V. F. Vil'danova, F. Kh. Mukminov, “Entropy solution for an equation with measure-valued potential in a hyperbolic space”, Sb. Math., 214:11 (2023), 1534

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$\begin{equation*} -\operatorname{div}_g(a(x,u,\nabla u))+b_0(x,u)+ b_1(x,u)\mu=f, \qquad f\in L_1(\mathbb{H}^n), \quad n\geqslant2, \end{equation*} \notag$

$\begin{equation*} -\operatorname{div}(a(x,\nabla u))=f(x,u), \qquad \sup_{|u|<c}|f(x,u)|\in L_{1,\mathrm{loc}}(\Omega), \quad c>0. \end{equation*} \notag$

$\begin{equation*} -\Delta u+\mu g(u)=\sigma, \qquad u|_{\partial\Omega}=0. \end{equation*} \notag$

$\begin{equation} \mathcal{L}u= -\operatorname{div}(a(x,u,\nabla u)+F(u))+\mu u=f, \qquad x\in\Omega, \quad u|_{\partial\Omega}=0, \end{equation} \tag{1.1}$

$\begin{equation*} -\operatorname{div}(a(x,\nabla u))+a_0(x,u)=\sigma \end{equation*} \notag$

$\begin{equation*} -\operatorname{div}(a(x,u,\nabla u))+ b(x,u,\nabla u) + |u|^{ p(x)-1} = \sigma \end{equation*} \notag$

$\begin{equation*} -\operatorname{div}a(x,u,\nabla u)+M'(x,u)+b(x,u,\nabla u)=\sigma,\qquad u|_{\partial \Omega}=0, \end{equation*} \notag$

$\begin{equation} g_{ij}(x)=\frac{4}{(1-|x|^2)^2}\delta_{ij}, \qquad x\in B_1, \quad i,j=1,\dots, n. \end{equation} \tag{2.1}$

$\begin{equation*} \partial B_1\equiv\partial_{\infty}\mathbb{H}^n=\{\infty\}. \end{equation*} \notag$

$\begin{equation} \rho(x)=\int_0^{|x|}\frac{2}{1-s^2}\,ds=\log \frac{1+|x|}{1-|x|}, \qquad x\in B_1\equiv \mathbb{H}^n, \end{equation} \tag{2.2}$

$\begin{equation} |x|=\operatorname{tanh}\biggl(\frac{\rho(x)}{2}\biggr). \end{equation} \tag{2.3}$

$\begin{equation} d\nu=\sqrt{g}\,dx_1\,dx_2\cdots dx_n=\frac{2^n}{(1-|x|^2)^n}\,dx, \end{equation} \tag{2.4}$

$\begin{equation} (\beta,\xi)_g\equiv(\beta,\xi)_{g,x}=\sum_{i,j=1}^n g_{ij}(x)\beta^i\xi^j; \end{equation} \tag{2.5}$

$\begin{equation*} |\beta|_g=\sqrt{(\beta,\beta)_g} \end{equation*} \notag$

$\begin{equation} (\nabla_g u)^i=\sum_{j=1}^ng^{ij}\frac{\partial u}{\partial x_j}, \qquad i=1,\dots ,n. \end{equation} \tag{2.6}$

$\begin{equation*} (df,X)=(\nabla f, X)_g=\sum_{i=1}^nX^i\frac{\partial f}{\partial x^i}. \end{equation*} \notag$

$\begin{equation*} \operatorname{div}_gX=\sum_{i=1}^n\frac{1}{\sqrt{g}}\,\frac{\partial }{\partial x^i}\bigl(X^i\sqrt{g}\bigr). \end{equation*} \notag$

Definition 1. Let $M(x,z)\colon \mathbb{H}^n \times\mathbb{R}\to\mathbb{R}_+$ be a function satisfying the following conditions:

1) $M(x,\,\cdot\,)$ is an $N$ -function in $z\in\mathbb{R}$ , that is, it is (downward) convex, nondecreasing, even and continuous, $M(x,0)=0$ for $\nu$ -almost all $x\in \mathbb{H}^n$ and $\inf_{x\in\mathbb{H}^n} M(x,z)>0$ for all $z\neq 0$ ; furthermore,

$\begin{equation} \lim_{z\to 0}\sup_{x\in\mathbb{H}^n} \frac{{M}(x,z)}{z}=0 \end{equation} \tag{2.7}$

and

$\begin{equation} \lim_{z\to \infty}\inf_{x\in\mathbb{H}^n}\frac{{M}(x,z)}{z}=\infty; \end{equation} \tag{2.8}$

it is obvious that the function

$M(x,z)/z$ is nondecreasing;

2) $M(\cdot,z)$ is a measurable function of $x\in \mathbb{H}^n$ for each $z\in\mathbb{R}$ .

Then $M(x,z)$ is called a Musielak-Orlicz function.

$\begin{equation*} \overline{M}(x,z)=\sup_{y\geqslant 0} ( yz-M(x,y)). \end{equation*} \notag$

$\begin{equation} |zy| \leqslant M(x,z)+ \overline{M}(x,y), \qquad z,y \in \mathbb{R}, \quad x \in \mathbb{H}^n. \end{equation} \tag{2.9}$

$\begin{equation} {M}(x,z),\overline{M}(x,z)\in L_{1,\mathrm{loc}}(\mathbb{H}^n), \qquad z>0. \end{equation} \tag{2.10}$

$\begin{equation*} M(x,2z)\leqslant CM(x,z)+G(x) \quad \forall\, z\in \mathbb{R}, \quad x\in \mathbb{H}^n. \end{equation*} \notag$

$\begin{equation*} \|u\|_{M}= \inf\{k>0\colon\varrho(k^{-1}u)\leqslant 1\}\quad\text{and} \quad \varrho(u)= \int_{\mathbb{H}^n}M(x,u)\,d\nu. \end{equation*} \notag$

$\begin{equation} \biggl|\int_{\mathbb{H}^n} u(x)v(x)\,d\nu\biggr| \leqslant 2\|u\|_{M}\|v\|_{\overline{M}}. \end{equation} \tag{2.11}$

$\begin{equation*} \|u\|_{M,1}=\|(|du|_g)\|_{M}=\|u\|_{V}. \end{equation*} \notag$

$\begin{equation*} \|u\|_{p^*,\mathbb{H}^n}\leqslant C\|\nabla_gu\|_{p,\mathbb{H}^n}\quad\text{and} \quad u\in C^\infty_0(\mathbb{H}^n), \quad p^*= \frac{np}{n-p}. \end{equation*} \notag$

$\begin{equation*} \mathfrak{B}_\sigma=\{x\in\mathbb{H}^n \mid \rho(x)<\sigma\} \end{equation*} \notag$

$\begin{equation*} B_r=\mathfrak{B}_{\ln((1+r)/(1-r))}. \end{equation*} \notag$

$\begin{equation} \|u\|_{W_p^1(\mathfrak{B}_\sigma)}\leqslant h(\sigma)\|u\|_{V}, \qquad \sigma>0, \end{equation} \tag{2.12}$

$\begin{equation*} \mathcal{B}u=b_0(x,u)+ b_1(x,u)\mu, \end{equation*} \notag$

$\begin{equation*} \langle\mathcal{B}u,v\rangle=\int_{\mathbb{H}^n } b_0(x,u)v\,d\nu + \int_{\mathbb{H}^n}b_1(x,u)v\,d\mu. \end{equation*} \notag$

$\begin{equation*} |B_r(x)|_\mu:=\int_{B_r(x)}\,d|\mu|\leqslant cr^{n(1-1/s)}, \qquad r>0, \quad x\in\Omega. \end{equation*} \notag$

$\begin{equation*} \int_{B_r(x)}\,d|\mu|\leqslant cr^\theta. \end{equation*} \notag$

$\begin{equation} \mu\in\mathbb{M}_s(B_r)\quad \forall\,r\in(0,1). \end{equation} \tag{2.13}$

$\begin{equation} -\operatorname{div}_g(a(x,u,d u))+\mathcal{B} u=f, \qquad f\in L_1(\mathbb{H}^n). \end{equation} \tag{2.14}$

$\begin{equation} \begin{gathered} \, \overline{M}(x,|a(x,r,y)|_g)\leqslant\operatorname{g}(r) \bigl(G(x)+M(x,|y|_g)\bigr), \qquad r\in\mathbb{R}, \quad y\in T_x^{*}\mathbb{H}^n; \end{gathered} \end{equation} \tag{2.15}$

$\begin{equation} (a(x,r,y),y)\geqslant c_0 M(x,|y|_g)-G(x), \qquad r\in\mathbb{R}, \quad c_0>0; \end{equation} \tag{2.16}$

$\begin{equation} (a(x,r,y)-a(x,r,z),y-z)>0, \qquad y\ne z, \ \ y,z\in T_x^{*}\mathbb{H}^n, \ \ r\in\mathbb{R}, \ \ x\in\mathbb{H}^n. \end{equation} \tag{2.17}$

$\begin{equation} |b_i(x,s)|\leqslant \operatorname{g}(r)\widetilde{G}_i(x), \qquad |s|\leqslant r, \quad\rho(x)\leqslant r, \quad i=0,1, \end{equation} \tag{2.18}$

$\begin{equation} b_i(x,r)r\geqslant 0, \qquad i=0,1. \end{equation} \tag{2.19}$

$\begin{equation} |b_1 (x,s)|>\widetilde{g}(r), \qquad s\geqslant r, \quad x\in \mathbb{H}^n. \end{equation} \tag{2.20}$

$\begin{equation*} T_k(r)= \begin{cases} k & \text{for }r>k, \\ r & \text{for }|r|\leqslant k, \\ -k & \text{for }r<-k. \end{cases} \end{equation*} \notag$

Definition 2. An entropy solution of the Dirichlet problem for equation (2.14) is a function $u\in \mathring{\mathcal{T}}_{M}^1(\mathbb{H}^n)$ such that for all $k>0$ and $\xi\in C_0^1(\mathbb{H}^n)$ the inequality

$\begin{equation} \int_{\mathbb{H}^n} \bigl((a(x,u, du),dT_k(u-\xi))-fT_k(u-\xi)\bigr)\,d\nu +\langle\mathcal{B}u,T_k(u-\xi)\rangle\leqslant 0 \end{equation} \tag{2.21}$

is well defined and valid, that is, all terms in it are finite.

Theorem 1. Assume that conditions (2.13) and (2.15)–(2.20) are met. Then the Dirichlet problem for equation (2.14) has an entropy solution.

Lemma 1. There exists $p\in(1,n)$ such that

$\begin{equation} \|u\|_{W_p^1(\mathfrak{B}_r)}\leqslant h(r)\|u\|_{V} \qquad \forall u\in V, \quad r>0, \end{equation} \tag{3.1}$

where

$h(r)$ is an increasing nonnegative function.

Proof. It follows from the

$\Delta_2$ -conditions on the functions

$M$ and

$\overline{M}$ (see the proof of Corollary 3.6.7 and Definition 2.1.2 in [11]) that there exist

$p\in(1,n)$ and

$\beta\in(0,1)$ such that for all

$\lambda>1$ the inequality

$\begin{equation} \beta\lambda^pM(x,r)<M(x,\lambda r) \end{equation} \tag{3.2}$

holds for all

$r>0$ and almost all

$x\in\mathbb{H}^n$ . It is obvious that

$\begin{equation*} \inf M(x,1)\int_{\mathfrak{B}_R} |du|_g^p\,d\nu\leqslant \int_{\mathfrak{B}_R} M(x,1)\,d\nu +\int_{\mathfrak{B}_R, |du|_g>1} M(x,1) |du|_g^p\,d\nu. \end{equation*} \notag$

Let

$\|u\|_{V}\leqslant1$ . Then by (3.2) we have

$\begin{equation} \beta\int_{\mathfrak{B}_R, |du|_g>1} M(x,1)|du|_g^p\,d\nu \leqslant\int_{\mathbb{H}^n} M(x,|du|_g)\,d\nu\leqslant1. \end{equation} \tag{3.3}$

Taking (2.10) into account we see that the integral

$\displaystyle\int_{\mathfrak{B}_R} |du|_g^p\,d\nu$ is bounded in the ball

$\|u\|_{V}\leqslant1$ . This yields the inequality

$\begin{equation} \||du|_g\|_{p,\mathfrak{B}_R}\leqslant C(R)\| u\|_{V}. \end{equation} \tag{3.4}$

Hence we have a continuous embedding

$\begin{equation*} V\hookrightarrow W^1_{p}({\mathfrak{B}_R}) \end{equation*} \notag$

and (3.1) holds. In fact, if the embedding operator is unbounded on

$C^\infty_0(\mathbb{H}^n)$ , then there exists a sequence of smooth functions

$v^k$ such that

$\begin{equation*} \|v^k\|_{W^1_{p}(\mathfrak{B}_R)}\geqslant k\| v^k\|_{V}. \end{equation*} \notag$

Multiplying both sides by an appropriate scalar we reduce this inequality to

$\begin{equation} 1\geqslant k\| v^k\|_{V}, \end{equation} \tag{3.5}$

where

$\| v^k\|_{W^1_{p}({\mathfrak{B}_R})}=1$ . By (3.5) we have

$\begin{equation*} \| v^k\|_{V}\to0. \end{equation*} \notag$

By Kondrashov’s theorem the

$v^k$ converge strongly in

$L_{p}(\mathfrak{B}_R)$ . Taking (3.4) into account we see that

$v^k\to C\ne0$ in the space

$W^1_{p}({\mathfrak{B}_R})$ . This contradicts inequality (3.5), which implies that

$v^k\to 0$ in

$V$ .

The proof is complete.

Lemma 2. Let $u(x)$ be a measurable function in $\mathbb{H}^n$ . Then $\{k\colon \operatorname{meas} \{x\in\mathbb{H}^n$ : $|u(x)|= k\}>0\}$ is finite or countable.

Proof. We select

$k_i$ such that

$\operatorname{meas}\{x\in\mathfrak{B}_{r}\colon |u(x)|= k_i\}>1/N$ . These sets are disjoint, so that

$\begin{equation*} \operatorname{meas} \{x\in\mathfrak{B}_{r}\colon |u(x)|= k_1\}+\operatorname{meas} \{x\in\mathfrak{B}_{r}\colon |u(x)|= k_2\}+\dotsb \leqslant \operatorname{meas} \mathfrak{B}_{r}. \end{equation*} \notag$

Hence there can be at most

$N \operatorname{meas} \mathfrak{B}_{r}$ such sets. Then the set

$\{k \colon \operatorname{meas}\{{x\in\mathfrak{B}_{r}}$ :

$|u(x)|= k\}>0\}$ is finite or countable. Now the required result follows easily.

Lemma 3. Let $v$ be a function such that $T_k(v)\in V$ for all $k>k_0$ and

$\begin{equation} \|T_k(v)\|_V^p\leqslant C k. \end{equation} \tag{3.6}$

Then

$\begin{equation} \operatorname{meas}\{\mathfrak{B}_{r} \colon |v|\geqslant k\}\leqslant \frac{C_1(r)}{k^{p^*(1-p^{-1})}}\quad\textit{for } k>k_0, \quad p^*=\frac{np}{n-p}. \end{equation} \tag{3.7}$

Proof. Using Sobolev’s inequality and (3.1) we obtain

$\begin{equation*} \|T_k(v)\|_{p^*,\mathfrak{B}_{r}}\leqslant C(r)\|T_k(v)\|_{W_p^1(\mathfrak{B}_{r})}\leqslant C(r)h(r)\|T_k(v)\|_V. \end{equation*} \notag$

For

$k_1\in(0,k]$ it is obvious that

$\begin{equation*} \operatorname{meas}\{\mathfrak{B}_{r} \colon |v|\geqslant k_1\} \leqslant \frac{\|T_k(v)\|_{p^*,\mathfrak{B}_{r}}^{p^*}}{k_1^{p^*}}\leqslant C_1(r) \frac{\|T_k(v)\|_V^{p^*}}{k_1^{p^*}}\leqslant C_1(r) \frac{Ck^{p^*/p}}{k_1^{p^*}}. \end{equation*} \notag$

Hence, setting

$k_1=k$ we obtain (3.7).

The proof is complete.

Lemma 4. Let $Q\subset \mathbb{H}^n$ , let $\{v^m\}_{m\in \mathbb{N}}$ be a bounded sequence in $L_M(Q)$ , let ${v\in L_M(Q)}$ , and assume that

$\begin{equation*} v^m\to v \quad\textit{$\nu$-a.e. in } Q, \qquad m\to\infty. \end{equation*} \notag$

Then

$\begin{equation*} v^m\rightharpoonup v \quad\textit{weakly in } L_M(Q), \qquad m\to\infty. \end{equation*} \notag$

Remark 1. In what follows, to avoid cumbersome reasoning, in place of statements of the form “from a sequence $u^m$ we can extract a subsequence converging $\nu$ -almost everywhere in $\mathbb{H}^n$ as $m\to \infty$ ” we will simply write “a sequence $u^m$ contains a subsequence converging $\nu$ -almost everywhere in $\mathbb{H}^n$ as $m\to \infty$ ”. We will also write “converges weakly along a subsequence” and so on, omitting a subscript indicating the subsequence.

Lemma 5. Let $v^j$ , $j\in \mathbb{N}$ , and $v$ be functions in $L_M(Q)$ such that

$\begin{equation*} v^j\to v \quad\textit{$\nu$-a.e. in } Q, \qquad j\to\infty, \end{equation*} \notag$

and

$\begin{equation*} M(x,v^j)\leqslant h\in L_1(Q), \qquad j\in \mathbb{N}. \end{equation*} \notag$

Then

$\begin{equation*} v^j\to v \quad\textit{strongly in } L_M(Q), \qquad j\to\infty. \end{equation*} \notag$

Lemma 6. Let $x\in \mathbb{H}^n$ be a point such that the function $a(x,\cdot\,{,}\,\cdot\,)$ is continuous for $v\in\mathbb{R}$ and $z\in T^*_x\mathbb{H}^n$ and the monotonicity condition (2.17) is met. Let $v_m\in\mathbb{R}$ and $z_m\in T^*_x\mathbb{H}^n$ be sequences such that $v_m\to v$ and

$\begin{equation*} \lim_{m\to\infty}\bigl((a(x,v_m,z_m)-a(x,v_m, z)), z_m- z\bigr)=0. \end{equation*} \notag$

Then

$z_m\to z$ .

Proof. Let

$s_0=|v|+|z|_g+1$ and

$\begin{equation*} \Lambda(x,r,y,z)=\bigl(a(x,r,y)-a(x,r,z),y-z), \qquad y,z\in T_x^{*}\mathbb{H}^n, \quad r\in\mathbb{R}. \end{equation*} \notag$

It follows from the hypotheses that

$\begin{equation} \Lambda(x,v_m,z_m,z)\to0, \qquad m\to \infty. \end{equation} \tag{3.8}$

The sequence

$v_{m}$ tends to

$v$ and

$|v|<s_0$ , so we can assume that

$|v_{m}|< s_0$ . The following inequality is easy to establish (see [14], formula (2.1)):

$\begin{equation} \Lambda(x,r,y,z)\geqslant|y-z|_g\Lambda\biggl(x,r,z+\frac{y-z}{|y-z|_g},z\biggr), \qquad |y-z|_g\geqslant2. \end{equation} \tag{3.9}$

Since at the point

$x$ the function

$a$ is continuous with respect to the other variables, the function

$\Lambda(x,r,y,z)$ is continuous in

$r,y$ and

$z$ . The function

$\Lambda(x,r,z+e,z)$ is continuous and positive and attains a positive minimum

$c(x)>0$ on the compact set

$|e|_g=1,|r|\leqslant s_0, |z|_g\leqslant s_0$ . Then from (3.9) we obtain

$\begin{equation} \Lambda(x,r,y,z)\geqslant|y-z|_gc(x), \qquad |y-z|_g\geqslant2, \quad|r|\leqslant s_0, \quad |z|_g\leqslant s_0. \end{equation} \tag{3.10}$

We claim that the sequence

$z_{m}$ is bounded. Otherwise a subsequence

$z_{m_k}\to \infty$ can be extracted and

$\Lambda(x,v_{m_k},z_{m_k},z)\geqslant|z_{m_k}-z|_gc(x)\to\infty$ by (3.10), which contradicts (3.8).

Thus, the sequence $z_{m}$ is bounded. We can extract from it a convergent subsequence $z_{m_k}\to z_0$ . Taking the limit we see that

$\begin{equation*} 0=\lim_{k\to\infty}\Lambda(x,v_{m_k},z_{m_k},z)= (a(t,x,v,z_0)-a(t,x,v, z))\cdot( z_0- z)=0. \end{equation*} \notag$

This is possible only for

$z_0= z$ . Then the whole sequence has the same limit. The proof is complete.

Lemma 7. Assume that conditions (2.15)–(2.17) are met in $\mathbb{H}^n$ and let $w^j\in V$ be a sequence such that

$\begin{equation} |w^j|\leqslant k \quad\textit{$\nu$-a.e. in } \mathbb{H}^n, \end{equation} \tag{3.11}$

$\begin{equation} dw^j\rightharpoonup dw \quad \textit{in } \mathbf{L}_M(\mathbb{H}^n), \qquad j\to \infty, \end{equation} \tag{3.12}$

$\begin{equation} w^j\to w \quad\textit{$\nu$-a.e. in } \mathbb{H}^n, \qquad j\to\infty, \end{equation} \tag{3.13}$

and

$\begin{equation} \lim_{j\to\infty}\int_{\mathfrak{B}_R}(a(x,w^j,dw^j)-a(x,w^j,d w),d(w^j-w))\,d\nu=0 \quad\forall\, R>0. \end{equation} \tag{3.14}$

Then

$\begin{equation} dw^j\to dw \quad \textit{strongly in } \mathbf{L}_{M,\mathrm{loc}}(\mathbb{H}^n), \qquad j\to \infty, \end{equation} \tag{3.15}$

along a subsequence.

Proof. Set

$\begin{equation*} q^j(x)=\Lambda(x,w^j,dw^j,dw), \qquad x\in \mathbb{H}^n, \quad j\in \mathbb{N}. \end{equation*} \notag$

By (2.17) the functions

$\{q^j(x)\}_{j\in \mathbb{N}}$ are nonnegative, and by (3.14)

$q^j\,{\to}\, 0$ in

$L_1(\mathfrak{B}_R)$ as

$j\to\infty$ . Hence

$\begin{equation*} q^j(x)\to 0 \quad \text{$\nu$-a.e. in } \mathfrak{B}_R \end{equation*} \notag$

along some subsequence. Using a diagonal procedure we can extract a subsequence such that

$\begin{equation} q^j(x)\to 0 \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad j\to \infty. \end{equation} \tag{3.16}$

From (3.12) we obtain

$\begin{equation} \|dw^j\|_{M}\leqslant C_3, \qquad j\in \mathbb{N}. \end{equation} \tag{3.17}$

Let $Q$ be the full-measure subset of points at which we have (3.16) and (3.13), inequalities (2.15)–(2.17) hold and

$\begin{equation*} |dw(x)|_g<\infty. \end{equation*} \notag$

Using Lemma 6 for $v_j= w^j$ , $z_j=d w^j$ and $z=dw$ we obtain the convergence

$\begin{equation} dw^j\to dw \end{equation} \tag{3.18}$

in the whole of

$Q$ .

It follows from (3.13), since $a(x,r,y)$ is continuous in $r$ and $y$ , that

$\begin{equation*} a(x,w^j,dw^j)\to a(x,w,dw) \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad j\to\infty. \end{equation*} \notag$

Furthermore, from (2.15) and (3.11) we obtain

$\begin{equation*} \overline{M}(x,|a(x,w^j,dw)|_g)\leqslant \operatorname{g}(k)( G(x))+M(x,|dw|_g))\in L_1(\mathbb{H}^n), \qquad j\in \mathbb{N}. \end{equation*} \notag$

Lemma 5 yields the convergence

$\begin{equation} a(x,w^j,dw)\to a(x,w,dw) \quad\text{strongly in } \mathbf L_{\overline{M}}(\mathbb{H}^n), \qquad j\to\infty. \end{equation} \tag{3.19}$

By (3.17) we have the weak convergence

$\begin{equation} a(x,w^j,dw^j)\rightharpoonup a(x,w,dw) \quad\text{weakly in } \mathbf L_{\overline{M}}(\mathbb{H}^n), \qquad j\to\infty. \end{equation} \tag{3.20}$

We introduce the notation $\lambda^j=(a(x,w^j,dw^j),dw^j)$ and $\lambda= (a(x,w,d w),dw)$ . Then it is easy to see that

$\begin{equation*} q^j=\lambda^j-(a(x,w^j,dw),dw^j)-(a(x,w^j,dw^j),dw)+(a(x,w^j,dw),dw). \end{equation*} \notag$

Using (3.12), (3.14) and (3.19) we can show that

$\begin{equation} \int_{\mathfrak{B}_R}\lambda^jd\nu\to \int_{\mathfrak{B}_R}\lambda \,d\nu, \qquad j\to\infty. \end{equation} \tag{3.21}$

It follows from (3.13) and (3.18) that

$\begin{equation} \lambda^j\to \lambda \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad j\to\infty. \end{equation} \tag{3.22}$

Using (2.16) and the

$\Delta_2$ -condition we obtain

$\begin{equation*} \begin{aligned} \, \lambda^j+\lambda &=(a(x,w^j,dw^j),dw^j)+ (a(x,w,d w),dw) \\ &\geqslant c_0 M(x,|d w^j|_g)+c_0 M(x,|dw|_g)-2G(x) \\ &\geqslant c_0 M\biggl(x,\frac{1}{2}| {d(w^j-w)}|_g\biggr)-2G(x) \\ &\geqslant \frac{c_0}{C} M(x,|d(w^j-w)|_g)-2G(x). \end{aligned} \end{equation*} \notag$

From Fatou’s lemma, (3.15) and (3.22) we deduce the inequality

$\begin{equation*} \int_{\mathfrak{B}_R}2\lambda \,d\nu\leqslant \lim_{j\to\infty}\inf \int_{\mathfrak{B}_R} \biggl(\lambda^j+\lambda-\frac{c_0}{C} M(x,|d(w^j-w)|_g)\biggr)d\nu. \end{equation*} \notag$

Taking (3.21) into account we conclude that

$\begin{equation*} 0\leqslant -\lim_{j\to\infty}\sup\int_{\mathfrak{B}_R} M(x,|d(w^j-w^j)|_g)\,d\nu. \end{equation*} \notag$

Hence

$\begin{equation*} 0\leqslant \lim_{j\to\infty}\inf\int_{\mathfrak{B}_R} M(x,|d(w^j-w)|_g)\,d\nu \leqslant\lim_{j\to\infty}\sup\int_{\mathfrak{B}_R} M(x,|d(w^j-w)|_g)\,d\nu\leqslant 0, \end{equation*} \notag$

so that

$\begin{equation*} \int_{\mathfrak{B}_R} M(x,|d(w^j-w)|_g)\,d\nu\to 0, \qquad j\to\infty, \end{equation*} \notag$

which yields the convergence (3.15).

Lemma 7 is proved.

Lemma 8. Let $\{v^j\}_{j\in \mathbb{N}}$ be a bounded sequence in $L_{\infty}(\mathbb{H}^n)$ , let $v\in L_{\infty}(\mathbb{H}^n)$ , and assume that

$\begin{equation*} v^j\to v \quad\textit{$\nu$-a.e. in } \mathbb{H}^n, \qquad j\to\infty. \end{equation*} \notag$

Then

$\begin{equation*} v^j\stackrel{*}\rightharpoonup v \quad\textit{weakly in } L_{\infty}(\mathbb{H}^n), \qquad j\to\infty. \end{equation*} \notag$

In addition, if $h^j$ , $j\in \mathbb{N}$ , and $h$ are functions in $L_M(\mathbb{H}^n)$ such that

$\begin{equation*} h^j \to h \quad\textit{strongly in } L_M(\mathbb{H}^n), \qquad j\to\infty, \end{equation*} \notag$

then

$\begin{equation*} v^jh^j\to v h \quad\textit{strongly in } L_M(\mathbb{H}^n), \qquad j\to\infty. \end{equation*} \notag$

Lemma 9. Let ( $S,\sum,\mu$ ) be a space with positive measure and $\{f_n\}$ be a nondecreasing sequence of nonnegative measurable, but not necessarily integrable functions. Then

$\begin{equation*} \lim_{n\to\infty}\int_{S}f_n(x)\, d\mu=\int_{S}\sup_{n}f_n(x)\,d\mu. \end{equation*} \notag$

$\begin{equation} \langle \widetilde{A}(u,v),v-w\rangle\geqslant\langle \widetilde{A}(u,w),v-w\rangle, \qquad u,v,w\in W. \end{equation} \tag{4.1}$

$\begin{equation} \langle \widetilde{A}(u,u),u\rangle\geqslant \alpha\|u\|_{W}^p-C(\|u\|_{W}+1),\qquad u\in W,\quad\alpha>0, \end{equation} \tag{4.2}$

$\begin{equation} \langle 2{K}(u),u\rangle+\alpha\|u\|_{W}^p\geqslant0, \qquad u\in W, \end{equation} \tag{4.3}$

Theorem 2. Assume that conditions (I) and (4.1)–(4.3) are met and ${K}$ is a weakly continuous operator. Also assume that for each sequence $u_j\rightharpoonup u$ converging weakly in $W$ the condition

$\begin{equation} \langle {K} (u_j),u_j-u\rangle\to0 \end{equation} \tag{4.4}$

is satisfied. Then

$A$ is a surjective operator.

Proof. That the equation

$A(u)=f$ is solvable for

$f\in W^*$ follows from the fact that

$A(u)= \widetilde{A}(u,u)+{K}(u)$ is a pseudomonotone coercive operator (see [16], Ch. 2, Theorem 2.7).

Verifying coercivity. From conditions (II) we obtain

$\begin{equation*} \begin{aligned} \, \langle {A}(u),u\rangle &=\langle \widetilde{A}(u,u)+{K}(u),u\rangle\geqslant \langle {K}(u),u\rangle+ \alpha\|u\|_{W}^p-C(\|u\|_{W}+1) \\ &\geqslant \frac{\alpha}{2} \|u\|_{W}^p-C(\|u\|_{W}+1). \end{aligned} \end{equation*} \notag$

Hence

$A$ is coercive.

Verifying pseudomonotonicity. Let $u_j\rightharpoonup u$ weakly in $W$ , and assume that

$\begin{equation} \limsup \langle A(u_j),u_j-u\rangle\leqslant0. \end{equation} \tag{4.5}$

We must prove that

$\begin{equation} \liminf\langle A(u_j),u_j-v\rangle\geqslant\langle A(u),u-v\rangle \end{equation} \tag{4.6}$

for all

$v\in W$ .

It follows from (4.4) and (4.5) that

$\begin{equation} \limsup \langle \widetilde{A}(u_j,u_j),u_j-u\rangle\leqslant0. \end{equation} \tag{4.7}$

By (4.1) we have

$\begin{equation} \langle \widetilde{A}(u_j,v),v-w\rangle\geqslant\langle\widetilde{A}(u_j,w),v-w\rangle \quad \forall\, v,w\in W. \end{equation} \tag{4.8}$

By condition (I) the sequence

$\widetilde{A}(u_j,v)$ converges strongly in

$W^*$ , and therefore

$\begin{equation} \langle \widetilde{A}(u_j,v),u_j-u\rangle\to0 \quad \forall\, v\in W. \end{equation} \tag{4.9}$

Hence by (4.8)

$\begin{equation*} \langle \widetilde{A}(u_j,u_j),u_j-u\rangle\geqslant\langle \widetilde{A}(u_j,u),u_j-u\rangle\to0. \end{equation*} \notag$

From this, taking (4.7) into account we obtain

$\begin{equation} \langle \widetilde{A}(u_j,u_j),u_j-u\rangle\to0. \end{equation} \tag{4.10}$

We substitute

$v=u_j$ and

$w=(1-\theta)u+\theta z$ ,

$\theta\in(0,1)$ , into (4.8). Then we have

$\begin{equation*} \begin{aligned} \, &\theta\langle \widetilde{A}(u_j,u_j),u-z\rangle \\ &\qquad\geqslant-\langle \widetilde{A}(u_j,u_j),u_j-u\rangle +\langle \widetilde{A}(u_j,w),u_j-u\rangle+\theta\langle \widetilde{A}(u_j,w),u-z\rangle. \end{aligned} \end{equation*} \notag$

Using (4.10) and (4.9) we derive the inequality

$\begin{equation*} \theta\liminf\langle \widetilde{A}(u_j,u_j),u-z\rangle\geqslant\theta\langle \widetilde{A}(u,w),u-z\rangle. \end{equation*} \notag$

Dividing by

$\theta$ and adding the result to (4.10) yields

$\begin{equation*} \liminf\langle \widetilde{A}(u_j,u_j),u_j-z\rangle\geqslant\langle \widetilde{A}(u,(1-\theta)u+\theta z),u-z\rangle. \end{equation*} \notag$

Letting

$\theta$ tend to zero and using the fact that

$\widetilde{A}$ is semicontinuous we obtain

$\begin{equation*} \liminf\langle \widetilde{A}(u_j,u_j),u_j-z\rangle\geqslant\langle \widetilde{A}(u,u),u-z\rangle. \end{equation*} \notag$

Using the fact that the operator

${K}$ is weakly continuous and (4.4) we can write

$\begin{equation*} \langle \widetilde{K}(u),u-z\rangle=\lim\langle \widetilde{K}(u_j),u-z\rangle=\lim\langle \widetilde{K}(u_j),u_j-z\rangle. \end{equation*} \notag$

From this we deduce (4.6).

Theorem 2 is proved.

$\begin{equation*} \langle \widetilde{A}(u,v),w\rangle=\int_{\mathbb{H}^n}(a^m(x,u,dv), d w) \,d\nu, \qquad u,v,w\in V. \end{equation*} \notag$

$\begin{equation*} \langle \widetilde{A}(u,u),u\rangle\geqslant c_0\varrho(|du|_g)-\int_{\mathbb{H}^n}G(x)\,d\nu. \end{equation*} \notag$

$\begin{equation} \begin{aligned} \, \notag \varrho(|du|_g) &=\int_{\mathbb{H}^n} M(x, |du|_g)\,d\nu =\int_{\mathbb{H}^n} M\biggl(x, k\frac{|du|_g}{k}\biggr)\,d\nu \\ &\geqslant\int_{\mathbb{H}^n} \beta k^p M\biggl(x,\frac{|du|_g}{k} \biggr)\,d\nu >\beta k^p = \widetilde{\beta}\|u\|_{V}^p. \end{aligned} \end{equation} \tag{4.11}$

$\begin{equation} \langle 2{K}(u),u\rangle+c_0\varrho(|du|_g)\geqslant0, \qquad u\in V, \end{equation} \tag{4.12}$

$\begin{equation} \langle K(u),v\rangle=\int_{\Omega}b(x,u)v(x)\,d\mu. \end{equation} \tag{4.13}$

$\begin{equation*} \int_{B_r(x)}\,d|\mu|\leqslant cr^\theta. \end{equation*} \notag$

$\begin{equation*} f\in L_q(\Omega\cap\{x^1=\dots =x^k=0\})\quad\text{and} \quad x'=(0,\dots ,0,x^{k+1},\dots ,x^{n}), \end{equation*} \notag$

$\begin{equation*} \int_{B_r(x_0)}|f(x)|\,dx'\leqslant\|f\|_q\biggl(\int_{B_r(x_0)\cap\{x^1=\dots =x^k=0\}}\,dx'\biggr)^{1-1/q} \leqslant cr^{(n-k)(1-1/q)}, \end{equation*} \notag$

$\begin{equation} W_p^1(\Omega)\hookrightarrow L_{q,\mu}(\Omega). \end{equation} \tag{4.14}$

$\begin{equation*} W_p^1(\Omega)\hookrightarrow L_{q_0}(\Omega) \end{equation*} \notag$

$\begin{equation} |b(x,r)|^{q'}\leqslant C(|r|^q+G_\mu(x)), \qquad r\in \mathbb{R}, \quad G_\mu\in L_{1,\mu}(\Omega). \end{equation} \tag{4.15}$

$\begin{equation} |\langle K(u),v\rangle|\leqslant C(\|u\|_{W_p^1(\Omega)}+C_1)\|v\|_{W_p^1(\Omega)}. \end{equation} \tag{4.16}$

$\begin{equation*} K\colon W_p^1(\mathfrak{B}_{\sigma})\to L_{q',\mu} (\mathfrak{B}_{\sigma}), \qquad q<\frac{n\theta}{n-p}, \quad \theta=n\biggl(1-\frac 1s\biggr), \end{equation*} \notag$

$\begin{equation} \langle K(u),v\rangle=\int_{\mathfrak{B}_{\sigma}}b(x,u(x))v(x)\,d\mu \end{equation} \tag{4.17}$

$\begin{equation*} \begin{gathered} \, f^m(x)=T_m(f(x))\chi_{m}(x), \\ \chi_{m}(x)= \begin{cases} 1 & \text{for } x\in \mathfrak{B}_m, \\ 0 & \text{for } x\notin \mathfrak{B}_m, \end{cases} \end{gathered} \end{equation*} \notag$

$\begin{equation*} b_i^m(x,r)=T_m(b_i(x,r))\chi_{m}(x), \qquad i=0,1. \end{equation*} \notag$

$\begin{equation} |b_i^m(x,r)|\leqslant |b_i(x,r)|\quad\text{and} \quad |b_i^m(x,r)|\leqslant m\chi_{m}(x), \qquad x\in \mathbb{H}^n. \end{equation} \tag{4.18}$

$\begin{equation} b_i^m(x,r)r\geqslant 0, \qquad x\in \mathbb{H}^n, \quad r\in \mathbb{R}. \end{equation} \tag{4.19}$

$\begin{equation} f^m\to f \quad \text{in } L_1(\mathbb{H}^n), \qquad m\to\infty, \end{equation} \tag{4.20}$

$\begin{equation} |f^m(x)|\leqslant |f(x)|\quad\text{and} \quad |f^m(x)|\leqslant m\chi_{m}(x), \qquad x\in \mathbb{H}^n, \quad m\in\mathbb{N}. \end{equation} \tag{4.21}$

$\begin{equation*} \langle\mathcal{B}_m u,v\rangle=\int_{\mathbb{H}^n}b_0^m(x,u)v\,d\nu +\int_{\mathbb{H}^n}b_1^m(x,u)v\,d\mu=\langle K_0 u,v\rangle+\langle K_1 u,v\rangle. \end{equation*} \notag$

$\begin{equation} -\operatorname{div}_g a^m(x,u,du)+\mathcal{B}_mu=f^m(x), \qquad x\in \mathbb{H}^n, \quad m\in\mathbb{N}, \end{equation} \tag{4.22}$

$\begin{equation} \int_{\mathbb{H}^n}(a(x,T_m(u^m),du^m),dv) \,d\nu + \langle\mathcal{B}_m u^m,v\rangle=\langle f^m,v\rangle \end{equation} \tag{4.23}$

Proof of Theorem 1. In (4.23) we set

$v=T_{k,h} (u^m)=T_k(u^m-T_h(u^m))$ . Taking (4.19) into account we obtain

$\begin{equation} \begin{aligned} \, &\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|<k+h\}}(a^m(x,u^m,du^m),du^m)\, d\nu \nonumber \\ &\qquad\qquad +k \int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_0^m(x,u^m)|\, d\nu+k \int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_1^m(x,u^m)|\,d\mu \nonumber \\ &\qquad\qquad +\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|< k+h\}}b_0^m(x,u^m)u^m\biggl(1-\frac{h}{|u_m|}\biggr) \,d\nu \nonumber \\ \nonumber &\qquad\qquad +\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|< k+h\}} b_1^m(x,u^m)u^m\biggl(1-\frac{h}{|u_m|}\biggr)\, d\mu \\ &\qquad\leqslant k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}|f^m|\,d\nu. \end{aligned} \end{equation} \tag{5.1}$

Taking (4.19) into account and using (4.21) and (2.16), from (5.1) we deduce

$\begin{equation} \begin{aligned} \, &\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|<k+h\}}((a^m(x,u^m,du^m),du^m)+G(x))\,d\nu \nonumber \\ &\qquad\qquad +k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_0^m(x,u^m)| \,d\nu+ k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_1^m(x,u^m)|\,d\mu \nonumber \\ &\qquad \leqslant\int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}(k|f|+|G|)\,d\nu, \qquad m\in\mathbb{N}. \end{aligned} \end{equation} \tag{5.2}$

Setting $h=0$ in (5.1) and using inequality (2.16) we obtain

$\begin{equation} \begin{aligned} \, &\int_{\{\mathbb{H}^n \colon |u^m|<k\}}c_0 M(x,|d u^m|_g) \,d\nu \nonumber \\ &\qquad\qquad+\int_{\{\mathbb{H}^n \colon |u^m|<k\}}b_0^m(x,u^m)u^m \,d\nu +k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k\}}|b_0^m(x,u^m)| \,d\nu \nonumber \\ &\qquad\qquad+\int_{\{\mathbb{H}^n \colon |u^m|<k\}}b_1^m(x,u^m)u^m\,d\mu +k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k\}}|b_1^m(x,u^m)|\,d\mu \nonumber \\ &\qquad\leqslant (k+1)C_1, \qquad m\in\mathbb{N}. \end{aligned} \end{equation} \tag{5.3}$

It follows from (5.3) that

$\begin{equation} \begin{aligned} \, \nonumber &\int_{\{\mathbb{H}^n \colon |u^m|<k\}}M(x,|du^m|_g)\,d\nu \\ &\qquad=\int_{\mathbb{H}^n}M(x,|dT_k (u^m)|_g)\,d\nu\leqslant C_1(k+1), \qquad k>1, \quad m\in\mathbb{N}. \end{aligned} \end{equation} \tag{5.4}$

Hence

$T_k (u^m)\in V$ for each

$k>1$ , and it follows from (4.11) that

$\begin{equation} \| T_k(u^m)\|_{V}^p\leqslant C_2k, \qquad m\in\mathbb{N}. \end{equation} \tag{5.5}$

Since

$V$ is reflexive, we can extract a subsequence

$\begin{equation} T_k (u^m)\rightharpoonup v_k\in V, \qquad m\to \infty, \end{equation} \tag{5.6}$

converging weakly in

$V$ . In view of (5.5) we can use Lemma 3, which yields the estimate

$\begin{equation} \operatorname{meas}\{x\in\mathfrak{B}_{R}\colon |u^m(x)|\geqslant k\}\leqslant \frac{C(R)}{k^{p^*(1-p^{-1})}}, \qquad m>k>k_0. \end{equation} \tag{5.7}$

Then, taking a sufficiently large

$R$ first and then selecting

$k$ , we obtain

$\begin{equation} \begin{aligned} \, \nonumber &\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k\}}(|f|+|G|)\,d\nu\leqslant\int_{\{\mathfrak{B}_R \colon |u^m|\geqslant k\}}(|f|+|G|)\,d\nu \\ &\qquad\qquad +\int_{\{\mathbb{H}^n \colon \rho(x)\geqslant R\}}(|f|+|G|)\,d\nu\leqslant\varepsilon(k), \qquad m>k, \end{aligned} \end{equation} \tag{5.8}$

where

$\varepsilon(k)\to0$ as

$k\to\infty$ .

By (2.16) the first integral in (5.2) is nonnegative, so that for $k=1$ we can write the inequality

$\begin{equation} \begin{gathered} \, \int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}|b_0^m(x,u^m)|\,d\nu+\int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}|b_1^m(x,u^m)|\,d\mu <C, \qquad m\in\mathbb{N}. \end{gathered} \end{equation} \tag{5.9}$

Using (2.20), for

$m> \widetilde{g}(h)$ we obtain

$\begin{equation*} \int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}\widetilde{g}(h)\chi_{m}(x)\,d\mu <C. \end{equation*} \notag$

Hence for each

$R>0$ we have

$\begin{equation} \operatorname{meas}_\mu\{x\in\mathfrak{B}_{R}\colon |u^m(x)|\geqslant h\}\leqslant \frac{C}{\widetilde{g}(h)}, \qquad m>\widetilde{g}(h)+R. \end{equation} \tag{5.10}$

Now we establish convergence with respect to a subsequence, namely,

$\begin{equation} u^m\to u \quad \nu\text{-a.e. and } \mu\text{-a.e. in } \mathbb{H}^n, \qquad m\to \infty. \end{equation} \tag{5.11}$

The sequence $T_s (u^m)$ is bounded in $V$ , and by (3.1) it is bounded in $W_p^1(\mathfrak{B}_{R})$ . By Kondrashov’s theorem we can extract a convergent subsequence such that $T_s (u^m) \to \widetilde{v}_{s}$ in $L_{p}(\mathfrak{B}_{R})$ as $m\to\infty$ . Hence $T_s(u^m)\to \widetilde{v}_{s}$ $\nu$ -almost everywhere in $\mathfrak{B}_{R}$ . By (5.6) we have $v_{s}=\widetilde{v}_{s}$ $\nu$ -almost everywhere in $\mathfrak{B}_{R}$ . Now using the diagonal procedure with respect to $R\in \mathbb{N}$ we find a subsequence such that $T_s(u^m)\to {v}_{s}$ $\nu$ -almost everywhere in $\mathbb{H}^n$ . Let $\Omega$ denote the set of points in $\mathbb{H}^n$ at which the sequence $u^m(x)$ has a finite limit. We denote this limit by $u(x)$ . Then for $x\in\Omega$ we have

$\begin{equation*} v_s(x)=\lim T_s(u^m(x))=T_s\lim u^m(x)=T_s(u). \end{equation*} \notag$

$\lim |T_s(u^m(x))|<s$ at some points

$x$ , then

$\lim T_s(u^m(x))=v_s(x)=\lim u^m(x)$ , that is,

$x\in\Omega$ . Thus, for

$\nu$ -almost all

$x\notin\Omega$ we have

$\begin{equation*} \lim T_s(u^m(x))=s\,\operatorname{sign}(v_s(x)) \end{equation*} \notag$

for each

$s>0$ . In particular,

$\begin{equation*} \lim T_{s+h}(u^m(x))=(s+h) \operatorname{sign}(v_{s+h}(x)). \end{equation*} \notag$

Then

$|u^m(x)|>s$ for large

$m$ , and therefore

$\lim |u^m(x)|=\infty$ . By (5.7) the set of such points in

$\mathfrak{B}_{R}$ has measure zero. Hence we conclude that the difference

$\mathbb{H}^n\setminus \Omega$ has measure zero, which proves the convergence (5.11) regarding the measure

$\nu$ . Then

$v_s(x)=T_s(u)$ for

$\nu$ -almost all

$x\in \mathbb{H}^n$ .

Also note that $T_s (u^m) \to v_{s}$ in $L_{q,\mu}(\mathfrak{B}_{R})$ , as follows from (4.14) and (3.1). Then $T_s(u^m)\to v_{s}$ $\mu$ -almost everywhere in $\mathfrak{B}_{R}$ (along a subsequence). Next, using a diagonal procedure with respect to $R\in \mathbb{N}$ we establish the following convergence along a subsequence:

$\begin{equation} T_s(u^m)\to T_s(u), \qquad m\to\infty, \end{equation} \tag{5.12}$

$\mu$ -almost everywhere in

$\mathbb{H}^n$ , so that we also prove (5.11).

Relation (5.6) can be written as

$\begin{equation} dT_k (u^m)\rightharpoonup dT_k(u) \quad \text{in } L_{M}(\mathbb{H}^n), \qquad m\to \infty. \end{equation} \tag{5.13}$

We show that for all $s>0$ and $R>0$

$\begin{equation} b_0^m(x,T_s(u^m)) \to b_0(x, T_s(u)) \quad\text{in } L_{1}(\mathfrak{B}_R), \qquad m\to \infty, \end{equation} \tag{5.14}$

and

$\begin{equation} b_1^m(x,T_s(u^m)) \to b_1(x, T_s(u)) \quad\text{in } L_{1,\mu}(\mathfrak{B}_R), \qquad m\to \infty. \end{equation} \tag{5.15}$

It follows from (5.11) that

$\begin{equation} b_0(x,T_s(u^m)) \to b_0(x, T_s(u)) \quad \nu\text{-a.e. in } \mathbb{H}^n, \qquad m\to \infty. \end{equation} \tag{5.16}$

By (2.18) we have

$|b_0^m(x,T_s(u^m))|\leqslant \operatorname{g}(s)\widetilde{G_0}(x)\in L_{1}(\mathfrak{B}_R)$ , so that (5.14) is a consequence of Lebesgue’s theorem. In a similar way we prove (5.15).

At this step we establish the strong convergence

$\begin{equation} dT_k(u^m)\to dT_k(u) \quad \text{in } L_{M,\mathrm{loc}}(\mathbb{H}^n), \qquad m\to \infty. \end{equation} \tag{5.17}$

From (5.4) and (2.15), for each $k>1$ we obtain the estimate

$\begin{equation} \|a(x,T_k(u^m),d T_k(u^m))\|_{ \overline{M},\mathbb{H}^n}\leqslant C_5(k), \qquad m\in\mathbb{N}. \end{equation} \tag{5.18}$

Let $k>0$ , $h>k+1$ and

$\begin{equation*} z^m=T_k(u^m)-T_k(u),\qquad m\in \mathbb{N}. \end{equation*} \notag$

Then by (5.12) we have

$\begin{equation} z^m\to 0 \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad m\to \infty, \end{equation} \tag{5.19}$

and also

$\begin{equation} |z^m|\leqslant 2k, \qquad m\in \mathbb{N}. \end{equation} \tag{5.20}$

We pointed out above that

$\begin{equation} \forall\, R>0\colon \ |z^m| \to 0\quad\text{in } L_{q,\mu}(\mathfrak{B}_{R}),\qquad m\to \infty. \end{equation} \tag{5.21}$

Using (5.19) and (5.20), from Lemma 8 we derive the convergence

$\begin{equation} |z^m| \stackrel{*}\rightharpoonup 0 \quad\text{in } L_{\infty}(\mathbb{H}^n), \qquad m\to \infty. \end{equation} \tag{5.22}$

By Lemma 5, for

$F\in L_M(\mathbb{H}^n)$ we have

$\begin{equation} Fz^m\to 0 \quad\text{in } L_M(\mathbb{H}^n), \qquad m\to \infty. \end{equation} \tag{5.23}$

Set $\eta_R(r)=\min(1,\max(0,R+1-r))$ . Substituting $z^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)$ into (4.23) as a test function we obtain

$\begin{equation} \begin{aligned} \, I_1^{mh} &=\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),d(\eta_R(\rho(x)) z^m\eta_{h-1}(|u^m|))\bigr)\,d\nu \nonumber \\ &=-\int_{\mathbb{H}^n} z^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)b_0^m(x,u^m)\,d\nu \nonumber \\ &\qquad - \int_{\mathbb{H}^n}b_1^m(x,u^m)z^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\mu \nonumber \\ &\qquad +\int_{\mathbb{H}^n} f^mz^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu \nonumber \\ &=I_2^{mh}+I_3^{mh}+I_4^{mh},\qquad m\in\mathbb{N}. \end{aligned} \end{equation} \tag{5.24}$

Estimates for the integrals $I_2^{mh}-I_4^{mh}$ . By the inequality

$\begin{equation*} \eta_R(\rho(x))\eta_{h-1}(|u^m|)|b_0^m(x,u^m)|\leqslant \operatorname{g}(h) \widetilde{G}_0(x), \end{equation*} \notag$

which follows from (2.18), and by the convergence (5.19), from Lebesgue’s theorem we obtain

$\begin{equation} |I_2^{mh}|\leqslant\int_{\mathfrak{B}_{R+1}}|z^m|\operatorname{g}(h) \widetilde{G}_0(x)\,d\nu=\varepsilon_{1}(m,h). \end{equation} \tag{5.25}$

Here and in what follows

$\begin{equation} \lim_{m\to\infty}\varepsilon_{i}(m,h)=0. \end{equation} \tag{5.26}$

In a quite similar way,

$\begin{equation} |I_3^{mh}|\leqslant\varepsilon_{2}(m,h). \end{equation} \tag{5.27}$

Using (4.21), (5.19) and (5.20) we obtain

$\begin{equation} |I_4^{mh}|\leqslant\int_{\mathbb{H}^n} |fz^m|\,d\nu=\varepsilon_{3}(m). \end{equation} \tag{5.28}$

It is obvious that $z^mu^m\geqslant0$ for $|u^m|\geqslant k$ , so for $|u^m|\geqslant h-1> k$ we have $z^m|u^m|=|z^m|u^m$ . Using this and (5.20) we obtain the estimate

$\begin{equation*} \begin{aligned} \, I_{12}^{mh} &=-\int_{\{\mathbb{H}^n\colon h-1\leqslant |u^m|<h\}}\bigl(a(x,T_h(u^m),dT_h(u^m)),d|u^m| \bigr) \eta_R(\rho(x)) z^m\,d\nu \\ &=-\int_{\{\mathbb{H}^n\colon h-1\leqslant |u^m|<h\}}\bigl((a(x,u^m,du^m),du^m )+G(x)\bigr) \eta_R(\rho(x)) |z^m|\,d\nu \\ &\qquad + \int_{\{\mathbb{H}^n\colon h-1\leqslant |u^m|<h\}}G(x) \eta_R(\rho(x)) |z^m|\,d\nu. \end{aligned} \end{equation*} \notag$

In view of (5.2) and (5.8) we see that

$\begin{equation} |I_{12}|^{mh}\leqslant\varepsilon(h), \qquad m\geqslant \widetilde{g}(h)+R, \end{equation} \tag{5.29}$

where

$\varepsilon(h)\to 0$ as

$h\to\infty$ .

Now, using (5.18) and the inequality $|d\eta_R(\rho(x))|_g\leqslant1$ we obtain

$\begin{equation} \begin{gathered} \, I^{mh}_{13} =\int_{\{\mathbb{H}^n\colon |u^m|<h\}}\bigl(a(x,T_h(u^m),d T_h(u^m)),d\eta_R(\rho(x))\bigr)\eta_{h-1}(|u^m|)z^m\,d\nu, \nonumber \\ |I^{mh}_{13}|\leqslant C_7(h)\|z^m\|_{M,\mathfrak{B}_{R+1}}=\varepsilon_5(m,h). \end{gathered} \end{equation} \tag{5.30}$

It is easy to show that $I^{mh}_1=I^{mh}_{11}+I^{mh}_{12}+I^{mh}_{13}$ , where

$\begin{equation*} I^{mh}_{11}=\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dz^m\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu. \end{equation*} \notag$

From (5.25)–(5.30) we can conclude that

$\begin{equation} |I^{mh}_{11}|\leqslant\varepsilon_6(m,h)+\varepsilon(h), \qquad m> \widetilde{g}(h)+R. \end{equation} \tag{5.31}$

After elementary transformations we arrive at the equalities

$\begin{equation*} \begin{aligned} \, I^{mh}_{11} &=\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dT_k(u^m)\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu \\ &\qquad -\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dT_k(u)\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu \\ &=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u^m)),dT_k(u^m)\bigr) \eta_R(\rho(x))\,d\nu \\ &\qquad -\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dT_k(u) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\bigr)\,d\nu \\ &=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),d T_k(u^m)),dz^m\bigr) \eta_R(\rho(x))\,d\nu \\ &\qquad +\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u^m)),dT_k(u)\bigr) \eta_R(\rho(x))\,d\nu \\ &\qquad -\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),d T_h(u^m)),dT_k(u)\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu. \end{aligned} \end{equation*} \notag$

The integrands in the last two integrals coincide for $|u^m|< k$ , so it is obvious that

$\begin{equation} \begin{aligned} \, \nonumber I^{mh}_{11} &=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u^m)),dz^m\bigr)\eta_R(\rho(x))\,d\nu \\ &\qquad +\int_{\{\mathbb{H}^n\colon |u^m|\geqslant k\}}\bigl(a(x,T_k(u^m),d T_k(u^m)) \nonumber \\ &\qquad\qquad-\eta_{h-1}a(x,T_h(u^m),dT_h(u^m)),dT_k(u)\bigr)\eta_R(\rho(x))\,d\nu \nonumber \\ &=I_{5}^m+I_{6}^{mh}. \end{aligned} \end{equation} \tag{5.32}$

In view of (5.18) we can establish the inequality

$\begin{equation} \begin{gathered} \, |I^{mh}_{6}|\leqslant C_{9}(h,k)\|\chi_{\{\mathfrak{B}_{R+1}\colon |u^m|\geqslant k\}}\, d T_k(u)\|_{M}. \end{gathered} \end{equation} \tag{5.33}$

Using Lemma 2 we select $k$ such that $\operatorname{meas}\{x\in\mathfrak{B}_{R+1}\colon |u(x)|= k\}=0$ . Then by (5.11) we have

$\begin{equation*} \chi_{\{\mathfrak{B}_{R+1}\colon |u^m|\geqslant k\}}\, dT_k(u)\to \chi_{\{\mathfrak{B}_{R+1}\colon |u|\geqslant k\}}\, dT_k(u)= 0 \quad \text{$\nu$-a.e. in } \mathfrak{B}_{R+1}, \quad m\to\infty. \end{equation*} \notag$

In view of (5.13) we have

$M(x,|d T_k(u)|_g)\in L_1(\mathbb{H}^n)$ , and from Lebesgue’s theorem we deduce that

$\begin{equation*} \chi_{\{\mathfrak{B}_{R+1}\colon |u^m|\geqslant k\}}\, dT_k(u)\to 0 \quad \text{in } L_M(\mathbb{H}^n), \qquad m\to\infty. \end{equation*} \notag$

Hence from (5.33) we obtain

$\begin{equation} I^{mh}_{6}=\varepsilon_8(m,h). \end{equation} \tag{5.34}$

It follows from (5.31), (5.32) and (5.34) that

$\begin{equation} |I^m_{5}|\leqslant \varepsilon_9(m,h)+\varepsilon(h). \end{equation} \tag{5.35}$

Using the notation

$\begin{equation*} q^m(x)=\Lambda\bigl(x,T_k(u^m),dT_k(u^m),dT_k(u)\bigr), \end{equation*} \notag$

from Lemma 6 we obtain

$\begin{equation*} \begin{aligned} \, 0 &\leqslant \int_{\mathbb{H}^n}q^m(x) \eta_R(\rho(x))\,d\nu=I^{m}_{5}-I_{54}^m \\ &=I^{m}_{5}-\int_{\mathbb{H}^n}\eta_R(\rho(x)) \bigl(a(x,T_k(u^m),dT_k(u)),d(T_k(u^m)-T_k(u))\bigr) \,d\nu. \end{aligned} \end{equation*} \notag$

Similarly to how we deduced (3.19), using (5.12) we can show that

$\begin{equation*} \begin{aligned} \, &\eta_R(\rho(x))a(x,T_k(u^m),dT_k (u)) \\ &\qquad \to \eta_R(\rho(x)) a(x,T_k(u),dT_k(u)) \quad\text{in } L_{\overline{M}}(\mathbb{H}^n), \qquad m\to \infty. \end{aligned} \end{equation*} \notag$

In combination with (5.13), this yields

$\begin{equation} I^m_{54}=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u)),d(T_k(u^m)-T_k(u))\bigr) \eta_R(\rho(x))\,d\nu=\varepsilon_{10}(m,h). \end{equation} \tag{5.36}$

From (5.35) and (5.36) we obtain

$\begin{equation*} \int_{\mathbb{H}^n}q^m(x)\eta_R(\rho(x)) \,d\nu\leqslant \varepsilon_{11}(m,h)+\varepsilon(h). \end{equation*} \notag$

Using (5.26) and taking the limit in the last inequality, first as

$m\to\infty$ and then as

$h\to\infty$ , we establish the relation

$\begin{equation*} \lim_{m\to \infty}\int_{\mathfrak{B}_R}q^m(x) \,d\nu=0, \end{equation*} \notag$

where positive

$R$ and

$k$ can be arbitrary.

By Lemma 7 for $w^j=T_{k}(u^j)$ and $w=T_{k}(u)$ , taking (5.12) into account we have (5.17). In view of (3.18),

$\begin{equation} dT_k(u^m)\to dT_k(u) \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad m\to\infty. \end{equation} \tag{5.37}$

To prove (2.21) consider the test function ${v = T_k(u^m-\xi)}$ in (4.23), where $\xi\in C_0^1(\mathbb{H}^n)$ ; then we obtain

$\begin{equation} \begin{aligned} \, \nonumber &\int_{\mathbb{H}^n} \bigl(a(x,T_m(u^m),du^m),dT_k(u^m-\xi)\bigr)\,d\nu \\ \nonumber &\qquad\qquad +\int_{\mathbb{H}^n} \bigl(b_0^m(x,u^m)-f^m\bigr)T_k(u^m-\xi)\,d\nu +\int_{\mathbb{H}^n} b_1^m(x,u^m)T_k(u^m-\xi)\,d\mu \\ &\qquad=I^m+J^m=0. \end{aligned} \end{equation} \tag{5.38}$

Set $\mathbf{M}=k+\|\xi\|_{\infty}$ . If $|u^m|\geqslant \mathbf{M}$ , then $|u^m-\xi|\geqslant |u^m|-\|\xi\|_{\infty}\geqslant k$ , and therefore $\{\mathbb{H}^n\colon |u^m-\xi|< k\}\subseteq \{\mathbb{H}^n\colon |u^m|< \mathbf{M}\}$ . Hence

$\begin{equation} \begin{aligned} \, I^m &=\int_{\mathbb{H}^n} \bigl(a(x,T_m(u^m),du^m),dT_k(u^m-\xi)\bigr)\,d\nu \nonumber \\ &=\int_{\mathbb{H}^n} \bigl(a(x,T_{\mathbf{M}}(u^m),d T_{\mathbf{M}}(u^m)),dT_k(u^m-\xi)\bigr)\,d\nu \nonumber \\ &=\int_{\mathbb{H}^n} \bigl(a(x,T_{\mathbf{M}}(u^m),dT_{\mathbf{M}}(u^m)),dT_{\mathbf{M}}(u^m)\,{-}\,d\xi\bigr) \chi_{\{\mathbb{H}^n\colon |u^m-\xi|<k\}}\,d\nu, \quad m\geqslant \mathbf{M}. \end{aligned} \end{equation} \tag{5.39}$

Let $w^m=u^m-\xi$ and $w=u-\xi$ . Since on the set where $|w^m|\to k$ as $m\to\infty$ we have $|w|= k$ , it follows that $d w=0$ . Therefore,

$\begin{equation} \begin{aligned} \, &dT_k(w^m)-dT_k(w) =\chi_{\{\mathbb{H}^n\colon |w^m|<k\}}(dw^m-d w) \nonumber \\ &\qquad +\bigl(\chi_{\{\mathbb{H}^n\colon |w^m|<k\}}-\chi_{\{\mathbb{H}^n\colon |w|<k\}}\bigr)\,dw\to 0 \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad m\to\infty. \end{aligned} \end{equation} \tag{5.40}$

Using inequality (2.9) and conditions (2.15) and (2.16) for

$\varepsilon\in (0,1)$ we find that

$\begin{equation*} \begin{aligned} \, &\bigl(a(x,T_{\mathbf{M}}(u^m),dT_{\mathbf{M}}(u^m)), dT_{\mathbf{M}}(u^m)-d\xi\bigr)\chi_{\{\mathbb{H}^n\colon |u^m-\xi|< k\}} \\ &\qquad \geqslant(c_0-\varepsilon \operatorname{g}(\mathbf{M}))M(x,|dT_{\mathbf{M}}(u^m)|_g)-C_1 G(x)-M(x,\varepsilon^{-1} |d\xi|_g). \end{aligned} \end{equation*} \notag$

Taking

$\varepsilon<c_0/\operatorname{g}(\mathbf{M})$ we obtain the inequality

$\begin{equation*} \begin{aligned} \, &\bigl(a(x,T_{\mathbf{M}}(u^m),dT_{\mathbf{M}}(u^m)),d( T_M(u^m)-\xi)\bigr)\chi_{\{\mathbb{H}^n\colon |u^m-\xi|<k\}} \\ &\qquad \geqslant -C_{10}(G(x)+M (x,|d\xi|_g))\in L_1(\mathbb{H}^n). \end{aligned} \end{equation*} \notag$

From (5.40), (5.12) and (5.37), since

$a(x,r,y)$ is continuous in

$(r,y)$ , by Fatou’s lemma we obtain

$\begin{equation} \begin{aligned} \, \lim_{m\to \infty}\inf I^m &\geqslant \int_{\mathbb{H}^n}\bigl(a(x,T_{\mathbf{M}}(u),dT_{\mathbf{M}}(u)),dT_k(u-\xi)\bigr)\,d\nu \nonumber \\ &=\int_{\mathbb{H}^n}\bigl(a(x,u,du),dT_k( u-\xi)\bigr)\,d\nu\geqslant C_I. \end{aligned} \end{equation} \tag{5.41}$

By (5.11), using Lemma 8 we have

$\begin{equation} T_k(u^m-\xi)\stackrel {*}\rightharpoonup T_k(u-\xi) \quad \text{in } L_{\infty}(\mathbb{H}^n), \qquad m\to \infty. \end{equation} \tag{5.42}$

We split the integral $J^m$ into two. The first term

$\begin{equation*} J^m_1=\int_{\mathbb{H}^n}b_0^m(x,u^m)T_k(u^m-\xi)\,d\nu+ \int_{\mathbb{H}^n}b_1^m(x,u^m)T_k(u^m-\xi)\,d\mu \end{equation*} \notag$

can be estimated as follows. Using (4.20) and (5.42) and taking the limit as

$m\to\infty$ we see that

$\begin{equation} J^m_2= \int_{\Omega}f^m T_k(u^m-\xi)\,dx\to \int_{\Omega}f T_k(u-\xi)\,dx=C_{J_2}. \end{equation} \tag{5.43}$

Now from (5.38) we obtain

$\begin{equation*} C_I+\liminf_{m\to \infty} J_1^m\leqslant C_{J_2}. \end{equation*} \notag$

Let $\operatorname{supp}\xi\subset \mathfrak{B}_{l_0}$ , $l\geqslant l_0$ , $\mathfrak{B}_{l,s}^m=\{x\in\mathfrak{B}_{l}\colon \ |u^m(x)|< s\}$ , $s\geqslant \mathbf{M}$ and $\mathfrak{B}_{l,s}=\{x\in\mathfrak{B}_{l}\colon \ |u(x)|< s\}$ . We choose $s$ so that $\operatorname{meas} \{x\in\mathfrak{B}_{l}\colon \ |u(x)|= s\}=0$ . Then taking (4.19) and the inequality $u^m(x)T_k(u^m-\xi)\geqslant0$ for $|u^m(x)|>\mathbf{M}$ into account we see that

$\begin{equation*} \begin{aligned} \, J^m_1 &=\int_{\mathbb{H}^n\backslash \mathfrak{B}_{l,s}^m}b_0^m(x,u^m)T_k(w^m)\,d\nu+\int_{\mathbb{H}^n\backslash \mathfrak{B}_{l,s}^m}b_1^m(x,u^m)T_k(w^m)\,d\mu \\ &\qquad +\int_{\mathfrak{B}_{l,s}^m}b_0^m(x,u^m)T_k(w^m)\,d\nu+ \int_{\mathfrak{B}_{l,s}^m}b_1^m(x,u^m)T_k(w^m)\,d\mu \\ &\geqslant\int_{\mathfrak{B}_{l,s}^m}b_0^m(x,T_s(u^m))T_k(w^m)\,d\nu+ \int_{\mathfrak{B}_{l,s}^m}b_1^m(x,T_s(u^m))T_k(w^m)\,d\mu= \overline{J}^{\,lm}_1. \end{aligned} \end{equation*} \notag$

Recalling (5.14), (5.15) and (5.42) and taking the limit as

$m\to \infty$ we obtain

$\begin{equation*} \begin{aligned} \, &\int_{\mathfrak{B}_{l,s}}b_1(x,T_s(u))T_k(u-\xi)\,d\mu +\int_{\mathfrak{B}_{l,s}}b_0(x,T_s(u))T_k(u-\xi)\,d\nu \\ &\qquad= \lim_{m\to \infty}\overline{J}^{\,lm}_1\leqslant \liminf_{m\to \infty} J_1^m. \end{aligned} \end{equation*} \notag$

Since by (2.19) we have

$\begin{equation*} \int_{\mathfrak{B}_{l,s}\setminus \mathfrak{B}_{l_0,s}}b_0(x,T_s(u))T_k(u-\xi)\,d\nu= \int_{\mathfrak{B}_{l,s}\setminus \mathfrak{B}_{l_0,s}}|b_0(x,T_s(u))T_k(u)|\,d\nu, \end{equation*} \notag$

Beppo Levi’s theorem shows that we can take the limit as

$l\to \infty$ . Setting

$\mathbb{H}_{s}=\{x\in\mathbb{H}^n\colon \ |u(x)|< s\}$ and taking the limit as

$l\to \infty$ we obtain

$\begin{equation*} \int_{\mathbb{H}_{s}}b_1(x,u)T_k(u-\xi)\,d\mu +\int_{\mathbb{H}_{s}}b_0(x,u)T_k(u-\xi)\,d\nu\leqslant \liminf_{m\to \infty} J_1^m. \end{equation*} \notag$

Since

$\begin{equation*} \int_{\mathbb{H}_{s}\setminus\mathbb{H}_{M}}b_1(x,u)T_k(u-\xi)\,d\mu= \int_{\mathbb{H}_{s}\setminus\mathbb{H}_{M}}|b_1(x,u)T_k(u-\xi)|\,d\mu \end{equation*} \notag$

by (2.19), we can take the limit as

$s\to \infty$ . As a result,

$\begin{equation} \int_{\mathbb{H}^n}b_1(x,u)T_k(u-\xi)\,d\mu +\int_{\mathbb{H}^n}b_0(x,u)T_k(u-\xi)\,d\nu\leqslant \liminf_{m\to \infty} J_1^m. \end{equation} \tag{5.44}$

Combining (5.38), (5.41), (5.44) and (5.43) we deduce (2.21).

Theorem 1 is proved.

\Bibitem{VilMuk23}

\by V.~F.~Vil'danova, F.~Kh.~Mukminov

\paper Entropy solution for an equation with measure-valued potential in a~hyperbolic space

\jour Sb. Math.

\yr 2023

\vol 214

\issue 11

\pages 1534--1559

\mathnet{http://mi.mathnet.ru/eng/sm9875}

\crossref{https://doi.org/10.4213/sm9875e}

\mathscinet{http://mathscinet.ams.org/mathscinet-getitem?mr=4720894}

\zmath{https://zbmath.org/?q=an:1540.35210}

\adsnasa{https://adsabs.harvard.edu/cgi-bin/bib_query?2023SbMat.214.1534V}

\isi{https://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=Publons&SrcAuth=Publons_CEL&DestLinkType=FullRecord&DestApp=WOS_CPL&KeyUT=001191951300002}

\scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-85188525946}

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