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<title>Localization of Nash-type equilibria <br />for systems with a partial variational structure: Localization of Nash-type equilibria <br />for systems with a partial variational structure</title>
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<h1>Localization of Nash-type equilibria <br />for systems with a partial variational structure</h1>
<p class="authors">
<span class="author">Andrei Stan\(^{\dag \ddag }\)</span>
</p>
<p class="date">September 4, 2023; accepted: November 6, 2023; published online: December 22, 2023.</p>
<p class="thanks">\(^{\dag }\)Faculty of Mathematics and Computer Science, Babeş-Bolyai University, 400084 Cluj-Napoca, Romania <br />\(^{\ddag }\)Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, P.O. Box 68-1, 400110 Cluj-Napoca, Romania</p>
</div>



<div class="abstract"> In this paper, we aim to generalize an existing result by obtaining localized solutions within bounded convex sets, while also relaxing specific initial assumptions. To achieve this, we employ an iterative scheme that combines a fixed-point argument based on the Minty-Browder Theorem with a modified version of the Ekeland variational principle for bounded sets. An application to a system of second-order differential equations with Dirichlet boundary conditions is presented. </div>
<p><b class="bfseries">MSC. </b>35B38, 47J25, 47J30, 47H10 </p>
<p><b class="bfseries">Keywords. </b>Nash equilibrium, iterative methods, Ekeland variational principle, monotone operator. </p>
<h1 id="sec1">1 Introduction</h1>
<p>  Numerous problems can be reduced to a fixed point equation \(N(u)=u\), where \(N\) is some operator. It is said that the equation has a variational form (or admits a variational structure) if it is equivalent with a critical point equation \(E'(u)=0.\) Also, in real word processes, one may need the solution to be positive and finite or limited in some sense. From a mathematical perspective, it means that the solution lies in a bounded subset of the positive cone of all states. </p>
<p>We aim to generalize a result from <span class="cite">
	[
	<a href="index.html#eu" >1</a>
	]
</span>, where a system of three equations </p>
<div class="displaymath" id="a0000000002">
  \begin{equation*} \begin{cases}  N_{1}(u,v,w)=u \\ N_{2}(u,v,w)=v \\ N_{3}(u,v,w)=w,\end{cases}\end{equation*}
</div>
<p>was studied, so that one of the equations lacked the variational structure. In <span class="cite">
	[
	<a href="index.html#eu" >1</a>
	]
</span>, sufficient conditions have been established for the system to admit a solution throughout the whole space, so that the second and third components of the solution represent a Nash equilibrium for the associated energy functionals. </p>
<p>In the present paper, a system with \(n\) equations is considered, </p>
<div class="equation" id="sistem">
<p>
  <div class="equation_content">
    \begin{equation} \label{sistem} \begin{cases}  N_1(u^1,\dots ,u^n)=u^1\\ \dots \\ N_p(u^1,\dots ,u^p,\dots ,u^n)=u^p \\ \dots \\ N_n(u^1,\dots ,u^n)=u^n \end{cases} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> where only the last \(n\)-\(p\) equations admit a variational structure. Our goal is to find a solution \((u^1_\ast ,\dots , u^p_\ast ,\dots ,u^n_\ast )\), such that \((u^{p+1}_\ast ,\dots ,u^n_\ast )\) is a Nash equilibrium for the corresponding energy functionals. Furthermore, this solution is intended to be localized in a bounded set of a positive cone. Localization is obtained in the Cartesian product of \(n\)-\(p\) bounded sets, as defined below. Specifically, </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  (u^{p+1}_\ast ,\dots ,u^n_\ast )\in K_{p+1}\times \dots \times K_n, \end{equation*}
</div>
<p> where \(K_i\) is a positive cone of some Hilbert space \(X_i\), with the corresponding inner product \((\cdot \,  , \,  \cdot )_i\) and norm \(|\cdot |_i\). Additionally, we require that </p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  r_q\leq l_q(u_\ast ^q)\,  \text{ ; }\,  |u_\ast ^q|_q \leq R_q \quad \left( q=\{ p+1,\dots , n\} \right), \end{equation*}
</div>
<p> for some positive real numbers \(r_q {\lt} R_q\), and a concave upper semicontinous functional \(l_q \colon K_q \to \mathbb {R}_+.\) </p>
<p>The study of systems with two equations, whose solutions represent a Nash-equilibrium localized in some conical sets, was previously made in <span class="cite">
	[
	<a href="index.html#bcp" >2</a>
	]
</span> or <span class="cite">
	[
	<a href="index.html#p1" >3</a>
	]
</span>. This was achieved imposing a Perov contraction condition and making use of Ekeland variational principle. In this paper, we use a monotony condition instead of Lipschitz one, inspired by <span class="cite">
	[
	<a href="index.html#p" >4</a>
	]
</span> and <span class="cite">
	[
	<a href="index.html#galewski" >5</a>
	]
</span>. </p>
<p>The idea of Nash equilibrium dates back to 1838, in a paper of Cournot <span class="cite">
	[
	<a href="index.html#cournot" >6</a>
	]
</span>, where the best output of a firm depending on the output of other firms was studied. Later, in 1951, the existence of a such equilibrium for any finite game was proved by John Forbes Nash <span class="cite">
	[
	<a href="index.html#nash_primul_articol" >7</a>
	]
</span>. From the physical point of view, a Nash equilibrium of a interconnected system with several entities, is that state in which the energy of each entity is minimal one with respect to the others. </p>
<p>The Nash equilibrium regards non-cooperative games, specifically those in which each player is unaware of how its change affects the output of the others. Heed that there are some other types of equilibrium, such as Pareto equilibrium for a cooperative game in which all parts are equal and collaborate to attain some (Pareto) optimal solution. If we are taking about a leading game, in which an individual (leader) takes the first step and all the others (followers) move accordingly, then we were dealing with a Stackelberg model and its equilibrium is called Stackelberg equilibrium. </p>
<p>The outline of this paper is as follows: We commence with <a href="index.html#sec2">section 2</a>, where several auxiliary results are presented. <a href="index.html#sec3">section 3</a> provides the main theoretical result while <a href="index.html#sec4">section 4</a> is dedicated to an application for a system of three second order differential equations subject to Dirichlet boundary conditions. Finally, we illustrate the application through a specific system. </p>
<h1 id="sec2">2 Preliminaries</h1>
<p>  Throughout this paper, \(\mathbb {R}^n\) is endowed with the usual scalar product denoted by \(x \cdot y=\sum _{i=1}^n x_iy_i\) and the Euclidian norm \(|x|=\left( \sum _{i=1}^n x_i^2\right)^{\frac{1}{2}}\). Also we consider the Hadamard product </p>
<div class="displaymath" id="a0000000005">
  \[  \circ \colon \mathbb {R}^{n}\times \mathbb {R}^{n}\rightarrow \mathbb {R}^{n}\text{ ; }x \circ y =(x_1y_1,\dots , x_n y_n)^T.  \]
</div>
<p> Unless otherwise noted, all vectors from \(\mathbb {R}^n\) are considered to be column. <div class="proposition_thmwrapper " id="propozitie_marginire_produs_hadamard">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">1</span>
  </div>
  <div class="proposition_thmcontent">
  <p> Let \(M=(m_{ij})_{1\leq i,j\leq n}\in \mathbb {M}_{n,n}\left(\mathbb {R}_+ \right)\) be a matrix with positive entries and let \(x=(x_i),y=(y_i),z=(z_i)\in \mathbb {R}^n_+\). If </p>
<div class="displaymath" id="a0000000006">
  \begin{equation*}  M x \circ y \leq z \end{equation*}
</div>
<p> then </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*}  Mx \cdot y\leq \sqrt{n} |z|. \end{equation*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000008">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Note that </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  \sum _{j=1}^n m_{ij}x_j y_i \leq z_i, \,  \text{ for }i=1,\dots , n. \end{equation*}
</div>
<p> Thus, </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  Mx \cdot y = \sum _{i=1}^n\sum _{j=1}^n m_{ij}x_j y_i \leq \sum _{i=1}^n z_i. \end{equation*}
</div>
<p> The conclusion immediately follows from the Cauchy-Schwartz inequality, as we have \(\sum _{i=1}^n z_i \leq \sqrt{n}|z|\). </p>
<p><div class="proof_wrapper" id="a0000000011">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> A square matrix of non-negative numbers \(A=[a_{i,j}]_{1\leq i,j\leq n}\in \mathbb {M}_{n,n}\left(\mathbb {R} \right)\) is said to be convergent to zero if \(A^k \to O_n\) as \(k \to \infty ,\) where \(O_n\) is the zero matrix. In the following, we outline a few characterizations of such matrices. <div class="theorem_thmwrapper " id="caracterizare_matrici_convergente_la_zero">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
    <span class="theorem_thmtitle">see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#galewski" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="index.html#nonnegative" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="index.html#role_of_matrices" >9</a>
	]
</span></span>
  </div>
  <div class="theorem_thmcontent">
  <p> The following statements are equivalent: </p>
<ol class="enumerate">
  <li><p>\(A\) is convergent to zero. </p>
</li>
  <li><p>\(\rho (A) {\lt} 1\), where \(\rho (A)\) represent the spectral radius of matrix \(A\). </p>
</li>
  <li><p>\(I - A\) is non-singular, and its inverse has nonnegative entries. </p>
</li>
  <li><p>There exist a positive diagonal matrix \(D=(d_{ii})_{1\leq i \leq n}\) such that </p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  D(I-A)x \,  \cdot \,  x {\gt} 0 ,\text{ for all }x\in \mathbb {R}^n\setminus \{ 0\} . \end{equation*}
</div>
</li>
</ol>

  </div>
</div> For any \(r\in \{ 1,\dots , n\} \), denote with \(A_{r}:=[a_{i,j}]_{1\leq i,j\leq r}\) a submatrix of \(A\). <div class="lemma_thmwrapper " id="submatrice_convergenta_la_zero">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p> If matrix \(A\) is convergent to zero, then \(A_r\) is also convergent to zero. <div class="proof_wrapper" id="a0000000013">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Note that </p>
<div class="displaymath" id="a0000000014">
  \begin{equation*}  A=\begin{bmatrix}  A_{r} 

&  B 

\\ C 

&  D 

\end{bmatrix}, \end{equation*}
</div>
<p> where \(B \in \mathbb {M}_{r,n-r}\left(\mathbb {R} \right)\), \(C \in \mathbb {M}_{n-r,r}\left(\mathbb {R} \right)\) and \(D\in \mathbb {M}_{n-r,n-r}\left(\mathbb {R} \right)\) are block matrices. Thus, </p>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  A^k=\begin{bmatrix}  A_{r}^k+P^1_{k-1} 

&  P^2_{k} 

\\ P^3_{k} 

&  P^4_{k} 

\end{bmatrix}\end{equation*}
</div>
<p> where \(P^1_m,P^2_m,P^3_m,P^4_m\) are some matrix polynomials of order \(m\) evaluated at \(A_{r},B,C,D\). Now, since \(A^k \to O_n\), clearly \( A_{r}^k+P^1_{n-1} \to O_n \). Thus \( A_{r}\) is convergent to zero. <div class="proof_wrapper" id="a0000000016">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> For the proof of our main result (<a href="index.html#teorema_principala">Theorem 7</a>), we need the following convergence lemma. <div class="lemma_thmwrapper " id="first_lemma">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">4</span>
    <span class="lemma_thmtitle">see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#eu" >1</a>
	, 
	Lemma 
	
	2.2
	]
</span></span>
  </div>
  <div class="lemma_thmcontent">
  <p>  Let \(\left( A_{k,p}\right) _{k\geq 1},\  \left( B_{k,p}\right) _{k\geq 1}\  \  \) be two sequences of vectors in \(\mathbb {R} _{+}^{n}\) (column vectors) depending on a parameter \(p,\) such that </p>
<div class="displaymath" id="a0000000017">
  \begin{equation*}  A_{k,p}\leq MA_{k-1,p}+B_{k,p} \end{equation*}
</div>
<p>for all \(k\) and \(p,\) where \(M\in \mathbb {M}_{n}(\mathbb {R}_{+})\) is a matrix with spectral radius less than one. If the sequence \(\left( A_{k,p}\right) _{k\geq 1}\) is bounded uniformly with respect to \(p\) and \(B_{k,p}\rightarrow O_{n}\) as \(k\rightarrow \infty \) uniformly with respect to \(p,\) then \(A_{k,p}\rightarrow O_{n}\) as \(k\rightarrow \infty \) uniformly with respect to \(p.\) </p>

  </div>
</div> </p>
<p>Let \(\left(X,|\cdot |_X\right)\) be a Hilbert space together with the scalar product \((\cdot \,  , \,  \cdot )_X\), \(X^\ast \) its dual and let \(\left\langle \cdot \,  , \,  \cdot \right\rangle ^\ast \) be the dual paring between \(X^\ast \) and \(X\). </p>
<p>For the convenience of the reader, we recall a theorem of Minty-Browder for operators from a Hilbert space to its dual. An operator \(T \colon X \to X^\ast \) is called <i class="itshape">strongly monotone </i> if there exists a real constant \(c{\gt}0\) such that </p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  \left\langle T(u)-T(v)\,  , \,  u-v\right\rangle ^\ast \geq c |u-v|_X^2, \text{ for all } u,v \in X. \end{equation*}
</div>
<p> <div class="theorem_thmwrapper " id="minty_browder">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
    <span class="theorem_thmtitle">Minty-Browder, see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#minty_browder" >10</a>
	, 
	Theorem
	
	9.14
	]
</span></span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(T \colon X \to X^\ast \) be a continuous and strongly monotone operator. Then \(T\) is bijective. </p>

  </div>
</div> </p>
<p>The following result is a variation of the Ekeland variational principle within a convex conical set (see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#p1" >3</a>
	]
</span>). Let \(K\subset X\) be a cone, and let \(l \colon K \to \mathbb {R}\) be an upper semicontinuous concave functional. Additionally, suppose we have an operator \(N \colon X \to X\) and a functional \(E \colon X \to \mathbb {R}\), such that \(E^\prime (u) = u - N(u)\) for all \(u \in X\). Consider two real numbers \(0 {\lt} r {\lt} R\), and let the convex conical set \(K_{r,R}\) be defined by </p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  K_{r,R}:=\{ u \in K \,  | \,  r\leq l(u) , \,  |u|_X\leq R\} . \end{equation*}
</div>
<p> <div class="lemma_thmwrapper " id="ekeland_in_conical_set">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">6</span>
    <span class="lemma_thmtitle">see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#p1" >3</a>
	, 
	Lemma
	
	2.1
	]
</span></span>
  </div>
  <div class="lemma_thmcontent">
  <p> Assume the following conditions are satisfied: </p>
<ol class="enumerate">
  <li><p>The functional \(E\) is bounded from below on \(K_{r,R}\), <i class="it">i.e.</i>, </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  m:=\inf _{K_{r,R}}E(\cdot ) {\gt} -\infty . \end{equation*}
</div>
</li>
  <li><p>There exists \(\varepsilon {\gt} 0\) such that for all \(u \in K_{r,R}\) satisfying both \(|u| = R\) and \(l(u) = r\), we have \(E(u) \geq m + \varepsilon \). </p>
</li>
  <li><p>\(l(N(u))\geq r\), for all \(u \in K_{r, R}.\) </p>
</li>
</ol>
<p> Then, there exists a sequence \((u_n)\in K_{r,R}\) such that </p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  E(u_n)\leq m+\tfrac {1}{n}, \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000022">
  \begin{equation*}  |E^\prime (u_n)+\lambda _nu_n|_X\leq \tfrac {1}{n}, \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000023">
  \begin{equation*}  \lambda _n=\begin{cases}  -\frac{1}{R^2}(E^\prime (u_n),u_n)_X, \text{ when }|u_n|_X=R \text{ and } (E^\prime (u_n),u_n)_X{\lt}0 \\ 0, \text{ otherwise}. \end{cases}\end{equation*}
</div>

  </div>
</div> </p>
<h1 id="sec3">3 Main result</h1>

<p>Let \(\left( X_i, \vert \cdot \vert _i\right)\) be Hilbert spaces identified with their duals \( \left(i \in \{  1, \dots , n\} \right)\). Denote \(X:=X_1 \times \dots \times X_n\) and \(X_{1,q}:=X_1\times \dots \times X_q\) (\(q\in \{ 1,\dots , n\} \)) together with the norms \(|u|_X=|u_1|_1+\dots +|u_n|_n\) and \(|u|_{X_{1,q}}:=|u_1|_1+\dots +|u_q|_q\), respectively. Similarly, \(\overline{X}q\) denotes the space obtained from \(X\) by excluding \(X_q\), <i class="it">i.e.</i>, </p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  \overline{X}_q:=X_1 \times \dots \times X_{q-1}\times X_{q+1}\times \dots \times X_n. \end{equation*}
</div>
<p> For simplicity, for any \(q\in \{ 1,\dots ,n\} \), we refer to </p>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  (u^1,\dots ,u^q)^T\text{ as }u^{1,q}, \, (u^{q+1},\dots ,u^n)^T \text{ as }u^{q+1,n} \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  (N_1(u),\dots ,N_q(u))^T\text{ as }N_{1,q}(u),\,  (N_{q+1}(u),\dots ,N_n(u))^T \text{ as }N_{q+1,n}(u). \end{equation*}
</div>
<p>With these notations, we have </p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  u=\left( u^{1,p},u^{p+1,n}\right)^T \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  \left(N_1(u),\dots ,N_n(u)\right)^T=\left(N_{1,q}(u),N_{q+1,n}(u) \right)^T. \end{equation*}
</div>
<p> On \(X_{1,q}\), we consider the vector-valued inner product </p>
<div class="displaymath" id="a0000000029">
  \begin{equation*}  \left\langle u, v\right\rangle :=\left( (u^1,v^1)_1,\dots , (u^q,v^q)_q\right)^T\in \mathbb {R}^q, \end{equation*}
</div>
<p> and vector valued norm </p>
<div class="displaymath" id="a0000000030">
  \begin{equation*}  \| u\| :=\left(|u^1|_1,\dots ,|u^q|_q \right)^T\in \mathbb {R}^q, \quad \end{equation*}
</div>
<p> for any \(u=(u^1,\dots ,u^q)\), \(v=(v^1,\dots ,v^q)\in X_{1,q}\). Note that these notations are consistent with respect to Hadamard product since \(\left\langle u\,  , \,  u \right\rangle =\|  u \|  \circ \|  u \| .\) </p>
<p>In the following, we assume that the operators \(N_1, \dots , N_n\) are continuous and that \(N_{p+1},\dots , N_n\) admit a variational structure, <i class="it">i.e.</i>, for each \(q \in \{  p+1,\dots ,n\} \), there exists a functional \(E_q \colon X \to \mathbb {R}\) such that for every \((u^1,\cdots ,u^{q-1},u^{q+1},\cdots ,u^n) \in \overline{X}_q\), the functional </p>
<div class="displaymath" id="a0000000031">
  \begin{equation*}  E_q(u^1,\cdots ,u^{q-1},\cdot ,u^{q+1},\cdots ,u^n) \end{equation*}
</div>
<p> has a Fréchet derivative \(E_{qq}\). Furthermore, this derivative is given by </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F1082721C0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{structura_variationala_N_q} E_{qq}(u)=u^q-N_q(u). \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p>For each \(q\in \{ p+1,\dots ,p \} \), let \(K_{q}\subset X_q\) be a cone and \(l_q \colon K_q \to \mathbb {R}_+\) be an upper semicontinous and concave functional such that \(l_q(0)=0\). Additionally, consider \(r_q,R_q \in \mathbb {R}_+\), and define \((K_q)_{r_q,R_q}\subset K_q\) as </p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  (K_q)_{r_q,R_q}:=\{ u^q \in K_q \,  : \,  r^q\leq l_q(u^q),\,  |u^q|_q \leq R_q \} . \end{equation*}
</div>
<p> Denote </p>
<div class="displaymath" id="a0000000033">
  \begin{equation*}  K:=(K_{p+1})_{r_{p+1},R_{p+1}} \times \dots \times (K_{n})_{r_n, R_n} \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  \overline{K}_q:= (K_{p+1})_{r_{p+1},R_{p+1}} \times \dots \times (K_{q-1})_{r_{q-1},R_{q-1}} \times (K_{q+1})_{r_{q+1},R_{q+1}}\times \dots \times (K_{n})_{r_{n},R_{n}}. \end{equation*}
</div>
<p><div class="theorem_thmwrapper " id="teorema_principala">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Under the above notations, let us assume the following: </p>
<ol class="enumerate">
  <li><p>There exists a matrix \(A=[a_{ij}]_{1\leq i,j \leq n}\) convergent to zero such that </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F1082908E0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{monotony_condition} \left\langle N_{1,n}\left( u\right)-N_{1,n}\left( v\right)\,  , \,  u-v \right\rangle \leq A\|  u-v \|  \circ \|  u-v \| , \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p> <i class="it">i.e.</i>, </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F108290DC0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{monotony_condition_pe_componente} \left\langle N_i(u)-N_i(v),u^i-v^i \right\rangle \leq \sum _{j=1}^n \left\vert u^i-v^i\right\vert _i \sum _{j=1}^n \left\vert u^j-v^j\right\vert _j, \, \left(i \in \{ 1,\dots ,n\} \right), \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p> for all \(u=(u^1,\dots ,u^n),v=(v^1,\dots ,v^n)\in X\). </p>
</li>
</ol>
<p> For each \(q \in \{  p+1,\dots , n\} ,\) one has </p>
<ol class="enumerate">
  <li><div class="displaymath" id="a0000000035">
  \begin{equation*} \begin{aligned} &  l_q\left( N_q(u)\right)\geq r_q \text{ for all } u\in X_{1,p} \times K , \\ &  N_q(u)-u^q-\lambda \,  u^q\neq 0 \text{ for all } \lambda {\gt}0 \text{ and } u\in X_{1,p}\times K \text{ with }|u^q|_q=R_q. \end{aligned}\end{equation*}
</div>
</li>
  <li><p>There exist \(m:= \inf _{u\in X_{1,p}\times K} E_q(u){\gt}-\infty \) and \(\varepsilon {\gt}0\) such that </p>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  E_q(u)\geq \inf _{(K_q)_{r_q,R_q}} E_q(u^1,\dots ,u^{q-1},\, \cdot \,  ,u^{q+1},\dots ,u^n)+\varepsilon , \end{equation*}
</div>
<p> for all \( (u^1,\dots ,u^{q-1},u^{q+1},u^n)\in X_{1,p}\times \overline{K}_q\) which satisfies \(l_q(u^q)=r_q\) and \(|u^q|_{q}=R_q\), simultaneously. </p>
</li>
  <li><p>The operator \(N_{q}\left( 0_{X_1},\dots ,0_{X_p},\cdots \right)\) is bounded on \(K\). </p>
</li>
</ol>
<p> Then there exists two points \(u_\ast ^{1,p} \in X_{1,p}\) and \(u_\ast ^{p+1,n} \in K\) such that </p>
<div class="displaymath" id="a0000000037">
  \begin{equation*}  u_\ast =(u_\ast ^{1,p},u_\ast ^{p+1,n}) \end{equation*}
</div>
<p> is a solution of \((\ref{sistem})\) and \(u_\ast ^{p+1,n}\) is a Nash equilibrium in K for the functionals \((E_{p+1},\dots ,E_n)\), <i class="it">i.e.</i>, </p>
<div class="displaymath" id="a0000000038">
  \begin{equation*}  E_{q}(u_\ast )=\inf _{\left( K_q\right)_{r_q,R_q}}E_q(u^{1,q-1}_\ast \, , \,  \cdot \,  , \,  u^{q+1,n}_\ast ) \quad \left( q=p+1, \dots , n\right). \end{equation*}
</div>
<p> <div class="proof_wrapper" id="a0000000039">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><b class="bfseries">Step 1: Construction of the approximation sequence.</b><br />Note that the submatrix \(A_p=[a_{ij}]_{1\leq i,j\leq p}\) converges to zero as a consequence of <a href="index.html#submatrice_convergenta_la_zero">Lemma 3</a>. Thus, <a href="index.html#caracterizare_matrici_convergente_la_zero">Theorem 2</a> guarantees the existence of a diagonal matrix \(D=(d_{ii})_{1\leq i\leq p}\) with \(d_{ii}{\gt} 0\) (\(i=1,\dots ,n\)) such that \(D(I-A_p)\) is positive definite, <i class="it">i.e.</i>, </p>
<div class="equation" id="c">
<p>
  <div class="equation_content">
    \begin{equation} \label{c} c:=\inf _{x\in R^p\setminus \{ 0\} } \frac{ D(I-A_p)x \,  \cdot \,  x}{|x|^2}>0. \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p> Let \(u^{p+1,n}_0 \in K\) be arbitrarily chosen and let \(T\colon X_{1,p} \to X_{1,p}^\ast \) be defined by </p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  T(u^{1,p})=D\left(u^{1,p}-N_{1,p}\left( u^{1,p}\,  , \,  u_0^{p+1,n}\right)\right)\text{, for all $u^{1,p}=(u^1,\dots ,u^p) \in X_{1,p}$}. \end{equation*}
</div>
<p> For any \(u^{1,p}\), \(\overline{u}^{1,p}\in X_{1,p}\), relations ?? yields </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F1082F1CA0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{monotony_equal} \begin{aligned} &  \left\langle T(u^{1,p})-T(\overline{u}^{1,p})\,  , \,  u^{1,p}-\overline{u}^{1,p}\right\rangle ^\ast = \\ & =\sum _{i=1}^p d_{ii} |u^i-\overline{u}^i|^2_i-\left( N_i(u^{1,p},u^{p+1,n}_0)-N_i(\overline{u}^{1,p},u^{p+1,n}_0)\,  , \,  u^i-\overline{u}^i\right)_i \\ & \geq \sum _{i=1}^p d_{ii}|u^i-\overline{u}^i|^2_i-d_{ii}\sum _{j=1}^p a_{ij}|u^i-\overline{u}^i|_i|u^j-\overline{u}^j|_j. \end{aligned} \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p> Thus, since </p>
<div class="displaymath" id="a0000000041">
  \begin{align*} &  \sum _{i=1}^p d_{ii}|u^i-\overline{u}^i|^2_i-d_{ii}\sum _{j=1}^p a_{ij}|u^q-\overline{u}^i|_i|u^j-\overline{u}^j|_j = \\ &  =D(I_p-A_p) \| u^{1,p}-\overline{u}^{1,p}\|  \cdot \| u^{1,p}-\overline{u}^{1,p}\| , \end{align*}
</div>
<p> from relation <a href="index.html#c" class="eqref">15</a> we deduce </p>
<div class="displaymath" id="a0000000042">
  \begin{equation*} \begin{aligned} &  \left\langle T(u^{1,p})-T(\overline{u}^{1,p})\,  , \,  u^{1,p}-\overline{u}^{1,p}\right\rangle \geq c|u^{1,p}-\overline{u}^{1,p}|^2_{X_{1,p}}. \end{aligned}\end{equation*}
</div>
<p> Consequently, \(T\) is strongly monotone. As \(T\) is clearly continuous, <a href="index.html#minty_browder">Theorem 5</a> guarantees that it is a bijection from \(X\) to \(X^\ast \). Therefore, there exist a unique \(u^{1,p}_1 \in X_{1,p}\) such that \(T(u^{1,p}_1)=0\), <i class="it">i.e.</i>, </p>
<div class="displaymath" id="a0000000043">
  \begin{equation*}  u_1^{1,p}=N_{1,p}\left( u_1^{1,p},u^{p+1,n}_0\right). \end{equation*}
</div>
<p>For each \(q \in \{ p+1,\dots , n\} \), we fix \(u^{1,q-1}_1\) previously determined and \(u^{q+1,n}_0\) initially set. Following <a href="index.html#ekeland_in_conical_set">Lemma 6</a>, we may find \(u^q_1\in (K_q)_{r_q,R_q}\) such that </p>
<div class="equation" id="a0000000044">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{aligned} \notag &  E_q\left(u_1^{1,q}\,  ,\,  u^{q+1,n}_0 \right)\leq \inf _{(K_q)_{r_q,R_q}}E_q\left(u_1^{1,q-1}\,  ,\,  \cdot \,  , \, u^{q+1,n}_0 \right)+1\,  , \\ &  \left\vert E_{qq}\left(u_1^{1,q}\,  ,\,  u^{q+1,n}_0 \right)+\lambda ^q_1 u^{q}_1 \right\vert _q\leq 1\,  , \end{aligned} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> where </p>
<div class="displaymath" id="a0000000045">
  \begin{equation*}  \lambda ^q_1:= \begin{cases}  \! -\frac{1}{R_q^2}\left(\! E_{qq}\left(\! u_1^{1,q},u_0^{q+1,n} \right)\! \! ,\!  u^q_1\right)_q, &  \! \! \text{if }|u^q_1|_q\! =\! R_q \text{ and} \left(\! E_{qq}\! \left(u_1^{1,q},u_0^{q+1,n} \right)\! , u^q_1\right)_q\! {\lt}\! 0 \\ 0, & \text{ otherwise } \end{cases}\end{equation*}
</div>
<p> Repeating the process for each step \(k\), we construct recursively a sequence </p>
<div class="displaymath" id="a0000000046">
  \begin{equation*}  u_k=(u^1_k,\dots , u^p_k,u^{p+1}_k,\dots , u^n_k)^T\in X_{1,p}\times K \end{equation*}
</div>
<p> such that </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F108358BE0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{punct_fix_u_k} u_k^{1,p}=N_{1,p}\left( u_k^{1,p},u^{p+1,n}_{k-1}\right). \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p> and </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F108358FA0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{inequality_on_derivative} \begin{aligned} &  E_q\left(u_k^{1,q}\,  ,\,  u^{q+1,n}_{k-1} \right)\leq \inf _{(K_q)_{r_q,R_q}}E_q\left(u_k^{1,q-1}\,  ,\,  \cdot \,  , \, u^{q+1,n}_{k-1} \right)+\tfrac {1}{k}\,  ,\\ &  \left\vert E_{qq}\left(u_{k}^{1,q}\,  ,\,  u^{q+1,n}_{k-1} \right)+\lambda ^q_k u^{q}_k \right\vert _q\leq \tfrac {1}{k} \,  , \end{aligned} \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> where </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F10836FD60>">
<p>
  <div class="equation_content">
    \begin{equation} \label{def_lambda}\notag \lambda ^q_k:= \begin{cases}  \! -\frac{1}{R_q^2}\left(\! E_{qq}\left(\! u_k^{1,q}\! ,u_{k-1}^{q+1,n} \right)\! , u^q_k\right)_q\! ,&  \text{if }|u^q_k|_q\! =\! R_q \text{ and} \left(\! E_{qq}\left(\! u_k^{1,q},\! u_{k-1}^{q+1,n} \right)\! , u^q_k\right)_q\! \! <\! 0 \\ 0, & \text{ otherwise.} \end{cases} \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p> for each \(q\in \{ p+1,\dots ,n\} \). <br /> <br /><b class="bfseries">Step 2: Boundedness of the sequence \((u^1_k,\dots ,u^p_k)\) . </b><br /></p>
<p>Condition \((h4)\) guarantees that the sequence \( N_{1,p}(0,u^{p+1,n}_{k-1})\) is uniformly bounded, <i class="it">i.e.</i>, there exists \(M{\gt}0\) such that </p>
<div class="displaymath" id="a0000000047">
  \begin{equation*}  M_0=\sup _{k\in \mathbb {N}}\left\vert N_{1,p}(0,u^{p+1,n}_{k-1})\right\vert _{X}. \end{equation*}
</div>
<p> Since \(u_k^{1,p}=N_{1,p}\left( u_k^{1,p},u^{p+1,n}_{k-1}\right)\), one has </p>
<div class="displaymath" id="a0000000048">
  \begin{align}  \notag \| u_k^{1,p}\|  \circ \| u_k^{1,p}\|  & = \left\langle u^{1,p}_k,N_{1,p}\left(u^{1,p}_k,u_{k-1}^{p+1,n}\right)\right\rangle \\ &  \leq \notag \left\langle u^{1,p}_k,N_{1,p}\left(u^{1,p}_k,u_{k-1}^{p+1,n}\right)-N_{1,p}\left(0,u_{k-1}^{p+1,n}\right)\right\rangle +M_0 \| u^{1,p}_k\| \\ &  \leq A_p\| u^{1,p}_k\| \circ \| u^{1,p}_k\| +M_0\| u^{1,p}_k\|  \label{ineq1}. \end{align}
</div>
<p> Multiplying both sides of <a href="index.html#ineq1" class="eqref">23</a> with the diagonal matrix \(D\), we obtain </p>
<div class="displaymath" id="a0000000049">
  \begin{equation*}  D(I_p-A_p)\| u_k^{1,p}\|  \circ \| u_k^{1,p}\|  \leq M_0 D \| u_k^{1,p}\| . \end{equation*}
</div>
<p> Therefore, relation <a href="index.html#c" class="eqref">15</a> together with <a href="index.html#propozitie_marginire_produs_hadamard">Proposition 1</a> leads to </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F1083A8EE0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{boundness_u^1_p}\notag \sqrt{n} M_0 |u^{1,p}|_{X_{1,p}} \geq D(I_p-A_p)\| u^{1,p}_k\|  \,  \cdot \,  \| u^{1,p}_k\|  \geq c|u^{1,p}|_{X_{1,p}}^2, \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p> which guarantees the boundedness of the sequence \(u^{1,p}_k\). <br /> <br /><b class="bfseries">Step 3: Convergence to zero of the real sequence \(\left(\lambda _k^q\right)\).</b> <br /></p>
<p>Let \(q\in \{  p+1, \dots ,n\} \) and assume </p>
<div class="displaymath" id="a0000000050">
  \begin{equation*}  \left( E_{qq}\left( u_{k}^{1,p},u_{k-1}^{p+1,n}\right)\,  , \,  u_{k}^{q} \right)_q{\lt}0. \end{equation*}
</div>
<p> Observe that </p>
<div class="displaymath" id="a0000000051">
  \begin{align*}  \lambda _{k}^{q}&  \leq -\tfrac {1}{R_{q}^{2}} \notag \left( E_{qq}\left( u_{k}^{1,p},u_{k-1}^{p+1,n}\right)\,  , \,  u_{k}^{q} \right)_q \\ &  =-1+\tfrac {1}{R_{q}^{2}} \left( u_{k}^{q},N_{q}\left( u_{k}^{1,p},u_{k-1}^{p+1,n}\right) \right). \end{align*}
</div>
<p> Thus, the integration of elementary computations with the monotonicity of \(N_q\) ensures that </p>
<div class="displaymath" id="a0000000052">
  \begin{align*}  \lambda _{k}^{q}\leq \tfrac {1}{R_{q}}M_0 + \tfrac {1}{R_{q}^{2}}\left\vert u_{k}^{q}\right\vert _q \sum _{i=1}^{p}a_{qi}\left\vert u_{k}^{p}\right\vert _q \end{align*}
</div>
<p> Now, the sequence \(\left(\lambda _k^q\right)\) is bounded as a consequence of <i class="itshape">Step 2</i>. Hence, eventually passing to a subsequence, \(\left(\lambda _k^q\right)\) converges to a non-negative real number \(\lambda ^q\). Further, since \(\left(u_k^{1,q}\right)\) and \(\left(u_{k-1}^{q+1,n}\right)\) are bounded, they have convergent subsequences whose limits are denoted by \(u^{1,p}\) and \(\underline{u}^{q+1,n}\), respectively. Now, if we take the limit in ??, we obtain</p>
<div class="displaymath" id="a0000000053">
  \begin{equation*}  E_{11}(u^{1,q},\underline{u}^{q+1,p})+\lambda ^q \,  (u^{1,q},\underline{u}^{q+1,p})=0, \end{equation*}
</div>
<p> where \(|u^q|_q=R_q\) if \(\lambda ^q\geq 0\). In this case, the Leray-Schauder boundary condition in the second relation of (h2) is contradicted, leading to the conclusion that \(\lambda ^q_k \to 0\) as \(k \to \infty \). <br /> <br /><b class="bfseries">Step 4: Convergence of the sequence \((u_k^1,\dots ,u_k^n)\).</b> <br /></p>
<p>For any \(q\in \{ 1,\dots , n\} \), let us denote </p>
<div class="displaymath" id="a0000000054">
  \begin{align*} &  x^q(k,m):=\left\vert u_{k+m}^q-u_k^q\right\vert _q , \\ &  c^q(k,m):=\begin{cases}  0,&  \text{ if }q\leq p \\ \left\vert E_{qq}\left( u_{k+m}^{1,q},u_{k+m-1}^{q+1,n} \right)-E_{qq}\left( u_{k}^{1,q},u_{k-1}^{q+1,n} \right) \right\vert _q,& \text{ if }p+1\leq q, \end{cases}\end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000055">
  \begin{align*} & x(k,m):=\left(x^1(k,m),\dots , x^n(k,m) \right)^T,\\ &  c(k,m):=\left(c^1(k,m),\dots , c^n(k,m) \right)^T. \end{align*}
</div>
<p> Hence, </p>
<div class="displaymath" id="a0000000056">
  \begin{align*}  x^q(k,m)^2& =\left( u^q_{k+m}-u^q_k\,  , \,  N_q(u_{k+m}^{1,q},u_{k+m-1}^{q+1,n}) -N_q(u_{k}^{1,q},u_{k-1}^{q+1,n}) \right)\\ &  \quad -\left( u^q_{k+m}-u^q_k\,  , \,  \notag E_{qq}(u_{k+m}^{1,q},u_{k+m-1}^{q+1,n}) -E_{qq}(u_{k}^{1,q},u_{k-1}^{q+1,n}) \right)_q. \end{align*}
</div>
<p> Using the monotonicity assumption of \(N_q\), we derive the subsequent inequality </p>
<div class="displaymath" id="xq">
  \begin{align} \label{xq} x^q(k,m)^2& \leq x^q(k,m)\left(\sum _{j=1}^q a_{q,j}\,  x^j(k,m) +\sum _{j=q+1}^n a_{q,j} \, x^j(k-1,m) \right) \\ &  \quad +x^q(k,m)c(k,m).\notag \end{align}
</div>
<p> We can put relations <a href="index.html#xq" class="eqref">24</a> in vector form as </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x000001F10863C280>">
  \begin{align} \label{x_k_patrat_marginire} x(k,m)\circ x(k,m) &  \leq A^\prime \,  x(k,m)\circ x(k,m)+A^{\prime \prime }\, x(k-1,m) \circ x(k,m)\\ & \quad +c(k,m) \circ x(k,m),\notag \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000057">
  \begin{eqnarray*}  ~ A^{\prime } & =& \left[ a_{ij}^{\prime }\right] _{1\leq i,j\leq n}~ \text{\  with }\quad a_{ij}^{\prime }= \begin{cases}  0,\,  \text{ if } 0\leq i\leq j \\ a_{ij},\,  \text{otherwise} \end{cases}, \\ A^{\prime \prime } & =& A-A^{\prime }. \end{eqnarray*}
</div>
<p> Observe that \(\eqref{x_k_patrat_marginire}\) is equivalent with </p>
<div class="displaymath" id="a0000000058">
  \begin{equation*}  x(k,m)\leq A^\prime x(k,m)+A^{\prime \prime }x(k-1,m)+c(k,m), \end{equation*}
</div>
<p> and consequently, </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F10863CAC0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{ineq_finala} x(k,m) \leq \left( I-A^{\prime }\right) ^{-1}A^{\prime \prime }x(k-1,m) +\left( I-A^{\prime }\right) ^{-1}c(k,m). \end{equation}
  </div>
  <span class="equation_label">28</span>
</p>
</div>
<p> In order to continue our proof, we need to show that \(c^q(k,m) \to 0\) uniformly with respect to \(m\). If \(q\leq p\) then there is nothing to prove. Let \( p+1\leq q \). Then </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F10866D1C0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{marginire_c_k_m} \begin{aligned}  c(k,m)& =\left\vert E_{qq}\left( u_{k+m}^{1,q},u_{k+m-1}^{q+1,n} \right)-E_{qq}\left( u_{k}^{1,q},u_{k-1}^{q+1,n} \right) \right\vert _q \\ &  \leq \left\vert E_{qq}\left( u_{k+m}^{1,q},u_{k+m-1}^{q+1,n} \right)+\lambda _{k+m}^q\, u_{k+m}^q\right\vert _q\\ &  \quad + \left\vert E_{qq}\left( u_{k}^{1,q},u_{k-1}^{q+1,n} \right)+\lambda _{k}^q\, u_{k}^q\right\vert _q\\ &  \quad + \left\vert \lambda _{k+m}^q\, u_{k+m}^q -\lambda _{k}^q\, u_{k}^q\right\vert _q. \end{aligned} \end{equation}
  </div>
  <span class="equation_label">29</span>
</p>
</div>
<p>Now, we conclude that \(c^q(k,m) \to 0\) uniformly with respect to \(m\) since each term of the right-hand side of ?? converges to zero as a consequence of ?? and <i class="itshape">Step 3</i>. </p>
<p>Following <span class="cite">
	[
	<a href="index.html#eu" >1</a>
	]
</span>, the matrix \(\left( I-A^{\prime }\right) ^{-1}A^{\prime \prime }\) is convergent to zero. Consequently, from <a href="index.html#first_lemma">Lemma 4</a> we deduce that \(x(k,m)\to 0\) uniformly with respect to \(m\), <i class="it">i.e.</i>, the sequences \( \left(u_k^1\right),\dots ,\left(u^n_k\right)\) are Cauchy. Denote with \(u^1_\ast ,\dots , u^n_\ast \) their limits. </p>
<p><b class="bfseries">Step 5: Passing to the limit.</b> </p>
<p>Passing to limit in \((\ref{punct_fix_u_k})\) and \((\ref{inequality_on_derivative})\) we deduce that \( u_\ast =\left( u^1_\ast ,\dots ,u^n_\ast \right)\) solves the system <a href="index.html#sistem" class="eqref">4</a>, <i class="it">i.e.</i>, </p>
<div class="displaymath" id="a0000000059">
  \begin{equation*}  u_\ast = N_{1,n}\left( u_\ast \right), \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000060">
  \begin{equation*}  E_q(u_\ast )=\inf _{(K_q)_{r_q,R_q}}E_q(u^1_\ast ,\dots , u^{q-1}_\ast ,\cdot ,u^{q+1}_\ast ,\dots ,u^n_\ast ), \end{equation*}
</div>
<p> for each \(q\in \{ p+1,\dots , n\} \). <div class="proof_wrapper" id="a0000000061">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000062">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">8</span>
    <span class="remark_thmtitle">Perov contraction condition</span>
  </div>
  <div class="remark_thmcontent">
  <p> If instead of condition (h1), we assume that \(N=(N_1,\dots , N_n)\) is a Perov contraction, then we can relax the spaces \(X_1,\dots ,X_p\) to complete metric spaces. In this particular case, the fixed point \(u^{1,p}_k\) from ?? can be derived directly since the operator </p>
<div class="displaymath" id="a0000000063">
  \begin{equation*}  N_q(u^1,\dots , u^{q-1},\,  \cdot \,  , u^{q+1},\dots , u^n) \end{equation*}
</div>
<p> is a Lipschitz contraction, for each \((u^1,\dots , u^{q-1}, u^{q+1},\dots , u^n) \in \overline{X}_q\). The above result generalizes the one provided in <span class="cite">
	[
	<a href="index.html#eu" >1</a>
	]
</span>, since the monotony condition ?? is weaker than a Perov contraction condition. For example, if \(n=1\), any decreasing function satisfies ??, but obviously not all of them are Lipschitz. </p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000064">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
    <span class="remark_thmtitle">Multiplicity solutions</span>
  </div>
  <div class="remark_thmcontent">
  <p> Note that if we choose </p>
<div class="displaymath" id="a0000000065">
  \begin{equation*}  r^q_1,r^q_2,\dots ,r^n_1,r^n_2 {\gt}0 \text{ and }R^q_1,R^q_2,\dots ,R^n_1,R^n_2 {\gt}0 \end{equation*}
</div>
<p> such that </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F1083B1940>">
<p>
  <div class="equation_content">
    \begin{equation} \label{multiplicity_cond_r_and_R} R^q_1< l_q (r^q_2), \quad \left( q=p+1,\dots ,n\right) \end{equation}
  </div>
  <span class="equation_label">30</span>
</p>
</div>
<p> then \(\left( K_q\right)_{r^q_1,R^q_1} \cap \left( K_q\right)_{r^q_2,R^q_2}=\emptyset \). Thus, multiple solutions localized in different conical sets can be found, if different pairs \((r^q_i,R^q_i)\) with property ?? are chosen. This localized solution may have the same components for the equations without a variational structure, but different for the others equations. </p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000066">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">10</span>
    <span class="remark_thmtitle">Limit cases</span>
  </div>
  <div class="remark_thmcontent">
  <p> In our theory, the constants \(r_q\) and \(R_q\) can approach their limit values, meaning \(r_q=0\) when we seek solutions within a ball, and \(R_q=\infty \) when we aim to find upper unbounded solutions. </p>

  </div>
</div> </p>
<h1 id="sec4">4 Application</h1>
<p>  We apply the results from <a href="index.html#sec3">section 3</a> to the Dirichlet problem </p>
<div class="equation" id="a0000000067">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{cases}  -u^{\prime \prime }(t)=\, f_{1}(t,u(t),v(t),w(t),u^\prime (t)) \\ - v^{\prime \prime }(t)=f_{2}(t,u(t),v(t),w(t)) \\ -w^{\prime \prime }(t)=f_{3}(t,u(t),v(t),w(t))\quad \text{ on } (0,T) \\ u(0)=u(T)=0\\ v(0)=v(T)=0\\ w(0)=w(T)=0 \end{cases} , \label{main_system} \end{equation}
  </div>
  <span class="equation_label">31</span>
</p>
</div>
<p>where the functions \(f_1\colon (0,T) \times \mathbb {R}^{4}\to \mathbb {R}_+\), \(f_2,f_3\colon (0,T) \times \mathbb {R}^3 \to \mathbb {R}_+\) are of Carathéodory type. We emphasize that the presence of \( u ^{\prime } \) on the right side of the first equation, as opposed to equations 2 and 3, makes it to lose the variational structure. Moreover, the Hilbert spaces \(X_1,X_2,X_3\) from <a href="index.html#teorema_principala">Theorem 7</a>, are the Sobolev space \(H_0^1(0,T)\) endowed with the inner product \(\left( u,v\right)_{H_0^1}=\displaystyle \int _0^T u^{\prime } v^{\prime }\) and the norm \(|u|_{H_0^1}=\Big( \displaystyle \int _0^T (u^{\prime })^2\Big)^{\frac{1}{2}}\). </p>
<p>Due to the continuous embeddings</p>
<div class="displaymath" id="a0000000068">
  \begin{equation*}  H_{0}^{1}\left( 0,T \right) \subset C[0,T] \subset L^{2}\left( 0,T \right) \subset \left(H_{0}^{1}(0,T)\right)^{\prime }, \end{equation*}
</div>
<p> the Poincaré inequality (see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#brezis" >11</a>
	]
</span>,<span class="cite">
	[
	<a href="index.html#rp" >12</a>
	, 
	Remark
	
	3.3
	]
</span>) holds with \(\lambda _1=\frac{\pi ^2}{T^2}\) (see, <span class="cite">
	[
	<a href="index.html#rp" >12</a>
	, 
	p.
	
	72
	]
</span>), <i class="it">i.e.</i>, </p>
<div class="displaymath" id="a0000000069">
  \begin{equation*}  |u|_{L^2}\leq \tfrac {1}{ \sqrt{ \lambda _1}}|u|_{H_0^1}=\tfrac {T}{\pi }|u|_{H_0^1}, \left( u\in H_0^1(0,T) \right), \end{equation*}
</div>
<p> where \(\lambda _{1}\) is the first eigenvalue of the Dirichlet problem \(-u^{\prime \prime }=\lambda u\), \(u(0)=u(T)=0\). Additionally, a positive constant \(c{\gt}0\) exists such that for all \(t\in (0,T)\) and \(u \in H_0^1(0,T)\), the following inequality holds true </p>
<div class="displaymath" id="a0000000070">
  \begin{equation*}  |u(t)|\leq c|u|_{H_0^1}. \end{equation*}
</div>
<p> Notably, \(c\) can be chosen as \(\sqrt{T}\). Indeed, from Cauchy-Schwartz inequality we immediately have </p>
<div class="displaymath" id="a0000000071">
  \begin{align*}  |u(t)|& =|u(t)-u(0)|\leq \int _0^t \left| u^\prime (x) \right| dx\\ & \leq \int _0^T |u^\prime (x)|dx\leq |u^\prime |_{L^2}|1|_{L^2}=\sqrt{T} |u|_{H_0^1} \text{, for all }t\in (0,T). \end{align*}
</div>
<p>Let \(\left( H_0^1(0,T)\right)^\prime \) be the dual space of \(H_0^1(0,T)\) and let \((\cdot \,  , \,  \cdot )^\prime \) be the dual pairing between \(\left( H_0^1(0,T)\right)^\prime \) and \(H_0^1(0,T)\), <i class="it">i.e.</i>, for any \(h\in \left( H_0^1(0,T)\right)^\prime \) and \(\phi \in H_0^1(0,T)\), \((h,\phi )^\prime \) stands for the value of the functional \(h\) evaluated at the element \(\phi \). From Riesz’s representation theorem, for each \(h\in \left( H_0^1(0,T)\right)^\prime \), there exists a unique \(u_h\in H_0^1(0,T)\) such that </p>
<div class="displaymath" id="a0000000072">
  \begin{equation*}  \text{$(h,\phi )^\prime =(u_h,\phi )_{H_0^1}$, for every $\phi \in H_0^1(0,T)$}. \end{equation*}
</div>
<p> Therefore, we can define the solution operator \(S\colon \left( H_0^1(0,T)\right)^\prime \to H_0^1(0,T)\), where \(S(h)=u_h\). If \(h \in L^2(0,T)\), then \(S(h)\) admits a representation via Green function \(G(t,s) \colon (0,T)^2 \to \mathbb {R}_+ \) (see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#green" >13</a>
	, 
	Example
	
	1.8.18
	]
</span>\()\), </p>
<div class="displaymath" id="a0000000073">
  \begin{equation*}  G(t,s)=\begin{cases}  s\left( 1-\frac{t}{T}\right),&  s\leq t \\ t\left( 1-\frac{s}{T}\right),&  s\geq t, \end{cases}\end{equation*}
</div>
<p> specifically \(S(h)(t)=\int _0^T G(t,s)h(s)ds\). </p>
<p>Let \(K:=K_2=K_3\) be the cone of nonegative functions from \(H_0^1(0,T)\) and \([a,b]\) be a fixed compact subinterval of \((0,T)\). Additionally, we consider the concave upper semicontinous functionals \(l_2,l_3 \colon K \to \mathbb {R}_+\), </p>
<div class="displaymath" id="a0000000074">
  \begin{equation*}  l_1(u)=l_2(u)=\min _{t\in [a,b]} u(t) \quad \left(u \in K \right) , \end{equation*}
</div>
<p> and the conical sets</p>
<div class="displaymath" id="a0000000075">
  \begin{equation*}  \left( K\right)_{r_j,R_j}=\{ u\in K_j\,  | \,  r_j\leq l_j(u),\,  |u|_{H_0^1}\leq R_j \}  \quad \left(j \in \{  2,3\}  \right), \end{equation*}
</div>
<p> where \(0{\lt}r_j{\lt}R_j\) are real numbers. </p>
<p>Note that the second and third equations from \((\ref{main_system})\) admit a variational form given by the energy functionals \(E_2, E_3 \colon H_0^1(0,T) \times K \times K \to \mathbb {R}\), </p>
<div class="displaymath" id="a0000000076">
  \begin{equation*} \begin{aligned} &  E_2(u,v,w):=\tfrac {1}{2}|v|_{H_0^1}^2-\int _{0}^T F_2(\cdot ,u,v,w)\\ &  E_3(u,v,w):=\tfrac {1}{2}|w|_{H_0^1}^2-\int _{0}^T F_3(\cdot ,u,v,w) \end{aligned}\end{equation*}
</div>
<p>where </p>
<div class="displaymath" id="a0000000077">
  \begin{equation*} \begin{aligned} &  F_2(x,u(x),v(x),w(x)):=\int _0^{v(x)} f_2(x,u(x),s,w(x))ds\\ &  F_3(x,u(x),v(x),w(x)):=\int _0^{w(x)} f_2(x,u(x),v(x),s)ds. \end{aligned}\end{equation*}
</div>
<p> If we identify \(H_0^1(0,T)\) with its dual, we deduce </p>
<div class="displaymath" id="a0000000078">
  \begin{align*} &  E_{22}(u,v,w)=v-Sf_2(u,v,w),\\ &  E_{33}(u,v,w)=w-Sf_3(u,v,w). \end{align*}
</div>
<p> Therefore, the system ?? is equivalent with the following fixed point equation </p>
<div class="displaymath" id="a0000000079">
  \begin{equation*} \begin{cases}  N_{1}(u,v,w)=u, \\ N_{2}(u,v,w)=v, \\ N_{3}(u,v,w)=w, \\ \end{cases}\end{equation*}
</div>
<p>where</p>
<div class="displaymath" id="a0000000080">
  \begin{equation*} \begin{cases}  N_{1}(u,v,w)=Sf_{1}(\cdot ,u,v,w,u^{\prime }) \\ N_{2}(u,v,w)=Sf_{2}(\cdot ,u,v,w) \\ N_{3}(u,v,w)=Sf_{3}(\cdot ,u,v,w). \end{cases}\end{equation*}
</div>
<p> Let us denote </p>
<div class="displaymath" id="a0000000081">
  \begin{equation*}  m:=\min _{t \in [a,b]}\int _0^T G(t,s)ds=\min _{t\in [a,b]}\tfrac {t(T-t)}{2}=\min \left\{ \tfrac {a(T-a)}{2},\tfrac {b(T-b)}{2}\right\} . \end{equation*}
</div>
<p> <div class="theorem_thmwrapper " id="application">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">11</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let the above assumptions be satisfied. Furthermore, assume the following</p>
<ol class="enumerate">
  <li><p>There exist \(a_{ij}, a_{14}{\gt}0\) \(\left( i,j\in \{ 1,3\} \right)\) such that for all real numbers \(x_1, \dots \, , x_4\) and \( \overline{x}_1, \dots \, , \overline{x}_4 \) we have </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F108721520>">
<p>
  <div class="equation_content">
    \begin{equation} \label{conditii_lipschitz} \begin{aligned} &  (x_1-\overline{x}_{1})\left(f_{1}(t,x_{1},...,x_{4})-f_{1}(t,\overline{x}_{1},...,\overline{x}_{4})\right)\leq |x_1-\overline{x}_{1}| \sum _{j=1}^{4}a_{1j}|x_{j}-\overline{x}_{j}|\\ &  (x_i-\overline{x}_{i})\left(f_{i}(t,x_{1},x_2,x_{3})-f_{i}(t,\overline{x}_{1},\overline{x}_2,\overline{x}_{3})\right)\leq |x_i-\overline{x}_{i}| \sum _{j=1}^{3}a_{ij}|x_{j}-\overline{x}_{j}|, \end{aligned} \end{equation}
  </div>
  <span class="equation_label">47</span>
</p>
</div>
<p> where \(i \in \{ 2,3\} \). </p>
</li>
  <li><p>The functions \(f_i(t,x,y,z)\), for \(i=2,3\), satisfy: </p>
<ol class="enumerate">
  <li><p>they are monotonically increasing with respect to the variables \(y\) and \(z\). </p>
</li>
  <li><div class="equation" id="<plasTeX.TeXFragment object at 0x000001F108726C40>">
<p>
  <div class="equation_content">
    \begin{equation} \label{marginire_inf_f_i} \begin{aligned} &  f_i(t,\cdot ,r_2,r_3) \geq \tfrac {r_i}{m (b-a)} \end{aligned} \end{equation}
  </div>
  <span class="equation_label">48</span>
</p>
</div>
<p> and </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F108726E20>">
<p>
  <div class="equation_content">
    \begin{equation} \label{norma_f_i} \begin{aligned} &  |f_i(t,\cdot ,0,0)|_{L^2}\leq \tfrac {\pi }{T}R_2 -\tfrac {T}{\pi }\left(a_{i2}R_2+a_{i3}R_3 \right) \end{aligned} \end{equation}
  </div>
  <span class="equation_label">49</span>
</p>
</div>
<p> for all \(t\in (0,T).\) </p>
</li>
  <li><p>there are real numbers \(M_1,M_2,M_3,M_4 {\gt}0\) such that </p>
<div class="displaymath" id="a0000000082">
  \begin{align*} &  f_2(t,\cdot ,cR_2,cR_3)\leq M_1\text{ ; }f_2(t,\cdot ,0,r_3)\geq M_2 \\ &  f_3(t,\cdot ,cR_2,cR_3)\leq M_3\text{ ; }f_3(t,\cdot ,r_2,0)\geq M_4 \end{align*}
</div>
<p> for every \(t\in (0,T)\) and </p>
<div class="displaymath" id="a0000000083">
  \begin{align*} &  TcR_2M_1-\tfrac {R_2^2}{2}{\lt}r_2(b-a)M_2, \\ &  TcR_3M_3-\tfrac {R_3^2}{2}{\lt}r_3(b-a)M_4. \end{align*}
</div>
</li>
</ol>
</li>
</ol>
<p> Then there exists \((u^\ast ,v^\ast ,w^\ast )\in H_0^1(0,T)\times (K_2)_{r_2,R_2} \times (K_3)_{r_3,R_3}\) a solution of the system ?? such that \((v^\ast ,w^\ast )\) is a Nash equilibrium for the energy functionals of the second and third equations. <div class="proof_wrapper" id="a0000000084">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> The proof entails checking that all condition<a href="index.html#teorema_principala">Theorem 7</a> are fulfilled. </p>
<p><i class="itshape">Verification of the condition (h1)</i>. Let \(u,\overline{u},v,\overline{v},w,\overline{w} \in H_0^1(0,T).\) Then, from ?? we have </p>
<div class="displaymath" id="a0000000085">
  \begin{align*} &  \left( N_1(u,v,w)-N_1(\overline{u},\overline{v},\overline{w})\,  , \,  u-\overline{u}\right)_{H_0^1}= \\ &  = \left(S\, f_{1}(\cdot ,u,v,w,u^\prime )-S\, f_{1}(\cdot ,\overline{u},\overline{v},\overline{w},\overline{u}^\prime ))\,  , \,  u-\overline{u}\right)_{H_0^1}\\ &  =\left(f_{1}(\cdot ,u,v,w,u^\prime )-f_{1}(\cdot ,\overline{u},\overline{v},\overline{w},\overline{u}^\prime )\,  , \,  u-\overline{u} \right)_{L^2} \\ &  \leq a_{11}|u-\overline{u}|^2_{L^2}+a_{12}|v-\overline{v}|_{L^2}|u-\overline{u}|_{L^2}+a_{13}|w-\overline{w}|_{L^2}|u-\overline{u}|_{L^2} \\ &  \quad +a_{14}|u^\prime -\overline{ u}^\prime |_{L^2}|u-\overline{u}|_{L^2} \\ &  \leq \left( a_{11}\tfrac {T^2}{\pi ^2}+a_{14}\tfrac {T}{\pi }\right)|u-\overline{u}|_{H_0^1} ^2 + a_{12}\tfrac {T^2}{\pi ^2}|u-\overline{u}|_{H_0^1}|v-\overline{v}|_{H_0^1}\\ &  \quad +a_{13}\tfrac {T^2}{\pi ^2}|u-\overline{u}|_{H_0^1}|w-\overline{w}|_{H_0^1}. \end{align*}
</div>
<p> Similarly </p>
<div class="displaymath" id="a0000000086">
  \begin{equation*} \begin{aligned} &  \left( N_2(u,v,w)-N_2(\overline{u},\overline{v},\overline{w})\,  , \,  v-\overline{v}\right)_{H_0^1} \leq \\ &  \leq \tfrac {T^2}{\pi ^2}|v-\overline{v}|_{H_0^1}\left[ a_{21}|u-\overline{u}|_{H_0^1}+a_{22}|v-\overline{v}|_{H_0^1}+a_{23}|w-\overline{w}|_{H_0^1}\right], \end{aligned}\end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000087">
  \begin{equation*} \begin{aligned} &  \left( N_3(u,v,w)-N_3(\overline{u},\overline{v},\overline{w})\,  , \,  w-\overline{w}\right)_{H_0^1} \leq \\ &  \leq \tfrac {T^2}{\pi ^2}|w-\overline{w}|_{H_0^1}\left[ a_{31}|u-\overline{u}|_{H_0^1}+a_{23}|v-\overline{v}|_{H_0^1}+a_{33}|w-\overline{w}|_{H_0^1}\right]. \end{aligned}\end{equation*}
</div>
<p> If the matrix </p>
<div class="equation" id="matrix">
<p>
  <div class="equation_content">
    \begin{equation}  A=\tfrac {T^2}{\pi ^2}\begin{bmatrix}  \left(a_{11} +\frac{\pi }{T}a_{41} \right) 

&  a_{12} 

&  a_{13} 

\\ a_{21} 

&  a_{22}

& a_{23}

\\ a_{31} 

& a_{32} 

& a_{33} 

\end{bmatrix} \label{matrix} \end{equation}
  </div>
  <span class="equation_label">50</span>
</p>
</div>
<p> is convergent to zero, then (h1) is fulfilled. </p>
<p>Let \(u\in H_0^1(0,T), v\in (K_2)_{r_2,R_2}\), \(w\in (K_3)_{r_3,R_3}\). Since \(f_2\) has positive values, using assumption (H2) we deduce </p>
<div class="displaymath" id="a0000000088">
  \begin{align*}  N_2(u(t),v(t),w(t))& =\int _0^T G(t,s)f_2(s,u(s),v(s),w(s))ds\\ & \geq \int _a^b G(t,s)f_2(s,u(s),v(s),w(s))ds \\ & \geq \int _a^b G(t,s)f_2(s,u(s),r_2,r_3)ds \\ &  \geq m(b-a) f_2(\xi ,u(\xi ), r_2,r_3) \geq r_2, \end{align*}
</div>
<p> for all \(t\in (0,T)\), where some \(\xi \in (a,b)\). This guarantees that \(l_2(N_2(u,v,w))\geq r_2\). Moreover, if \(|v|_{H_0^1}=R_2\), it follows that \(N_2(u,v,w)\neq (1+\lambda )v\) for every \(\lambda {\gt}0\). Indeed, if we assume the contrary, then </p>
<div class="displaymath" id="a0000000089">
  \begin{align*}  (1+\lambda )|v|_{H_0^1}^2& =(1+\lambda )R_2^2 =(N_2(u,v,w),v)_{H_0^1} \\ &  =(f_2(u,v,w),v)_{L^2} = \int _0^T v(s)f_2(s,u(s),v(s),w(s))ds \\ & = \int _0^T v(s)\left[f_2(s,u(s),v(s),w(s))-f_2(s,u(s),0,0)\right]ds\\ &  \quad +\int _0^T v(s)f_2(s,u(s),0,0)ds. \end{align*}
</div>
<p> Using the monotonicity conditions of \(f_2\), we easily deduce </p>
<div class="displaymath" id="a0000000090">
  \begin{align*}  (1+\lambda ) |v|_{H_0^1}^2& \leq \int _0^T v(s) \left[a_{22}v(s)+a_{23}w(s) \right]ds+|v|_{L^2}|f_2(\cdot ,u(\cdot ),0,0)|_{L^2} \\ &  \leq \tfrac {T^2}{\pi ^2}\left[ a_{22}R_2^2+a_{23}R_2 R_3\right]+\tfrac {T}{\pi }R_2|f_2(\cdot ,u(\cdot ),0,0)|_{L^2}. \end{align*}
</div>
<p> Hence, relation ?? yields </p>
<div class="displaymath" id="a0000000091">
  \begin{equation*}  (1+\lambda ) |v|_{H_0^1}^2\leq R_2^2, \end{equation*}
</div>
<p> which is a contradiction. Applying the same reasoning, a similar relation also holds for \(f_3\), and therefore condition (h2) is verified. </p>
<p><i class="itshape">Verification of conditions (h3), (h4)</i>. One easily sees that for any \(v \in (K_2)_{r_2,R_2}\) and \(w\in (K_3)_{r_3,R_3}\), one has </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F10880C700>">
<p>
  <div class="equation_content">
    \begin{equation} \label{marginire_v_w} 0\leq v(t) \leq cR_2 \text{ ; }0\leq w(t) \leq cR_3. \end{equation}
  </div>
  <span class="equation_label">51</span>
</p>
</div>
<p> Consequently,</p>
<div class="displaymath" id="a0000000092">
  \begin{equation*} \begin{aligned}  |N_1(0,v,w)|& = \int _0^T G(t,s)f_1(s,0,v(s),w(s),0)ds \\ &  \leq \int _0^TG(t,s)f_1(s,0,cR_2,cR_3,0) ds \\ &  \leq T|G|_\infty |f_1(\theta ,0,cR_2,cR_3,0)|_\infty , \end{aligned}\end{equation*}
</div>
<p> for some \(\theta \in (0,T)\). Hence, condition (h4) holds true. </p>
<p>Since </p>
<div class="displaymath" id="a0000000093">
  \begin{equation*}  F_2(t,u(t),v(t),w(t)) \leq cR_2 f_2\left(t,u(t),cR_2,cR_3\right), \text{ for all }t\in (0,T), \end{equation*}
</div>
<p> we can deduce </p>
<div class="displaymath" id="a0000000094">
  \begin{align*}  E_2(u,v,w)&  \geq -\int _0^T cR_2 f_2(x,u(x),cR_2,cR_3)dx \\ &  \geq -cR_2 \int _0^T\left[f_2(x,u(x),cR_2,cR_3)-f_2(x,u(x),0,0) \right]dx\\ &  \quad -cR_2|f(\cdot ,u(\cdot ),0,0)|_{L^1} \\ &  \geq -cR^2_2T \left( a_{22}cR_2+a_{23}cR_3\right)-cR_2|f(\cdot ,u(\cdot ),0,0)|_{L^1}{\gt}-\infty , \end{align*}
</div>
<p> which guarantees that \(E_2\) is lower bounded on \(H_0^1(0,T) \times (K_2)_{r_2,R_2}\times (K_3)_{r_3,R_3}\). Furthermore, when \(l_2(v)=r_2\) and \(|v|_{H_0^1}=R_2\) simultaneously, it follows that </p>
<div class="displaymath" id="a0000000095">
  \begin{align}  \notag E_2(u,v,w)& =\tfrac {|v|_{H_0^1}^2}{2}-\int _0^T\int _0^{v(x)} f_2(x,u(x),v(x),w(x)) ds dx \\ &  \geq \tfrac {R_2^2}{2}-\int _0^T \notag \int _0^{cR_2}f_2(x,u(x),s,cR_3) ds dx \\ &  \geq \notag \tfrac {R_2^2}{2}-\int _0^T cR_2 f_2(x,u(x),cR_2,cR_3)dx \\ & \geq \tfrac {R_2^2}{2} -cR_2TM_1 \label{marginire_inferioara_E2} \geq -r_2(b-a)M_2+\varepsilon . \end{align}
</div>
<p> On the other hand, </p>
<div class="displaymath" id="a0000000096">
  \begin{align}  \notag E_2(u,r_2,w) & = -\int _0^T \int _0^r f_2(x,u(x),s,w(x))dsdx \\ &  \leq -\int _a^b \notag \int _0^{r_2}f_2(x,u(x),0,r_3)ds dx \\ &  =-\int _a^b r_2 f_2(x,u(x),0,r_3)dx \leq -r_2 (b-a)M_2. \label{marginire_superiara_E2} \end{align}
</div>
<p> From ?? and ??, we easily deduce </p>
<div class="displaymath" id="a0000000097">
  \begin{equation*}  E_2(u,v,w) \geq -r_2 (b-a)M_2+\varepsilon \geq E_2(u,r_2,w)+\varepsilon . \end{equation*}
</div>
<p> Following the same reasoning for \(E_3\) and \(f_3\) we obtain </p>
<div class="displaymath" id="a0000000098">
  \begin{equation*}  \inf _{ H_0^1(0,T)\times (K)_{r_2,R_2}\times (K)_{r_3,R_3}} E_3(\cdot ){\gt}-\infty \end{equation*}
</div>
<p> and \( E_3(u,v,w)\geq E_3(u,v,w)+\varepsilon \) whenever \(l_3(w)=r_3\) and \(|w|_{H_0^1}=R_3\) simultaneously. Thus conditions \((h3)\) is satisfied. </p>
<p>As all the assumptions of <a href="index.html#teorema_principala">Theorem 7</a> are satisfied, there exists </p>
<div class="displaymath" id="a0000000099">
  \begin{equation*}  (u^\ast ,v^\ast ,w^\ast )\in H_0^1(0,T) \times (K_2)_{r_2,R_2}\times (K_3)_{r_3,R_3} \end{equation*}
</div>
<p> a solution to system ??, such that \((v^\ast ,w^\ast )\) is a Nash equilibrium for the energy functionals \((E_2,E_3)\). <div class="proof_wrapper" id="a0000000100">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="a0000000101">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">12</span>
  </div>
  <div class="example_thmcontent">
  <p>Let the system</p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F108874F40>">
<p>
  <div class="equation_content">
    \begin{equation} \label{sistem_aplicatie_localizare} \begin{cases}  -u^{\prime \prime }(t)=\bar{a}_1 \left(e^{-u^2(t)}+e^{-\left(u^{\prime }(t)\right)^2}+e^{-v^2(t)}+e^{-w^2(t)}\right) \\ -v^{\prime \prime }(t)=\bar{a}_2\left(e^{-u^2(t)}+\arctan (v(t)+2w(t))+\frac{\pi }{2}\right)\\ -w^{\prime \prime }(t)=\bar{a}_2\left(e^{-u^2(t)}+\arctan (2v(t)+w(t))+\frac{\pi }{2}\right) \end{cases} \text{on} (0,3), \end{equation}
  </div>
  <span class="equation_label">54</span>
</p>
</div>
<p> with Dirichlet boundary conditions </p>
<div class="displaymath" id="a0000000102">
  \begin{equation*} \begin{cases}  u(0)=u(3)=0\\ v(0)=v(3)=0\\ w(0)=w(3)=0, \end{cases}\end{equation*}
</div>
<p> where \(\bar{a}_i\,  (i=1,3)\) are positive real numbers. </p>
<p>We apply the results from <a href="index.html#application">Theorem 11</a> with, </p>
<div class="displaymath" id="a0000000103">
  \[  \begin{aligned} &  f_1\left(x_1, x_2, x_3, x_4\right)=\bar{a}_1\left(e^{-x_1^2}+e^{-x_2^2}+e^{-x_3^2}+e^{-x_4^2}\right) \\ &  f_2\left(x_1, x_2, x_3\right)=\bar{a}_2\left(e^{-x_1^2}+\arctan \left(x_2+x_3\right)+\tfrac {\pi }{2}\right) \\ &  f_3\left(x_1, x_2, x_3\right)=\bar{a}_3\left(e^{-x_2^2}+\arctan \left(x_2+x_3\right)+\tfrac {\pi }{2}\right) \end{aligned}  \]
</div>
<p> Here, we choose \(c = \sqrt{3}\) and set \(R_1 = R_2=\infty \). The value of \(r\) is determined in such a way that, for each \(i=2,3\), there exist suitable constants \(\bar{a}_2\) and \(\bar{a}_3\) such that the following inequality has solutions: </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F10880EB80>">
<p>
  <div class="equation_content">
    \begin{equation} \label{alegere_constanta_r} \bar{a}_i\left(\arctan 2r +\tfrac {\pi }{2}\right) \geq r. \end{equation}
  </div>
  <span class="equation_label">61</span>
</p>
</div>
<p> The closed interval \([a, b]\) is selected to be [1, 2]. As a result </p>
<div class="displaymath" id="a0000000104">
  \begin{equation*}  m=\min \left\{ \tfrac {1(3-1)}{2},\tfrac {2(3-2)}{2}\right\} =1. \end{equation*}
</div>
<p> If the matrix </p>
<div class="displaymath" id="a0000000105">
  \begin{equation*}  A= \tfrac {9}{\pi ^2}\begin{bmatrix}  \bar{a}_1\left( \frac{\pi }{3}+1\right) 

&  \bar{a}_1 

&  \bar{a}_1 

\\ \bar{a}_{2} 

&  \bar{a}_{2} 

&  \bar{a}_{2} 

\\ \bar{a}_{2} 

&  \bar{a}_{3} 

&  \bar{a}_{3} 

\end{bmatrix}\end{equation*}
</div>
<p> is convergent to zero, then the system ?? has a solution \((u^\ast , v^\ast , w^\ast )\) such that \((v^\ast , w^\ast )\) is a Nash equilibrium on \((K)_{r,R}\times (K)_{r,R}\) for the energy functionals associated with the second and third equations. </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000106">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We will demonstrate that all the conditions outlined in <a href="index.html#application">Theorem 11</a> are satisfied. It is clear that the functions \(f_1, f_2, f_3\) are all nonegative. Since these functions are constructed from Lipschitz functions, each with a Lipschitz constant not exceeding 1, the coefficients \(a_{ij}\) are: </p>
<div class="displaymath" id="a0000000107">
  \[  a_{ij} = \bar{a}_i \quad (i=1,3) \text{ and } a_{14} = \overline{a}_1.  \]
</div>
<p> Therefore, condition (H1) is satisfied, given that the matrix \(A\) converges to zero. On the condition (\(H_2\)), the first relation (i) follows immediately, as the function \(\arctan \) is increasing. </p>
<p><i class="itshape">Check of (ii)</i>. For each \(i \in \{ 2, 3\} \), one has: </p>
<div class="displaymath" id="a0000000108">
  \[  f_i(x_1, r, r) \geqslant \bar{a}_i \arctan 2r \geq r  \]
</div>
<p> As we sought solutions without upper bounds, it is not necessary to verify the second conditions from (ii) and condition (iii). Thus, all the assumptions of <a href="index.html#application">Theorem 11</a> are satisfied, which concludes our proof. <div class="proof_wrapper" id="a0000000109">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>In order the straighten our theory, below we present a numerical example for a particular choice of constants \(\bar{a}_i\). <div class="example_thmwrapper " id="a0000000110">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">13</span>
  </div>
  <div class="example_thmcontent">
  <p>Let the system </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x000001F1088C11C0>">
<p>
  <div class="equation_content">
    \begin{equation} \label{sistem_aplicatie_numeric} \begin{cases}  -u^{\prime \prime }(t)=0.1 \left(e^{-u^2(t)}+e^{-\left(u^{\prime }(t)\right)^2}+e^{-v^2(t)}+e^{-w^2(t)}\right) \\ -v^{\prime \prime }(t)=0.40\left(e^{-u^2(t)}+\arctan (v(t)+w(t))\right). \\ -w^{\prime \prime }(t)=0.45\left(e^{-u^2(t)}+\arctan (v(t)+w(t))\right) \end{cases}, \end{equation}
  </div>
  <span class="equation_label">62</span>
</p>
</div>
<p> together with the boundary conditions </p>
<div class="displaymath" id="a0000000111">
  \begin{equation*} \begin{cases}  u(0)=u(3)=0\\ v(0)=v(3)=0\\ w(0)=w(3)=0 \end{cases}. \end{equation*}
</div>
<p> Here, we chose \(r:=\min \{ r_2,r_3\} \) where \(r_2\) and \(r_3\) are the solutions of the equations </p>
<div class="displaymath" id="a0000000112">
  \begin{equation*}  0.4 \arctan \left(2x+\tfrac {\pi }{2}\right)=x \text{ and } 0.45 \arctan \left(2x+\tfrac {\pi }{2}\right)=x. \end{equation*}
</div>
<p>After straightforward calculations, we obtain an approximate solution of \(r\approx 0.478\). </p>
<figure id="fig.1">
  <div class="centered"> <img src="images/img-0001.jpg" alt="\includegraphics[height=4in, width=5.0776in]{figure-Stan.jpg}" style="height:4in; width:5.0776in" />
 </div>
<figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Plot of the solutions</span> 
</figcaption>


</figure>
<p>In <a href="index.html#fig.1">figure 1</a>, you can observe the approximated solutions of the system ??. It is clear that the minimum values of the second and third solutions over the interval \([1, 2]\) surpass the threshold of \(r=0.478\), that is, </p>
<div class="displaymath" id="a0000000113">
  \begin{equation*}  \min _{t\in [1,2]}\{ v(t), w(t)\} \geq r. \end{equation*}
</div>
<p> However, the first equation does not satisfy this requirement since we did not request localization. </p>

  </div>
</div> </p>
<p><small class="footnotesize"><dl class="bibliography">
  <dt><a name="eu">1</a></dt>
  <dd><p><a href ="http://doi.org/10.24193/subbmath.2021.2.14"> <i class="sc">A. Stan</i>, <i class="it">Nonlinear systems with a partial Nash type equilibrium</i>, Stud. Univ. Babeş-Bolyai Math., <b class="bf">66</b> (2021) no. 2, pp. 397–408, <a href="http://doi.org/10.24193/subbmath.2021.2.14">http://doi.org/10.24193/subbmath.2021.2.14</a>. </a> </p>
</dd>
  <dt><a name="bcp">2</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s11784-021-00852-6"> <i class="sc">I. Benedetti, T. Cardinali</i>, <i class="sc">R. Precup</i>, <i class="it">Fixed point-critical point hybrid theorems and application to systems with partial variational structure</i>, J. Fixed Point Theory Appl., <b class="bfseries">23</b> (2021) art. no. 63, <a href="https://doi.org/10.1007/s11784-021-00852-6">https://doi.org/10.1007/s11784-021-00852-6</a>. </a> </p>
</dd>
  <dt><a name="p1">3</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.jmaa.2018.03.035"> <i class="sc">R. Precup</i>,  <i class="it">A critical point theorem in bounded convex sets and localization of Nash-type equilibria of nonvariational systems</i>, J. Math. Anal. Appl., <b class="bfseries">463</b> (2018), pp. 412–431, <a href="https://doi.org/10.1016/j.jmaa.2018.03.035">https://doi.org/10.1016/j.jmaa.2018.03.035</a>. </a> </p>
</dd>
  <dt><a name="p">4</a></dt>
  <dd><p><a href ="https://doi.org/10.1515/anona-2014-0006"> <i class="sc">R. Precup</i>, <em>Nash-type equilibria and periodic solutions to nonvariational systems</em>, Adv. Nonlinear Anal., <b class="bfseries">3</b> (2014) no. 4, pp. 197–207, <a href="https://doi.org/10.1515/anona-2014-0006">https://doi.org/10.1515/anona-2014-0006</a>. </a> </p>
</dd>
  <dt><a name="galewski">5</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.amc.2020.125456"> <i class="sc">M. Beldinski</i>, <i class="sc">M. Galewski</i>, <em>Nash type equilibria for systems of non-potential equations</em>, Appl. Math. Comput., <b class="bfseries">385</b> (2020), pp. 125456, <a href="https://doi.org/10.1016/j.amc.2020.125456">https://doi.org/10.1016/j.amc.2020.125456</a>. </a> </p>
</dd>
  <dt><a name="cournot">6</a></dt>
  <dd><p><i class="sc">A. Cournot</i>, <i class="it">The mathematical principles of the theory of wealth</i>, Economic J., 1838. </p>
</dd>
  <dt><a name="nash_primul_articol">7</a></dt>
  <dd><p><a href ="https://doi.org/10.2307/1969529"> <i class="sc">J. Nash</i>, <em>Non-cooperative games</em>, Ann. Math., <b class="bfseries">54</b> (1951) no. 2, pp. 286–295, <a href="https://doi.org/10.2307/1969529">https://doi.org/10.2307/1969529</a>. </a> </p>
</dd>
  <dt><a name="nonnegative">8</a></dt>
  <dd><p><i class="sc">A. Berman</i>,  <i class="sc">R.J. Plemmons</i>, <em>Nonnegative Matrices in the Mathematical Sciences</em>, Academic Press, 1997. </p>
</dd>
  <dt><a name="role_of_matrices">9</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.mcm.2008.04.006"><i class="sc">R. Precup</i>, <em>The role of matrices that are convergent to zero in the study of semilinear operator systems</em>, Math. Comput. Modelling, <b class="bfseries">49</b> (2009) no. 3, pp. 703–708, <a href="https://doi.org/10.1016/j.mcm.2008.04.006">https://doi.org/10.1016/j.mcm.2008.04.006</a>. </a> </p>
</dd>
  <dt><a name="minty_browder">10</a></dt>
  <dd><p><i class="sc">G. Ciarlet</i>, <em>Linear and Nonlinear Functional Analysis with Applications</em>, SIAM, Philadelphia, 2013. </p>
</dd>
  <dt><a name="brezis">11</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/978-0-387-70914-7"> <i class="sc">H. Brezis</i>, <em>Functional Analysis, Sobolev Spaces and Partial Differential Equations</em>, Springer, New York, 2011, <a href="https://doi.org/10.1007/978-0-387-70914-7">https://doi.org/10.1007/978-0-387-70914-7</a>. </a> </p>
</dd>
  <dt><a name="rp">12</a></dt>
  <dd><p><i class="sc">R. Precup</i>, <em>Linear and Semilinear Partial Differential Equations</em>, De Gruyter, Berlin, 2013. </p>
</dd>
  <dt><a name="green">13</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/978-1-4614-9506-2"> <i class="sc">A. Cabada</i>, <em>Green’s Functions in the Theory of Ordinary Differential Equations</em>, Springer, New York, 2014, <a href="https://doi.org/10.1007/978-1-4614-9506-2">https://doi.org/10.1007/978-1-4614-9506-2</a>. </a> </p>
</dd>
</dl>
</small></p>

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