Upper Bound F550F7

1. **Problem statement:** We have the autonomous system of differential equations: $$\begin{cases} x' = -x + xy \cos(x), \\ y' = -y + yz \sin(y), \\ z' = -z, \end{cases}$$ with initial conditions $x(0) = x_0$, $y(0) = y_0$, $z(0) = z_0$. We want to prove that for all $t \geq 0$: $$|x(t)| \leq |x_0| e^{|y_0| e^{|z_0|}} e^{-t}.$$ 2. **Step 1: Solve for $z(t)$** The equation for $z$ is independent: $$z' = -z, \quad z(0) = z_0.$$ This is a simple linear ODE with solution: $$z(t) = z_0 e^{-t}.$$ Hence, $$|z(t)| = |z_0| e^{-t} \leq |z_0|.$$ 3. **Step 2: Analyze $y(t)$** The equation for $y$ is: $$y' = -y + y z \sin(y) = y(-1 + z \sin(y)).$$ Taking absolute values and using $|\sin(y)| \leq 1$: $$|y'| \leq |y| (1 + |z|).$$ More precisely, consider the differential inequality for $|y|$: $$\frac{d}{dt} |y| \leq -|y| + |y| |z| = |y|(-1 + |z|).$$ Since $|z(t)| = |z_0| e^{-t}$, we have: $$\frac{d}{dt} |y| \leq |y|(-1 + |z_0| e^{-t}).$$ 4. **Step 3: Solve the inequality for $|y|$** Rewrite as: $$\frac{d}{dt} |y| \leq |y|(-1 + |z_0| e^{-t}).$$ Divide both sides by $|y|$ (assuming $|y| > 0$): $$\frac{d}{dt} \ln |y| \leq -1 + |z_0| e^{-t}.$$ Integrate from 0 to $t$: $$\ln |y(t)| - \ln |y_0| \leq \int_0^t (-1 + |z_0| e^{-s}) ds = -t + |z_0|(1 - e^{-t}).$$ Exponentiate: $$|y(t)| \leq |y_0| e^{-t + |z_0|(1 - e^{-t})} = |y_0| e^{|z_0|} e^{-t - |z_0| e^{-t}} \leq |y_0| e^{|z_0|} e^{-t}.$$ 5. **Step 4: Analyze $x(t)$** The equation for $x$ is: $$x' = -x + x y \cos(x) = x(-1 + y \cos(x)).$$ Taking absolute values and using $|\cos(x)| \leq 1$: $$\frac{d}{dt} |x| \leq |x|(-1 + |y|).$$ 6. **Step 5: Use the bound on $|y(t)|$ in the inequality for $|x|$** We have: $$\frac{d}{dt} |x| \leq |x|(-1 + |y_0| e^{|z_0|} e^{-t}).$$ Divide both sides by $|x|$ (assuming $|x| > 0$): $$\frac{d}{dt} \ln |x| \leq -1 + |y_0| e^{|z_0|} e^{-t}.$$ Integrate from 0 to $t$: $$\ln |x(t)| - \ln |x_0| \leq \int_0^t (-1 + |y_0| e^{|z_0|} e^{-s}) ds = -t + |y_0| e^{|z_0|} (1 - e^{-t}).$$ Exponentiate: $$|x(t)| \leq |x_0| e^{-t + |y_0| e^{|z_0|} (1 - e^{-t})} = |x_0| e^{|y_0| e^{|z_0|}} e^{-t - |y_0| e^{|z_0|} e^{-t}} \leq |x_0| e^{|y_0| e^{|z_0|}} e^{-t}.$$ 7. **Conclusion:** We have shown that for all $t \geq 0$: $$|x(t)| \leq |x_0| e^{|y_0| e^{|z_0|}} e^{-t},$$ which is the desired inequality. This completes the proof.