20211108 微分跟踪器
微分跟蹤器
引理 1. 設 z(t)z(t)z(t) 是 [0,∞)[0, \infty)[0,∞) 上的連續函數, 且lim?t→∞z(t)=0,\lim _{t \rightarrow \infty} z(t)=0, t→∞lim?z(t)=0,若令x(t)=z(Rt),R>0x(t)=z(Rt), R>0 x(t)=z(Rt),R>0則對任意給定的 T>0T>0T>0, 有 lim?R→∞∫0T∣x(t)∣dt=0.\lim _{R \rightarrow \infty} \int_{0}^{T}|x(t)| dt=0. R→∞lim?∫0T?∣x(t)∣dt=0.
證明：
lim?R→∞∫0T∣x(t)∣dt=lim?R→∞∫0T∣z(Rt)∣dt=lim?R→∞1R∫0T∣z(Rt)∣dRt=lim?R→∞1R∫0RT∣z(t)∣dt=0.\lim _{R \rightarrow \infty} \int_{0}^{T}|x(t)| dt =\lim _{R \rightarrow \infty} \int_{0}^{T}|z(Rt)| dt =\lim _{R \rightarrow \infty} \frac{1}{R} \int_{0}^{T}|z(Rt)| d{Rt} =\lim _{R \rightarrow \infty} \frac{1}{R} \int_{0}^{RT}|z(t)| d{t}=0. R→∞lim?∫0T?∣x(t)∣dt=R→∞lim?∫0T?∣z(Rt)∣dt=R→∞lim?R1?∫0T?∣z(Rt)∣dRt=R→∞lim?R1?∫0RT?∣z(t)∣dt=0.
根據引理 1 及變換:
{s=tRx1(s)=z1(t)+cx2(s)=Rz2(t)\left\{\begin{array}{l} s=\frac{t}{R} \\ x_{1}(s)=z_{1}(t)+c \\ x_{2}(s)=R z_{2}(t) \end{array}\right. ????s=Rt?x1?(s)=z1?(t)+cx2?(s)=Rz2?(t)?
引理 2. 若系統 {z˙1=z2,z˙2=f(z1,z2)\left\{\begin{array}{l} \dot{{z}}_{1}=z_{2}, \\ \dot{z}_{2}=f\left(z_{1}, z_{2}\right) \end{array}\right. {z˙1?=z2?,z˙2?=f(z1?,z2?)?的任意解滿足: z1(t)→0,z2(t)→0(t→∞)z_{1}(t) \rightarrow 0, z_{2}(t) \rightarrow 0 (t \rightarrow \infty)z1?(t)→0,z2?(t)→0(t→∞), 則對任意固定的常數 ccc, 系統 {x˙1=x2x˙2=R2f(x1?c,x2R)\left\{\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}\right. {x˙1?=x2?x˙2?=R2f(x1??c,Rx2??)?的解 x1(t)x_{1}(t)x1?(t) 對于任意 T>0T>0T>0，有
lim?R→∞∫0T∣x1(t)?c∣dt=0\lim _{R \rightarrow \infty} \int_{0}^{T}\left|x_{1}(t)-c\right| d t=0 R→∞lim?∫0T?∣x1?(t)?c∣dt=0
證明：
dx1(s)ds=dz1(t)dtR=Rz˙1(t)=Rz2(t)=x2(s)\frac{\mathrmozvdkddzhkzd x_1(s)}{\mathrmozvdkddzhkzd s} = \frac{\mathrmozvdkddzhkzd z_1(t)}{\mathrmozvdkddzhkzd \frac{t}{R}}=R\dot z_1(t)=R z_2(t) = x_2(s)dsdx1?(s)?=dRt?dz1?(t)?=Rz˙1?(t)=Rz2?(t)=x2?(s)
dx2(s)ds=Rdz2(t)dtR=R2z˙2(t)=R2f(z1,z2)=R2f(x1(s)?c,x2(s)R)\frac{\mathrmozvdkddzhkzd x_2(s)}{\mathrmozvdkddzhkzd s} = \frac{R\mathrmozvdkddzhkzd z_2(t)}{\mathrmozvdkddzhkzd \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) dsdx2?(s)?=dRt?Rdz2?(t)?=R2z˙2?(t)=R2f(z1?,z2?)=R2f(x1?(s)?c,Rx2?(s)?)
因此，系統等價變換成立。
同時，因為有z1(t)→0當t→∞z_1(t) \rightarrow 0 當 t \rightarrow \inftyz1?(t)→0當t→∞，且z1(t)z_1(t)z1?(t)可導
lim?R→∞∫0T∣z1(t)∣dt=0\lim _{R \rightarrow \infty} \int_{0}^{T}\left|z_{1}(t)\right| d t=0 R→∞lim?∫0T?∣z1?(t)∣dt=0
即可得
lim?R→∞∫0T∣x1(t)?c∣dt=0\lim _{R \rightarrow \infty} \int_{0}^{T}\left|x_{1}(t)-c\right| d t=0 R→∞lim?∫0T?∣x1?(t)?c∣dt=0
參考文獻：https://wenku.baidu.com/view/e1ed0cf8aef8941ea76e05e9.html
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