Second-order integro-differential parabolic variational inequalities arising from the valuation of American option

Abstract

This article deals with the existence and uniqueness of solution to the ${\mathbb{R}}_{d_1}$-valued parabolic variational inequalities with integro-differential terms which arise from the valuation of American option. The authors use the penalty method to construct a sequence of approximation parabolic problem and hence obtain the existence and uniqueness of solution to the approximation problem by using fixed point theory. Then the solution of parabolic variational inequalities is obtained by showing that the solution of this penalty problem converges to the variational inequalities. The uniqueness of the solution is also proven.

MSC:35B40, 35K35.

1 Introduction

In this paper, we are concerned with the existence and uniqueness of solution to an ${\mathbb{R}}_{d_1}$-valued parabolic variational inequality with the integro-differential terms

$$\{\begin{array}{ll}{(u^k)}_t-L_{k}u-F_k(x,t,u,\mathrm{\nabla }u)\ge 0,& \text{in }{Q}_T,\\ u^k(x,t)\ge u_0^k(x),& \text{in }{Q}_T,\\ ({(u^k)}_t-L_{k}u-F_k(x,t,u,\mathrm{\nabla }u))(u^k-u_0^k(x))=0,& \text{in }{Q}_T,\\ u^k(x,0)=u_0^k(x),& \text{on }\mathrm{\Omega },\\ u^k(x,t)=g^k(x,t),& \text{on }\partial \mathrm{\Omega }\times (0,T),\end{array}$$

(1)

where $\mathrm{\Omega }\in {\mathbb{R}}_d$ is an open set with a smooth boundary ∂ Ω, $Q_T=\mathrm{\Omega }\times [0,T]$ is a parabolic domain for some $T>0$. $L=(L_1,L_2,\dots ,L_{d_1})$, and $L_k=L_k(x,t)$ is a divergence form second-order elliptic operator satisfying

$$L_{k}u=\sum _{r=1}^{d_1}\sum _{i,j=1}^d\frac{\partial }{\partial x_j}(a_{kr}^{ij}(x,t)\frac{\partial }{\partial x_i}{u}^r)-\sum _{r=1}^{d_1}\sum _{i=1}^d{b}_{kr}^i(x,t)\frac{\partial }{\partial x_i}{u}^r-\sum _{r=1}^{d_1}{c}_{kr}(x,t)u^r.$$

Moreover, the integro-differential operator $F_k$ is defined by

$$F_k(x,t,u,\mathrm{\Delta }u)={\int }_{\mathrm{\Omega }}{f}_k(x,t,y,u(x,t),\mathrm{\nabla }u(x,t))\mathrm{d}y$$

which is a continuous integral operator as defined in [1].

The concrete motivations of studying (1) can be easily found in the literature. If $d_1=1$, $F_k(x,t,u,\mathrm{\Delta }u)=0$, and the operator $L_k$ degenerates to

$$Lu={\partial }_{t}u-\frac{{\sigma }^2}{2}\frac{{\partial }^2}{\partial x^2}u+(\frac{{\sigma }^2}{2}-r)\frac{\partial }{\partial x}u+ru,$$

(1) becomes the linear variational inequality based on the famous Black-Scholes equation (see [2]). Such a variational inequality arises in many applications of American option pricing (see [3, 4]). To deal with this problem, some scholars often introduce a free and moving boundary problem. By adding a certain penalty term to the Black-Scholes equation, the solution of this variational inequality is extended to a fixed domain. Furthermore, this penalty term forces the solution to stay above the payoff function at expiry. Throughout the last decade, a number of papers addressing penalty schemes for American options have been published (see, for instance, [5–9] and references therein).

However, several Black-Scholes models proposed in recent years, such as the model found in [8, 9], have allowed the risk assets driven by levy processes. In this model, the risk asset S as in the Black-Scholes model, follows the stochastic process (assume $d_1=1$)

$$\mathrm{d}{S}_t=\mu S_t\mathrm{d}t+\sigma S_t\mathrm{d}{z}_t,$$

from which we obtain the following PIDE:

$${\partial }_{t}u=\frac{{\sigma }^2}{2}[{\partial }_{xx}u-{\partial }_{x}u]+\int [u(x+y,t)-u(x,t)-(e^y-1){\partial }_{x}u]v(\mathrm{d}x),$$

(2)

where $z_t$ is a kind of levy process ( for details, see [10, 11]). The authors in [9] generalize (2) and prove the existence and uniqueness of a classical solution to a more general problem in the parabolic domain $Q_T=\mathrm{\Omega }\times [0,T]$, where Ω is an open, unbounded subset of ${\mathbb{R}}_d$, with a smooth boundary ∂ Ω. A related work in the context of quantum mechanics has been studied in [12, 13].

Therefore, the authors of this paper intend to study a more complex variational inequality involved in American option based on the more complicated PIDE type Black-Scholes equation than (2). And we will consider ${\mathbb{R}}_{d_1}$-valued parabolic variational inequality (1) using the penalty method. The rest of this article is as follows. In Section 2, we state the main result. Section 3 discusses the penalty problem which will be used to prove our main result. In Section 4, we show the proof of our main result.

2 Main result

In this section, we present some notations and lemmas which are important to be used to prove our main result. Define ${\mathbb{L}}_p(\mathrm{\Omega })=L_p(\mathrm{\Omega };{\mathbb{R}}_{d_1})$ as the space of all ${\mathbb{R}}_{d_1}$-value functions $u=(u^1,u^2,\dots ,u^{d_1})$ satisfying

$${\parallel u\parallel }_{{\mathbb{L}}_p(\mathrm{\Omega })}^p=\sum _{k=1}^{d_1}{\parallel u^k\parallel }_{L_p({\mathbb{R}}_d)}^p<\mathrm{\infty }.$$

Let $1\le p\le \mathrm{\infty }$ and $k\in \mathbb{N}$. By $W_p^k$, we mean that the ℝ-value functions space satisfies

$$W_p^k(\mathrm{\Omega })=\{u\in L^p(\mathrm{\Omega };\mathbb{R})|D^{\alpha }u\in L^p(\mathrm{\Omega };\mathbb{R}),1\le \alpha \le k\}$$

which is Banach spaces with the norm

$${\parallel v\parallel }_{W_p^k(\mathrm{\Omega })}=\sum _{0\le |\alpha |\le k}{\parallel D^{\alpha }v\parallel }_{L^p(\mathrm{\Omega })}.$$

Further, we use the following Sobolev spaces which are modified from ℝ-value version to ${\mathbb{R}}_{d_1}$-value version

$${\mathbb{W}}_p^k(\mathrm{\Omega })=\{u\in L^p(\mathrm{\Omega };{\mathbb{R}}_{d_1})|D^{\alpha }u\in L^p(\mathrm{\Omega };{\mathbb{R}}_{d_1}),1\le \alpha \le k\}$$

with the norm

$${\parallel v\parallel }_{{\mathbb{W}}_p^k(\mathrm{\Omega })}=\sum _{0\le |\alpha |\le k}{\parallel D^{\alpha }v\parallel }_{{\mathbb{L}}^p(\mathrm{\Omega })}.$$

As mentioned in [14], we shall use $C^{2k+\delta ,k+\delta /2}({\overline{Q}}_T)$ to denote the Holder space which satisfies

$$C^{2k+\delta ,k+\delta /2}({\overline{Q}}_T)=\{u|D^{\alpha }{\partial }_t^{\rho }\in C^{\delta ,\delta /2}({\overline{Q}}_T),|\alpha |+2\rho \le 2k\}.$$

For $0<T<\mathrm{\infty }$, we also use some following common notation from PDEs without mentioning:

$$\begin{array}{c}{H}^k(\mathrm{\Omega })=W_2^k(\mathrm{\Omega }),{\parallel \cdot \parallel }_{H^k(\mathrm{\Omega })}={\parallel \cdot \parallel }_{W_2^k(\mathrm{\Omega })},\hfill \\ {\mathbb{H}}^k(\mathrm{\Omega })={\mathbb{W}}_2^k(\mathrm{\Omega }),{\parallel \cdot \parallel }_{{\mathbb{H}}^k(\mathrm{\Omega })}={\parallel \cdot \parallel }_{{\mathbb{W}}_2^k(\mathrm{\Omega })},\hfill \\ L^p(0,T;H^k(\mathrm{\Omega }))=\{v(\cdot ,t)|v(\cdot ,t)\in H^k(\mathrm{\Omega })\text{ in }(0,T);{\parallel v(\cdot ,t)\parallel }_{H^k(\mathrm{\Omega })}\in L^p(0,T)\},\hfill \\ {\mathbb{L}}^p(0,T;{\mathbb{H}}^k(\mathrm{\Omega }))=\{v(\cdot ,t)|v(\cdot ,t)\in {\mathbb{H}}^k(\mathrm{\Omega })\text{ in }(0,T);{\parallel v(\cdot ,t)\parallel }_{{\mathbb{H}}^k(\mathrm{\Omega })}\in L^p(0,T)\},\hfill \\ {\parallel u\parallel }_{L^p(0,T;H^k(\mathrm{\Omega }))}^p={\int }_0^T{\parallel u(\cdot ,t)\parallel }_{H^k(\mathrm{\Omega })}^p\mathrm{d}t,{\parallel u\parallel }_{L^p(0,T;{\mathbb{H}}^k(\mathrm{\Omega }))}^p={\int }_0^T{\parallel u(\cdot ,t)\parallel }_{{\mathbb{H}}^k(\mathrm{\Omega })}^p\mathrm{d}t,\hfill \end{array}$$

where $0<T<\mathrm{\infty }$.

Moreover, we use the following lemmas to show the existence and uniqueness of the solution for the penalty approximation of our main problem.

Definition 2.1 Assume that X is a real Banach space, the space $C([0,T];X)$ consists of all continuous functions $u:[0,T]\to X$ with

$${\parallel u\parallel }_{C([0,T];X)}=\underset{0\le t\le T}{max}{\parallel u\parallel }_X<\mathrm{\infty }.$$

Then $C([0,T];X)$ is a Banach space endowed with the norm ${\parallel u\parallel }_{C([0,T];X)}$.

Definition 2.2 A mapping $A:X\to X$ is called to be compact if and only if the sequence ${\{A[u]\}}_{k=1}^{\mathrm{\infty }}$ is precompact for each bounded sequence ${\{u\}}_{k=1}^{\mathrm{\infty }}$, that is, there exists a subsequence ${\{u_{k_j}\}}_{j=1}^{\mathrm{\infty }}$ such that ${\{A[u_{k_j}]\}}_{j=1}^{\mathrm{\infty }}$ converges in X.

Lemma 2.3 (Schaefer’s fixed point theorem)

Suppose that $A:X\to X$ is a continuous and compact mapping. Further, assume that the set

$$\{u\in X:u=\lambda A[u],\mathit{\text{ for some }}0\le \lambda \le 1\}$$

is bounded. Then A has a fixed point.

The following two lemmas from the linear theory of parabolic partial differential equations can be found in [15].

Lemma 2.4 (Energy estimates)

Consider the problem

$$\{\begin{array}{ll}\frac{\partial v^k}{\partial t}-L_{k}v=f_k(x,t),& \mathit{\text{in }}{Q}_T,\\ v^k(x,0)=0,& \mathit{\text{on }}\mathrm{\Omega },\\ v^k(x,t)=0,& \mathit{\text{on }}\partial \mathrm{\Omega }\times (0,T)\end{array}$$

(3)

with $f_k\in L^2(0,T;L^2(\mathrm{\Omega }))$. Then there exists a unique solution $v\in L^2(0,T;{\mathcal{H}}_0^1(B_R))\cap C([0,T];{\mathcal{L}}^2(B_R))$ to problem (9) that satisfies

$$\underset{0<t<T}{max}{\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}+{\parallel v(t)\parallel }_{L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))}+{\parallel \frac{\partial v}{\partial t}\parallel }_{L^2(0,T;{\mathbb{H}}_0^{-1}(\mathrm{\Omega }))}\le C{\parallel f\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))},$$

where C is a positive constant depending only on Ω, T and the operator L.

Lemma 2.5 (Improved regularity)

There exists a unique weak solution

$$u\in L^2(0,T;{\mathbb{H}}_0^1(B_R))\cap C([0,T];{\mathbb{L}}^2(B_R))$$

to problem (3), with $\frac{\partial u}{\partial t}\in L^2(0,T;{\mathbb{H}}^{-1}(\mathrm{\Omega }))$. Moreover,

$$u\in L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))\cap L^{\mathrm{\infty }}(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega })),\frac{\partial u}{\partial t}\in L^2(0,T;{\mathbb{H}}^{-1}(\mathrm{\Omega })).$$

We also have the estimate

$$\begin{array}{r}\underset{0<t<T}{ess\hspace{0.17em}sup}{\parallel u(t)\parallel }_{{\mathbb{H}}_0^1(\mathrm{\Omega })}+{\parallel u(t)\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}+{\parallel \frac{\partial u}{\partial t}\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\\ \le C{\parallel f\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))},\end{array}$$

(4)

where C is a positive constant depending only on Ω, T and the operator L.

Throughout this section, we impose the following assumptions:

(A1) The coefficients $a_{kr}^{ij}(x,t)$, $b_{kr}^i(x,t)$, $c_{kr}(x,t)$ belong to the Holder space $C^{\delta ,\delta /2}({\overline{Q}}_T)$.

(A2) The operator $\frac{\partial }{\partial t}-L_k$ is (uniformly) parabolic, that is, there exists a constant $\theta >0$ such that

$$\theta \sum _{k=1}^{d_1}{\left|\mathrm{\nabla }{v}^k\right|}^2\le \sum _{k,r=1}^{d_1}\sum _{i,j=1}^d{a}_{kr}^{ij}(x,t){\partial }_i{v}^r\cdot {\partial }_j{v}^k\text{for all }(x,t)\in Q_{T}v\in {\mathbb{R}}_d.$$

(A3) There exists a positive constant B such that for all $v\in H_0^1(\mathrm{\Omega })$ and $(x,t)\in Q_T$, we have

$$B\sum _{k=1}^{d_1}{\partial }_i{v}^k\cdot v^k\le \sum _{k,r=1}^{d_1}{b}_{kr}^i(x,t){\partial }_i{v}^r\cdot v^k,c_{kr}(x,t)>0,k,r=1,2,\dots ,d_1.$$

(A4) $u_0^k(x)$ and $g_k(x,t)$ belong to the Holder spaces $C^{2+\delta }(R^d)$ and $C^{2+\delta ,1+\delta /2}({\overline{Q}}_T)$, respectively. Moreover, for all $(x,t)\in Q_T$, we have that $u_0^k(x)<g_k(x,t)$, $k=1,2,\dots ,d_1$.

(A5) The following two consistency conditions

$$g_k(x,0)=u_0^k(x)\text{and}{g}_k(x,0)-L_k(x,0)u_0^k(x)=0$$

hold for all $x\in \partial \mathrm{\Omega }$, $k=1,2,\dots ,d_1$.

(A6) $f_k(x,t,y,z,p)$ is nonnegative and belongs to $C^1(Q_T\times \mathrm{\Omega }\times {\mathbb{R}}_{d_1}\times {\mathbb{R}}_{d_1})$, $k=1,2,\dots ,d_1$, and there exists $0<\epsilon <1$ satisfying

$$|f_k(\cdot ,\cdot ,\cdot ,z_1,p_1)-f_k(\cdot ,\cdot ,\cdot ,z_2,p_2)|\le \epsilon |z_1-z_2|+\epsilon |p_1-p_2|.$$

(A7) For some $C_0>0$, $f=(f_1,f_2,\dots ,f_{d_1})$ satisfies the estimate

$$|f_k(x,t,y,z,p)|\le C_0(f_{0,k}+\sum _{r=1}^{d_1}\left|z^r\right|+\sum _{r=1}^{d_1}\left|p^r\right|)$$

for all $(x,t,y,z,p)\in Q_T\times \mathrm{\Omega }\times R_{d_1}\times R_{d_1}$, where $C_0$ is a positive constant independent of $f,f_{0,k}\in {\mathbb{L}}^2(\mathrm{\Omega })$.

(A8) If $u\in {\mathbb{H}}_{\mathrm{loc}}^2({\mathbb{R}}_d)$, then $F(x,t,u,\mathrm{\nabla }u)\in L^2(0,T;{\mathbb{L}}_{\mathrm{loc}}^2(R))$. If $w_n\to w$ in $L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))$, then $F_k(x,t,w_n,\mathrm{\Delta }{w}_n)\to F_k(x,t,w,\mathrm{\Delta }w)$ in $L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))$.

Using these assumptions, we will elaborate our main result.

Theorem 2.6 Under hypotheses (A1)-(A8), there exists a unique solution $u\in L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))\cap C([0,T];{\mathcal{L}}^2(\mathrm{\Omega }))$ to variational inequality (1) satisfying

$${\parallel u\parallel }_{L^2(0,T;{\mathbb{H}}_0^2(\mathrm{\Omega }))}\le N({\parallel u_0\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}+{\parallel g\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+{\parallel F(\cdot ,\cdot ,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}).$$

The proof of Theorem 2.6 will be given in Section 4.

3 The penalty problem

In order to prove the existence and uniqueness of the solution, we consider the following penalty approximation of problem (1)

$$\{\begin{array}{ll}{(u_{\epsilon }^k)}_t-L_k{u}_{\epsilon }-F_k(x,t,u_{\epsilon },\mathrm{\nabla }{u}_{\epsilon })+{\beta }_{\epsilon }(u_{\epsilon }^k-u_0^k(x))=0,& \text{in }{Q}_T,\\ u_{\epsilon }^k(x,0)=u_0^k(x),& \text{on }\mathrm{\Omega },\\ u_{\epsilon }^k(x,t)=g^k(x,t),& \text{on }\partial \mathrm{\Omega }\times (0,T),\end{array}$$

(5)

where ${\beta }_{\epsilon ,k}(\cdot )$ is the penalty function satisfying

$$\begin{array}{r}0<\epsilon \le 1,{\beta }_{\epsilon ,k}(x)\in C^2(R),{\beta }_{\epsilon ,k}(x)\le 0,{\beta }_{\epsilon ,k}^{\prime }(x)\ge 0,{\beta }_{\epsilon ,k}^{″}(x)\le 0,\\ \underset{\epsilon \to 0}{lim}{\beta }_{\epsilon ,k}(x)=\{\begin{array}{ll}0,& x>0,\\ -\mathrm{\infty },& x<0.\end{array}\end{array}$$

(6)

Next, we introduce a change of variables

$$v(x,t)=u_{\epsilon }(x,t)-\phi (x,t)$$

to transform problem (5) into the zero boundary condition of the form

$$\{\begin{array}{ll}{(v^k)}_t-L_{k}v=F_k(x,t,v+\phi ,\mathrm{\nabla }(v+\phi ))-{\beta }_{\epsilon }(v^k+{\phi }^k-u_0^k(x)),& \text{in }{Q}_T,\\ v^k(x,0)=0,& \text{on }\mathrm{\Omega },\\ v^k(x,t)={\phi }^k(x,t)=0,& \text{on }\partial \mathrm{\Omega }\times (0,T).\end{array}$$

(7)

Here $\phi =({\phi }^1,{\phi }^2,\dots ,{\phi }^{d_1})\in C^{2+\delta ,1+\delta /2}({\overline{Q}}_T)$ is the unique solution to the problem

$$\{\begin{array}{ll}{(u^k)}_t-L_{k}u=0,& \text{in }{Q}_T,\\ u^k(x,0)=u_0^k(x),& \text{on }\mathrm{\Omega },\\ u^k(x,t)=g_k(x,t),& \text{on }\partial \mathrm{\Omega }\times (0,T)\end{array}$$

(8)

satisfying

$${\parallel \phi \parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}\le N({\parallel u_0\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}+{\parallel g\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}).$$

For further details, see Theorem 10.4.1 in [15].

Definition 3.1 v is said to be a weak solution to problem (7) if

$$v\in L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega })),\frac{\partial v}{\partial t}\in L^2(0,T;{\mathbb{H}}^{-1}(\mathrm{\Omega }))$$

and

$$\begin{array}{r}{\int }_{\mathrm{\Omega }}(\frac{\partial v^k}{\partial t}g+\sum _{r=1}^{d_1}\sum _{i,j=1}^d{a}_{kr}^{ij}(x,t){\partial }_i{v}^r\cdot {\partial }_{j}g)\mathrm{d}x\\ +{\int }_{\mathrm{\Omega }}(\sum _{r=1}^{d_1}\sum _{i=1}^d{b}_{kr}^i(x,t){\partial }_i{v}^r\cdot g+\sum _{r=1}^{d_1}{c}_{kr}(x,t)v^r\cdot g)\mathrm{d}x\\ ={\int }_{\mathrm{\Omega }}{F}_k(x,t,v+\phi ,\mathrm{\nabla }(v+\phi ))g\mathrm{d}x-{\int }_{\mathrm{\Omega }}{\beta }_{\epsilon }(v^k+{\phi }^k-u_0^k)g\mathrm{d}x\end{array}$$

(9)

for all $g\in {\mathbb{H}}_0^1(\mathrm{\Omega })$.

Lemma 3.2 If $v\in L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))$ and $\frac{\partial v}{\partial t}\in L^2(0,T;{\mathbb{H}}^{-1}(\mathrm{\Omega }))$, then

$$v\in C([0,T];{\mathbb{L}}^2(\mathrm{\Omega })).$$

The mapping $t\to {\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2$ is absolutely continuous with

$$\frac{\mathrm{d}}{\mathrm{d}t}{\parallel v^k(t)\parallel }_{L^2(\mathrm{\Omega })}^2=2{\int }_{\mathrm{\Omega }}\frac{\partial v^k}{\partial t}{v}^k\mathrm{d}t,\frac{\mathrm{d}}{\mathrm{d}t}{\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2=2\sum _{k=1}^{d_1}{\int }_{\mathrm{\Omega }}\frac{\partial v^k}{\partial t}\cdot v^k\mathrm{d}t$$

for any $0\le t\le T$. The proof is quite standard and can be found in [15].

Lemma 3.3 If v is a weak solution to problem (7), then there exists a positive constant C such that

$${\parallel v(t)\parallel }_{L^2(0,T;{\mathcal{H}}_0^1(\mathrm{\Omega }))}\le C,$$

where C is independent of v.

Proof Choosing $v\in {\mathbb{H}}_0^1(\mathrm{\Omega })$ as the test function in (9), we obtain

$$\begin{array}{r}{\int }_{\mathrm{\Omega }}(\frac{\partial v^k}{\partial t}{v}^k+\sum _{r=1}^{d_1}\sum _{i,j=1}^d{a}_{kr}^{ij}(x,t){\partial }_i{v}^r\cdot {\partial }_j{v}^k)\mathrm{d}x\\ +{\int }_{\mathrm{\Omega }}(\sum _{r=1}^{d_1}\sum _{i,j=1}^d{b}_{kr}^i(x,t){\partial }_i{v}^r\cdot v^k+\sum _{r=1}^{d_1}{c}_{kr}(x,t)v^r{v}^k)\mathrm{d}x\\ ={\int }_{\mathrm{\Omega }}{F}_k(x,t,v+\phi ,\mathrm{\nabla }(v+\phi ))v\mathrm{d}x-{\int }_{\mathrm{\Omega }}{\beta }_{\epsilon }(v^k+{\phi }^k-u_0^k)v^k\mathrm{d}x.\end{array}$$

(10)

From (1), we easily have that

$$v^k+{\phi }^k\ge u_0^k\text{for all }(x,t)\in Q_T.$$

This and (6) lead to

$$\left|{\int }_{\mathrm{\Omega }}{\beta }_{\epsilon }(v^k+{\phi }^k-u_0^k)v^k\mathrm{d}x\right|\le |{\beta }_{\epsilon }(0)|\cdot {\int }_{\mathrm{\Omega }}\left|v^k\right|\mathrm{d}x.$$

(11)

By (A7), so that we get

$$\begin{array}{r}\left|F_k(x,t,v+\phi ,\mathrm{\nabla }(v+\phi ))\right|\\ \le \left|{\int }_{\mathrm{\Omega }}{f}_k(x,t,y,v+\phi ,\mathrm{\nabla }(v+\phi ))\mathrm{d}y\right|\\ \le C_0{\int }_{\mathrm{\Omega }}(f_{0,k}+\sum _{k=1}^{d_1}\left|v^k\right|+\sum _{k=1}^{d_1}\left|\mathrm{\nabla }{v}^k\right|+\sum _{k=1}^{d_1}\left|{\phi }^k\right|+\sum _{k=1}^{d_1}\left|\mathrm{\nabla }{\phi }^k\right|)\mathrm{d}x,\end{array}$$

(12)

where $C_0$ is a positive constant which depends only on the region Ω. It follows by (A2) and (A3) that

$$\theta \sum _{k=1}^{d_1}{\left|\mathrm{\nabla }{v}^k\right|}^2\le \sum _{k,r=1}^{d_1}\sum _{i,j=1}^d{a}_{kr}^{ij}{\partial }_i{v}^r\cdot {\partial }_j{v}^k,$$

(13)

$$\begin{array}{rl}{\int }_{\mathrm{\Omega }}\sum _{k,r=1}^{d_1}\sum _{i=1}^d{b}_{kr}^i(x,t){\partial }_i{v}^r\cdot v^k\mathrm{d}x& \ge B\sum _{k=1}^{d_1}\sum _{i=1}^d{\int }_{\mathrm{\Omega }}{\partial }_i{v}^k\cdot v^k\mathrm{d}x\\ =\frac{B}{2}\sum _{k=1}^{d_1}\sum _{i=1}^d{\int }_{\mathrm{\Omega }}{\partial }_i{\left(v^k\right)}^2\mathrm{d}x=0\end{array}$$

(14)

and

$${\int }_{\mathrm{\Omega }}{c}_{kr}(x,t)v^2\mathrm{d}x\ge 0.$$

(15)

Now, we use Theorem 10.4.1 in [15] and substitute (11)-(15) into (10) to arrive at

$$\begin{array}{r}\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}{\parallel v^k\parallel }_{{\mathcal{L}}^2(\mathrm{\Omega })}^2+\theta \sum _{k=1}^{d_1}{\int }_{\mathrm{\Omega }}{\left|\mathrm{\nabla }{v}^k\right|}^2\mathrm{d}x\\ \le C_0{\int }_{\mathrm{\Omega }}(f_{0,k}+\sum _{k=1}^{d_1}\left|v^k\right|+\sum _{k=1}^{d_1}\left|\mathrm{\nabla }{v}^k\right|+\sum _{k=1}^{d_1}\left|{\phi }^k\right|+\sum _{k=1}^{d_1}\left|\mathrm{\nabla }{\phi }^k\right|)\cdot \sum _{k=1}^{d_1}\left|v^k\right|\mathrm{d}x\\ +|{\beta }_{\epsilon }(0)|\cdot {\int }_{\mathrm{\Omega }}\sum _{k=1}^{d_1}\left|v^k\right|\mathrm{d}x.\end{array}$$

(16)

It follows, by using the Cauchy inequality with $\omega >0$, that

$$\begin{array}{r}\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}{\parallel v^k\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2+\theta \sum _{k=1}^{d_1}{\int }_{\mathrm{\Omega }}{\left|\mathrm{\nabla }{v}^k\right|}^2\mathrm{d}x\\ \le C_0(\omega {\int }_{\mathrm{\Omega }}{f}_{0,k}^2\mathrm{d}x+\frac{1}{4\omega }{\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|v^k\right|\right)}^2\mathrm{d}x)+C_0{\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|v^k\right|\right)}^2\mathrm{d}x\\ +C_0(\omega {\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|\mathrm{\nabla }{v}^k\right|\right)}^2\mathrm{d}x+\frac{1}{4\omega }{\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|v^k\right|\right)}^2\mathrm{d}x)\\ +C_0(\omega {\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|{\phi }^k\right|\right)}^2\mathrm{d}x+\frac{1}{4\omega }{\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|v^k\right|\right)}^2\mathrm{d}x)\\ +C_0(\omega {\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|\mathrm{\nabla }{\phi }^k\right|\right)}^2\mathrm{d}x+\frac{1}{4\omega }{\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|v^k\right|\right)}^2\mathrm{d}x)\\ +|{\beta }_{\epsilon }(0)|\cdot (\omega {\int }_{\mathrm{\Omega }}\mathrm{d}x+\frac{1}{4\omega }{\int }_{\mathrm{\Omega }}{\left(\sum _{k=1}^{d_1}\left|v^k\right|\right)}^2\mathrm{d}x).\end{array}$$

Next, we choose $0<\omega \ll 1$ to arrive at

$$\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}{\parallel v^k\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2+C_1{\parallel v\parallel }_{{\mathbb{H}}_0^1(\mathrm{\Omega })}^2\le C_2{\parallel v\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2+\tilde{C}(t),$$

where $C_1$ and $C_2$ are positive constants and

$$\tilde{C}(t)=C_0\omega {\int }_{\mathrm{\Omega }}(f_{0,k}^2+{\left(\sum _{k=1}^{d_1}\left|{\phi }^k\right|\right)}^2+{\left(\sum _{k=1}^{d_1}\left|\mathrm{\nabla }{\phi }^k\right|\right)}^2)\mathrm{d}x+\omega |{\beta }_{\epsilon }(0)|\cdot {\int }_{\mathrm{\Omega }}1\mathrm{d}x.$$

Note that ${\phi }_k\in C^{2+\delta ,1+\delta /2}({\overline{Q}}_T)$. Therefore we choose a positive constant $C_3={sup}_{0<t<T}\tilde{C}(t)$ to obtain

$$\frac{1}{2}\frac{d}{dt}{\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2+C_1{\parallel v\parallel }_{{\mathbb{H}}_0^1(\mathrm{\Omega })}^2\le C_2{\parallel v\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2+C_3.$$

(17)

On the one hand, letting $\eta (t)={\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2$, (17) gives

$${\eta }^{\prime }(t)\le 2C_2\eta (t)+2C_3.$$

Using the differential form of Gronwall inequality with $\eta (0)={\parallel v(0)\parallel }_{{\mathcal{L}}^2(\mathrm{\Omega })}^2=0$, we get

$${\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2\le 2e^{2C_{2}t}{C}_{3}t$$

and

$$\underset{0<t<T}{max}{\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2\le C_4,$$

(18)

where $C_4$ is a positive constant independent of v.

On the other hand, an integration of (17) from 0 to T with ${\parallel v(0)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2=0$ yields

$$C_1{\int }_0^T{\parallel v\parallel }_{{\mathbb{H}}_0^1(\mathrm{\Omega })}^2\mathrm{d}t\le C_2{\int }_0^T{\parallel v\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2\mathrm{d}t+C_{3}T+\underset{t\to T^{-}}{lim}{\parallel v(t)\parallel }_{{\mathbb{L}}^2(\mathrm{\Omega })}^2.$$

Using inequality (18), we obtain

$${\parallel v(t)\parallel }_{L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))}\le \frac{C_2{C}_4}{C_1}T+\frac{C_3}{C_1}T+\frac{C_4}{C_1}.$$

Thus, the proof is ended by letting $C=\frac{C_2{C}_4}{C_1}T+\frac{C_3}{C_1}T+\frac{C_4}{C_1}>0$. □

Lemma 3.4 The solution to problem (7) satisfies

$$u^k(x,t)\ge u_0^k(x).$$

Proof Using (A8), there exists a positive constant M satisfying

$$|F_k(x,t,u,\mathrm{\nabla }u)|\le M.$$

Here we plan to finish the proof by using contradiction. Assume that ${\bigcup }_{k=1}^{d_1}\{u^k(x,t)<u_0^k(x)\}$ is not empty, that is, at least there exists one k satisfying $u^k<u_0^k$. Let ${\beta }_{\epsilon }(0)=-M$, from (16), one gets

$$u_t-Lu=F_k(x,t,u,\mathrm{\nabla }u)-{\beta }_{\epsilon }(u^k-u_0^k(x))>F_k(x,t,u,\mathrm{\nabla }u)-{\beta }_{\epsilon }(0)\ge 0.$$

Using the standard maximum principle and (A4), one gets

$$u^k(x,t)\ge u_0^k(x).$$

This is obviously contradictory. Therefore, we conclude that $u^k(x,t)\ge u_0^k(x)$ for any $(x,t)\in Q_T$. □

Lemma 3.5 If (A1)-(A8) are satisfied, there exists a unique weak solution $v\in L^2(0,T;{\mathbb{H}}_0^2(\mathrm{\Omega }))\cap C([0,T];{\mathbb{L}}^2(\mathrm{\Omega }))$ to problem (7) satisfying

$${\parallel v\parallel }_{L^2(0,T;{\mathbb{H}}_0^2(\mathrm{\Omega }))}\le N{\parallel F(\cdot ,\cdot ,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+N|{\beta }_{\epsilon }(0)|T,$$

(19)

where N is a positive constant independent of v.

Proof Given $w\in L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))$, set

$$f_w=(f_{1,w},f_{2,w},\dots ,f_{d_1,w}),f_{k,w}(x,t)=F_k(x,t,w+\phi ,\mathrm{\nabla }(w+\phi ))-{\beta }_{\epsilon }(w^k+{\phi }^k-u_0^k).$$

The use of (A8) leads to

$$f_w(x,t)\in L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega })).$$

(20)

By Lemma 2.4, there exists a unique solution $v\in L^2(0,T;{\mathbb{H}}_0^2(\mathrm{\Omega }))\cap C([0,T];{\mathbb{L}}^2(\mathrm{\Omega }))$ to

$$\{\begin{array}{ll}{(v^k)}_t-L_{k}v=f_{k,w}(x,t)& \text{in }{Q}_T,\\ v^k(x,0)=0,& \text{on }\mathrm{\Omega },\\ v^k(x,t)=0,& \text{on }\partial \mathrm{\Omega }\times (0,T).\end{array}$$

(21)

Define the mapping

$$M:L^2(0,T;{\mathbb{H}}_0^2(\mathrm{\Omega }))\to L^2(0,T;{\mathbb{H}}_0^2(\mathrm{\Omega })),w\mapsto M(w)=v,$$

where v is derived from w via (21).

Here we plan to prove the existence and uniqueness by Schaefer’s fixed point theorem. So that we need to present the continuity and compactness of the mapping M. In this proof we only prove the continuity of the mapping M. The compactness can be obtained by following similar arguments, so we omit it here.

Let ${\{w_n\}}_n\in L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))$ be a sequence such that

$$w_k\to w\text{in }{L}^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega })).$$

(22)

By the improved regularity (4), we obtain

$${\parallel v_n\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}\le {\parallel f_{w_k}\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\text{for }{v}_n=M[w_n],n=1,2,\dots .$$

By using (A8) and Lemma 3.4 with ${\beta }_{\epsilon }(x)\le 0$, ${\beta }_{\epsilon }^{\prime }(x)\ge 0$, we have

$$\begin{array}{r}{\parallel f_{w_k}\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\\ \le {\parallel F_k(x,t,w+\phi ,\mathrm{\nabla }w+\mathrm{\nabla }\phi )\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+|{\beta }_{\epsilon }(0)|d_{1}T\\ \le {\parallel F_k(x,t,w+\phi ,\mathrm{\nabla }w+\mathrm{\nabla }\phi )-F_k(x,t,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\\ +{\parallel F_k(x,t,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+|{\beta }_{\epsilon }(0)|d_{1}T\\ \le \alpha {\parallel w\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+\alpha {\parallel \mathrm{\nabla }w\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\\ +{\parallel F_k(x,t,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+|{\beta }_{\epsilon }(0)|d_{1}T\\ \le \alpha {\parallel w\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}+{\parallel F_k(x,t,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+|{\beta }_{\epsilon }(0)|d_{1}T.\end{array}$$

The two inequalities above lead to

$$\begin{array}{r}{\parallel v_n\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}\\ \le \alpha {\parallel w_n\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}+{\parallel F_k(x,t,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+|{\beta }_{\epsilon }(0)|d_{1}T.\end{array}$$

(23)

Next, we pay our attention to the sequence ${\{{\parallel f_{w_n}\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\}}_n$. Since $w_n\to w$ in $L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))$, from (A8) we have

$${\parallel f_{w_n}(x,t)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\to {\parallel f_w(x,t)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}.$$

This and (20) lead to the fact that the sequence ${\{{\parallel f_{w_n}\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\}}_n$ is bounded, that is,

$${\parallel f_{w_n}\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}\le C_6.$$

(24)

Combing (23) with (24), the sequence ${\{v_n\}}_n$ is bounded uniformly in $L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))$. In a similar way, ${\{\frac{\partial v_n}{\partial t}\}}_n$is uniformly bounded in $L^2(0,T;{\mathbb{H}}^{-1}(\mathrm{\Omega }))$. By using Rellich’s theorem (see [16]), there exist a subsequence ${\{v_j\}}_j\in L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))$ and a function $v\in L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))$ which satisfies

$$v_{n_j}\to v\text{in }{L}^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))\text{ as }j\to \mathrm{\infty }$$

such that

$${\int }_{\mathrm{\Omega }}(\frac{\partial v_{n_j}^k}{\partial t}\varphi -L_k{v}_{n_j}\varphi )\mathrm{d}x={\int }_{\mathrm{\Omega }}{f}_{k,w_{n_j}}(x,t)\varphi \mathrm{d}x\text{for each }\varphi \in H_0^1(\mathrm{\Omega }).$$

(25)

We combine (22) with (25) to arrive at

$${\int }_{\mathrm{\Omega }}(\frac{\partial v^k}{\partial t}\varphi -L_{k}v\varphi )\mathrm{d}x={\int }_{\mathrm{\Omega }}{f}_{k,w}(x,t)\varphi \mathrm{d}x.$$

Therefore,

$$v=A[w],A[w_n]\to A[w]\text{in }{L}^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega })).$$

Further, by Lemma 3.3, we have that $\{w\in L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega })):w=A[w]\}$ is bounded. Hence the existence and uniqueness of this theorem are proven by using Lemma 2.3 with $\lambda =1$.

Finally, we pay our attention to the estimate (19). Letting $n\to \mathrm{\infty }$ in (23), we obtain

$${\parallel v\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}\le \alpha {\parallel v\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}+{\parallel F_k(x,t,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+|{\beta }_{\epsilon }(0)|d_{1}T.$$

It follows by $0<\alpha <1$ that

$${\parallel v\parallel }_{L^2(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))}\le \frac{1}{1-\alpha }{\parallel F_k(x,t,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+\frac{1}{1-\alpha }|{\beta }_{\epsilon }(0)|d_{1}T.$$

Therefore, the proof is complete. □

4 The proof of the main result

In this section, we prove that the solution to problem (5) converges to that of problem (1) when $\epsilon \to 0$. From Lemma 3.3 and Lemma 3.5, we conclude that $u_{\epsilon }$ in (7) exists a unique solution satisfying

$$\begin{array}{c}{\parallel u_{\epsilon }\parallel }_{L^2(0,T;{\mathbb{H}}_0^1(\mathrm{\Omega }))}\le C,\hfill \\ {\parallel u_{\epsilon }\parallel }_{L^2(0,T;{\mathbb{H}}_0^2(\mathrm{\Omega }))}\le N{\parallel F(\cdot ,\cdot ,0,0)\parallel }_{L^2(0,T;{\mathbb{L}}^2(\mathrm{\Omega }))}+N|{\beta }_{\epsilon }(0)|d_{1}T.\hfill \end{array}$$

Thus, there exists a subsequence of $\{u_{\epsilon }\}$ still denoted by itself for convenience, and $u\in L(0,T;{\mathbb{H}}^2(\mathrm{\Omega }))$ such that

$$\begin{array}{c}{u}_{\epsilon }^k\stackrel{\mathrm{weak}}{\to }{u}^k\text{in }L(0,T;H^2(\mathrm{\Omega })),k=1,2,\dots ,d_1,\hfill \\ u_{\epsilon }^k\stackrel{\mathrm{uniformly}}{\to }{u}^k\text{in }C({\mathrm{\Omega }}_T),k=1,2,\dots ,d_1.\hfill \end{array}$$

Letting $\epsilon \to 0$ in (5) and using the maximum principle, we arrive at

$$\{\begin{array}{l}{(u^k)}_t-L_{k}u-F_k(x,t,u,\mathrm{\nabla }u)\ge 0,\\ u^k(x,0)=u_0^k(x),\\ u^k(x,t)=g_k(x).\end{array}$$

(26)

Comparing (1) with (26), we only need to prove that

$$[{\left(u^k\right)}_t-L_{k}u-F_k(x,t,u,\mathrm{\nabla }u)](u^k(x,t)-u_0^k(x))=0$$

holds. Note that $u^k(x,0)\ge u_0^k(x)$ derived from Lemma 3.3. Thus, we can end the proof by showing

$${\left(u^k\right)}_t-L_{k}u-F_k(x,t,u,\mathrm{\nabla }u)=0\text{when }{u}^k(x,t)\ge u_0^k(x).$$

(27)

In fact, there exists a positive constant such that for all $(x_0,t_0)\in \{u^k(x,t)>u_0^k(x)\}$,

$$u_{\epsilon }^k(x_0,t_0)>\alpha +u_0^k(x_0)$$

holds when ε is sufficiently small. Further, from (15) we have that

$$0\ge {\beta }_{\epsilon }(u_{\epsilon }^k-u_0^k)\ge {\beta }_{\epsilon }(\alpha )\to 0.$$

(28)

Therefore, we conclude that

$${\left(u^k\right)}_t-L_{k}u-F_k(x,t,u,\mathrm{\nabla }u)=0\text{in }L(0,T;H^2(\mathrm{\Omega })).$$

Now, we prove the uniqueness by contradiction. Assume that $u_1=(u_1^1,u_1^2,\dots ,u_1^{d_1})$ and $u_2=(u_2^1,u_2^2,\dots ,u_2^{d_1})$ are the solutions of (1) and $u_1\ne u_2$. That is, there exists at least one k satisfying $u_1^k\ne u_2^k$. For simplicity, we assume that $\{u_1^k>u_2^k\}$ is not empty (if not, we assume $\{u_1^k<u_2^k\}$ is not empty), so that we have

$$u_1^k>u_2^k\ge u_0^k(x)\text{in }\{u_1^k>u_2^k\}.$$

This and (27) lead to

$$\begin{array}{c}{\left(u_1^k\right)}_t-L_k{u}_1-F_k(x,t,u_1,\mathrm{\Delta }{u}_1)=0\text{in }\{u_1^k>u_2^k\},\hfill \\ {\left(u_2^k\right)}_t-L_k{u}_2-F_k(x,t,u_2,\mathrm{\Delta }{u}_2)\ge 0\text{in }\{u_1^k>u_2^k\},\hfill \\ {(u_1^k-u_2^k)}_t-L_k(u_1-u_2)-(F_k(x,t,u_1,\mathrm{\Delta }{u}_1)-F_k(x,t,u_2,\mathrm{\Delta }{u}_2))\le 0.\hfill \end{array}$$

By the maximum principle and (A8), we have that

$$u_1^k-u_2^k\le 0\text{in }\{u_1^k>u_2^k\}.$$

This is obviously contradictory. Therefore, we conclude that problem (1) has a unique solution. Moreover the estimate can be easily obtained by Lemma 3.5 and (28) with $\epsilon \to 0$.