拉普拉斯变换的性质
贡献者： FFjet
前面我们介绍了拉普拉斯变换，那么这个变换到底给我们带来了什么特别的好处呢？我们下面就来看一下它到底具有哪些性质。
1. 线性性质
若 $\mathscr L[f_1(t)] = \bar f_1(p)$，$\mathscr L[f_2(t)] = \bar f_2(p)$，则
\begin{equation}
\mathscr L[c_1f_1(t)+c_2f_2(t)] = c_1\bar f_1(p) + c_2\bar f_2(p)~,
\end{equation}
证明起来也十分容易。
\begin{equation}
\begin{aligned} \mathscr L[c_{1} f_{1}(t)+c_{2} f_{2}(t)] & = \int_{0}^{\infty}\left[c_{1} f_{1}(t)+c_{2} f_{2}(t)\right] \mathrm{e}^{-\rho t} \mathrm{d} t \\ &=\int_{0}^{\infty} c_{1} f_{1}(t) \mathrm{e}^{-p t} \mathrm{d} t+\int_{0}^{\infty} c_{2} f_{2}(t) \mathrm{e}^{-\rho t} \mathrm{d} t \\ &=c_{1} \bar{f}_{1}(p)+c_{2} \bar{f}_{2}(p) ~.\end{aligned}
\end{equation}
它能让我们干什么呢？我们可以把未知的函数拆成已知函数的线性组合，相加它们的拉普拉斯变换来得到所要的未知函数的拉普拉斯变换。例如：
2. 导数定理
拉普拉斯变换的导数定理是说：
\begin{equation}
\mathscr L[f'(t)]=p\bar f(p)-f(0)~,
\end{equation}
证明同样直接积分就可以了。
\begin{equation}
\begin{aligned} \mathscr L[f^{\prime}(t)] & = \int_{0}^{\infty} f^{\prime}(t) \mathrm{e}^{-p t} \mathrm{d} t=\int_{0}^{\infty} \mathrm{e}^{-p t} \mathrm{d} f \\ &=\left[\mathrm{e}^{-p t} f(t)\right]_{0}^{\infty}-\int_{0}^{\infty} f(t) \mathrm{d}\left(\mathrm{e}^{-p t}\right) \end{aligned}~.
\end{equation}
\begin{equation}
\begin{aligned} \mathscr L[f^{\prime}(t)] & = -f(0)-\int_{0}^{\infty} f(t) \mathrm{d}\left(\mathrm{e}^{-\rho t}\right)=p \int_{0}^{\infty} f(t) \mathrm{e}^{-p t} \mathrm{d} t-f(0) \\ &=p \bar{f}(p)-f(0) \quad\left(\operatorname{Re} p>\sigma_{0}\right) \end{aligned}~.
\end{equation}
我们还可以推广到高阶导数的情形：
\begin{equation}
\mathscr L[f^{(n)}(t)] = p^{n} f(p)-p^{n-1} f(0)-p^{n-2} f^{\prime}(0)-\cdots-p f^{(n-2)}(0)-f^{(n-1)}(0)~.
\end{equation}
有了导数定理，让我们来想想导数的逆运算——积分！有没有积分定理呢？有！
3. 积分定理
\begin{equation}
\mathscr L\left[\int_{0}^{t} \psi(\tau) \mathrm{d} \tau \right]=\frac{1}{p} \mathscr{L}[\psi(t)]~,
\end{equation}
证明其实也一样简单。设 $\displaystyle f(t)=\int_{0}^{t} \psi(\tau) \mathrm{d} \tau$，对 $f(t)$ 应用导数定理，可知：
\begin{equation}
\mathscr[f^{\prime}(t)]=p \mathscr{L}[f(t)]-f(0)=p \mathscr{L}[f(t)]~,
\end{equation}
\begin{equation}
\frac{1}{p} \mathscr{L}[\psi(t)]=\mathscr{L}[f(t)]=\mathscr{L}\left[\int_{0}^{t} \psi(\tau) \mathrm{d} \tau\right]~,
\end{equation}
4. 相似性定理
\begin{equation}
\mathscr{L}[f(at)]=\frac{1}{a} \bar{f}\left(\frac{p}{a}\right)~.
\end{equation}
5. 位移定理
\begin{equation}
\mathscr{L}[\mathrm{e}^{-\lambda t} f(t)]= \bar{f}(p+\lambda)~.
\end{equation}
这个证明也十分简单，留作练习。
6. 延迟定理
\begin{equation}
\mathscr{L}\left [f\left(t-t_{0}\right)\right] =\mathrm{e}^{-p t_{0}} \bar f(p)~,
\end{equation}
用换元积分即可证明。具体来说：
\begin{equation}
\mathscr L\left [f\left(t-t_{0}\right)\right] =\int_{0}^{\infty} f\left(t-t_{0}\right) \mathrm{e}^{-p t} \mathrm{d} t~.
\end{equation}
\begin{equation}
\mathscr L\left [f\left(t-t_{0}\right)\right] =\int_{t_0}^{\infty} f\left(t-t_{0}\right) \mathrm{e}^{-p t} \mathrm{d} t~,
\end{equation}
\begin{equation}
\begin{aligned} \mathscr L\left[f\left(t-t_{0}\right)\right] &=\int_{0}^{\infty} f(\xi) \mathrm{e}^{-p\left(\xi+t_{0}\right)} \mathrm{d} \xi=\mathrm{e}^{-p t_{0}} \int_{0}^{\infty} f(\xi) \mathrm{e}^{-p \xi} \mathrm{d} \xi \\ &=\mathrm{e}^{-p t_{0}} \bar{f}(p) \end{aligned}~.
\end{equation}
7. 卷积定理
卷积定理是说，若 $\mathscr L[f_1(t)] = \bar f_1(p), \mathscr L[f_2(t)] = \bar f_2(p)$，则
\begin{equation}
\mathscr L[f_{1}(t) * f_{2}(t) ]= \bar{f}_{1}(p) \bar{f}_{2}(p)~,
\end{equation}
下面我们来证明它。这个证明稍复杂一些，毕竟它是二重积分：
\begin{equation}
\begin{aligned} & \mathscr{L}\left[f_{1}(t) * f_{2}(t)\right] \\=& \int_{0}^{\infty} f_{1}(t) * f_{2}(t) \mathrm{e}^{-p t} \mathrm{d} t \\=& \int_{0}^{\infty}\left[\int_{0}^{t} f_{1}(\tau) f_{2}(t-\tau) \mathrm{d} \tau\right] \mathrm{e}^{-p t} \mathrm{d} t \end{aligned}~.
\end{equation}
图 1：积分限
我们需要交换积分次序，首先我们来看一下积分区域，它就是图 1 中的绿色区域。由此可见，现在改变积分次序，就是说
\begin{equation}
\begin{aligned} & \mathscr{L}\left[f_{1}(t) * f_{2}(t)\right] \\=& \int_{0}^{\infty}\left[\int_{\tau}^{\infty} f_{2}(t-\tau) \mathrm{e}^{-p t} \mathrm{d} t\right] f_{1}(\tau) \mathrm{d} \tau \end{aligned}~.
\end{equation}
\begin{equation}
\begin{aligned} \mathscr{L}\left[f_{1}(t) * f_{2}(t)\right] &=\int_{0}^{\infty}\left[\int_{0}^{\infty} f_{2}(\xi) \mathrm{e}^{-p \xi} \mathrm{d} \xi\right] f_{1}(\tau) \mathrm{e}^{-p \tau} \mathrm{d} \tau \\ &=\int_{0}^{\infty} f_{1}(\tau) \mathrm{e}^{-p \tau} \mathrm{d} \tau \int_{0}^{\infty} f_{2}(\xi) \mathrm{e}^{-p \xi} \mathrm{d} \xi \\ &=\bar{f}_{1}(p) \bar{f}_{2}(p) \end{aligned}~.
\end{equation}