in case of the Laplace operator ($p_i=2$) to Jannelli–Solimini [@JanSol], where the decay estimate has been proved to hold true for solutions of with right-hand side $f{\left(}x,u{\right)}=\sum_{i=1}^Na_i{\left(}x{\right)}\left|u\right|^{q_i^*-2}u$, where $q_i^*:=2^*{\left(}1-1/q_i{\right)}$, $q_i\in{\left(}n/2,\infty{\right]}$, $\left|a_i{\left(}x{\right)}\right|={\text{O}}\big(\left|x\right|^{-n/q_i}\big)$ for large $\left|x\right|$, and $a_i$ belongs to the Marcinkiewicz space $M^{q_i}{\left(}{\mathbb{R}}^n{\right)}$ for all $i=1,\dotsc,N$.

The next results concern the case $p_+\ge p_*$, namely $p_i\ge p_*$ for some index $i$. In particular, we are now exclusively in the case where the exponents $p_i$ are not all equal.

In the limit case $p_+=p_*$, we prove the following result.

\[Th2\] Assume that $p_+=p_*$. Let $f:{\mathbb{R}}^n\times{\mathbb{R}}\to{\mathbb{R}}$ be a Caratheodory function such that holds true and $u$ be a solution of . Then for any $q>p_*$, there exists a constant $C_q=C{\left(}n,\operatorname{\overrightarrow{p}},\Lambda,u,q{\right)}$ such that $$\label{Th2Eq}
\left|u{\left(}x{\right)}\right|^{q}+\sum_{i=1}^n\left|\partial_{x_i}u{\left(}x{\right)}\right|^{p_i}\le C_q\bigg(1+\sum_{i=1}^n\left|x_i\right|^{\frac{qp_i}{q-p_i}}\bigg)^{-1}\quad\text{for a.e. }x\in{\mathbb{R}}^n.$$

Beyond this limit case, namely when $p_*<p_+<p^*$, we find the following result.

\[Th3\] Assume that $p_*<p_+<p^*$. Let $f:{\mathbb{R}}^n\times{\mathbb{R}}\to{\mathbb{R}}$ be a Caratheodory function such that holds true and $u$ be a solution of . Then there exist a real number $q_0=q_0{\left(}n,\operatorname{\overrightarrow{p}}{\right)}<p_+$ such that the two following assertions hold true.

1.  There exists a constant $R_0=R_0{\left(}n,\operatorname{\overrightarrow{p}},\Lambda,u{\right)}$ such that $$\label{Th3Eq1}
    u{\left(}x{\right)}=0\quad\text{for all }x\in{\mathbb{R}}^n\text{ such that }\sum_{i\in\mathcal{I}_0}\left|x_i\right|\ge R_0\,,$$ where $\mathcal{I}_0$ is the set of all indices $i$ such that $p_i>q_0$. 