On the competition of two conflicting messages

Abstract

There are plenty of conflicting messages in online social networks. This paper addresses the competition of two conflicting messages. Based on a novel individual-level competing spreading model (the generic UABU model), three criteria for one or two messages to terminate are presented. These criteria manifest the influence of the two message-spreading networks on the evolution of the two messages. Extensive computer simulations show that when a message terminates, the dynamics of a simplified UABU model (the linear UABU model) fits well with the expected evolutionary process of the message. These findings help in understanding the competing spreading process of two conflicting messages.

Appendices

Appendix A: Proof of Lemma 1

Given a sufficiently small time interval $\Delta t >0$, it follows from the total probability formula that

$$\begin{aligned} \begin{aligned}&A_i(t+\Delta t)\\&= \left( 1 - A_i(t) - B_i(t)\right) \Pr \{X_{i}(t+\Delta t)=1 \mid X_{i}(t)=0\} \\&\quad +\, A_i(t) \Pr \{X_{i}(t+\Delta t)=1 \mid X_{i}(t)=1\} \\&\quad +\, B_i(t) \Pr \{X_{i}(t+\Delta t)=1 \mid X_{i}(t)=2\}, 1 \le i \le N. \end{aligned} \end{aligned}$$

(A.1)

By the conditional total probability formula and in view of the model (2), we get that

$$\begin{aligned}&\Pr \{X_{i}(t+\Delta t)=1 \mid X_{i}(t)=0\}\nonumber \\&\quad = \sum _{\mathbf {x} \in \{0, 1, 2\}^N, x_i = 0} \Pr \{X_{i}(t+\Delta t)=1 \mid X_{i}(t)=0, \nonumber \\&\quad \quad \mathbf {X}(t) = \mathbf {x}\} \cdot \Pr \{\mathbf {X}(t) = \mathbf {x} \mid X_{i}(t)=0\}\nonumber \\&\quad = \frac{\Delta t}{1 - A_i(t) - B_i(t)} \cdot \sum _{\mathbf {x} \in \{0, 1, 2\}^N, x_i = 0} \sum _{j = 1}^N \beta _{ij}^{UA} 1_{\{x_j=1\}}\nonumber \\&\quad \quad \cdot \Pr \{\mathbf {X}(t) = \mathbf {x}\} + o(\Delta t)\nonumber \\&\quad = \frac{\Delta t}{1 - A_i(t) - B_i(t)} \cdot \sum _{j = 1}^N \beta _{ij}^{UA} \sum _{\mathbf {x} \in \{0, 1, 2\}^N} 1_{\{x_i = 0, x_j=1\}}\nonumber \\&\quad \quad \cdot \Pr \{\mathbf {X}(t) = \mathbf {x}\} + o(\Delta t) \nonumber \\&\quad = \frac{\Delta t}{1 - A_i(t) - B_i(t)} \sum _{j = 1}^N \beta _{ij}^{UA} \Pr \{X_i(t) = 0,\nonumber \\&\quad \quad X_j(t) = 1\} + o(\Delta t), \quad 1 \le i \le N. \end{aligned}$$

(A.2)

Similarly, we can derive that

$$\begin{aligned}&\Pr \{X_{i}(t+\Delta t)=2 \mid X_{i}(t)=1\} \nonumber \\&\quad = \frac{\Delta t}{A_i(t)} \cdot \sum _{j = 1}^N \beta _{ij}^{AB} \Pr \{X_i(t) = 1, X_j(t) = 2\}\nonumber \\&\qquad +\, o(\Delta t), \quad 1 \le i \le N \end{aligned}$$

(A.3)

and that

$$\begin{aligned}&\Pr \{X_{i}(t+\Delta t) = 0 \mid X_{i}(t)=1\} = \delta _i^A \Delta t+o(\Delta t),\nonumber \\&\quad 1 \le i \le N. \end{aligned}$$

(A.4)

It follows that

$$\begin{aligned}&\Pr \{X_{i}(t+\Delta t)=1 \mid X_{i}(t)=1\}\nonumber \\&\quad = 1-\frac{\Delta t}{A_i(t)} \cdot \sum _{j = 1}^N \beta _{ij}^{AB} \Pr \{X_i(t) = 1, X_j(t) = 2\} \nonumber \\&\qquad -\,\delta _i^A \Delta t + o(\Delta t), \quad 1 \le i \le N. \end{aligned}$$

(A.5)

Besides, we have

$$\begin{aligned}&\Pr \{X_{i}(t+\Delta t)=1 \mid X_{i}(t)=2\} \nonumber \\&\quad = \frac{\Delta t}{B_i(t)} \cdot \sum _{j = 1}^N \beta _{ij}^{BA} \Pr \{X_i(t) = 2, X_j(t) = 1\} \nonumber \\&\qquad +\, o(\Delta t), \quad 1 \le i \le N. \end{aligned}$$

(A.6)

Substituting these equations into Eq. (A.1), rearranging the terms, dividing both sides by $\Delta t$ and letting $\Delta t \rightarrow 0$, we get that

$$\begin{aligned} \frac{\text {d}A_i(t)}{\text {d}t}= & {} \sum _{j = 1}^N \beta _{ij}^{UA} \Pr \{X_i(t) = 0, X_j(t) = 1\}\nonumber \\&+\sum _{j = 1}^N \beta _{ij}^{BA} \Pr \{X_i(t) = 2, X_j(t) = 1\} \nonumber \\&- \sum _{j = 1}^N \beta _{ij}^{AB} \Pr \{X_i(t) = 1, X_j(t) = 2\} \nonumber \\&- \delta _i^A A_i(t), \quad 1 \le i \le N. \end{aligned}$$

(A.7)

The last N equations in Lemma 1 can be derived in an analogous way. The proof is complete.

Appendix B: Proof of Theorem 1

(a) Suppose the model (10) admits a A-dominant equilibrium $\mathbf {E}=(A_1,\ldots ,A_N,0,\ldots , 0)^T$. Let $\mathbf {A}=(A_1,\ldots ,A_N)^T$. We show that $\mathbf {0}< \mathbf {A} < \mathbf {1}$. It follows from the model that

$$\begin{aligned} A_i=\frac{f_i^{UA}(\mathbf {A})}{\delta _i^A+f_i^{UA}(\mathbf {A})}< 1, \quad 1 \le i \le N. \end{aligned}$$

(B.1)

Hence, $\mathbf {A} < \mathbf {1}$. On the contrary, suppose that some $A_k = 0$. It follows from the model (10) that $f_k^{UA}(\mathbf {A})=0$. As $G_A$ is strongly connected, we get that some $\beta _{kl}^{UA} >0$, implying that $A_l = 0$. Repeating this argument, we finally get that $\mathbf {A} = \mathbf {0}$, contradicting the assumption that $\mathbf {E}$ is a A-dominant equilibrium. Hence, $\mathbf {A} > \mathbf {0}$.

Define a continuous mapping $\mathbf {H}=(H_1,\ldots ,H_N)^T\!: (0,1]^N \rightarrow (0,1]^N$ by

$$\begin{aligned} H_i(\mathbf {x})=\frac{f_i^{UA}(\mathbf {x})}{\delta _i^A+f_i^{UA}(\mathbf {x})}, \;\, \mathbf {x}=(x_1,\ldots ,x_N)^T{\in }(0,1]^N. \end{aligned}$$

(B.2)

It suffices to show that $\mathbf {H}$ has a unique fixed point. Let $\mathbf {B}(t) \equiv \mathbf {0}$ and rewrite the model (10) as

$$\begin{aligned} \frac{\text {d}\mathbf {A}(t)}{\text {d}t}=\mathbf {C}_A\mathbf {A}(t)+\mathbf {G}(\mathbf {A}(t)), \end{aligned}$$

(B.3)

where $\mathbf {G}(\mathbf {A}(t))=o(\Vert \mathbf {A}(t)\Vert )$. By Lemma 3, $\mathbf {C}_A$ has a positive eigenvector $\mathbf {v}=(v_1,\ldots ,v_N)^T$ belonging to the eigenvalue $s(\mathbf {C}_A)$. As $s(\mathbf {C}_A) > 0$, we have $\mathbf {C}_A\mathbf {v}=s(\mathbf {C}_A)\mathbf {v} > \mathbf {0}.$ Hence, there is a small $\varepsilon > 0$ such that

$$\begin{aligned} \mathbf {C}_A \cdot (\varepsilon \mathbf {v})+\mathbf {G}(\varepsilon \mathbf {v})=\varepsilon s(\mathbf {C}_A)\mathbf {v}+\mathbf {G}(\varepsilon \mathbf {v})\ge \mathbf {0}, \end{aligned}$$

(B.4)

which is equivalent to $\mathbf {H}(\varepsilon \mathbf {v})\ge \varepsilon \mathbf {v}$. On the other hand, it is easily verified that $\mathbf {H}$ is monotonically increasing, i.e., $\mathbf {u} \ge \mathbf {w}$ implies $\mathbf {H}(\mathbf {u}) \ge \mathbf {H}(\mathbf {w})$. Define a compact convex set as $K=\prod _{i=1}^N[\varepsilon v_i,1]$. Then, $\mathbf {H}|_{K}$ maps K into K. It follows from Lemma 9 that $\mathbf {H}$ has a fixed point in K. Denote this fixed point by $\mathbf {A}^*=(A_1^*,\ldots ,A_N^*)^T$.

Suppose $\mathbf {H}$ has a fixed point $\mathbf {A}^{**}=(A_1^{**},\ldots ,A_N^{**})^T$ other than $\mathbf {A}^{*}$. Let

$$\begin{aligned} \theta =\max _{1 \le i \le N}\frac{A_i^*}{A_i^{**}}, \quad i_0=\arg \max _{1 \le i \le N}\frac{A_i^*}{A_i^{**}}. \end{aligned}$$

(B.5)

Without loss of generality, assume $\theta >1$. It follows that

$$\begin{aligned} \begin{aligned} A_{i_0}^{*}&= H_{i_0}(\mathbf {A}^*)\le H_{i_0}(\theta \mathbf {A}^{**}) \frac{f_{i_0}^{UA}(\theta \mathbf {A}^{**})}{\delta _{i_0}^A +f_{i_0}^{UA}(\theta \mathbf {A}^{**})}\\&\quad <\, \frac{f_{i_0}^{UA}(\theta \mathbf {A}^{**})}{\delta _{i_0}^A+f_{i_0}^{UA}(\mathbf {A}^{**})} \le \frac{\theta f_{i_0}^{UA}(\mathbf {A}^{**})}{\delta _{i_0}^A+f_{i_0}^{UA}(\mathbf {A}^{**})}\\&=\theta H_{i_0}(\mathbf {A}^{**})=\theta A_{i_0}^{**}, \end{aligned} \end{aligned}$$

(B.6)

where $f_{i_0}^{UA}(\theta \mathbf {A}^{**}) \le \theta f_{i_0}^{UA} (\mathbf {A}^{**})$ follows from the concavity of $f_{i_0}^{UA}$. This contradicts the assumption that $A_{i_0}^{*}=\theta A_{i_0}^{**}$. Hence, $\mathbf {A}^{*}$ is the unique fixed point of $\mathbf {H}$. The proof is complete.

(b) The argument is analogous to that for Claim (a) and hence is omitted.

Appendix C: Proof of Theorem 2

Let $(\mathbf {A}(t)^T, \mathbf {B}(t)^T)^T$ be a solution to the model (10). It follows from the first N equations of the model (13), which is an equivalent form of the model (10), that

$$\begin{aligned} \frac{\text {d}\mathbf {A}(t)}{\text {d}t}\le (\mathbf {I}_N-diag\mathbf {A}(t)) \mathbf {f}_{UA} (\mathbf {A}(t))-\mathbf {D}_A \mathbf {A}(t). \end{aligned}$$

(C.1)

Consider the comparison system

$$\begin{aligned} \frac{\text {d}\mathbf {u}(t)}{\text {d}t}=(\mathbf {I}_N-diag\mathbf {u}(t))\mathbf {f}_{UA} (\mathbf {u}(t))-\mathbf {D}_A \mathbf {u}(t) \end{aligned}$$

(C.2)

with $\mathbf {u}(0) = \mathbf {A}(0)$. This system admits the trivial equilibrium $\mathbf {0}$. Moreover, it follows from Lemma 7 that $\mathbf {u}(t) \ge \mathbf {A}(t) \ge \mathbf {0}$. We proceed by distinguishing two possibilities.

Case 1 $s(\mathbf {C}_A)<0$. By Lemma 5, there is a positive definite diagonal matrix $\mathbf {P}_1$ such that $\mathbf {C}_A^T\mathbf {P}_1+\mathbf {P}_1\mathbf {C}_A$ is negative definite. Let $\mathbf {u} = (u_1,\ldots ,u_N)^T$, and define a positive definite function as

$$\begin{aligned} V_1(\mathbf {u})=\mathbf {u}^T \mathbf {P}_1 \mathbf {u}. \end{aligned}$$

(C.3)

By calculations, we get that

$$\begin{aligned} \begin{aligned}&\frac{\text {d}V_1(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)}= 2\mathbf {u}(t)^T \mathbf {P}_1 \frac{\text {d}\mathbf {u}(t)}{\text {d}t}\\&\quad \le 2\mathbf {u}(t)^T \mathbf {P}_1\left[ \mathbf {f}_{UA}(\mathbf {u}(t)) - \mathbf {D}_R\mathbf {u}(t) \right] \\&\quad \le 2\mathbf {u}(t)^T \mathbf {P}_1\mathbf {C}_A \mathbf {u}(t)\\&= \mathbf {u}(t)^T[\mathbf {C}_A^T\mathbf {P}_1+\mathbf {P}_1 \mathbf {C}_A]\mathbf {u}(t) \le 0. \end{aligned} \end{aligned}$$

(C.4)

Here, the second inequality follows from the concavity of $\mathbf {f}_{UA}(\mathbf {x}) - \mathbf {D}_A\mathbf {x}$. Furthermore, $\frac{\text {d}V_1(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)}=0$ if and only if $\mathbf {u}(t)=\mathbf {0}$. According to the LaSalle invariance principle (Corollary 4.1 in [52]), the trivial equilibrium $\mathbf {0}$ of the system (C.2) is asymptotically stable for $[0,1]^N$.

Case 2: $s(\mathbf {C}_A)=0$. By Lemma 6, there is a positive definite diagonal matrix $\mathbf {P}_2$ such that $\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A$ is negative semi-definite. Define a positive definite function as

$$\begin{aligned} V_2(\mathbf {u})=\mathbf {u}^T \mathbf {P}_2 \mathbf {u}. \end{aligned}$$

(C.5)

Similarly, we have

$$\begin{aligned} \frac{\text {d}V_2(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)} \le \mathbf {u}(t)^T[\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A]\mathbf {u}(t) \le 0. \end{aligned}$$

(C.6)

If $\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A$ is negative definite, the subsequent argument is analogous to that for Case 1. Now, assume $\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A$ is not negative definite, which implies

$$\begin{aligned} s(\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A)=0. \end{aligned}$$

(C.7)

As $\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A$ is Metzler and irreducible, it follows from Lemma 3 that (a) 0 is a simple eigenvalue of $\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A$, and (b) up to scalar multiple, $\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A$ has a positive eigenvector belonging to eigenvalue 0. Obviously, $\frac{\text {d}V_2(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)}=0$ if $\mathbf {u}(t)=\mathbf {0}$. On the contrary, suppose $\frac{\text {d}V_2(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)}=0$ for some $\mathbf {u}(t)\ge \mathbf {0}$. If $\mathbf {u}(t) > \mathbf {0}$, then $\mathbf {f}_{UA}(\mathbf {u}(t))>\mathbf {0}$, implying $\frac{\text {d}V_2(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)}<0$, a contradiction. If $\mathbf {u}(t)$ has a zero component, then $\mathbf {u}(t)$ is not an eigenvector of $\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A$ belonging to eigenvalue 0. It follows from the Rayleigh formula (Theorem 4.2.2 in [47]) that

$$\begin{aligned} \mathbf {u}(t)^T[\mathbf {C}_A^T\mathbf {P}_2+\mathbf {P}_2\mathbf {C}_A]\mathbf {u}(t)<0, \end{aligned}$$

(C.8)

implying $\frac{\text {d}V_2(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)}<0$, again a contradiction. Hence, $\mathbf {u}(t)=\mathbf {0}$ if $\frac{\text {d}V_2(\mathbf {u}(t))}{\text {d}t}\mid _{(C.2)}=0$. It follows from the LaSalle invariance principle that the trivial equilibrium $\mathbf {0}$ of the system (C.2) is asymptotically stable with respect to $[0,1]^N$.

Combining Cases 1 and 2, we get $\mathbf {u}(t) \rightarrow \mathbf {0}$ as $t \rightarrow \infty$. According to Lemma 7, we get $\mathbf {A}(t)\le \mathbf {u}(t)$, which implies $\mathbf {R}(t) \rightarrow \mathbf {0}$ as $t \rightarrow \infty$.

Similarly, we can derive that $\mathbf {B}(t) \rightarrow \mathbf {0}$ as $t \rightarrow \infty$. The proof is complete.

Appendix D: Proof of Corollary 1

• (a) We first show $s(\mathbf {C}_A) < 0$. As $\mathbf {C}_A\mathbf {D}_A^{-1}$ is Metzler and irreducible, and it follows from Lemma 3 that $\mathbf {C}_A\mathbf {D}_A^{-1}$ has a positive eigenvector $\mathbf {x}$ belonging to eigenvalue $s(\mathbf {C}_A\mathbf {D}_A^{-1})$. So,

  $$\begin{aligned} \left( \mathbf {C}_A\mathbf {D}_A^{-1}+\mathbf {I}_N\right) \mathbf {x}= \left[ s(\mathbf {C}_A\mathbf {D}_A^{-1})+1\right] \mathbf {x}. \end{aligned}$$

  (D.1)

  That is, $\mathbf {x}$ is an eigenvector of $\mathbf {C}_A\mathbf {D}_A^{-1}+\mathbf {I}_N$ belonging to eigenvalue $s(\mathbf {C}_A\mathbf {D}_A^{-1})+1$. It follows from Lemma 2 that

  $$\begin{aligned} s(\mathbf {C}_A\mathbf {D}_A^{-1}) = \rho (\mathbf {C}_A\mathbf {D}_A^{-1}+\mathbf {I}_N)-1 < 0. \end{aligned}$$

  (D.2)

  By Lemma 5, there is a positive definite diagonal matrix $\mathbf {D}$ such that the matrix

  $$\begin{aligned} \mathbf {P}=(\mathbf {C}_A \mathbf {D}_A^{-1})^T\mathbf {D}+\mathbf {D}(\mathbf {C}_A\mathbf {D}_A^{-1}) \end{aligned}$$

  (D.3)

  is negative definite. Direct calculations give

  $$\begin{aligned} \left[ \mathbf {D}_A^{\frac{1}{2}}\mathbf {C}_A\mathbf {D}_A^{-\frac{1}{2}}\right] ^T \mathbf {D}+\mathbf {D}\left[ \mathbf {D}_A^{\frac{1}{2}}\mathbf {C}_A \mathbf {D}_A^{-\frac{1}{2}}\right] =\mathbf {D}_A^{\frac{1}{2}}\mathbf {P}\mathbf {D}_A^{\frac{1}{2}}. \end{aligned}$$

  (D.4)

  As $\mathbf {D}_A^{\frac{1}{2}}\mathbf {P}\mathbf {D}_A^{\frac{1}{2}}$ is negative definite, $\mathbf {D}_A^{\frac{1}{2}}\mathbf {C}_A\mathbf {D}_A^{-\frac{1}{2}}$ is diagonally stable and hence Hurwitz. It follows that

  $$\begin{aligned} s(\mathbf {C}_A) = s(\mathbf {D}_A^{\frac{1}{2}}\mathbf {C}_A\mathbf {D}_A^{-\frac{1}{2}}) < 0. \end{aligned}$$

  (D.5)

  Similarly, we have $s(\mathbf {C}_B) < 0$. The declared result follows from Theorem 2.

• (b) By the concavity of $f_i^{UA}(\mathbf {x})$, we have $\frac{\partial f_i^{UA}(\mathbf {0})}{\partial x_j}\le \beta _{ij}^{UA}$. That is, $\mathbf {C}_A + \mathbf {D}_A \le \mathbf {M}_{UA}$. Hence,

  $$\begin{aligned} \rho (\mathbf {C}_A\mathbf {D}_A^{-1}+\mathbf {I}_N) \le \rho (\mathbf {M}_{UA}\mathbf {D}_A^{-1}) < 1. \end{aligned}$$

  (D.6)

  Similarly, we have $\rho (\mathbf {C}_B\mathbf {D}_B^{-1}+\mathbf {I}_N) < 1$. The claim follows from Claim (a) of this corollary.

• (c) The claim follows from Claim (b) of this corollary and the well-known inequality $\rho (\mathbf {M}) \le ||\mathbf {M}||_1$.

• (d) The claim follows from Claim (b) of this corollary and the well-known inequality $\rho (\mathbf {M}) \le ||\mathbf {M}||_{\infty }$.

Appendix E: Proof of Theorem 3

Let $(\mathbf {A}(t)^T, \mathbf {B}(t)^T)^T$ be a solution to the model (10) with $\mathbf {A}(0)\ne \mathbf {0}$. It follows from the last N equations of the model (9) that

$$\begin{aligned} \frac{\text {d}\mathbf {B}(t)}{\text {d}t}\le (\mathbf {I}_N-diag(\mathbf {B}(t)))\mathbf {f}_{UB} (\mathbf {B}(t))-\mathbf {D}_B\mathbf {B}(t). \end{aligned}$$

(E.1)

By an argument analogous to that for Theorem 2, we get $\mathbf {B}(t)\rightarrow \mathbf {0}$. Consider the following limit system of model (5).

$$\begin{aligned} \frac{\text {d}\mathbf {u}(t)}{\text {d}t}=(\mathbf {I}_N-diag\mathbf {u}(t))\mathbf {f}_{UA} (\mathbf {u}(t))-\mathbf {D}_A \mathbf {u}(t) \end{aligned}$$

(E.2)

with $\mathbf {u}(0) = \mathbf {A}(0)$. Theorem 1 confirms that the system admits a unique nonzero equilibrium $\mathbf {A}^*=(A_1^*,\ldots ,A_N^*)$. By Lemma 8, it suffices to show that $\mathbf {A}^*$ is asymptotically stable for $(0,1]^N$. Given a solution $\mathbf {u}(t) = (u_1(t), \ldots , u_N(t))^T$ to the system (E.2) with $\mathbf {u}(0) > \mathbf {0}$. First, let us show the following claim.

Claim 1

$\mathbf {u}(t) > \mathbf {0}$ for all $t > 0$.

Proof of Claim 1

On the contrary, suppose there is $t_0 > 0$ such that (a) $\mathbf {u}(t) > \mathbf {0}$, $0< t < t_0$, and (b) $u_i(t_0)=0$ for some i. According to the smoothness of $\mathbf {u}(t)$, we get $\frac{du_{i}(t_0)}{\text {d}t}=0$, implying that $f_{i}^{UA}(\mathbf {u}(t_0))=0$. As $G_A$ is strongly connected, there is j such that $\beta _{ij}^{UA}>0$, which implies $u_{j}(t_0)=0$. Working inductively, we conclude that $\mathbf {u}(t_0)=0$. This contradicts the uniqueness of the solution to the system (E.2) with given initial condition. Claim 1 is proven. $\square$

For $t > 0$, let

$$\begin{aligned} Z(\mathbf {u}(t))=\max _{1 \le k \le N} \frac{u_k(t)}{A_k^*}, \quad z(\mathbf {u}(t))=\min _{1 \le k \le N} \frac{u_k(t)}{A_k^*}. \end{aligned}$$

(E.3)

Define a function $V_3$ as

$$\begin{aligned} V_3(\mathbf {u}(t))=\max \{Z(\mathbf {u}(t))-1,0\}+\max \{1-z(\mathbf {u}(t)),0\}. \end{aligned}$$

(E.4)

It is easily verified that $V_3$ is positive definite with respect to $\mathbf {A}^*$, i.e., (a) $V_3(\mathbf {u}(t))\ge 0$, and (b) $V_3(\mathbf {u}(t))=0$ if and only if $\mathbf {u}(t)=\mathbf {A}^{*}$. Next, let us show that $D^+V_3(\mathbf {u}(t)) \le 0$, where $D^+$ stands for the upper right Dini derivative. To this end, we need to show the following two claims for $t > 0$.

Claim 2

$D^+Z(\mathbf {u}(t))\le 0$ if $Z(\mathbf {u}(t))\ge 1$. Moreover, $D^+Z(\mathbf {u}(t))<0$ if $Z(\mathbf {u}(t))>1$.

Claim 3

$D_+z(\mathbf {u}(t))\ge 0$ if $z(\mathbf {u}(t))\le 1$. Moreover, $D_+z(\mathbf {u}(t))> 0$ if $z(\mathbf {u}(t))<1$. Here, $D_+$ stands for the lower right Dini derivative.

Proof of Claim 2

Choose $k_0$ such that

$$\begin{aligned} Z(\mathbf {u}(t))=\frac{u_{k_0}(t)}{A_{k_0}^*}, \quad D^+Z(\mathbf {u}(t))=\frac{u_{k_0}^{'}(t)}{A_{k_0}^*}. \end{aligned}$$

(E.5)

Then,

$$\begin{aligned} \frac{A_{k_0}^{*}}{u_{k_0}(t)}u_{k_0}^{'}(t) {=}\left( 1-u_{k_0}(t)\right) \frac{A_{k_0}^{*}}{u_{k_0}(t)}f_{k_0}^{UA}(\mathbf {u}(t))-\delta _{k_0}^A A_{k_0}^{*}. \end{aligned}$$

(E.6)

If $f_{k_0}^{UA}(\mathbf {u}(t))=0$, then $\frac{A_{k_0}^{*}}{u_{k_0}(t)}u_{k_0}^{'}(t) <0$, which implies $D^+Z(\mathbf {u}(t))<0$. Now assume $f_{k_0}^{UA}(\mathbf {u}(t))>0$, then

$$\begin{aligned} \begin{aligned} \frac{A_{k_0}^{*}}{u_{k_0}(t)}u_{k_0}^{'}(t)&\le (1-A_{k_0}^{*})\frac{A_{k_0}^{*}}{u_{k_0}(t)}f_{k_0}^{UA}(\mathbf {u}(t))-\delta _{k_0}^A A_{k_0}^{*}\\&\le (1-A_{k_0}^{*})f_{k_0}^{UA}\left( \frac{A_{k_0}^{*}}{u_{k_0}(t)}\mathbf {u}(t)\right) -\delta _{k_0}^A A_{k_0}^{*} \\&\le (1-A_{k_0}^{*})f_{k_0}^{UA}\left( \mathbf {A}^*\right) -\delta _{k_0}^A A_{k_0}^{*}=0, \end{aligned} \end{aligned}$$

(E.7)

where the second inequality follows from the concavity of $f_{k_0}^{UA}$, and the third inequality follows from the monotonicity of $f_{k_0}^{UA}$. This implies $D^+Z(\mathbf {u}(t))\le 0$. Noting that the first inequality is strict if $Z(\mathbf {u}(t))>1$, we get that $D^+Z(\mathbf {u}(t))<0$ if $Z(\mathbf {u}(t))>1$. Claim 2 is proven. $\square$

The argument for Claim 3 is analogous to that for Claim 2 and hence is omitted. Next, consider three possibilities.

• Case 1: $Z(\mathbf {u}(t)) < 1$. Then, $z(\mathbf {u}(t)) < 1$ and $V_3(\mathbf {u}(t)) = 1 - z(\mathbf {u}(t))$. Hence, $D^+V_3(\mathbf {u}(t)) = -D_+z(\mathbf {u}(t)) < 0$.

• Case 2: $z(\mathbf {u}(t)) > 1$. Then, $Z(\mathbf {u}(t)) > 1$ and $V_3(\mathbf {u}(t)) = Z(\mathbf {u}(t)) - 1$. Hence, $D^+V_3(\mathbf {u}(t)) = D^+Z(\mathbf {u}(t)) < 0$.

• Case 3 If $Z(\mathbf {u}(t)) \ge 1$, $z(\mathbf {u}(t)) \le 1$. Then, $V_3(\mathbf {u}(t)) = Z(\mathbf {u}(t)) - z(\mathbf {u}(t))$ and $D^+V_3(\mathbf {u}(t)) = D^+Z(\mathbf {u}(t)) - D_+z(\mathbf {u}(t)) \le 0$. Moreover, the equality holds if and only if $\mathbf {u}(t) = \mathbf {A}^*$.

The declared result follows from the LaSalle invariance principle.