A Selection Procedure for Extracting the Unique Feller Weak Solution of Degenerate Diffusions

Abstract

In this work, we show that for the martingale problem for a class of degenerate diffusions with bounded continuous drift and diffusion coefficients, the small noise limit of non-degenerate approximations leads to a unique Feller limit. The proof uses the theory of viscosity solutions applied to the associated backward Kolmogorov equations. Under appropriate conditions on drift and diffusion coefficients, we will establish a comparison principle and a one-one correspondence between Feller solutions to the martingale problem and continuous viscosity solutions of the associated Kolmogorov equation. This work can be considered as an extension to the work in Borkar and Kumar in (J Theor Probab 23(3): 729–747, 2010).

Appendix

Lemma 5

If u and v are viscosity subsolutions of (2.1) and v is $C^{1,2}((0,T)\times {{\mathbb {R}}}^n)$, then $u+v$ is also a subsolution.

Proof

From the definition of viscosity subsolution, we know that when ${{\mathcal {P}}}_{{\mathcal {O}}}^{2,+} u(x,t) \ne \phi$,

$$\begin{aligned}&a- {{\mathcal {L}}}(X,p,x)\le 0, \text { for} \,(t,x)\in (0,T)\times {{\mathbb {R}}}^n \,\text {and} \,(a,p,X)\in {{\mathcal {P}}}_{{\mathcal {O}}}^{2,+} u(x,t), \end{aligned}$$

(A.1)

$$\begin{aligned}&b- {{\mathcal {L}}}(Y,q,x)\le 0, \text { for} \,(t,x)\in (0,T)\times {{\mathbb {R}}}^n\, \text {and}\, (b,q,Y)\in {{\mathcal {P}}}_{{\mathcal {O}}}^{2,+} v(x,t). \end{aligned}$$

(A.2)

From the differentiability of v and definition of ${{\mathcal {P}}}_{{\mathcal {O}}}^{2,+}$, it is clear that ${{\mathcal {P}}}_{{\mathcal {O}}}^{2,+} v(x,t)\ne \emptyset$, for every $(x,t)\in (0,T)\times {{\mathbb {R}}}^n$ and more importantly, is a singleton. Using Taylor’s theorem and the definition of ${{\mathcal {P}}}_{{\mathcal {O}}}^{2,+} v(x,t)$ (on Page 5), we have

$$\begin{aligned} v(y,s)&= v(x,t) + b (s-t) + q.(y-x)+ \frac{1}{2} (y-x)^\dagger Y(y-x) + o\left( |t-s| +\Vert x-y\Vert ^2\right) \end{aligned}$$

(A.3)

Note that the ’$\le$’ in the definition of ${{\mathcal {P}}}_{{\mathcal {O}}}^{2,+}$ became ’$=$’ because of the differentiability of v. Let $(c,r,Z)\in {{\mathcal {P}}}_{{\mathcal {O}}}^{2,+}(u+v)(x,t)$. Then we have $(c-b, r-q, Z-Y)\in {{\mathcal {P}}}_{{\mathcal {O}}}^{2,+} u(x,t)$. To see this, we use the definition of ${{\mathcal {P}}}_{{\mathcal {O}}}^{2,+}(u+v)(x,t)$:

$$\begin{aligned}{} & {} u(y,s)+v(y,s)\le u (x,t)+ v(x,t) + c (s-t) + r.(y-x)+ \frac{1}{2} (y-x)^\dagger Z(y-x) \\ {}{} & {} \quad + o\left( |t-s| +\Vert x-y\Vert ^2\right) . \end{aligned}$$

Now subtracting Equation (A.3) from the above inequality gives us the following:

$$\begin{aligned}{} & {} u(y,s)\le u (x,t) + (c-b) (s-t) + (r-q).(y-x)+ \frac{1}{2} (y-x)^\dagger (Z-Y)(y-x) \\ {}{} & {} \quad + o\left( |t-s| +\Vert x-y\Vert ^2\right) \end{aligned}$$

and this impies that $(c-b, r-q, Z-Y)\in {{\mathcal {P}}}_{{\mathcal {O}}}^{2,+} u(x,t)$. Since u is a subsolution, we have

$$\begin{aligned} c-b-{{\mathcal {L}}}(Z-Y,r-q,x)\le 0 \ \Longrightarrow \\ c-{{\mathcal {L}}}(Z,r,x)\le b-{{\mathcal {L}}}(Y,q,x)\le 0, \end{aligned}$$

from (2.6) and the linearity of ${{\mathcal {L}}}$. Finally, from the definition of viscosity subsolution, we have the result. $\square$

In the rest of the appendix, we assume the conditions of the statement of Lemma 2. Let $M_{\alpha , \beta , \rho }$ be as defined in (4.2) and $({\hat{x}},{\hat{y}}, {\hat{t}})$ is the maximizer of $M_{\alpha , \beta , \rho }$ on ${{{\mathbb {R}}}^n}\times {{{\mathbb {R}}}^n}\times [0,T]$.

Lemma 6

[25, Pg. 29] Suppose (4.1) holds. Then there exists $\beta _0$ and $\rho _0$ such that for every $\forall \alpha >0$, $\beta <\beta _0$ and $\rho <\rho _0$, we have the following:

$$\begin{aligned} \sup _{{{{\mathbb {R}}}^n}\times {{{\mathbb {R}}}^n}\times [0,T]}M_{\alpha , \rho ,\beta }>\eta >0. \end{aligned}$$

Proof

Let

$$\begin{aligned} \gamma \doteq \lim _{r\rightarrow 0}\sup \left\{ u(x,t)-v(y,t): \Vert x-y\Vert <r, (x,y,t) \in {{\mathbb {R}}}^n\times {{\mathbb {R}}}^n \times [0,T]\ \right\} . \end{aligned}$$

From (3.2), clearly we have,

$$\begin{aligned} \gamma \ge \sup _{ {{\mathbb {R}}}^n\times [0,T]}\left[ u(x,t)-v(x,t)\right] > \sup _{{{\mathbb {R}}}^n}\left\{ \left[ u(x,0)-v(x,0)\right] \vee 0\right\} \doteq M_b. \end{aligned}$$

For $\epsilon >0$, there is a $r_0$ such that for $r<r_0$, we have

$$\begin{aligned} \sup \left\{ u(x,t)-v(y,t): \Vert x-y\Vert <r, ( x,y,t) \in {{\mathbb {R}}}^n\times {{\mathbb {R}}}^n \times [0,T]\ \right\} >\gamma -\epsilon . \end{aligned}$$

To ensure that $\alpha \Vert x_0-y_0\Vert ^2$ is small, choose $r< \min \{\sqrt{\frac{\varepsilon }{2\alpha }},r_0\}$. Now there exists $(x_0,y_0,t_0)\in {{\mathbb {R}}}^n\times {{\mathbb {R}}}^n \times [0,T]$ such that

$$\begin{aligned} u(x_0,t_0)-v(y_0,t_0)+\epsilon >\gamma -\epsilon \text { and } \alpha \Vert x_0-y_0\Vert ^2<\varepsilon . \end{aligned}$$

It is also clear that there exists $\beta _0$ and $\rho _0$ such that $\frac{\rho }{T-t_0}<\varepsilon$ and $\beta \Vert x_0\Vert ^2<\varepsilon$, for every $\beta <\beta _0$ and $\rho <\rho _0$. To summarize, we have shown that

$$\begin{aligned} M_{\alpha , \beta , \rho }(x_0,y_0,t_0)&= u(x_0,t_0)-v(y_0,t_0)-\alpha \Vert x_0-y_0\Vert ^2-\frac{\rho }{T-t_0}-\beta \Vert x_0\Vert ^2 \end{aligned}$$

(A.4)

$$\begin{aligned}&> \gamma -5\varepsilon \end{aligned}$$

(A.5)

$$\begin{aligned}&>M_b, \text { for small enough}\, \varepsilon . \end{aligned}$$

(A.6)

This concludes the proof. $\square$

Note that ${\hat{x}}$, ${\hat{y}}$ and ${\hat{t}}$ depend on $\alpha$, $\beta$ and $\rho$. We then have:

Lemma 7

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } \varlimsup _{\beta ,\rho \rightarrow 0} \alpha \Vert {\hat{x}} -{\hat{y}}\Vert ^2 =0. \end{aligned}$$

Proof

From Lemma 6, $\sup _{{{\mathbb {R}}}^n\times {{\mathbb {R}}}^n \times [0,T]}M_{\alpha ,\beta ,\rho }>0$ and this along with the boundedness of u, v means that

$$\begin{aligned} \alpha \Vert {\hat{x}}-{\hat{y}}\Vert ^2 +\beta \Vert {\hat{x}}\Vert ^2 + \frac{\rho }{T-{\hat{t}}}\le \Vert u\Vert _\infty +\Vert v\Vert _\infty \doteq M. \end{aligned}$$

This immediately gives us

$$\begin{aligned} \Vert {\hat{x}}-{\hat{y}}\Vert \le \sqrt{\frac{M}{\alpha }}, \; \beta \Vert {\hat{x}}\Vert ^2\le M \text { and } \frac{\rho }{T-{\hat{t}}}\le M, \end{aligned}$$

then

$$\begin{aligned} \sup _{{{\mathbb {R}}}^n\times {{\mathbb {R}}}^n\times [0,T]}M_{\alpha ,\beta ,\rho }-\sup _{{{\mathbb {R}}}^n\times {{\mathbb {R}}}^n\times [0,T]}M_{\frac{\alpha }{2},\frac{\beta }{2},\frac{\rho }{2}}&\le M_{\alpha ,\beta ,\rho }({\hat{x}},{\hat{y}}, {\hat{t}})- M_{\frac{\alpha }{2},\frac{\beta }{2},\frac{\rho }{2}}({\hat{x}},{\hat{y}},{\hat{t}})\\&= -\frac{\alpha }{2}\Vert {\hat{x}}-{\hat{y}}\Vert ^2 -\frac{\beta }{2}\Vert {\hat{x}}\Vert ^2 -\frac{\rho }{2(T-{\hat{t}})} \end{aligned}$$

From Lemma 6, we know that $M_{\alpha ,\beta ,\rho }$ is bounded away from zero from below uniformly for small enough $\beta$ and $\rho$ and all $\alpha$. From the monotone convergence theorem,

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } \lim _{\beta ,\rho \rightarrow 0}\left( \alpha \Vert {\hat{x}} -{\hat{y}}\Vert ^2 + \beta \Vert {\hat{x}}\Vert ^2 + \frac{\rho }{T-{\hat{t}}}\right) =0 \end{aligned}$$

and we are done. $\square$

Lemma 8

For u, v as in Lemma 2, there exists $\alpha _0$, $\beta _0$ and $\rho _0$ such that for every $\forall \alpha >\alpha _0$, $\beta <\beta _0$ and $\rho <\rho _0$, we have the following:

$$\begin{aligned} \sup _{{{{\mathbb {R}}}^n}\times {{{\mathbb {R}}}^n}\times [0,T]}M_{\alpha , \rho ,\beta } > \sup _{{{\mathbb {R}}}^n}\left\{ [u(x,0)-v(x,0)]\vee 0\right\} . \end{aligned}$$

Proof

Suppose v is uniformly continuous. Assume the contrary to the statement of the lemma. Then there exists a sequence $(\alpha _n,\beta _n,\rho _n) \rightarrow (\infty , 0 ,0)$ such that the corresponding ${\hat{t}}=0$ (as ${\hat{t}}\ne T$ because of the term involving $\rho$). It is easy to see from definition of $M_{\alpha ,\beta ,\rho }$, Lemmas 6 and 7 that

$$\begin{aligned} M_b<M_{\alpha _n,\beta _n,\rho _n}({\hat{x}},{\hat{y}},0)&\le u({\hat{x}},0)-v({\hat{y}},0)\\&\le v({\hat{x}},0)-v({\hat{y}},0)+ \sup _{{{\mathbb {R}}}^n}\left\{ [u(\hat{x},0)-v(\hat{x},0)]\vee 0\right\} . \end{aligned}$$

Letting $n \rightarrow \infty$ and using uniform continuity of v, we have $M_b < M_b$, a contradiction.$\square$