20211108 differential tracker

Differential tracker

lemma 1. set up $z(t)$ yes $[0,\infty )$ Continuous functions on , And $$\underset{t\to \infty }{\lim }z(t)=0,$$ If order $$x(t)=z(Rt),R>0$$ For any given $T>0$, Yes $$\underset{R\to \infty }{\lim }{\int }_0^T|x(t)|dt=0.$$

prove ：
$$\underset{R\to \infty }{\lim }{\int }_0^T|x(t)|dt=\underset{R\to \infty }{\lim }{\int }_0^T|z(Rt)|dt=\underset{R\to \infty }{\lim }\frac{1}{R}{\int }_0^T|z(Rt)|dRt=\underset{R\to \infty }{\lim }\frac{1}{R}{\int }_0^{RT}|z(t)|dt=0.$$

According to lemma 1 And transformation :
$$\{\begin{array}{l}s=\frac{t}{R}\\ x_1(s)=z_1(t)+c\\ x_2(s)=R{z}_2(t)\end{array}$$

lemma 2. If system $$\{\begin{array}{l}{\dot{z}}_1=z_2,\\ {\dot{z}}_2=f\left(z_1,z_2\right)\end{array}$$ The arbitrary solution of satisfies : $z_1(t)\to 0,z_2(t)\to 0(t\to \infty )$, Then for any fixed constant $c$, System $$\{\begin{array}{l}{\dot{x}}_1=x_2\\ {\dot{x}}_2=R^{2}f\left(x_1-c,\frac{x_2}{R}\right)\end{array}$$ Solution $x_1(t)$ For any $T>0$, Yes
$$\underset{R\to \infty }{\lim }{\int }_0^T\left|x_1(t)-c\right|dt=0$$

prove ：
$$\frac{\mathrm{d}{x}_1(s)}{\mathrm{d}s}=\frac{\mathrm{d}{z}_1(t)}{\mathrm{d}\frac{t}{R}}=R{\dot{z}}_1(t)=R{z}_2(t)=x_2(s)$$

$$\frac{\mathrm{d}{x}_2(s)}{\mathrm{d}s}=\frac{R\mathrm{d}{z}_2(t)}{\mathrm{d}\frac{t}{R}}=R^2{\dot{z}}_2(t)=R^{2}f\left(z_1,z_2\right)=R^{2}f\left(x_1(s)-c,\frac{x_2(s)}{R}\right)$$

therefore , The system equivalence transformation holds .

meanwhile , Because there is $z_1(t)\to 0Whent\to \infty$, And $z_1(t)$ Derivable

$$\underset{R\to \infty }{\lim }{\int }_0^T\left|z_1(t)\right|dt=0$$

You can get
$$\underset{R\to \infty }{\lim }{\int }_0^T\left|x_1(t)-c\right|dt=0$$

reference ：https://wenku.baidu.com/view/e1ed0cf8aef8941ea76e05e9.html