I have the following system, a special case of Kuramoto model:
$$ \frac{d\theta_i}{dt}=\frac{\kappa}{N} \sum_{j=1}^{N} \sin(\theta_j - \theta_i) ~~~~~ (i=1, \dots, N)$$
Here $\kappa$ is a positive constant. My goal is to prove the following exponential decay:
$$\max_{1\leq i, j\leq N} | \theta_i(0) - \theta_j(0)| < \frac \pi 2 \implies |\theta_i(t)-\theta_j(t)| \leq Ae^{-ct}, ~\forall t \geq 0 $$ for some positive constants $c$ and $A$ depending on $i$ and $j$.
Could anyone check my proof or suggest an alternative approach?
My attempt:
Claim 1. We may assume that $\sum_{i=1}^{N} \theta_i=0$
proof. Define $\theta_c=\frac{1}{N}\sum_{i=1}^{N} \theta_i$ and set $\tilde{\theta}_i=\theta_i - \theta_c$. Then $$\frac{d}{dt}\theta_c=\frac{1}{N} \sum_{l=1}^{N} \frac{d}{dt} \theta_l = \frac{\kappa}{N} \sum_{i,j=1}^{N} \sin (\theta_j - \theta_i)=0$$ Therefore, $$\frac{d\tilde{\theta_i}}{dt}=\frac{d\theta_i}{dt}=\frac{\kappa}{N} \sum_{j=1}^{N} \sin(\theta_j - \theta_i)=\frac{\kappa}{N} \sum_{j=1}^{N} \sin(\tilde{\theta}_j - \tilde{\theta}_i)$$Moreover $$\max_{i,j} | \theta_i(0) - \theta_j(0) | = \max_{i, j} | \tilde{\theta}_i(0) - \tilde{\theta}_j(0) | $$ $$ | \theta_i(t) - \theta_j(t) | = | \tilde{\theta}_i(t) - \tilde{\theta}_j(t) |, ~~ \forall t \geq 0 $$
Finally, $$\sum_{i=1}^{N} \tilde{\theta}_i = \sum_{i=1}^{N} \left( \theta_i - \frac1N \sum_{j=1}^{N} \theta_j \right) = 0 $$
$\blacksquare$
Claim 2. $$\max_{1\leq i, j\leq N} | \theta_i(0) - \theta_j(0)| < \frac \pi 2 \implies \max_{1\leq i, j\leq N} | \theta_i(t) - \theta_j(t)| < \frac \pi 2, ~~ \forall t \geq 0$$ proof. Pick $C>0$ such that $$\max_{1\leq i, j\leq N} | \theta_i(0) - \theta_j(0)| \leq C< \frac \pi 2$$ and define $$S = \{ T \geq 0 : \max_{i,j} | \theta_i(t) - \theta_j(t)|\leq C, ~~\forall t \in [0, T] \}$$ Then $0 \in S \neq \emptyset$. We will show that $\sup S = \infty$. Otherwise, $\sup S =: T < \infty$. Then there exist $T_1, T_2, \dots $ such that $T- \frac 1n < T_n \leq T $ with $T_n \in S$. Consider $g : t \mapsto \max_{i, j} | \theta_i(t) - \theta_j(t)|$, which is a continuous function. Since $0 \leq g(T_n) \leq C$ for each $n$, we can find a subsequence $\left(T_{n_j}\right)_{j=1}^{\infty}$ such that $g(T_{n_j})$ converges, by the Weierstrass theorem. By continuity of $g$, $g(T_{n_j})\to g(T) \leq C$. Hence $T \in S$. To see this, let $s\in [0, T]$ be given. If $s=T$, then $g(s) \leq C$. If $s<T$, we can pick some $T_m$ such that $s<T_m \in S$. Now consider $g(T)=\max_{i,j} | \theta_i(T) - \theta_j(T) | = \theta_\alpha(T) - \theta_\beta(T)$, where $\theta_\alpha(T) = \max_{i} \theta_i(T)$ and $\theta_\beta(Y)= \min_{j} \theta_j(T)$ ($1 \leq \alpha, \beta \leq N$). If $\theta_\alpha(T) = \theta_\beta(T)$, then $g(T)=0$. In this case, $T+\delta \in S$ for some $\delta >0$ by the continuity of $g$. Otherwise, $$\frac{d\theta_\alpha}{dt}(T)=\frac{\kappa}{N}\sum_{j=1}^{N} \sin \left( \theta_j(T) - \theta_\alpha(T) \right) <0 $$ $$\frac{d\theta_\beta}{dt}(T)=\frac{\kappa}{N}\sum_{j=1}^{N} \sin \left( \theta_j(T) - \theta_\beta(T) \right) >0 $$
Note that $\theta_j(T) - \theta_\alpha(T) \in (-\frac \pi 2, 0] $ and $\theta_j(T) - \theta_\beta(T) \in [0, \frac \pi 2]$. As a result, there exists $\delta > 0 $ such that $g(T+t) \leq g(T)$ for all $t \in [0, \delta]$. $\blacksquare$
Claim 3. $\theta_j - \theta_i $ has a constant sign for every $t \geq 0$.
proof. We will use the global existence-uniqueness theorem from the ODE theory. First, if $\theta_i = \theta_j$ for all $t \geq 0$ ($i \neq j$), then we can rename $\theta_1, \dots, \theta_N$ with $\theta_i \to \theta_1$, $\theta_j \to \theta_N$. Then we can consider a system of $(N-1)$ differential equations:
$$ \frac{d}{dt}\theta_i = \frac{\kappa}{N} \sum_{j=2}^{N} \sin(\theta_j - \theta_i) + \frac{\kappa}{N} \sin(\theta_N - \theta_i) ~~ (i=2,3, \dots, N)$$
This system satisfies the condition of global existence-uniqueness theorem, independent of initial data. Let's see this. Define $g : \mathbb R \times \mathbb R^{N-1} \to \mathbb R^{N-1}$ by $g(t, \theta_2, \dots, \theta_N) = (\frac{d}{dt}\theta_2, \dots, \frac{d}{dt}\theta_N)$, and write $\theta = (\theta_2, \dots, \theta_N)$. Then \begin{align} \lVert g(t, \theta) \rVert^2 = \sum_{i=2}^{N} \left( \frac{\kappa}{N} \sum_{j=2}^{N} \sin(\theta_j - \theta_i) + \frac{\kappa}{N} \sin(\theta_N - \theta_i) \right)^2 \leq \kappa^2 (N-1)\end{align} \begin{align} \lVert g(t, \theta^1) - g(t, \theta^2) \rVert^2 = \sum_{i=2}^{N} \left( \frac{\kappa}{N} \sum_{j=1}^{N} \left( \sin(\theta^1_j - \theta^1_i) - \sin(\theta^2_j - \theta^2_i) \right) \right)^2 \end{align}
At this point we need some estimations. First, $$ \sin(\theta^1_j - \theta^1_i) - \sin(\theta^2_j - \theta^2_i) = 2\cos \left( \frac{\theta^1_j -\theta^1_i + \theta^2_j - \theta^2_i}{2} \right) \sin \left( \frac{(\theta^1_j -\theta^1_i) - (\theta^2_j - \theta^2_i)}{2} \right) $$
Thus \begin{align} \left| \sum_{j=1}^{N} \left( \sin(\theta^1_j - \theta^1_i) - \sin(\theta^2_j - \theta^2_i) \right) \right|^2 &= 4 \left| \sum_{j=1}^{N} \cos \left( \frac{\theta^1_j -\theta^1_i + \theta^2_j - \theta^2_i}{2} \right) \sin \left( \frac{(\theta^1_j -\theta^1_i) - (\theta^2_j - \theta^2_i)}{2} \right) \right|^2 \\ & \leq 4 \left| \sum_{j=1}^{N} \cos^2 \left( \frac{\theta^1_j -\theta^1_i + \theta^2_j - \theta^2_i}{2} \right) \right| \left| \sum_{j=1}^{N} \sin^2 \left( \frac{(\theta^1_j -\theta^1_i) - (\theta^2_j - \theta^2_i)}{2} \right) \right|\\ &\leq 4N \sum_{j=1}^{N} \sin^2 \left( \frac{(\theta^1_j -\theta^1_i) - (\theta^2_j - \theta^2_i)}{2} \right) \\ & \leq 4N\sum_{j=1}^{N} \left| \frac{(\theta^1_j -\theta^1_i) - (\theta^2_j - \theta^2_i)}{2} \right|^2 \\ & \leq N \left( |\theta^1_j - \theta^2_j| + |\theta^1_i - \theta^2_j| \right)^2 \end{align}
Combining this with the CBS inequality, we get $$\sum_{i=1}^{N} \left( \sum_{j=1}^{N} \left( \sin(\theta^1_j - \theta^1_i) - \sin(\theta^2_j - \theta^2_i) \right) \right)^2 \leq 4N^2 \sum_{l=1}^N | \theta^1_l - \theta^2_l |^2$$ To sum up, $\lVert g(t, \theta^1) - g(t, \theta^2) \rVert^2 \leq 8\kappa^2 \lVert \theta^1 - \theta^2 \rVert^2 $
Similarly, come back to our original Kuramoto model and define $f : \mathbb R \times \mathbb R^N \to \mathbb R^N$ by $f(t, \theta) = (\frac{d}{dt}\theta_1, \dots, \frac{d}{dt}\theta_N)$. Then same method yields $\lVert f(t, \theta) \rVert^2 \leq \kappa^2 N$ and $\lVert f(t, \theta^1) - f(t, \theta^2) \rVert^2 \leq 4\kappa^2 \lVert \theta^1 -\theta^2 \rVert ^2$.
Now we can conclude that $\theta_i = \theta_j$ at some $t\geq 0$, then in fact $\theta_i(0) = \theta_j(0)$. $\blacksquare$
Claim 4. $$| \theta_i(0) - \theta_j(0)| < \frac \pi 2 \implies |\theta_i(t)-\theta_j(t)| \leq Ae^{-ct}, ~\forall t \geq 0 $$ for some positive constants $c$ and $A$ depending on $i$ and $j$.
proof. If $\theta_i(0) = \theta_j(0)$, then $\theta_i = \theta_j$ so there is nothing to prove. Suppose that $\theta_i(0)>\theta_j(0)$. Then $\theta_i(t) > \theta_j(t)$ for all $t \geq 0$. Now \begin{align} \frac{d}{dt}(\theta_i - \theta_j) &= \frac{2\kappa}{N} \sin\left( \frac{\theta_j - \theta_i}{2}\right) \sum_{l=1}^{N} \cos \left( \theta_l - \frac{\theta_j + \theta_i}{2}\right) \\ & \leq 2\kappa \sin \left( \frac{\theta_j - \theta_i}{2}\right) \\ &\leq 2\kappa (\theta_j - \theta_i) \\ &= -\kappa(\theta_i - \theta_j) \end{align}
By Gronwall's lemma, \begin{align} (\theta_i - \theta_j)(t) &\leq (\theta_i(0) - \theta_j)(0) \exp \left( \int_{0}^{t} -\kappa ds \right) \\ &= \left(\theta_i(0) - \theta_j(0) \right) e^{-\kappa t}\end{align} $\blacksquare$
I briefly read your argument. It seems that Claim 1 and 2 are not used afterwards. (If not, it would be better to specify the usage of these claims.) Moreover, Claim 3 and 4 yields Claim 2.
Here is an alternative argument using Claim 2, which makes it possible to avoid using Claim 3. Suppose $\max_{1\leq i, j\leq N} | \theta_i(0) - \theta_j(0)| \leq C < \frac \pi 2 $. Then $\max_{1\leq i, j\leq N} | \theta_i(t) - \theta_j(t)| \leq C< \frac \pi 2, ~~ \forall t \geq 0$ by Claim 2.
Now \begin{align} \frac{d}{dt} (\theta_i - \theta_j)^2 &=2(\theta_i - \theta_j)\left(\ \frac{d\theta_i}{dt} - \frac{d\theta_j}{dt} \right) \\ &=2(\theta_i - \theta_j)\frac{2\kappa}{N}\sin\left( \frac{\theta_j - \theta_i}{2}\right) \sum_{i=1}^{N} \cos \left( \theta_l - \frac{\theta_i + \theta_j}{2}\right)\end{align}
Note that $$(\theta_i - \theta_j)\sin \left(\frac{\theta_j - \theta_i}{2} \right) \leq -\frac{1}{\pi}(\theta_i - \theta_j)^2 \leq 0$$ and $$\cos \left( \theta_l - \frac{\theta_i + \theta_j}{2} \right) = \cos\left( \frac{(\theta_l - \theta_i) + (\theta_l-\theta_j)}{2} \right)\geq \cos C >0 $$holds
Thus \begin{align} \frac{d}{dt} (\theta_i - \theta_j)^2 \leq -\frac{4\kappa\cos C}{\pi }(\theta_i - \theta_j)^2 \end{align} Now apply the Gronwall's lemma.