Proof. Let $u \in W^{1,2}_{\Gamma_{0}}(\Omega)$. Since $\beta \ge 0$, we have

(2.2)\begin{equation} \mathfrak{b}_M(u,u) \ge \int_{\Omega}\langle \nabla u,A\nabla u\rangle+\int_{\Omega}u \bigg(\langle \vec b,\nabla u\rangle+uc+uM\bigg)-\int_{\Omega}\int_{\Omega}uK(x,y)u(y)dy. \end{equation}

We will obtain estimates for each of the above terms. Since $\mathcal{L}$ is uniformly elliptic in Ω, we get

(2.3)\begin{equation} \int_{\Omega} \langle \nabla u, A \nabla u \rangle \ge \theta \int_{\Omega} \langle \nabla u, \nabla u \rangle = \theta \| \nabla u \|^{2}_{L^{2}(\Omega)}. \end{equation}

Moreover, by using Young’s inequality, we find that

(2.4)\begin{equation} \int_{\Omega} \langle \vec b, \nabla u \rangle u \ge -\frac{\|\vec b\|_{L^{\infty}(\Omega)}}{2} \left( \epsilon \| u \|_{L^{2}(\Omega)}^2 + \frac{\|\nabla u \|_{L^{2}(\Omega)}^2}{\epsilon} \right), \end{equation}

where ϵ is a positive constant. Furthermore,

(2.5)\begin{equation} \int_{\Omega} (uc + uM)u \ge (M + \inf_{\Omega} c) \int_{\Omega} u^2 = (M + \inf_{\Omega} c) \| u \|^{2}_{L^{2}(\Omega)}. \end{equation}

Finally, there exists a positive constant $C_1 \gt 0$ such that

(2.6)\begin{equation} -\int_{\Omega} \int_{\Omega} u K(x, y) u(y) \, dy \ge -\| K \|_{L^{\infty}(\Omega\times\Omega)} C_{1} \| u \|^2_{L^{2}(\Omega)}. \end{equation}

Consequently, substituting (2.3), (2.4), (2.5), and (2.6) into (2.2), we have that

\begin{align*}\begin{array}{ll} \mathfrak{b}_M(u, u) &\ge \left( \theta - \frac{\|\vec b\|_{L^{\infty}(\Omega)}}{2 \epsilon} \right) \|\nabla u \|^{2}_{L^{2}(\Omega)} \\ &+ \left( M + \inf_{\Omega} c - \frac{\|\vec b\|_{L^{\infty}(\Omega)}}{2} \epsilon - \| K \|_{L^{\infty}(\Omega\times\Omega)} C_{1} \right) \| u \|^{2}_{L^{2}(\Omega)}. \end{array}\end{align*}

Choose ϵ > 0 such that $\theta - \|\vec b\|_{L^{\infty}(\Omega)}/2 \epsilon = \theta/2 \gt 0,$ and define $M_{0} := -\inf_{\Omega} c + \| \vec b \|_{L^\infty(\Omega)}\epsilon/2 + \| K \|_{L^{\infty}(\Omega\times\Omega)} C_{1} + \theta/2.$ Then, for all $M \ge M_{0}$, we have that

\begin{equation*} \mathfrak{b}_M(u, u) \ge \frac{\theta}{2} \left( \|\nabla u \|^{2}_{L^{2}(\Omega)} + \| u \|^{2}_{L^{2}(\Omega)} \right) = \frac{\theta}{2} \| u \|^2_{W^{1,2}_{\Gamma_{0}}(\Omega)}. \end{equation*}

This concludes the proof.

Proof. First, note that since Ω is of class C ², by [Reference López-Gómez22, Lemma 2.1], there exist $\psi \in C^{2}(\bar{\Omega})$ and γ > 0 such that

\begin{equation*}\frac{\partial\psi}{\partial\eta}(x)\ge\gamma, \quad\forall x\in\Gamma_{1}.\end{equation*}

Defining $h(x)= e^{\omega \psi(x)}$, where ω > 0 is a constant, we observe that $h(x) \gt 0$ for all $ x \in \Omega $. Moreover, on Γ₁ we find

\begin{equation*} \mathfrak{B}h=\frac{\partial h}{\partial \eta}+\beta h=\omega h\frac{\partial\psi}{\partial\eta}+\beta h=h(\omega\frac{\partial\psi}{\partial\eta}+\beta)\ge h(\omega\gamma+\beta) \gt 0, \end{equation*}

for all $\omega \gt -\inf_{\Gamma_1} \beta/\gamma=:\omega_0$.

On the other hand, due to the regularity of the coefficients of $\mathcal{L_I}$, we have $\mathcal{L_I}h \in L^\infty(\Omega)$. Consequently,

\begin{equation*}(\mathcal{L_I}+M)h \ge -\|\mathcal{L_I}h\|_{L^{\infty}(\Omega)} + M \inf h \gt 0 \quad \mbox{in }\Omega,\end{equation*}

whenever $M \gt \Arrowvert{\mathcal L}_{\mathcal I}h{\Arrowvert_{L^\infty(\Omega)}/\inf h}=:\widetilde M$. Hence, h satisfies the required condition.

Proof. Consider $\mathfrak{b}_0, \mathfrak{b}_{0h}:W^{1,2}_{\Gamma_{0}}(\Omega)\times W^{1,2}_{\Gamma_{0}}(\Omega)\longrightarrow \mathbb{R}$, the bilinear forms given by

\begin{equation*}\mathfrak{b}_0(v,w):=\int_{\Omega}\langle \nabla w,A\nabla v\rangle+\int_{\Omega}w \bigg(\langle \vec b,\nabla v\rangle+vc+vM\bigg)+\int_{\Gamma_{1}}w\beta vdS\end{equation*}

and

\begin{equation*}\mathfrak{b}_{0h}(v,w):=\int_{\Omega}\langle\nabla w,A\nabla v\rangle+\int_{\Omega}w \bigg(\langle \vec b_h,\nabla v\rangle+vc_h+vM\bigg) +\int_{\Gamma_{1}}w\beta_h vdS.\end{equation*}

By [Reference López-Gómez22, Lemma 4.2], it holds for all $v,w\in W^{1,2}_{\Gamma_{0}}(\Omega)$,

\begin{equation*}\mathfrak{b}_0(hv,\frac{w}{h})=\mathfrak{b}_{0h}(v,w).\end{equation*}

Thus

\begin{equation*} \mathfrak{b}_M(hv,\frac{w}{h})= \mathfrak{b}_{0h}(v,w)-\int_{\Omega}\int_{\Omega}wK_h(x,y)v(y)dy =\mathfrak{b}_h(v,w). \end{equation*}

Proof. Suppose that $v\in W^{1,2}_{\Gamma_0}(\Omega)$ is a weak solution of (2.8). Defining u = hv, by Lemma 2.4, it holds that

\begin{equation*}\mathfrak{b}_M(u,\frac{\phi}{h})=\mathfrak{b}_M(hv,\frac{\phi}{h})=\mathfrak{b}_h(v,\phi)=\langle \frac{g}{h},\phi\rangle_{L^{2}(\Omega)}, \quad\forall \phi\in C^{\infty}_{\Gamma_{0}}(\Omega).\end{equation*}

Since $C^{\infty}_{\Gamma_0}(\Omega)$ is dense in $W^{1,2}_{\Gamma_0}(\Omega)$, it follows that

\begin{equation*} \mathfrak{b}_M(u,\psi)=\langle g,\psi\rangle_{L^{2}(\Omega)}, \quad \forall \psi\in C^{\infty}_{\Gamma_{0}}(\Omega), \end{equation*}

which shows that $u\in W^{1,2}_{\Gamma_0}(\Omega)$ is a weak solution of (2.1).

Conversely, suppose $u\in W^{1,2}_{\Gamma_0}(\Omega)$ is a weak solution of (2.1) and define $v:=u/h$. For all $\phi\in C^{\infty}_{\Gamma_{0}}(\Omega),$ we have $\phi/h\in C^{2}_{\Gamma_0}(\Omega)\subset W^{1,2}_{\Gamma_0}(\Omega).$ Using again the density of $C^{\infty}_{\Gamma_0}(\Omega)$ in $W^{1,2}_{\Gamma_0}(\Omega)$, there exists a sequence $\psi_n \in C^{\infty}_{\Gamma_0}(\Omega)$ such that

\begin{equation*}\psi_n\to \frac{\phi}{h}\text{ in }W^{1,2}_{\Gamma_0}(\Omega).\end{equation*}

Thus, by definition of weak solution, u verifies

\begin{equation*} \mathfrak{b}_M(u,\psi_n) = \mathfrak{b}_M(hv,\psi_n) = \langle u, \psi_n\rangle_{L^2(\Omega)}. \end{equation*}

Letting $n\to \infty$ in above equality, it becomes apparent that

\begin{equation*} \mathfrak{b}_M(hv,\frac{\phi}{h})=\langle g,\frac{\phi}{h}\rangle_{L^{2}(\Omega)}=\langle \frac{g}{h},\phi\rangle_{L^{2}(\Omega)},\quad \forall \phi\in C^{\infty}_{\Gamma_{0}}(\Omega). \end{equation*}

By Lemma 2.4, this is equivalent to

\begin{equation*}\mathfrak{b}_h(v,\phi)=\langle \frac{g}{h},\phi\rangle_{L^{2}(\Omega)}, \quad \forall \phi\in C^{\infty}_{\Gamma_{0}}(\Omega).\end{equation*}

Therefore, v is a weak solution of (2.8).

Proof. Let $g \in L^2(\Omega)$. Since $\beta_h\ge0$, it follows from Proposition 2.1 that there exists a unique weak solution, $v\in W^{1,2}_{\Gamma_0}(\Omega)$, to the problem (2.8). Now, by Proposition 2.2, we have that $u:=vh$ is the unique weak solution to the problem (2.1), where h is as in Lemma 2.3.

Proof. First, observe that $h\ge0$ is also a supersolution of $(\mathcal{L},\mathfrak B,\Omega),$ since $\mathcal{L}h\ge \mathcal{L_I}h\ge0$ in Ω.

Now, let $u\in W^{2,p}(\Omega)$, p > N, be a supersolution of $(\mathcal{L_I},\mathfrak B,\Omega)$. If $u\ge0$, then u is also a supersolution of $(\mathcal{L},\mathfrak B,\Omega)$, and we can apply Theorem 2.4 from [Reference López-Gómez22] to conclude the result.

Suppose instead that there exists some $x_-\in\Omega$ such that $u(x_-) \lt 0.$ Define

\begin{equation*}\Lambda:=\{\lambda \gt 0; v_{\lambda}(x):=u(x)+\lambda h(x)\ge0,\,\, x\in\Omega\}\end{equation*}

and $\mu:=\inf\Lambda \gt 0. $ Then $v_\mu\ge0$ satisfies

\begin{equation*} \left\{ \begin{array}{ll} \mathcal{L}v_\mu\geq\mathcal{L_I}v_\mu=\mathcal{L_I}u+\mu\mathcal{L_I}h\ge0&\text{ in } \Omega,\\ \mathfrak{B}v_\mu=\mathfrak B u+\mu\mathfrak B h\ge0&\text{ on }\partial \Omega. \end{array}\right. \end{equation*}

Thus, v_µ is a supersolution of $(\mathcal{L},\mathfrak B,\Omega)$. By following the same steps as in the proof of [Reference López-Gómez22, Theorem 2.4], we conclude that $v_\mu = 0$, which ensures that alternative A3 holds.

Proof. We begin by showing that T_M is well-defined and compact. Let $g\in C^{1}_{\mathfrak{B}}(\overline{\Omega})$. By Theorem 2.5, there exists a unique $u\in W_{\Gamma_0}^{1,2}(\Omega)$ which is the weak solution of problem (2.1). Moreover, since $g\in C^{1}_{\mathfrak{B}}(\overline{\Omega})$, it follows that $g\in L^\infty(\Omega)$. Define $U(x)=\int_{\Omega}K(x,y)u(y)dy$. Then, $U\in L^{\infty}(\Omega)$ and, consequently, u is also a weak solution of

\begin{equation*} \left\{\begin{array}{ll} (\mathcal{L}+M)u=g(x)+U(x)&\text{ in }\Omega,\\ \mathfrak{B}u=0&\text{ on }\partial\Omega. \end{array} \right. \end{equation*}

By elliptic regularity, $u\in W^{2,p}(\Omega)$ for all $p\ge1$. The compact Sobolev embedding theorem then implies that $u\in C^{1}_{\mathfrak{B}}(\overline{\Omega})$. This establishes that T_M is well-defined and compact. The linearity of T_M follows from the linearity of $\mathcal{L_I}+M$.

It remains to prove that T_M is strongly positive. To this end, consider $g\in \mathcal P\setminus \{0\}.$ Then $u=T_Mg$ satisfies

\begin{equation*} \left\{\begin{array}{ll} (\mathcal{L_I}+M)u=g \gt 0 &\text{ in }\Omega,\\ \mathfrak{B}u=0 &\text{ on }\partial\Omega. \end{array} \right. \end{equation*}

Thus, u ≠ 0 is a supersolution of $(\mathcal{L_I} + M, \mathfrak{B}, \Omega)$. By Lemma 2.3, there exists a supersolution $h \in C^{2}(\overline{\Omega})$ such that $h(x) \gt 0$ in $\overline{\Omega}$. Then we are in the conditions of Proposition 2.3 and u satisfies one of the alternatives $A1$, $A2$, or $A3$. Since u ≠ 0, alternative $A1$ cannot occur. If $A3$ were true, then $u=\rho h$, for some ρ < 0 constant. In particular, this would imply

\begin{equation*}0 \lt g=(\mathcal{L_I}+M)u=\rho(\mathcal{L_I}+M)h \lt 0 \quad \mbox{in }\Omega,\end{equation*}

which is a contradiction. Therefore u must satisfy alternative $A2,$ meaning $u\in int(\mathcal P)$. This shows that T_M is strongly positive.

Proof. By Proposition 2.4, we can apply the Krein Rutman Theorem (see [Reference López-Gómez22, Theorem 6.3] or [Reference Daners and Koch Medina12, Th. 12.3]) to obtain that $r(T_M) \gt 0$ is the unique eigenvalue of T_M with an associated eigenfunction in $\mathrm{int} (\mathcal P)$. Additionally, this eigenvalue is algebraically simple and satisfies $r(T_M)\ge \tau$ for any other eigenvalue τ of T_M. Since $r(T_M)$ is an eigenvalue of T_M if and only if $\sigma_0:={1}/{r(T_M)}-M$ is an eigenvalue of problem (EP), it follows that σ ₀ satisfies the conditions of the theorem.

Proof. To show that (i) implies (ii), it suffices to observe that if $\sigma_0 \gt 0$, then the principal eigenfunction $\varphi_0 \gt 0$ associated with σ ₀ is a positive strict supersolution of $(\mathcal{L_{I}},\mathfrak{B},\Omega)$.

Suppose that $(\mathcal{L_{I}},\mathfrak{B},\Omega)$ admits a positive strict supersolution $h \in W^{2,p}(\Omega)$, p > N. By Corollary 2.3, any supersolution $u \in W^{2,p}(\Omega)$, p > N, of $(\mathcal{L_I}, \mathfrak{B}, \Omega),$ must satisfy one of the alternatives $A1, A2$, or $A3$. Since u ≠ 0, Alternative $A1$ cannot hold. Moreover, as h is a positive strict supersolution, u cannot satisfy $A3$ because this would imply $u = \rho h$ with ρ < 0. Hence,

\begin{equation*}0 \leq \mathcal{L_I}u = \rho \mathcal{L_I}h \leq 0 \quad \text{ in } \Omega \quad \mbox{and} \quad 0 \leq \mathfrak{B}u = \rho \mathfrak{B}h \leq 0 \quad \text{ on }\partial\Omega,\end{equation*}

which cannot occur because h is a strict supersolution of $(\mathcal{L_I},\mathfrak{B},\Omega)$. Thus, u must satisfy $A2$. Therefore, (ii) implies (iii).

That (iii) implies that (iv) is immediate.

To show that (iv) implies (i), we argue as follows. Assume that $(\mathcal{L_{I}},\mathfrak{B},\Omega)$ satisfies the maximum principle, and suppose, for contradiction, that $\sigma_0 \leq 0$. Then, a principal eigenfunction $\varphi_0 \gt 0$ associated with σ ₀ satisfies

\begin{equation*} \mathcal{L_{I}}(-\varphi_{0}) = -\sigma_0\varphi_{0} \ge 0 \text{ in } \Omega \quad \mbox{and} \quad \mathfrak{B}(-\varphi_{0}) = 0 \mbox{ on } \partial\Omega.\end{equation*}

By (iv), this implies $-\varphi_{0} \ge 0$, contradiction $\varphi_{0} \gt 0$. Thus $\sigma_0 \gt 0$.

Now, let us prove that (i) implies (v). Let $g \in C^{1}_{\mathfrak{B}}(\overline{\Omega})$, and let $u \in W^{2,p}(\Omega)$, p > N, be a solution of (2.9). Choose $M \gt \max\{M_{0},\widetilde{M}\}$, where M ₀ and $\widetilde{M}$ are as in Lemmas 2.2 and 2.3. Then,

\begin{equation*} (\mathcal{L_{I}}+M)u = M u + g \quad \text{ in } \Omega. \end{equation*}

This implies

\begin{equation*} \left(\frac{1}{M}-T_M\right)u = \frac{1}{M}T_Mg. \end{equation*}

Since $\sigma_0 \gt 0$, it becomes apparent that $ r(T_M) \lt 1/M$. By the Krein Rutman Theorem (see [Reference López-Gómez22, Th. 6.3] or [Reference Daners and Koch Medina12, Th. 12.3]), the operator

\begin{equation*} \left(\frac{1}{M}-T_M\right)^{-1} \in \mathfrak{L}(C^{1}_{\mathfrak{B}}(\overline{\Omega})) \end{equation*}

is well-defined and strongly positive. Therefore,

\begin{equation*} u = \frac{1}{M}\left(\frac{1}{M}-T_M\right)^{-1}T_Mg =: S_0 g. \end{equation*}

where S ₀ is strongly positive as the composition of strongly positive operators. This concludes the implication of (i) in (v).

To finish, suppose that S ₀ is well-defined and strongly positive. Then, given $g \in C^{1}_{\mathfrak{B}}(\overline{\Omega})$, g > 0, the unique solution $u\in W^{2,p}_{\Gamma_0}(\Omega),$ $p \gt N,$ of (2.9) satisfies $u\gg0$. Thus u is a positive strict supersolution of $(\mathcal{L_{I}},\mathfrak{B},\Omega)$.

Proof. To establish (i), we need to show that

\begin{equation*} \sigma_1[\mathcal{L_{I}}-\sigma_0,\mathfrak{D},\Omega] \gt 0. \end{equation*}

By Theorem 2.8, it suffices to find a positive strict supersolution for $(\mathcal{L_{I}}-\sigma_0,\mathfrak{B},\Omega)$. Indeed, it follows that the positive eigenfunction of $\sigma_1[\mathcal{L_{I}},\mathfrak{B},\Omega]$ satisfies this property.

To prove (ii), let $\varphi_0 \gt 0$ be an eigenfunction associated with $\sigma_0 = \sigma_1[\mathcal{L_{I}},\mathfrak{B},\Omega]$. Consequently, within $\Omega_*$, we have

\begin{equation*}\begin{array}{ll} (\mathcal{L}_{*\mathcal{I}}-\sigma_0)\varphi_0&=\mathcal{L}\varphi_0-\int_{\Omega_*}K(x,y)\varphi_0(y)dy-\sigma_0 \varphi_0\\ &\ge \mathcal{L}\varphi_0-\int_{\Omega_*}K(x,y)\varphi_0(y)dy-\int_{\Omega\setminus\Omega_*}K(x,y)\varphi_0(y)dy-\sigma_0 \varphi_0\\ &= (\mathcal{L_{I}}-\sigma_0)\varphi_0 = 0. \end{array}\end{equation*}

Furthermore, as $\Omega_*$ is a proper subdomain of Ω, we have $\partial\Omega_*\cap\Omega\neq\emptyset$ and

\begin{equation*}\mathfrak{B}[\Omega_*]\varphi_0=\left\{ \begin{array}{ll} \varphi_0(x) \gt 0 & \text{on }\partial\Omega_*\cap\Omega,\\ \varphi_0(x)=0 & \text{on }\partial\Omega_*\cap\Gamma_0,\\ \frac{\partial\varphi_0}{\partial\eta}(x)+\beta(x)\varphi_0(x)=0 &\text{on }\partial\Omega_*\cap\Gamma_{1}. \end{array} \right. \end{equation*}

Therefore, φ ₀ is a positive strict supersolution of $(\mathcal{L}_{\mathcal{I}_*}-\sigma_0,\mathfrak{B}[\Omega_*],\Omega_*)$. By Theorem 2.8, we have the result.

The proof of item (iii) is analogous to the local case.

Finally, to prove (iv), we argue as follows. Let $\varphi_2 \gt 0$ be an eigenfunction associated with $\sigma_1[\mathcal{L_I}_2,\mathfrak{B},\Omega]$. In Ω, we have

\begin{equation*} \mathcal{L}\varphi_2-\sigma_1[\mathcal{L_I}_2,\mathfrak{B},\Omega]\varphi_2=\int_{\Omega}K_2(x,y)\varphi_2(y)dy. \end{equation*}

Thus,

\begin{equation*}\begin{array}{ll} (\mathcal{L_I}_1-\sigma_1[\mathcal{L_I}_2,\mathfrak{B},\Omega])\varphi_2&=\mathcal{L}\varphi_2-\int_{\Omega}K_1(x,y)\varphi_2(y)dy-\sigma_1[\mathcal{L_I}_2,\mathfrak{B},\Omega])\varphi_2\\ &=\int_{\Omega}(K_2(x,y)-K_1(x,y))\varphi_2(y)dy \gt 0 \quad \text{ in }\Omega, \end{array}\end{equation*}

where the last inequality holds due to $K_1 \lt K_2$ and the fact that $\varphi_2(x) \gt 0$ for all $x \in \Omega$. Furthermore, $\mathfrak{B}\varphi_2=0$ on $\partial\Omega$. Therefore, $\varphi_2 \gt 0$ is a positive strict supersolution of $(\mathcal{L_I}_1-\sigma_1[\mathcal{L_I}_2,\mathfrak{B},\Omega],\mathfrak{B},\Omega)$. By Theorem 2.8, we conclude that $\sigma_1[\mathcal{L_I}_1-\sigma_1[\mathcal{L_I}_2,\mathfrak{B},\Omega],\mathfrak{B},\Omega] \gt 0$, which proves the result.

Proof. The proof of (i) is analogous to the local case and will be omitted.

To prove (ii), we proceed as follows. For simplicity, let us denote $\sigma_n := \sigma_1[\mathcal{L_I}_n,\mathfrak{B},\Omega]$. Since $K_n \to K$ in $L^{\infty}(\Omega \times \Omega)$, for every ϵ > 0, there exists $n_0 \in \mathbb{N}$ such that

\begin{equation*} 0 \leq K_n \leq K + \epsilon, \quad \text{a.e. in } \Omega, \end{equation*}

whenever $n \geq n_0$. By Proposition 2.5 (iv), it follows that

\begin{equation*} \sigma_1[\mathcal{L_{I+\epsilon}},\mathfrak{B},\Omega] \leq \sigma_n \leq \sigma_1[\mathcal{L},\mathfrak{B},\Omega], \end{equation*}

where

\begin{equation*} \mathcal{L_{I+\epsilon}}u := \mathcal{L}u - \int_{\Omega} (K(x, y) + \epsilon) u(y) \, dy. \end{equation*}

Since the sequence $\{\sigma_n\}_{n \geq 1}$ is bounded, there exists $\sigma^* \in \mathbb{R}$ such that, along some subsequence relabeled by n,

\begin{equation*} \sigma_n \to \sigma^*. \end{equation*}

Furthermore, arguing as in the proof of Lemma 2.3 and noting that $\{K_n\}_{n\geq1}$ is bounded in $L^{\infty}(\Omega \times \Omega)$, there exists $\widetilde{M}_1 \gt 0$ (independent of n) such that, for any $M \gt \widetilde{M}_1$, the family of operators $(\mathcal{L_I}_n + M, \mathfrak{B}_h, \Omega)$, $n \geq 1$, admits a strict supersolution (also independent of n), $h \in C^2(\overline{\Omega})$, with $h(x) \gt 0$ for all $x \in \overline{\Omega}$.

Let $\varphi_n \gt 0$ denote the eigenfunction associated with σ_n such that $\|\varphi_n/h\|_{L^2(\Omega)} = 1$. Using (2.7), it follows that $v_n := \varphi_n/h$ satisfies, for $M \gt \widetilde{M}_1$,

(2.11)\begin{equation} \begin{cases} (\mathcal{L_{I}}_{hn} + M)v_n = (\sigma_n + M)v_n & \text{ in } \Omega, \\ \mathfrak{B}_h v_n = 0 & \text{ on } \partial\Omega, \end{cases} \end{equation}

where

\begin{equation*} \mathcal{L_I}_{hn}u := \mathcal{L}_h u - \int_{\Omega} K_{hn}(x, y) u(y) \, dy, \quad \text{and} \quad K_{hn}(x, y) := K_n(x, y) \frac{h(y)}{h(x)}. \end{equation*}

Utilizing the argument presented in the proof of Lemma 2.2, there exists a constant $M^* \gt 0$ (independent of n) such that the bilinear form $\mathfrak{b}_{hn}$ associated with (2.11) is coercive whenever $M \gt M^*$. Consequently, we fix $M \gt \max\{\widetilde{M}_1, M^*\}$.

Note that v_n satisfies

\begin{equation*} \frac{\theta}{2}\|v_n\|^2_{W^{1,2}_{\Gamma_0}(\Omega)} \leq \mathfrak{b}_{hn}(v_n, v_n) = \sigma_n \|v_n\|^2_{L^2(\Omega)} \leq \sigma_1[\mathcal{L},\mathfrak{B},\Omega]. \end{equation*}

Hence, v_n is bounded in $W^{1,2}_{\Gamma_0}(\Omega)$, and therefore there exists $v^* \in W^{1,2}_{\Gamma_0}(\Omega)$ such that, along some subsequence relabeled by n,

(2.12)\begin{equation} v_n \rightharpoonup v^* \text{ in } W^{1,2}_{\Gamma_0}(\Omega) \quad \text{and} \quad v_n \to v^* \text{ in } L^2(\Omega). \end{equation}

Given that v_n is a solution of (2.11), by definition, we have

(2.13)\begin{equation} \mathfrak{b}_{hn}(v_n, \psi) = \sigma_n \int_\Omega v_n \psi, \quad \forall \psi \in C^{\infty}_{\Gamma_0}(\Omega). \end{equation}

Combining (2.12) with the Dominated Convergence Theorem and letting $n \to +\infty$ in (2.13) yields

\begin{equation*} \mathfrak{b}_{h}(v^*, \psi) = \sigma^* \int_\Omega v^* \psi, \quad \forall \psi \in C^{\infty}_{\Gamma_0}(\Omega). \end{equation*}

Since $v_n \geq 0$ and $\|v_n\|_{L^2(\Omega)} = 1$, it follows that $v^* \geq 0$ and $v^*\neq 0$. By Lemma 2.4, $v^*h \gt 0$ is the eigenfunction associated with $\sigma^*$ and, therefore, $\sigma^* = \sigma_1[\mathcal{L_I}, \mathfrak{B}, \Omega]$. This concludes the proof.

Proof. Fix $\lambda \lt \sigma_0=\sigma_1[\mathcal{L_I},\mathfrak{B},\Omega]$. Then $\sigma_1[\mathcal{L_I}-\lambda,\mathfrak{B},\Omega] = \sigma_0-\lambda \gt 0$. By Theorem 2.8, there exists $w_1 \in C_\mathfrak{B}^1(\overline{\Omega}) \cap W^{2,p}(\Omega)$, p > N, a unique solution of

\begin{equation*}\begin{cases} (\mathcal{L_{I}}-\lambda)u=1&\text{ in }\Omega,\\ \mathfrak{B}u=0&\text{on }\partial\Omega. \end{cases} \end{equation*}

Moreover, $w_1\gg0$, and therefore $\mathfrak{P}\neq\emptyset$. Thus, for every $\lambda \lt \sigma_0$, we have

\begin{equation*} \lambda\le\inf_{x\in\Omega}\frac{\mathcal{L_{I}}w_1(x)}{w_1(x)}\le\sup_{\psi\in\mathfrak{P}}\inf_{x\in\Omega}\frac{\mathcal{L_{I}}\psi(x)}{\psi(x)}. \end{equation*}

Consequently,

\begin{equation*}\sigma_0\le\sup_{\psi\in\mathfrak{P}}\inf_{x\in\Omega}\frac{\mathcal{L_{I}}\psi(x)}{\psi(x)}.\end{equation*}

Assume, by contradiction, that the above inequality is strict. Then there exist ϵ > 0 and $w_2\in\mathfrak{P}$ such that

\begin{equation*}\sigma_0+\epsilon \lt \frac{\mathcal{L_I }w_2}{w_2} \quad\text{ in }\Omega.\end{equation*}

Therefore w ₂ is a positive strict supersolution of $(\mathcal{L_I}-\sigma_0-\epsilon,\mathfrak{B},\Omega)$. By Theorem 2.8, we have

\begin{equation*}0 \lt \sigma_1[\mathcal{L_I}-\sigma_0-\epsilon,\mathfrak{B},\Omega]=-\epsilon,\end{equation*}

which is a contradiction.

Proof. Consider the function $G: W^{2,p}(\Omega)\rightarrow C(\overline{\Omega})$, p > N, given by

\begin{equation*}G(u)=(\mathcal{L}-c)u+c+R(u),\end{equation*}

where $R(u):=-\sum_{i,j=1}^{N}a_{ij}\partial_i u\partial_j u$. By the proof of [Reference López-Gómez22, Th. 8.3], we have that G is a concave function and satisfies

\begin{equation*}G(\log \psi)=\frac{\mathcal{L}\psi}{\psi},\end{equation*}

for all $\psi\in \mathfrak{P}$. Now define $H:W^{2,p}(\Omega)\rightarrow C(\overline{\Omega})$, $p \gt N,$ given by

\begin{equation*}H(u)=-\int_{\Omega} K(x,y)e^{u(y)-u(x)}dy.\end{equation*}

Let us show that H is concave. Indeed, for each $x,y\in \Omega$, $u_1,u_2\in W^{2,p}(\Omega)$, $p \gt N,$ we have

\begin{equation*}\begin{array}{ll} H(tu_1+(1-t)u_2)&=-\int_{\Omega} K(x,y)e^{t(u_1(y)+u_1(x))+(1-t)(u_2(y)-u_2(x))}dy\\ &\ge -\int_{\Omega} K(x,y)\left[te^{u_1(y)+u_1(x)}+(1-t)e^{u_2(y)-u_2(x)}\right]dy\\ &=t H(u_1)+(1-t)H(u_2). \end{array}\end{equation*}

Since both G and H are concave, it follows that $G_{\mathcal{I}}(u):=G(u)+H(u)$ is also concave. Furthermore, for all $\psi\in \mathfrak{P}$, we have

\begin{equation*} G_\mathcal{I}(\log\psi)=G(\log\psi)+H(\log\psi) =\frac{\mathcal{L}\psi}{\psi}-\int_{\Omega} K(x,y)\frac{\psi(y)}{\psi(x)}dy =\frac{\mathcal{L_I}\psi}{\psi}. \end{equation*}

Now let $\psi_1,\psi_2\in \mathfrak{P}$, $P_1,P_2\in L^{\infty}(\Omega)$. Since $\phi\in\mathfrak{P}$, it follows that $1/\phi\in L^{\infty}(\Omega)$ and $\nabla\phi\in L^{\infty}(\Omega,\mathbb{R}^N)$. Consequently, $\psi_1^t\psi_2^{1-t}\in \mathfrak{P}$. Therefore,

\begin{equation*}\begin{array}{ll} \frac{[\mathcal{L_I}+tP_1+(1-t)P_2](\psi_1^t\psi_2^{1-t})}{\psi_1^t\psi_2^{1-t}} &=tP_1+(1-t)P_2+G_\mathcal{I}(\log(\psi_1^t\psi_2^{1-t}))\\ &\ge tP_1+(1-t)P_2+tG_\mathcal{I}(\log(\psi_1))+(1-t)G_\mathcal{I}(\log(\psi_2))\\ &=t\frac{(\mathcal{L_I}+P_1)\psi_1}{\psi_1}+(1-t)\frac{(\mathcal{L_I}+P_2)\psi_2}{\psi_2}. \end{array}\end{equation*}

Since this last inequality holds for any $\psi_1,\psi_2\in\mathfrak{P}$ and in view of (2.14), we obtain the result.

Proof. To establish that $\Sigma_0(\lambda)$ is real analytic, we consider the family of operators $\mathcal{T}(\lambda): W^{2,p}(\Omega_0)\rightarrow L^2(\Omega_0)$ given by

\begin{equation*} \mathcal{T}(\lambda)u:=\mathcal{L_I}u-\lambda m u. \end{equation*}

For each $u\in W^{2,p}(\Omega_0)$, p > N, and $v\in L^2(\Omega_0)$, the mapping $ \lambda \in \mathbb{R} \mapsto \langle v,\mathcal{L_I}(\lambda)u\rangle_{L^2(\Omega_0)} $ is real analytic. Hence, $\mathcal{T}(\lambda)$ is a real holomorphic family of type (A), as described in [Reference Kato21]. Consequently, Theorems 1.7 and 1.8 of [Reference Kato21, $\S$ VII.1.3] imply that $ \Sigma_0(\lambda) $ is real analytic.

The concavity of $\Sigma_0(\lambda)$ follows directly from Proposition 2.7.

To prove (i), note that the continuity of m ensures the existence of $r_1 \gt 0$ such that $\inf_{B} m \gt 0$, where $B:=B_{r_1}(x_+)$ is such that $\overline{B} \subset \Omega_0$. Using Proposition 2.5 (ii), we conclude that

\begin{equation*}\begin{array}{ll} \Sigma_0(\lambda) & \lt \sigma_1[\mathcal{L} - \int_{B}K(x,y) \cdot dy-\lambda m,\mathfrak{D},B] \\ & \lt \sigma_1[\mathcal{L}- \int_{B}K(x,y) \cdot dy,\mathfrak{D},B]-\lambda \inf_{B} m. \end{array}\end{equation*}

Letting $\lambda \to +\infty$ in the above inequality yields the desired result. The proof of (ii) follows similarly by considering a point $x_-\in\Omega_0$, where $m(x_-) \lt 0$.

Proof. By assumption (m ₁) and using Proposition 3.1 (i), we have

\begin{equation*} \lim_{\lambda\to +\infty}\Sigma_0(\lambda)=-\infty. \end{equation*}

On the other hand, by hypothesis (Λ ₀), we get $\Sigma_0(\lambda_0) \gt 0$. Since $\Sigma_0\neq 0$ is real analytic and concave to ensure that there exists a unique $\lambda_1^{\mathfrak B_0}(\Omega_0) \gt \lambda_0$, $\Sigma_0(\lambda_1^{\mathfrak B_0}(\Omega_0))=0$.

Proof. By Proposition 3.1, the mappings defined in (3.1) are real analytic and concave. Since $b\geq0$, Proposition 2.5 (iii) ensures that the mapping $\mu\in[0,+\infty)\mapsto\Sigma(\cdot,\mu)$ is strictly increasing. Thus,

(3.4)\begin{equation} \Sigma(\lambda)=\Sigma(\lambda,0)=\sigma_1[\mathcal{L_{I}}-\lambda m,\mathfrak{B},\Omega] \lt \sigma_1[\mathcal{L_{I}}-\lambda m+\mu b,\mathfrak{B},\Omega]=\Sigma(\lambda,\mu). \end{equation}

Using Proposition 2.5 (ii), we also obtain

(3.5)\begin{equation} \Sigma(\lambda,\mu)=\sigma_1[\mathcal{L_{I}}-\lambda m+\mu b,\mathfrak{B},\Omega] \lt \sigma_1[\mathcal{L_{I}}_0-\lambda m,\mathfrak{B}_0,\Omega_0]=\Sigma_0(\lambda), \end{equation}

for all $\lambda\in\mathbb R, \mu\geq0$. Combining (3.4) and (3.5) gives (3.2). Finally, (3.3) follows directly from [Reference López-Gómez22, Th. 9.6].

Proof. By hypothesis (Λ ₀) and equation (3.3), there exists $\bar\mu \gt 0$, such that $\Sigma(\lambda_0,\mu) \gt 0$, for all $\mu\ge\bar\mu$. By hypothesis (m ₁) and Proposition 3.1 (ii), for each $\mu \ge \bar \mu$, we have

\begin{equation*} \lim_{\lambda \to +\infty} \Sigma(\lambda,\mu) =-\infty. \end{equation*}

Thus, for a fixed $\mu \ge\bar \mu$, the monotonicity properties of the continuous mapping $\lambda \mapsto \Sigma(\lambda,\mu)$, guarantee the existence of a unique $\lambda_1^{\mathfrak B}(\mu,\Omega) \gt \lambda_0$, such that

\begin{equation*} \Sigma(\lambda_1^{\mathfrak B}(\mu,\Omega), \mu)=0. \end{equation*}

Hence, $\lambda_1^{\mathfrak B}(\mu,\Omega)$ is an principal eigenvalue of (E_λ,μ).

Next, we prove that the mapping $\mu \in [\bar\mu,+\infty)\mapsto \lambda_1^{\mathfrak B}(\mu,\Omega)$ is strictly increasing. Consider $\bar\mu\le\mu_1 \lt \mu_2.$ Then, there exist unique real numbers $\lambda_1^{\mathfrak B}(\mu_1,\Omega)$ and $\lambda_1^{\mathfrak B}(\mu_2,\Omega)$ both greater than $\lambda_0,$ such that

\begin{equation*}\Sigma(\lambda_1^{\mathfrak B}(\mu_1,\Omega),\mu_1)=0 \quad\text{and}\quad\Sigma(\lambda_1^{\mathfrak B}(\mu_2,\Omega),\mu_2)=0.\end{equation*}

Since the mapping $\mu\mapsto\Sigma(\lambda,\mu)$ is increasing for all $\lambda\in\mathbb R$, we have

\begin{equation*} \Sigma(\lambda_1^{\mathfrak B}(\mu_1,\Omega),\mu_2) \gt \Sigma(\lambda_1^{\mathfrak B}(\mu_1,\Omega),\mu_1)=0. \end{equation*}

Arguing as in the proof of Theorem 3.2, there exists a unique $\lambda^* \gt \lambda_1^{\mathfrak B}(\mu_1,\Omega)$ such that $\Sigma(\lambda^*,\mu_2)=0.$ By uniqueness, we must have $\lambda^*=\lambda_1^{\mathfrak B}(\mu_2,\Omega).$ This proves that $\lambda_1^\mathfrak{B}(\mu,\Omega)$ is strictly increasing with respect to µ.

To prove continuity, let $\{\mu_n\}_{n\geq1}$ be an increasing sequence such that $\mu_n\to\mu^*$, with $\bar\mu\le\mu_n,$ for all $n\in\mathbb{N}$. By Proposition 2.5 (ii), we have

\begin{equation*} \Sigma(\lambda,\mu) \lt \Sigma_0(\lambda),\quad\forall\mu\geq0, \end{equation*}

which implies that

(3.7)\begin{equation} \lambda_1^{\mathfrak{B}}(\mu,\Omega) \lt \lambda_1^{\mathfrak{B}_0}(\Omega_0),\quad\forall \mu\geq0. \end{equation}

Therefore, $\left\{\lambda_1^{\mathfrak{B}}(\mu_n,\Omega)\right\}_{n\geq1}$ is an increasing and bounded sequence, which implies the existence of $\bar{\lambda} \in \mathbb{R}$ such that

\begin{equation*} \lim_{n\to\infty}\lambda_1^{\mathfrak{B}}(\mu_n,\Omega) = \bar{\lambda}. \end{equation*}

Moreover, for each $n \in \mathbb{N}$,

\begin{equation*} \Sigma(\lambda_1^{\mathfrak{B}}(\mu_n,\Omega), \mu_n) = 0. \end{equation*}

By the continuity of $(\lambda, \mu) \mapsto \Sigma(\lambda, \mu) $,

\begin{equation*} 0 = \lim_{n\to\infty}\Sigma(\lambda_1^{\mathfrak{B}}(\mu_n,\Omega), \mu_n) = \Sigma\left(\lim_{n\to\infty}\lambda_1^{\mathfrak{B}}(\mu_n,\Omega), \lim_{n\to\infty}\mu_n\right) = \Sigma(\bar{\lambda}, \mu^*). \end{equation*}

By arguments similar to those in Theorem 3.2, the map $\lambda \in (\lambda_0,+\infty) \mapsto \Sigma(\lambda, \mu^*)$ has a unique zero. Therefore, we must have $\bar{\lambda} = \lambda_1^{\mathfrak{B}}(\mu^*, \Omega)$.

Finally, we prove (3.6). From (3.7), it follows that $\lambda_1^{\mathfrak{B}}(\mu, \Omega)$ is bounded for all $\mu \geq \bar\mu$. Since $\lambda_1^\mathfrak{B}(\mu,\Omega)$ is strictly increasing with respect to µ, we can define

\begin{equation*} \lambda_0 \lt \lambda_{\infty} := \lim_{\mu\to\infty}\lambda_1^{\mathfrak{B}}(\mu, \Omega). \end{equation*}

Consequently, for any ϵ > 0, there exists $\mu(\epsilon) \geq \bar\mu$ such that

\begin{equation*} -\lambda_\infty m(x) - \epsilon \le -\lambda_1^{\mathfrak{B}}(\mu, \Omega)m(x) \le -\lambda_\infty m(x) + \epsilon, \end{equation*}

for all $\mu \ge \mu(\epsilon)$. By Proposition 2.5 (iii), for $\mu \ge \mu(\epsilon)$, we have

\begin{equation*}\begin{array}{ll} \sigma_1[\mathcal{L_{I}} - \lambda_\infty m(x) + \mu b(x), \mathfrak{B}, \Omega] - \epsilon &\le \sigma_1[\mathcal{L_{I}} - \lambda_1^{\mathfrak{B}}(\mu, \Omega)m(x) + \mu b(x), \mathfrak{B}, \Omega] \\ & \lt \sigma_1[\mathcal{L_{I}}_0 - \lambda_1^{\mathfrak{B}}(\mu, \Omega)m(x), \mathfrak{B}_0, \Omega_0]. \end{array}\end{equation*}

Letting $\mu\to\infty$ in the above inequalities and applying [Reference López-Gómez22, Th. 9.6], along with the continuity of the principal eigenvalue with respect to the potential, we obtain that, for every ϵ > 0,

\begin{equation*}\begin{array}{ll} \sigma_1[\mathcal{L_{I}}_0-\lambda_\infty m(x),\mathfrak{B}_0,\Omega_0]-\epsilon&\le\lim_{\mu\to\infty}\sigma_1[\mathcal{L_{I}}-\lambda_1^{\mathfrak{B}}(\mu,\Omega)m(x)+\mu b(x),\mathfrak{B},\Omega]\\ &\le\sigma_1[\mathcal{L_{I}}_0-\lambda_\infty m(x),\mathfrak{B}_0,\Omega_0]. \end{array}\end{equation*}

Letting $\epsilon\to 0^+$ yields

\begin{equation*}\lim_{\mu\to\infty}\sigma_1[\mathcal{L_{I}}-\lambda_1^{\mathfrak{B}}(\mu,\Omega)m(x)+\mu b(x),\mathfrak{B},\Omega] =\sigma_1[\mathcal{L_{I}}_0-\lambda_\infty m(x),\mathfrak{B}_0,\Omega_0].\end{equation*}

On the other hand, for any $\mu\geq\bar\mu$, we have

\begin{equation*} 0=\Sigma(\lambda_1^{\mathfrak{B}}(\mu,\Omega),\mu)=\sigma_1[\mathcal{L_{I}}-\lambda_1^{\mathfrak{B}}(\mu,\Omega)m(x)+\mu b(x),\mathfrak{B},\Omega], \end{equation*}

and thus

\begin{equation*}\begin{array}{ll} 0=\lim_{\mu\to\infty}\Sigma(\lambda_1^{\mathfrak{B}}(\mu,\Omega),\mu)&=\lim_{\mu\to\infty}\sigma_1[\mathcal{L_{I}}-\lambda_1^{\mathfrak{B}}(\mu,\Omega)m(x)+\mu b(x),\mathfrak{B},\Omega]\\ &=\sigma_1[\mathcal{L_{I}}_0-\lambda_\infty m(x),\mathfrak{B}_0,\Omega_0]\\ &=\Sigma_0(\lambda_{\infty}). \end{array}\end{equation*}

By the uniqueness of the zero of the mapping $\lambda\in (\lambda_0,+\infty)\mapsto\Sigma_0(\lambda)$, we conclude that

\begin{equation*}\lambda_{\infty}=\lambda_1^{\mathfrak{B}_0}(\Omega_0).\end{equation*}

Proof. Since $\Sigma(\lambda,0)=\Sigma(\lambda)$, the mapping $\mu\mapsto\Sigma(\lambda,\mu)$ is strictly increasing, and (Λ′₀) is satisfied, we obtain (3.8). The remaining assertions follow from the previous corollary.

Proof. Consider, for all $ u,v \in W^{1,2}_{\Gamma_0}(\Omega) $, the bilinear form associated with the operator $ \mathcal{L_I} $:

(3.12)\begin{equation} \mathfrak{b}(u,v)=\int_{\Omega}\langle A\nabla u,\nabla v\rangle+\int_{\Omega}\left(\langle \vec b, \nabla u\rangle +uc\right)v+\int_{\Gamma_1}\beta uv dS-\int_{\Omega}\int_{\Omega}vK(x,y)u(y)dy, \end{equation}

and we define

\begin{equation*} \mathfrak{b^+}(u,v):=\mathfrak{b}(u,v)+\int_{\Gamma_1}\beta^-uvdS, \end{equation*}

where $ \beta^-(x):=\max\{-\beta(x),0\} $. By Lemma 2.2, for M > 0 sufficiently large, we find that

\begin{equation*} \mathfrak{b}^+_M(u,v):=\mathfrak{b}^+(u,v)+M\int_\Omega uv \end{equation*}

is coercive. To simplify the notation, let us denote $ \lambda_\mu=\lambda^{\mathfrak{B}}_1(\mu,\Omega) $. Thus, φ_µ satisfies

\begin{equation*} (\mathcal{L_{I}}+M)\varphi_\mu+\mu b\varphi_\mu=\lambda_\mu\varphi_\mu +M\varphi_\mu \quad \text{in } \Omega. \end{equation*}

Multiplying this equation by φ_µ, integrating over Ω, and applying the integration by parts, we obtain

(3.13)\begin{equation} \mathfrak{b}^+_M(\varphi_\mu,\varphi_\mu)=\lambda_\mu \int_\Omega m\varphi^2_\mu+M\int_\Omega\varphi^2_\mu+\int_{\Gamma_1}\beta^-\varphi_\mu^2-\mu \int_\Omega b\varphi^2_\mu. \end{equation}

Since $ b_M^+ $ is coercive, there exists a constant $ C_0 \gt 0 $ such that

\begin{align*}\begin{array}{lc}C_0\Arrowvert\varphi_\mu\Arrowvert_{W_{\Gamma_0}^{1,2}(\Omega)}^2&\leq\lambda_\mu\int_\Omega m\varphi_\mu^2+M\int_\Omega\varphi_\mu^2+\int_{\Gamma_1}\beta^-\varphi_\mu^2-\mu\int_\Omega b\varphi_\mu^2\\&\leq\lambda_\infty\Arrowvert m\Arrowvert_{L^\infty(\Omega)}+M+\Arrowvert\beta\Arrowvert_{L^\infty(\partial\Omega)}\int_{\Gamma_1}\varphi_\mu^2.\end{array}\end{align*}

By [Reference Afrouzi and Brown2, Lemma 1], for each ϵ > 0, there exists $ C(\epsilon) \gt 0 $ such that

\begin{equation*}\begin{array}{ll} \int_{\Gamma_1}\varphi_\mu^2 &\le \int_{\partial\Omega}\varphi_\mu^2 \le \epsilon\int_{\Omega}|\nabla\varphi_\mu|^2+C(\epsilon)\int_{\Omega}\varphi_\mu^2\\ &\le \epsilon\left(\int_{\Omega}|\nabla\varphi_\mu|^2+\int_{\Omega}\varphi_\mu^2\right)+C(\epsilon). \end{array}\end{equation*}

Thus,

\begin{equation*} C_0\|\varphi_\mu\|^2_{W^{1,2}_{\Gamma_{0}}(\Omega)} \le \lambda_{\infty}\|m\|_{L^\infty(\Omega)}+M+\|\beta\|_{L^\infty(\partial\Omega)}\left(\epsilon\|\varphi_\mu\|^2_{W^{1,2}_{\Gamma_{0}}(\Omega)}+C(\epsilon)\right). \end{equation*}

Choosing ϵ > 0 sufficiently small such that $ \|\beta\|_{L^\infty(\partial\Omega)}\epsilon \lt C_0 $, we have

\begin{equation*} \|\varphi_\mu\|^2_{W^{1,2}_{\Gamma_{0}}(\Omega)} \le \sqrt{\frac{\lambda_{\infty}\|m\|_{L^\infty(\Omega)}+M+\|\beta\|_{L^\infty(\partial\Omega)}C(\epsilon)}{C_0-\epsilon\|\beta\|_{L^\infty(\partial\Omega)}}} =:\tilde{C}. \end{equation*}

Thus, $ \{\varphi_\mu\}_\mu $ is bounded in $ W^{1,2}_{\Gamma_{0}}(\Omega)$. Therefore, for any increasing sequence $ \{\mu_n\}_n $ such that $ \lim_{n \to \infty} \mu_n =\infty $, we can extract a subsequence, relabelled by n, and a function $ \psi\in W^{1,2}_{\Gamma_{0}}(\Omega) $, such that

\begin{equation*} \varphi_{\mu_n}\rightharpoonup\psi\text{ in } W^{1,2}_{\Gamma_{0}}(\Omega) \quad \mbox{and} \quad \varphi_{\mu_n}\to\psi\text{ in } L^{2}(\Omega). \end{equation*}

For simplicity, let

\begin{equation*}\varphi_n:=\varphi_{\mu_n}\ \text{and}\ \lambda_n:=\lambda_{\mu_n}.\end{equation*}

On the other hand, it follows from (3.13) that

\begin{equation*}\begin{array}{ll} \mu_n \int_\Omega b\varphi^2_n &= \lambda_n\int_\Omega\varphi^2_{n}-\mathfrak{b}^+_M(u,v)+M\int_\Omega \varphi^2_{n}+\int_{\Gamma_1}\beta^-\varphi_{n}^2\\ &\le \lambda_\infty m\int_\Omega\varphi^2_{n}+M\int_\Omega\varphi^2_{n}+\int_{\Gamma_1}\beta^-\varphi_{n}^2\\ &\le\lambda_{\infty}\|m\|_{L^\infty(\Omega)}+M+\|\beta^-\|_{L^\infty(\Omega)}\left( \epsilon\|\varphi_{n}\|^2_{W^{1,2}_{\Gamma_{0}}(\Omega)}+C(\epsilon)\right)\\ &\le\lambda_{\infty}\|m\|_{L^\infty(\Omega)}+M+\|\beta^-\|_{L^\infty(\Omega)}\left( \epsilon \tilde{C}+C(\epsilon)\right) \lt +\infty. \end{array}\end{equation*}

Consequently,

\begin{equation*} \lim_{n\to\infty}\int_{\Omega\setminus\overline{\Omega}_0}b\varphi^2_n=0. \end{equation*}

Since $ \{\varphi_{n}\}_n $ converges in $ L^2(\Omega) $, the above limit implies that ψ = 0 on $ \Omega\setminus\ \overline{\Omega}_{0} $. In particular, by [Reference Cano-Casanova and López-Gómez8, Th. 6.4], we conclude that $ \psi \in W_{\partial \Omega_0 \cap \Omega}^{1,2}(\Omega_0) $.

Now, we show that

\begin{equation*} \lim_{n\to\infty}\|\varphi_{n} -\psi\|_{W^{1,2}_{\Gamma_{0}}(\Omega)}=0. \end{equation*}

Since φ_n is an eigenfunction associated with $ \lambda_n = \lambda_1^{\mathfrak{B}}(\mu_n,\Omega) $, we obtain

\begin{align*} &\mathfrak{b}(\varphi_n,\varphi_n)-2\mathfrak{b}(\varphi_n,\varphi_k)+\mathfrak{b}(\varphi_k,\varphi_k)\\ &\quad=\lambda_n\int_{\Omega}m(\varphi_n-\varphi_k)^2+(\lambda_k-\lambda_n)\int_{\Omega}m\varphi_k^2+(\mu_n-\mu_k)\int_{\Omega}b\varphi_k^2. \end{align*}

Thus, since $ \{\varphi_n\}_n $ and $ \{\lambda_n\}_n $ are Cauchy sequences, given ϵ > 0, there exists $ n_0\in \mathbb{N} $ such that

(3.14)\begin{align} &\mathfrak{b}(\varphi_n,\varphi_n-\varphi_k)+\mathfrak{b}(\varphi_k-\varphi_n,\varphi_k)\nonumber\\ &\quad=\lambda_n\int_{\Omega}m(\varphi_n-\varphi_k)^2+(\lambda_k-\lambda_n)\int_{\Omega}m\varphi_k^2\nonumber\\ &\qquad+(\mu_n-\mu_k)\int_{\Omega}b\varphi_k^2\nonumber\\ &\quad\le\lambda_n\int_{\Omega}m(\varphi_n-\varphi_k)^2+(\lambda_k-\lambda_n)\int_{\Omega}m\varphi_k^2 \lt \epsilon, \end{align}

whenever $ n,k\ge n_0 $. On the other hand. using the definition of the bilinear form (3.12), we have

(3.15)\begin{align} &\mathfrak{b}(\varphi_n,\varphi_n-\varphi_k)+\mathfrak{b}(\varphi_k-\varphi_n,\varphi_k)\nonumber\\ &\quad=\int_{\Omega}\langle A\nabla (\varphi_n-\varphi_k),\nabla (\varphi_n-\varphi_k)\rangle+\int_{\Omega}\langle \vec b, \nabla \varphi_n\rangle(\varphi_n-\varphi_k)\nonumber\\ &\qquad+\int_{\Omega}\langle \vec b,\nabla(\varphi_k-\varphi_n)\rangle\varphi_n + \int_{\Omega}c(\varphi_n-\varphi_k)^2\nonumber\\ &\qquad+\int_{\Gamma_1}\beta (\varphi_n-\varphi_k)^2 dS-\int_{\Omega}\int_{\Omega}(\varphi_n-\varphi_k) K(x,y)\varphi_n(y)dy\nonumber\\ &\qquad-\int_{\Omega}\int_{\Omega}\varphi_k K(x,y)(\varphi_k-\varphi_n)(y)dy. \end{align}

Next, note that

(3.16)\begin{align} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\theta\|\varphi_n-\varphi_k\|_{W_{\Gamma_{0}}^{1,2}(\Omega)}\le\int_{\Omega}\langle A\nabla (\varphi_n-\varphi_k),\nabla (\varphi_n-\varphi_k)\rangle; \end{align}

(3.17)\begin{align} \int_{\Omega}\langle \vec b,\nabla(\varphi_k-\varphi_n)\rangle\varphi_n&\le\|\vec b\|_{L^{\infty}(\Omega)}\int_{\Omega}|\nabla(\varphi_k-\varphi_n)||\varphi_k|\nonumber\\ &=\|\vec b\|_{L^{\infty}(\Omega)}\left(\int_{\Omega}|\varphi_k-\varphi_n|\nabla|\varphi_k|+\int_{\Gamma_1}\beta|\varphi_k||\varphi_k-\varphi_n|\right)\nonumber\\ &\le \tilde{C} \|\vec b\|_{L^{\infty}(\Omega)}\|\varphi_k-\varphi_n\|_{L^2(\Omega)}. \end{align}

Moreover, all other terms in (3.15) are Cauchy sequences. Thus, by (3.14), (3.16), and (3.17), we obtain that $\{\varphi_n\}_{n}$ is a Cauchy sequence in $W^{1,2}_{\Gamma_{0}}(\Omega)$ and, therefore, convergent. Since φ_n converges to ψ in $L^2(\Omega)$, we obtain (3.10).

Finally, we prove that ψ > 0 is a solution of

(3.18)\begin{equation}\left\{\begin{array}{lc}{{\mathcal L}_{\mathcal I}}_0\psi=\lambda m(x)\psi&\text{in\ }\Omega_0,\\{\mathfrak B}_0\psi=0&\text{on}\ \partial\Omega_0.\end{array}\right.\end{equation}

To do this, pick any test function $\xi \in C^{\infty}_{\Gamma_0}(\Omega_{0})$. Multiplying the equation of φ_n by ξ and integrating by parts, we have

\begin{align*} \lambda_n\int_{\Omega_0} m\varphi_n\xi&= \int_{\Omega_0}\langle A\nabla \varphi_{n},\nabla \xi\rangle+\int_{\Omega_0}\left(\langle \vec b, \nabla \varphi_{n}\rangle +\varphi_{n} c\right)\xi+\int_{\Gamma_1}\beta \xi \varphi_{n} dS\\ &\qquad-\int_{\Omega_0}\int_{\Omega_0}\xi K(x,y)\varphi_{n}(y)dy. \end{align*}

Letting $n \to \infty$ and using the convergence properties of $\{\varphi_n\}_n$ combined with the Dominated Convergence Theorem of Lebesgue, we conclude that

\begin{align*} \lambda_\infty\int_{\Omega_0} m\psi\xi &= \int_{\Omega_0}\langle A\nabla \psi,\nabla \xi\rangle+\int_{\Omega_0}\left(\langle \vec b, \nabla \psi\rangle +\psi c\right)\xi \\ &\qquad+\int_{\Gamma_1}\beta \xi \varphi_\mu dS-\int_{\Omega_0}\int_{\Omega_0}\xi K(x,y)\psi (y)dy, \end{align*}

for all $\xi\in C^\infty_{\Gamma_0}(\Omega_{0})$. This shows that ψ > 0 is a weak solution of (3.18). By the uniqueness of the principal eigenvalue, it follows that $\psi=\varphi_{0}$ in Ω₀.

To obtain (3.11), we proceed as follows. If N = 1, by [Reference Gilbarg and Neil18, Theorem 7.26], it holds that $W^{1,2}(\Omega)\hookrightarrow C^{0,\frac{1}{2}}(\overline{\Omega})$, which shows the result.

Now, assume $a_{ij}\in C^2(\Omega_0), b_j, c, m, K(\cdot,y)\in C^1(\Omega_0),$ for all $y\in \Omega.$ Note that

\begin{equation*} \mathcal{L_I}_0(\varphi_\mu-\psi)=\lambda_\mu m\varphi_\mu-\lambda^{\mathfrak{B}_0}_1(\Omega_0) m\psi \quad \text{in }\Omega_0, \end{equation*}

which is equivalent to

\begin{equation*} \mathcal{L}(\varphi_\mu-\psi)=\lambda_\mu m(\varphi_\mu-\psi)+(\lambda_\mu-\lambda^{\mathfrak{B}_0}_1(\Omega_0))m\psi+\int_{\Omega_0}K(x,y)(\varphi_\mu-\psi)(y)dy \quad \text{in }\Omega_0. \end{equation*}

Thus, defining

\begin{equation*} f_\mu:=\lambda_\mu m(\varphi_\mu-\psi)+(\lambda_\mu-\lambda^{\mathfrak{B}_0}_1(\Omega_0))m\psi+\int_{\Omega_0}K(x,y)(\varphi_\mu-\psi)(y)dy, \end{equation*}

and combining $m\in C(\overline{\Omega}),$ with (3.9) and (3.10), we obtain

(3.19)\begin{equation} \begin{array}{ll} \lim_{\mu\to\infty}\|f_\mu\|_{W^{1,2}(\Omega_0)}=0. \end{array} \end{equation}

By [Reference Gilbarg and Neil18, Theorem 8.10], for every proper subdomain $\Omega_1\subset\Omega_0,$ there exists a constant $C_1 \gt 0$ such that

\begin{equation*} \|\varphi_\mu-\psi\|_{W^{3,2}(\Omega_1)}\le C_1\left( \|\varphi_\mu-\psi\|_{W^{1,2}(\Omega_0)}+\|f_\mu\|_{W^{1,2}(\Omega_0)}\right). \end{equation*}

Thus, taking $\mu \to \infty$, using (3.19) and (3.10), we have

(3.20)\begin{equation} \begin{array}{ll} \lim_{\mu\to\infty}\|\varphi_\mu-\psi\|_{W^{3,2}(\Omega_1)}=0. \end{array} \end{equation}

If $N \lt 6,$ by Sobolev embeddings (see [Reference Gilbarg and Neil18, Th. 7.26])

\begin{equation*} \lim_{\mu\to\infty}\|\varphi_\mu-\psi\|_{C(\overline{\Omega}_1)}=0. \end{equation*}

This concludes the proof in this case.

Finally, assume $a_{ij} \in C^4(\Omega_0), b_j, c, m, K(\cdot,y) \in C^3(\Omega_0),$ for all $y \in \Omega.$ Using (3.20), we have

\begin{equation*} \lim_{\mu\to\infty}\|f_\mu\|_{W^{3,2}(\Omega_0)}=0. \end{equation*}

Thus, by [Reference Gilbarg and Neil18, Th. 8.10], there exists a constant $C_2 \gt 0$ such that, for every proper subdomain $\Omega_2 \subset \Omega_1$, we have

\begin{equation*} \|\varphi_\mu-\psi\|_{W^{5,2}(\Omega_1)} \le C_2\left( \|\varphi_\mu-\psi\|_{W^{1,2}(\Omega_0)} + \|f_\mu\|_{W^{3,2}(\Omega_0)} \right). \end{equation*}

Therefore, if N < 10, arguing as above, we obtain the result for this case.

By an inductive argument, we obtain for each $N \ge 1$, the existence of $r = r(N) \ge 1$ that satisfies the desired result.