1$) it seems hopeless to obtain such results without further assumptions. Therefore, the structure condition (H3), often called the angle condition, is assumed in [1] and [2]. In this note we assume that the matrix $\sigma$ satisfies another type of assumption \vskip .3 true cm \noindent (H3)' (sign condition) For all $i=1,\hdots,m$, $$ \sigma_i(x,u,F)\cdot F_i\ge 0\,. $$ Here $\sigma_i$ and $F_i$ are respectively the $i$-th row vectors of $\sigma$, $F$, and the dot denotes the inner product in $\Bbb R^n$. This indicates that system (1.1) is weakly-coupled. Using (H3)' to replace (H3) and assuming (H0)-(H2), we can prove all the results in [1] and [2]. To make this note short we are only concerned with system (1.1). In [1], a notion of solution where the weak derivative $Du$ is replaced by the approximate derivative $\ap Du$ was introduced. This kind of notion is useful when one considers the case $1

0$ and $|D\eta(y)|\le C(1+|y|)^{-1}$ for some constant $C<\infty$. \endproclaim We will prove the following conclusion. \proclaim{Theorem 2} Suppose (H0), (H1), (H2) and (H3)' hold, and that one of the following conditions is satisfied. \parindent=25pt \item{(i)} $F\mapsto\sigma(x,u,F)$ is a $C^1$ function. \item{(ii)} There exists a function $W: \varOmega\times \Bbb R^m \times \Bbb M^{m\times n}\to \Bbb R$ such that $\sigma(x,u,F)=\frac{\partial W}{\partial F}(x,u,F)$ and $F\mapsto W(x,u,F)$ is convex and $C^1$. \item{(iii)} $\sigma$ is strictly monotone with respect to $F$, i.e., $\sigma$ is monotone and $\big(\sigma(x,u,F)-\sigma(x,u,G)\big):(F-G)=0$ implies $F=G$. \parindent=12.0pt \noindent Also assume that $\mu$ is an $\Bbb R^m$-valued Radon measure on $\varOmega$ with finite mass. Then Problem (1.1)-(1.2) has a solution in the sense of Definition 1, which satisfies the weak Lebesgue space estimate $$ \|u\|^*_{L^{s^*,\infty}(\varOmega)}+\|\ap Du\|^*_{L^{s,\infty} (\varOmega)}\le C(c_1,c_2, \|\mu\|_{\Cal M},\meas\varOmega)\, $$ where $$ s=\frac n{n-1}(p-1),\quad s^*=\frac{ns}{n-s}=\frac{n}{n-p}(p-1)\,. $$ \endproclaim Theorem 2 remains true if we assume that (H0), (H2), (H3)' and the matrix $\sigma$ is strictly $p$-quasi-monotone, see Corollary 4 in [1]. We will follow the same procedure as in [1] to prove Theorem 2. Instead of writing all details of the proof, we will only prove Lemmas 10 and 11 in [1] where the structure condition (H3) has been used. Actually we will only show how to prove the essential parts of Lemmas 10 and 11 when we use (H3)' to replace (H3). Our main idea is to choose slightly different test functions. The relationship between (H3) and (H3)' will also be established in this section. In the scalar case $m=1$, (H3) or (H3)' is redundant. In the vector case $m=2$, (H3) implies (H3)' by choosing the vector $a=(0,1)\in\Bbb R^2$ or $a=(1,0)\in\Bbb R^2$ in (H3). Nevertheless there seems no direct implication between (H3) and (H3)' when $m\ge 3$. To our knowledge, there are two known examples satisfying (H0)-(H3). \proclaim{Example 3} $$\sigma(x,u,F)=a(u)|F|^{p-2}F$$ where $a:\Bbb R^m\rightarrow \Bbb R$ is a continuous function, which is bounded from above and below by positive constants. \endproclaim \proclaim{Example 4} $$\sigma(x,u,F)=\big((FA):F\big)^{(p-2)/2}FA$$ where $A=A(x,u):\varOmega\times \Bbb R^m\to\Bbb M^{n\times n}$ is a symmetric matrix satisfying the ellipticity condition $\nu_1|\zeta|^2\le\sum^n_{i,j=1}A_{ij}(x,u)\zeta_i\zeta_j\le \nu_2 |\zeta|^2 $ with constants $\nu_1, \nu_2>0$. \endproclaim It is not difficult to verify that Examples 3 and 4 satisfy the sign condition (H3)'. Next, we present a linear elliptic system which satisfies (H0)-(H2) and (H3)', but does not satisfy (H3). To some extent, this example shows the restriction of the structure condition (H3). Since the sign condition (H3)' requires that the components of the matrix $\sigma$ can be only weakly-coupled, this shows the limitation of (H3)'. Therefore, it would be interesting to prove the same results in [1] and [2] under a weaker condition than (H3) and (H3)'. \proclaim{Example 5} Let $$F=\pmatrix F_{11}&F_{12}\\ F_{21}&F_{22}\endpmatrix \in \Bbb M^{2\times 2}\,,\quad \sigma(F)=\pmatrix \varepsilon F_{11}&\varepsilon^{\alpha} F_{12}\\ F_{21}&F_{22}\endpmatrix\in \Bbb M^{2\times 2} $$ with $1\leq \alpha $, $0<\varepsilon\le 1/5$. \endproclaim It is obvious that $\sigma(F)$ satisfy (H0), (H1), (H2) and (H3)'. We will show that $\sigma(F)$ does not satisfy (H3). Choosing $a=(\varepsilon^{1/2},(1-\varepsilon)^{1/2})\in \Bbb R^2$, we have $|a|=1 $ and $$ M=\Id-a\otimes a=\pmatrix 1-\varepsilon &-\varepsilon^{1/2} (1-\varepsilon)^{1/2}\\ -\varepsilon^{1/2}(1-\varepsilon)^{1/2}& \varepsilon\endpmatrix\in \Bbb M^{2\times 2}. $$ Now choosing $F=\pmatrix 1&1\\1&1\endpmatrix$, we have $\sigma(F)=\pmatrix \varepsilon&\varepsilon^{\alpha}\\1&1\endpmatrix$ and $$ MF=\pmatrix 1-\varepsilon -\varepsilon^{1/2}(1-\varepsilon)^{1/2}& 1-\varepsilon -\varepsilon^{1/2}(1-\varepsilon)^{1/2}\\ \varepsilon -\varepsilon^{1/2}(1-\varepsilon)^{1/2}& \varepsilon-\varepsilon^{1/2}(1-\varepsilon)^{1/2}\endpmatrix. $$ Therefore, $$ \align \sigma(F):MF &=(\varepsilon+\varepsilon^{\alpha}) \big(1-\varepsilon -\varepsilon^{1/2}(1 -\varepsilon)^{1/2}\big) +2\big( \varepsilon -\varepsilon^{1/2}(1-\varepsilon)^{1/2}\big)\\ &<\varepsilon+\varepsilon^{\alpha}+2\varepsilon -2\varepsilon^{1/2}(1 -\varepsilon)^{1/2} \\ &< 4\varepsilon -2\varepsilon^{1/2}(1 -\varepsilon)^{1/2}\\ & <2\varepsilon^{1/2}\big(2\varepsilon^{1/2}- (1 -\varepsilon)^{1/2}\big) \le 0\,, \endalign $$ which implies that (H3) does not hold for this example. \head 2. Key Lemmas \endhead In [1], the following {\it spherical truncation function} $$ T_{\alpha}(y)=\min\{1,\frac \alpha{|y|}\}y \tag 2.1 $$ is used with $\alpha>0$ and $y=(y_1,\hdots,y_m)\in \Bbb R^m$. Roughly speaking, $T_{\alpha}(u)$ is used as a test function for system (1.1) in [1]. Direct calculation shows that $$ DT_{\alpha}(y)=\cases \Id, &\text{for $|y|\le \alpha$},\\ \frac{\alpha}{|y|}(\Id-\frac y{|y|}\otimes \frac y{|y|}), &\text{for $|y|> \alpha$}. \endcases $$ From this viewpoint, it is easy to understand why the structure condition (H3) is assumed in [1] and [2]. In this note we introduce the following {\it cubic truncation function} $$ \align {\Cal T}_{\alpha}(y)&=\big(T_{\alpha}(y_1),\hdots,T_{\alpha}(y_m)\big)\\ &=\big(\max\{-\alpha,\min\{\alpha,y_1\}\},\hdots, \max\{-\alpha,\min\{\alpha,y_m\}\}\big) \tag 2.2 \endalign $$ to build up some simpler test functions when the sign condition (H3)' is satisfied. To prove Theorem 2, we only need to prove Lemmas 10 and 11 in [1] as we explained in the introduction. First we recall \proclaim{Lemma 10} Let $\varOmega\subset \Bbb R^n$ be an open set and $f\in L^1(\varOmega;\Bbb R^m)$. Assume that $\sigma$ satisfies (H2) and (H3)' with $p=q$ and that $u\in W^{1,p}_0(\varOmega;\Bbb R^m)$ is a solution of $$ -\dive \sigma\big(x,u(x),Du(x)\big)=f(x) \tag 2.3 $$ in the sense of distributions. Then $$ u\in L^{s^*,\infty}(\varOmega;\Bbb R^m),\quad Du\in L^{s,\infty}(\varOmega;\Bbb M^{m\times n}) $$ where $$ s=\frac n{n-1}(p-1),\quad s^*=\frac{ns}{n-s}=\frac{n}{n-p}(p-1). $$ Moreover, $$ \|u\|^*_{L^{s^*,\infty}(\varOmega)}+\|Du\|^*_{L^{s,\infty} (\varOmega)}\le C(c_1,c_2, \|f\|_{L^1}, \meas\varOmega).\tag 2.4 $$ \endproclaim \demo{Proof} In the weak formulation of (2.3) we use the test function ${\Cal T}_ {\alpha}(u)$ to replace $T_{\alpha}(u)$ in [1]. Then we have $$ \sum^m_{i=1}\int_{|u_i|\le \alpha} \sigma_i(x,u,Du)\cdot Du_i \,dx = \int_{\varOmega}f\cdot{\Cal T}_{\alpha}(u)\,dx\le m\alpha\|f\|_{L^1}. $$ By (H2) and (H3)', we obtain $$ \aligned \int_{\max\{|u_1|,\hdots,|u_m|\} \le \alpha} (c_1|Du|^p-c_2) \,dx & \le \int_{\max\{|u_1|,\hdots,|u_m|\} \le \alpha} \sigma(x,u,Du): Du \,dx\\ &\le \sum^m_{i=1}\int_{|u_i|\le \alpha} \sigma_i(x,u,Du)\cdot Du_i \,dx, \endaligned $$ which implies $$ \int_{|u|\le \alpha} |Du|^p \,dx \le \int_{\max\{|u_1|,\hdots,|u_m|\} \le \alpha} |Du|^p \,dx \le C(\alpha\|f\|_{L^1}+\meas\varOmega).\tag 2.5 $$ Actually (2.5) is exactly (4.5) in [1], which is the starting point for proving (2.4). Now following the same procedure as in [1], we obtain (2.4). \enddemo And now we prove Lemma 11 in [1]. It contains a div-curl inequality, which is the crucial ingredient for the argument in [1]. Let $f^k(x):\varOmega\to\Bbb R^m$ denote a bounded sequence in $L^1(\varOmega)$. Suppose that $u^k\in W^{1,1}(\varOmega;\Bbb R^m)$ is the weak solution of the system $$ -\dive \sigma\big(x,u^k(x),\,Du^k(x)\big)=f^k(x)\tag 2.6 $$ with $Du^k\in L_{\loc}^r(\varOmega; \Bbb M^{m\times n})$ and $\sigma^k(x)=\sigma \big(x,u^k(x),Du^k(x)\big)\in L_{\loc}^{r'} (\varOmega; \Bbb M^{m\times n})$ for some $r\in (1,\infty)$. Suppose further that (1) there exist an $s>0$ such that $\int_{\varOmega}|Du^k|^s\,dx\le C$ uniformly in $k$, (2) $\sigma^k$ is equi-integrable, (3) $u^k\to u $ in measure and $u$ is almost everywhere approximately differentiable. And we may assume that $\{Du^k\}$ generates a Young measure $\nu$ (see [3]) such that $\nu_x$ is a probability measure for almost every $x\in \varOmega$, see Theorem 5 (iii) in [1]. Now we recall \proclaim{Lemma 11} Suppose the sequence $\{u^k\}$ is constructed as above. Then (after passage to a subsequence) the sequence $\sigma^k(x)$ converges weakly to $\bar \sigma(x)$ in $ L^1(\varOmega)$ where $$ \bar \sigma(x)=\langle \nu_x,\sigma(x,u(x),\cdot)\rangle= \int_{\Bbb M^{m\times n}}\sigma(x,u(x),\lambda)\,d\nu_x(\lambda)\,. $$ Moreover, the following inequality holds, $$ \int_{\Bbb M^{m\times n}}\sigma(x,u(x),\lambda):\lambda\,d\nu_x(\lambda)\le \bar\sigma(x): \ap Du(x)\quad \text{for a.e. } x\in \varOmega\,. \tag 2.7 $$ \endproclaim \demo{Proof} First we choose $\varphi_1\in C^{\infty}_0(\varOmega;\Bbb R)$ with $\varphi_1\ge 0$ and $\int_{\varOmega}\varphi_1\,dx=1$. Choosing the test function ${\Cal T}_1(u^k-v)\varphi_1$ in (2.6) where $v \in C^1(\varOmega;\Bbb R^m)$ is a suitable comparison function as in [1], we have $$ \int_{\varOmega}\sigma^k:D\big({\Cal T}_1(u^k-v)\varphi_1\big)\,dx= \int_{\varOmega}f^k\cdot {\Cal T}_1(u^k-v)\varphi_1\,dx\,.\tag 2.8 $$ Let $h^k =\sigma^k:D\big({\Cal T}_1(u^k-v)\varphi_1\big)$. Then $$ \align h^k &=\sigma^k:D{\Cal T}_1(u^k-v)D(u^k-v) \varphi_1+ \sigma^k:{\Cal T}_1(u^k-v)\otimes D\varphi_1\\ &=\sum^m_{i=1}\sigma^k_i\cdot Du^k_i \chi_{\{|u^k_i-v_i|\le 1\}} \varphi_1- \sum^m_{i=1}\sigma^k_i\cdot Dv_i \chi_{\{|u^k_i-v_i|\le 1\}} \varphi_1\\ &\quad+ \sigma^k:{\Cal T}_1(u^k-v)\otimes D\varphi_1. \endalign $$ In view of (H3)', the first term on the right hand side of the last equality is nonnegative. Recalling our assumptions, we conclude that $(h^k)^{-}$ is equi-integrable. By Theorem 5, Lemma 6 in [1] and the equi-integrability of $\sigma^k(x)$ together with the convergence of $u^k(x)$ in measure we obtain $$ \align \liminf_{k\to\infty}\int_{\varOmega} h^k\,dx &\ge \int_{\varOmega}\varphi_1\int_{\Bbb M^{m\times n}}\sigma\big(x,u(x),\lambda \big):D{\Cal T}_1(u-v)\big(\lambda-Dv(x)\big)\,d\nu_x(\lambda)\,dx\\ &\quad +\int_{\varOmega}\bar\sigma:{\Cal T}_1(u-v)\otimes D\varphi_1\,dx\,. \endalign $$ The right hand side of (2.8) may be estimated as in [1]. And then we use the blow-up method in [1] to prove (2.7) with some minor changes. \enddemo Now the lemmas involving (H3) in [1] have been proved if (H3) is replaced by (H3)'. Therefore, Theorem 2 and other results in [1] hold. \medskip \noindent {\bf Acknowledgements:} The author would like to thank the anonymous referee for reading an early version of this article and offering his/her valuable suggestions. \Refs \widestnumber\key{3} % \ref \key 1 \by G. Dolzmann, N. Hungerb\"uhler \& S. M\"uller \paper Non-linear elliptic systems with measure-valued right hand side \jour Math. Z. \vol 226(4) \yr 1997 \pages 545--574 \endref % \ref \key 2 \by R. Landes \paper On the existence of weak solutions of perturbated systems with critical growth \jour J. reine angew. Math. \vol 393 \yr 1989 \pages 21--38 \endref % \ref \key 3 \by D. Kinderlehrer \& P. Pedregal \paper Gradient Young measures generated by sequences in Sobolev spaces \jour J. Geom. 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