Proof. Since we have a positive solution pair $(u,\,v)$ in the annulus, and both functions start positive and increasing, a critical point must exist for both $u$ and $v$ by Rolle's theorem.

The uniqueness of $\tau _u$ follows from the fact that, since $v$ is positive, any critical point of $u$ is a strict local maximum; likewise for $\tau _v$ and $v$.

Proof. We revisit the moving planes method as performed in [Reference dos Santos and Nornberg6] in order to treat the general range $pq>1$. The lack of $C^1$ or even Lipschitz continuity on the nonlinearities is allowed there, namely when either $p<1$ or $q<1$. Accordingly to the notation in [Reference dos Santos and Nornberg6], for the annulus we have

\[ \Lambda_1=\Lambda_2=\frac{1}{2}(\frak a +\frak b), \]

see § 2 in [Reference dos Santos and Nornberg6] for the corresponding definitions.

For any direction $\gamma >0$ (as positive axis $\{x_1=0\}$) it follows as in [Reference dos Santos and Nornberg6, Step 1 of the proof of proposition 3.1, p.4183] that $\gamma \cdot Du <0$ in the maximal cap $\Sigma _{\Lambda _1}$. The union of these maximal caps, originated by all directions $\gamma =\frac {x}{|x|}$, $x\neq 0$, produces the half annulus $A_{\Lambda _1,\frak b}$. In particular, no critical points exist in the open annulus $A_{\Lambda _1,\frak b}$.

If we had $\partial _r u (x_0)=0$ or $\partial _r v (x_0)=0$ for some $x_0$ with $|x_0|=\Lambda _1$, then the Hopf lemma in [Reference dos Santos and Nornberg6, lemma 3.3] would give us $U^{\Lambda _1}\equiv 0$ or $V^{\Lambda _1}\equiv 0$ in $\Sigma _{\Lambda _1}$. Since the system is strongly coupled, this means $U^{\Lambda _1}=V^{\Lambda _1}\equiv 0$ in $\Sigma _{\Lambda _1}$. So $u(x^{\Lambda _1})=v(x^{\Lambda _1})=0$ for all $x\in \Sigma _{\Lambda _1}\cap \partial A_{\frak a, \frak b}$ due to the boundary condition on $|x|=\frak b$, but this is impossible since the solution is positive.

Proof. We recall that in $[\frak a,\, \tau _* ]$ we have $u^\prime,\, v^\prime \ge 0$, $u^{\prime \prime }\le 0$, and

\begin{align*} \textstyle u^{\prime\prime}v^\prime +\frac{v^p v^\prime}{\lambda}={-} \frac{(\tilde{N}_-{-}1)u^\prime v^\prime}{ r}, \quad v^{\prime\prime}u^\prime +\frac{u^q u^\prime}{\lambda}={-} \frac{(\tilde{N}_-{-}1)u^\prime v^\prime}{ r}. \end{align*}

In $[\tau ^*,\, \frak b)$ we have $u^\prime,\, v^\prime <0$ and

\[ u^{\prime\prime}v^\prime +\frac{v^p v^\prime}{\Lambda} \ge u^{\prime\prime}v^\prime +\frac{v^p v^\prime}{\sigma}={-}\frac{(\hat{N}-1)u^\prime v^\prime}{ r} \ge - \frac{(\tilde{N}_-{-}1)u^\prime v^\prime}{ r}, \]

where $(\sigma,\, \hat {N})$ is either $(\lambda,\, N)$ or $(\Lambda,\, \tilde {N}_+)$, and analogously $v^{\prime \prime }u^\prime +\frac {u^q u^\prime }{\Lambda } \ge - \frac {(\tilde {N}_--1)u^\prime v^\prime }{ r}$. Thus, for $\kappa =2(\tilde {N}_- -1)$, we obtain in $[\tau ^*,\, \frak b)$,

\begin{align*} ({E}^{\Lambda}_1)^\prime (r) & = \kappa r^{\kappa -1} \left\{ u^\prime v^\prime +\frac{v^{p+1}}{\Lambda (p+1)}+\frac{u^{q+1}}{\Lambda (q+1)} \right\}\\ & \quad+r^\kappa\left\{ u^{\prime\prime} v^\prime +u^\prime v^{\prime\prime} +\frac{v^{p}v^\prime}{\Lambda}+\frac{u^{q}u^\prime}{\Lambda} \right\} \ge 0, \end{align*}

while in $(\frak a,\, \tau _*]$ it yields

\begin{align*} ({E}^{\lambda}_1)^\prime (r) & = \kappa r^{\kappa -1} \left\{ u^\prime v^\prime +\frac{v^{p+1}}{\lambda (p+1)}+\frac{u^{q+1}}{\lambda (q+1)} \right\}\\ & \quad +r^\kappa\left\{ u^{\prime\prime} v^\prime +u^\prime v^{\prime\prime} +\frac{v^{p}v^\prime}{\lambda}+\frac{u^{q}u^\prime}{\lambda} \right\} \ge 0. \end{align*}

On the other hand, in $[\frak a,\, \tau _* ]$ one writes

\[ u^{\prime\prime}v^\prime +\frac{v^p v^\prime}{\lambda} \le - \frac{(\tilde{N}_+{-}1)u^\prime v^\prime}{ r}, \quad v^{\prime\prime}u^\prime +\frac{u^q u^\prime}{\lambda} \le - \frac{(\tilde{N}_+{-}1)u^\prime v^\prime}{r}, \]

while in $[\tau ^*,\, \frak b)$,

\[ u^{\prime\prime}v^\prime +\frac{v^p v^\prime}{\lambda}\le u^{\prime\prime}v^\prime +\frac{v^p v^\prime}{\sigma} \le - \frac{(\tilde{N}_+{-}1) u^\prime v^\prime}{ r}, \quad v^{\prime\prime}u^\prime +\frac{u^q u^\prime}{\lambda} \le - \frac{(\tilde{N}_+{-}1)u^\prime v^\prime}{ r}, \]

so anyways $\mathcal {E}_\lambda ^\prime (r) \le - \frac {2(\tilde {N}_+-1)}{ r}u^\prime v^\prime \le 0$.

Let us now analyse the interval $[\tau _* ,\, \tau ^*]$; to fix the ideas say $\tau _*=\tau _v$ and $\tau ^*=\tau _u$. Then, in $[\tau _v,\,\tau _u]$ we have $u^\prime >0$ and $v^\prime <0$ (recall that $\tau ^*\le \frak b$). Hence

\begin{align*} & u^{\prime\prime}v^\prime +\frac{v^p v^\prime}{\lambda} \ge - \frac{(\tilde{N}_+{-}1)u^\prime v^\prime}{ r},\\ & v^{\prime\prime}u^\prime +\frac{u^q u^\prime}{\lambda}\ge v^{\prime\prime} u^\prime +\frac{u^q u^\prime}{\sigma}={-} \frac{(\hat{N}-1)u^\prime v^\prime}{ r} \ge - \frac{(\tilde{N}_+{-}1)u^\prime v^\prime}{ r}. \end{align*}

Thus, for $\kappa =2(\tilde {N}_+-1)$ we get $\mathcal {E}_{\lambda }^\prime (r) \ge - \frac {\kappa }{ r} \,u^\prime v^\prime \ge 0$. The reasoning is analogous when instead $\tau _*=\tau _u\,$ and $\tau ^*=\tau _v$.

Proof. By proposition 2.3 we have $\mathcal {E}_\lambda (r) \leq \mathcal {E}_\lambda (\frak a)$ for all $r\le \tau _*$, that is,

(2.15)\begin{equation} \frac{1}{p+1}v^{p+1}(r)+\frac{1}{q+1}u^{q+1}(r) \le \lambda \,\delta \mu \quad \textrm{for all}\ [\frak a, \tau_*], \end{equation}

since $u^\prime v^\prime \ge 0$ in $[\frak a,\, \tau _*]$. Observe that (2.15) implies

(2.16)\begin{align} u^q(r) \le C (\delta \mu)^{\frac{q}{q+1}} \quad \textrm{for all}\, r\in [\frak a, \tau_v], \quad v^p(r) \le C (\delta \mu)^{\frac{p}{p+1}}\quad \textrm{for all}\, r\in [\frak a, \tau_u], \end{align}

since $\tau _u$ (resp. $\tau _v$) is the maximum point for $u$ (resp. $v$) in $[\frak a,\, \rho _u]$ (resp. in $[\frak a,\, \rho _v]$).

Next we write the equation for $v$ in $[\frak a,\, \tau _v]$ as $(v^\prime r^{\tilde {N}_- -1})^\prime = - \frac {u^q}{\lambda }r^{\tilde N_- -1}$, and so

(2.17)\begin{equation} 0=v^\prime (\tau_v)\, \tau_v^{\tilde N_-{-}1}=\mu \, {\frak a}^{\tilde N_-{-}1}- \frac{1}{ \lambda }\int_{\frak a}^{\tau_v} r^{\tilde N_-{-}1} {u^q}(r) \,{\rm d}r . \end{equation}

By combining the estimate for $u$ in (2.16) and equality (2.17) we obtain

\[ \mu =\frac{1}{\lambda {\frak a}^{\tilde N_-{-}1}} \int_{\frak a}^{\tau_v} r^{\tilde N_-{-}1} u^q(r) \,{\rm d}r \le \frac{C}{\tilde{N}_-} (\delta \mu)^{\frac{q}{q+1}} \, \tau_v^{\tilde N_-}, \]

from which we derive (2.13).

Analogously, in $[\frak a,\, \tau _u]$ one writes $(u^\prime r^{\tilde {N}_- -1})^\prime = - \frac {v^p}{\lambda }r^{\tilde N_- -1}$ and so

(2.18)\begin{equation} 0=u^\prime (\tau_u)\, \tau_u^{\tilde N_-{-}1} ={\delta}\,{\frak a}^{\tilde N_-{-}1} - \frac{1}{ \lambda }\int_{\frak a}^{\tau_u} r^{\tilde N_-{-}1} {v^p}(r)\, {\rm d}r . \end{equation}

Thus, using the estimate for $v$ in (2.16) into (2.18) one reaches (2.14) out.

Proof. Set $C_1=({\frak {b}^{\tilde N_-}}{/C_0})^{q+1}$ and $C_2=({\frak {b}^{\tilde N_-}}{/C_0})^{p+1}$. By (2.13) and (2.14) we derive

\[ \mu\le C_1 \,\delta^{q} \quad and \quad \delta\le C_2\, \mu^{p}. \]

The combination of these two estimates then implies

\[ \delta^{pq-1} \ge \frac{1}{C_1^p C_2} , \quad \mu^{pq-1} \ge \frac{1}{C_1 C_2^q}, \]

which gives us the lower bound (2.20).

Proof. We fix the annulus $A_{\frak a , \frak b}$ with $0<\frak a <\frak b <+\infty$. Assume by contradiction that there exists a sequence of shooting parameters $(\delta _k ,\, \mu _k)$ with respective solutions $(u_k,\,v_k)$ of (2.2) in $A_{\frak a , \frak b}$ such that at least one of them converges to infinity, that is $\delta _k\to +\infty$ or $\mu _k\to +\infty$. The first step is to show that both of them approach infinity in this case. Step (1) $\delta _k\to +\infty$ and $\mu _k \to +\infty$.

Assume on the contrary that either $\delta _k\to \infty$ or $\mu _k\to \infty$, and the other one is bounded. To fix the ideas we suppose $\delta _k\to +\infty$ and $\mu _k\le C$ for all $k$. Then, by (2.14) we obtain $\frak b \ge \tau _u \to +\infty$, which is impossible since the annulus $A_{\frak a, \frak b}$ is fixed. Analogously, if $\mu _k\to +\infty$ and $\delta _k\le C$ for all $k$, one finds the absurdity $\frak b \ge \tau _v \to +\infty$ by (2.13).

We set

\[ w_k (r):=\frac{1}{p+1}\,v_k^{p+1} (r)+\frac{1}{q+1}\,u_k^{q+1}(r). \]

Step (2) $w_k(\tau _*^k)\to +\infty$ and $w_k(\tau ^*_k)\to +\infty$. We already know that the energy $E_1^{\lambda }$ is increasing in $[\frak a,\, \tau _*^k]$ by proposition 2.3, and the annulus is fixed so that $\frak a\le \tau ^k_*\le \frak b$ for all $k$. Thus,

(3.2)\begin{equation} w_k(\tau_*^k)\ge C_0\,\delta_k\mu_k\ {\rm for\ all}\ k,\; {\rm where}\ C_0=C_0(\frak a, \frak b, N, \lambda, \Lambda, p, q). \end{equation}

On the other hand, $w_k(\tau ^*_k) \ge w_k(\tau _*^k)$ by proposition 2.3, since the energy $\mathcal {E}_\lambda$ is increasing in $[\tau _*^k,\, \tau ^*_k]$. Now, by Step 1 we have $\delta _k\mu _k \to +\infty$. This proves Step 2. Step (3) $\|u_k\|_\infty \to +\infty$ and $\|v_k\|_\infty \to +\infty$. By Step 2 we know that at least one of the norms sequences satisfies $\|u_k\|_\infty \to +\infty$ or $\|v_k\|_\infty \to +\infty$. Without loss we assume $\|u_k\|_\infty \to +\infty$.

Suppose by contradiction that $\|v_k\|_{L^\infty (A)} \leq C$ is bounded in the annulus $A=A_{\frak a, \frak b}$. Recall that $u_k$ solves $-\mathcal {M}^\pm (D^2 u_k)=v_k^p$ in $A$, with $u_k=0$ on $\partial A$. Now we are going to use the Alexandrov-Bakelman-Pucci estimate (ABP), which can be found for instance in [Reference Caffarelli and Cabré3]. By ABP we then get $u_k\le C$ in $A_{\frak a, \frak b}$, which is impossible. Thus, $\|v_k\|_\infty \to +\infty$.

Step (4) $\lim _{k \to \infty } {\tau ^*_k}= \frak b$. Otherwise we may write $\frak b>(1+\epsilon )\tau ^*_k$ for all $k$, up to a subsequence, for some $\epsilon >0$. In particular, $u_k,\, v_k$ are both positive and decreasing in the interval $[\tau _k^*,\,(1+\epsilon )\tau _k^*]$.

We consider the annulus $A_k=A_{\tau _k^*, r}$. Then $U_k=t_ku_k$ and $v_k$ solve

\[ -\mathcal{M}^\pm (D^2 U_k) \ge t_k v_k^p, \; -\mathcal{M}^\pm (D^2 v_k) \ge u_k^q \ge t_k\, U_k^{1/p} \; {\rm in}\ A_k, \; U_k,v_k>0\ {\rm in}\ A_k; \]

while $u_k$ and $V_k=s_kv_k$ satisfy

\[ -\mathcal{M}^\pm (D^2 u_k) \ge v_k^{p}\ge s_k \, V_k^{1/q},\quad -\mathcal{M}^\pm (D^2 V_k) \ge s_k u_k^q\quad {\rm in}\ A_k,\quad u_k,V_k> 0 in A_k, \]

where

\[ t_k=\min_{A_k} u_k ^{\frac{pq-1}{p+1}}=u_k ^{\frac{pq-1}{p+1}} (r), \quad s_k=\min_{A_k} v_k ^{\frac{pq-1}{q+1}}=v_k^{\frac{pq-1}{q+1}}(r). \]

Hence, by the definition of first eigenvalue $\lambda _1^+(\mathcal {D})=\lambda _1^+(\mathcal {M}^\pm,\, \mathcal {M}^\pm,\, \mathcal {D})$ for the fully nonlinear Lane-Emden systems in [Reference Moreira dos Santos, Nornberg, Schiera and Tavares13], we have

(3.3)\begin{equation} u_k^{\frac{pq-1}{p+1}} (r),\, v_k^{\frac{pq-1}{q+1}} (r) \le \lambda_1^+( A_k)\le \lambda_1^+(A_{\tau_k^*, i_k}) \le\frac{1}{{\frak a }^2} \lambda_1^+\left(A_{1,1+\frac{\epsilon}{2}}\right) =:C_1, \end{equation}

for all $r\in I_k=[i_k,\, j_k]$, where $i_k:=\left (1+\frac {\epsilon }{2}\right )\tau _k^*$  and $j_k:=(1+\epsilon )\tau _k^*$, since $\epsilon >0$ is fixed. Recall the energy $E_1^{\Lambda }$ is increasing in $[\tau ^*_k,\,\frak b]$ by proposition 2.3. Thus, ${E}_1^{\Lambda }(\tau _k^*)\le {E}_1^{\Lambda } (r)$ for all $r\in I_k$. Hence, this, (3.2), and (3.3) give us for $r\in I_k$

(3.4)\begin{equation} \textstyle (u_k^\prime v_k^\prime)(r) \ge \frac{1}{\Lambda} \left( \frac{\frak a}{\frak b}\right)^{2(\tilde{N}_-{-}1)} w_k(\tau_k^*)-\frac{1}{\Lambda} w_k(r)\ge c_0 \,\delta_k\mu_k \end{equation}

for large $k$. Recall that $u_k^\prime <0$ and $v_k^\prime <0$ in $I_k$. In particular, $u_k^\prime (r)$ or $v_k^\prime (r)$ goes to $-\infty$ as $k\to \infty$, for all $r\in I_k$.

Observe that $(u_k^\prime )^2 (r) +(v_k^\prime )^2 (r)\ge 2 (u_k^\prime v_k^\prime )(r)\ge 2c_0\, \delta _k \mu _k$ for all $r\in I_k$. Then for each $k$ we have either

\[ |\mathcal{C}_k|=| \, \{ r\in I_k : (u_k^\prime)^2 (r) \ge c_0\, \delta_k \mu_k \}\, |\ge \frac{1}{2} |I_k| \]

or

\[ |\mathcal{D}_k|=| \, \{ r\in I_k : (v_k^\prime)^2 (r) \ge c_0\, \delta_k \mu_k \}\, |\ge \frac{1}{2} |I_k|. \]

So we can extract a subsequence for which one of the two situations occurs, namely the first one $|\mathcal {C}_k|\ge \frac {1}{2} |I_k|$.

Next, the Fundamental Theorem of Calculus and the Lebesgue integration for this subsequence imply

\begin{align*} u_k(i_k) & \ge u_k(i_k)-u_k(\frak b)= \int^{\frak b}_{i_k} ({-}u^\prime_k)\,\geq\int_{\mathcal{C}_k} ({-}u^\prime_k) \\ & = |\mathcal{C}_k|\, (c_0 \, \delta_k \mu_k)^{1/2} \ge {\frac{1}{2}}|I_k|\, (c_0 \, \delta_k \mu_k)^{1/2} \ge {\frac{\epsilon}{4}}\frak a\, (c_0 \, \delta_k \mu_k)^{1/2}\to +\infty \end{align*}

as $k\to +\infty$ by using the fact that $\epsilon >0$ is fixed fulfilling $|I_k|=\frac {\varepsilon }{2}\tau _k^*\ge \frac {\varepsilon }{2}\frak a$. Hence we reach a contradiction with the estimate (3.3). The case when $|\mathcal {D}_k|\ge \frac {1}{2} |I_k|$ is analogous.

Step (5) Conclusion

We reach a contradiction by putting together theorem 2.2 with Step 4, since $\frak b >\frak a$.

Proof. Let $(u,\,v)$ be a solution pair of (2.2) and denote $A=A_{\frak a ,\frak b}$. We set $U=tu$, with $t>0$, and write

\[ \begin{cases} -\mathcal{M}^\pm (D^2 U) \le t \,v^p\\ \;-\mathcal{M}^\pm (D^2 v)\le u^{q-\frac{1}{p}}\,u^{\frac{1}{p}} \le t^{-\frac{1}{p}}\, \|u\|_\infty^{\frac{pq-1}{p}} U^{\frac{1}{p}} = t \, U^{\frac{1}{p}} \end{cases} \]

since $pq>1$, as long as we choose $t=\|u\|_{L^\infty ( A)}^{\frac {pq-1}{p+1}}$. Hence, by applying the ABP estimate in the domain $A$ for each of the scalar PDE inequalities above we obtain

\[ \sup_A U \le C\, t\, \sup_A v^p \,|A|^{1/N} , \quad \sup_A v \le C\, t\, \sup_A U^{\frac{1}{p}} \,|A|^{1/N}. \]

Then by taking the $1/p$ power of the inequality above for $U$, and replacing it into the inequality satisfied by $v$, one finds

\[ \sup_A v\le C^{\frac{p+1}{p}} t^{\frac{p+1}{p}} \sup_A v^+ \; |A|^\frac{p+1}{Np} \quad \Rightarrow \quad t \ge \frac{1}{C |A|^{1/p}} \to \infty \;\; \textrm{ as }\, |A|\to 0. \]

On the other hand, by writing $v^p=v^{p-\frac {1}{q}}\, v^{\frac {1}{q}}\le \|v\|_\infty ^{\frac {pq-1}{q}}v^{\frac {1}{q}}$ and arguing similarly with the pair $(u,\,sV)$, where $V=sv$ for $s=\|v^+\|_{L^\infty (A)}^{\frac {pq-1}{q+1}}$ we get $v(\tau _v)=\|v^+\|_{L^\infty (A)} \to \infty$ as $|A|\to 0$ as well.

Proof. Proof of the existence in the annulus via degree theory

We consider $X=C(\bar {A}_{\frak a , \frak b})\times C(\bar {A}_{\frak a , \frak b})$, with the norm $\|(u,\,v)\|:=\max \{ \|u\|_{L^\infty (A_{\frak a , \frak b})},\, \, \|v\|_{L^\infty (A_{\frak a , \frak b})} \}$.

Let $K=\{ (u,\,v)\in X : u,\,v\ge 0 \}$, and denote $\mathcal {B}_{\frak s} =\{ (u,\,v)\in K : \|(u,\,v)\| < \frak s\}$.

For any $(u,\,v)\in K$ and $t\ge 0$ we define the operator $F(t,\,u,\,v)=(U,\,V)$, with $U=U_t$ and $V=V_t$ , as the unique solution of the problem

\[ -\mathcal{M}^\pm (D^2 U)= (v+t)^p, \quad -\mathcal{M}^\pm (D^2 V)= (u+t)^q \textrm{ in } A_{\frak a , \frak b}, \quad U,V=0 \textrm{ on } \partial A_{\frak a , \frak b}. \]

In particular, $(U,\,V)\in K$ by the maximum principle for scalar equations. We denote $\Phi (\cdot )=F(0,\, \cdot )$. Our goal is to show that $\Phi$ has a positive fixed point $(u,\,v)$.

Let us verify the hypotheses in proposition 4.1.

1. (i) We take $(u,\,v)\in K$ such that $\|(u,\,v)\|=r$, for some $r>0$ to be chosen, and $(u,\,v)=\theta \Phi (u,\,v)$, $\theta \in (0,\,1]$. In particular, $\|u\|_\infty,\, \, \|v\|_\infty \le r$. As before, we set $\tilde u=\kappa u$ and write

   \[ \begin{cases} -\mathcal{M}^\pm (D^2 \tilde u) = \theta \kappa v^p \le \kappa v^p \\ \;-\mathcal{M}^\pm (D^2 v)= \, \theta u^{q-\frac{1}{p}}\,u^{\frac{1}{p}} \le \kappa^{-\frac{1}{p}}\, \|u\|_\infty^{\frac{pq-1}{p}} (\tilde u)^{\frac{1}{p}} = \kappa (\tilde u)^{\frac{1}{p}} \end{cases} \]
   since $pq>1$, as long as $\kappa :=\|u\|_{L^\infty ( A_{\frak a , \frak b})}^{\frac {pq-1}{p+1}} \le r^{\frac {pq-1}{p+1}}$. Then we choose $r>0$ small enough such that $r^{\frac {pq-1}{p+1}}<\lambda _1^+ (\mathcal {M}^\pm,\, \mathcal {M}^\pm,\,A_{\frak a , \frak b})$. Since $u,\,v=0$ on $\partial A_{\frak a , \frak b}$, then by the maximum principle for the Lane-Emden system for fully nonlinear operators with weights in [Reference Moreira dos Santos, Nornberg, Schiera and Tavares13] we get $u,\,v\le 0$ in $A_{\frak a , \frak b}$. Since $(u,\,v)\in K$ then $u,\,v\equiv 0$ in $A_{\frak a , \frak b}$, but this contradicts the fact that $\|(u,\,v)\|=r>0$.

2. (ii) Case 1: $t\ge t_0$

If $F_t$ has a fixed point $(u_t,\,v_t)$ then $\tilde u_t=\kappa u_t$ and $v_t$ solve

\[ \begin{cases} -\mathcal{M}^\pm (D^2 \tilde u_t)=\kappa (v_t+t)^p \ge \kappa v_t^p \\ -\mathcal{M}^\pm (D^2 v_t)= \, (u_t+t)^{q-\frac{1}{p}}\,(u_t+t)^{\frac{1}{p}} \ge t_0^{\frac{pq-1}{p}} \kappa^{-\frac{1}{p}} (\tilde u_t)^{\frac{1}{p}} = \kappa (\tilde u_t)^{\frac{1}{p}} \end{cases} \]

with $\tilde u_t,\,v_t>0$ in $A_{\frak a, \frak b}$ , where

\[ \kappa= t_0 ^{\frac{pq-1}{p+1}}. \]

Now, the definition of first eigenvalue $\lambda _1^+(A_{\frak a , \frak b})=\lambda _1^+(\mathcal {M}^\pm,\, \mathcal {M}^\pm,\,A_{\frak a , \frak b})$ for the fully nonlinear weighted Lane-Emden system in [Reference Moreira dos Santos, Nornberg, Schiera and Tavares13] yields

\[ \kappa \le \lambda_1^+( A_{\frak a, \frak b}). \]

Thus we choose $t_0$ large enough such that $\kappa =2 \lambda _1^+( A_{\frak a, \frak b})$ in order to derive a contradiction.

Case 2: $t\le t_0$

In this case we infer that lemma 3.1 immediately produces a priori bounds for the fixed points of $F(t,\,\cdot )$ in bounded intervals of $t$, that is, for each fixed $t_0>0$ it will give $\|(u_t ,\,v_t)\|\le C(t_0)$ for all solutions $u=u_t$, $v=v_t$ of $F(t,\,u,\,v)=(u ,\,v)$ with $t\in [0,\,t_0]$. Indeed, we define the function

\[ w_t (r):=\frac{1}{p+1}\,|v_t +t|^{p+1} (r)+\frac{1}{q+1}\,|u_t +t|^{q+1}(r)\quad {\rm for}\ t\ge 0. \]

Then Step 1, Step 2 hold for $t>0$ exactly as in the case $t=0$. Moreover, the symmetry result in theorem 2.2 applies as well (and so Step 5) since we maintain the zero boundary condition $u_t=v_t=0$ on $\partial A_{\frak a, \frak b}$. On the other hand, a positive solution $(u_t,\, v_t)$ of

\[ -\mathcal{M}^\pm (D^2 u_t)= (v_t+t)^p, \quad -\mathcal{M}^\pm (D^2 v_t)= (u_t+t)^q \textrm{ in } A_{\frak a , \frak b}, \quad u_t , v_t=0 \textrm{ on } \partial A_{\frak a , \frak b} \]

produces a positive solution $(\tilde u_t ,\, \tilde v_ t)$, with $\tilde {u}_t =u_t +t$ and $\tilde {v}_t = v_t +t$, of

\[ -\mathcal{M}^\pm (D^2 \tilde{u}_t)= \tilde{v}_t^p, \quad -\mathcal{M}^\pm (D^2 \tilde{v}_t ) = \tilde{u}_t^q \textrm{ in } A_{\frak a , \frak b}, \quad \tilde{u}_t , \tilde{v}_t= t \textrm{ on } \partial A_{\frak a , \frak b} . \]

Thus, the proof of Step 4 in lemma 3.1 is unchangeable for $(\tilde u_t,\, \tilde v_t)$ in place of $(u,\,v)$, since we only used in such a proof that the solution is nonnegative at $\frak b$.

Therefore, it is enough to choose $R=2C(t_0)$ in order to conclude that $F_t$ does not have fixed points satisfying $\|(u_t,\,v_t)\|= R$ whenever $t\le t_0$. The complementary case $\|(u_t,\,v_t)\|= R$ with $t\ge t_0$ is automatically fulfilled by Case 1.