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Solutions for linear conservation laws with gradient constraints

José Francisco Rodrigues    Lisa Santos
Abstract

We consider variational inequality solutions with prescribed gradient constraints for first order linear boundary value problems. For operators with coefficients only in L2L^{2}, we show the existence and uniqueness of the solution by using a combination of parabolic regularization with a penalization in the nonlinear diffusion coefficient. We also prove the continuous dependence of the solution with respect to the data, as well as, in a coercive case, the asymptotic stabilization as time t+t\rightarrow+\infty towards the stationary solution. In a particular situation, motivated by the transported sandpile problem, we give sufficient conditions for the equivalence of the first order problem with gradient constraint with a two obstacles problem, the obstacles being the signed distances to the boundary. This equivalence, in special conditions, illustrates also the possible stabilization of the solution in finite time.


Dedicado a João Paulo Dias, no seu ativo septuagésimo aniversário!


1 Introduction

Several works have developed solutions u=u(x,t)u=u(x,t) to the linear equation of first order

tu+𝒃u+cu=f,\partial_{t}u+\boldsymbol{b}\cdot\nabla u+cu=f, (1)

for t>0t>0 and xx in an open subset Ω\Omega of N\mathbb{R}^{N}, where 𝒃=𝒃(x,t)\boldsymbol{b}=\boldsymbol{b}(x,t) is a given vector field and c=c(x,t)c=c(x,t) and f=f(x,t)f=f(x,t) are given functions.

The well-known DiPerna and Lions theory of renormalized solutions, when 𝒃\boldsymbol{b} is given in Sobolev spaces, has been extended by Ambrosio to BV coefficients for the Cauchy problem and has found several applications in the study of hyperbolic systems of multidimensional conservation laws (see, for instance [1], for an introduction and references). The initial-boundary value problem for (1) with a C1C^{1} vector field 𝒃\boldsymbol{b} has been studied in the pioneer work of Bardos [2] using essentially a L2L^{2} approach for the transport operator. This method also holds for Lipschitz vector fields, as observed in [8], and was extended by Boyer [5] for solenoidal vector fields in Sobolev spaces that do not need to be tangential to the boundary of Ω\Omega, i.e. 𝒃𝒏0\boldsymbol{b}\cdot\boldsymbol{n}\neq 0 on Ω\partial\Omega for t>0t>0.

The delicate point is then to prescribe the boundary data to the normal trace of 𝒃\boldsymbol{b} on the portion of the space-time boundary ΓΩ×(0,T)\Gamma_{-}\subset\partial\Omega\times(0,T) where the characteristics are entering the domain QT=Ω×(0,T)Q_{T}=\Omega\times(0,T). In the case when Γ\Gamma_{-} does not vary with tt, Besson and Pousin [3] have treated the initial-inflow problems for the continuity equation (1) with 𝑳\boldsymbol{L}^{\infty} velocity fields 𝒃\boldsymbol{b} with c=𝒃=div𝒃c=\nabla\cdot\boldsymbol{b}=\text{div}\,\boldsymbol{b} also in L(QT)L^{\infty}(Q_{T}). Recently Crippa et al. [7] have also considered this problem without that restriction on Γ\Gamma_{-} and with similar assumptions on 𝒃\boldsymbol{b} in BV.

Here we are interested in the initial-boundary value problem for (1) under the additional gradient constraint

|u(x,t)|g(x,t),(x,t)QT,|\nabla u(x,t)|\leq g(x,t),\quad(x,t)\in Q_{T}, (2)

where g=g(x,t)g=g(x,t) is a given strictly positive and bounded function. This problem was already considered in [20] in the framework of a quasilinear continuity equation

tu+𝚽(u)=F(u)\partial_{t}u+\nabla\cdot\boldsymbol{\Phi}(u)=F(u) (3)

and a Lipschitz semilinear lower order term F=F(x,t,u)F=F(x,t,u), with a gradient bound in (2) that may depend also on the solution but not on time. As observed in [20], in the linear transport equation (1), corresponding to

𝚽(u)=𝒃uandF(u)=f+(𝒃c)u\boldsymbol{\Phi}(u)=\boldsymbol{b}u\quad\text{and}\quad F(u)=f+\big{(}\nabla\cdot\boldsymbol{b}-c\big{)}u

with regular coefficients and g=g(x)g=g(x) independent of tt, the problem is well-posed in terms of a first order variational inequality with the convex set

𝕂g={vH01(Ω):|v(x)|g(x) a.e. xΩ}.\mathbb{K}_{g}=\big{\{}v\in H^{1}_{0}(\Omega):|\nabla v(x)|\leq g(x)\text{ a.e. }x\in\Omega\big{\}}. (4)

In [20] it is also proved the existence and asymptotic behaviour of quasivariational solutions for positive nonlinear gradient constraints g=g(x,u)g=g(x,u) depending continuously on the solution u=u(x,t)u=u(x,t). Here H01(Ω)H^{1}_{0}(\Omega) denotes the usual Sobolev space of functions vanishing on the boundary Ω\partial\Omega, as the gradient bound allows to prescribe values on the whole boundary. Moreover, it allows also to consider the data 𝒃\boldsymbol{b}, cc and ff only in L2(QT)L^{2}(Q_{T}), provided c12𝒃c-\frac{1}{2}\nabla\cdot\boldsymbol{b} is bounded from below.

A motivation for the constraint (2) applied to the equation (1) is the “transported sandpile” problem. Following Prigozhin [14, 15], the gradient of the shape of a growing pile of grains z=z(x,t)z=z(x,t) characterized by its angle of repose α>0\alpha>0 is constrained by its surface slope, i.e. g=arctanαg=\arctan\alpha. A general conservation of mass, in the form (3) with 𝚽=μu+𝒃u\boldsymbol{\Phi}=-\mu\nabla u+\boldsymbol{b}u and source density FF, with transport directed by 𝒃\boldsymbol{b} and dropping flow directed to the steepest descent μu-\mu\nabla u, should be then subjected to the unilateral conditions

μ0,|u|gand|u|<gμ=0.\mu\geq 0,\quad|\nabla u|\leq g\quad\text{and}\quad|\nabla u|<g\Rightarrow\mu=0.

We illustrate this problem with the interesting example of the one dimensional special case announced in [19]: Ω=(0,1)\Omega=(0,1), 𝒃=1=g\boldsymbol{b}=1=g, i.e. α=π4\alpha=\frac{\pi}{4} and f(x,t)=tf(x,t)=t. Taking as initial condition the parabola z0(x)=12x2z_{0}(x)=-\frac{1}{2}x^{2}, up to the point ξ0=31\xi_{0}=\sqrt{3}-1, and the straight line z0(x)=x1z_{0}(x)=x-1, for ξ0x1\xi_{0}\leq x\leq 1, the profile of the “transported sandpile” growth attains a steady state exactly at t=54t=\frac{5}{4}. This happens with the first free boundary point ξ(t)\xi(t) increasing from ξ0\xi_{0} up to t=12t=\frac{1}{2}, touching then the boundary x=1x=1, and decreasing till the midpoint x=12x=\frac{1}{2}. At this point, the free boundary ξ(t)\xi(t) meets a second increasing free boundary ζ(t)=2(t1)\zeta(t)=2(t-1), that appears at t=1t=1 and increases up to the final stabilization at t=54.t=\frac{5}{4}.

Refer to caption Refer to caption Refer to caption Refer to caption \begin{array}[]{ccc}\begin{minipage}{113.81102pt} \includegraphics[height=106.69783pt]{C-Ex-RS2014-0.pdf} \end{minipage}\begin{minipage}{113.81102pt} \includegraphics[height=106.69783pt]{C-Ex-RS2014-1.pdf} \end{minipage}\begin{minipage}{113.81102pt} \includegraphics[height=106.69783pt]{C-Ex-RS2014-2.pdf} \end{minipage}\begin{minipage}{113.81102pt} \includegraphics[height=106.69783pt]{C-Ex-RS2014-3.pdf} \end{minipage}\end{array}
Figure 1: The free boundary of the transported sandpile problem at t=0,3/4,9/8t=0,3/4,9/8 and 5/45/4.

The explicit sandpile profile is given by

z(x,t)={tx12x2 if 0xξ(t) and 0t1,x1 if ξ(t)<x1 and 0t12,1x if ξ(t)<x1 and 12<t1,x if 0xζ(t) and 1<t54,tx12x2 if ζ(t)<xξ(t) and 1<t54,x1 if ξ(t)<x1 and 1<t54,12|x12| if t>54,z(x,t)=\left\{\begin{array}[]{ll}tx-\frac{1}{2}x^{2}&\mbox{ if }0\leq x\leq\xi(t)\mbox{ and }0\leq t\leq 1,\vskip 8.53581pt\\ x-1&\mbox{ if }\xi(t)<x\leq 1\mbox{ and }0\leq t\leq\frac{1}{2},\vskip 8.53581pt\\ 1-x&\mbox{ if }\xi(t)<x\leq 1\mbox{ and }\frac{1}{2}<t\leq 1,\vskip 8.53581pt\\ x&\mbox{ if }0\leq x\leq\zeta(t)\mbox{ and }1<t\leq\frac{5}{4},\vskip 8.53581pt\\ tx-\frac{1}{2}x^{2}&\mbox{ if }\zeta(t)<x\leq\xi(t)\mbox{ and }1<t\leq\frac{5}{4},\vskip 8.53581pt\\ x-1&\mbox{ if }\xi(t)<x\leq 1\mbox{ and }1<t\leq\frac{5}{4},\vskip 8.53581pt\\ \frac{1}{2}-|x-\frac{1}{2}|&\mbox{ if }t>\frac{5}{4},\end{array}\right.

where ξ(t)=t1+(1t)2+2\xi(t)=t-1+\sqrt{(1-t)^{2}+2}, if 0t120\leq t\leq\frac{1}{2}, and ξ(t)=t+1(t+1)22\xi(t)=t+1-\sqrt{(t+1)^{2}-2}, if 12<t54\frac{1}{2}<t\leq\frac{5}{4}.

It is clear that z(t)𝕂1H01(Ω)z(t)\in\mathbb{K}_{1}\subset H^{1}_{0}(\Omega).

We introduce the function d(x)=12|x12|d(x)=\frac{1}{2}-|x-\frac{1}{2}| and the convex set

𝕂={vH01(Ω):d(x)v(x)d(x) a.e. x(0,1)}𝕂1.\mathbb{K}_{\vee}^{\wedge}=\big{\{}v\in H^{1}_{0}(\Omega):-d(x)\leq v(x)\leq d(x)\text{ a.e. }x\in(0,1)\big{\}}\supset\mathbb{K}_{1}.

Since tz+xz=t\partial_{t}z+\partial_{x}z=t in A={(x,t)QT:|xz(x,t)|<1}A=\big{\{}(x,t)\in Q_{T}:|\partial_{x}z(x,t)|<1\big{\}}, by simple computation and integration in QTQ_{T}, we easily conclude that zz, which (using =sup\vee=sup and =inf\wedge=inf) can be written as

z(x,t)=(d(x))((tx12x2)d(x)),z(x,t)=\big{(}-d(x)\big{)}\vee\big{(}(tx-\frac{1}{2}x^{2})\wedge d(x)\big{)},

is then the unique solution in 𝕂\mathbb{K}_{\vee}^{\wedge} of the variational inequality

QT(tu+xut)(wu)0w(t)𝕂, 0<t<T,u(0)=z0.\int_{Q_{T}}\big{(}\partial_{t}u+\partial_{x}u-t\big{)}(w-u)\geq 0\quad\forall\,w(t)\in\mathbb{K}_{\vee}^{\wedge},\,0<t<T,\qquad u(0)=z_{0}. (5)

But since z0,z(t)𝕂1z_{0},z(t)\in\mathbb{K}_{1}, zz is also the solution of the variational inequality (5) with w(t)𝕂1𝕂w(t)\in\mathbb{K}_{1}\subset\mathbb{K}_{\vee}^{\wedge}, which has at most one solution also in the convex set 𝕂1\mathbb{K}_{1}, defined as in (4) with g1g\equiv 1.

In Section 2 we establish the existence and the uniqueness of the solution of the first order variational inequality associated with the general linear equation (1) in a family of time dependent convex sets with gradient constraints of the type (4) with g=g(x,t)g=g(x,t). We improve the results of [20] under general square integrability assumptions on the coefficients and on the data, by direct estimates in the parabolic-penalized problem and passage to the limit, first in the penalization parameter ε\varepsilon, and afterwards in the regularization parameter δ\delta. The continuous dependence of the solution with respect to the gradient constraint variations in LL^{\infty}, to the coefficients of the operator and the data in L1L^{1}, is proven in Section 3 under the weak coercive condition (7), as well as the asymptotic convergence towards the unique stationary solution under the stronger coercive assumption (23).

Finaly, in Section 4, we consider the special case of a constant vector 𝒃\boldsymbol{b}, with g=1g=1 and f=f(t)f=f(t) bounded, to show the equivalence of the variational inequalities with the gradient constraint and with the two obstacles, i.e. with the signed distances to the boundary constraints on the solution. This is a first result of this type for first order variational inequalities, similar to the elliptic well-known case of the elastoplastic torsion problem (see, for instance, [16] and its references) and to the parabolic case without convection considered in [21, 22], where it was shown that this equivalence is not always possible in the general case. With additional conditions, that include the above one dimensional transported sand pile problem, we establish the finite time stabilization of the solution. This extends to the convective problem a similar result by Cannarsa et al. [6] and raises the interesting open question of establishing more general conditions on the finite time stabilization of evolutionary problems with gradient constraints.

2 Existence and uniqueness of the variational solution

Let Ω\Omega be a bounded open subset of N\mathbb{R}^{N} with a Lipschitz boundary Ω\partial\Omega and, for any T>0T>0, denote QT=Ω×(0,T)Q_{T}=\Omega\times(0,T).

Assume that

𝒃𝑳2(QT)andcL2(QT),\boldsymbol{b}\in\boldsymbol{L}^{2}(Q_{T})\quad\text{and}\quad c\in L^{2}(Q_{T}), (6)

and there exists ll\in\mathbb{R} such that

c12𝒃linQT,c-\frac{1}{2}\nabla\cdot\boldsymbol{b}\geq l\quad\text{in}\quad Q_{T}, (7)

being this inequality satisfied in the distributional sense, since 𝒃\nabla\cdot\boldsymbol{b} does not need to be a function.

In addition we also suppose given

fL2(QT)andu0𝕂g(0),f\in L^{2}(Q_{T})\quad\text{and}\quad u_{0}\in\mathbb{K}_{g(0)}, (8)

with

gW1,(0,T;L(Ω)),gm>0.g\in W^{1,\infty}\big{(}0,T;L^{\infty}(\Omega)\big{)},\quad g\geq m>0. (9)

As in (4), we define, for t0,t\geq 0,

𝕂g(t)={vH01(Ω):|v(x)|g(x,t) for a.e. in xΩ}.\mathbb{K}_{g(t)}=\big{\{}v\in H^{1}_{0}(\Omega):|\nabla v(x)|\leq g(x,t)\mbox{ for a.e. in }x\in\Omega\big{\}}.

Consider the following variational inequality problem: To find uu, in an appropriate space, such that

{u(t)𝕂g(t) for a.e. t(0,T),u(0)=u0,Ωtu(t)(vu(t))+Ω𝒃(t)u(t)(vu(t))+Ωc(t)u(t)(vu(t))Ωf(t)(vu(t)),v𝕂g(t), for a.e. t(0,T).\begin{array}[]{l}\left\{\begin{array}[]{l}u(t)\in\mathbb{K}_{g(t)}\mbox{ for a.e. }t\in(0,T),\qquad u(0)=u_{0},\\ \\ \displaystyle\int_{\Omega}\partial_{t}u(t)(v-u(t))+\displaystyle\int_{\Omega}\boldsymbol{b}(t)\cdot\nabla u(t)(v-u(t))+\displaystyle\int_{\Omega}c(t)\,u(t)(v-u(t))\geq\displaystyle\int_{\Omega}f(t)(v-u(t)),\\ \\ \hfill{\forall\,v\in\mathbb{K}_{g(t)},\mbox{ for a.e. }t\in(0,T)}.\end{array}\right.\end{array} (10)
Theorem 2.1

With the assumptions (6)-(9), problem (10) has a unique solution

uL(0,T;W01,(Ω))𝒞(Q¯T),tuL2(QT).u\in L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)}\cap\mathscr{C}(\overline{Q}_{T}),\qquad\partial_{t}u\in L^{2}(Q_{T}).

Proof  To prove the uniqueness of the solution we assume there exist two solutions u1u_{1} and u2u_{2}. Using u2=u2(t)u_{2}=u_{2}(t) as test function in (10) for the variational inequality of u1u_{1} and reciprocally, setting u¯=u1u2\bar{u}=u_{1}-u_{2} at a.e. t>0t>0, we obtain

Ωtu¯(t)u¯(t)+Ω𝒃(t)u¯(t)u¯(t)+Ωc(t)u¯2(t)0.\int_{\Omega}\partial_{t}\bar{u}(t)\bar{u}(t)+\int_{\Omega}\boldsymbol{b}(t)\cdot\nabla\bar{u}(t)\,\bar{u}(t)+\int_{\Omega}c(t)\bar{u}^{2}(t)\leq 0.

Using (7), for any v𝒞0(Ω)v\in\mathscr{C}^{\infty}_{0}(\Omega), we have

12Ω𝒃(t)v2+Ωc(t)v2lΩv2\frac{1}{2}\int_{\Omega}\boldsymbol{b}(t)\cdot\nabla v^{2}+\int_{\Omega}c(t)v^{2}\geq l\int_{\Omega}v^{2}

and, by approximation in H01(Ω)H^{1}_{0}(\Omega) of u¯(t)\bar{u}(t), we obtain,

ddtΩ|u¯(t)|2+2lΩ|u¯(t)|20.\frac{d\ }{dt}\int_{\Omega}|\bar{u}(t)|^{2}+2l\int_{\Omega}|\bar{u}(t)|^{2}\leq 0.

By Gronwall’s inequality, we conclude u¯0\bar{u}\equiv 0 from u¯(0)=0\bar{u}(0)=0.

To prove the existence of a solution, we consider a family of approximating quasilinear parabolic problems for uεδu^{\varepsilon\delta}, with ε,δ(0,1)\varepsilon,\delta\in(0,1), defined as follows

{tuεδδ(kε(|uεδ|2g2)uεδ)+𝒃δuεδ+cδuεδ=fδ in QT,uεδ=0 on Ω×(0,T),uεδ(0)=u0ε in Ω0,\left\{\begin{array}[]{l}\partial_{t}u^{\varepsilon\delta}-\delta\nabla\cdot(k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\nabla u^{\varepsilon\delta})+\boldsymbol{b}^{\delta}\cdot\nabla u^{\varepsilon\delta}+c^{\delta}\,u^{\varepsilon\delta}=f^{\delta}\mbox{ in }Q_{T},\vskip 8.53581pt\\ u^{\varepsilon\delta}=0\mbox{ on }\partial\Omega\times(0,T),\vskip 8.53581pt\\ u^{\varepsilon\delta}(0)={u_{0}^{\varepsilon}}\mbox{ in }\Omega_{0},\end{array}\right. (11)

where 𝒃δ\boldsymbol{b}^{\delta}, cδc^{\delta}, fδf^{\delta} and u0εu_{0}^{\varepsilon} are 𝒞\mathscr{C}^{\infty} appropriate regularizations of 𝒃\boldsymbol{b}, cc, ff and u0u_{0}, respectively, with |u0ε|g(0)|\nabla u_{0}^{\varepsilon}|\leq g(0) and kεk_{\varepsilon} is a smooth real function such that kε(s)=1k_{\varepsilon}(s)=1 if s0s\leq 0 and kε(s)=esεk_{\varepsilon}(s)=e^{\frac{s}{\varepsilon}} if sεs\geq\varepsilon. Notice that this problem has a unique solution uεδH1(0,T;L2(Ω))L(0,T;H01(Ω))𝒞(Q¯T)u^{\varepsilon\delta}\in H^{1}\big{(}0,T;L^{2}(\Omega)\big{)}\cap L^{\infty}\big{(}0,T;H^{1}_{0}(\Omega)\big{)}\cap\mathscr{C}(\overline{Q}_{T}), by the classical theory of parabolic quasilinear problems (see, for instance, [10]).

We prove first several a priori estimates.

Estimate 1

kε(|uεδ|2g2)L1(QT)1δC1,\|k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\|_{L^{1}(Q_{T})}\leq\frac{1}{\delta}C_{1}, (12)

for some constant C1C_{1} dependent only on mm, |l||l|, fL2(QT)\|f\|_{L^{2}(Q_{T})}, gL2(QT)\|g\|_{L^{2}(Q_{T})} and u0L2(Ω)\|u_{0}\|_{L^{2}(\Omega)}.

Multiplying the equation of the problem (11) by uεδu^{\varepsilon\delta} and integrating over Qt=Ω×]0,t[Q_{t}=\Omega\times]0,t[, we have

12Ω|uεδ(t)|2+δQtkε(|uεδ|2g2)|uεδ|2+Qt(𝒃δuεδ)uεδ+Qtcδ|uεδ|2=Qtfδuεδ+12Ω|u0ε|2.\frac{1}{2}\int_{\Omega}|u^{\varepsilon\delta}(t)|^{2}+\delta\int_{Q_{t}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})|\nabla u^{\varepsilon\delta}|^{2}+\int_{Q_{t}}\big{(}\boldsymbol{b}^{\delta}\cdot\nabla u^{\varepsilon\delta}\big{)}\,u^{\varepsilon\delta}+\int_{Q_{t}}c^{\delta}\,|u^{\varepsilon\delta}|^{2}=\int_{Q_{t}}f^{\delta}u^{\varepsilon\delta}+\frac{1}{2}\int_{\Omega}|{u_{0}^{\varepsilon}}|^{2}.

Observing that

Qt(𝒃δuεδ)uεδ=12Qt(𝒃δ)(uεδ)2,\int_{Q_{t}}\big{(}\boldsymbol{b}^{\delta}\cdot\nabla u^{\varepsilon\delta}\big{)}\,u^{\varepsilon\delta}=-\frac{1}{2}\int_{Q_{t}}\big{(}\nabla\cdot\boldsymbol{b}^{\delta}\big{)}\,\big{(}u^{\varepsilon\delta}\big{)}^{2},

and using the coercive inequality for the regularized coefficients

cδ12𝒃δ=(c12𝒃)ρδlρδ=l,c^{\delta}-\tfrac{1}{2}\nabla\cdot\boldsymbol{b}^{\delta}=(c-\tfrac{1}{2}\nabla\cdot\boldsymbol{b})*\rho_{\delta}\geq l*\rho_{\delta}=l,

we have

12Ω|uεδ(t)|2+δQtkε(|uεδ|2g2)|uεδ|2fδL2(QT)uεδL2(Qt)+12u0L2(Ω)2+|l|uεδL2(Qt)2.\frac{1}{2}\int_{\Omega}|u^{\varepsilon\delta}(t)|^{2}+\delta\int_{Q_{t}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})|\nabla u^{\varepsilon\delta}|^{2}\leq\|f^{\delta}\|_{L^{2}(Q_{T})}\|u^{\varepsilon\delta}\|_{L^{2}(Q_{t})}+\frac{1}{2}\|u_{0}\|_{L^{2}(\Omega)}^{2}+|l|\|u^{\varepsilon\delta}\|_{L^{2}(Q_{t})}^{2}.

Hence, by the integral Gronwall’s inequality, there exists a positive constant CTC_{T}, independent of ε\varepsilon and δ\delta, such that

uεδL2(QT)CT\|u^{\varepsilon\delta}\|_{L^{2}(Q_{T})}\leq C_{T}

and so

δQtkε(|uεδ|2g2)|uεδ|2C,\delta\int_{Q_{t}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})|\nabla u^{\varepsilon\delta}|^{2}\leq C^{\prime},

where C=C(fL2(QT),u0L2(Ω)),|l|)C^{\prime}=C^{\prime}(\|f\|_{L^{2}(Q_{T})},\|u_{0}\|_{L^{2}(\Omega))},|l|).

On the other hand, we observe that

QTkε(|uεδ|2g2)|uεδ|2=QTkε(|uεδ|2g2)(|uεδ|2g2)+QTkε(|uεδ|2g2)g2.\int_{Q_{T}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})|\nabla u^{\varepsilon\delta}|^{2}=\int_{Q_{T}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\big{(}|\nabla u^{\varepsilon\delta}|^{2}-g^{2}\big{)}+\int_{Q_{T}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})g^{2}. (13)

Since kε(s)=1k_{\varepsilon}(s)=1 for s0s\leq 0 and kε(s)s0k_{\varepsilon}(s)s\geq 0, for all s0s\geq 0, then

Qtkε(|uεδ|2g2)(|uεδ|2g2)={|uεδ|2g2}kε(|uεδ|2g2)(|uεδ|2g2)+{|uεδ|2>g2}kε(|uεδ|2g2)(|uεδ|2g2)QTg2.\int_{Q_{t}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\big{(}|\nabla u^{\varepsilon\delta}|^{2}-g^{2}\big{)}=\int_{\{|\nabla u^{\varepsilon\delta}|^{2}\leq g^{2}\}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\big{(}|\nabla u^{\varepsilon\delta}|^{2}-g^{2}\big{)}\\ +\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}\}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\big{(}|\nabla u^{\varepsilon\delta}|^{2}-g^{2}\big{)}\geq-\int_{Q_{T}}g^{2}. (14)

From (13) and (14) we obtain

Qtkε(|w|2g2)1m2(QTkε(|w|2g2)|w|2+QTg2)1m2(1δC+gL2(QT)2)1δC1,\int_{Q_{t}}k_{\varepsilon}(|\nabla w|^{2}-g^{2})\leq\frac{1}{m^{2}}\left(\int_{Q_{T}}k_{\varepsilon}(|\nabla w|^{2}-g^{2})|\nabla w|^{2}+\int_{Q_{T}}g^{2}\right)\leq\frac{1}{m^{2}}\Big{(}\frac{1}{\delta}C^{\prime}+\|g\|^{2}_{L^{2}(Q_{T})}\Big{)}\leq\frac{1}{\delta}C_{1},

where C1=C1(m,fL2(QT),gL2(QT),u0L2(Ω),|l|)C_{1}=C_{1}(m,\|f\|_{L^{2}(Q_{T})},\|g\|_{L^{2}(Q_{T})},\|u_{0}\|_{L^{2}(\Omega)},|l|).

Estimate 2

uεδ𝑳p(QT)Dδ,\|\nabla u^{\varepsilon\delta}\|_{\boldsymbol{L}^{p}(Q_{T})}\leq D_{\delta}, (15)

where, for any δ>0\delta>0 and any 1p<1\leq p<\infty, the constant DδD_{\delta} depends only on pp, mm, |l||l|, fL2(QT)\|f\|_{L^{2}(Q_{T})}, gL2(QT)\|g\|_{L^{2}(Q_{T})}, u0L2(Ω)\|u_{0}\|_{L^{2}(\Omega)} and a negative power of δ\delta.

From (12) we know that

QTkε(|uεδ|2g2)1δC1,\int_{Q_{T}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\leq\frac{1}{\delta}C_{1},

where C1C_{1} is the positive constant of Estimate 1. So,

1δC1{|uεδ|2>g2+ε}kε(|uεδ|2g2)={|uεδ|2>g2+ε}e|uεδ|2g2ε\frac{1}{\delta}C_{1}\geq\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}+\varepsilon\}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})=\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}+\varepsilon\}}e^{\frac{|\nabla u^{\varepsilon\delta}|^{2}-g^{2}}{\varepsilon}}

and recalling that, for all s>0s>0 and all jj\in\mathbb{N}, essjj!e^{s}\geq\frac{s^{j}}{j!}, we get, for any jj\in\mathbb{N},

{|uεδ|2>g2+ε}(|uεδ|2g2)jj!εj{|uεδ|2>g2+ε}e|uεδ|2g2ε1δj!εjC1.\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}+\varepsilon\}}\left(|\nabla u^{\varepsilon\delta}|^{2}-g^{2}\right)^{j}\leq j!\varepsilon^{j}\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}+\varepsilon\}}e^{\frac{|\nabla u^{\varepsilon\delta}|^{2}-g^{2}}{\varepsilon}}\leq\frac{1}{\delta}j!\varepsilon^{j}C_{1}.

Given 1p<1\leq p<\infty, we have

QT|uεδ|2p={|uεδ|2g2+ε}|uεδ|2p+{|uεδ|2>g2+ε}|uεδ|2p\int_{Q_{T}}|\nabla u^{\varepsilon\delta}|^{2p}=\int_{\{|\nabla u^{\varepsilon\delta}|^{2}\leq g^{2}+\varepsilon\}}|\nabla u^{\varepsilon\delta}|^{2p}+\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}+\varepsilon\}}|\nabla u^{\varepsilon\delta}|^{2p} (16)

and, since gg is bounded, we can estimate, for any pp\in\mathbb{N}, the second integral in the second term of (16) as follows,

{|uεδ|2>g2+ε}|uεδ|2p{|uεδ|2>g2+ε}j=0p(pj)gL(QT)2p2j(|uεδ|2g2)j1δj=0p(pj)gL(QT)2p2jj!εjC1.\displaystyle\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}+\varepsilon\}}|\nabla u^{\varepsilon\delta}|^{2p}\leq\displaystyle{\int_{\{|\nabla u^{\varepsilon\delta}|^{2}>g^{2}+\varepsilon\}}\sum_{j=0}^{p}}\binom{p}{j}\displaystyle{\|g\|_{L^{\infty}(Q_{T})}^{2p-2j}\left(|\nabla u^{\varepsilon\delta}|^{2}-g^{2}\right)^{j}}\\ \displaystyle{\leq\frac{1}{\delta}\sum_{j=0}^{p}}\binom{p}{j}\|g\|_{L^{\infty}(Q_{T})}^{2p-2j}j!\varepsilon^{j}C_{1}.

The first integral in the second term of (16) is clearly bounded since

{|uεδ|2g2+ε}|uεδ|2pQT(g2+1)p\int_{\{|\nabla u^{\varepsilon\delta}|^{2}\leq g^{2}+\varepsilon\}}|\nabla u^{\varepsilon\delta}|^{2p}\leq\int_{Q_{T}}\left(g^{2}+1\right)^{p}

and the conclusion follows easily, first for 2p2p\in\mathbb{N} and afterwards for any 1p<1\leq p<\infty.

Estimate 3

tuεδL2(QT)24(𝒃δ𝑳s(QT)2+C3cδ𝑳s(QT)2)uεδ𝑳q(QT)2+C4,\|\partial_{t}u^{\varepsilon\delta}\|_{L^{2}(Q_{T})}^{2}\leq 4\big{(}\|\boldsymbol{b}^{\delta}\|^{2}_{\boldsymbol{L}^{s}(Q_{T})}+C_{3}\|c^{\delta}\|^{2}_{\boldsymbol{L}^{s}(Q_{T})}\big{)}\|\nabla u^{\varepsilon\delta}\|^{2}_{\boldsymbol{L}^{q}(Q_{T})}+C_{4}, (17)

where, for 2<s<2NN22<s<\frac{2N}{N-2}, C3C_{3} is an upper bound, independent of q=2ss2q=\frac{2s}{s-2}, of the Poincaré constant for W01,q(Ω)W^{1,q}_{0}(\Omega) and C4C_{4} is a positive constant depending only on mm, |l||l|, fL2(QT)\|f\|_{L^{2}(Q_{T})}, gW1,(0,T;L(Ω))2\|g\|^{2}_{W^{1,\infty}(0,T;L^{\infty}(\Omega))} and u0L2(Ω)\|u_{0}\|_{L^{2}(\Omega)}.

We multiply the equation of problem (11) by tuεδ\partial_{t}u^{\varepsilon\delta} and we integrate over QtQ_{t}, noting that tuεδ=0\partial_{t}u^{\varepsilon\delta}=0 on Ω×(0,T)\partial\Omega\times(0,T). Denoting ϕε(s)=0skε(τ)𝑑τ\phi_{\varepsilon}(s)=\displaystyle{\int_{0}^{s}k_{\varepsilon}(\tau)d\tau}, we have

Qt|tuεδ|2+δ2Qtddt(ϕε(|uεδ|2g2))+δQtkε(|uεδ|2g2)gtg+Qt(𝒃δuεδ)tuεδ+Qtcδuεδtuεδ=Qtfδtuεδ.\int_{Q_{t}}|\partial_{t}u^{\varepsilon\delta}|^{2}+\frac{\delta}{2}\int_{Q_{t}}\frac{d\ }{dt}\left(\phi_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\right)+\delta\int_{Q_{t}}k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})g\partial_{t}g\\ +\int_{Q_{t}}(\boldsymbol{b}^{\delta}\cdot\nabla u^{\varepsilon\delta})\,\partial_{t}u^{\varepsilon\delta}+\int_{Q_{t}}c^{\delta}\,u^{\varepsilon\delta}\,\partial_{t}u^{\varepsilon\delta}=\int_{Q_{t}}f^{\delta}\partial_{t}u^{\varepsilon\delta}.

We choose 2<s<2NN22<s<\frac{2N}{N-2} and q=2ss2q=\frac{2s}{s-2}, and so we have 1s+1q+12=1\frac{1}{s}+\frac{1}{q}+\frac{1}{2}=1. Then

|Qt(𝒃δuεδ)tuεδ|𝒃δLs(QT)uεδLq(QT)tuεδL2(QT)𝒃δLs(QT)2uεδLq(QT)2+14tuεδL2(QT)2,\Big{|}\int_{Q_{t}}(\boldsymbol{b}^{\delta}\cdot\nabla u^{\varepsilon\delta})\,\partial_{t}u^{\varepsilon\delta}\Big{|}\leq\|\boldsymbol{b}^{\delta}\|_{L^{s}(Q_{T})}\|\nabla u^{\varepsilon\delta}\|_{L^{q}(Q_{T})}\|\partial_{t}u^{\varepsilon\delta}\|_{L^{2}(Q_{T})}\leq\|\boldsymbol{b}^{\delta}\|_{L^{s}(Q_{T})}^{2}\|\nabla u^{\varepsilon\delta}\|_{L^{q}(Q_{T})}^{2}+\frac{1}{4}\|\partial_{t}u^{\varepsilon\delta}\|^{2}_{L^{2}(Q_{T})},

and

|Qtcδuεδtuεδ|cδLs(QT)uεδLq(QT)tuεδL2(QT)cδLs(QT)2uεδLq(QT)2+14tuεδL2(QT)2.\Big{|}\int_{Q_{t}}c^{\delta}\,u^{\varepsilon\delta}\,\partial_{t}u^{\varepsilon\delta}\Big{|}\leq\|c^{\delta}\|_{L^{s}(Q_{T})}\|u^{\varepsilon\delta}\|_{L^{q}(Q_{T})}\|\partial_{t}u^{\varepsilon\delta}\|_{L^{2}(Q_{T})}\leq\|c^{\delta}\|_{L^{s}(Q_{T})}^{2}\|u^{\varepsilon\delta}\|_{L^{q}(Q_{T})}^{2}+\frac{1}{4}\|\partial_{t}u^{\varepsilon\delta}\|^{2}_{L^{2}(Q_{T})}.

So

14QT|tuεδ|2\displaystyle\frac{1}{4}\int_{Q_{T}}|\partial_{t}u^{\varepsilon\delta}|^{2}\leq fδL2(QT)2+δ2Ωϕε(|uεδ(0)|2g2(0))δ2Ωϕε(|uεδ(t)|2g2(t))\displaystyle\|f^{\delta}\|_{L^{2}(Q_{T})}^{2}+\frac{\delta}{2}\int_{\Omega}\phi_{\varepsilon}\big{(}|\nabla u^{\varepsilon\delta}(0)|^{2}-g^{2}(0)\big{)}-\frac{\delta}{2}\int_{\Omega}\phi_{\varepsilon}\big{(}|\nabla u^{\varepsilon\delta}(t)|^{2}-g^{2}(t)\big{)}
+C1gL(QT)tgL(QT)+(𝒃δLs(QT)2+CqcδLs(QT)2)uεδLq(QT)2,\displaystyle+C_{1}\|g\|_{L^{\infty}(Q_{T})}\|\partial_{t}g\|_{L^{\infty}(Q_{T})}+\big{(}\|\boldsymbol{b}^{\delta}\|_{L^{s}(Q_{T})}^{2}+C_{q}\|c^{\delta}\|_{L^{s}(Q_{T})}^{2}\big{)}\|\nabla u^{\varepsilon\delta}\|_{L^{q}(Q_{T})}^{2},

being CqC_{q} a Poincaré constant. Observe that, since Ω\Omega is bounded we may find a positive upper bound C3C_{3} of CqC_{q}, independently of qq\leq\infty.

On one hand

Ωϕε(|uεδ(0)|2g2(0))0,\displaystyle\int_{\Omega}\phi_{\varepsilon}\big{(}|\nabla u^{\varepsilon\delta}(0)|^{2}-g^{2}(0)\big{)}\leq 0,

because |uεδ(0)|=|u0ε|g(0)|\nabla u^{\varepsilon\delta}(0)|=|\nabla{u_{0}^{\varepsilon}}|\leq g(0). On the other hand, if we set Λ={(x,t)QT:|uεδ(x,t)|<g(x,t)}\Lambda=\{(x,t)\in Q_{T}:|\nabla u^{\varepsilon\delta}(x,t)|<g(x,t)\}, we have

ϕε(|uεδ(x,t)|2g2(x,t))=|uεδ(x,t)|2g2(x,t)g2(x,t)for a.e. (x,t)Λ,\displaystyle\qquad\phi_{\varepsilon}(|\nabla u^{\varepsilon\delta}(x,t)|^{2}-g^{2}(x,t))=|\nabla u^{\varepsilon\delta}(x,t)|^{2}-g^{2}(x,t)\geq-g^{2}(x,t)\qquad\mbox{for a.e. }(x,t)\in\Lambda,
ϕε(|uεδ(x,t)|2g2(x,t))0g2(x,t)for a.e. (x,t)QTΛ.\displaystyle\qquad\phi_{\varepsilon}(|\nabla u^{\varepsilon\delta}(x,t)|^{2}-g^{2}(x,t))\geq 0\geq-g^{2}(x,t)\qquad\mbox{for a.e. }(x,t)\in Q_{T}\setminus\Lambda.

Consequently, for a.e. t(0,T)t\in(0,T),

Ωϕε(|uεδ(t)|2g2(t))gL(0,T;L2(Ω))2.-\int_{\Omega}\phi_{\varepsilon}(|\nabla u^{\varepsilon\delta}(t)|^{2}-g^{2}(t))\leq\|g\|^{2}_{L^{\infty}(0,T;L^{2}(\Omega))}.

So,

14tuεδL2(QT)2fδL2(QT)2+δ2gL(0,T;L2(Ω))2+C1gL(QT)tgL(QT)+(𝒃δLs(QT)2+CqcδLs(QT)2)uεδLq(QT)2,\frac{1}{4}\|\partial_{t}u^{\varepsilon\delta}\big{\|}^{2}_{L^{2}(Q_{T})}\leq\|f^{\delta}\|^{2}_{L^{2}(Q_{T})}+\frac{\delta}{2}\|g\|^{2}_{L^{\infty}(0,T;L^{2}(\Omega))}\\ +C_{1}\|g\|_{L^{\infty}(Q_{T})}\|\partial_{t}g\|_{L^{\infty}(Q_{T})}+\big{(}\|\boldsymbol{b}^{\delta}\|_{L^{s}(Q_{T})}^{2}+C_{q}\|c^{\delta}\|_{L^{s}(Q_{T})}^{2}\big{)}\|\nabla u^{\varepsilon\delta}\|_{L^{q}(Q_{T})}^{2},

and the proof of Estimate 3 is concluded.

By (15) and (17), we know there exist constants DδD_{\delta}, CδC_{\delta} and C4C_{4}, independent of ε\varepsilon, such that, for each N<p<N<p<\infty,

uεδLp(0,T;W01,p(Ω))Dδ,tuεδL2(QT)(𝒃δ𝑳s(QT)+cδLs(QT))Cδ+C4.\|u^{\varepsilon\delta}\|_{L^{p}(0,T;W^{1,p}_{0}(\Omega))}\leq D_{\delta},\qquad\|\partial_{t}u^{\varepsilon\delta}\|_{L^{2}(Q_{T})}\leq\big{(}\|\boldsymbol{b}^{\delta}\|_{\boldsymbol{L}^{s}(Q_{T})}+\|c^{\delta}\|_{L^{s}(Q_{T})}\big{)}C_{\delta}+C_{4}.

Since uεδu^{\varepsilon\delta} is bounded in H1(0,T;L2(Ω))Lp(0,T;W01,p(Ω)𝒞0,1N/p(Ω¯))H^{1}\big{(}0,T;L^{2}(\Omega)\big{)}\cap L^{p}\big{(}0,T;W^{1,p}_{0}(\Omega)\cap\mathscr{C}^{0,1-N/p}(\overline{\Omega})\big{)}, independently of ε(0,1)\varepsilon\in(0,1), for p>Np>N, by a known compactness theorem ([23], page 84), {uεδ}ε\{u^{\varepsilon\delta}\}_{\varepsilon} is relatively compact in 𝒞([0,T];𝒞(Ω¯))\mathscr{C}\big{(}[0,T];\mathscr{C}(\overline{\Omega})\big{)}. Then, at least for a subsequence,

uεδε0uδ in 𝒞(Q¯T).u^{\varepsilon\delta}\underset{\varepsilon\rightarrow 0}{\longrightarrow}u^{\delta}\qquad\mbox{ in }\mathscr{C}(\overline{Q}_{T}).

The above estimates also imply that we may choose, always with fixed δ\delta,

uεδ–⇀ε0uδ weakly in Lp(0,T;W01,pΩ)), 1p<,tuεδ–⇀ε0tuδ weakly in L2(QT).u^{\varepsilon\delta}\underset{\varepsilon\rightarrow 0}{\relbar\joinrel\relbar\joinrel\rightharpoonup}u^{\delta}\quad\mbox{ weakly in }L^{p}(0,T;W^{1,p}_{0}\Omega)),\ 1\leq p<\infty,\qquad\partial_{t}u^{\varepsilon\delta}\underset{\varepsilon\rightarrow 0}{\relbar\joinrel\relbar\joinrel\rightharpoonup}\partial_{t}u^{\delta}\quad\mbox{ weakly in }L^{2}(Q_{T}).

Given vL(0,T;H01(Ω))v\in L^{\infty}(0,T;H^{1}_{0}(\Omega)) such that v(t)𝕂g(t)v(t)\in\mathbb{K}_{g(t)} for a.e. t(0,T)t\in(0,T), we multiply the equation of problem (11) by v(t)uεδ(t)v(t)-u^{\varepsilon\delta}(t), we use the monotonicity of kεk_{\varepsilon} and we integrate over Ω×(s,t)\Omega\times(s,t), 0s<tT0\leq s<t\leq T, to conclude that

stΩtuεδ(vuεδ)+δstΩv(vuεδ)+stΩ𝒃δuεδ(vuεδ)+stΩcuεδ(vuεδ)stΩfδ(vuεδ).\int_{s}^{t}\int_{\Omega}\partial_{t}u^{\varepsilon\delta}(v-u^{\varepsilon\delta})+\delta\int_{s}^{t}\int_{\Omega}\nabla v\cdot\nabla(v-u^{\varepsilon\delta})\\ +\int_{s}^{t}\int_{\Omega}\boldsymbol{b}^{\delta}\cdot\nabla u^{\varepsilon\delta}(v-u^{\varepsilon\delta})+\int_{s}^{t}\int_{\Omega}cu^{\varepsilon\delta}(v-u^{\varepsilon\delta})\geq\int_{s}^{t}\int_{\Omega}f^{\delta}(v-u^{\varepsilon\delta}).

Letting ε0\varepsilon\rightarrow 0, since ss and tt are arbitrary, we obtain that for a.e. t(0,T)t\in(0,T)

Ωtuδ(t)(v(t)uδ(t))+δΩv(t)(v(t)uδ(t))+Ω𝒃δ(t)uδ(t)(v(t)uδ(t))+Ωcδ(t)uδ(t)(v(t)uδ(t))Ωfδ(t)(v(t)uδ(t)),\int_{\Omega}\partial_{t}u^{\delta}(t)(v(t)-u^{\delta}(t))+\delta\int_{\Omega}\nabla v(t)\cdot\nabla(v(t)-u^{\delta}(t))\\ +\int_{\Omega}\boldsymbol{b}^{\delta}(t)\cdot\nabla u^{\delta}(t)\big{(}v(t)-u^{\delta}(t)\big{)}+\int_{\Omega}c^{\delta}(t)\,u^{\delta}(t)\big{(}v(t)-u^{\delta}(t)\big{)}\geq\int_{\Omega}f^{\delta}(t)\big{(}v(t)-u^{\delta}(t)\big{)},

for all vL(0,T;H01(Ω))v\in L^{\infty}(0,T;H^{1}_{0}(\Omega)), such that, v(t)𝕂g(t)v(t)\in\mathbb{K}_{g(t)} for a.e. t(0,T)t\in(0,T).

Set Aε={(x,t)QT:|uεδ(x,t)|2g2(x,t)ε}A_{\varepsilon}=\{(x,t)\in Q_{T}:|\nabla u^{\varepsilon\delta}(x,t)|^{2}-g^{2}(x,t)\geq\sqrt{\varepsilon}\}. Since kε(|uεδ|2g2)e1εk_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})\geq\displaystyle e^{\frac{1}{\sqrt{\varepsilon}}} in AεA_{\varepsilon}, then we have

|Aε|=Aε1Aεkε(|uεδ|2g2)e1εC1δe1ε,\left|A_{\varepsilon}\right|=\int_{A_{\varepsilon}}1\leq\int_{A_{\varepsilon}}\displaystyle{\frac{k_{\varepsilon}(|\nabla u^{\varepsilon\delta}|^{2}-g^{2})}{e^{\frac{1}{\sqrt{\varepsilon}}}}}\leq\frac{C_{1}}{\delta}e^{-\frac{1}{\sqrt{\varepsilon}}},

by (12), being C1C_{1} a constant independent of ε\varepsilon as we have seen. So we have

QT(|uδ|2g2)+lim infε0QT(|uεδ|2g2ε)+=lim infε0Aε(|uεδ|2g2ε)limε0Mδ|Aε|12=0,\int_{Q_{T}}\left(|\nabla u^{\delta}|^{2}-g^{2}\right)^{+}\leq\liminf_{\varepsilon\rightarrow 0}\int_{Q_{T}}\left(|\nabla u^{\varepsilon\delta}|^{2}-g^{2}-\sqrt{\varepsilon}\right)^{+}\\ =\liminf_{\varepsilon\rightarrow 0}\int_{A_{\varepsilon}}\left(|\nabla u^{\varepsilon\delta}|^{2}-g^{2}-\sqrt{\varepsilon}\right)\leq\lim_{\varepsilon\rightarrow 0}\,M_{\delta}\,\left|A_{\varepsilon}\right|^{\frac{1}{2}}=0, (18)

where MδM_{\delta} is an upper bound of |uεδ|2g2εL2(QT)\|\,|\nabla u^{\varepsilon\delta}|^{2}-g^{2}-\sqrt{\varepsilon}\,\|_{L^{2}(Q_{T})}, independent of ε\varepsilon. Consequently,

|uδ|g a.e. in QT|\nabla u^{\delta}|\leq g\qquad\mbox{ a.e. in }Q_{T}

and so uδ(t)𝕂g(t)u^{\delta}(t)\in\mathbb{K}_{g(t)} for a.e. t(0,T)t\in(0,T). Let zL(0,T;H01(Ω))z\in L^{\infty}(0,T;H^{1}_{0}(\Omega)) be such that z(t)𝕂g(t)z(t)\in\mathbb{K}_{g(t)}. Defining v=u+θ(zu)v=u+\theta(z-u), θ(0,1]\theta\in(0,1], then v(t)𝕂g(t)v(t)\in\mathbb{K}_{g(t)}. Using v(t)v(t) as test function in (10) and dividing both sides of the inequality by θ\theta, we get

Ωtuδ(t)(z(t)uδ(t))+δΩuδ(t)(z(t)uδ(t))+δθΩ|(z(t)uδ(t))|2Ω𝒃δ(t)uδ(t)(z(t)uδ(t))+Ωcδ(t)uδ(t)(z(t)uδ(t))Ωfδ(t)(z(t)uδ(t))\int_{\Omega}\partial_{t}u^{\delta}(t)(z(t)-u^{\delta}(t))+\delta\int_{\Omega}\nabla u^{\delta}(t)\cdot\nabla(z(t)-u^{\delta}(t))+\delta\,\theta\int_{\Omega}|\nabla(z(t)-u^{\delta}(t))|^{2}\\ \int_{\Omega}\boldsymbol{b}^{\delta}(t)\cdot\nabla u^{\delta}(t)(z(t)-u^{\delta}(t))+\int_{\Omega}c^{\delta}(t)\,u^{\delta}(t)(z(t)-u^{\delta}(t))\geq\int_{\Omega}f^{\delta}(t)(z(t)-u^{\delta}(t))

and, letting θ0\theta\rightarrow 0, we conclude that uδu^{\delta} solves the following variational inequality

{uδ(t)𝕂g(t) for a.e. t(0,T),uδ(0)=u0,Ωtuδ(t)(vuδ(t))+δΩuδ(t)(vuδ(t))+Ω𝒃δ(t)uδ(t)(vuδ(t))+Ωcδ(t)uδ(t)(vuδ(t))Ωfδ(t)(vuδ(t)),v𝕂g(t), for a.e. t(0,T).\begin{array}[]{l}\left\{\begin{array}[]{l}u^{\delta}(t)\in\mathbb{K}_{g(t)}\mbox{ for a.e. }t\in(0,T),\qquad u^{\delta}(0)=u_{0},\\ \\ \displaystyle\int_{\Omega}\partial_{t}u^{\delta}(t)(v-u^{\delta}(t))+\delta\int_{\Omega}\nabla u^{\delta}(t)\cdot\nabla(v-u^{\delta}(t))\\ \\ \hskip 28.45274pt+\displaystyle\int_{\Omega}\boldsymbol{b}^{\delta}(t)\cdot\nabla u^{\delta}(t)(v-u^{\delta}(t))+\displaystyle\int_{\Omega}c^{\delta}(t)\,u^{\delta}(t)(v-u^{\delta}(t))\geq\displaystyle\int_{\Omega}f^{\delta}(t)(v-u^{\delta}(t)),\\ \\ \hskip 227.62204pt{\forall\,v\in\mathbb{K}_{g(t)},\mbox{ for a.e. }t\in(0,T)}.\end{array}\right.\end{array} (19)

Recalling the Estimate 3 we have

tuεδL2(QT)2\displaystyle\|\partial_{t}u^{\varepsilon\delta}\|_{L^{2}(Q_{T})}^{2} (𝒃δ𝑳s(QT)2+C3cδLs(QT)2)uεδ𝑳2ss2(QT)2+C4\displaystyle\leq\big{(}\|\boldsymbol{b}^{\delta}\|^{2}_{\boldsymbol{L}^{s}(Q_{T})}+C_{3}\|c^{\delta}\|^{2}_{L^{s}(Q_{T})}\big{)}\|\nabla u^{\varepsilon\delta}\|^{2}_{\boldsymbol{L}^{\frac{2s}{s-2}}(Q_{T})}+C_{4}
(𝒃δ𝑳s(QT)2+C3cδLs(QT)2)(QT(|uεδ|2ss2g2ss2ε)++QT(g2ss2+ε))s2s+C4.\displaystyle\leq\big{(}\|\boldsymbol{b}^{\delta}\|^{2}_{\boldsymbol{L}^{s}(Q_{T})}+C_{3}\|c^{\delta}\|^{2}_{L^{s}(Q_{T})}\big{)}\Big{(}\int_{Q_{T}}(|\nabla u^{\varepsilon\delta}|^{\frac{2s}{s-2}}-g^{\frac{2s}{s-2}}-\sqrt{\varepsilon})^{+}+\int_{Q_{T}}(g^{\frac{2s}{s-2}}+\sqrt{\varepsilon})\Big{)}^{\frac{s-2}{s}}+C_{4}.

Passing to the lim inf\liminf when ε0\varepsilon\rightarrow 0 and arguing as in (18), we conclude that

lim infε0QT(|uεδ|2ss2g2ss2ε)+=0\liminf_{\varepsilon\rightarrow 0}\int_{Q_{T}}(|\nabla u^{\varepsilon\delta}|^{\frac{2s}{s-2}}-g^{\frac{2s}{s-2}}-\sqrt{\varepsilon})^{+}=0

and, consequently,

tuδL2(QT)4(𝒃δ𝑳s(QT)2+C3cδLs(QT)2)gL2ss2(QT)2+C4.\|\partial_{t}u^{\delta}\|_{L^{2}(Q_{T})}\leq 4\big{(}\|\boldsymbol{b}^{\delta}\|^{2}_{\boldsymbol{L}^{s}(Q_{T})}+C_{3}\|c^{\delta}\|^{2}_{L^{s}(Q_{T})}\big{)}\|g\|_{L^{\frac{2s}{s-2}}(Q_{T})}^{2}+C_{4}.

Observing that

(𝒃δ𝑳s(QT)2+C3cδLs(QT)2)gL2ss2(QT)2s2(𝒃δ𝑳2(QT)2+C3cδL2(QT))gL(QT)2,\big{(}\|\boldsymbol{b}^{\delta}\|^{2}_{\boldsymbol{L}^{s}(Q_{T})}+C_{3}\|c^{\delta}\|^{2}_{L^{s}(Q_{T})}\big{)}\|g\|_{L^{\frac{2s}{s-2}}(Q_{T})}^{2}\underset{s\rightarrow 2}{\longrightarrow}\big{(}\|\boldsymbol{b}^{\delta}\|^{2}_{\boldsymbol{L}^{2}(Q_{T})}+C_{3}\|c^{\delta}\|_{L^{2}(Q_{T})}\big{)}\|g\|_{L^{\infty}(Q_{T})}^{2},

we have the sequence {tuδ}δ\{\partial_{t}u^{\delta}\}_{\delta} uniformly bounded in L2(QT)L^{2}(Q_{T}).

Moreover, the sequence {uδ}δ\{u^{\delta}\}_{\delta} is uniformly bounded in L(0,T;W01,(Ω))L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)}, independently of δ\delta, since each uδ(t)u^{\delta}(t) belongs to 𝕂g(t)\mathbb{K}_{g(t)}. So, there exists a function uL(0,T;W01,(Ω))H1(0,T;L2(Ω))𝒞(Q¯T)u\in L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)}\cap H^{1}\big{(}0,T;L^{2}(\Omega)\big{)}\cap\mathscr{C}(\overline{Q}_{T}) and, at least for a subsequence,

uδδ0u in 𝒞(Q¯T),u^{\delta}\underset{\delta\rightarrow 0}{\longrightarrow}u\qquad\mbox{ in }\mathscr{C}(\overline{Q}_{T}),
uδ–⇀δ0u weakly in Lp(0,T;W01,p(Ω)), 1p<,tuδ–⇀δ0tu weakly in L2(QT).u^{\delta}\underset{\delta\rightarrow 0}{\relbar\joinrel\relbar\joinrel\rightharpoonup}u\quad\mbox{ weakly in }L^{p}\big{(}0,T;W^{1,p}_{0}(\Omega)\big{)},\ 1\leq p<\infty,\qquad\partial_{t}u^{\delta}\underset{\delta\rightarrow 0}{\relbar\joinrel\relbar\joinrel\rightharpoonup}\partial_{t}u\quad\mbox{ weakly in }L^{2}(Q_{T}).

Integrating in (19) between ss and tt, for 0s<tT0\leq s<t\leq T, and passing to the limit when δ0\delta\rightarrow 0, we get

stΩtu(vu)+stΩ𝒃u(vu)+stΩcu(vu)stΩf(vu),\int_{s}^{t}\int_{\Omega}\partial_{t}u(v-u)+\int_{s}^{t}\int_{\Omega}\boldsymbol{b}\cdot\nabla u(v-u)+\int_{s}^{t}\int_{\Omega}c\,u(v-u)\geq\int_{s}^{t}\int_{\Omega}f(v-u),

for all vv such that v(t)𝕂g(t)v(t)\in\mathbb{K}_{g(t)} for a.e. t(0,T)t\in(0,T). Since ss and tt are arbitrary, we can drop the integration in time. Since uδ(t)𝕂g(t)u^{\delta}(t)\in\mathbb{K}_{g(t)} for a.e. t(0,T)t\in(0,T), the same holds for u(t)u(t), concluding that uu solves the variational inequality (10). \square

Remark 2.2

We observe that in the proof of the uniqueness of the solution it is sufficient to assume only

𝒃𝑳1(QT)andcL1(QT),\boldsymbol{b}\in\boldsymbol{L}^{1}(Q_{T})\quad\text{and}\quad c\in L^{1}(Q_{T}),

instead of (6).

Similarly, we may replace (7) by the different weak coercive assumption by assuming the existence of rr\in\mathbb{R}, such that, in the sense of distributions,

c𝒃r in QT,c-\nabla\cdot\boldsymbol{b}\geq r\quad\text{ in }Q_{T},

in order to have also the uniqueness of the solution to the variational inequality (10).

In fact, assuming that there are two solutions u1u_{1} and u2u_{2}, we may choose for test function v=u1+ζ2sζ(u2u1)v=u_{1}+\zeta^{2}s_{\zeta}(u_{2}-u_{1}) in the variational inequality for u1u_{1}, where sζ:s_{\zeta}:\mathbb{R}\rightarrow\mathbb{R} is a sequence of C1C^{1} increasing odd functions approximating pointwise the sign function sgn0\operatorname{sgn}^{0} and ζ\zeta is sufficient small. Then, choosing also v=u2+ζ2sζ(u1u2)v=u_{2}+\zeta^{2}s_{\zeta}(u_{1}-u_{2}) in the variational inequality for u2u_{2}, we get

Ωtu¯(t)sζ(u¯(t))+Ω𝒃(t)u¯(t)sζ(u¯(t))+Ωc(t)u¯(t)sζ(u¯(t))0\int_{\Omega}\partial_{t}\bar{u}(t)\,s_{\zeta}(\bar{u}(t))+\int_{\Omega}\boldsymbol{b}(t)\cdot\nabla\bar{u}(t)\,s_{\zeta}(\bar{u}(t))+\int_{\Omega}c(t)\,\bar{u}(t)\,s_{\zeta}(\bar{u}(t))\leq 0

Noting Sζ(s)=0ssζ(τ)𝑑τζ0|s|S_{\zeta}(s)=\displaystyle\int_{0}^{s}s_{\zeta}(\tau)\,d\tau\underset{\zeta\rightarrow 0}{\longrightarrow}|s| and τsζ(τ)ζ0|τ|\tau s_{\zeta}(\tau)\underset{\zeta\rightarrow 0}{\longrightarrow}|\tau|, by the dominated convergence theorem, we have

ddtΩ|u¯(t)|+Ω𝒃(t)|u¯(t)|+Ωc(t)|u¯(t)|0\frac{d}{dt}\int_{\Omega}|\bar{u}(t)|+\int_{\Omega}\boldsymbol{b}(t)\cdot\nabla|\bar{u}(t)|+\int_{\Omega}c(t)\,|\bar{u}(t)|\leq 0

and so

ddtΩ|u¯(t)|+rΩ|u¯(t)|0.\frac{d}{dt}\int_{\Omega}|\bar{u}(t)|+r\int_{\Omega}|\bar{u}(t)|\leq 0.

Since u¯(0)=0\bar{u}(0)=0, by the Gronwall’s inequality, we conclude the uniqueness from

Ω|u¯(t)|ertΩ|u¯(0)|=0.\int_{\Omega}|\bar{u}(t)|\leq e^{rt}\int_{\Omega}|\bar{u}(0)|=0.

3 Stability and asymptotic behaviour in time

In this section, the stability of the solutions of the variational inequality (10), as well as its asymptotic limit when t+t\rightarrow+\infty is based in the following Lemma, which is due essentially to [22].

Lemma 3.1

For i=1,2i=1,2, let gig_{i} belong to L(QT)L^{\infty}(Q_{T}). If v1Lq(0,T;W01,p(Ω))v_{1}\in L^{q}\big{(}0,T;W^{1,p}_{0}(\Omega)\big{)}, 1p,q1\leq p,q\leq\infty, is such that v1(t)𝕂g1(t)v_{1}(t)\in\mathbb{K}_{g_{1}(t)} for a.e. t(0,T)t\in(0,T) then there exists v^2Lq(0,T;W01,p(Ω))\widehat{v}_{2}\in L^{q}\big{(}0,T;W^{1,p}_{0}(\Omega)\big{)} such that v^2(t)𝕂g2(t)\widehat{v}_{2}(t)\in\mathbb{K}_{g_{2}(t)} for a.e. t(0,T)t\in(0,T) and a positive constant CC such that

v1v^2Lq(0,T;W01,p(Ω))Cg1g2L(QT).\|v_{1}-\widehat{v}_{2}\|_{L^{q}(0,T;W^{1,p}_{0}(\Omega))}\leq C\|g_{1}-g_{2}\|_{L^{\infty}(Q_{T})}.

Proof  Let α(t)=g1(t)g2(t)L(Ω)\alpha(t)=\|g_{1}(t)-g_{2}(t)\|_{L^{\infty}(\Omega)}. Define ψ(t)=1+α(t)m\displaystyle{\psi(t)=1+\frac{\alpha(t)}{m}} and v^2(t)=1ψ(t)v1(t)\displaystyle{\widehat{v}_{2}(t)=\frac{1}{\psi(t)}\,v_{1}(t)}.

Since

|v^2(t)|=1ψ(t)|v1(t)|1ψ(t)g1(t)|\nabla\widehat{v}_{2}(t)|=\frac{1}{\psi(t)}|\nabla v_{1}(t)|\leq\frac{1}{\psi(t)}g_{1}(t)

and

g1(t)ψ(t)=mm+α(t)g1(t)g2(t)\frac{g_{1}(t)}{\psi(t)}=\frac{m}{m+\alpha(t)}g_{1}(t)\leq g_{2}(t)

then v^2(t)𝕂g2(t)\widehat{v}_{2}(t)\in\mathbb{K}_{g_{2}(t)} for a.e. t(0,T)t\in(0,T). The conclusion follows immediately from

|(v1v^2)|=|11ψ(t)||v1||v1|mg1g2L(QT).|\nabla(v_{1}-\widehat{v}_{2})|=\Big{|}1-\frac{1}{\psi(t)}\Big{|}|\nabla v_{1}|\leq\frac{|\nabla v_{1}|}{m}\,\|g_{1}-g_{2}\|_{L^{\infty}(Q_{T})}.

\square

The continuous dependence result is a consequence of the boundedness of the solution and of its gradient, when we impose the weakly coercive assumption (7).

Theorem 3.2

For i=1,2i=1,2, let uiu_{i} denote the solution of the variational inequality (10) with data (𝐛i,ci,fi,(\boldsymbol{b}_{i},c_{i},f_{i}, gi,u0i)g_{i},u_{0i}) satisfying assumptions (6)-(9). Then there exists a positive constant C=C(T)C=C(T), depending on TT, such that

u1u2L(0,T;L2(Ω))2C(u01u02L2(Ω)2+𝒃1𝒃2𝑳1(QT)+c1c2L1(QT)+f1f2L1(QT)+g1g2L(QT)).\|u_{1}-u_{2}\|_{L^{\infty}(0,T;L^{2}(\Omega))}^{2}\leq C\big{(}\|u_{01}-u_{02}\|_{L^{2}(\Omega)}^{2}+\|\boldsymbol{b}_{1}-\boldsymbol{b}_{2}\|_{\boldsymbol{L}^{1}(Q_{T})}+\|c_{1}-c_{2}\|_{L^{1}(Q_{T})}\\ +\|f_{1}-f_{2}\|_{L^{1}(Q_{T})}+\|g_{1}-g_{2}\|_{L^{\infty}(Q_{T})}\big{)}.

Proof  Let u^2\widehat{u}_{2} be defined as in Lemma 3.1, for the solution u1u_{1} and u^1\widehat{u}_{1} be the corresponding function for u2u_{2}. Using u^1\widehat{u}_{1} as test function in the variational inequality (10), we obtain

Ωtu1(t)(u1(t)u2(t))+Ω𝒃1(t)u1(t)(u1(t)u2(t))+Ωc1(t)u1(t)(u1(t)u2(t))Ωf1(t)(u1(t)u2(t))+Ω(tu1(t)+𝒃1(t)u1(t)+c1(t)u1(t)f1(t))(u^1(t)u2(t))\int_{\Omega}\partial_{t}u_{1}(t)\big{(}u_{1}(t)-u_{2}(t)\big{)}+\int_{\Omega}\boldsymbol{b}_{1}(t)\cdot\nabla u_{1}(t)\big{(}u_{1}(t)-u_{2}(t)\big{)}+\int_{\Omega}c_{1}(t)u_{1}(t)\big{(}u_{1}(t)-u_{2}(t)\big{)}\\ \leq\int_{\Omega}f_{1}(t)\big{(}u_{1}(t)-u_{2}(t)\big{)}+\int_{\Omega}\big{(}\partial_{t}u_{1}(t)+\boldsymbol{b}_{1}(t)\cdot\nabla u_{1}(t)+c_{1}(t)u_{1}(t)-f_{1}(t)\big{)}\,\big{(}\widehat{u}_{1}(t)-u_{2}(t)\big{)}

and a similar inequality is true using the variational inequality of u2u_{2}, by replacing the data f1,𝒃1,c1f_{1},\boldsymbol{b}_{1},c_{1} by f2,𝒃2,c2f_{2},\boldsymbol{b}_{2},c_{2} and u^1\widehat{u}_{1} by u^2\widehat{u}_{2}. Then we have

Ωt(u1(t)u2(t))(u1(t)u2(t))+Ω𝒃1(t)(u1(t)u2(t))(u1(t)u2(t))+Ωc1(t)(u1(t)u2(t))2Θ(t),\int_{\Omega}\partial_{t}\big{(}u_{1}(t)-u_{2}(t)\big{)}\,\big{(}u_{1}(t)-u_{2}(t)\big{)}+\int_{\Omega}\boldsymbol{b}_{1}(t)\cdot\nabla\big{(}u_{1}(t)-u_{2}(t)\big{)}\,\big{(}u_{1}(t)-u_{2}(t)\big{)}\\ +\int_{\Omega}c_{1}(t)\,\big{(}u_{1}(t)-u_{2}(t)\big{)}^{2}\leq\Theta(t), (20)

with

Θ(t)\displaystyle\Theta(t) =Ω(tu1(t)+𝒃1(t)u1(t)+c1(t)u1(t)f1(t))(u^1(t)u2(t))\displaystyle=\int_{\Omega}\big{(}\partial_{t}u_{1}(t)+\boldsymbol{b}_{1}(t)\cdot\nabla u_{1}(t)+c_{1}(t)u_{1}(t)-f_{1}(t)\big{)}\,\big{(}\widehat{u}_{1}(t)-u_{2}(t)\big{)}\hskip 56.9055pt
+Ω(tu2(t)+𝒃(t)u2(t)+c(t)u2(t)f2(t))(u^2(t)u1(t))\displaystyle\ \ +\int_{\Omega}\big{(}\partial_{t}u_{2}(t)+\boldsymbol{b}(t)\cdot\nabla u_{2}(t)+c(t)u_{2}(t)-f_{2}(t)\big{)}\,\big{(}\widehat{u}_{2}(t)-u_{1}(t)\big{)}
+Ω(𝒃1(t)𝒃2(t))u2(t)(u1(t)u2(t))\displaystyle\ \ +\int_{\Omega}\big{(}\boldsymbol{b}_{1}(t)-\boldsymbol{b}_{2}(t)\big{)}\cdot\nabla u_{2}(t)\,\big{(}u_{1}(t)-u_{2}(t)\big{)}
+Ω[(c1(t)c2(t))u2(t)+(f1(t)f2(t))](u1(t)u2(t)).\displaystyle\ \ +\int_{\Omega}\big{[}\big{(}c_{1}(t)-c_{2}(t)\big{)}\,u_{2}(t)+\big{(}f_{1}(t)-f_{2}(t)\big{)}\big{]}\,\big{(}u_{1}(t)-u_{2}(t)\big{)}.

Using the boundedness of the solutions uiu_{i}, i=1,2,i=1,2, and their gradients and recalling the L2(QT)L^{2}(Q_{T}) estimates of tui\partial_{t}u_{i}, we have

0TΘ(τ)𝑑τ\displaystyle\int_{0}^{T}\Theta(\tau)d\tau CM(g1g2L(QT)+𝒃1𝒃2𝑳1(QT)+c1c2L1(QT)+f1f2L1(QT)),\displaystyle\leq C_{M}\big{(}\|g_{1}-g_{2}\|_{L^{\infty}(Q_{T})}+\|\boldsymbol{b}_{1}-\boldsymbol{b}_{2}\|_{\boldsymbol{L}^{1}(Q_{T})}+\|c_{1}-c_{2}\|_{L^{1}(Q_{T})}+\|f_{1}-f_{2}\|_{L^{1}(Q_{T})}\big{)},

where CMC_{M} is a positive constant depending on TT, on the norms of the solutions and their derivatives (which can be bounded in terms of the data) and on the constant CC of Lemma 3.1.

Setting w=u1u2w=u_{1}-u_{2} in the inequality (20), we obtain using (7),

ddtΩ|w(t)|22|ł|Ω|w(t)|2+2Θ(t).\frac{d\ }{dt}\int_{\Omega}|w(t)|^{2}\leq 2|\l |\int_{\Omega}|w(t)|^{2}+2\Theta(t).

Applying Gronwall’s inequality, we conclude

Ω|u1(t)u2(t)|2e2|ł|T(u10u20L2(Ω)2+2CM(g1g2L(QT)+𝒃1𝒃2𝑳1(QT)+c1c2L1(QT)+f1f2L1(QT))).\int_{\Omega}\big{|}u_{1}(t)-u_{2}(t)\big{|}^{2}\leq e^{2|\l |T}\Big{(}\|u_{10}-u_{20}\|_{L^{2}(\Omega)}^{2}+2C_{M}\big{(}\|g_{1}-g_{2}\|_{L^{\infty}(Q_{T})}\\ +\|\boldsymbol{b}_{1}-\boldsymbol{b}_{2}\|_{\boldsymbol{L}^{1}(Q_{T})}+\|c_{1}-c_{2}\|_{L^{1}(Q_{T})}+\|f_{1}-f_{2}\|_{L^{1}(Q_{T})}\big{)}\Big{)}.

\square

In order to consider the corresponding time independent solution to the first order variational inequality, we give stationary data f,g,𝒃,cf_{\infty},g_{\infty},\boldsymbol{b}_{\infty},c_{\infty} satisfying the assumptions

gL(Ω),gm>0,fL1(Ω),g_{\infty}\in L^{\infty}(\Omega),\ g_{\infty}\geq m>0,\qquad f_{\infty}\in L^{1}(\Omega), (21)
𝒃𝑳1(Ω),cL1(Ω),\boldsymbol{b}_{\infty}\in\boldsymbol{L}^{1}(\Omega),\quad c_{\infty}\in L^{1}(\Omega), (22)
c12𝒃λ>0inΩ\ c_{\infty}-\frac{1}{2}\nabla\cdot\boldsymbol{b}_{\infty}\geq\lambda>0\quad\text{in}\quad\Omega (23)

in the distributional sense, where we set accordingly

𝕂g={wH01(Ω):|w|g a.e. in Ω}.\mathbb{K}_{g_{\infty}}=\big{\{}w\in H^{1}_{0}(\Omega):|\nabla w|\leq g_{\infty}\mbox{ a.e. in }\Omega\big{\}}. (24)

Then, the stationary problem can be written as

u𝕂g:Ω𝒃u(wu)+Ωcu(wu)Ωf(wu),w𝕂g.u_{\infty}\in\mathbb{K}_{g_{\infty}}:\quad\int_{\Omega}\boldsymbol{b}_{\infty}\cdot\nabla u_{\infty}(w-u_{\infty})+\int_{\Omega}c_{\infty}\,u_{\infty}(w-u_{\infty})\geq\int_{\Omega}f_{\infty}(w-u_{\infty}),\qquad\forall\,w\in\mathbb{K}_{g_{\infty}}. (25)

Since the convex set 𝕂g\mathbb{K}_{g_{\infty}} is bounded in H01(Ω)H^{1}_{0}(\Omega) and the first order linear operator in the left hand side of (25) is pseudo-monotone, by the classical theory (see, for instance, [12]) it has a solution, which is unique by the strict coerciveness induced by the condition λ>0\lambda>0 in (23).

In order to study the asymptotic convergence of the solution of the variational inequality (10) to the stationary solution of (25), we consider solutions global in time. This is easily obtained if we assume that (6)-(8) are satisfied for any T>0T>0 and replace (9) by

gW1,(0,;L(Ω)),gm>0.\displaystyle g\in W^{1,\infty}\big{(}0,\infty;L^{\infty}(\Omega)\big{)},\ g\geq m>0. (26)

We need an auxiliary lemma.

Lemma 3.3

([9], pg. 286) Let φ:(0,)\varphi:(0,\infty)\rightarrow\mathbb{R} be a nonnegative function, absolutely continuous in any compact subinterval of (0,)(0,\infty), ΦLloc1(0,)\Phi\in L^{1}_{loc}(0,\infty) a nonnegative function and μ\mu a positive constant, such that,

φ(t)+μφ(t)Φ(t),t>0.\varphi^{\prime}(t)+\mu\varphi(t)\leq\Phi(t),\qquad\qquad\forall\,\ t>0. (27)

Then, for any s,t>0s,t>0,

φ(t+s)eμt+11eμ[supτsττ+1Φ(ξ)𝑑ξ].\varphi(t+s)\leq e^{-\mu t}+\frac{1}{1-e^{-\mu}}\left[\sup_{\tau\geq s}\int_{\tau}^{\tau+1}\Phi(\xi)d\xi\right].

\square

In order to apply this Lemma to

φ(t)=Ω|u(t)u|2,t>0,\varphi(t)=\int_{\Omega}\big{|}u(t)-u_{\infty}\big{|}^{2},\qquad\ t>0, (28)

we shall require the additional assumptions on the coefficients and on the data

𝒃L(0,;𝑳2(Ω))andc,fL(0,;L2(Ω)).\boldsymbol{b}\in L^{\infty}\big{(}0,\infty;\boldsymbol{L}^{2}(\Omega)\big{)}\quad\text{and}\quad c,f\in L^{\infty}\big{(}0,\infty;L^{2}(\Omega)\big{)}. (29)
Theorem 3.4

Assume that f,g,𝐛,c,u0f,g,\boldsymbol{b},c,u_{0} satisfy the assumptions (6)-(8), (26), (29) and f,g,𝐛,cf_{\infty},g_{\infty},\boldsymbol{b}_{\infty},c_{\infty} satisfy the assumption (21),(22) and (23). Suppose, in addition, that

tt+1Ω|f(τ)f|𝑑τ𝑑xt0,tt+1Ω|𝒃(τ)𝒃|𝑑τ𝑑xt0,tt+1Ω(c(τ)c)𝑑τ𝑑xt0\int_{t}^{t+1}\int_{\Omega}\big{|}f(\tau)-f_{\infty}\big{|}d\tau dx\underset{t\rightarrow\infty}{\longrightarrow}0,\quad\int_{t}^{t+1}\int_{\Omega}\big{|}\boldsymbol{b}(\tau)-\boldsymbol{b}_{\infty}\big{|}d\tau dx\underset{t\rightarrow\infty}{\longrightarrow}0,\quad\int_{t}^{t+1}\int_{\Omega}\big{(}c(\tau)-c_{\infty}\big{)}d\tau dx\underset{t\rightarrow\infty}{\longrightarrow}0

and there exists γ>12\gamma>\frac{1}{2}, such that, for some constant D>0D>0,

g(t)gL(Ω)Dtγ,t>0.\|g(t)-g_{\infty}\|_{L^{\infty}(\Omega)}\leq\frac{D}{t^{\gamma}},\qquad t>0. (30)

If uu and uu_{\infty} are, respectively, the unique solutions of the variational inequalities (10) and (25) then, for every α\alpha, 0<α<10<\alpha<1,

u(t)tu in 𝒞0,α(Ω¯)u(t)\underset{t\rightarrow\infty}{\longrightarrow}u_{\infty}\quad\mbox{ in }\mathscr{C}^{0,\alpha}(\overline{\Omega})

Proof  First we need to return to the estimate (17) of the existence proof in order to prove that, under the additional assumptions of this theorem, there are positive constants A,BA,B, independent of TT, such that,

tuL2(Ω×(0,T))AT+B.\|\partial_{t}u\|_{L^{2}(\Omega\times(0,T))}\leq A\sqrt{T}+B.

Since |u(x,t)|g(x,t)|\nabla u(x,t)|\leq g(x,t) for a.e. (x,t)Q=Ω×(0,)(x,t)\in Q_{\infty}=\Omega\times(0,\infty) and gL(Q)g\in L^{\infty}(Q_{\infty}), we have now uL(0,;W1,(Ω))u\in L^{\infty}(0,\infty;W^{1,\infty}(\Omega)). This yields the estimate

uL2(QT)2=QT|u|2cgT\|u\|^{2}_{L^{2}(Q_{T})}=\int_{Q_{T}}|u|^{2}\leq c_{g}T

where the constant cg>0c_{g}>0 is independent of TT. Using similar estimates for fL2(QT)2\|f\|^{2}_{L^{2}(Q_{T})} with the constant cgc_{g} replaced by cf=fL2(0,;L2(Ω))2c_{f}=\|f\|^{2}_{L^{2}(0,\infty;L^{2}(\Omega))}, as well as for c𝒃=𝒃𝑳2(0,;L2(Ω))2c_{\boldsymbol{b}}=\|\boldsymbol{b}\|^{2}_{\boldsymbol{L}^{2}(0,\infty;L^{2}(\Omega))} and cc=cL2(0,;L2(Ω))2c_{c}=\|c\|^{2}_{L^{2}(0,\infty;L^{2}(\Omega))}, we may conclude that the constant C1=C1(T)C_{1}=C_{1}(T) of (12), in the Estimate 1, grows also linearly with TT, i.e. C1c0+c1TC_{1}\leq c_{0}+c_{1}T, where c0c_{0} depends only on u0u_{0} and c1c_{1} depends on mm, cfc_{f}, cgc_{g}, c𝒃c_{\boldsymbol{b}} and ccc_{c}. Using this fact in the Estimate 3, we may now easily deduce (3) from (17), with s=2s=2 and q=q=\infty, since C4C_{4}, depending on ff and on C1C_{1} grows also linearly with TT.

Using Lemma 3.1, we choose u^𝕂g(t)\widehat{u}_{\infty}\in\mathbb{K}_{g(t)}, for a.e. t(0,T)t\in(0,T), as test function in (10). Then

Ωtu(t)(u(t)u)+Ω𝒃(t)u(t)(u(t)u)+Ωc(t)u(t)(u(t)u)Ωf(t)(u(t)u)+Ω(tu(t)+𝒃(t)u(t)+c(t)u(t)f(t))(u^u).\int_{\Omega}\partial_{t}u(t)\big{(}u(t)-u_{\infty}\big{)}+\int_{\Omega}\boldsymbol{b}(t)\cdot\nabla u(t)\big{(}u(t)-u_{\infty}\big{)}+\int_{\Omega}c(t)u(t)\big{(}u(t)-u_{\infty}\big{)}\\ \leq\int_{\Omega}f(t)\big{(}u(t)-u_{\infty}\big{)}+\int_{\Omega}\big{(}\partial_{t}u(t)+\boldsymbol{b}(t)\cdot\nabla u(t)+c(t)u(t)-f(t)\big{)}\,\big{(}\widehat{u}_{\infty}-u_{\infty}\big{)}.

Analogously, with u^(t)𝕂g\widehat{u}(t)\in\mathbb{K}_{g_{\infty}}, for a.e. t(0,T)t\in(0,T), we obtain the inequality

Ω𝒃u(u(t)u)+Ωcu(u(t)u)Ωf(u(t)u)+Ω(𝒃u+cuf)(u(t)u^(t)).\int_{\Omega}\boldsymbol{b}_{\infty}\cdot\nabla u_{\infty}\big{(}u(t)-u_{\infty}\big{)}+\int_{\Omega}c_{\infty}u_{\infty}\big{(}u(t)-u_{\infty}\big{)}\\ \geq\int_{\Omega}f_{\infty}\big{(}u(t)-u_{\infty}\big{)}+\int_{\Omega}\big{(}\boldsymbol{b}_{\infty}\cdot\nabla u_{\infty}+c_{\infty}u_{\infty}-f_{\infty}\big{)}\,\big{(}u(t)-\widehat{u}(t)\big{)}.

Then, simple algebraic manipulations lead to

Ωt(u(t)u)(u(t)u)+Ω𝒃(u(t)u)(u(t)u)+Ωc(u(t)u)(u(t)u)Θ(t),\int_{\Omega}\partial_{t}\big{(}u(t)-u_{\infty}\big{)}\big{(}u(t)-u_{\infty}\big{)}+\int_{\Omega}\boldsymbol{b}_{\infty}\cdot\nabla\big{(}u(t)-u_{\infty}\big{)}\big{(}u(t)-u_{\infty}\big{)}+\int_{\Omega}c_{\infty}\big{(}u(t)-u_{\infty}\big{)}\big{(}u(t)-u_{\infty}\big{)}\\ \leq\Theta(t), (31)

where

Θ(t)=Ω(tu(t)+𝒃(t)u(t)+c(t)u(t)f(t))(u^u)+Ω(𝒃u+cuf)(u^(t)u(t))+Ω(𝒃(t)𝒃)u(t)(uu(t))+Ω(c(t)c)u(t)(uu(t))+Ω(f(t)f)(u(t)u).\Theta(t)=\int_{\Omega}\big{(}\partial_{t}u(t)+\boldsymbol{b}(t)\cdot\nabla u(t)+c(t)u(t)-f(t)\big{)}\,\big{(}\widehat{u}_{\infty}-u_{\infty}\big{)}+\int_{\Omega}\big{(}\boldsymbol{b}_{\infty}\cdot\nabla u_{\infty}+c_{\infty}u_{\infty}-f_{\infty}\big{)}\,\big{(}\widehat{u}(t)-u(t)\big{)}\\ +\int_{\Omega}(\boldsymbol{b}(t)-\boldsymbol{b}_{\infty})\cdot\nabla u(t)\big{(}u_{\infty}-u(t)\big{)}+\int_{\Omega}(c(t)-c_{\infty})u(t)\big{(}u_{\infty}-u(t)\big{)}+\int_{\Omega}\big{(}f(t)-f_{\infty}\big{)}\big{(}u(t)-u_{\infty}\big{)}.

Using (23) and the definition (28), from (31), we obtain the differential inequality with μ=2λ\mu=2\lambda and where, taking into account (3), we may choose Φ(t)2|Θ(t)|\Phi(t)\geq 2|\Theta(t)| given by

Φ(t)=C((At+B+C)g(t)gL(Ω)+𝒃(t)𝒃𝑳1(Ω)+c(t)cL1(Ω)+f(t)fL1(Ω)).\Phi(t)=C\big{(}(A\sqrt{t}+B+C)\|g(t)-g_{\infty}\|_{L^{\infty}(\Omega)}+\|\boldsymbol{b}(t)-\boldsymbol{b}_{\infty}\|_{\boldsymbol{L}^{1}(\Omega)}+\|c(t)-c_{\infty}\|_{L^{1}(\Omega)}+\|f(t)-f_{\infty}\|_{L^{1}(\Omega)}\big{)}.

Then, using the assumptions and observing that the number γ\gamma in (30) is greater than 12\frac{1}{2}, we have

tt+1Φ(τ)𝑑τCtt+1(f(τ)fL1(Ω)+𝒃(τ)𝒃L1(Ω)+c(τ)cL1(Ω))𝑑τ+Ctt+1(τ12+1)g(τ)gL(Ω)𝑑τt+0.\int_{t}^{t+1}\Phi(\tau)d\tau\leq C\int_{t}^{t+1}\big{(}\|f(\tau)-f_{\infty}\|_{L^{1}(\Omega)}+\|\boldsymbol{b}(\tau)-\boldsymbol{b}_{\infty}\|_{L^{1}(\Omega)}+\|c(\tau)-c_{\infty}\|_{L^{1}(\Omega)}\big{)}d\tau\\ +C^{\prime}\int_{t}^{t+1}(\tau^{\frac{1}{2}}+1)\|g(\tau)-g_{\infty}\|_{L^{\infty}(\Omega)}d\tau\underset{t\rightarrow+\infty}{\longrightarrow}0.

Therefore, by Lemma 3.3, u(t)t+uu(t)\underset{t\rightarrow+\infty}{\longrightarrow}u_{\infty} in L2(Ω)L^{2}(\Omega). Since uu belongs to L(0,;W1,(Ω))L^{\infty}\big{(}0,\infty;W^{1,\infty}(\Omega)\big{)}, the compact inclusion of W1,(Ω)W^{1,\infty}(\Omega) in C0,α(Ω¯)C^{0,\alpha}(\overline{\Omega}) implies, first for a subsequence, and after for the whole sequence, that u(t)t+uu(t)\underset{t\rightarrow+\infty}{\longrightarrow}u_{\infty} in C0,α(Ω¯)C^{0,\alpha}(\overline{\Omega}), concluding the proof.

\square

4 Finite time stabilization in a special case

In this section we assume that Ω\partial\Omega is of class 𝒞2\mathscr{C}^{2} and

𝒃N,c0,g1,z0𝕂1andfL(0,T).\boldsymbol{b}\in\mathbb{R}^{N},\quad c\equiv 0,\quad g\equiv 1,\quad z_{0}\in\mathbb{K}_{1}\quad\text{and}\quad f\in L^{\infty}(0,T). (32)

We consider the following two obstacles problem

{z(t)𝕂 for a.e. t(0,T),z(0)=u0,Ωtz(t)(vz(t))+Ω𝒃z(t)(vz(t))Ωf(t)(vz(t)),v𝕂, for a.e. t(0,T),\begin{array}[]{l}\left\{\begin{array}[]{l}z(t)\in\mathbb{K}_{\vee}^{\wedge}\mbox{ for a.e. }t\in(0,T),\qquad z(0)=u_{0},\\ \\ \displaystyle\int_{\Omega}\partial_{t}z(t)(v-z(t))+\displaystyle\int_{\Omega}\boldsymbol{b}\cdot\nabla z(t)(v-z(t))\geq\displaystyle\int_{\Omega}f(t)(v-z(t)),\ \forall\,v\in\mathbb{K}_{\vee}^{\wedge},\mbox{ for a.e. }t\in(0,T),\end{array}\right.\end{array} (33)

where

𝕂={vH01(Ω):d(x)v(x)d(x) for a.e. xΩ}.\mathbb{K}_{\vee}^{\wedge}=\{v\in H^{1}_{0}(\Omega):-d(x)\leq v(x)\leq d(x)\text{ for a.e. }x\in\Omega\}.

Here d(x)=d(x,Ω)d(x)=d(x,\partial\Omega) is the distance function to the boundary Ω\partial\Omega. Notice that dW01,(Ω)d\in W^{1,\infty}_{0}(\Omega), |d(x)|1|\nabla d(x)|\leq 1, a.e. xΩx\in\Omega and ΔdC\Delta d\leq C for some constant C=C(Ω)>0C=C(\Omega)>0. Observe that z0𝕂1𝕂z_{0}\in\mathbb{K}_{1}\subset\mathbb{K}_{\vee}^{\wedge}.

Theorem 4.1

Under the assumptions (32), the inequality (33) has a unique solution

zL(0,T;W01,(Ω))H1(0,T;L2(Ω))𝒞(Q¯T),z\in L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)}\cap H^{1}\big{(}0,T;L^{2}(\Omega)\big{)}\cap\mathscr{C}(\overline{Q}_{T}),

which satisfies |z|1|\nabla z|\leq 1 a.e. in QTQ_{T} and is the unique solution of the variational inequality (10).

Proof  For ε,δ(0,1)\varepsilon,\delta\in(0,1), we consider the following family of penalized problems for zεδz^{\varepsilon\delta},

{tzεδδΔzεδ+𝒃zεδ+δε(zεδ(zεδd)(d))=fδin QT,zεδ(0)=z0εon Ω,zεδ=0on Ω×(0,T),\left\{\begin{array}[]{l}\partial_{t}z^{\varepsilon\delta}-\delta\Delta z^{\varepsilon\delta}+\boldsymbol{b}\cdot\nabla z^{\varepsilon\delta}+\frac{\delta}{\varepsilon}\big{(}z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d)\big{)}=f^{\delta}\ \text{in }Q_{T},\vskip 8.53581pt\\ z^{\varepsilon\delta}(0)=z_{0}^{\varepsilon}\ \text{on }\Omega,\qquad z^{\varepsilon\delta}=0\ \text{on }\partial\Omega\times(0,T),\end{array}\right. (34)

where fδf^{\delta} and z0εz_{0}^{\varepsilon} are regularizations of the functions ff and z0z_{0}, with |z0ε|1|\nabla{z_{0}^{\varepsilon}}|\leq 1. This problem has a unique solution zεδH2,1(QT)z^{\varepsilon\delta}\in H^{2,1}(Q_{T}), since the operator

Pεv,w=δεΩ(v(vd)(d))w\langle P_{\varepsilon}v,w\rangle=\frac{\delta}{\varepsilon}\int_{\Omega}\big{(}v-(v\wedge d)\vee(-d)\big{)}w (35)

is monotone (see, for instance, [12]).

We obtain firstly an estimate of |zεδ||\nabla z^{\varepsilon\delta}| on Ω×(0,T)\partial\Omega\times(0,T). Since Ω\partial\Omega is of class 𝒞2\mathscr{C}^{2}, there exists r>0r>0 such that, if Br(x)B_{r}(x) denotes the ball with centre in xx and radius rr, then for all x0Ωx_{0}\in\partial\Omega there exists y0Ny_{0}\in\mathbb{R}^{N} such that B¯r(y0)Ω¯={x0}\overline{B}_{r}(y_{0})\cap\overline{\Omega}=\{x_{0}\}. Placing the origin of the coordinates in the point y0y_{0}, let ηε(s)=esε\eta_{\varepsilon}(s)=e^{-\frac{s}{\sqrt{\varepsilon}}} and

φ¯(x)=d(x)+Mε(1ηε(|x|r)),φ¯(x)=d(x)Mε(1ηε(|x|r)),\overline{\varphi}(x)=d(x)+M\varepsilon\big{(}1-\eta_{\varepsilon}(|x|-r)\big{)},\qquad\underline{\varphi}(x)=-d(x)-M\varepsilon\big{(}1-\eta_{\varepsilon}(|x|-r)\big{)},

where MM is a positive constant, depending on δ\delta, to be chosen later. We show that φ¯\overline{\varphi} is a supersolution of (34). Analogously, it can be verified that φ¯\underline{\varphi} is a subsolution. We start by observing that

φ¯(x0)=0=zεδ(x0,t)andφ¯0=zεδon Ω×(0,T).\overline{\varphi}(x_{0})=0=z^{\varepsilon\delta}(x_{0},t)\qquad\ \text{and}\qquad\overline{\varphi}\geq 0=z^{\varepsilon\delta}\ \text{on }\partial\Omega\times(0,T).

Since z0ε𝕂z_{0}^{\varepsilon}\in\mathbb{K}_{\vee}^{\wedge}, then

φ¯(x)d(x)z0ε(x).\overline{\varphi}(x)\geq d(x)\geq z_{0}^{\varepsilon}(x).

We compute

xiφ¯(x)=xid(x)+Mεηε(|x|r)xi|x|and|φ¯|1+Mε,\partial_{x_{i}}\overline{\varphi}(x)=\partial_{x_{i}}d(x)+M\sqrt{\varepsilon}\eta_{\varepsilon}(|x|-r)\tfrac{x_{i}}{|x|}\quad\text{and}\quad|\nabla\overline{\varphi}|\leq 1+M\sqrt{\varepsilon}, (36)
xi2φ¯(x)=xi2d(x)Mηε(|x|r)xi2|x|2+Mεηε(|x|r)(1|x|xi2|x|3)\partial^{2}_{x_{i}}\overline{\varphi}(x)=\partial^{2}_{x_{i}}d(x)-M\eta_{\varepsilon}(|x|-r)\,\tfrac{x_{i}^{2}}{|x|^{2}}+M\sqrt{\varepsilon}\eta_{\varepsilon}(|x|-r)\Big{(}\tfrac{1}{|x|}-\tfrac{x_{i}^{2}}{|x|^{3}}\Big{)}

and

Δφ¯(x)=Δd(x)+Mηε(|x|r)(1+εN1|x|).\Delta\overline{\varphi}(x)=\Delta d(x)+M\eta_{\varepsilon}(|x|-r)\Big{(}-1+\sqrt{\varepsilon}\,\tfrac{N-1}{|x|}\Big{)}.

Let

Lw=twδΔw+𝒃w+δε(w(wd)(d)).Lw=\partial_{t}w-\delta\Delta w+\boldsymbol{b}\cdot\nabla w+\frac{\delta}{\varepsilon}(w-(w\wedge d)\vee(-d)).

Then, recalling that there exists a positive constant CC such that ΔdC\Delta d\leq C and choosing ε\varepsilon sufficiently small, such that, 1εN1|x|1εN1r121-\sqrt{\varepsilon}\,\tfrac{N-1}{|x|}\geq 1-\sqrt{\varepsilon}\,\frac{N-1}{r}\geq\frac{1}{2} we have

Lφ¯f\displaystyle L\overline{\varphi}-f =δΔd+Mδηε(|x|r)(1εN1|x|)+𝒃(d+Mεηε(|x|r)x|x|)+Mδ(1ηε(|x|r))f\displaystyle=-\delta\Delta d+M\,\delta\,\eta_{\varepsilon}(|x|-r)\Big{(}1-\sqrt{\varepsilon}\,\tfrac{N-1}{|x|}\Big{)}+\boldsymbol{b}\cdot\Big{(}\nabla d+M\sqrt{\varepsilon}\,\eta_{\varepsilon}(|x|-r)\,\tfrac{x}{|x|}\Big{)}+M\,\delta\,\big{(}1-\eta_{\varepsilon}(|x|-r)\big{)}-f
δC+Mδ2ηε(|x|R)|𝒃||𝒃|Mεηε(|x|R)+Mδ(1ηε(|x|r))fL(0,T)\displaystyle\geq-\delta\,C+M\tfrac{\delta}{2}\,\eta_{\varepsilon}(|x|-R)-|\boldsymbol{b}|-|\boldsymbol{b}|\,M\,\sqrt{\varepsilon}\,\eta_{\varepsilon}(|x|-R)+M\,\delta\big{(}1-\eta_{\varepsilon}(|x|-r)\big{)}-\|f\|_{L^{\infty}(0,T)}
=M(δ+(δ2|𝒃|εδ)ηε(|x|r))δC|𝒃|fL(0,T).\displaystyle=M\left(\delta+(\tfrac{\delta}{2}-|\boldsymbol{b}|\,\sqrt{\varepsilon}-\delta)\,\eta_{\varepsilon}(|x|-r)\right)-\delta\,C-|\boldsymbol{b}|-\|f\|_{L^{\infty}(0,T)}. (37)

Observe now that the term δ2|𝒃|εδ\tfrac{\delta}{2}-|\boldsymbol{b}|\,\sqrt{\varepsilon}-\delta is negative and, since ηε(|x|r)1\eta_{\varepsilon}(|x|-r)\leq 1, we have the following inequality

M(δ+(δ2|𝒃|εδ)ηε(|x|r))M(δ2|𝒃|ε).M\left(\delta+(\tfrac{\delta}{2}-|\boldsymbol{b}|\,\sqrt{\varepsilon}-\delta)\,\eta_{\varepsilon}(|x|-r)\right)\geq M\big{(}\tfrac{\delta}{2}-|\boldsymbol{b}|\,\sqrt{\varepsilon}\big{)}.

We can fix ε0\varepsilon_{0} such that, for 0<εε00<\varepsilon\leq\varepsilon_{0}, we have |𝒃|εδ4|\boldsymbol{b}|\,\sqrt{\varepsilon}\leq\frac{\delta}{4}. From (4), we obtain then

Lφ¯fMδ4δC|𝒃|fL(0,T)=0,L\overline{\varphi}-f\geq M\,\tfrac{\delta}{4}-\delta\,C-|\boldsymbol{b}|-\|f\|_{L^{\infty}(0,T)}=0,

provided

M=C1δ,C1=4(δC+|𝒃|+fL(0,T)),M=\frac{C_{1}}{\delta},\qquad C_{1}=4(\delta\,C+|\boldsymbol{b}|+\|f\|_{L^{\infty}(0,T)}), (38)

concluding then that φ¯\overline{\varphi} is a supersolution of (34). Analogously, φ¯\underline{\varphi} is a subsolution of (34) and so we have

φ¯zεδφ¯inQTandzεδ(x0,t)=φ¯(x0)=φ¯(x0).\underline{\varphi}\leq z^{\varepsilon\delta}\leq\overline{\varphi}\quad\text{in}\,\,Q_{T}\qquad\text{and}\qquad z^{\varepsilon\delta}(x_{0},t)=\overline{\varphi}(x_{0})=\underline{\varphi}(x_{0}). (39)

Observe that, from (36), we obtain

|zεδ(x0,t)|max{|φ¯(x0)|,|φ¯(x0)|}1+C1δε|\nabla z^{\varepsilon\delta}(x_{0},t)|\leq\max\{|\nabla\overline{\varphi}(x_{0})|,|\nabla\underline{\varphi}(x_{0})|\}\leq 1+\tfrac{C_{1}}{\delta}\,\sqrt{\varepsilon}

for an arbitrary point x0Ωx_{0}\in\partial\Omega at any t(0,T)t\in(0,T). We wish to prove that this estimate is true a.e. in QTQ_{T}. Differentiate the first equation of (34) with respect to xkx_{k}, multiply it by zxkεδz^{\varepsilon\delta}_{x_{k}} and sum over kk. Setting v=|zεδ|2v=|\nabla z^{\varepsilon\delta}|^{2} and noticing that zxkεδΔzxkεδ=12Δv(zxkxkεδ)2z^{\varepsilon\delta}_{x_{k}}\Delta z^{\varepsilon\delta}_{x_{k}}=\frac{1}{2}\Delta v-(z^{\varepsilon\delta}_{x_{k}x_{k}})^{2} we get

12tvδ2Δv+12𝒃v+δε(vz~εδzεδ)0,\frac{1}{2}\partial_{t}v-\frac{\delta}{2}\Delta v+\frac{1}{2}\boldsymbol{b}\cdot\nabla v+\frac{\delta}{\varepsilon}\big{(}v-\nabla\tilde{z}^{\varepsilon\delta}\cdot\nabla z^{\varepsilon\delta}\big{)}\leq 0,

being z~εδ=zεδ(zεδd)(d)\tilde{z}^{\varepsilon\delta}=z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d). Using the Cauchy-Schwartz inequality, we obtain

tvδΔv+𝒃v+2δε(v|z~εδ|v12)0.\partial_{t}v-\delta\Delta v+\boldsymbol{b}\cdot\nabla v+\frac{2\delta}{\varepsilon}\big{(}v-|\nabla\tilde{z}^{\varepsilon\delta}|v^{\frac{1}{2}}\big{)}\leq 0.

Multiplying the above inequality by (v(1+Mε)2)+(v-(1+M\sqrt{\varepsilon})^{2})^{+} and integrating over QtQ_{t}, we have

12Ω|(v(t)(1+Mε)2)+|2+δQt|(v(1+Mε)2)+|2+Qt𝒃(v(1+Mε)2)+(v(1+Mε)2)++2δεQt(v|z~εδ|v12)(v(1+Mε)2)+0.\frac{1}{2}\int_{\Omega}\big{|}(v(t)-(1+M\sqrt{\varepsilon})^{2})^{+}\big{|}^{2}+\delta\int_{Q_{t}}|\nabla(v-(1+M\sqrt{\varepsilon})^{2})^{+}|^{2}\\ +\int_{Q_{t}}\boldsymbol{b}\cdot\nabla(v-(1+M\sqrt{\varepsilon})^{2})^{+}\,(v-(1+M\sqrt{\varepsilon})^{2})^{+}+\frac{2\delta}{\varepsilon}\int_{Q_{t}}\big{(}v-|\nabla\tilde{z}^{\varepsilon\delta}|v^{\frac{1}{2}}\big{)}(v-(1+M\sqrt{\varepsilon})^{2})^{+}\leq 0. (40)

Since

Qt𝒃(v(1+Mε)2)+(v(1+Mε)2)+=0\displaystyle\int_{Q_{t}}\boldsymbol{b}\cdot\nabla(v-(1+M\sqrt{\varepsilon})^{2})^{+}\,(v-(1+M\sqrt{\varepsilon})^{2})^{+}=0

and

Qt(v|z~εδ|v12)(v(1+Mε)2)+={zεδ>d}(vv12)(v(1+Mε)2)++{zεδ<d}(vv12)(v(1+Mε)2)+0,\int_{Q_{t}}\big{(}v-|\nabla\tilde{z}^{\varepsilon\delta}|v^{\frac{1}{2}}\big{)}(v-(1+M\sqrt{\varepsilon})^{2})^{+}\\ =\int_{\{z^{\varepsilon\delta}>d\}}\big{(}v-v^{\frac{1}{2}}\big{)}(v-(1+M\sqrt{\varepsilon})^{2})^{+}+\int_{\{z^{\varepsilon\delta}<-d\}}\big{(}v-v^{\frac{1}{2}}\big{)}(v-(1+M\sqrt{\varepsilon})^{2})^{+}\geq 0,

from (40) we conclude that (v(1+Mε)2)+0(v-(1+M\sqrt{\varepsilon})^{2})^{+}\equiv 0.

Then, recalling the choice of MM done in (38), we have

|zεδ|2=v1+C1δε a.e. in QT,|\nabla z^{\varepsilon\delta}|^{2}=v\leq 1+\tfrac{C_{1}}{\delta}\sqrt{\varepsilon}\quad\text{ a.e. in }Q_{T}, (41)

and {zεδ}ε\{z^{\varepsilon\delta}\}_{\varepsilon} is uniformly bounded in L(0,T;W01,(Ω))L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)}. Using (39), it is easy see that

C1δε(zεδ(zεδd)(d))C1.-C_{1}\leq\frac{\delta}{\varepsilon}(z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d))\leq C_{1}.

In fact, in the set {zεδ>d}\{z^{\varepsilon\delta}>d\} we have

δε(zεδ(zεδd)(d))=δε(zεδd)δε(φ¯d)C1,\frac{\delta}{\varepsilon}(z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d))=\frac{\delta}{\varepsilon}(z^{\varepsilon\delta}-d)\leq\frac{\delta}{\varepsilon}(\overline{\varphi}-d)\leq C_{1},

in the set {dzεδd}\{-d\leq z^{\varepsilon\delta}\leq d\} we have zεδ(zεδd)(d)=0z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d)=0 and in the set {zεδ<d}\{z^{\varepsilon\delta}<-d\} we have

δε(zεδ(zεδd)(d))=δε(zεδ+d)δε(φ¯+d)C1.\frac{\delta}{\varepsilon}(z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d))=\frac{\delta}{\varepsilon}(z^{\varepsilon\delta}+d)\geq\frac{\delta}{\varepsilon}(\underline{\varphi}+d)\geq-C_{1}.

Multiplying the first equation of (34) by tzεδ\partial_{t}z^{\varepsilon\delta}, we obtain

Qt|tzεδ|2+δQtzεδtzεδ+QT𝒃zεδtzεδ+δεQt(zεδ(zεδd)(d))tzεδ=Qtftzεδ\int_{Q_{t}}|\partial_{t}z^{\varepsilon\delta}|^{2}+\delta\int_{Q_{t}}\nabla z^{\varepsilon\delta}\cdot\nabla\partial_{t}z^{\varepsilon\delta}+\int_{Q_{T}}\boldsymbol{b}\cdot\nabla z^{\varepsilon\delta}\partial_{t}z^{\varepsilon\delta}+\frac{\delta}{\varepsilon}\int_{Q_{t}}\big{(}z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d)\big{)}\partial_{t}z^{\varepsilon\delta}=\int_{Q_{t}}f\partial_{t}z^{\varepsilon\delta}

and so

Qt|tzεδ|2\displaystyle\int_{Q_{t}}|\partial_{t}z^{\varepsilon\delta}|^{2} +δ2Ω|zεδ(t)|2\displaystyle+\frac{\delta}{2}\int_{\Omega}|\nabla z^{\varepsilon\delta}(t)|^{2}
δ2Ω|u0ε|2+(|𝒃|zεδ𝑳2(QT)+δε(zεδ(zεδd)(d)L(QT)\displaystyle\leq\frac{\delta}{2}\int_{\Omega}|\nabla u^{\varepsilon}_{0}|^{2}+\big{(}|\boldsymbol{b}|\,\|\nabla z^{\varepsilon\delta}\|_{\boldsymbol{L}^{2}(Q_{T})}+\|\frac{\delta}{\varepsilon}(z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d)\|_{L^{\infty}(Q_{T})}
+fL2(0,T))tzεδL2(QT)\displaystyle\quad\quad+\|f\|_{L^{2}(0,T)}\big{)}\|\partial_{t}z^{\varepsilon\delta}\|_{L^{2}(Q_{T})}
δ2Ω|u0ε|2+(|𝒃|(1+C1δε+C1)|QT|12+fL2(0,T))tzεδL2(QT),\displaystyle\leq\frac{\delta}{2}\int_{\Omega}|\nabla u_{0}^{\varepsilon}|^{2}+\big{(}|\boldsymbol{b}|(1+\tfrac{C_{1}}{\delta}\,\sqrt{\varepsilon}+C_{1})|Q_{T}|^{\frac{1}{2}}+\|f\|_{L^{2}(0,T)}\big{)}\|\partial_{t}z^{\varepsilon\delta}\|_{L^{2}(Q_{T})},
δ2Ω|u0ε|2+12(|𝒃|(1+C1δε+C1)|QT|12+fL2(0,T))2+12tzεδL2(QT)2\displaystyle\leq\frac{\delta}{2}\int_{\Omega}|\nabla u_{0}^{\varepsilon}|^{2}+\frac{1}{2}\big{(}|\boldsymbol{b}|(1+\tfrac{C_{1}}{\delta}\,\sqrt{\varepsilon}+C_{1})|Q_{T}|^{\frac{1}{2}}+\|f\|_{L^{2}(0,T)}\big{)}^{2}+\frac{1}{2}\|\partial_{t}z^{\varepsilon\delta}\|_{L^{2}(Q_{T})}^{2}

where |QT||Q_{T}| denotes the Lebesgue measure of QTQ_{T}. So, for δ\delta fixed,

tzεδL2(QT)2δΩ|u0ε|2+(|𝒃|(1+C1δε+C1)|QT|12+fL2(0,T))2.\|\partial_{t}z^{\varepsilon\delta}\|_{L^{2}(Q_{T})}^{2}\leq\delta\int_{\Omega}|\nabla u_{0}^{\varepsilon}|^{2}+\big{(}|\boldsymbol{b}|(1+\tfrac{C_{1}}{\delta}\,\sqrt{\varepsilon}+C_{1})|Q_{T}|^{\frac{1}{2}}+\|f\|_{L^{2}(0,T)}\big{)}^{2}. (42)

Then, there exists zδL(0,T;W01,(Ω))H1(0,T;L2(Ω))z^{\delta}\in L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)}\cap H^{1}\big{(}0,T;L^{2}(\Omega)\big{)} such that

zεδ–⇀ε0zδ in L(0,T;W01,(Ω))-weakandtzεδ–⇀ε0tzδ in L2(QT).z^{\varepsilon\delta}\underset{\varepsilon\rightarrow 0}{\relbar\joinrel\relbar\joinrel\rightharpoonup}z^{\delta}\text{ in }L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)}\text{-weak}*\quad\text{and}\quad\partial_{t}z^{\varepsilon\delta}\underset{\varepsilon\rightarrow 0}{\relbar\joinrel\relbar\joinrel\rightharpoonup}\partial_{t}z^{\delta}\text{ in }L^{2}(Q_{T}).

Multiplying the first equation of the problem (34) by vzεδ(t)v-z^{\varepsilon\delta}(t), where v𝕂v\in\mathbb{K}_{\vee}^{\wedge} and integrating over Ω×(s,t)\Omega\times(s,t), 0s<tT0\leq s<t\leq T, we obtain

stΩtzεδ(vzεδ)+δstΩzεδ(vzεδ)+stΩ𝒃zεδ(vzεδ)+δεΩ(zεδ(zεδ(t)d)(d))(vzεδ)=stΩfδ(vzεδ).\int_{s}^{t}\int_{\Omega}\partial_{t}z^{\varepsilon\delta}(v-z^{\varepsilon\delta})+\delta\int_{s}^{t}\int_{\Omega}\nabla z^{\varepsilon\delta}\cdot\nabla(v-z^{\varepsilon\delta})+\int_{s}^{t}\int_{\Omega}\boldsymbol{b}\cdot\nabla z^{\varepsilon\delta}\,(v-z^{\varepsilon\delta})\\ +\frac{\delta}{\varepsilon}\int_{\Omega}\big{(}z^{\varepsilon\delta}-(z^{\varepsilon\delta}(t)\wedge d)\vee(-d)\big{)}(v-z^{\varepsilon\delta})=\int_{s}^{t}\int_{\Omega}f^{\delta}(v-z^{\varepsilon\delta}).

For v𝕂v\in\mathbb{K}_{\vee}^{\wedge}, the operator PεP_{\varepsilon} defined in (35) is monotone, we have

stΩ(zεδ(zεδd)(d))(vzεδ)0.\int_{s}^{t}\int_{\Omega}\big{(}z^{\varepsilon\delta}-(z^{\varepsilon\delta}\wedge d)\vee(-d)\big{)}(v-z^{\varepsilon\delta})\leq 0.

So, letting ε0\varepsilon\rightarrow 0, we obtain

stΩtzδ(vzδ)+δstΩzδ(vzδ)+stΩ𝒃zδ(vzδ)stΩfδ(vzδ).\int_{s}^{t}\int_{\Omega}\partial_{t}z^{\delta}(v-z^{\delta})+\delta\int_{s}^{t}\int_{\Omega}\nabla z^{\delta}\cdot(v-z^{\delta})+\int_{s}^{t}\int_{\Omega}\boldsymbol{b}\cdot\nabla z^{\delta}\,(v-z^{\delta})\geq\int_{s}^{t}\int_{\Omega}f^{\delta}(v-z^{\delta}). (43)

By (39), the function zδz^{\delta} is such that zδ(t)𝕂z^{\delta}(t)\in\mathbb{K}_{\vee}^{\wedge}, for a.e. t(0,T)t\in(0,T).

To prove that {tzδ}δ\{\partial_{t}z^{\delta}\}_{\delta} is bounded in L2(QT)L^{2}(Q_{T}), let ε0\varepsilon\rightarrow 0 in (42), obtaining

Qt|tzδ|2δΩ|u0|2+(|𝒃|(1+C1)|QT|12+fL2(0,T))2.\int_{Q_{t}}|\partial_{t}z^{\delta}|^{2}\leq\delta\int_{\Omega}|\nabla u_{0}|^{2}+\big{(}|\boldsymbol{b}|(1+C_{1})|Q_{T}|^{\frac{1}{2}}+\|f\|_{L^{2}(0,T)}\big{)}^{2}.

Analogously, letting ε0\varepsilon\rightarrow 0 in (41), we obtain

|zδ|1 a.e. in QT.|\nabla z^{\delta}|\leq 1\quad\text{ a.e. in }Q_{T}.

We can now pass easily to the limit when δ0\delta\rightarrow 0 in inequality (43). Observing that zδz^{\delta} converges to some function zz weakly* in L(0,T;W01,(Ω))L^{\infty}\big{(}0,T;W^{1,\infty}_{0}(\Omega)\big{)} and tzδ\partial_{t}z^{\delta} converges weakly in L2(QT)L^{2}(Q_{T}) to tz\partial_{t}z, we find for all 0<s<t<T0<s<t<T

stΩtz(vz)+stΩ𝒃z(vz)stΩf(vz)\int_{s}^{t}\int_{\Omega}\partial_{t}z(v-z)+\int_{s}^{t}\int_{\Omega}\boldsymbol{b}\cdot\nabla z\,(v-z)\geq\int_{s}^{t}\int_{\Omega}f(v-z)

and it follows also

Ωtz(t)(vz(t))+Ω𝒃z(t)(vz(t))Ωf(t)(vz(t))for a.e.t(0,T).\int_{\Omega}\partial_{t}z(t)(v-z(t))+\int_{\Omega}\boldsymbol{b}\cdot\nabla z(t)\,(v-z(t))\geq\int_{\Omega}f(t)(v-z(t))\,\,\,\,\text{for a.e.}\,\,t\in(0,T).

Since zδ(t)𝕂z^{\delta}(t)\in\mathbb{K}_{\vee}^{\wedge} for a.e. t(0,T)t\in(0,T), we also have z(t)𝕂z(t)\in\mathbb{K}_{\vee}^{\wedge} and the proof of existence of solution for the variational inequality (33) is complete. The uniqueness is also clear.

The inclusion 𝕂1𝕂\mathbb{K}_{1}\subset\mathbb{K}_{\vee}^{\wedge} and the fact z(t)𝕂1z(t)\in\mathbb{K}_{1} for a.e. t(0,T)t\in(0,T) implies that the function zz also solves the problem (10). \square

Remark 4.2

The first order variational inequalities of obstacle type have been introduced by Bensoussan and Lions in [4] and have been studied in [13] and in [17], for general linear operators and general obstacles, and extended to a quasilinear two obstacles problem in [11]. In all those cases the notion of solution is less regular and the boundary data can only be prescribed on part of the boundary. In addition, the solution cannot have a gradient in L2L^{2} and the best that can be expected in general is the operator tu+𝐛u+cuL2\partial_{t}u+\boldsymbol{b}\cdot\nabla u+c\,u\in L^{2}, as a consequence of Lewy-Stampacchia inequalities. These estimates can be obtained from the regularized parabolic inequality (43) and, as in [18], it allows the passage to the limit δ0\delta\rightarrow 0 without the estimates on the gradient and on the time derivative. It is an open question to establish the equivalence of the first order obstacle problem with the variational inequality with gradient constraint for more general first order linear operators.

Theorem 4.3

In addition to the assumptions (32), suppose

𝒃z0f(t) in {xΩ:d(x)<z0(x)}fort>0,\boldsymbol{b}\cdot\nabla z_{0}\leq f(t)\text{ in }\big{\{}x\in\Omega:-d(x)<z_{0}(x)\big{\}}\,\,for\,\,t>0, (44)
f=f(t)is increasing and nonnegative,f=f(t)\,\,\text{is increasing and nonnegative}, (45)
lim inftf(t)>|𝒃|+2D,\liminf_{t\rightarrow\infty}f(t)>|\boldsymbol{b}|+2D, (46)

where D=dL(Ω)=maxxΩ¯d(x,Ω)D=\|d\|_{L^{\infty}(\Omega)}=\displaystyle\max_{x\in\overline{\Omega}}d(x,\partial\Omega). Then there exists T<T_{*}<\infty such that the solution zz of the variational inequality (10), or equivalently of (33), satisfies

z(t)=dfor alltT.z(t)=d\quad\text{for all}\,\,t\geq T_{*}.

Proof  We consider zz as the solution of the variational inequality (33).

Step 1: z0z(t) for all t>0.z_{0}\leq z(t)\quad\text{ for all }t>0.

Let v(t)=z(t)+(z0z(t))+v(t)=z(t)+(z_{0}-z(t))^{+} and note that v(t)𝕂v(t)\in\mathbb{K}_{\vee}^{\wedge}. Then

Ωtz(t)(z0u(t))++Ω𝒃z(t)(z0z(t))+Ωf(t)(z0z(t))+.\int_{\Omega}\partial_{t}z(t)(z_{0}-u(t))^{+}+\int_{\Omega}\boldsymbol{b}\cdot\nabla z(t)(z_{0}-z(t))^{+}\geq\int_{\Omega}f(t)(z_{0}-z(t))^{+}. (47)

On the other hand, by (44), we have

𝒃z0(z0z(t))+f(t)(z0z(t))+ in {d<z0},\boldsymbol{b}\cdot\nabla z_{0}(z_{0}-z(t))^{+}\leq f(t)(z_{0}-z(t))^{+}\quad\text{ in }\quad\{-d<z_{0}\},

and also on {d=z0}\{-d=z_{0}\} since, in this last set, (z0z(t))+0(z_{0}-z(t))^{+}\equiv 0 (recall that f0f\geq 0). Then

Ωtz0(z0z(t))++Ω𝒃z0(z0z(t))+Ωf(t)(z0z(t))+.\int_{\Omega}\partial_{t}z_{0}(z_{0}-z(t))^{+}+\int_{\Omega}\boldsymbol{b}\cdot\nabla z_{0}(z_{0}-z(t))^{+}\leq\int_{\Omega}f(t)(z_{0}-z(t))^{+}. (48)

From (47) and (48) we get

Ωt(u0z(t))(z0z(t))++Ω𝒃(z0z(t))(z0z(t))+0.\int_{\Omega}\partial_{t}(u_{0}-z(t))(z_{0}-z(t))^{+}+\int_{\Omega}\boldsymbol{b}\cdot\nabla(z_{0}-z(t))(z_{0}-z(t))^{+}\leq 0.

But

Ω𝒃(z0z(t))(z0z(t))+=12Ω𝒃((z0z(t))+)2=12Ω𝒃((z0z(t))+)2=0\int_{\Omega}\boldsymbol{b}\cdot\nabla(z_{0}-z(t))(z_{0}-z(t))^{+}=\frac{1}{2}\int_{\Omega}\boldsymbol{b}\cdot\nabla\big{(}(z_{0}-z(t))^{+}\big{)}^{2}=-\frac{1}{2}\int_{\Omega}\nabla\cdot\boldsymbol{b}\,\big{(}(z_{0}-z(t))^{+}\big{)}^{2}=0

and so

12Ω|(z0z(t))+|212Ω|(z0z(0))+|2=0,\frac{1}{2}\int_{\Omega}\big{|}(z_{0}-z(t))^{+}\big{|}^{2}\leq\frac{1}{2}\int_{\Omega}\big{|}(z_{0}-z(0))^{+}\big{|}^{2}=0,

which implies that z0z(t)z_{0}\leq z(t), for all t 0t\>0.

Step 2: z(t)z(t+h) for all t,h>0.z(t)\leq z(t+h)\quad\text{ for all }t,h>0.

Observe that v(t)=z(t+h)(z(t)z(t+h))𝕂v(t)=z(t+h)-(z(t)-z(t+h))^{-}\in\mathbb{K}_{\vee}^{\wedge}, so we can choose v(t)v(t) as test function in (33). Noting that

v(t)z(t)=z(t+h)z(t)(z(t)z(t+h))=(z(t)z(t+h))+v(t)-z(t)=z(t+h)-z(t)-\big{(}z(t)-z(t+h)\big{)}^{-}=-\big{(}z(t)-z(t+h)\big{)}^{+}

we get

Ωtz(t)(z(t)z(t+h))+Ω𝒃z(t)(z(t)z(t+h))+Ωf(t)(z(t)z(t+h))+.-\int_{\Omega}\partial_{t}z(t)\big{(}z(t)-z(t+h)\big{)}^{+}-\int_{\Omega}\boldsymbol{b}\cdot\nabla z(t)\big{(}z(t)-z(t+h)\big{)}^{+}\geq-\int_{\Omega}f(t)\big{(}z(t)-z(t+h)\big{)}^{+}. (49)

Choosing v(t)=z(t+h)+(z(t)z(t+h))+v(t)=z(t+h)+\big{(}z(t)-z(t+h)\big{)}^{+} as test function in (33) in the instant t+ht+h and observing that

v(t)z(t)=z(t+h)z(t)+(z(t)z(t+h))+=(z(t)z(t+h)),v(t)-z(t)=z(t+h)-z(t)+\big{(}z(t)-z(t+h)\big{)}^{+}=\big{(}z(t)-z(t+h)\big{)}^{-},

we have

Ωtz(t+h)(z(t)z(t+h))+Ω𝒃z(t+h)(z(t)z(t+h))Ωf(t+h)(z(t)z(t+h)).\int_{\Omega}\partial_{t}z(t+h)\big{(}z(t)-z(t+h)\big{)}^{-}+\int_{\Omega}\boldsymbol{b}\cdot\nabla z(t+h)\big{(}z(t)-z(t+h)\big{)}^{-}\geq\int_{\Omega}f(t+h)\big{(}z(t)-z(t+h)\big{)}^{-}. (50)

From (49) and (50) we get

Ωt(z(t)z(t+h))(z(t)z(t+h))+Ω𝒃(z(t)z(t+h))(z(t)z(t+h))Ω(f(t)f(t+h))(z(t)z(t+h))0,\int_{\Omega}\partial_{t}(z(t)-z(t+h))\big{(}z(t)-z(t+h)\big{)}^{-}+\int_{\Omega}\boldsymbol{b}\cdot\nabla(z(t)-z(t+h))\,\big{(}z(t)-z(t+h)\big{)}^{-}\\ \leq\int_{\Omega}(f(t)-f(t+h))\big{(}z(t)-z(t+h)\big{)}^{-}\leq 0,

because f(t)f(t+h)f(t)\leq f(t+h), by assumption (45) and (z(t)z(t+h))0\big{(}z(t)-z(t+h)\big{)}^{-}\geq 0. As

Ω𝒃(z(t)z(t+h))(z(t)z(t+h))=0,\int_{\Omega}\boldsymbol{b}\cdot\nabla(z(t)-z(t+h))\,\big{(}z(t)-z(t+h)\big{)}^{-}=0,

we obtain

12Ω(|(z(t)z(t+h))|212Ω(|(z(0)z(h))|20,\frac{1}{2}\int_{\Omega}\Big{(}\big{|}(z(t)-z(t+h)\big{)}^{-}\Big{|}^{2}\leq\frac{1}{2}\int_{\Omega}\Big{(}\big{|}(z(0)-z(h)\big{)}^{-}\Big{|}^{2}\leq 0,

using Step 1. So z(t)z(t+h)z(t)\leq z(t+h), for all t,h>0t,h>0.

Step 3: There exists z𝒞(Ω¯)z_{\infty}\in\mathscr{C}(\overline{\Omega}) such that limt+z(x,t)=z(x),\displaystyle\lim_{t\rightarrow+\infty}z(x,t)=z_{\infty}(x), uniformly in xΩ¯x\in\overline{\Omega}.

Since the sequence of continuous functions {z(t)}t>0{\{z(t)\}}_{t>0} is increasing in tt and is bounded from above by dd, this conclusion follows immediatly.

However, in this special case we have a finite time stabilization.

First we prove that the function zz_{\infty} coincides with dd. We recall that tzL2(QT)\partial_{t}z\in L^{2}(Q_{T}), for any T>0T>0, and we set ψ(t)=Ωz(t)\psi(t)=\displaystyle\int_{\Omega}z(t). Observe that ψL(0,)|Ω|D\|\psi\|_{L^{\infty}(0,\infty)}\leq|\Omega|D, where |Ω||\Omega| denotes the Lebesgue measure of Ω\Omega. Since {z(t)}t>0{\{z(t)\}}_{t>0} is increasing, then tz0\partial_{t}z\geq 0 and

ψ(t)t+Ωz,ψ(t)0 for a.e. t>0.\psi(t)\underset{t\rightarrow+\infty}{\longrightarrow}\int_{\Omega}z_{\infty},\qquad\psi^{\prime}(t)\geq 0\quad\text{ for a.e. }t>0.

This implies that

lim infttz(t)=0in L1(Ω).\liminf_{t\rightarrow\infty}\partial_{t}z(t)=0\quad\text{in }L^{1}(\Omega).

Choosing v=dv=d as test function in (33) we obtain, for a.e. t(0,)t\in(0,\infty),

Ωtz(t)(dz(t))+Ω𝒃z(t)(dz(t))Ωf(t)(dz(t))\int_{\Omega}\partial_{t}z(t)(d-z(t))+\int_{\Omega}\boldsymbol{b}\cdot\nabla z(t)(d-z(t))\geq\int_{\Omega}f(t)(d-z(t))

and so

Ωtz(t)(dz(t))+|𝒃|Ω(dz(t))f(t)Ω(dz(t)).\int_{\Omega}\partial_{t}z(t)(d-z(t))+|\boldsymbol{b}|\int_{\Omega}(d-z(t))\geq f(t)\int_{\Omega}(d-z(t)).

Since dz(t)d\geq z(t), taking lim inft\displaystyle\liminf_{t\rightarrow\infty} to both sides of the inequality and using the assumption (46), we obtain

|𝒃|Ω(dz)(|𝒃|+2D)Ω(dz),|\boldsymbol{b}|\int_{\Omega}(d-z_{\infty})\geq(|\boldsymbol{b}|+2D)\int_{\Omega}(d-z_{\infty}),

which is a contradiction unless z=dz_{\infty}=d.

Consider the following subsets of Q=Ω×(0,)Q_{\infty}=\Omega\times(0,\infty)

Λ={d<z<d},I+={z=d},I={z=d}.\Lambda=\big{\{}-d<z<d\big{\}},\quad I^{+}=\big{\{}z=d\big{\}},\quad I^{-}=\big{\{}z=-d\big{\}}.

Since zz solves the two obstacle problem (33), it is well known that the following inequalities are verified a.e. in QQ_{\infty}:

tz+𝒃z=finΛ,tz+𝒃zfinI+,tu+𝒃ufinI.\displaystyle\partial_{t}z+\boldsymbol{b}\cdot\nabla z=f\quad\text{in}\quad\Lambda,\quad\partial_{t}z+\boldsymbol{b}\cdot\nabla z\leq f\quad\text{in}\quad I^{+},\quad\partial_{t}u+\boldsymbol{b}\cdot\nabla u\geq f\quad\text{in}\quad I^{-}.

If there is no finite time stabilization of the solution, since z(t)z(t) is increasing in time, we may find a point (x0,t0)(x_{0},t_{0}) and an open subset ω0\omega_{0} of Ω\Omega with x0ω0x_{0}\in\omega_{0}, such that, (x,t)ΛI(x,t)\in\Lambda\cup I^{-} for t>t0t>t_{0}. So,

f(t)tz(x,t)+𝒃z(x,t) for a.e. (x,t)ω0×[t0,+).f(t)\leq\partial_{t}z(x,t)+\boldsymbol{b}\cdot\nabla z(x,t)\quad\text{ for a.e. }(x,t)\in\omega_{0}\times[t_{0},+\infty).

Then, for any tt0t\geq t_{0} and any open set ωω0\omega\subset\omega_{0}, we have

tt+1f(τ)\displaystyle\int_{t}^{t+1}f(\tau) 1|ω|tt+1ω(tz(x,τ)+𝒃z(x,τ))\displaystyle\leq\frac{1}{|\omega|}\int_{t}^{t+1}\int_{\omega}\big{(}\partial_{t}z(x,\tau)+\boldsymbol{b}\cdot\nabla z(x,\tau)\big{)}
1|ω|ω(z(x,t+1)z(x,t))+|𝒃|(2D+|𝒃|).\displaystyle\leq\frac{1}{|\omega|}\int_{\omega}\big{(}z(x,t+1)-z(x,t)\big{)}+|\boldsymbol{b}|\leq\big{(}2D+|\boldsymbol{b}|\big{)}.

As a consequence,

lim inftf(t)lim infttt+1f(τ)𝑑τ2D+|𝒃|\liminf_{t\rightarrow\infty}f(t)\leq\liminf_{t\rightarrow\infty}\int_{t}^{t+1}f(\tau)d\tau\leq 2D+|\boldsymbol{b}|

and this is a contradiction with (46). So z(t)z(t) must stabilize in finite time. \square

Acknowledgments

This research was partially supported by CMAT - “Centro de Matemática da Universidade do Minho”, financed by FEDER Funds through “Programa Operacional Factores de Competitividade - COMPETE” and by Portuguese Funds through FCT, “Fundação para a Ciência e a Tecnologia”, within the Project PEst-OE/MAT/UI0013/2014.

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José Francisco Rodrigues            Lisa Santos
jfrodrigues@ciencias.ulisboa.pt            lisa@math.uminho.pt
CMAF-FC_ULisboa,            CMAT / Dept. of Mathematics and Applications
Av. Prof. Gama Pinto, 2,            Campus de Gualtar,
1649-003 Lisboa, Portugal            4710-057 Braga, Portugal