The Laplace Transform: The Inverse Transforms

Let $\mathcal{L}^{-1}\{f(s)\}$ denote a function whose Laplace transform is $f(s)$ i.e. if $\mathcal{L}\{F(t)\}=f(s)$ then $F(t)=\mathcal{L}^{-1}\{f(s)\}$. For example,
$$\mathcal{L}^{-1}\left\{\frac{1}{s-k}\right\}=e^{kt},\ \mathcal{L}^{-1}\left\{\frac{k}{s^2+k^2}\right\}=\sin kt$$
$F(t)$ is called the inverse transform of $f(s)$. There is a question that must be addressed. Is the inverse transform of $f(s)$ unique? The answer is not really. For example, we know that $F_1(t)=e^{kt}$ is an inverse transform of $\frac{1}{s-k}$. Define
$$F_2(t)=\left\{\begin{aligned}
e^{kt},\ &0<t<2,\ t>2\\
1,\ &t=2
\end{aligned}\right.$$
The transform of $F_2(t)$ is
\begin{align*}
\mathcal{L}\{F_2(t)\}&=\int_0^\infty e^{-st}F_2(t)dt\\
&=\int_0^2 e^{-st}e^{kt}dt+\int_2^\infty e^{-st}e^{kt}dt\\
&=\frac{1}{s-k}
\end{align*}
So are we in trouble then? Not really. We introduce the following theorem without proof.

Theorem. If two functions $F_1(t)$ and $F_2(t)$ have the same Laplace transform, then $F_2(t)=F_1(t)+N(t)$ where $N(t)$ satisfies
$$\int_0^T N(t)dt=0$$
for every positive $T$.

Such a function $N(t)$ is called a null function. In the above example,
$$N(t)=\left\{\begin{aligned}
1-e^{2k},\ &=t=2\\
0,\ &t\ne 2
\end{aligned}\right.$$
In view of this theorem the inverse transform is essentially unique because a null function is not important in the applications. For this reason this theorem is called the uniqueness of the inverse transform.

Due to the linearity of $\mathcal{L}$, we have
\begin{align*}
\mathcal{L}\{AF(t)+BG(t)\}&=A\mathcal{L}\{F(t)\}+B\mathcal{L}\{G(t)\}\\
&=Af(s)+Bg(s)
\end{align*}
Now
\begin{align*}
\mathcal{L}^{-1}\{Af(s)+Bg(s)\}&=AF(t)+BG(t)\\
&=A\mathcal{L}^{-1}\{f(s)\}+B\mathcal{L}^{-1}\{g(s)\}
\end{align*}
so $\mathcal{L}^{-1}$ is also linear.

Suppose that the Laplace transform of $F(t)$ converges when $s>k$.
$$f(s)=\int_0^\infty e^{-st}F(t)dt\ (s>k)$$
Substitute $s-a$ for $s$. Then
\begin{align*}
f(s-a)&=\int_0^\infty e^{-(s-a)t}F(t)dt\\
&=\int_0^\infty e^{-st}e^{at}F(t)dt\\
&=\mathcal{L}\{e^{at}F(t)\}
\end{align*}
Therefore
\begin{equation}
\mathcal{L}\{e^{at}F(t)\}=f(s-a),\ s>a+k
\end{equation}

Example.
$$\mathcal{L}\{e^{at}t^n\}=\frac{n!}{(s-a)^{n+1}},\ s>a$$

Example.
$$\mathcal{L}\{e^{-at}\cos kt\}=\frac{s+a}{(s+a)^2+k^2},\ s>-a$$

Partial fractions often play a useful role in finding $\mathcal{L}^{-1}$ as seen in the following examples.

Example. Find $\mathcal{L}^{-1}\left\{\frac{s+1}{s^2+2s}\right\}$

Solution. $s^2+2s=s(s+1)$ so let
$$\frac{s+1}{s^2+2s}=\frac{A}{s}+\frac{B}{s+2}$$
Then we have
\begin{align*}
s+1&=A(s+2)+Bs\\
&=(A+B)s+2A
\end{align*}
This means $A+B=1$ and $2A=1$ i.e. $A=B=\frac{1}{2}$. Hence
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{s+1}{s^2+2s}\right\}&=\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{2}\right\}+\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}\\
&=\frac{1}{2}+\frac{1}{2}e^{-2t}
\end{align*}

Example. Find $\mathcal{L}^{-1}\left\{\frac{a^2}{s(s+a)^2}\right\}$ ($a\ne 0$).

Solution. Let
$$\frac{a^2}{s(s+a)^2}=\frac{A}{s}+\frac{B}{s+a}+\frac{C}{(s+a)^2}$$
Then we have
$$a^2=A(s+a)^2+Bs(s+a)+Cs$$
For $s=0$, $a^2=Aa^2$ i.e. $A=1$. For $s=-a$, $a^2=-Ca$ i.e. $C=-a$. For $s=a$, $a^2=3a^2+2a^2B$ which is reduced to $-2a^2=2a^2B$ i.e. $B=-1$. Hence
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{a^2}{s(s+a)^2}\right\}&=\mathcal{L}^{-1}\left\{\frac{1}{2}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\}-a\mathcal{L}^{-1}\left\{\frac{1}{(s+a)^2}\right\}\\
&=1-e^{-at}-ate^{-at}
\end{align*}

Example. Find $\mathcal{L}^{-1}\left\{\frac{s}{(s^2+a^2)(s^2+b^2}\right\}$ ($a^2\ne b^2$).

Solution. As usual one can start with
$$\frac{s}{(s^2+a^2)(s^2+b^2)}=\frac{As+B}{s^2+a^2}+\frac{Cs+D}{s^2+b^2}$$
However one can write
\begin{align*}
\frac{s}{(s^2+a^2)(s^2+b^2)}&=\frac{s}{a^2-b^2}\left[\frac{s^2+a^2-(s^2+b^2)}{(s^2+a^2)(s^2+b^2)}\right]\\
&=\frac{1}{b^2-a^2}\left[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}\right]
\end{align*}
Hence
$$\mathcal{L}^{-1}\left\{\frac{s}{(s^2+a^2)(s^2+b^2)}\right\}=\frac{1}{b^2-a^2}(\cos at-\cos bt)$$

Example. Find $F(t)$ is $f(s)=\frac{5s+3}{(s-1)(s^2+2s+5)}$.

Solution. Let
$$\frac{5s+3}{(s-1)(s^2+2s+5)}=\frac{A}{s-1}+\frac{Bs+C}{s^2+2s+5}$$
Then we obtain $A=1$, $B=-1$ and $C=2$.
\begin{align*}
f(s)&=\frac{1}{s-1}-\frac{s-2}{s^2+2s+5}\\
&=\frac{1}{s-1}-\frac{s-2}{(s+1)^2+4}\\
&=\frac{1}{s-1}-\frac{s+1-3}{(s+1)^2+2^2}\\
&=\frac{1}{s-1}-\frac{s+1}{(s+1)^2+2^2}+\frac{3}{2}\frac{2}{(s+1)^2+2^2}
\end{align*}
Hence
$$F(t)=e^t-e^{-t}\cos 2t+\frac{3}{2}e^{-t}\sin 2t$$

One thought on “The Laplace Transform: The Inverse Transforms

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