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

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