Existence of an Effective Burning Velocity in a Cellular Flow for the Curvature G-Equation Proved Using a Game Analysis

Abstract

G-equation is a popular level set model in turbulent combustion, and becomes an advective mean curvature type evolution equation when curvature of a moving flame in a fluid flow is considered:

$$\begin{aligned} G_t + \left( 1-d\, \text {div}\left( \frac{DG}{|DG|}\right) \right) _+|DG|+V(x)\cdot DG=0. \end{aligned}$$

Here $d>0$ is the Markstein number and the positive part $()_+$ is imposed to avoid a non-physical negative laminar flame speed. For simplicity of presentation, we focus mainly on the case when $V:{{\mathbb {R}}}^2\rightarrow {{\mathbb {R}}}^2$ is the two dimensional cellular flow with Hamiltonian $H = \sin x_1 \, \sin x_2$ and amplitude A. Our main result is that for any unit vector $p\in {{\mathbb {R}}}^2$, there exists a positive number ${\overline{H}}(p)$ such that if $G(x,0)=p\cdot x$, then

$$\begin{aligned} \left| G(x,t)-p\cdot x+{\overline{H}}(p)t\right| \le C \quad \text {in}\, {{\mathbb {R}}}^2\times [0,\infty ) \end{aligned}$$

for a constant C depending only on on the Markstein number d and the cellular flow amplitude A. The number ${\overline{H}}(p)$ corresponds to the effective burning velocity in the physics literature. The non-coercivity encountered here is one of the major difficulties for homogenization of the mean curvature-type equations. To overcome it, we introduce a new approach that combines PDE methods with a dynamical analysis of the Kohn-Serfaty deterministic game characterization of the curvature G-equation utilizing the streamline structure of cellular flows. Extension to general two-dimensional incompressible flows is also discussed. In three dimensional incompressible flows, the existence of ${\overline{H}}(p)$ might fail when the flow intensity exceeds a bifurcation value even for simple shear flows Mitake et al. (Bifurcation of homogenization and nonhomogenization of the curvature G-equation with shear flows (2023)).

Appendix

It is well known that the front propagation under the viscosity solution framework is consistent with the classical meaning when the smooth solutions exist. See section 6 in [16] for instance. The following conclusion is a special case in our context, which is needed to derive a reachability property. For the reader’s convenience, we present its proof here in 2D, which is sufficient for our purpose.

Throughout this section, we only assume that $V\in W^{1,\infty }({{\mathbb {R}}}^2)$ and

$$\begin{aligned} V(x)\cdot (1,0)=0 \quad \text {for}\, x\in \{0\}\times [0,1]. \end{aligned}$$

(5.1)

Let $S=(0,1)^2$ and

$$\begin{aligned} g_S(x)={\left\{ \begin{array}{ll} -{2\over \pi }\arctan (d(x,\partial S)) \quad \text {for}\, x\in S\\ {2\over \pi }\arctan (d(x,\partial S)) \quad \text {otherwise}. \end{array}\right. } \end{aligned}$$

Here $d(x,\partial S)$ is the distance from x to $\partial S$.

Lemma 5.1

Suppose that $G\in C({{\mathbb {R}}}^2\times [0,\infty ))$ is the unique viscosity solution to equation (1.2) subject to

$$\begin{aligned} G(x,0)=g_S(x). \end{aligned}$$

Then for a given $\delta \in (0,{1\over 2})$, there exists $t_{\delta }>0$ depending only on d, V and $\delta$ such that

$$\begin{aligned} G((0,\theta ), t)<0 \quad \text {for}\, (\theta , t)\in [\delta , 1-\delta ]\times (0, t_\delta ]. \end{aligned}$$

Proof

Intuitively, this conclusion is obvious since the speed along the normal direction $\vec {n}=(-1,0)$ at the point $(0, \theta )$ is

$$\begin{aligned} v_{\vec {n}}=1-d\kappa +V(x)\cdot \vec {n}=1. \end{aligned}$$

To make this rigorous, we need to build smooth supersolutions and employ comparison principle. It suffices to prove this at a fixed $\theta \in [\delta ,1-\delta ]$.

Step 1: Choose an ellipse. Consider the ellipse

$$\begin{aligned} E_{\theta }(t): \frac{(x_1-a_0 - \nu )^2}{a^2(t)} + \frac{(x_2-\theta )^2}{b^2(t)} = 1. \end{aligned}$$

Here $\nu \in (0,a_0)$ is added for technical convenience and will be sent to zero later (See Fig. 16).

Fig. 16

Propagation of the ellipse

Let $b_0={\delta \over 2}$. Then choose $a_0\in (0, {b_0\over 4})$ small enough such that

$$\begin{aligned} |V(x)\cdot (1,0)|<{1\over 8} \quad \text {if}\, x\in [-4a_0, 4a_0]\times [0,1] \end{aligned}$$

and

$$\begin{aligned} \frac{d64a_0}{b_{0}^2}+4M_0\sqrt{3}\frac{a_0}{b_{0}}<{1\over 8}. \end{aligned}$$

Here $M_0=\max _{x\in {{\mathbb {R}}}^2}|V(x)|$.

Then we define $a(t) = a_0 + \frac{1}{2}t$ and $b(t) = b_{0} - Lt$ for $L>0$ satisfying

$$\begin{aligned} {1\over 2}-{3La_0\over 4b_0}<1-{db_0\over {a_0}^{2}}-M_0. \end{aligned}$$

Hereafter we require $0 \le t \le t_\delta$ for $t_\delta = \min \left\{ 2a_0, \frac{b_{0}}{2\,L}\right\}$, which implies that

$$\begin{aligned} (a(t), b(t))\in [a_0,\ 2a_0]\times \left[ \frac{b_0}{2},\ b_0\right] \end{aligned}$$

and

$$\begin{aligned} E_\theta (t)\subset R_0=[-4a_0, 4a_0]\times [0,1] \end{aligned}$$

for $t\in [0,t_\delta ]$. Also, $4a_0<b_0$ leads to

$$\begin{aligned} a(t)< b(t) \quad \text {for all}\, t\in [0,t_\delta ]. \end{aligned}$$

For convenience, we drop the dependence of a and b on t. Hereafter $t\in [0,t_\delta ]$ unless specified otherwise. Let $\vec {n}$ be the outward unit normal vector along $E_{\theta }(t)$ that has the following parameterization: for $\phi \in [0, 2\pi ]$

$$\begin{aligned} {\left\{ \begin{array}{ll} x_1=a_0+\nu +a\cos \phi \\ x_2=\theta +b\sin \phi . \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} {V} \cdot \vec {n} = {V} \cdot \left( \frac{b\cos \phi }{\sqrt{a^2\sin ^2\phi + b^2\cos ^2\phi }}, \frac{a\sin \phi }{\sqrt{a^2\sin ^2\phi + b^2\cos ^2\phi }}\right) . \end{aligned}$$

If $|\sin \phi |<{\sqrt{3}\over 2}$,

$$\begin{aligned} \begin{array}{ll} |{V} \cdot \vec {n}| &{}\le |V(x)\cdot (1,0)| + M_0 \cdot \frac{a\sin \phi }{\sqrt{a^2\sin ^2\phi + b^2\cos ^2\phi }}\\ &{}\le |V(x)\cdot (1,0)| + M_0 \cdot \frac{a}{b} \cdot |\tan \phi |\\ &{} \le |V(x)\cdot (1,0)| + M_0 \cdot \frac{2a_0}{b_0/2} \cdot \sqrt{3} \\ &{}= |V(x)\cdot (1,0)|+ 4\sqrt{3}M_0 \cdot \frac{a_0}{b_0}< {1\over 8}+{1\over 8}={1\over 4}. \end{array} \end{aligned}$$

(5.2)

Step 2: Evolution of an elliptic boundary. Let us recall some basic facts. Given a $C^1$ function f(x, t) and the family of level curves

$$\begin{aligned} C(t)=\{x\in {{\mathbb {R}}}^2\ | f(x,t)=0\}, \end{aligned}$$

if $D_xf\not =0$, the propagation speed of C(t) along the outward normal direction $\vec {n}={D_xf\over |D_xf|}$ is given by

$$\begin{aligned} v_{\vec n}=-{f_t\over |D_xf|}, \end{aligned}$$

which can be easily derived through the chain rule. Moreover, the corresponding mean curvature along C(t) is

$$\begin{aligned} \kappa =\textrm{div}_x(\vec {n}). \end{aligned}$$

Now let us verify that for $t\in [0,t_\delta ]$, the propagation of $E_{\theta }(t)$ obeys the following inequality:

$$\begin{aligned} v_{\vec n}< 1 - d\kappa + {V} \cdot \vec {n}, \end{aligned}$$

(5.3)

which will be used to construct a supersolution. Fix any $a, b > 0$ and an ellipse

$$\begin{aligned} \frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} = 1, \end{aligned}$$

direct compuations show that its curvature at point $P(a\cos \phi , b\sin \phi )$, $0 \le \phi < 2\pi$, is

$$\begin{aligned} \kappa = \frac{ab}{\left( a^2\sin ^2\phi + b^2\cos ^2\phi \right) ^{\frac{3}{2}}} \end{aligned}$$

and the normal velocity at $P(a\cos \phi , b\sin \phi )$ is

$$\begin{aligned} v_{\vec {n}} = \frac{a^{\prime }b\cos ^2\phi + ab^{\prime }\sin ^2\phi }{\sqrt{a^2\sin ^2\phi + b^2\cos ^2\phi }}. \end{aligned}$$

Case 1. If $|\sin \phi |\le {\sqrt{3}\over 2}$, then

$$\begin{aligned} d\kappa = \frac{dab}{\left( a^2\sin ^2\phi + b^2\cos ^2\phi \right) ^{\frac{3}{2}}} \le \frac{dab}{\left( b^2\cos ^2\phi \right) ^{\frac{3}{2}}} \le \frac{dab}{(b/2)^3}\le \frac{d64a_0}{b_{0}^2} <{1\over 8}. \end{aligned}$$

Note that

$$\begin{aligned} v_{\vec n}= \frac{\frac{b}{2}\cos ^2\phi - La \sin ^2\phi }{\sqrt{a^2 \sin ^2\phi + b^2\cos ^2\phi }} \le {{b\over 2}\cos ^2\phi \over \sqrt{b^2\cos ^2\phi }}\le \frac{1}{2}. \end{aligned}$$

Since for $t\in [0,t_\delta ]$,

$$\begin{aligned} E_{\theta }(t)\subset [-4a_0, 4a_0]\times [0,1], \end{aligned}$$

(5.2) implies that $v_{\vec {n}} \le {1\over 2}< 1-{1\over 8}-{1\over 4}< 1 -d \kappa + {V} \cdot \vec {n}$.

Case 2. If $|\sin \phi |\ge {\sqrt{3}\over 2}$, then

$$\begin{aligned} v_{\vec n}= & {} \frac{\frac{b}{2}\cos ^2\phi - La \sin ^2\phi }{\sqrt{a^2 \sin ^2\phi + b^2\cos ^2\phi }} \le {{b\over 2}\cos ^2\phi \over \sqrt{b^2\cos ^2\phi }}-{La\sin ^2\phi \over b}\\\le & {} {1\over 2}-{3La\over 4b}\le {1\over 2}-{3La_0\over 4b_0}. \end{aligned}$$

Meanwhile,

$$\begin{aligned} d\kappa = \frac{dab}{\left( a^2\sin ^2\phi + b^2\cos ^2\phi \right) ^{\frac{3}{2}}}\le {db\over a^2}\le {db_0\over {a_0}^{2}}. \end{aligned}$$

Therefore, due to the choice of L, we have $v_{\vec n}< 1 - d\kappa + {V} \cdot \vec {n}$.

Combining case 1 and case 2, we see that (5.3) holds, i.e., the evolution of $E_{\theta }(t)$ satisfies

$$\begin{aligned} v_{\vec n} < 1 -d \kappa + {V} \cdot \vec {n},&\text {for}&0 \le t \le t_\delta . \end{aligned}$$

Step 3: Comparison. Let

$$\begin{aligned} \Psi (x,t)=\frac{(x_1-a_0 - \nu )^2}{a^2(t)} + \frac{(x_2-\theta )^2}{b^2(t)} -1. \end{aligned}$$

Owing to (5.3), we may choose $\mu _0\in (0, {1\over 2})$ such that (5.3) also holds along the curve $\{x\in {{\mathbb {R}}}^2|\ \Psi (x,t)=\mu \}$ for all $\mu \in [-\mu _0,\mu _0]$ and $t\in [0,t_\delta ]$. Equivalently, for

$$\begin{aligned} D_{\theta }= & {} \left\{ (x,t)\in {{\mathbb {R}}}^2\times [0,t_\delta ]\ |-\mu _0\le \Psi (x,t) \le \mu _0\right\} ,\nonumber \\{} & {} \Psi _t + \left( 1-d\, \, \textrm{div}\left( {D\Psi \over |D\Psi |}\right) \right) |D\Psi |+V(x)\cdot D\Psi \ge 0 \quad \text {on}\, D_{\theta }. \end{aligned}$$

(5.4)

Let $\{h_k\}_{k\ge 1}\in C^{\infty }({{\mathbb {R}}})$ be a sequence of functions such that

$$\begin{aligned} 0<h_{k}^{'}\le 1 \quad \text {in}\, I_k=\left( -\mu _0+{1\over k},\ \mu _0-{1\over k}\right) ,\quad h_{k}^{'}=0 \quad \text {in}\, {{\mathbb {R}}}\backslash I_k \end{aligned}$$

and $\lim _{k\rightarrow +\infty }h_k(s)=h(s)$ uniformly in ${{\mathbb {R}}}$, where

$$\begin{aligned} h(s)={\left\{ \begin{array}{ll} \mu _0 \quad \text {for}\, s\ge \mu _0\\ s\quad \text {for}\, s\in [-\mu _0, \mu _0]\\ -\mu _0 \quad \text {for}\, s\le -\mu _0. \end{array}\right. } \end{aligned}$$

Apparently, $\Psi _k(x,t)=h_k(\Psi (x,t))$ satisfies

$$\begin{aligned} {\partial \Psi _{k}\over \partial t} + \left( 1-d\, \, \textrm{div}\left( {D\Psi _k\over |D\Psi _k|}\right) \right) |D\Psi _k|+V(x)\cdot D\Psi _k\ge 0 \quad \text {on}\, {{\mathbb {R}}}^2\times (0,t_\delta ). \end{aligned}$$

(5.5)

By stability, we have that $G_1(x,t)=h(\Psi (x,t))\in W^{1,\infty }({{\mathbb {R}}}^2\times [0,\infty ))$ is a viscosity supersolution of

$$\begin{aligned} {\partial G_{1}\over \partial t }+ \left( 1-d\, \, \textrm{div}\left( {DG_1\over |DG_1|}\right) \right) |DG_1|+V(x)\cdot DG_1\ge 0 \quad \text {on}\, {{\mathbb {R}}}^2\times (0,t_\delta ). \end{aligned}$$

(5.6)

Since $(a)_+\ge a$, $G_1=G_1(x,t)$ is also a viscosity supersolution of equation (1.2) on ${{\mathbb {R}}}^2\times (0,t_\delta )$.

Because for fixed $\nu >0$

$$\begin{aligned} \{G_1(x,0)\le 0\}=\{\Psi (x,0)\le 0\}\subset S, \end{aligned}$$

we can choose a function $\xi \in C^{\infty }({{\mathbb {R}}})$ such that $\dot{\xi }>0$, $\xi (0)=0$, $\sup _{s\in {{\mathbb {R}}}}|\xi (s)|<\infty$ and

$$\begin{aligned} g(x)\le \xi (G_1(x,0)). \end{aligned}$$

Since $\xi (G_1(x,0))$ is also a viscosity supersolution of equation (1.2) on ${{\mathbb {R}}}^2\times (0,t_\delta )$, thanks to Theorem 2.1, we have that

$$\begin{aligned} G(x,t)\le \xi (G_1(x,t)) \quad \text {for}\, (x,t)\in {{\mathbb {R}}}^2\times [0,t_\delta ]. \end{aligned}$$

In particular, this implies that

$$\begin{aligned} \{x\in {{\mathbb {R}}}^2|\ G_1(x,t)<0\}=\{x\in {{\mathbb {R}}}^2|\ \xi (G_1(x,t))<0\}\subset \{x\in {{\mathbb {R}}}^2|\ G(x,t)<0\}. \end{aligned}$$

Note that $\{x\in {{\mathbb {R}}}^2|\ G_1(x,t)<0\}=\{x\in {{\mathbb {R}}}^2|\ \Psi (x,t)<0\}$. Sending $\nu \rightarrow 0$, we have that for $t\in [0,t_\delta ]$,

$$\begin{aligned} \left\{ x=(x_1,x_2)\in {{\mathbb {R}}}^2|\ \frac{(x_1-a_0)^2}{a^2(t)} + \frac{(x_2-\theta )^2}{b^2(t)}<1\right\} \subset \{x\in {{\mathbb {R}}}^2|\ G(x,t)<0\}. \end{aligned}$$

Then

$$\begin{aligned} G((0,\theta ),t)<0 \quad \text {for}\, t\in (0, t_\delta ], \end{aligned}$$

which finishes the proof. Note that $t_\delta$ only depends on $\delta$ and V. $\square$

Let $\Omega \subset {{\mathbb {R}}}^2$ be an open convex set. Denote by $G_{\Omega }(x,t)$ the unique viscosity solution to equation (1.2) subject to $G_\Omega (x,0)=g_{\Omega }(x)$ where

$$\begin{aligned} g_{\Omega }(x)={\left\{ \begin{array}{ll} -{2\over \pi }\arctan (d(x, \partial \Omega )) \quad \text {for}\, x\in \Omega \\ {2\over \pi }\arctan (d(x,\partial \Omega ) )\quad \text {otherwise}. \end{array}\right. } \end{aligned}$$

Given two sets $E_1$ and $E_2$, their Hausdorff distance

$$\begin{aligned} d_H(E_1, E_2)=\max \{ \max _{x\in E_1}d(x, E_2), \ \max _{x\in E_2}d(x, E_1)\}. \end{aligned}$$

Also, for $\alpha >0$ and $\delta \in (0, {1\over 2})$, we write

$$\begin{aligned} W_{\alpha , \delta }=[-\alpha , \alpha ]\times [\delta , 1-\delta ]. \end{aligned}$$

Lemma 5.2

Let $S=(0,1)^2$. For given $\delta \in (0,{1\over 2})$ and $n\in {{\mathbb {N}}}$, there exists $\sigma _{\delta , n}>0$ such that if

$$\begin{aligned} d_H(S, \Omega )\le \sigma _{\delta ,n} \quad \textrm{and} \quad \alpha \le \sigma _{\delta ,n}, \end{aligned}$$

then

$$\begin{aligned} G_\Omega (x,t)<0 \quad \text {for}\, (x,t)\in W_{\alpha ,\delta }\times \left[ {t_{\delta }\over n},\ t_{\delta }\right] . \end{aligned}$$

Here $t_\delta$ is from the previous Lemma 5.1.

Proof

We argue by contradiction. If not, then there exist a sequence of convex open sets $\{\Omega _m\}_{m\ge 1}$ such that

$$\begin{aligned} d_H(S, \Omega _m)\le {1\over m} \end{aligned}$$

and for some $(x_m,t_m)\in W_{{1\over m},\delta }\times \left[ {t_{\delta }\over n},\ t_{\delta }\right]$

$$\begin{aligned} G_{\Omega _m}(x_m,t_m)\ge 0. \end{aligned}$$

Fig. 17

Reach $\Omega$ from $W_{\alpha ,\delta }$

Since

$$\begin{aligned} \lim _{m\rightarrow +\infty }g_{\Omega _m}(x)=g_S(x) \quad \text {uniformly on}\, {{\mathbb {R}}}^2, \end{aligned}$$

due to Remark 2.1, the uniqueness of viscosity solutions, we have that

$$\begin{aligned} \lim _{m\rightarrow +\infty }G_{\Omega _m}(x,t)=G(x,t) \quad \text {locally uniformly on}\, {{\mathbb {R}}}^2\times [0,\infty ). \end{aligned}$$

Here G is from Lemma 5.1. The proof is similar to that of (2.4). Also, up to a subsequence if necessary, we may assume that $\lim _{m\rightarrow +\infty }(x_m,t_m)=((0,\theta ), {\bar{t}})$ for $(\theta , {\bar{t}})\in [\delta , 1-\delta ]\times \left[ {t_{\delta }\over n},\ t_{\delta }\right]$. Then we have that

$$\begin{aligned} G((0,\theta ), {\bar{t}})=\lim _{m\rightarrow +\infty }G_{\Omega _m}(x_m,t_m)\ge 0. \end{aligned}$$

This is a contradiction. $\square$

As an immediate corollary, we have the following reachability.

Lemma 5.3

Consider the game in section 2.2. Under the assumption of Lemma 5.2, every point on $W_{\alpha ,\delta }$ can reach $\Omega$ within time $t_{\delta }\over n$. Also, it is easy to see that if we replace S by an arbitrary rectangle, all previous results are still true in the corresponding forms (See Fig. 17).

Remark 5.1

Due to the hidden stochastic nature of the game trajectory, it is not clear to us how to use pure game dynamics to prove the above reachability conclusion. An interesting analog in random walk (or Brownian motion) is to use strong maximum principle of the Laplace equation to show that a particle has a positive probability to exit from any small window of the boundary.