comparison result on a system of linearly coupled HJBs

We will prove comparison result on a system of linearly coupled HJB equations. Such a system usually arises from stochastic control problem with regime-switching diffusion.

Consider a system of Dirichlet PDE given by

$\displaystyle \begin{array}{ll} F_i(x, u_i(x), Du_i(x), D^2 u_i(x)) - \sum_{j\in \mathcal{M}} q_{ij} u_i(x) = 0, \quad (x,i) \in \Omega \times \mathcal{M}. \end{array} \ \ \ \ \ (1)$

In the above, ${\Omega\subset \mathbb{R}^d}$ is bounded open domain, and ${\mathcal{M}=\{1,\dots,m\}}$, and the matrix ${Q=(q_{ij}) \in \mathbb{R}^{m\times m}}$ is a generator of a continuous Markov chain, i.e.

$\displaystyle q_{ij} \ge 0, \ \forall i \neq j \in \mathcal{M}; \quad q_{ii} = - \sum_{j\neq i} q_{ij}, \forall i\in \mathcal{M}. \ \ \ \ \ (2)$

Solution of the system is a function set ${u = \{u_i: i\in \mathcal{M}\}}$. Our goal is to show comparison result in viscosity sense. Note that, since (1) is not proper by (2), we can not directly apply the result of [CIL92].

Definition 1 A continuous function ${u: O \times \mathcal{M} \mapsto \mathbb{R}}$ is a viscosity subsolution of (1), if, for each ${u_i}$ of ${i\in \mathcal{M}}$, for every ${x_0 \in O}$ and every smooth test function ${\varphi:O \rightarrow \mathbb{R}}$ satisfying

$\displaystyle (u_i-\varphi)(x) \le (u_i - \varphi) (x_0) =0, \ \forall x\in \Omega$

we have

$\displaystyle F_{i}(x_0, u_{i}(x_0), D \varphi(x_0), D^2 \varphi(x_0)) - \sum_{j\in \mathcal{M}} q_{ij} u_{i} (x_0) \le 0.$

Similarly, define a viscosity supersolution.

Following two assumptions will be imposed for ${F_i = F}$, which is satisfied by most HJB equations.

Assumption 1 ${F: \mathbb{R}^d \times \mathbb{R} \times \mathbb{R}^d \times \mathcal{S}(d) \rightarrow \mathbb{R}}$ satisfies, for any ${ r\ge s}$ and ${(x,p,X) \in \bar \Omega \times \mathbb{R}^d \times \mathcal{S}(d)}$

$\displaystyle \gamma (r - s) \le F(x,r,p,X) - F(x, s, p, X) \ \ \ \ \ (3)$

for some ${\gamma>0}$.

Assumption 2 ${F: \mathbb{R}^d \times \mathbb{R} \times \mathbb{R}^d \times \mathcal{S}(d) \rightarrow \mathbb{R}}$ satisfies

$\displaystyle F(y, r, \alpha (x-y), Y) - F(x,r, \alpha (x-y),X) \le \omega(\alpha |x-y|^2 + |x-y|), \ \ \ \ \ (4)$

for some function ${\omega: [0,\infty] \rightarrow [0,\infty]}$ with ${\omega(0+) = 0}$, whenever ${x, y \in \Omega, r\in \mathbb{R}, X, Y\in \mathcal{S}(d)}$, and

$\displaystyle - 3 \alpha \left[ \begin{array}{cc} I & 0 \\ 0 & I \end{array} \right] \le \left[ \begin{array}{cc} X & 0 \\ 0 & -Y \end{array} \right] \le 3 \alpha \left[ \begin{array}{cc} I & -I \\ -I & I \end{array} \right]; \ \ \ \ \ (5)$

Theorem 2 (Comparison result) Let ${\Omega}$ be bounded open subset of ${\mathbb{R}^d}$, ${F_i \in C(\Omega \times \mathbb{R} \times \mathbb{R}^d \times \mathcal{S}(d))}$ be proper and satisfy Assumption 1 and Assumption 2 for each ${i\in \mathcal{M}}$. Let ${u\in USC(\bar \Omega \times \mathcal{M})}$ (respectively ${v\in LSC(\bar \Omega \times \mathcal{M})}$) be a subsolution (respectively supersolution) of (1) and ${u \le v}$ on ${\partial \Omega \times \mathcal{M}}$. Then ${u\le v}$ in ${\bar \Omega \times \mathcal{M}}$.

Proof: Let ${(\hat x, \ell) \in \bar\Omega \times \mathcal{M}}$ be maximizer of

$\displaystyle (u_\ell- v_\ell)(\hat x) = \max_{\bar \Omega \times \mathcal{M}} (u_i - v_i) (x) := \delta. \ \ \ \ \ (6)$

Such a maximizer exists due to USC property of ${(u_i - v_i)}$ and finiteness of ${\mathcal{M}}$. It is enough to show ${\delta\le 0}$. To to contrary, if ${\delta>0}$, then maximizer must be interior point, i.e. ${(\hat x, \ell) \in \Omega \times \mathcal{M}}$. Define

$\displaystyle \Phi^\alpha(x,y) = u_\ell(x) - v_\ell(y) - \frac \alpha 2 |x-y|^2, \ (x,y) \in \bar \Omega^2.$

Let the maximizer of ${\Phi^\alpha}$ (exists) be ${(x_\alpha, y_\alpha)}$, writing

$\displaystyle M^\alpha = \Phi^\alpha(x_\alpha, y_\alpha) = \max_{x,y} \Phi^\alpha(x,y).$

Then, ${(x_\alpha, y_\alpha)}$ satisfies

$\displaystyle \lim_{\alpha \rightarrow \infty} x_\alpha = \hat x, \quad \lim_{\alpha \rightarrow \infty} M_\alpha = \delta, \quad \lim_{\alpha \rightarrow \infty} \alpha |x_\alpha - y_\alpha|^2 = 0. \ \ \ \ \ (7)$

By Ishii’s lemma, there also exists ${X, Y \in \mathcal{S}(d)}$, such that

1. ${(\alpha(x_\alpha - y_\alpha), X) \in \bar D^{2,+} u_\ell(x_\alpha)}$,
2. ${(\alpha(x_\alpha - y_\alpha), Y) \in \bar D^{2,-} v_\ell(x_\alpha)}$,
3. ${X}$ and ${Y}$ satisfies (5), in particular, ${X\le Y}$.

By the above first two properties, we can write

$\displaystyle F_\ell(x_\alpha, u_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), X) - \sum_{j} q_{\ell j} u_j(x_\alpha) \le 0$

and

$\displaystyle F_\ell (x_\alpha, v_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), Y) - \sum_{j} q_{\ell j} v_j(x_\alpha) \ge 0.$

If we write ${\Delta_i := u_i - v_i}$, then above two inequalities yield

$\displaystyle F_\ell(x_\alpha, u_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), X) - F_\ell (x_\alpha, v_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), Y) \le \sum_{j} q_{\ell j} \Delta_j(x_\alpha) \ \ \ \ \ (8)$

Now, we are ready to find a contradiction as follows, by the fact of (7), (8), Assumption 1, and Assumption 2,

$\displaystyle \begin{array}{ll} \gamma \delta & \le \gamma (u_\ell (x_\alpha) - v_\ell(y_\alpha)) \\ & \le F_\ell(x_\alpha, u_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), X) - F_\ell(x_\alpha, v_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), X) \\ & \le F_\ell(x_\alpha, u_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), X) - F_\ell(x_\alpha, v_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), Y) + \\ & \quad \quad \quad F_\ell(x_\alpha, v_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), Y) - F_\ell(x_\alpha, v_\ell(x_\alpha), \alpha(x_\alpha - y_\alpha), X) \\ & \le \sum_{j} q_{\ell j} \Delta_j(x_\alpha) + \omega(\alpha |x_\alpha - y_\alpha|^2 + |x_\alpha -y_\alpha|) \end{array}$

Since as ${\alpha \rightarrow \infty}$,

1. ${ \lim\sup \sum_{j} q_{\ell j} \Delta_j(x_\alpha) = \sum_{j} q_{\ell j} \Delta_j(\hat x) \le 0}$ by the choice of (6), property of (2), and USC property of ${\Delta_j}$.
2. ${ \omega(\alpha |x_\alpha - y_\alpha|^2 + |x_\alpha -y_\alpha|)\rightarrow 0}$ due to (7).

Therefore, ${\delta \le 0}$ if we force the limsup ${\alpha \rightarrow \infty}$, which leads to contradiction to ${\delta>0}$. $\Box$