Appendix
In this appendix, we prove the existence and uniqueness of the solution for problem (1.1).
A.1 The existence of solutions for problem (1.1)
In the first part, we give the existence of solutions for problem (1.1).
Theorem 3
Let
f
satisfy (f1) and the Keller-Osserman condition (2.7), and
b
satisfy (b1). Then problem (1.1) has at least one solution
\(u\in W^{1,p}_{\mathrm{loc}}(\overline{\Omega})\)
satisfying
$$ u(x)\geq\psi\bigl(\overline{v}(x)\bigr), \quad \forall x\in\Omega. $$
(A.1)
Furthermore, if
f
satisfies
\(\int^{1}_{0}f^{-\frac{1}{p-1}}(s)\, ds=\infty\), then
$$ u>0, \quad \forall x\in\Omega, $$
(A.2)
where
ψ
is the solution of problem (1.4) and
\(\overline {v}\in W^{1,p}_{0}(\Omega)\)
is the unique solution for problem
$$ -\Delta_{p}\overline{v}=b(x),\qquad \overline{v}(x)>0,\quad x\in \Omega,\qquad \overline {v}|_{\partial\Omega}=0. $$
(A.3)
Remark 4
By Lemma 2(3), we can see that f satisfies the Keller-Osserman condition under our hypotheses on f in Theorem 1.
Remark 5
By the similar argument in [46], we show that (f1) and the Keller-Osserman condition imply (f2). Indeed, if we can prove that there exist two positive numbers ρ and M such that
$$ \frac{f^{p'-1}(s)}{s}\geq\rho^{p'} \quad \mbox{for } s\geq M, $$
(A.4)
then it will be done since
$$F(s)= \int^{s}_{0}f(t)\, dt\leq sf(s)\leq \frac{f^{p'}(s)}{\rho^{p'}} \quad \mbox{for } s\geq M, $$
which, in turn, yields \([F(s)]^{-\frac{1}{p}}\geq\frac{\rho^{\frac{1}{p-1}}}{f^{\frac{1}{p-1}}(s)}\), so that the Keller-Osserman condition implies (A.4). Then, we will prove (A.4) by contradiction. Assume that there exists an increasing sequence \({s_{j}}\) of real numbers such that \(\lim_{j\rightarrow\infty}s_{j}=\infty\) and \(\frac{f(s_{j})}{s_{j}}<\frac{1}{j}\) for all j. Since f is increasing, we have \(f(s)\leq f(s_{j})\) for all \(s\in[0,s_{j} ]\), which, in turn, produces \(F(s)\leq sf(s)\leq sf(s_{j})\rightarrow\infty\) for \(s\in[0,s_{j}]\). Hence,
$$\begin{aligned} \int^{s_{j}}_{s_{1}}\bigl[F(s)\bigr]^{-\frac{1}{p}}\, ds &\geq \int ^{s_{j}}_{s_{1}}\bigl[F(s)\bigr]^{-\frac{1}{p}}\, ds \geq \biggl[\frac{j}{s_{j}} \biggr]^{\frac{1}{p'}} \int^{s_{j}}_{s_{1}}s^{-\frac{1}{p}}\, ds \\ &= \frac{p}{p-1}j^{\frac{1}{p'}} \biggl(1- \biggl(\frac{s_{1}}{s_{j}} \biggr) \biggr)\rightarrow\infty \end{aligned}$$
as \(j\rightarrow\infty\), which contradicts the Keller-Osserman condition (2.7). Thus, (A.4) must be true, and then (f2) holds.
Proof
Let
$$ v=\phi(u)= \int^{\infty}_{u}f^{-\frac{1}{p-1}}(v)\, dv, \quad u>0. $$
(A.5)
We see that problem (1.1) is equivalent to the following problem:
$$ -\Delta_{p}v+g(v)|\nabla v|^{p}=b(x), \quad v>0, x\in \Omega, \qquad \overline {v}|_{\partial\Omega}=0, $$
(A.6)
where \(g(v)=f'(\psi(v))\), and ϕ is also the inverse function of ψ.
Now let \(v\in W^{1,p}_{\mathrm{loc}}(\Omega)\cap L^{\infty}_{\mathrm{loc}}(\Omega)\) be any solution of problem (A.6). We claim that
$$ v(x)\leq \overline{v}(x), \quad \forall x\in\Omega. $$
(A.7)
Indeed, assume on the contrary that \(\{x\in\Omega:v(x)>\overline{v}(x)\}\neq\emptyset\). Then, on its arbitrary connected component D, we have \(-\Delta _{p}(v-\overline{v})(x)\leq0\), \(x\in D\), since \(g(v)\geq0\). It follows by \((v-\overline{v})|_{\partial\Omega}=0\) and the maximum principle that \(v(x)\leq\overline{v}(x)\) for all \(x\in D\). This is a contradiction. Thus, (A.7) holds, i.e., any weak solution u of problem (1.1) satisfies (A.1). Moreover, by the definition of ψ and the condition \(\int^{1}_{0}\frac{ds}{f(s)}=\infty\) we see that \(\psi(\overline{v}(x))>0\), \(\forall x\in\Omega\), and (A.2) holds. Next, we consider the perturbed problem
$$ \Delta_{p}u=b(x)f(u), \quad x\in\Omega,\qquad u|_{\partial\Omega}=m\in N. $$
(A.8)
By (b1) and (f1) we see that \(\overline{u}_{m}=m\) is a supersolution of problem (A.8). To construct a subsolution \(\underline{u}_{1}\) of problem (A.8), we let \(\overline{v}_{1}\in C^{2+\alpha}(\overline{\Omega})\) be the unique solution of the problem
$$ -\Delta_{p}\overline{v}_{1}=b(x), \qquad \overline{v}(x)>0, \quad x\in\Omega, \qquad \overline{v}_{1}|_{\partial\Omega}= \int^{\infty }_{1}f^{-\frac{1}{p-1}}(s)\, ds, $$
(A.9)
and \(\overline{u}_{1}=\psi(\overline{v}_{1})\). Then we see that \(\underline{u}_{1}|_{\partial\Omega}=1\leq m\) and
$$-\Delta _{p}\overline{v}_{1}=\frac{\Delta_{p}\underline {u}_{1}}{f(\underline{u}_{1})} - \frac{f'(\underline{u}_{1})}{f^{2}(\underline{u}_{1})}|\nabla\underline {u}_{1}|^{p}=b(x), \quad x\in\Omega, $$
which yields
$$\Delta_{p}\underline{u}_{1}\geq b(x)f(\underline{u}_{1}), \quad x\in\Omega, $$
i.e., \(\underline{u}_{1}\) is a subsolution of problem (A.8). Moreover, \(\underline{u}_{1}\leq1\leq m\), \(x\in\Omega\), due to the maximum principle. Thus, problem (A.8) has one solution \(u_{m}\in W^{1,p}_{0}\) in the order interval \([u_{1},m]\), and the maximum principle again yields that the map \(m\rightarrow u_{m}\) is increasing. On the other hand, the classical Keller-Osserman condition guarantees that the problem
$$ \Delta_{p}u=b_{0}f(u), \quad x\in\Omega_{0}, \qquad u|_{\partial\Omega _{0}}=\infty, $$
(A.10)
has one solution \(u_{\Omega_{0}}\in W^{1,p}_{\mathrm{loc}}(\Omega)\) for each \(\Omega_{0}\subset\subset\Omega\), where \(b_{0}=\min_{x\in\overline{\Omega}_{0}}b(x)\). By the maximum principle we have \(u_{m}\leq u_{\Omega_{0}}(x)\), \(x\in\Omega_{0}\), and \(u(x):=\lim_{m\rightarrow\infty}u_{m}(x)\) exists for \(x\in\Omega_{0}\). Thus, u is the desired solution for problem (1.1) by the standard bootstrap argument, the arbitrariness of \(\Omega_{0}\), and (A.1). □
A.2 The uniqueness of the solution for problem (1.1)
In the second part, we prove the uniqueness of the solution for problem (1.1). The method is similar to the idea in [40, 58].
Theorem 4
Under the hypotheses in Theorem
1, problem (1.1) admits a unique solution.
Proof
Since \(\frac{f(s)}{s^{q}}\) is increasing in \([S_{0},\infty)\) for some \(q>p-1\) and \(S_{0}\) large enough, by Lemma 2(2) we have
$$ \frac{f(s)}{s} \mbox{ is also increasing in } [S_{0}, \infty). $$
(A.11)
Let \(u_{0}\) be the minimal solution for problem (1.3), and u be another solution for problem (1.1). We prove that \(u = u_{0}\) in Ω. In fact, by the maximum principle we have
$$ u_{0}\leq u \quad \mbox{in } \Omega. $$
(A.12)
Moreover, by the asymptotic behavior (1.3) we deduce that
$$ \lim_{d(x)\rightarrow0}\frac{u_{0}(x)}{u(x)}=1. $$
(A.13)
For any \(\epsilon>0\), setting \(w=(1+\epsilon)u_{0}\), we have
$$ \lim_{d(x)\rightarrow0}\bigl(w(x)-u(x)\bigr) =\lim_{d(x)\rightarrow0} \biggl(\frac{(1+\epsilon)u_{0}(s)}{u(s)}-1 \biggr)=+\infty. $$
(A.14)
Now, for small \(\epsilon> 0\), we define the open set
$$ D_{\epsilon}=\bigl\{ x\in\Omega:w(x)< u(x)\bigr\} . $$
(A.15)
We may assume that \(D_{\epsilon}\) is nonempty for ϵ small enough; otherwise, there is nothing to prove. Indeed, notice that \(D_{\epsilon}\) increases as \(\epsilon\rightarrow0\). Moreover, we may also assume that \(D_{\epsilon}\rightarrow\Omega\) as \(\epsilon\rightarrow0\); if there exists \(x_{0}\in\Omega\) and a sequence \(\epsilon_{n}\rightarrow0\) such that \(x_{0}\in D_{\epsilon_{n}}\) for all n, then we have \((1+\epsilon_{n})u_{0}(x_{0})\geq u(x_{0})\). Then the strong maximum principle yields \(u\equiv u_{0}\) in Ω. Finally, we have \(D_{\epsilon}\subset\Omega\) by (A.13).
Next, we choose \(\eta>0\) such that \(u_{0}\geq S_{0}\) in \(\Omega_{\eta}\) and define \(D_{\epsilon,\eta}=D_{\epsilon}\cap D_{\eta}\). Notice that \(D_{\epsilon,\eta}\) is a nonempty open set for small ϵ. Moreover, by (A.11) we have
$$ \Delta _{p}w=(1+\epsilon)b(x)f(u_{0})\leq b(x)f(w),\quad x\in D_{\epsilon,\eta}. $$
(A.16)
It follows by (f1) that
$$ \Delta _{p}(u-w)\geq b(x) \bigl(f(u)-f(w)\bigr)\geq0, \quad x\in D_{\epsilon,\eta }. $$
(A.17)
Thus, by the maximum principle we obtain
$$ u(x)-w(x)\leq\max_{\partial D_{\epsilon,\eta}}(u-w), \quad x\in D_{\epsilon,\eta}. $$
(A.18)
Since \(\partial D_{\epsilon,\eta}=(\partial D_{\epsilon}\cap D_{\eta})\cup( D_{\epsilon}\cap\partial D_{\eta})\), \(D_{\epsilon}\cap\partial\Omega=\emptyset\), and \((u-w)|_{\partial D_{\epsilon}}=0\), we see that the maximum of \(u-w\) is achieved on \(D_{\epsilon}\cap\partial D_{\eta}=( D_{\epsilon}\cap {x:d(x)=\eta})\). Hence,
$$ u(x)-w(x)\leq\max_{( D_{\epsilon}\cap {x:d(x)=\eta})}(u-w), \quad x\in D_{\epsilon,\eta}. $$
(A.19)
Letting \(\epsilon\rightarrow0\) in (A.19), we obtain
$$ u-u_{0}\leq\max_{d(x)=\eta}(u-u_{0}):=\theta \quad \mbox{in } \Omega _{\eta}. $$
(A.20)
On the other hand, by (A.12) and (f1) we have
$$ \Delta _{p}(u-u_{0})=b(x) \bigl(f(u)-f(u_{0}) \bigr)\geq0, \quad x\in\Omega^{\eta }={x\in\Omega\mbox{: }d(x)> \eta}. $$
(A.21)
The maximum principle implies that \(u-u_{0}\leq\theta\) in \(\Omega^{\eta}\), and hence \(u-u_{0}\leq\theta\) in the whole Ω. Then the strong maximum principle gives \(u-u_{0}\equiv\theta\). We obtain that \(f(u)=f(u+\theta)\) in Ω, which can only hold if \(\theta=0\). Thus, \(u=u_{0}\), which shows the uniqueness. □