Nash equilibria for quasi-linear parabolic problems

Abstract

This paper discusses the application of the Nash strategy to a quasi-linear parabolic equation and a quasi-linear parabolic equation with a semilinear term. First, we demonstrate the existence of a Nash quasi-equilibrium for both equations using the fixed point method. Subsequently, we establish that the functionals are convex, ensuring that the Nash quasi-equilibrium is, in fact, a Nash equilibrium. Additionally, in conjunction with the theoretical results, numerical techniques are developed for each problem. To describe the iterative algorithms, the Newton’s Method is utilized in combination with the Finite Element Method and Finite Difference Method. All algorithms are implemented using the Freefem++ software, and various results are presented through figures and comparative graphics.

Appendix A

In this section we will deal with the well-posedness of differential equations (1.1) and (1.2). To facilitate the exposition we consider on the right side of each equation a function f and g, respectively.

Theorem 5

Under the hypotheses $({\textbf{H}}1)$, $({\textbf{H}}2)$, $u_0 \in H^1_0(\Omega ) \cap H^2(\Omega )$, $f \in L^2(\Omega \times (0,T))$, there exist a positive constant $\epsilon _0$ such that, if $u_0$ satisfies $(\Vert u_0 \Vert + a_1 \Vert u_0\Vert + \vert f \vert _{L^2(Q)}) < e_0$, then the problem (1.1) admits a unique solution $u: Q \longrightarrow \mathbb {R}$, satisfying

$$\begin{aligned} u \in L^{\infty }(0,T;H^{1}_{0}(\Omega )\cap L^2(\Omega )) \cap L^{2}(0,T;H^1_{0}(\Omega )\cap H^2(\Omega )), u_t \in L^2(0,T;L^2(\Omega )). \end{aligned}$$

Theorem 6

Under the hypotheses $({\textbf{H}}1)$, $({\textbf{H}}2)$, $u_0 \in H^1_0(\Omega ) \cap H^2(\Omega )$, $g \in L^2(\Omega \times (0,T))$, there exist a positive constant $\epsilon _0$ such that, if $u_0$ satisfies $(\Vert u_0 \Vert + a_1 \Vert u_0\Vert + \vert g \vert _{L^2(Q)}) < e_0$ then the problem (1.2) admits a unique solution $u: Q \rightarrow \mathbb {R}$, satisfying

$$\begin{aligned}u \in L^{\infty }(0,T;H^{1}_{0}(\Omega )\cap L^2(\Omega )) \cap L^{2}(0,T;H^1_{0}(\Omega )\cap H^2(\Omega )), u_t \in L^2(0,T;L^2(\Omega )). \end{aligned}$$

Proof of Theorem 5

To prove the theorem, we employ Galerkin Method with the Hilbertian basis from $H^1_0(\Omega )$, given by the eigenvectors $(w_j)$ of the spectral problem: $(w_j,v) = \lambda _j (w_j, v)$ for all $v \in V = H^1_0(\Omega ) \cap H^2(\Omega )$ and $j = 1, 2, \ldots$. We represent by $V_m$ the subspace of V generated by vectors ${w_1, w_2,\ldots , w_m }$. We propose the following approximate problem: Determine $u_m \in V_m$, so that

$$\begin{aligned} (3) \, \left\{ \begin{array}{ll} (u'_m,v)+ (a(u_m)\nabla u_m,\nabla v) = (f,v) \\ begin{eqnarray*}8pt] u_m(0)=u_{0m}\rightarrow u_0 ~in~ H^{1}_0(\Omega ) \end{array} \right. \end{aligned}$$

Existence

The system of ordinary differential equations (3) has a local solution $u_m = u_m (x,t)$ in the interval $(0, T_m)$. The estimates that follow permit to extend the solution $u_m (x,t)$ to interval [0, T[ for all $T > 0$ and to take the limit in (3).

Estimate I Taking $v = u_m (t)$ in equation $(3)_1$

$$\begin{aligned} (u'_m,u_m)+ (a(u_m)\nabla u_m,\nabla u_m) = (f,u_m) \end{aligned}$$

from hypothesis (H1), we have the following inequality,

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\vert u_m \vert ^2 + a_0 \Vert u_m \Vert ^{2}\le & {} \int _{\Omega } \vert f \vert \vert u_m \vert \\\le & {} \vert f \vert _{L^2} \vert u_m \vert _{L^2} \\\le & {} \frac{1}{2} \vert f \vert ^2_{L^2} + \frac{1}{2}\vert u_m \vert ^2_{L^2} \,, \end{aligned}$$

and integrating over (0, t), we obtain:

$$\begin{aligned} \frac{1}{2}\vert u_m(t) \vert ^2_{L^2} + a_0 \int _0^t \Vert u_m \Vert ^{2}\le & {} \frac{1}{2}\vert u_m(0)\vert ^2_{L^2} + \frac{1}{2} \int _{0}^T \vert f \vert ^2_{L^2} + \frac{1}{2}\int _{0}^t \vert u_m \vert ^2_{L^2} \\\le & {} \frac{1}{2}\vert u_0\vert ^2_{L^2} + \frac{1}{2}\vert f \vert ^2_{L^2(Q)} + \frac{1}{2} \int _{0}^t \vert u_m \vert ^2_{L^2} \\ \vert u_m \vert ^2 + \int _0^t \Vert u_m \Vert ^{2}\le & {} C_1(\vert u_0\vert ^2_{L^2} + \vert f \vert ^2_{L^2(Q)}) + C_2 \int _{0}^t \vert u_m \vert ^2_{L^2} \,. \end{aligned}$$

Therefore, applying the Gronwall’s inequality, we obtain the following estimate:

$$\begin{aligned} \vert u_m\vert _{L^{\infty }(0,T,L^2)} + \vert u_m \vert _{L^2(0,T,H^1_0)} \le C(\vert u_0\vert _{L^2} + \vert f \vert _{L^2(Q)})e^{TC_2} \,. \end{aligned}$$

Estimate II Taking $v = u'_m (t)$ in equation $(3)_1$, we obtain

$$\begin{aligned} (u'_m,u'_m)+ (a(u_m)\nabla u_m,\nabla u'_m) = (f,u'_m) \end{aligned}$$

since

$$\begin{aligned} \frac{d}{dt}\int _{\Omega }a(u_m)\vert \nabla u_m \vert ^2=\int _{\Omega }\frac{da}{du}(u_m)u'_m\vert \nabla u_m \vert ^2 + 2 \int _{\Omega }a(u_m)\nabla u_m \nabla u'_m \end{aligned}$$

we get

$$\begin{aligned} \vert u'_m \vert ^2_{L^2} + \frac{1}{2}\frac{d}{dt}\int _{\Omega }a(u_m)\vert \nabla u_m \vert ^2= & {} \frac{1}{2}\int _{\Omega }\frac{da}{du}(u_m)u'_m\vert \nabla u_m \vert ^2 + (f,u'_m) \\= & {} I_1 + I_2 \,. \end{aligned}$$

On the other hand, from hypothesis (H2), we have the following inequality,

$$\begin{aligned} \vert I_1 \vert\le & {} \frac{1}{2} M C_0 \int _{\Omega }\vert u'_m \vert \vert \nabla u_m \vert \vert \nabla u_m \vert \\\le & {} C\vert u'_m \vert _{L^2}\vert \nabla u_m \vert _{L^4} \vert \nabla u_m \vert _{L^4} \\\le & {} C\vert u'_m \vert _{L^2} \vert \nabla u_m \vert _{H^1} \vert \Delta u_m \vert _{L^2} \\\le & {} \epsilon \vert \Delta u_m \vert ^2_{L^2} + C(\epsilon )\vert u'_m \vert ^2_{L^2} \vert \nabla u_m \vert ^2_{L^2} \\ \vert I_2 \vert\le & {} \int _{\Omega } \vert f \vert \vert u'_m \vert \\\le & {} \vert f \vert _{L^2} \vert u'_m \vert _{L^2} \\\le & {} \epsilon _2 \vert u'_m \vert ^2_{L^2} + C(\epsilon _2) \vert f \vert ^2_{L^2} \,. \end{aligned}$$

Substituting into the right-hand side of (6.1), we get

$$\begin{aligned}{} & {} \vert u'_m \vert ^2_{L^2} + \frac{1}{2}\frac{d}{dt}\int _{\Omega }a(u_m)\vert \nabla u_m \vert ^2\nonumber \\{} & {} \quad \le \epsilon \vert \Delta u_m \vert ^2_{L^2} + C(\epsilon )\vert u'_m \vert ^2_{L^2} \vert \nabla u_m \vert ^2_{L^2} + \epsilon _2 \vert u'_m \vert ^2_{L^2} + C(\epsilon _2) \vert f \vert ^2_{L^2}. \end{aligned}$$

(6.1)

Now taking $v=-\Delta u_m$ in equation $(3)_1$, we obtain

$$\begin{aligned} (u'_m,-\Delta u_m)+ (a(u_m)\nabla u_m,\nabla (-\Delta u_m)) = (f,-\Delta u_m) \end{aligned}$$

since

$$\begin{aligned}\int _{\Omega } u'_m (-\Delta u_m)= \int _{\Omega } \nabla u'_m \nabla u_m = \frac{1}{2} \frac{d}{dt}\int _{\Omega } \vert \nabla u_m \vert ^2 = \frac{1}{2} \frac{d}{dt}\Vert u_m \Vert ^2\end{aligned}$$

and

$$\begin{aligned} \int _{\Omega } a(u_m)\nabla u_m \nabla (-\Delta u_m)= & {} - \int _{\Omega } \nabla \cdot (a(u_m)\nabla u_m)(-\Delta u_m) \\= & {} -\int _{\Omega } (\frac{da}{du}(u_m)\nabla u_m \nabla u_m + a(u_m) \Delta u_m)(-\Delta u_m) \\= & {} \int _{\Omega } \frac{da}{du}(u_m)\vert \nabla u_m \vert ^2 \Delta u_m + a(u_m)\vert \Delta u_m)\vert ^2 \end{aligned}$$

substituting

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert u_m \Vert ^2 + \int _{\Omega } a(u_m)\vert \Delta u_m\vert ^2= & {} - \int _{\Omega } \frac{da}{du}(u_m)\vert \nabla u_m \vert ^2 \Delta u_m + (f,-\Delta u_m) \\= & {} I_1 + I_2 \,. \end{aligned}$$

On the other hand, from hypothesis (H2), we have the following inequality:

$$\begin{aligned} \vert I_1 \vert\le & {} M \int _{\Omega }\vert \nabla u_m \vert \vert \nabla u_m \vert \vert \Delta u_m \vert \nonumber \\\le & {} C\vert \nabla u_m \vert _{L^4} \vert \nabla u_m \vert _{L^4} \vert \Delta u_m \vert _{L^2} \nonumber \\\le & {} C\vert \nabla u_m \vert _{L^2} \vert \nabla u_m \vert _{H^1} \vert \Delta u_m \vert _{L^2} \nonumber \\\le & {} C\vert \nabla u_m \vert _{L^2} \vert u_m \vert _{H^2} \vert \Delta u_m \vert _{L^2} \nonumber \\\le & {} C\vert \nabla u_m \vert _{L^2} \vert \Delta u_m \vert _{L^2} \vert \Delta u_m \vert _{L^2} \nonumber \\\le & {} \epsilon \vert \Delta u_m \vert ^2_{L^2} + C(\epsilon )\vert \nabla u_m \vert ^2_{L^2} \vert \Delta u_m \vert ^2_{L^2} \,, \nonumber \\ \vert I_2 \vert\le & {} \int _{\Omega } \vert f \vert \vert \Delta u_m \vert \nonumber \\\le & {} \vert f \vert _{L^2} \vert \Delta u_m \vert _{L^2} \nonumber \\\le & {} \epsilon \vert \Delta u_m \vert ^2_{L^2} + C(\epsilon ) \vert f \vert ^2_{L^2} \,, \end{aligned}$$

(6.2)

substituting

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert u_m \Vert ^2 + a_0 \vert \Delta u_m\vert ^2_{L^2} \le 2\epsilon \vert \Delta u_m \vert ^2_{L^2} + C(\epsilon )\vert \nabla u_m \vert ^2_{L^2} \vert \Delta u_m \vert ^2_{L^2} + C(\epsilon ) \vert f \vert ^2_{L^2} \,. \end{aligned}$$

(6.3)

From the estimates (6.1) and (6.3), we obtain

$$\begin{aligned} \begin{aligned} \displaystyle \frac{1}{2}\frac{d}{dt}\displaystyle&\left( \int _{\Omega }a(u_m)\vert \nabla u_m \vert ^2 + \Vert u_m \Vert ^2 \displaystyle \right) + \vert u'_m \vert ^2_{L^2} + a_0\vert \Delta u_m \vert ^2_{L^2} \\&\le \displaystyle 3\epsilon \vert \Delta u_m \vert ^2_{L^2} + C(\epsilon )\vert \nabla u_m \vert ^2_{L^2} \vert \Delta u_m \vert ^2_{L^2} + C \vert f \vert ^2_{L^2} + C(\epsilon )\vert u'_m \vert ^2_{L^2} \vert \nabla u_m \vert ^2_{L^2} + \epsilon _2 \vert u'_m \vert ^2_{L^2} \end{aligned} \end{aligned}$$

(6.4)

and consequently,

$$\begin{aligned} \begin{aligned} \displaystyle \frac{1}{2}\frac{d}{dt}\displaystyle \left( \int _{\Omega }a(u_m)\vert \nabla u_m \vert ^2 + \Vert u_m \Vert ^2 \displaystyle \right)&+ \frac{1}{2}\vert u'_m \vert ^2_{L^2} + \frac{a_0}{2}\vert \Delta u_m \vert ^2_{L^2} \\&\le \displaystyle C\vert \nabla u_m \vert ^2_{L^2} \vert \Delta u_m \vert ^2_{L^2} + C\vert u'_m \vert ^2_{L^2} \vert \nabla u_m \vert ^2_{L^2} + C \vert f \vert ^2_{L^2} \end{aligned} \end{aligned}$$

(6.5)

$$\begin{aligned} \begin{aligned} \displaystyle \frac{1}{2}\frac{d}{dt}\displaystyle \left( \int _{\Omega }a(u_m)\vert \nabla u_m \vert ^2 + \Vert u_m \Vert ^2 \displaystyle \right)&+ \frac{1}{4}\vert u'_m \vert ^2_{L^2} + \frac{a_0}{4}\vert \Delta u_m \vert ^2_{L^2} + \vert u'_m \vert ^2_{L^2}(\frac{1}{4} - C\Vert u_m \Vert ^2) \\&\displaystyle + \vert \Delta u_m \vert ^2 _{L^2}(\frac{a_0}{4}- C\Vert u_m \Vert ^2) \le C \vert f \vert ^2_{L^2} \,. \end{aligned} \end{aligned}$$

(6.6)

Now, if the initial data are sufficiently small, we obtain

$$\begin{aligned} \vert u'_m \vert ^2_{L^2}\Big {(}\frac{1}{4} - C\Vert u_m \Vert ^2\Big {)}>0 ~,~ \vert \Delta u_m \vert ^2 _{L^2}\Big {(}\frac{a_0}{4}- C\Vert u_m \Vert ^2 \Big {)}>0 \,. \end{aligned}$$

Integrating over (0, t), we obtain

$$\begin{aligned}{} & {} \begin{aligned} \displaystyle \frac{1}{2}\int _{\Omega }a(u_m(t))\vert \nabla u_m(t) \vert ^2&+ \frac{1}{2}\Vert u_m(t)\Vert ^2 + \frac{1}{4}\int ^t_0 \vert u'_m \vert ^2_{L^2} + \frac{a_0}{4}\int ^t_0 \vert \Delta u_m \vert ^2_{L^2} \\&\le \displaystyle \frac{1}{2}\int _{\Omega }a(u_m(0))\vert \nabla u_m(0) \vert ^2 + \frac{1}{2}\Vert u_m(0)\Vert ^2 + C\int ^T_0\vert f \vert ^2_{L^2} \end{aligned} \nonumber \\{} & {} \frac{a_0}{2}\Vert u_m(t) \Vert ^2 + \frac{1}{2}\Vert u_m(t)\Vert ^2 + \frac{1}{4}\int ^t_0 \vert u'_m \vert ^2_{L^2} + \frac{a_0}{4}\int ^t_0 \vert \Delta u_m \vert ^2_{L^2} \le \frac{a_1}{2}\Vert u_0 \Vert ^2 + \frac{1}{2}\Vert u_0 \Vert ^2 + C\int ^T_0\vert f \vert ^2_{L^2} \nonumber \\{} & {} \vert u_m \vert _{L^{\infty }(0,T;H^1_0)} + \vert u'_m \vert _{L^2(0,T;L^2)} + \vert u_m \vert _{L^2(0,T;H^2)} \le C( \Vert u_0 \Vert + \vert f \vert _{L^2(Q)}). \end{aligned}$$

(6.7)

$\square$