Existence of Solutions for Nonlinear Choquard Equations with (p,q)-Laplacian on Finite Weighted Lattice Graphs

Lemma 7.

Under the conditions of Theorem 1,

(i): there exists $ρ > 0$ , $\forall u \in N_{p, q}$ , and one has $∥ u ∥ \geq ρ$ ;
(ii): $m_{p, q} > 0$ ;
(iii): $\forall u \in X ∖ 0$ , and there exists a unique $t_{u} > 0$ such that $t_{u} u \in N_{p, q}$ .

Proof. (i) For any

u \in N_{p, q}

with

∥ u ∥ \leq 1

, from (32) and (12), we deduce that

${∥ u ∥}_{p, V}^{p} + {∥ u ∥}_{q, V}^{q} = \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{{| u (y) |}^{r}}{d {(x, y)}^{N - α}}) {| u |}^{r} d μ \leq C {∥ u ∥}^{2 r} .$

(33)

Since

{∥ u ∥}_{p, V}, {∥ u ∥}_{q, V} < 1

, it follows from (13) that

${∥ u ∥}_{p, V}^{p} + {∥ u ∥}_{q, V}^{q} \geq {∥ u ∥}_{p, V}^{q} + {∥ u ∥}_{q, V}^{q} \geq c_{q} {(∥ u ∥}_{p, V} + {∥ u ∥}_{q, V})^{q} = c_{q} {∥ u ∥}^{q} .$

(34)

Combining (33), (34) and $2 r > q$ , we conclude that there exists $1 > ρ > 0$ such that $∥ u ∥ \geq ρ$ , $\forall u \in N_{p, q}$ .

(ii) We first prove the following inequality:

${∥ u ∥}_{p, V} \leq C {∥ u ∥}_{q, V} .$

(35)

In fact, by using the H

\ddot{o}

lder inequality, one has

$\begin{matrix} \int_{K^{N}} {| \nabla u |}^{p} d μ & \leq {(\int_{K^{N}} {| \nabla u |}^{q} d μ)}^{\frac{p}{q}} \cdot {(\int_{K^{N}} 1^{\frac{q}{q - p}} d μ)}^{\frac{q - p}{q}} \\ = {(\sum_{x \in K^{N}} μ (x))}^{\frac{q - p}{q}} {(\int_{K^{N}} {| \nabla u |}^{q} d μ)}^{\frac{p}{q}} = C_{4} {(\int_{K^{N}} {| \nabla u |}^{q} d μ)}^{\frac{p}{q}}, \end{matrix}$

(36)

and

$\begin{matrix} \int_{K^{N}} V (x) {| u |}^{p} d μ & = \int_{K^{N}} V {(x)}^{\frac{q - p}{q}} \cdot V {(x)}^{\frac{p}{q}} {| u |}^{p} d μ \\ \leq {(\int_{K^{N}} V (x) {| u |}^{q} d μ)}^{\frac{p}{q}} \cdot {(\int_{K^{N}} V (x) d μ)}^{\frac{q - p}{q}} \leq {(V_{m} \sum_{x \in K^{N}} μ (x))}^{\frac{q - p}{q}} \cdot {(\int_{K^{N}} V (x) {| u |}^{q} d μ)}^{\frac{p}{q}} \\ = C_{5} {(\int_{K^{N}} V (x) {| u |}^{q} d μ)}^{\frac{p}{q}}, \end{matrix}$

(37)

where $V_{m} = {max}_{x \in K^{N}} V (x) < + \infty$ . Thus,

$\int_{K^{N}} {(| \nabla u |}^{p} + {V (x) | u |}^{p}) d μ \leq C_{6} [{(\int_{K^{N}} {| \nabla u |}^{q} d μ)}^{\frac{p}{q}} + {(\int_{K^{N}} V (x) {| u |}^{q} d μ)}^{\frac{p}{q}}] .$

(38)

From (13), we have

${(x + y)}^{r} \leq C_{r} (x^{r} + y^{r}), x, y \geq 0, r > 1,$

i.e.,

$x + y \leq C_{7} {(x^{r} + y^{r})}^{\frac{1}{r}}, x, y \geq 0, r > 1 .$

(39)

In (39), let

$x = {(\int_{K^{N}} {| \nabla u |}^{q} d μ)}^{\frac{p}{q}}, y = {(\int_{K^{N}} V (x) {| u |}^{q} d μ)}^{\frac{p}{q}}, r = \frac{q}{p} > 1,$

and we can derive that

${(\int_{K^{N}} {| \nabla u |}^{q} d μ)}^{\frac{p}{q}} + {(\int_{K^{N}} V (x) {| u |}^{q} d μ)}^{\frac{p}{q}} \leq C_{7} {(\int_{K^{N}} {| \nabla u |}^{q} d μ + \int_{K^{N}} V (x) {| u |}^{q} d μ)}^{\frac{p}{q}} .$

(40)

Substituting (40) into (38), we deduce that (35) holds. It follows from (35) that

${(1 + C) ∥ u ∥}_{q, V} \geq {∥ u ∥}_{p, V} + {∥ u ∥}_{q, V} = ∥ u ∥ \geq ρ,$

which implies that ${∥ u ∥}_{q, V} \geq C_{8}$ . Thus, we have by (13) that

$\begin{matrix} {∥ u ∥}_{p, V}^{p} & + {∥ u ∥}_{q, V}^{q} = {∥ u ∥}_{p, V}^{p} + {∥ u ∥}_{q, V}^{p} \cdot {∥ u ∥}_{q, V}^{q - p} \geq {∥ u ∥}_{p, V}^{p} + C_{8}^{q - p} {∥ u ∥}_{q, V}^{p} \\ \geq C_{9} {(∥ u ∥}_{p, V}^{p} + {∥ u ∥}_{q, V}^{p}) \geq C_{10} {(∥ u ∥}_{p, V} + {∥ u ∥}_{q, V})^{p} = C_{10} {∥ u ∥}^{p} \\ \geq C_{10} ρ^{p} . \end{matrix}$

(41)

In view of (32) and (41), we derive that

$m_{p, q} = inf_{u \in N_{p, q}} (\frac{1}{p} {∥ u ∥}_{p, V}^{p} + \frac{1}{q} {∥ u ∥}_{q, V}^{q} - \frac{1}{2 r} \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{{| u (y) |}^{r}}{d {(x, y)}^{N - α}}) {| u |}^{r} d μ)$

$= inf_{u \in N_{p, q}} [(\frac{1}{p} - \frac{1}{2 r}) {∥ u ∥}_{p, V}^{p} + (\frac{1}{q} - \frac{1}{2 r}) {∥ u ∥}_{q, V}^{q}]$

$\geq (\frac{1}{q} - \frac{1}{2 r}) inf_{u \in N_{p, q}} {(∥ u ∥}_{p, V}^{p} + {∥ u ∥}_{q, V}^{q}) \geq (\frac{1}{q} - \frac{1}{2 r}) C_{10} ρ^{p} > 0 .$

(iii) For each fixed

u \in X ∖ 0

and

\forall t \geq 0

, we set

w_{u} (t) = J_{p, q} (t u)

, namely

$w_{u} (t) = \frac{t^{p}}{p} {∥ u ∥}_{p, V}^{p} + \frac{t^{q}}{q} {∥ u ∥}_{q, V}^{q} - \frac{t^{2 r}}{2 r} \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{{| u |}^{r}}{d {(x, y)}^{N - α}}) {| u |}^{r} d μ .$

(42)

Taking the derivative of

w_{u} (t)

, we obtain

$\begin{matrix} w_{u}^{'} (t) & = t^{p - 1} {∥ u ∥}_{p, V}^{p} + t^{q - 1} {∥ u ∥}_{q, V}^{q} - t^{2 r - 1} \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{{| u |}^{r}}{d {(x, y)}^{N - α}}) {| u |}^{r} d μ \\ = t^{p - 1} [{∥ u ∥}_{p, V}^{p} - t^{q - p} (t^{2 r - q} \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{{| u |}^{r}}{d {(x, y)}^{N - α}}) {| u |}^{r} d μ - {∥ u ∥}_{q, V}^{q})] . \end{matrix}$

(43)

It is easy to see that the function

$k_{u} (t) = t^{q - p} (t^{2 r - q} \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{{| u |}^{r}}{d {(x, y)}^{N - α}}) {| u |}^{r} d μ - {∥ u ∥}_{q, V}^{q})$

(44)

is strictly increasing on $(t_{u}^{*}, + \infty)$ , and $k_{u} (t_{u}^{*}) = 0$ , where

$t_{u}^{*} = {(\frac{{∥ u ∥}_{q, V}^{q}}{\int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{{| u |}^{r}}{d {(x, y)}^{N - α}}) {| u |}^{r} d μ})}^{\frac{1}{2 r - q}} .$

It follows from (43) and (44) that there exists a unique $t_{u} > t_{u}^{*}$ such that $w_{u}^{'} (t) > 0$ on $(t_{u}^{*}, t_{u})$ and $w_{u}^{'} (t) < 0$ on $(t_{u}, + \infty)$ . Moreover, if $t \in (0, t_{u}^{*}]$ , then $w_{u}^{'} (t) > 0$ . Thus, ${max}_{t \geq 0} w_{u} (t) = w_{u} (t_{u})$ and $w_{u}^{'} (t_{u}) = 0$ , i.e., $t_{u} u \in N_{p, q}$ . □

Lemma 8.

Set maps $t : X ∖ 0 \to (0, + \infty) : u \mapsto t (u) : = t_{u}$ , and $λ : X ∖ 0 \to N_{p, q} : u \mapsto λ (u) : = t_{u} u$ . Under the conditions of Theorem 1,

(i): t and λ are continuous;
(ii): ${λ |}_{S}$ (S represents the unit sphere in X) is a homeomorphism between S and $N_{p, q}$ .

Proof.

Assume that $u_{n} \to u$ in $X ∖ 0$ as $n \to + \infty$ , and it suffices to show that $t_{u_{n}} \to t_{u}$ . Set $t_{u_{n}} = t_{n}$ ; then, $λ (u_{n}) = t_{n} u_{n}$ . Since $λ (s u) = λ (u)$ for any $s > 0$ , we may assume that $∥ u_{n} ∥ = 1$ . In the following, we will prove that there exist $δ > 0$ , $C_{S} > 0$ , such that $δ \leq t_{n} \leq C_{S}$ , $\forall n \in N$ . Indeed, $t_{n}$ is the unique root for

$w_{u_{n}}^{'} (t_{n}) = t_{n}^{p - 1} θ_{u_{n}} (t_{n}) = 0$ ,

where

$θ_{u_{n}} (t) = ∥ u_{n} ∥_{p, V}^{p} - t^{2 r - p} \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{| u_{n} |^{r}}{d {(x, y)}^{N - α}}) | u_{n} |^{r} d μ + t^{q - p} {∥ u_{n} ∥}_{q, V}^{q}$

$\geq ∥ u_{n} ∥_{p, V}^{p} - t^{2 r - p} \int_{K^{N}} (\sum_{y \in K^{N}, y \neq x} \frac{| u_{n} |^{r}}{d {(x, y)}^{N - α}}) {| u_{n} |}^{r} d μ$

$\geq ∥ u_{n} ∥_{p, V}^{p} - C t^{2 r - p} {∥ u_{n} ∥}^{2 r}$

$= ∥ u_{n} ∥_{p, V}^{p} - C t^{2 r - p} .$

Taking

δ > 0

small enough, one has

$θ_{u_{n}} (t) > 0, \forall t \in (0, δ) .$

Hence,

$w_{u_{n}}^{'} (t) = t^{p - 1} θ_{u_{n}} (t) > 0, \forall t \in (0, δ),$

and then $t_{n} > δ > 0$ . On the other hand, we will prove that there exists $C_{S} > 0$ such that $t_{n} \leq C_{S}$ . We show this by contradiction. Assume that $t_{n} \to + \infty$ , $n \to + \infty$ . Observe that

$w_{u_{n}} (t_{n}) = max_{t \geq 0} w_{u_{n}} (t) = max_{t \geq 0} J_{p, q} (t u_{n}) > 0 .$

Nevertheless, by (15), we can easily find that $w_{u_{n}} (t_{n}) = J_{p, q} (t_{n} u_{n}) \to - \infty$ as $n \to + \infty$ , which is a contradiction. Thus, we may assume that $t_{n} \to \bar{t} > 0$ after passing to the subsequence. Then, $λ (u_{n}) = t_{n} u_{n} \to \bar{t} u$ as $n \to + \infty$ . Thus, we simply need to demonstrate that $\bar{t} = t_{u}$ . Since $N_{p, q}$ is closed and $t_{n} u_{n} \in N_{p, q}$ , $\bar{t} u \in N_{p, q}$ . According to the uniqueness of $t_{u}$ , one has $\bar{t} = t_{u}$ and $λ (u_{n}) \to \bar{t} u = t_{u} u = λ (u)$ . Hence, the map t and $λ$ are continuous.

(ii) Let $u_{i} \in S$ , $λ (u_{i}) = t_{i} u_{i}$ , $i = 1, 2$ . If $λ (u_{1}) = λ (u_{2})$ , i.e., $t_{1} u_{1} = t_{2} u_{2}$ , one has $t_{1} = t_{2} > δ > 0$ after taking the norm on both sides. Thus, we have $u_{1} = u_{2}$ and ${λ |}_{S}$ is injective. For each $u \in N_{p, q}$ , let $\bar{u} = \frac{u}{∥ u ∥}$ ; then, $\bar{u} \in S$ . Owing to $u = ∥ u ∥ \bar{u}$ and $t_{\bar{u}}$ being unique, we have $t_{\bar{u}} = ∥ u ∥$ . Thus, $λ (\bar{u}) = t_{\bar{u}} \bar{u} = u = N_{p, q}$ , i.e., ${λ |}_{S}$ is surjective. Therefore, ${λ |}_{S} : S \to N_{p, q}$ is a homeomorphism. We complete the proof of Lemma 8. □

In what follows, we will prove that $c_{1}$ and $c_{2}$ are equal.

Proof.

Firstly, we prove

c_{1} = m_{p, q}

. From Lemma 7, we understand that a unique

t_{u} > 0

exists for which

$max_{t \geq 0} w_{u} (t) = max_{t \geq 0} J_{p, q} (t u) = w_{u} (t_{u}) = J_{p, q} (t_{u} u),$

where $w_{u} (t)$ is defined by (42). Thus, one has

$c_{1} = inf_{u \in X ∖ 0} max_{t \geq 0} J_{p, q} (t u) = inf_{u \in X ∖ 0} J_{p, q} (t_{u} u) = inf_{u \in N_{p, q}} J_{p, q} (u) = m_{p, q} .$

(46)

Secondly, we prove

c_{1} \geq c_{2}

. It follows from (15) that there is a large enough

t_{0} > 0

such that

w_{u} (t_{0}) < 0

for each

u \in X ∖ 0

. Define

γ_{0} : [0, 1] \to X : t \mapsto t_{0} t u

. Since

γ_{0} (0) = 0

J_{p, q} (γ_{0} (1)) < 0

, which implies that

γ_{0} \in Γ_{2}

. Thus, for each

u \in X ∖ 0

, we have

$max_{t \geq 0} J_{p, q} (t u) \geq max_{t \in [0, 1]} J_{p, q} (t_{0} t u) = max_{t \in [0, 1]} J_{p, q} (γ_{0} (t)) \geq inf_{γ \in Γ_{2}} max_{t \in [0, 1]} J_{p, q} (γ (t)),$

which implies

$c_{1} = inf_{u \in X ∖ 0} max_{t \geq 0} J_{p, q} (t u) \geq inf_{γ \in Γ_{2}} max_{t \in [0, 1]} J_{p, q} (γ (t)) = c_{2} .$

(47)

Finally, we prove

c_{2} \geq m_{p, q}

. By Lemma 7, for each

u \in X ∖ 0

, there exists a unique

t_{u} > 0

such that

t_{u} u \in N_{p, q}

. We can subsequently split X into two components, namely

X = X_{1} \cup X_{2}

, where

X_{1} = u \in X : t_{u} \geq 1

and

X_{2} = u \in X : t_{u} < 1

. We now assert that every

γ \in Γ_{2}

must cross

N_{p, q}

. Indeed, if t is small enough, we can easily find that

γ (t)

and 0 belong to the component

X_{1}

. Consider the function

w_{γ (1)} (t) = J_{p, q} (t γ (1))

\forall t \in [0, + \infty)

, and we have

w_{γ (1)} (0) = 0

and

w_{γ (1)} (1) < 0

. From (12), it holds that

$w_{γ (1)} (t) \geq \frac{t^{p}}{p} {∥ γ (1) ∥}_{p, V}^{p} + \frac{t^{q}}{q} {∥ γ (1) ∥}_{q, V}^{q} - C_{11} \frac{t^{2 r}}{2 r} {∥ γ (1) ∥}^{2 r},$

which suggests that $w_{γ (1)} (t) > 0$ for small enough $t > 0$ . Thus, there exist $t_{γ (1)} \in (0, 1)$ such that ${max}_{t \geq 0} w_{γ (1)} (t) = J_{p, q} (t_{γ (1)} γ (1))$ . Thus, $t_{γ (1)} < 1$ and $γ (1) \in X_{2}$ . Furthermore, according to Lemma 8, the map $u \mapsto t_{u}$ is continuous, which means that every $γ \in Γ_{2}$ has to cross $N_{p, q}$ . Hence, $\forall γ \in Γ_{2}$ , and there exists $t_{0} \in (0, 1)$ such that $γ (t_{0}) \in N_{p, q}$ . Thus, we have

$inf_{u \in N_{p, q}} J_{p, q} (u) \leq J_{p, q} (γ (t_{0})) \leq max_{t \in [0, 1]} J_{p, q} (γ (t)),$

which implies that

$m_{p, q} = inf_{u \in N_{p, q}} J_{p, q} (u) \leq inf_{γ \in Γ_{2}} max_{t \in [0, 1]} J_{p, q} (γ (t)) = c_{2} .$

(48)

Combining (46), (47), (48) and Lemma 7, we find that $c_{1} = c_{2} = m_{p, q} > 0$ holds, which gives the desired result. □

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Dandan Yang www.mdpi.com

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