《常微分方程教程》习题2-2,4:一个跟踪问题

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\title{\textbf{《常微分方程教程》习题2-2,4\\[2em]一个跟踪问题}} 

\author{\small{叶卢庆}\\{\small{杭州师范大学理学院,学号:1002011005}}\\{\small{Email:h5411167@gmail.com}}} % Institution
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\begin{exercise}[2-2,4]
跟踪:设某 $A$ 从 $Oxy$  平面的原点出发,沿 $x$ 轴正方向前进;同时某 $B$
从点 $(0,b)$ 开始跟踪 $A$,即 $B$ 的运动方向永远指向 $A$ 并与 $A$ 保持
等距 $b$.试求 $B$ 的光滑运动轨迹.
\end{exercise}
\begin{proof}[解]
设在时刻 $t$ 的时候 $A$ 位于 $(f(t),0)$.其中 $f(0)=0$,且 $f(t)$ 是关于
$t$ 的严格单调增函数.设在时刻 $t$ 的
$B$ 位于 $(P(t),Q(t))$,其中 $P(0)=0,Q(0)=b$.不妨设 $b\neq 0$,否则 $B$
的运动将与 $A$ 重合,这是没什么意思的,再根据对称性不妨设 $b>0$.且由于 $B$ 的路径光滑,因此关于
$t$ 的函数 $P,Q$ 都是连续可微的.由于 $B$ 的方向一直指向 $A$,因此
\begin{equation}
  \label{eq:10.51}
  (P'(t),Q'(t))=k(f(t)-P(t),-Q(t)).
\end{equation}
其中 $k>0$.由于 $A,B$ 间距始终为 $b$,因此
\begin{equation}
  \label{eq:10.52}
  [P(t)-f(t)]^2+Q(t)^2=b^2.
\end{equation}
当 $Q(t)\neq 0$ 时,$Q'(t)$ 也不为0.此时 将(1) 代入 (2) 可得
\begin{equation}
  \label{eq:11.02}
  (P'(t))^2+(Q'(t))^2=b^2k^2=b^2\frac{Q'(t)^{2}}{Q(t)^{2}}.
\end{equation}
于是我们就得到了微分方程
\begin{equation}
  \label{eq:11.54}
  (\frac{P'(t)}{Q'(t)})^2+1=\frac{b^2}{Q(t)^2}.
\end{equation}
也就是
$$
(\frac{dP(t)}{dQ(t)})^2+1=\frac{b^2}{Q(t)^2}.
$$
也即
$$
\frac{dx}{dy}=-\sqrt{(\frac{b}{y})^2-1}.
$$
令 $\frac{b}{y}=\cosh a$.其中 $a\in \mathbf{R}^{+}$,于是,
$$
\frac{dy}{da}=\frac{-b\tanh a}{\cosh a}.
$$
且
$$
\frac{dx}{dy}=-\sinh a.
$$
因此,
$$
\frac{dx}{da}=b(\tanh a)^2=b-b\tanh'a.
$$
因此,
$$
x=ba-b\tanh a+C.
$$
因此,
$$
x=b\cosh^{-1}\frac{b}{y}-b\tanh(\cosh^{-1}\frac{b}{y})+C.
$$
将初始条件 $x=0,y=b$ 代入,解得 $C=0$.于是 $B$ 的光滑轨迹为
$$
x=b\cosh^{-1}\frac{b}{y}-b\tanh(\cosh^{-1}\frac{b}{y}).
$$
通过这个方程,我们发现 $B$ 的运动轨迹和 $A$ 的运动无关!\\

当 $Q(t)=0$ 时,易得 $B$ 已经和 $A$ 同在 $x$ 轴上运动.
\end{proof}
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