Uniqueness of Renormalized Solutions forNonlinear Degenerate Problems

早稲田大学大学院・理工学研究科 高木 悟 (Satoru Takagi )Graduate School of Science and Engineering,

Waseda University

1 Introduction

Let $\Omega$ be abounded domain in $\mathbb{R}^{N},$ $N\geq 1$ , and let $T>0$ . When $N\geq 2$ weassume that $\Omega$ has aLipschitz boundary an. We consider the initial-boundaryvalue problem

(E) $\{$

$\frac{\partial g(u)}{\partial t}-\triangle b(u)+\mathrm{d}\mathrm{i}\mathrm{v}\phi(u)$ $=$ $f$ in $Q=(0, T)\cross\Omega$ ,

$b(u)$ $=$ $0$ on $\Sigma=(0, T)\cross\partial\Omega$ ,

$g(u)(0, \cdot)$ $=g(u_{0})$ in $\Omega$ ,

where

(H1) $g,$$b:\mathbb{R}arrow \mathbb{R}$ are continuous and nondecreasing functions satisfying the

normalization conditions $g(0)=b(0)=0$ , and $\phi:\mathbb{R}arrow \mathbb{R}^{N}$ is acontinu-ous $N$-dimensional vector-valued function satisfying $\phi(0)=0$ .

(H2) $f\in L^{1}(Q)$ and $u_{0}$ : $\Omegaarrow\overline{\mathbb{R}}$ is measurable with $g(u_{0})\in L^{1}(\Omega)$ , where$\overline{\mathbb{R}}=[-\infty, \infty]$ .

(H3) For any measurable functions $u,$ $v:Qarrow \mathbb{R}$

$((\nabla b(u)-\phi(u))-(\nabla b(v)-\phi(v)))\cdot(\nabla u-\nabla v)$

$+C(u, v)(1+|\nabla b(u)-\phi(u)|^{2}+|\nabla b(v)-\phi(v)|^{2})|u-v|\geq 0$ ,

where $C$ : $\mathbb{R}\mathrm{x}\mathbb{R}arrow \mathbb{R}^{+}$ is continuous.

数理解析研究所講究録 1323巻 2003年 59-75

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Many authors have considered the problems like (E) as well as the sta-tionary problems under various assumptions on the vector field and have in-troduced several different notions of solutions for these problems in order toprove existence and uniqueness of such solutions, see $[1]-[3],$ $[6],$ $[10]$ and [14],for example.

Due to the possible degeneracy of $b$ and $g$ , in general, we are not ableto expect that solution in the sense of distribution for (E) is unique. Wethus consider the problem (E) adopting the notion of renormalized solutions.The notion of renormalized solutions was introduced by DiPerna and Lions intheir papers [8] and [9] dealing with existence of asolution for the Boltzmannequation. We can also treat the problem in the case of large data in asense byutilizing this theory. In this report we shall prove uniqueness and acomparisonresult of renormalized solutions for the problem (E) with no growth conditionapplying the method of doubling variables both in space and time introducedby Kruzhkov [12]. As to some studies of renormalized solutions, see [4], [5],[7], [11], [13], [15] and [16], for example.

We shall mention the notations and definitions. For $k>\mathrm{O}$ we define atruncate function $T_{k}$ by

$T_{k}(u)=\{$

$k$ if $u>k$$u$ if $|u|\leq k$

$-k$ if $u<-k$

as usual. We introduce the following functions

$S(r)=\{$1if $r>0$

$[0, 1]$ if $r=0$

0if $r<0$

and

$S_{0}(r)=\{$1if $r>0$0if $r\leq 0$ ’

and also define nonnegative functions $r^{+}$ and $r^{-}$ by $r^{+}= \max(r, 0)$ and $r^{-}=$

$- \min(r, 0)$ , respectively.We now define arenormalized solution as in [7].

Definition 1.1. A renormalized solution of (E) is a rneasurable function$u$ : $Qarrow \mathbb{R}$ satisfying

(R1) $g(u)\in L^{1}(Q)_{J}$

(R2) $T_{k}(u)\in L^{2}(0,T;H_{0}^{1}(\Omega))$ for any $k>0$ ,

(R3) $b(T_{k}(u))\in L^{2}(0, T;H_{0}^{1}(\Omega))$ for any $k>0_{r}$

(R4) $\phi(T_{k}(u))\in L^{2}(Q)^{N}$ for any $k>0_{1}$

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(R5) for all $h\in C_{0}^{1}(\mathbb{R})$ and $\xi\in C_{0}^{\infty}([0, T)\cross\Omega)_{J}$

$\int_{Q}\xi_{t}\int_{u_{0}}^{u}h(r)dg(r)dxdt+\int_{Q}\xi fh(u)$ dxdt

$=$ $\int_{Q}(\nabla b(u)-\phi(u))\cdot\nabla(h(u)\xi)dxdt$, (1.1)

$moreover_{J}$

$\int_{Q\cap\{n\leq|u|\leq n+1\}}\nabla b(u)\cdot\nabla udxdtarrow 0$ as $narrow\infty$ . (1.2)

Remark 1.2. Note that each integral in (1.1) and (1.2) is well-defined. Infact, the right-hand side of (1.1) is identified with

$\int_{Q\cap\{|u|<k\}}(\nabla b(T_{k}(u))-\phi(T_{k}(u)))\cdot\nabla(h(T_{k}(u))\xi)dxdt$

for $k>\mathrm{O}$ such that $\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}h\subset(-k, k)$ . Similarly, the integral in (1.2) has to beunderstood as

$\int_{Q\cap\{n\leq|u|\leq n+1\}}\nabla b(T_{n+1}(u))\cdot\nabla T_{n+1}(u)dxdt$ .

2Main theorem

We obtain the following comparison result.Theorem 2.1. Suppose that (H1) and (H3) hold. Let $u_{0i}$ : $\Omegaarrow\overline{\mathbb{R}}$ be mea-surable with $g(u_{0i})\in L^{1}(\Omega),$ $f_{i}\in L^{1}(Q)$ and let $u.\cdot$ be a renormalized solutionof $(\mathrm{E}_{i})$ for $i=1,2$ , where

$(\mathrm{E}_{j})$ $\{$

$\frac{\partial g(u_{i})}{\partial t}-\triangle b(u_{i})+\mathrm{d}\mathrm{i}\mathrm{v}\phi(u_{i})$ $=$ $f_{i}$ in $Q=(0, T)\cross\Omega$ ,

$b(u\dot{.})$ $=$ $0$ on $\Sigma=(0, T)\cross\partial\Omega$ ,

$g(u.\cdot)(0, \cdot)$ $=g(u_{0i})$ in $\Omega$ .Then there exists $\kappa\in S(u_{1}-u_{2})$ such that for $a.e$ . $\tau\in(0, T)$ ,

$\int_{\Omega}(g(u_{1})(\tau, x)-g(u_{2})(\tau, x))^{+}dx$

$\leq$ $\int_{\Omega}(g(u_{01})(x)-g(u_{02})(x))^{+}dx+\int_{0}^{\tau}\int_{\Omega}\kappa(f_{1}(t, x)-f_{2}(t, x))$dxdt.

(2.1)

Moreover; for any $u_{0}$ satisfying (H2) there exists a unique solution for (E).

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In order to prove this theorem, we start with the following lemma.

Lemma 2.2. Let $u$ be a renormalized solution of (E), Then

$\int_{Q}S_{0}(u-k)((h(u)(\nabla b(u)-\phi(u))+h(k)\phi(k))\cdot\nabla\xi$

$- \xi fh(u)-\xi_{t}\int_{k}^{u}h(r)dg(r)+\xi h’(u)(\nabla b(u)-\phi(u))\cdot\nabla u)$ dxdt

$\leq$ $\int_{\Omega}\xi(0,x)S_{0}(u_{0}-k)\int_{k}^{u0}h(r)dg(r)dx$ (2.2)

and

$\int_{Q}S_{0}(-k-u)((h(u)(\nabla b(u)-\phi(u))+h(-k)\phi(-k))\cdot$ V4

$- \xi fh(u)-\xi_{t}\int_{-k}^{u}h(r)dg(r)+\xi h’(u)(\nabla b(u)-\phi(u))\cdot\nabla u)$ dxdt

$\geq$ $\int_{\Omega}\xi(0, x)S_{0}(-k-u_{0})\int_{-k}^{u0}h(r)dg(r)dx$ (2.3)

for any $h\in C_{0}^{1}(\mathbb{R})^{+}$ and for any pair $(k, \xi)$ satisfying

$(k, \xi)\in \mathbb{R}\cross C_{0}^{\infty}([0, T)\cross\Omega)^{+}$ or $(k, \xi)\in \mathbb{R}^{+}\cross C_{0}^{\infty}([0, T)\mathrm{x}\overline{\Omega})^{+}$ , (2.4)

where $\mathbb{R}^{+}=[0, \infty)$ and $X^{+}$ denotes all nonnegative functions which belong to$X$ with $X=C_{0}^{1}(\mathbb{R}),$ $C_{0}^{\infty}([0, T)\cross\Omega)$ or $C_{0}^{\infty}([0, T)\cross\overline{\Omega})$ .

Remark 2.3. Note that if $u$ is a renormalized solution of (E), then $-u$ isa renormalized solution of the problem associated with the equation $\tilde{g}(v)_{t}-$

$\triangle\tilde{b}(v)+\mathrm{d}\mathrm{i}\mathrm{v}\tilde{\phi}(v)=\tilde{f,}$ where $\tilde{g}(r)=-g(-r),$ $\tilde{b}(r)=-b(-r)_{\rangle}\tilde{\phi}(r)=-\phi(-r)$ ,$\tilde{f}=-f$ and initial data $\tilde{u_{0}}=-u_{0}$ .

Sketch of the proof of Lemma 2.2. Due to Remark 2.3 it is sufficient to show(2.2). Let $h\in C_{0}^{1}(\mathbb{R})^{+}$ . For $\epsilon>0$ let $N_{\epsilon}\in W^{1,\infty}(\mathbb{R})$ be defined by $N_{\epsilon}(r)=$

$\inf(r^{+}/\epsilon, 1)$ . For $\epsilon>0$ we see that $N_{\epsilon}(u-k)\xi\in L^{2}(0,T;H_{0}^{1}(\Omega))\cap L^{\infty}(Q)$ forany pair $(k, \xi)$ satisfying (2.4). Since $u$ is arenormalized solution we find

$G_{h}(u)$ $:=$ $\int_{0}^{u}h(r)dg(r)\in L^{1}(Q)$ ,

$\frac{\partial G_{h}(u)}{\partial t}$$\in$ $L^{2}(0, T;H^{-1}(\Omega))+L^{1}(Q)$

and $G_{h}(u)(0, \cdot)$ $=$ $\int_{0}^{u\mathrm{o}}h(r)dg(r)\in H^{-1}(\Omega)+L^{1}(\Omega)$ .

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Therefore we have that

$- \int_{0}^{T}\langle G_{h}(u)_{t}, N_{\epsilon}(u-k)\xi\rangle dt$

$= \int_{Q}\xi_{t}\int_{u\mathrm{o}}^{u}N_{\epsilon}(r-k)dG_{h}(r)$ dxdt

$= \int_{Q}\xi_{t}\int_{u\mathrm{o}}^{u}N_{\epsilon}(r-k)h(r)dg(r)$ dxdt.

Letting $\epsilonarrow 0$ on the right we obtain

$\int_{Q}\xi_{t}\int_{u\mathrm{o}}^{u}S_{0}(r-k)h(r)dg(r)$ dxdt

$=$ $\int_{\Omega}\xi(0, x)S_{0}(u_{0}-k)\int_{k}^{u_{0}}h(r)dg(r)dx$

$+ \int_{Q}\xi_{t}S_{0}(u-k)\int_{k}^{u}h(r)dg(r)$ dxdt.

On the other hand we have

$- \int_{0}^{T}\langle$ $G_{h}(u)_{t}$ , N\’e $(u-k)\xi\rangle$ $dt$

$=$ $\int_{Q}(N_{\epsilon}(u-k)\xi)_{t}\int_{u\mathrm{o}}^{u}h(r)dg(r)$ dxdt

$=$ $- \int_{Q}fh(u)N_{e}(u-k)\xi dxdt$

$+ \int_{Q}(\nabla b(u)-\phi(u))\cdot\nabla(h(u)N_{\epsilon}(u-k)\xi)dxdt$

and since $fh(u)N_{\epsilon}(u-k)\xi\in L^{1}(Q)$ from the Lebesgue convergence theoremit follows that

$\lim_{\epsilonarrow 0}(-\int_{Q}fh(u)N_{\epsilon}(u-k)\xi dxdt)=-\int_{Q}fh(u)S_{0}(u-k)\xi dxdt$ .

As to the second integral we find

$\int_{Q}(\nabla b(u)-\phi(u))\cdot\nabla(h(u)N_{\epsilon}(u-k)\xi)dxdt$

$= \int_{Q}N_{\epsilon}(u-k)(\xi h’(u)(\nabla b(u)-\phi(u))\cdot\nabla u+h(u)(\nabla b(u)-\phi(u))\cdot\nabla\xi)$ dxdt

$+ \frac{1}{\epsilon}\int_{Q\cap\{0<u-k<\epsilon\}}\xi h(u)(\nabla b(u)-\phi(u))\cdot\nabla udxdt$

$arrow\int_{Q}S_{0}(u-k)(\xi h’(u)(\nabla b(u)-\phi(u))\cdot\nabla u+h(u)(\nabla b(u)-\phi(u))\cdot\nabla\xi)$ dxdt

$+ \lim_{\epsilonarrow 0}\frac{1}{\epsilon}\int_{Q\cap\{0<u-k<\epsilon\}}\xi h(u)(\nabla b(u)-\phi(u))\cdot\nabla udxdt$ as $\epsilonarrow 0$ .

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Due to the divergence theorem we obtain

0 $=$ $\int_{Q}\mathrm{d}\mathrm{i}\mathrm{v}(\xi\int_{0}^{N_{e}(u-k)}h(\epsilon r+k)(\nabla b(\epsilon r+k)-\phi(\epsilon r+k))dr)$ dxdt

$=$ $\int_{Q}\int_{0}^{N_{\epsilon}(u-k)}h(\epsilon r+k)(\nabla b(\epsilon r+k)-\phi(\epsilon r+k))\cdot\nabla\xi$ drdxdt

$+ \frac{1}{\epsilon}\int_{Q\cap\{0<u-k<\epsilon\}}\xi h(u)(\nabla b(u)-\phi(u))\cdot\nabla udxdt$

whenever the pair $(k, \xi)$ satisfies (2.4), hence

$\lim_{\epsilonarrow}\inf_{0}\frac{1}{\epsilon}\int_{Q\cap\{0<u-k<\epsilon\}}\xi h(u)(\nabla b(u)-\phi(u))\cdot\nabla udxdt$

$\geq$ $\int_{Q}S_{0}(u-k)h(k)\phi(k)\cdot\nabla\xi$ .

Combining these estimates above we finally obtain (2.2). $\square$

We next prove the following renormalized Kato inequality.

Lemma 2.4. Let $u_{0:}$ : $\Omegaarrow\overline{\mathbb{R}}$ be measurable with $g(u_{0\dot{\iota}})\in L^{1}(\Omega),$ $f_{i}\in L^{1}(Q)$

and let $u_{i}$ be a renormalized solution of $(\mathrm{E}_{i})$ for $i=1,2$ . Then there exists$\kappa\in S(u_{1}-u_{2})$ such that for $a.e$ . $t\in(0, T)$ ,

$- \int_{Q\cap\{u_{1}>u_{2}\}}\xi_{t}\int_{u_{2}}^{u_{1}}h(r)dg(r)$ dxdt

$- \int_{\Omega\cap\{u_{01}>u_{02}\}}\xi(0, x)\int_{u_{02}}^{u_{01}}h(r)dg(r)dx$

$+ \int_{Q\cap\{u_{1}>u_{2}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))$

$-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))\cdot\nabla\xi$ dxdt

$+ \int_{Q\cap\{u_{1}>u_{2}\}}\xi(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt

$\leq$ $\int_{Q}\xi\kappa(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$ (2.5)

for all $h\in C_{0}^{1}(\mathbb{R})^{+}$ and all $\xi\in C_{0}^{\infty}([0, T)\cross\overline{\Omega})^{+}$ .

Sketch of the proof of Lemma 2.4. We adopt the method of doubling variablesintroduced by Kruzhkov. Thus we choose two different pairs of variables $(s, y)$

and $(t, x)$ in $Q$ and consider $u_{1},$ $f_{1}$ as functions in $(s, y),$ $u_{2},$ $f_{2}$ in $(t, x)$ . Let$\xi\in C_{0}^{\infty}([0,T)\cross \mathbb{R}^{N})^{+}$ be such that

$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\xi\cap([0,T)\cross \mathbb{R}^{N})\subset([0,T)\cross B)$

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where $B$ is aball for which

either $B\cap\partial\Omega=\emptyset$ or $B\subset\subset B’$ and$B’\cap\partial\Omega$ is apart of the graph of aLipschitz continuous function.

(2.6)

Then there exists asequence of mollifiers $\sigma_{l}$ defined in $\mathbb{R}$ with $\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\sigma\iota\subset$

$(-2/l, 0)$ and there exists asequence of mollifiers $\rho_{n}$ in $\mathbb{R}^{N}$ s $\mathrm{u}\mathrm{c}$h that $x\vdasharrow$

$\rho_{n}(x-y)\in C_{0}^{\infty}(\Omega)$ for any $y\in B\cap\Omega$ ,

$\mu_{n}(x)=\int_{\Omega}\rho_{n}(x-y)dy$

is an increasing sequence for any $x\in B$ and $\mu_{n}(x)=1$ for any $x\in B$ with$d(x, \mathbb{R}^{N}\backslash \Omega)>c/n$, where $c$ is apositive constant depending on $B$ . Further,for sufficiently large 1and $n$ , the function $\xi^{(l,n)}$ defined by

$\xi^{(l,n)}(t, x,s, y)=\xi(t,x)\rho_{n}(x-y)\sigma_{l}(t-s)$

satisfies

$(s,y)\mapsto\succ\xi^{(l,n)}(t,x, s,y)\in C_{0}^{\infty}([0, T)\cross\overline{\Omega})$ for any $(t, x)\in Q$ ,$(t, x)\vdash+\xi^{(l,n)}(t,x, s, y)\in C_{0}^{\infty}([0, T)\cross\Omega)$ for any $(s, y)\in Q$ ,

and the function $\xi^{(n)}$ defined by

$\xi^{(n)}=\int_{Q}\xi^{(l,n)}(t, x, s, y)dyds=\xi\mu_{n}$

satisfies

$\xi^{(n)}\in C_{0}^{\infty}([0, T)\cross\Omega)$ , $0\leq\xi^{(m)}\leq\xi^{(n)}\leq\xi$ for any $m\leq n$ .

We thus apply Lemma 2.2 with $u=u_{1},$ $k=0,$ $f=f_{1},$ $\xi=\xi^{(l,n)}(t, x, \cdot)$ and$h(\cdot)N_{\epsilon}(\cdot-u_{2}^{+})$ in the place of $h$ , and we have

$\int_{Q}(\xi^{(l,n)})_{\theta}\int_{u_{2}^{+}}^{u_{1}^{+}}h(r)N_{\epsilon}(r-u_{2}^{+})dg(r)$ dyds

$+ \int_{\Omega}\xi^{(l,n)}(t, x,0, y)\int_{u_{2}^{+}}^{u_{01}^{+}}h(r)N_{\epsilon}(r-u_{2}^{+})dg(r)dy$

$+ \int_{Q}f_{1}h(u_{1})N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)}dyds$

$\geq$ $\int_{Q}(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla_{y}(h(u_{1})N_{e}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)})$dyds (2.7)

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and since $u_{2}$ is arenormalized solution of $(\mathrm{E}_{2})$ we obtain from (1.1) that

$\int_{Q}(\xi^{(l,n)})_{t}\int_{u_{1}^{+}}^{u_{2}}h(r)N_{\epsilon}(u_{1}^{+}-r^{+})dg(r)$ dxdt

$+ \int_{\Omega}\xi^{(l,n)}(0, x, s, y)\int_{u_{1}^{+}}^{u_{02}}h(r)N_{\text{\’{e}}}(u_{1}^{+}-r^{+})dg(r)dx$

$+ \int_{Q}f_{2}h(u_{2})N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)}dxdt$

$=$ $\int_{Q}(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla_{x}(h(u_{2})N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)})dxdt$ . (2.8)

Integrating (2.7) in $(t, x)$ and (2.8) in $(s,y)$ , respectively, over $Q$ and takingtheir difference we obtain

$\int_{Q\cross Q}((\xi^{(l,n)})_{s}\int_{u_{2}^{+}}^{u_{1}^{+}}h(r)N_{\epsilon}(r-u_{2}^{+})dg(r)$

$-( \xi^{(l,n)})_{t}\int_{u_{1}^{+}}^{u_{2}}h(r)N_{\epsilon}(u_{1}^{+}-r^{+})dg(r))$ dydsdxdt

$+( \int_{Q\mathrm{X}\Omega}\xi^{(l,n)}(t, x, 0, y)\int_{u_{2}^{+}}^{u_{01}^{+}}h(r)N_{\epsilon}(r-u_{2}^{+})dg(r)$ dydxdt

$- \int_{\Omega \mathrm{x}Q}\xi^{(l,n)}(0, x,s, y)\int_{u_{1}^{+}}^{u_{02}}h(r)N_{\epsilon}(u_{1}^{+}-r^{+})dg(r)dydsdx)$

$+ \int_{Q\mathrm{x}Q}(f_{1}h(u_{1})-f_{2}h(u_{2}))N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)}$ dydsdxdt

$\geq$ $\int_{Q\mathrm{x}Q}((\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla_{y}(h(u_{1})N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)})$

$-(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla_{x}(h(u_{2})N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)}))$ dydsdxdt.

We shall denote the three integrals on the left by $J_{1},$ $J_{2}$ and $J_{3}$ , the integralon the right by $J_{4}$ , respectively. We begin with the first term $J_{1}$ .

$\lim_{\epsilonarrow 0}J_{1}$$=$ $\int_{Q\cross Q}((\xi^{(l,n)})_{s}S_{0}(u_{1}^{+}-u_{2}^{+})\int_{u_{2}^{+}}^{u_{1}^{+}}h(r)dg(r)$

$-( \xi^{(l,n)})_{t}S_{0}(u_{1}^{+}-u_{2}^{+})\int_{u_{1}^{+}}^{u_{2}}h(r)dg(r))$ dydsdxdt

$=$ $\int_{Q\mathrm{X}Q}\xi_{t}\rho_{n}\sigma_{l}S_{0}(u_{1}^{+}-u_{2}^{+})\int_{u_{2}^{+}}^{u_{1}^{+}}h(r)dg(r)dydsdxdt$

$- \mathit{1}_{\mathrm{x}Q\cap\{u_{1}>0\}\cap\{u_{2}<0\}}(\xi^{(l,n)})_{t}\int_{0}^{u_{2}}h(r)dg(r)dydsdxdt$ . (2.9)

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As to $J_{2}$ we see from $\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\sigma_{l}\subset(-2/l, 0)$ that

$\lim_{\epsilonarrow 0}J_{2}$

$=$ $\int_{\Omega \mathrm{x}(0,2/l)\mathrm{x}\Omega}\xi^{(l,n)}(0, x, s, y)S_{0}(u_{1}^{+}-u_{02}^{+})\int_{u_{02}}^{u_{1}^{+}}h(r)dg(r)$dydsdx

$=$ $\int_{\Omega\cross(0,2/l)\cross\Omega}\xi^{(l,n)}(0, x, s, y)S_{0}(u_{1}^{+}-u_{02}^{+})\int_{u_{02}^{+}}^{u_{1}^{+}}h(r)dg(r)$ dydsdx

$+ \int_{\Omega \mathrm{x}(0,2/l)\mathrm{x}\Omega\cap\{u_{1}>0\}\cap\{u_{02}<0\}}\xi^{(l,n)}(0,x, s, y)\int_{u_{02}}^{0}h(r)dg(r)$dydsdx. (2.10)

In the third term we deduce that

$\lim_{\epsilonarrow 0}J_{3}=\int_{Q\mathrm{x}Q\cap\{u_{1}^{+}>u_{2}^{+}\}}(f_{1}h(u_{1})-f_{2}h(u_{2}))\xi^{(l,n)}$ dydsdxdt. (2.11)

It remainds to consider $J_{4}$ . In terms of the divergence theorem we have

$J_{4}= \int_{Q\cross Q}\xi^{(l,n)}N_{\epsilon}(u_{1}-u_{2}^{+})h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}dydsdxdt$

$+ \int_{Q\cross Q}h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla_{y}(N_{\epsilon}(u_{1}-u_{2}^{+})\xi^{(l,n)})$ dydsdxdt

$- \int_{Q\mathrm{x}Q}(\nabla b(u_{2}^{+})-\phi(u_{2}^{+}))\cdot\nabla_{x}(h(u_{2}^{+})N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)})dydsdxdt$

$+ \int_{Q\cross Q\cap\{u_{2}<0\}}(\nabla b(u_{2}^{+})-\phi(u_{2}^{+}))\cdot\nabla_{x}(h(u_{2}^{+})N_{\epsilon}(u_{1}^{+}-u_{2}^{+})\xi^{(l,n)})$ dydsdxdt

$- \int_{Q\cross Q\cap\{u_{2}<0\}}(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla_{x}(h(u_{2})N_{\epsilon}(u_{1}^{+}-u_{2})\xi^{(l,n)})$ dydsdxdt

$= \int_{Q\cross Q}\xi^{(l,n)}N_{\epsilon}(u_{1}-u_{2}^{+})(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2}^{+})(\nabla b(u_{2}^{+})-\phi(u_{2}^{+}))\cdot\nabla u_{2}^{+})$ dydsdxdt

$+ \int_{Q\mathrm{x}Q}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))-h(u_{2}^{+})(\nabla b(u_{2}^{+})-\phi(u_{2}^{+})))$

. $(\nabla_{x}+\nabla_{y})(N_{\epsilon}(u_{1}-u_{2}^{+})\xi^{(l,n)})$ dydsdxdt

$- \int_{Q\cross Q\cap\{u_{2}<0\}}N_{\epsilon}(u_{1}^{+})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla_{x}(h(u_{2})\xi^{(l,n)})dydsdxdt$

Then we find

$\lim_{\epsilonarrow}\inf_{0}\int_{Q_{\mathrm{X}}Q}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))-h(u_{2}^{+})(\nabla b(u_{2}^{+})-\phi(u_{2}^{+})))$

$.(\nabla_{x}+\nabla_{y})(N_{\epsilon}(u_{1}-u_{2}^{+})\xi^{(l,n)})$ dydsdxdt

$\geq$ $\int_{Q\mathrm{x}Q\cap\{u_{1}>u_{2}^{+}\}}\rho_{n}\sigma\iota(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))$

$-h(u_{2}^{+})(\nabla b(u_{2}^{+})-\phi(u_{2}^{+})))\cdot\nabla\xi dydsdxdt$ (2.12)

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$\lim_{\epsilonarrow}\inf_{0}\int_{Q\cross Q}N_{\epsilon}(u_{1}-u_{2}^{+})\xi^{(l,n)}(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dydsdxdt

$\geq$ $\int_{Q\cross Q\cap\{u_{1}>u_{2}^{+}\}}\xi^{(l,n)}(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2}^{+})(\nabla b(u_{2}^{+})-\phi(u_{2}^{+}))\cdot\nabla u_{2})dydsdxdt$. (2.13)

As to the remaining term we obtain from Lemma 2.2 that

$\int_{Q\cap\{u_{2}<0\}}((\xi^{(l,n)})_{t}\int_{u02}^{u_{2}}h(r)dg(r)+\xi^{(l,n)}f_{2}h(u_{2})$

$-(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla(h(u_{2})\xi^{(l,n)}))$ dxdt $\leq 0$ .

Since $1-N_{\epsilon}(u_{1})\geq 0$ , multiplying $(1-N_{\epsilon}(u_{1}))$ to the previous inequality andintegrating in $(s, y)$ over $Q$ we have

$- \int_{Q\mathrm{x}Q\cap\{u_{2}<0\}}N_{\epsilon}(u_{1})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla_{x}(h(u_{2})\xi^{(l,n)})dydsdxdt$

$\geq$ $\int_{Q\mathrm{x}Q\cap\{u_{2}<0\}}((\xi^{(l,n)})_{t}\int_{u_{02}}^{u_{2}}h(r)dg(r)+\xi^{(l,n)}f_{2}h(u_{2})$

$-(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla_{x}(h(u_{2})\xi^{(l,n)}))$ dydsdxdt

$- \int_{Q\mathrm{x}Q\cap\{u_{2}<0\}}N_{\epsilon}(u_{1})(\xi^{(l,n)})_{t}\int_{u_{02}}^{u_{2}}h(r)dg(r)$ dydsdxdt

$- \int_{Q\mathrm{x}Q\cap\{u_{2}<0\}}N_{\epsilon}(u_{1})\xi^{(l,n)}f_{2}h(u_{2})dydsdxdt$

$arrow$ $\int_{Q\mathrm{x}Q\cap\{u_{2}<0\}}((\xi^{(l,n)})_{t}\int_{u_{02}}^{u_{2}}h(r)dg(r)+\xi^{(l,n)}f_{2}h(u_{2})$

$-(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla_{x}(h(u_{2})\xi^{(l,n)}))$ dydsdxdt

$- \int_{Q\mathrm{x}Q\cap\{u_{1}>0\}\cap\{u_{2}<0\}}(\xi^{(l,n)})_{t}\int_{u_{02}}^{u_{2}}h(r)dg(r)dydsdxdt$

$- \int_{Q\mathrm{x}Q\cap\{u_{1}>0\}\cap\{u_{2}<0\}}\xi^{(l,n)}f_{2}h(u_{2})$ dydsdxdt as $\epsilonarrow 0$ . (2.14)

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Combining these estimates (2.9) - (2.14) we deduce that

$\int_{Q}\xi_{t}S_{0}(u_{1}^{+}-u_{2}^{+})\int_{2}^{u_{1}^{+}}+h(r)dg(r)$ dxdt

$+ \int_{\Omega}\xi(0, x)S_{0}(u_{01}^{+}-u_{02}^{+})\int_{u_{02}^{+}}^{u_{01}^{+}}h(r)dg(r)dx$

$+ \int_{Q}\xi\kappa_{+}S_{0}(u_{1})(f_{1}h(u_{1})-(1-S_{0}(-u_{2}))f_{2}h(u_{2}))$ dxdt

$\geq$ $\int_{Q\cap\{u_{1}>u_{2}^{+}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))-h(u_{2}^{+})(\nabla b(u_{2}^{+})-\phi(u_{2}^{+})))\cdot\nabla\xi$ dxdt

$+ \int_{Q\cap\{u_{1}>u_{2}^{+}\}}\xi(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2}^{+})(\nabla b(u_{2}^{+})-\phi(u_{2}^{+}))\cdot\nabla u_{2})$ dxdt

$+ \lim_{narrow\infty}\int_{Q\cap\{u_{2}<0\}}((\xi^{(n)})_{t}\int_{u_{02}}^{u_{2}}h(r)dg(r)+f_{2}h(u_{2})\xi^{(n)}$

$-(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla(h(u_{2})\xi^{(n)}))$ dxdt

for any $\xi\in C_{0}^{\infty}([0, T)\cross B)^{+}$ , where $\kappa_{+}\in S(u_{1}^{+}-u_{2}^{+})$ . We also obtain fromRemark 2.3 that there exists $\kappa_{-}\in S(u_{2}^{-}-u_{1}^{-})$ such that

$\int_{Q}\xi_{t}S_{0}(u_{2}^{-}-u_{1}^{-})\int_{-u_{2}^{-}}^{-u_{1}^{-}}h(r)dg(r)$ dxdt

$+ \int_{\Omega}\xi(0, x)S_{0}(u_{02}^{-}-u_{01}^{-})\int_{-u_{02}^{-}}^{-u_{01}^{-}}h(r)dg(r)dx$

$+ \int_{Q}\xi\kappa_{-}S_{0}(u_{2}^{-})((1-S_{0}(u_{1}^{+}))f_{1}h(u_{1})-f_{2}h(u_{2}))$ dxdt

$\geq$ $\int_{Q\cap\{u_{2}^{-}>u_{1}^{-}\}}(h(u_{1})(\nabla b(-u_{1}^{-})-\phi(-u_{1}^{-}))$

$-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))\cdot\nabla\xi$ dxdt

$+ \int_{Q\cap\{u_{2}^{-}>u_{1}^{-}\}}\xi(h’(u_{1})(\nabla b(-u_{1}^{-})-\phi(-u_{1}^{-}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt

- $\lim_{narrow\infty}\int_{Q\cap\{u_{1}>0\}}((\xi^{(n)})_{t}\int_{u_{01}}^{u_{1}}h(r)dg(r)+f_{1}h(u_{1})\xi^{(n)}$

$-(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla(h(u_{1})\xi^{(n)}))$ dxdt

for any $\xi\in C_{0}^{\infty}([0,T)\cross B)^{+}$ . Since $\tilde{\kappa}=(1-S_{0}(u_{1}^{+}))S_{0}(-u_{2}^{+})\kappa_{-}+S_{0}(u_{1}^{+})\kappa_{+}=$

$(1-S_{0}(-u_{2}))S_{0}(u_{1})\kappa_{+}+S_{0}(-u_{2})\kappa_{-}\in S(u_{1}-u_{2})$ , summing up the previous

69

two inequalities we have

$\int_{Q}\xi_{t}S_{0}(u_{1}-u_{2})\int_{u_{2}}^{u_{1}}h(r)dg(r)$ dxdt

$+ \int_{\Omega}\xi(0, x)S_{0}(u_{01}-u_{02})\int_{u_{02}}^{u_{01}}h(r)dg(r)dx$

$+ \int_{Q}\xi\tilde{\kappa}(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$

$- \int_{Q\cap\{u_{1}>u_{2}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))\cdot\nabla\xi dxdt$

$- \int_{Q\cap\{u_{1}>u_{2}\}}\xi(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt

$\geq$ $\lim_{narrow\infty}\int_{Q\cap\{u_{2}<0\}}((\xi^{(n)})_{t}\int_{u_{02}}^{u_{2}}h(r)dg(r)+\xi^{(n)}f_{2}h(u_{2})$

$-(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla(h(u_{2})\xi^{(n)}))$ dxdt

- $\lim_{narrow\infty}\int_{Q\cap\{u_{1}>0\}}((\xi^{(n)})_{t}\int_{u_{01}}^{u_{1}}h(r)dg(r)+\xi^{(n)}f_{1}h(u_{1})$

$-(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla(h(u_{1})\xi^{(n)}))$ clxdt (2.15)

for any $\xi\in C_{0}^{\infty}([0, T)\cross B)^{+}$ .Now let $\xi\in C_{0}^{\infty}([0, T)\cross B)^{+}$ . Then $\xi^{(m)}=\xi\mu_{m}\in C_{0}^{\infty}([0, T)\cross\Omega)$ and we

see that

$- \int_{Q\cap\{u_{1}>u_{2}\}}\xi_{t}^{(m)}\int_{u_{2}}^{u_{1}}h(r)dg(r)$ dxdt

$- \int_{\Omega\cap\{u_{01}>u_{02}\}}\xi^{(m)}(0, x)\int_{u02}^{u_{01}}h(r)dg(r)dx$

$+ \int_{Q\cap\{u_{1}>u_{2}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))$

$-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))\cdot\nabla\xi^{(m)}$ dxdt

$+ \int_{Q\cap\{u\iota>u_{2}\}}\xi^{(m)}(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt

$\leq$ $\int_{Q}\xi^{(m)}\kappa(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$ .

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$\mathrm{n}\mathrm{c}\mathrm{e}\xi^{(m)}=\xi-\xi(1-\mu_{m})$ we obtain from (2.15) that

$\int_{Q\cap\{u_{1}>u_{2}\}}\xi_{\mathrm{t}}\int_{u_{2}}^{u_{1}}h(r)dg(r)$ dxdt

$+ \int_{\Omega\cap\{u_{01}>u_{02}\}}\xi(0, x)\int_{u_{02}}^{u_{01}}h(r)dg(r)dx$

$+ \int_{Q}\xi\overline{\kappa}(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$

$+ \int_{Q}\xi(\kappa-\tilde{\kappa})(f_{1}h(u_{1})-f_{2}h(u_{2}))$ dxdt

$- \int_{Q\cap\{u_{1}>u_{2}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))$

$-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))$ .V4 dxdt

$- \int_{Q\cap\{u_{1}>u_{2}\}}\xi(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt

$\geq$ $\int_{Q\cap\{u_{1}>u_{2}\}}(\xi(1-\mu_{m}))_{t}\int_{u_{2}}^{u_{1}}h(r)dg(r)$ dxdt

$+ \mathit{1}_{\cap\{u_{01}>u_{02}\}}^{(\xi(1-\mu_{m}))(0,x)}\int_{u_{02}}^{u_{01}}h(r)dg(r)dx$

$+ \int_{Q}(\xi(1-\mu_{m}))\tilde{\kappa}(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$

$+ \int_{Q}(\xi(1-\mu_{m}))(\kappa-\tilde{\kappa})(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$

$- \int_{Q\cap\{u_{1}>u_{2}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))$

$-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))\cdot\nabla(\xi(1-\mu_{m}))$ dxdt

$- \int_{Q\cap\{u_{1}>u_{2}\}}(\xi(1-\mu_{m}))(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt

$\geq\lim_{narrow\infty}\int_{Q\cap\{u_{2}<0\}}((\xi(1-\mu_{m})\mu_{n})_{t}\int_{u_{02}}^{u_{2}}h(r)dg(r)+(\xi(1-\mu_{m})\mu_{n})f_{2}h(u_{2})$

$-(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla(h(u_{2})\xi(1-\mu_{m})\mu_{n}))dxdt$

$- \mathrm{h}.\mathrm{m}narrow\infty\int_{Q\cap\{u_{1}>0\}}((\xi(1-\mu_{m})\mu_{n})_{t}\int_{u_{01}}^{u_{1}}h(r)dg(r)+(\xi(1-\mu_{m})\mu_{r\iota})f_{1}h(u_{1})$

$-(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla(h(u_{1})\xi(1-\mu_{m})\mu_{n}))$ dxdt

$+[_{\cap}(\xi(1-\mu_{m}))(\kappa-\tilde{\kappa})(f_{1}h(u_{1})-f_{2}h(u_{2}))$ dxdt.

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In the last term on the right it is clear that the integral converges to 0asm $arrow\infty$ . Since if n is large enough then $\mu_{n}=1$ on $\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\mu_{m}$ we find that$(1-\mu_{m})\mu_{n}=\mu_{n}-\mu_{m}$ . Therefore the remaining terms tend to 0as m $arrow\infty$ .It implies that

$\int_{Q}\xi_{t}S_{0}(u_{1}-u_{2})\int_{u_{2}}^{u_{1}}h(r)dg(r)$ dxdt

$+ \int_{\Omega}\xi(0, x)S_{0}(u_{01}-u_{02})\int_{u_{02}}^{u_{01}}h(r)dg(r)dx$

$+ \int_{Q}\xi\kappa(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$

$- \int_{Q\cap\{u_{1}>u_{2}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))\cdot\nabla\xi dxdt$

$- \int_{Q\cap\{u_{1}>u_{2}\}}\xi(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt $\geq$ 0,

with $\kappa\in S(u_{1}-u_{2})$ .

To this end, let $B_{0}\subset\subset\Omega$ be such that $\bigcup_{i=0}^{n}B_{j}$ is acovering of $\Omega$ , where $B_{j}$

is aball satisfing (2.6) for $i=1,$ $\cdots,$ $n$ . Let $\{\nu\dot{.}\}_{i=0}^{n}$ be such that $\nu.\cdot\in C_{0}^{\infty}(B:)$

for $i=0,1,$ $\cdots,$ $n$ and let $\xi\in C_{0}^{\infty}([0, T)\cross\overline{\Omega})^{+}$ . Then for $i=0,1,$ $\cdots,$ $n$ wehave

$\int_{Q}(\xi\nu_{i})_{t}S_{0}(u_{1}-u_{2})\int_{u_{2}}^{u_{1}}h(r)dg(r)$ dxdt

$+ \int_{\Omega}(\xi\nu.\cdot)(0, x)S_{0}(u_{01}-u_{02})\int_{u_{02}}^{u_{01}}h(r)dg(r)dx$

$+ \int_{Q}(\xi\nu\dot{.})\kappa(f_{1}h(u_{1})-f_{2}h(u_{2}))$ dxdt

$- \int_{Q\cap\{u_{1}>u_{2}\}}(h(u_{1})(\nabla b(u_{1})-\phi(u_{1}))-h(u_{2})(\nabla b(u_{2})-\phi(u_{2})))\cdot\nabla(\xi\nu\dot{.})$ dxdt

$- \int_{Q\cap\{u_{1}>u_{2}\}}(\xi\nu\dot{.})(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt $\geq$ 0.

Since $\xi=\sum_{=0}^{n}\dot{.}(\xi\nu_{i})$ we obtain (2.5) for any $\xi\in C_{0}^{\infty}([0, T)\cross\overline{\Omega})^{+}$ . $\square$

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3Proof of the main theorem

We finally give the proof of our main result.

Proof of Theorem 2.1. Let $u_{i}$ be arenormalized solution of (Ej) for $i=1,2$ .Choosing $\xi=\alpha\otimes 1$ with $\alpha\in C_{0}^{\infty}([0, T))$ in (2.5) there exists $\kappa\in S(u_{1}-u_{2})$

$- \int_{Q}\alpha_{t}(S_{0}(u_{1}-u_{2})\int_{u_{2}}^{u_{1}}h(r)dg(r)-S_{0}(u_{01}-u_{02})\int_{u_{02}}^{u_{01}}h(r)dg(r))$ dxdt

$+ \int_{Q\cap\{u_{1}>u_{2}\}}\alpha(h’(u_{1})(\nabla b(u_{1})-\phi(u_{1}))\cdot\nabla u_{1}$

$-h’(u_{2})(\nabla b(u_{2})-\phi(u_{2}))\cdot\nabla u_{2})$ dxdt

$\leq$ $\int_{Q}\alpha\kappa(f_{1}h(u_{1})-f_{2}h(u_{2}))dxdt$ (3.1)

for any $h\in W^{1,\infty}(\mathbb{R})$ with compact support. We now define the function$h_{n}\in W^{1,\infty}(\mathbb{R})$ by $h_{n}(r)= \inf((n+1-|r|)^{+}, 1)$ and replace $h$ by $h_{n}$ in (3.1).As to the second integral on the left we divide as

$\int_{Q\cap\{u_{1}>u_{2}\}}\alpha h_{n}’(u_{1})\nabla b(u_{1})\cdot\nabla u_{1}dxdt-\int_{Q\cap\{u_{1}>u_{2}\}}\alpha h_{n}’(u_{2})\nabla b(u_{2})\cdot\nabla u_{2}dxdt$

$- \int_{Q\cap\{u_{1}>u_{2}\}}\alpha(h_{n}’(u_{1})\phi(u_{1})\cdot\nabla u_{1}-h_{n}’(u_{2})\phi(u_{2})\cdot\nabla u_{2})$dxdt.

Since $u_{1},$ $u_{2}$ are renormalized solutions we see from (1.2) that the first twointegrals on the right tend to 0as $narrow\infty$ . Moreover, thanks to the divergencetheorem we have

$- \int_{Q\cap\{u_{1}>u_{2}\}}\alpha(h_{n}’(u_{1})\phi(u_{1})\cdot\nabla u_{1}-h_{n}’(u_{2})\phi(u_{2})\cdot\nabla u_{2})dxdt$

$=$ $\int_{Q}\alpha \mathrm{d}\mathrm{i}\mathrm{v}(-\int_{\inf(u_{1},u_{2})}^{u_{1}}h_{n}’(r)\phi(r)dr)$ $dxdt=0$ .

Therefore the second integral on the left in (3.1) converges to 0as $h=h_{n}arrow 1$

and it implies that

$- \int_{Q}\alpha_{t}((g(u_{1})(t,x)-g(u_{2})(t, x))^{+}-(g(u_{01})(x)-g(u_{02})(x))^{+})$ dxdt

$\leq$ $\int_{Q}\alpha\kappa(f_{1}(t, x)-f_{2}(t,x))dxdt$

for all $\alpha\in C_{0}^{\infty}([0,T))$ . We thus conclude the proof of our main theorem. 0

73

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