It is worthy to note that, for any smooth function \(f_{n}\), there exists at least one solution to the obstacle problem (6). Indeed, one can proceed exactly as in [1, 11] to obtain \(W^{1,p}\)-solutions due to assumptions (2)–(4) on a and \(r-1< p\). These solutions, in particular, are also entropy solutions. In this section, using the method of [8] and [19], we establish several auxiliary results on convergence of sequences of entropy solutions when \(f_{n}\rightarrow f\) in \(L^{1}(\varOmega )\).
Lemma 2
Let
\(v_{0}\in K_{g,\psi }\cap L^{\infty }(\varOmega )\), and let
u
be an entropy solution of the obstacle problem associated with
\((f,\psi ,g)\). Then we have
$$\begin{aligned} \int _{\{ \vert u \vert < t\}}\frac{ \vert \nabla u \vert ^{p}}{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \leq C \bigl(1+t^{r}\bigr), \quad \forall t>0, \end{aligned}$$
where
C
is a positive constant depending only on
α, β, p, r, b, \(\Vert j \Vert _{p'}\), \(\Vert \nabla v_{0} \Vert _{p}\), \(\Vert v_{0} \Vert _{\infty }\), and
\(\Vert f \Vert _{1}\).
Proof
Take \(v_{0}\) as a test function in (7). For t large enough such that \(t- \Vert v_{0} \Vert _{\infty }>0\), we get
$$\begin{aligned} \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla u}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x\leq{}& \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \\ & {} + \int _{\varOmega }\bigl(f-b \vert u \vert ^{r-2}u\bigr) T_{t}(u-v_{0})\,\mathrm{d}x. \end{aligned}$$
(11)
We estimate each integration in the right-hand side of (11). It follows from (3) and Young’s inequality with \(\varepsilon >0\) that
$$\begin{aligned}& \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x\leq \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta ( \vert j \vert + \vert \nabla u \vert ^{p-1})\cdot \vert \nabla v_{0} \vert }{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x}\leq \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta \varepsilon ( \vert j \vert ^{p'}+ \vert \nabla u \vert ^{p})}{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x}\quad {} + \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta C(\varepsilon ) \vert \nabla v_{0} \vert ^{p}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x} \leq \varepsilon \int _{\{ \vert v_{0}-u \vert < t\}}\frac{ \vert \nabla u \vert ^{p}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x}\quad {} +C\bigl( \Vert j \Vert _{p'}^{p'}+ \Vert \nabla v_{0} \Vert _{p}^{p}\bigr), \end{aligned}$$
(12)
$$\begin{aligned}& \begin{aligned}[b] - \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x={}&- \int _{\{ \vert u-v_{0} \vert \leq t\}} b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x \\ & {} - \int _{\{ \vert u-v_{0} \vert > t\}}b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x. \end{aligned} \end{aligned}$$
(13)
Note that on the set \(\{ \vert u-v_{0} \vert \leq t\}\),
$$\begin{aligned} \bigl\vert \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \bigr\vert \leq t \bigl\vert t+ \Vert v_{0} \Vert _{\infty } \bigr\vert ^{r-1}\leq C\bigl(1+t ^{r} \bigr), \end{aligned}$$
(14)
where C is a constant depending only on r, \(\Vert v_{0} \Vert _{\infty }\).
On the set \(\{ \vert u-v_{0} \vert > t\}\), we have \(\vert u \vert \geq t- \Vert v_{0} \Vert _{\infty } >0\), thus u and \(T_{t}(u-v_{0})\) have the same sign. It follows
$$\begin{aligned} - \int _{\{ \vert u-v_{0} \vert > t\}} b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x\leq 0. \end{aligned}$$
(15)
Combining (13)–(15), we get
$$\begin{aligned}& - \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x\leq C\bigl(1+t^{r}\bigr), \end{aligned}$$
(16)
$$\begin{aligned}& \begin{aligned}[b] \int _{\{ \vert v_{0}-u \vert < t\}}\frac{ \vert \nabla u \vert ^{p}}{(1+ \vert u \vert )^{\theta (p-1)}} \,\mathrm{d}x\leq{}& C\bigl( \Vert j \Vert _{p'}^{p'}+ \Vert \nabla v_{0} \Vert _{p}^{p}+t \Vert f \Vert _{1}+1+t ^{r}\bigr) \\ \leq{}& C\bigl(1+t^{r}\bigr). \end{aligned} \end{aligned}$$
(17)
Replacing t with \(t+ \Vert v_{0} \Vert _{\infty }\) in (17) and noting that \(\{ \vert u \vert < t\}\subset \{ \vert v_{0}-u \vert < t+ \Vert v_{0} \Vert _{\infty }\}\), one may obtain the desired result. □
In the rest of this section, let \(\{u_{n}\}\) be a sequence of entropy solutions of the obstacle problem associated with \((f_{n},\psi ,g)\) and assume that
$$\begin{aligned} f_{n}\rightarrow f \quad \text{in } L^{1}(\varOmega ) \quad \text{and} \quad \Vert f_{n} \Vert _{1}\leq \Vert f \Vert _{1}+1. \end{aligned}$$
Lemma 3
There exists a measurable function
u
such that
\(u_{n}\rightarrow u \)
in measure, and
\(T_{k}(u_{n})\rightharpoonup T_{k}(u)\)
weakly in
\(W^{1,p}(\varOmega )\)
for any
\(k>0\). Thus
\(T_{k}(u_{n})\rightarrow T _{k}(u) \)
strongly in
\(L^{p}(\varOmega )\)
and a.e. in Ω.
Proof
Let s, t, and ε be positive numbers. One may verify that, for every \(m,n\geq 1\),
$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n}-u_{m} \vert >s\bigr\} \bigr)\leq{}& \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t \bigr\} \bigr)+ \mathcal{L}^{N}\bigl( \bigl\{ \vert u_{m} \vert >t\bigr\} \bigr) \\ & {} +\mathcal{L}^{N}\bigl(\bigl\{ \bigl\vert T_{k}(u_{n})-T_{k}(u_{m}) \bigr\vert >s\bigr\} \bigr), \end{aligned}$$
(18)
and
$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t \bigr\} \bigr)=\frac{1}{t^{p}} \int _{\{ \vert u_{n} \vert >t\}}t ^{p}\,\mathrm{d}x\leq \frac{1}{t^{p}} \int _{\varOmega } \bigl\vert T_{t}(u_{n}) \bigr\vert ^{p} \,\mathrm{d}x. \end{aligned}$$
(19)
Due to \(v_{0}=g+(\psi -g)^{+}\in K_{g,\psi }\cap L^{\infty }(\varOmega )\), by Lemma 2, one has
$$\begin{aligned} \int _{\varOmega } \bigl\vert \nabla T_{t}(u_{n}) \bigr\vert ^{p}\,\mathrm{d}x= \int _{\{ \vert u_{n} \vert < t\}} \vert \nabla u_{n} \vert ^{p}\,\mathrm{d}x \leq C(1+t)^{\theta (p-1)}\bigl(1+t^{r} \bigr). \end{aligned}$$
(20)
Note that \(T_{t}(u_{n})-T_{t}(g)\in W^{1,p}_{0}(\varOmega )\). By (19), (20), and Poincaré’s inequality, for every \(t> \Vert g \Vert _{\infty }\) and for some positive constant C independent of n and t, there holds
$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t \bigr\} \bigr)\leq{}& \frac{1}{t^{p}} \int _{\varOmega } \bigl\vert T _{t}(u_{n}) \bigr\vert ^{p}\,\mathrm{d}x \\ \leq{}& \frac{2^{p-1}}{t^{p}} \int _{\varOmega } \bigl\vert T_{t}(u_{n})-T_{t}(g) \bigr\vert ^{p} \,\mathrm{d}x+\frac{2^{p-1}}{t^{p}} \Vert g \Vert _{p}^{p} \\ \leq{}& \frac{C}{t^{p}} \int _{\varOmega } \bigl\vert \nabla T_{t}(u_{n})- \nabla T _{t}(g) \bigr\vert ^{p}\,\mathrm{d}x+ \frac{2^{p-1}}{t^{p}} \Vert g \Vert _{p}^{p} \\ \leq{}&\frac{C}{t^{p}} \int _{\varOmega } \bigl\vert \nabla T_{t}(u_{n}) \bigr\vert ^{p}\,\mathrm{d}x+\frac{C}{t ^{p}} \Vert g \Vert _{1,p}^{p} \\ \leq{}& \frac{C(1+t^{r+\theta (p-1)})}{t^{p}}. \end{aligned}$$
Since \(0\leq \theta <\frac{p-r}{p-1}\), there exists \(t_{\varepsilon }>0\) such that
$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t \bigr\} \bigr)< \frac{\varepsilon }{3}, \quad \forall t\geq t_{\varepsilon }, \forall n\geq 1. \end{aligned}$$
(21)
Now we have as in (19)
$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \bigl\vert T_{t_{\varepsilon }}(u_{n})-T_{t_{\varepsilon }}(u _{m}) \bigr\vert >s\bigr\} \bigr) &=\frac{1}{s^{p}} \int _{\{ \vert T_{t_{\varepsilon }} (u_{n})-T_{t_{\varepsilon }}(u_{m}) \vert >s \}} s^{p}\,\mathrm{d}x \\ &\leq \frac{1}{s^{p}} \int _{\varOmega } \bigl\vert T_{t_{\varepsilon }}(u_{n})-T _{t_{\varepsilon }}(u_{m}) \bigr\vert ^{p}\,\mathrm{d}x. \end{aligned}$$
(22)
Using (20) and the fact that \(T_{t}(u_{n})-T_{t}(g)\in W^{1,p} _{0}(\varOmega )\) again, we see that \(\{T_{t_{\varepsilon }}(u_{n})\}\) is a bounded sequence in \(W^{1,p}(\varOmega )\). Thus, up to a subsequence, \(\{T_{t_{\varepsilon }}(u_{n})\}\) converges strongly in \(L^{p}( \varOmega )\). Taking into account (22), there exists \(n_{0}=n_{0}(t _{\varepsilon },s)\geq 1\) such that
$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \bigl\vert T_{t_{\varepsilon }}(u_{n})-T_{t_{\varepsilon }}(u _{m}) \bigr\vert >s\bigr\} \bigr)< \frac{\varepsilon }{3}, \quad \forall n,m\geq n_{0}. \end{aligned}$$
(23)
Combining (18), (21), and (23), we obtain
$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n}-u_{m} \vert >s\bigr\} \bigr)< \varepsilon , \quad \forall n,m\geq n_{0}. \end{aligned}$$
Hence \(\{u_{n}\}\) is a Cauchy sequence in measure, and therefore there exists a measurable function u such that \(u_{n}\rightarrow u\) in measure. The remainder of the lemma is a consequence of the fact that \(\{T_{k}(u_{n})\}\) is a bounded sequence in \(W^{1,p}(\varOmega )\). □
Proposition 4
There exist a subsequence of
\(\{u_{n}\}\)
and a measurable function
u
such that, for each
q
given in (8), we have
$$\begin{aligned} u_{n}\rightarrow u \quad \textit{strongly in } W^{1,q}( \varOmega ). \end{aligned}$$
Furthermore, if
\(0\leq \theta < \min \{ \frac{1}{N-p+1},\frac{N}{N-1}-\frac{1}{p-1},\frac{p-r}{p-1} \}\), then
$$\begin{aligned} \frac{a(x, \nabla u_{n})}{(1+ \vert u_{n} \vert )^{\theta (p-1)}} \rightarrow \frac{a(x, \nabla u)}{(1+ \vert u \vert )^{\theta (p-1)}} \quad \textit{strongly in } \bigl(L^{1}(\varOmega )\bigr)^{N}. \end{aligned}$$
To prove Proposition 4, we need two preliminary lemmas.
Lemma 5
There exist a subsequence of
\(\{u_{n}\}\)
and a measurable function
u
such that, for each
q
given in (8), we have
\(u_{n}\rightharpoonup u \)
weakly in
\(W^{1,q}(\varOmega )\), and
\(u_{n}\rightarrow u \)
strongly in
\(L^{q}(\varOmega ) \).
Proof
Let \(k>0\) and \(n\geq 1\). Define \(D_{k}=\{ \vert u_{n} \vert \leq k\}\) and \(B_{k}=\{k\leq \vert u_{n} \vert < k+1\}\). Using Lemma 2 with \(v_{0}=g+(\psi -g)^{+}\), we get
$$\begin{aligned} \int _{D_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{\theta (p-1)}} \,\mathrm{d}x\leq C \bigl(1+k^{r}\bigr), \end{aligned}$$
(24)
where C is a positive constant depending only on α, β, b, p, r, \(\Vert j \Vert _{p'}\), \(\Vert f \Vert _{1}\), \(\Vert \nabla v_{0} \Vert _{p}\), and \(\Vert v_{0} \Vert _{\infty }\).
Using the function \(T_{k}(u_{n})\) for \(k> \{ \Vert g \Vert _{\infty }, \Vert \psi \Vert _{\infty }\}\), as a test function for the problem associated with \((f_{n},\psi ,g)\), we obtain
$$\begin{aligned} & \int _{\varOmega }\frac{a(x,\nabla u_{n})\cdot \nabla (T_{1}(u_{n}-T_{k}(u _{n})))}{(1+ \vert u_{n} \vert )^{\theta (p-1)}}\,\mathrm{d}x+ \int _{\varOmega }b \vert u_{n} \vert ^{r-2}u _{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \\ & \quad \leq \int _{\varOmega }f_{n}T_{1} \bigl(u_{n}-T_{k}(u_{n})\bigr)\, \mathrm{d}x, \end{aligned}$$
which and (2) give
$$\begin{aligned} \int _{B_{k}} \frac{\alpha \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{\theta (p-1)}}\,\mathrm{d}x+ \int _{\varOmega }b \vert u_{n} \vert ^{r-2} u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \leq \Vert f_{n} \Vert _{1}\leq \Vert f \Vert _{1}+1. \end{aligned}$$
Note that on the set \(\{ \vert u_{n} \vert \geq k+1\} \), \(u_{n}\) and \(T_{1}(u_{n}-T _{k}(u_{n}))\) have the same sign. Then
$$\begin{aligned} \int _{\varOmega } \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x ={}& \int _{D_{k}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \\ & {} + \int _{B_{k}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \\ & {} + \int _{\{ \vert u_{n} \vert \geq k+1\}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u _{n})\bigr)\,\mathrm{d}x \\ \geq{}& \int _{B_{k}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr) \,\mathrm{d}x. \end{aligned}$$
Thus we have
$$\begin{aligned} \int _{B_{k}} \frac{\alpha \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{\theta (p-1)}}\,\mathrm{d}x+ \leq{}& \Vert f \Vert _{1}+1- \int _{B_{k}}b \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u _{n})\bigr)\,\mathrm{d}x \\ \leq{}& \Vert f \Vert _{1}+1+ \int _{B_{k}}b \vert u_{n} \vert ^{r-1}\,\mathrm{d}x \\ \leq{}& C \biggl(1+ \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{\frac{r-1}{q ^{*}}} \vert B_{k} \vert ^{1-\frac{r-1}{q^{*}}} \biggr) , \end{aligned}$$
(25)
where q is given in (8) and \(q^{*}=\frac{Nq}{N-q} \).
Let \(s=\frac{q\theta (p-1)}{p}\). Note that \(q< p\) and \(\frac{ps}{p-q}< q ^{*}\). For \(\forall k>0\), we estimate \(\int _{B_{k}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x\) as follows:
$$\begin{aligned} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x={}& \int _{B_{k}}\frac{ \vert \nabla u _{n} \vert ^{q}}{(1+ \vert u_{n} \vert )^{s}}\cdot \bigl(1+ \vert u_{n} \vert \bigr)^{s}\,\mathrm{d}x \\ \leq{}& \biggl( \int _{B_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{ \theta (p-1)}}\,\mathrm{d}x \biggr)^{\frac{q}{p}} \biggl( \int _{B_{k}}\bigl(1+ \vert u _{n} \vert \bigr)^{\frac{ps}{p-q}}\,\mathrm{d}x \biggr)^{\frac{p-q}{p}} \\ \leq{}& C \biggl( \int _{B_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{ \theta (p-1)}}\,\mathrm{d}x \biggr)^{\frac{q}{p}} \biggl( \vert B_{k} \vert ^{ \frac{p-q}{p}}+ \biggl( \int _{B_{k}} \vert u_{n} \vert ^{\frac{ps}{p-q}}\,\mathrm{d}x \biggr)^{\frac{p-q}{p}} \biggr) \\ \leq{}& C \biggl( \int _{B_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{ \theta (p-1)}}\,\mathrm{d}x \biggr)^{\frac{q}{p}} \biggl( \vert B_{k} \vert ^{ \frac{p-q}{p}} + \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{\frac{s}{q ^{*}}} \vert B_{k} \vert ^{\frac{p-q}{p}-\frac{s}{q^{*}}} \biggr) \\ \leq{}& C \biggl( \vert B_{k} \vert ^{\frac{p-q}{p}}+ \vert B_{k} \vert ^{\frac{p-q}{p}-\frac{s}{q ^{*}}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{ \frac{s}{q^{*}}}+ \vert B_{k} \vert ^{1-p_{1}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{p_{1}} \\ & {} + \vert B_{k} \vert ^{1-p_{2}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{p _{2}} \biggr) \quad \text{by (25)} \\ ={}& C \biggl( \vert B_{k} \vert ^{\frac{p-q}{p}}+ \vert B_{k} \vert ^{\frac{p-q}{p}-\frac{s}{q ^{*}}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{\frac{s}{q^{*}}} \\ & {} + \vert B_{k} \vert ^{1-p_{1}-C_{1}} \vert B_{k} \vert ^{C_{1}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}}\,\mathrm{d}x \biggr)^{p_{1}} \\ & {} + \vert B_{k} \vert ^{1-p_{2}-C_{2}} \vert B_{k} \vert ^{C_{2}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}}\,\mathrm{d}x \biggr)^{p_{2}} \biggr), \end{aligned}$$
where \(p_{1}=\frac{q}{p}\frac{r-1}{q^{*}}\), \(p_{2}=\frac{s}{q^{*}}+ \frac{q}{p}\frac{r-1}{q^{*}}\), \(C_{1}\) and \(C_{2}\) are positive constants to be chosen later.
Note that \(\theta <\frac{p-r}{p-1}\), it follows
$$\begin{aligned} \frac{\theta (p-1)}{p}+\frac{r-1}{p}< \frac{p-1}{p}< 1- \frac{1}{N}=1- \frac{1}{q}+\frac{1}{q^{*}}. \end{aligned}$$
Thus
$$\begin{aligned} \frac{\theta q(p-1)}{p}+\frac{q(r-1)}{p}+1< q+\frac{q}{q^{*}} \quad &\Leftrightarrow\quad s+\frac{q(r-1)}{p}+1< q+\frac{q}{q^{*}} \\ & \Leftrightarrow\quad p_{2}+\frac{1-p _{2}}{q^{*}+1}< \frac{q}{q^{*}}. \end{aligned}$$
Note that \(p_{1}< p_{2}<1\). Then, for \(i=1,2\), we always have
$$\begin{aligned} p_{i}+\frac{1-p_{i}}{q^{*}+1}< \frac{q}{q^{*}}< 1. \end{aligned}$$
From this, we may find positive \(C_{i}\) (\(i=1,2\)) such that
$$\begin{aligned} p_{i}+\frac{1-p_{i}}{q^{*}+1}< p_{i}+C_{i}< \frac{q}{q^{*}}< 1, \quad i=1,2. \end{aligned}$$
(26)
It follows
$$\begin{aligned} \frac{1-p_{i}}{q^{*}+1}< C_{i}\quad \Leftrightarrow \quad 1-p_{i}-C_{i}< C_{i}q^{*}, \quad i=1,2, \end{aligned}$$
which implies
$$\begin{aligned} C_{i}\alpha _{i}q^{*}= \frac{C_{i}q^{*}}{1-p_{i}-C_{i}}>1, \quad i=1,2, \end{aligned}$$
(27)
with \(\alpha _{i}=\frac{1}{1-p_{i}-C_{i}}>1\), \(i=1,2\). Let \(\beta _{i}=\frac{1}{p _{i}+C_{i}}>1\), \(i=1,2\). Then we have \(\frac{1}{\alpha _{i}}+\frac{1}{ \beta _{i}}=1\) (\(i=1,2\)).
Since \(\vert B_{k} \vert \leq \frac{1}{k^{q^{*}}}\int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \), \(\vert B_{k} \vert ^{1-p_{1}-C_{1}} \leq \vert \varOmega \vert ^{1-p_{1}-C_{1}}\), and \(\vert B_{k} \vert ^{1-p_{2}-C_{2}} \leq \vert \varOmega \vert ^{1-p_{2}-C_{2}}\), we have, for \(k\geq k_{0}\geq 1\),
$$\begin{aligned} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq{}&\frac{C}{k^{q^{*} ( \frac{p-q}{p}-\frac{s}{q^{*}} )}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{\frac{p-q}{p}} \\ & {} +\frac{C}{k^{q^{*}C_{1}}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{p_{1}+C_{1}}+\frac{C}{k^{q^{*}C_{2}}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}}\,\mathrm{d}x \biggr)^{p_{2}+C_{2}}. \end{aligned}$$
Summing up from \(k=k_{0}\) to \(k=K\) and using Hölder’s inequality, one has
$$\begin{aligned} \sum_{k=k_{0}}^{K} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq{}& C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{s}{q ^{*}})\frac{p}{q}}} \Biggr)^{\frac{q}{p}}\cdot \Biggl(\sum _{k=k_{0}} ^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{\frac{p-q}{p}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{1}\alpha _{1}}} \Biggr)^{\frac{1}{ \alpha _{1}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}} \,\mathrm{d}x \biggr)^{\beta _{1}(p_{1}+C_{1})} \Biggr)^{\frac{1}{\beta _{1}}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{2}\alpha _{2}}} \Biggr)^{\frac{1}{ \alpha _{2}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}} \,\mathrm{d}x \biggr)^{\beta _{2}(p_{2}+C_{2})} \Biggr)^{\frac{1}{\beta _{2}}} \\ ={}& C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{s}{q ^{*}})\frac{p}{q}}} \Biggr)^{\frac{q}{p}}\cdot \Biggl(\sum _{k=k_{0}} ^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{\frac{p-q}{p}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{1}\alpha _{1}}} \Biggr)^{\frac{1}{ \alpha _{1}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{p_{1}+C_{1}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{2}\alpha _{2}}} \Biggr)^{\frac{1}{ \alpha _{2}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{p_{2}+C_{2}}. \end{aligned}$$
(28)
Note that
$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x= \int _{D_{k_{0}}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x+\sum_{k=k_{0}}^{K} \int _{B_{k}} \vert \nabla u _{n} \vert ^{q} \,\mathrm{d}x. \end{aligned}$$
(29)
To estimate the first integral in the right-hand side of (29), we compute by using Hölder’s inequality and (24), obtaining
$$\begin{aligned} \int _{D_{k_{0}}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x \leq{}& \biggl( \int _{D_{k _{0}}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{\theta (p-1)}}\,\mathrm{d}x \biggr)^{\frac{q}{p}} \biggl( \int _{D_{k_{0}}}\bigl(1+ \vert u_{n} \vert \bigr)^{ \frac{ps}{p-q}} \,\mathrm{d}x \biggr)^{\frac{p-q}{p}} \\ \leq{}& C, \end{aligned}$$
(30)
where C depends only on α, β, b, p, θ, \(\Vert j \Vert _{p'}\), \(\Vert f \Vert _{1}\), \(\Vert \nabla v_{0} \Vert _{p}\), \(\Vert v_{0} \Vert _{\infty }\), and \(k_{0}\).
Note that \(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{s}{q ^{*}})\frac{p}{q}}}\) and \(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C _{i}\alpha _{i}}}\) converge as \(K\rightarrow \infty \) due to the fact that \(q^{*}(\frac{p-q}{p}-\frac{s}{q^{*}})\frac{p}{q}>1\) and \(q^{*}C_{i} \alpha _{i}>1\) by (27), respectively. Combining (28)–(30), we get for \(k_{0}\) large enough
$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq{}& C+C \biggl( \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{ \frac{p-q}{p}} \\ & {} + C \biggl( \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{p _{1}+C_{1}} \\ & {} +C \biggl( \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{p _{2}+C_{2}}. \end{aligned}$$
(31)
Since \(p>q\), \(T_{K}(u_{n})\in W^{1,q}(\varOmega )\), \(T_{K}(g)=g\in W^{1,q}( \varOmega )\) for \(K> \Vert g \Vert _{\infty }\). Hence \(T_{K}(u_{n})-g\in W^{1,q} _{0}(\varOmega )\). Using the Sobolev embedding \(W^{1,q}_{0}(\varOmega ) \subset L^{q^{*}}(\varOmega )\) and Poincaré’s inequality, we obtain
$$\begin{aligned} \bigl\Vert T_{K}(u_{n}) \bigr\Vert ^{q}_{q^{*}}\leq{}& 2^{q-1}\bigl( \bigl\Vert T_{K}(u_{n})-g \bigr\Vert ^{q}_{q ^{*}}+ \Vert g \Vert ^{q}_{q^{*}}\bigr) \\ \leq{}& C\bigl( \bigl\Vert \nabla \bigl(T_{K}(u_{n})-g \bigr) \bigr\Vert ^{q}_{q}+ \Vert g \Vert ^{q}_{q^{*}}\bigr) \\ \leq{}& C \bigl( \bigl\Vert \nabla T_{K}(u_{n}) \bigr\Vert ^{q}_{q}+ \Vert \nabla g \Vert ^{q}_{q}+ \Vert g \Vert ^{q}_{q^{*}} \bigr) \\ \leq{}& C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x \biggr). \end{aligned}$$
(32)
Using the fact that
$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x\leq \int _{\{ \vert u_{n} \vert \leq K\}} \bigl\vert T_{K}( u_{n}) \bigr\vert ^{q^{*}} \,\mathrm{d}x\leq \bigl\Vert T _{K}( u_{n}) \bigr\Vert ^{q^{*}}_{q^{*}}, \end{aligned}$$
(33)
we obtain from (31)–(32)
$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq{}& C+C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x \biggr)^{\frac{q ^{*}}{q}\frac{p-q}{p}} \\ & {} +C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x \biggr)^{(p_{1}+C_{1})\frac{q^{*}}{q}} \\ & {} +C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x \biggr)^{(p_{2}+C_{2})\frac{q^{*}}{q}} . \end{aligned}$$
(34)
Note that \(p< N\Leftrightarrow \frac{q^{*}}{q}\frac{p-q}{p}<1\) and \((p_{i}+C_{i})\frac{q^{*}}{q}<1 \) by (26). It follows from (34) that, for \(k_{0}\) large enough, \(\int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x \) is bounded independently of n and K. Using (32) and (33), we deduce that \(\int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x\) is also bounded independently of n and K. Letting \(K\rightarrow \infty \), we deduce that \(\Vert \nabla u_{n} \Vert _{q}\) and \(\Vert u_{n} \Vert _{q^{*}}\) are uniformly bounded independently of n. Particularly, \(u_{n}\) is bounded in \(W^{1,q}(\varOmega )\). Therefore, there exist a subsequence of \(\{u_{n}\}\) and a function \(v\in W^{1,q}(\varOmega )\) such that \(u_{n}\rightharpoonup v\) weakly in \(W^{1,q}(\varOmega )\), \(u_{n}\rightarrow v\) strongly in \(L^{q}(\varOmega )\) and a.e. in Ω. By Lemma 3, \(u_{n}\rightarrow u\) in measure in Ω, we conclude that \(u=v\) and \(u\in W^{1,q}(\varOmega )\). □
Lemma 6
There exist a subsequence of
\(\{u_{n}\}\)
and a measurable function
u
such that
\(\nabla u_{n}\)
converges almost everywhere in
Ω
to ∇u.
Proof
Define \(A(x,u,\xi )=\frac{a(x,\xi )}{(1+ \vert u \vert )^{\theta (p-1)}}\) (for the sake of simplicity, we omit the dependence of \(A(x,u,\xi )\) on x). Let \(h>0\), \(k> \max \{ \Vert g \Vert _{\infty }, \Vert \psi \Vert _{\infty }\}\), and \(n\geq h+k\). Take \(T_{k}(u)\) as a test function for (7), obtaining
$$\begin{aligned} I_{7}(n,k,h)\leq \int _{\varOmega }f_{n}T_{h} \bigl(u_{n}-T_{k}(u)\bigr)\,\mathrm{d}x+ \int _{\varOmega }b \vert u_{n} \vert ^{r-2}u_{n}T_{h}\bigl(u_{n}-T_{k}(u) \bigr)\,\mathrm{d}x, \end{aligned}$$
where
$$\begin{aligned} I_{7}(n,k,h)= \int _{\varOmega }A(u_{n},\nabla u_{n}) \cdot \nabla T_{h}\bigl(u_{n}-T_{k}(u) \bigr)\,\mathrm{d}x. \end{aligned}$$
Note that \(r-1< q^{*}\), and \(\int _{\varOmega } \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x\) is uniformly bounded (see the proof of Lemma 5), thus \(\vert u_{n} \vert \) converges strongly in \(L^{1}(\varOmega )\). Therefore we have
$$\begin{aligned} \lim_{n\rightarrow \infty } \int _{\varOmega } \vert u_{n} \vert ^{r-2}u_{n}T_{h}\bigl(u_{n}-T_{k}(u) \bigr)\,\mathrm{d}x= \int _{\varOmega } \vert u \vert ^{r-2}uT_{h} \bigl(u-T_{k}(u)\bigr)\,\mathrm{d}x. \end{aligned}$$
Then, using the strong convergence of \(f_{n}\) in \(L^{1}(\varOmega )\), one has
$$\begin{aligned} \lim_{n\rightarrow \infty }I_{7}(n,k,h) \leq \int _{\varOmega }-fT_{h}\bigl(u-T_{k}(u) \bigr)\,\mathrm{d}x+ \int _{\varOmega }b \vert u \vert ^{r-2}uT_{h} \bigl(u-T_{k}(u)\bigr)\,\mathrm{d}x. \end{aligned}$$
It follows
$$\begin{aligned} \lim_{k\rightarrow \infty }\lim_{n\rightarrow \infty }I_{7}(n,k,h) \leq 0. \end{aligned}$$
Thanks to Lemma 3 and Lemma 5, we can proceed exactly as [19, Lemma 6] to conclude that, up to subsequence, \(\nabla u_{n}\rightarrow \nabla u\)
a.e. □
Proof of Proposition 4
We shall prove that \(\nabla u_{n}\) converges strongly to ∇u in \(L^{q}(\varOmega )\) for each q being given by (8). To do that,we will apply Vitali’s theorem, using the fact that by Lemma 5, \(\nabla u_{n}\) is bounded in \(L^{q}(\varOmega )\) for each q given by (8). So let \(s\in (q,\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)})\) and \(E\subset \varOmega \) be a measurable set. Then we have by Hölder’s inequality
$$\begin{aligned} \int _{E} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq \biggl( \int _{E} \vert \nabla u_{n} \vert ^{r} \,\mathrm{d}x \biggr)^{\frac{q}{s}}\cdot \vert E \vert ^{\frac{s-q}{s}}\leq C \vert E \vert ^{ \frac{s-q}{s}}\rightarrow 0 \end{aligned}$$
uniformly in n, as \(\vert E \vert \rightarrow 0\). From this and from Lemma 6, we deduce that \(\nabla u_{n}\) converges strongly to ∇u in \(L^{q}(\varOmega )\).
Now assume that \(0\leq \theta < \min \{\frac{1}{N-p+1},\frac{N}{N-1}- \frac{1}{p-1},\frac{p-r}{p-1}\}\). Note that since \(\nabla u_{n}\) converges to ∇u a.e. in Ω, to prove the convergence
$$\begin{aligned} \frac{a(x, \nabla u_{n})}{(1+ \vert u_{n} \vert )^{\theta (p-1)}} \rightarrow \frac{a(x, \nabla u)}{(1+ \vert u \vert )^{\theta (p-1)}} \quad \text{strongly in } \bigl(L^{1}(\varOmega )\bigr)^{N}, \end{aligned}$$
it suffices, thanks to Vitali’s theorem, to show that, for every measurable subset \(E\subset \varOmega \), \(\int _{E} \vert \frac{a(x, \nabla u_{n})}{(1+ \vert u_{n} \vert )^{\theta (p-1)}} \vert \,\mathrm{d}x\) converges to 0 uniformly in n, as \(\vert E \vert \rightarrow 0\). Note that \(p-1<\frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)})\) by assumptions. For any \(q\in (p-1, \frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} )\), we deduce by Hölder’s inequality
$$\begin{aligned} \int _{E} \biggl\vert \frac{a(x, \nabla u_{n})}{(1+ \vert u_{n} \vert )^{\theta (p-1)}} \biggr\vert \,\mathrm{d}x\leq{}& \beta \int _{E}\bigl(j+ \vert \nabla u_{n} \vert ^{p-1}\bigr)\,\mathrm{d}x \\ \leq{}& \beta \Vert j \Vert _{p'} \vert E \vert ^{\frac{1}{p}}+\beta \biggl( \int _{E} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x \biggr)^{\frac{p-1}{q}} \vert E \vert ^{\frac{q-p+1}{q}} \\ \rightarrow &0 \quad \text{uniformly in } n \text{ as } \vert E \vert \rightarrow 0. \end{aligned}$$
□
Lemma 7
There exists a subsequence of
\(\{u_{n}\}\)
such that, for all
\(k>0\),
$$\begin{aligned} \frac{a(x, \nabla T_{k}(u_{n}))}{(1+ \vert T_{k}(u_{n}) \vert )^{\theta (p-1)}} \rightarrow \frac{a(x, \nabla T_{k}(u))}{(1+ \vert T_{k}(u) \vert )^{\theta (p-1)}} \quad \textit{strongly in } \bigl(L^{1}(\varOmega )\bigr)^{N}. \end{aligned}$$
Proof
See the proof of [19, Lemma 7]. □