【高数+复变函数】Laplace变换的性质

通过上一节【高数+复变函数】Laplace变换的学习，我们知道了Laplace的基本概念：
F ( s ) = ∫ 0 + ∞ f ( t ) e − s t d t F(s)=\int_0^{+\infty} f(t) \mathrm{e}^{-s t} \mathrm{~d} t F(s)=∫0+∞f(t)e−st dt
这一节我们学习Laplace变换的一些常用性质。

一、性质

1. 线性性质

L [ α f 1 ( t ) + β f 2 ( t ) ] = α L [ f 1 ( t ) ] + β L [ f 2 ( t ) ] , \mathscr{L}\left[\alpha f_1(t)+\beta f_2(t)\right]=\alpha \mathscr{L}\left[f_1(t)\right]+\beta \mathscr{L}\left[f_2(t)\right] \text {, } L[αf1(t)+βf2(t)]=αL[f1(t)]+βL[f2(t)],

它的证朋只需根据定义,利用积分性质就可推出

例：求 f ( t ) = s i n h t f(t)=sinht f(t)=sinht的Laplace变换
L [ f ( t ) ] = L [ e t − e − t 2 ] = 1 2 [ L [ e t ] + L [ e − t ] ] = 1 2 [ 1 s − 1 + 1 s + 1 ] = 1 s 2 + 1 \mathscr{L}[f(t)]=\mathscr{L}[\frac{e^t-e^{-t}}{2}]=\frac{1}{2}[\mathscr{L}[e^t]+\mathscr{L}[e^{-t}]]=\frac{1}{2}[\frac{1}{s-1}+\frac{1}{s+1}]=\frac{1}{s^2+1} L[f(t)]=L[2et−e−t]=21[L[et]+L[e−t]]=21[s−11+s+11]=s2+11

2. 微分性质

L [ f ′ ( t ) ] = s F ( s ) − f ( 0 ) . \mathscr{L}\left[f^{\prime}(t)\right]=s F(s)-f(0) . L[f′(t)]=sF(s)−f(0).

它的证明只需根据定义,利用分部积分性质就可推出

推广：
S [ f ′ ′ ( t ) ] = s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 ) \mathscr{S}\left[f^{\prime \prime}(t)\right]=s^2 F(s)-s f(0)-f^{\prime}(0) S[f′′(t)]=s2F(s)−sf(0)−f′(0)
L [ f ( n ) ( t ) ] = s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ′ ( 0 ) − ⋯ − f ( n − 1 ) ( 0 ) \mathscr{L}\left[f^{(n)}(t)\right] = s^n F(s)-s^{n-1} f(0)-s^{n-2} f^{\prime}(0)-\cdots-f^{(n-1)}(0) L[f(n)(t)]=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−f(n−1)(0)
可以正反两用，求 E [ f ( n ) ( t ) ] \mathscr{E}\left[f^{(n)}(t)\right] E[f(n)(t)]或者 F ( s ) F(s) F(s)

例1 求函数 f ( t ) = t m f(t)=t^m f(t)=tm 的 Laplace 变换, 其中 m m m 是正整数

由于 f ( 0 ) = f ′ ( 0 ) = ⋯ = f ( m − 1 ) ( 0 ) = 0 f(0)=f^{\prime}(0)=\cdots=f^{(m-1)}(0)=0 f(0)=f′(0)=⋯=f(m−1)(0)=0, 而 f ( m ) ( t ) = m f^{(m)}(t)=m f(m)(t)=m !所以：
L [ m ! ] = s n F ( s ) \mathscr{L}[m!]=s^nF(s) L[m!]=snF(s)
而 L [ m ! ] = m ! L [ 1 ] = m ! s \mathscr{L}[m!]=m!\mathscr{L}[1]=\frac{m!}{s} L[m!]=m!L[1]=sm!，其中 L [ 1 ] \mathscr{L}[1] L[1]可理解成 f ( t ) = 1 f(t)=1 f(t)=1

所以
F ( s ) = m ! s n + 1 ( 由 L [ 1 ] 产生 R e s > 0 ) F(s)=\frac{m!}{s^{n+1}}(由\mathscr{L}[1]产生Res>0) F(s)=sn+1m!(由L[1]产生Res>0)
例2 求函数 f ( t ) = cos ⁡ k t f(t)=\cos k t f(t)=coskt 的 Laplace 变换.

解由于 f ( 0 ) = 1 , f ′ ( 0 ) = 0 , f ′ ′ ( t ) = − k 2 cos ⁡ k t f(0)=1, f^{\prime}(0)=0, f^{\prime \prime}(t)=-k^2 \cos k t f(0)=1,f′(0)=0,f′′(t)=−k2coskt, 则有
即
− k 2 L [ cos ⁡ k t ] = s 2 L [ cos ⁡ k t ] − s , -k^2 \mathscr{L}[\cos k t]=s^2 \mathscr{L}[\cos k t]-s, −k2L[coskt]=s2L[coskt]−s,
移项化简得
L [ cos ⁡ k t ] = s s 2 + k 2 ( Re ⁡ ( s ) > 0 ) \mathscr{L}[\cos k t]=\frac{s}{s^2+k^2} \quad(\operatorname{Re}(s)>0) L[coskt]=s2+k2s(Re(s)>0)

利用了cos二阶导的不变性

3. 像函数的微分性质

F ′ ( s ) = − L [ t f ( t ) ] F^{\prime}(s)=-\mathscr{L}[t f(t)] F′(s)=−L[tf(t)]

推广:
F ( n ) ( s ) = ( − 1 ) n L [ t n f ( t ) ] F^{(n)}(s)=(-1)^n \mathscr{L}\left[t^n f(t)\right] F(n)(s)=(−1)nL[tnf(t)]
例3 求函数 f ( t ) = t sin ⁡ k t f(t)=t \sin k t f(t)=tsinkt 的 Laplace 变换.

令 g ( t ) = s i n k t g(t)=sinkt g(t)=sinkt
L [ t g ( t ) ] = − F ′ ( g ( t ) ) = − d d s ( k s 2 + k 2 ) = 2 k s ( s 2 + k 2 ) 2 , Re ⁡ ( s ) > 0 \mathscr{L}[tg(t)]=-F^{'}(g(t))=-\frac{d}{d s}\left(\frac{k}{s^2+k^2}\right)=\frac{2 k s}{\left(s^2+k^2\right)^2}, \quad \operatorname{Re}(s)>0 L[tg(t)]=−F′(g(t))=−dsd(s2+k2k)=(s2+k2)22ks,Re(s)>0

4. 积分性质

若 L [ f ( t ) ] = F ( s ) \mathscr{L}[f(t)]=F(s) L[f(t)]=F(s), 则
L [ ∫ 0 t f ( t ) d t ] = 1 s F ( s ) . \mathscr{L}\left[\int_0^t f(t) \mathrm{d} t\right]=\frac{1}{s} F(s) . L[∫0tf(t)dt]=s1F(s).
利用 L [ f ( t ) ] \mathscr{L}[f(t)] L[f(t)]作为中介进行转换。

推广：
L [ ∫ 0 t d t ∫ 0 t d t ⋯ ∫ 0 t f ( t ) d t ] = 1 s n F ( s ) . \mathfrak{L}\left[\int_0^t \mathrm{~d} t \int_0^t \mathrm{~d} t \cdots \int_0^t f(t) \mathrm{d} t\right]=\frac{1}{s^n} F(s) . L[∫0t dt∫0t dt⋯∫0tf(t)dt]=sn1F(s).

5. 象函数的积分性质

L [ f ( t ) t ] = ∫ s ∞ F ( s ) d s \mathscr{L}\left[\frac{f(t)}{t}\right]=\int_s^{\infty} F(s) \mathrm{d} s L[tf(t)]=∫s∞F(s)ds

证明：
∫ s ∞ F ( u ) d u = ∫ s ∞ d u ∫ 0 + ∞ f ( t ) e − u t d t = ∫ 0 + ∞ f ( t ) d t ∫ s ∞ e − u t d u = ∫ 0 + ∞ f ( t ) t e − s t d t = L [ f ( t ) t ] . \begin{aligned} \int_s^{\infty} F(u) \mathrm{d} u & =\int_s^{\infty} \mathrm{d} u \int_0^{+\infty} f(t) e^{-u t} \mathrm{~d} t \\ & =\int_0^{+\infty} f(t) \mathrm{d} t \int_s^{\infty} e^{-u t} \mathrm{~d} u \\ & =\int_0^{+\infty} \frac{f(t)}{t} e^{-s t} \mathrm{~d} t=\mathfrak{L}\left[\frac{f(t)}{t}\right] . \end{aligned} ∫s∞F(u)du=∫s∞du∫0+∞f(t)e−ut dt=∫0+∞f(t)dt∫s∞e−ut du=∫0+∞tf(t)e−st dt=L[tf(t)].
令 s = 0 s=0 s=0可得
∫ 0 + ∞ f ( t ) t d t = ∫ 0 ∞ F ( s ) d s \int_0^{+\infty} \frac{f(t)}{t} \mathrm{~d} t=\int_0^{\infty} F(s) \mathrm{d} s ∫0+∞tf(t) dt=∫0∞F(s)ds
例4 求函数 f ( t ) = sinh ⁡ t t f(t)=\frac{\sinh t}{t} f(t)=tsinht 的 Laplace 变换

前面已经证得 L [ sinh ⁡ t ] = 1 s 2 − 1 \mathscr{L}[\sinh t]=\frac{1}{s^2-1} L[sinht]=s2−11,所以
L [ sinh ⁡ t t ] = ∫ 1 ∞ L [ sinh ⁡ t ] d s = ∫ s ∞ 1 s 2 − 1 d s = 1 2 ln ⁡ s − 1 s + 1 ∣ 0 ∞ = 1 2 ln ⁡ s + 1 s − 1 \mathscr{L}\left[\frac{\sinh t}{t}\right] =\int_1^{\infty} \mathscr{L}[\sinh t] \mathrm{d} s=\int_s^{\infty} \frac{1}{s^2-1} \mathrm{~d} s =\left.\frac{1}{2} \ln \frac{s-1}{s+1}\right|^{\infty}_0=\frac{1}{2} \ln \frac{s+1}{s-1} L[tsinht]=∫1∞L[sinht]ds=∫s∞s2−11 ds=21lns+1s−1 0∞=21lns−1s+1

通常我们还可以运用此积分性质计算一些复杂积分：

如果积分 ∫ 0 + x f ( t ) t d t \int_0^{+x} \frac{f(t)}{t} \mathrm{~d} t ∫0+xtf(t) dt 存在, 取 s = 0 s=0 s=0, 则有
∫ 0 + ∞ f ( t ) t d t = ∫ 0 + ∞ F ( s ) d s ∴ 狄利克雷积分 ∫ 0 + ∞ sin ⁡ t t d t = ∫ 0 ∞ 1 s 2 + 1 d s = arctan ⁡ s ∣ 0 ∞ = π 2 . \int_0^{+\infty} \frac{f(t)}{t} \mathrm{~d} t=\int_0^{+\infty} F(s) \mathrm{d} s\\ \therefore 狄利克雷积分\int_0^{+\infty} \frac{\sin t}{t} \mathrm{~d} t=\int_0^{\infty} \frac{1}{s^2+1} \mathrm{~d} s=\left.\arctan s\right|_0 ^{\infty}=\frac{\pi}{2} . ∫0+∞tf(t) dt=∫0+∞F(s)ds∴狄利克雷积分∫0+∞tsint dt=∫0∞s2+11 ds=arctans∣0∞=2π.

6. 位移性质

L [ e a t f ( t ) ] = F ( s − a ) ( Re ⁡ ( s − a ) > c ) ∵ L [ e a t f ( t ) ] = ∫ 0 + ∞ e a t f ( t ) e − s t d t = ∫ 0 + ∞ f ( t ) e − ( s − a ) t d t \mathscr{L}\left[\mathrm{e}^{at} f(t)\right]=F(s-a) \quad(\operatorname{Re}(s-a)>c) \\\because \mathscr{L}\left[\mathrm{e}^{a t} f(t)\right]=\int_0^{+\infty} \mathrm{e}^{a t} f(t) \mathrm{e}^{-s t} \mathrm{~d} t=\int_0^{+\infty} f(t) \mathrm{e}^{-(s-a) t} \mathrm{~d} t L[eatf(t)]=F(s−a)(Re(s−a)>c)∵L[eatf(t)]=∫0+∞eatf(t)e−st dt=∫0+∞f(t)e−(s−a)t dt

这个性质表朋了一个象原函数乘函数 e a t \mathrm{e}^{at} eat 的 Laplace 变换等于其象函数作位移 a a a.

7. 延迟性质

若 L [ f ( t ) ] = F ( s ) \mathscr{L}[f(t)]=F(s) L[f(t)]=F(s), 又 t < 0 t<0 t<0 时 f ( t ) = 0 f(t)=0 f(t)=0, 则对于任一非负实数 τ \tau τ, 有
L [ f ( t − τ ) ] = e − s τ F ( s ) 变量代换证明 L [ f ( t − τ ) u ( t − τ ) ] = e − s τ F ( s ) 看图理解，只有 t > τ 时有用 \mathfrak{L}[f(t-\tau)]=e^{-s \tau} F(s)变量代换证明\\ \mathfrak{L}[f(t-\tau) u(t-\tau)]=e^{-s \tau} F(s)看图理解，只有t>\tau时有用 L[f(t−τ)]=e−sτF(s)变量代换证明L[f(t−τ)u(t−τ)]=e−sτF(s)看图理解，只有t>τ时有用

Notice：when t<0, f(t) = 0

8. 相似性质

设 F(s)=L[f(t)], 则 L[f(at)]=1aF(sa)(a>0), 其中 Re⁡(s)>as0.
证明：
L [ f ( a t ) ] = ∫ 0 + ∞ f ( a t ) e − s t d t u = a t = 1 a ∫ 0 + ∞ f ( u ) e − s a u d u = 1 a F ( s a ) \begin{aligned} \mathfrak{L}[f(a t)] & =\int_0^{+\infty} f(a t) e^{-s t} \mathrm{~d} t \quad u=a t \\ & =\frac{1}{a} \int_0^{+\infty} f(u) e^{-\frac{s}{a} u} \mathrm{~d} u \\ & =\frac{1}{a} F\left(\frac{s}{a}\right) \end{aligned} L[f(at)]=∫0+∞f(at)e−st dtu=at=a1∫0+∞f(u)e−asu du=a1F(as)