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The comparison lemma (from Khalil's Nonlinear Systems book) is stated as follows:

Lemma 3.4

Consider the scalar differential equation

$$\dot{u} = f(t,u), \quad u(t_0) = u_0 $$

where $f(t,u)$ is continuous in $t$ and locally Lipschitz in $u$ for all $t \ge0$. Let $[t_0,T)$ be the maximimal interval of existence for the solution $u(t)$. Let $v(t)$ be a continuous function that satisfies the differential inequality

$$\dot{v}(t) \le f(t,v(t)), \quad v(t_0)\le u_0, \quad \forall t\in [t_0,T) $$

Then, $$ v(t) \le u(t), \quad \forall t\in [t_0,T).$$

Question

I have an inequality such as this, but $f(t,u)$ is nontrivial to solve analytically, but I believe I can obtain an asymptotic solution, say

$$ u(t) \sim \tilde{u}(t),\quad t \rightarrow \infty.$$

For, say, $T = \infty$, are there any existing theorems such that it possible to conclude that

$$ v(t) \le \tilde{u}(t), \quad \forall t\in [T_0,\infty),$$

for some $T_0$ sufficiently large?

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1 Answer 1

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Consider the differential inequality:

$$\dot{x}\leq-kx^3$$ where $k$ is a positive constant. This inequality describes the behavior of a system with a negative feedback loop, where the rate of change of the system state $x$ is proportional to the cube of the negative of the system state. This type of system is known as a "cubic system".

To analyze the stability of this system, we can use the Comparison Lemma with a Lyapunov function that is non-negative definite. One such Lyapunov function is: $$V(x)=\frac{1}{2}x^2$$ This function is non-negative definite, because $V(x) = 0$ if and only if $x = 0$, and $V(x) > 0$ for all $x \neq 0$. We can also show that the time derivative of this function along the trajectory of the system is:$$\dot{V}(x)=x\dot{x}\leq-kx^4\leq0$$ where we have used the differential inequality and the fact that $x^4$ is non-negative. Thus, we have shown that the Lyapunov function is decreasing along the trajectory of the system.

Now, let's consider a second system described by the differential equation:$\dot{y}=\sqrt{ky^3}$ with a Lyapunov function:$$W(y)=\frac12y^2\quad\text{}$$ which is positive definite, because $W(y) = 0$ if and only if $y = 0$, and $W(y) > 0$ for all $y \neq 0$. We can also show that the time derivative of this function along the trajectory of the system is:$$\dot W(y)=y\dot y=-ky^4\le0\quad\text{}$$ Thus, we have shown that the Lyapunov function is decreasing along the trajectory of the system.

Now, if we can find a function $z(t)$ that satisfies $x(t) \leq z(t) \leq y(t)$ for all $t$, then we can use the Comparison Lemma to conclude that the first system is stable. One such function is:$$z(t)=\sqrt{\frac{2}{k}}\frac{y(t)^{\frac{3}{2}}}{\sqrt{y(t)^{\frac{3}{2}}+x_0^{\frac{3}{2}}}}$$ where $x_0$ is the initial condition of the first system. We can show that $z(t)$ satisfies the differential inequality and that $x(t) \leq z(t) \leq y(t)$ for all $t$. Therefore, by the Comparison Lemma, we can conclude that the first system is stable, since we have shown that the Lyapunov function of the first system is bounded by the Lyapunov function of the second system.

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