Consider the following system of coupled differential equations \begin{align} \dot{x}_{1}&= -b_1\sin(x_{1})+c(\sin(x_{2})-\sin(x_{3})) \\ \dot{x}_{2}&= a-2c\sin(x_{2})+b_1\sin(x_{1})-b_4\sin(x_{4}) \\ \dot{x}_{3}&=a-2c\sin(x_{3})-b_1\sin(x_{1})+b_4\sin(x_{4}) \\ \dot{x}_{4}&= -b_4\sin(x_{4})-c(\sin(x_{2})-\sin(x_{3})), \\ \end{align} where $x_3(t):=x_{1}(t)+x_{2}(t)-x_{4}(t)$, $a,b_1,b_4,c>0$ and $x_{i}(0):=x_{i,0}$, $i=1,2,3,4$. Let $\varepsilon(t) := b_1\sin(x_{1})-b_4\sin(x_{4})$ and suppose that $a>2c+|\varepsilon(t)|$ for all $t\ge 0$. I'm interested in the behavior of $$ f(x_{2},x_{3}):=\sin(x_{2}-x_{3}). $$ Of course, by using the relation $x_3(t):=x_{1}(t)+x_{2}(t)-x_{4}(t)$, we have $|f(x_{2},x_{3})|=|\sin(x_{1}-x_{4})|\le |x_1|+|x_4|$. I would like however to derive a "tighter" bound on $|f(x_{2},x_{3})|$ which possibly depends on parameter $a$. So far, my approach has been to explicitly write the solution $x_{2}(t)$ and $x_{3}(t)$ according to the accepted answer to this question, namely (after some straightforward manipulations) \begin{align} x_2(t)&= x_{2,0} + at + \int_{x_{2,0}}^{x_2(t)}\frac{2c\sin(s)-\varepsilon}{a-2c\sin(s)+\varepsilon}\mathrm{d}s,\\ x_3(t)&= x_{3,0} + at + \int_{x_{3,0}}^{x_3(t)}\frac{2c\sin(s)+\varepsilon}{a-2c\sin(s)-\varepsilon}\mathrm{d}s. \end{align} Then, after plugging the latter expressions into $f(x_2,x_3)$, we get $$ f(x_2,x_3)=\sin\left(\int_{x_{2,0}}^{x_2(t)}\frac{2c\sin(s)-\varepsilon}{a-2c\sin(s)+\varepsilon}\mathrm{d}s-\int_{x_{3,0}}^{x_3(t)}\frac{2c\sin(s)+\varepsilon}{a-2c\sin(s)-\varepsilon}\mathrm{d}s-(x_{2,0}-x_{3,0})\right). $$
I'm stuck at this point. In particular, it seems to me that the latter expression does not vanish as $|x_1|+|x_4|$ goes to zero (so contradicting a remark above). Indeed there is always a constant term depending on the initial conditions $x_{2,0}$, $x_{3,0}$ that does not cancel out as $|x_1|+|x_4|\to 0$. Perhaps I did a mistake at some point but I couldn't figure it out. So, any help for resolving this contradiction is really appreciated!
