As was told in the introduction, Laplace transform can handle e.g. impulse or periodic function as the driving function. We are going to cover some more tools and operations here.

We have experience that solving DE may led to solution which has to be prefixed by $t,\ t^2,\ \dots$. For example $y^{(4)} - 3y''+2y' = 0$ has solution $y_c = c_1+c_2\color{red}{e^{t}} + c_3\color{red}{te^t} -c_4e^{-2t}$.

That might happen also to solutions based on $\sin(), \cos()$ as well. Such functions might appear not only within the solution, but also on the right side of DE.

It was told that ${\scr L}[e^{at}t^n] = n!/(s-a)^{n+1}$ and the general rule is

$$
\begin{align}
\text{If}\ F(s)={\scr L}[f(t)],\ \text{then}\nonumber \\
\color{red}{{\scr L}[t^n\cdot f(t)] = (-1)^n \frac{d^n}{ds^n}F(s)} \label{ref:lapla_transf_tn}
\end{align}
$$

Find ${\scr L}[t\cdot e^t]$ and ${\scr L}[t^2\cdot e^t]$ using the rule $(\ref{ref:lapla_transf_tn})$.

If $f(t) = \color{red}{e^t}$ then the Laplace transform is $F(s) = 1/(s-1)$. That is easily solved or the tables can be used. Now let us find ${\scr L}[t\cdot \color{red}{e^t}]$ and ${\scr L}[t^2\cdot \color{red}{e^t}]$. For that purpose, let us differentiate $F(s)$ once and twice.

$$ \begin{align} \underline{\underline{{\scr L}\left[t\cdot e^t\right]}} &= (-1)\frac{d}{ds}\cdot\frac{1}{s-1} = \underline{\underline{\frac {1}{(s-1)^2}}} \nonumber \\ \underline{\underline{{\scr L}\left[t^2\cdot e^t\right]}} &= (-1)^2\frac{d^2}{ds^2}\cdot\frac{1}{(s-1)^2} = \underline{\underline{2\frac {1}{(s-1)^3}}} \nonumber \end{align} $$Solving DE by Laplace transform sometimes leads to such **fractions which are not
easily solved by partial fraction decomposition. We would like to take the
inverse way of such transforms.**

For example $\frac{1}{s(s^2+1)}$ might be not obvious how to be solved and then convolution theorem is a handy tool.

The convolution is defined by the following integral.

$$ \begin{align} \color{red}{f \ast g = g \ast f = \int_{\tau = 0}^t f(\tau)\ g(t-\tau)\ d\tau}. \label{ref:lapl_convol} \\ \end{align} $$The convolution theorem states that

$$ {\scr L}\left[ f(t) \right]\cdot{\scr L}\left[ g(t) \right] = F(s)\cdot G(s) = {\scr L}\left[ f \ast g\right ], \\ $$
and we are more interested into the inverse form of the theorem:
$$
\begin{equation}
\color{red}{{\scr L}^{-1}\left[F(s)\cdot G(s)\right] = f \ast g } \label{ref:lapl_convol1}
\end{equation}
$$

Use the convolution theorem to find inverse transform of $F(s) = \frac{1}{s(s^2+1)}$.

We can detect two fractions, which are products of $f(t) = \sin(t) $ and $g(t) = 1$. But this time they are multiplied together and how to separate them into two fractions? If we do not know, we may feed them according to $(\ref{ref:lapl_convol1})$ into $(\ref{ref:lapl_convol})$.

$$ {\scr L}^{-1}\left[ \frac{1}{s^2+1} \cdot \frac{1}{s}\right] = \begin{vmatrix} f(t) = \sin t \\ g(t) = 1 \end{vmatrix} = f \ast g\\ \int_{\tau = 0}^t \sin \tau \cdot 1\ d\tau = \left.-\cos \tau \right|_0^t = -\cos t + \cos 0 \implies \\ \implies \underline{\underline{{\scr L}^{-1}\left[\frac{1}{s(s^2+1)}\right] = 1-\cos t}} $$Note that the result could be also achieved by partial fraction decomposition if one knows all the rules.

$$ \begin{align} F(s) = \frac{1}{s(s^2+1)} &= \frac{A}{s} + \frac{Bs + C}{s^2 + 1} \nonumber \\ 1 &= A(s^2 + 1) + Bs^2 + Cs \nonumber \\ 1 &= s^2(A+B) +s(C) + 1(A) \implies A=1,\ B=-1,\ C=0 \implies \nonumber \\ \implies \frac{1}{s(s^2+1)} &= \frac{1}{s} - \frac{s}{s^2 + 1} \nonumber \\ \end{align} $$And so we have approached the same result of inverse transform

$$ \underline{\underline{{\scr L}^{-1}\left[F(s)\right] = 1 - \cos t}}. $$In many applications the nonhomogeneous term in a linear DE is a periodic function. For example $\sin(t)$ and $\cos(t)$ are periodic function with period of $T=2\pi$. The Laplace transform of periodic function is

$$ \begin{equation} \color{red}{ {\scr L}[f(t)] = \frac{{\scr L}[f_T(t)]}{1-e^{-sT}}} \hskip2em \text{where}\\ f_T(t) = \left\{ \begin{aligned} f(t) & \hskip2em 0 \leq t \leq T, \\ 0 & \hskip2em t \gt T \end{aligned} \right. \label{ref:lapl_perod} \end{equation} $$So it is Laplace transform of the first period divided by $1-e^{-sT}$

Determine the Laplace transform of function

$$ f(t) = \left\{ \begin{aligned} 1 \hskip2em & 0 \leq t \leq T/2 \\ 0 \hskip2em & T/2 \lt t \lt T \\ \end{aligned} \right. \hskip2em \left. f(t+T) = f(t),\ t \geq 0 \right. $$The function is periodic with a period $T$. We have to evaluate Laplace transform of the first period and then use theorem $(\ref{ref:lapl_perod})$.

$$ \begin{align} {\scr L}[f_T(t)] &= \int_0^T f(t)e^{-st}\ dt = \int_0^{T/2}1\cdot e^{-st}\ dt = \nonumber \\ &=\left.-\frac{1}{s}e^{-st}\right|_{t=0}^{T/2} = -\frac 1{s}e^{-s{T}/{2}} - (-\frac 1 s e^0) = \frac{1-e^{-s{T}/{2}}}{s} \nonumber \end{align} $$The Laplace transform of the first period was evaluated by means of integral. Now substitute to $(\ref{ref:lapl_perod})$:

$$ {\scr L}[f(t)] = \frac{\frac{1-e^{-sT/2}}{s}}{1-e^{-s{T}}} = \frac{1-e^{-s{T}/{2}}}{s(1-e^{-sT})} \\ \underline{\underline{{\scr L}[f(t)] =\frac{1}{s(1+e^{-s{T}/{2}})}}}, \hskip2em s\gt 0 $$The last simplification comes from $\frac{1-a}{1-a^2} = \frac{1}{1+a}$.

The other useful property of Laplace transform is that it can apply an impulse as a driving function $f(t)$. It can be either force, voltage or some other physical phenomena.

We want to have a function which in general has a value of zero for any $t$
except the peak within requested short timeframe. As we have e.g. unit time
function already, we would like if such "peak" function is *unit* also, namely
the area below the function is 1. Such function $\delta$ was introduced by Paul
Dirac in 1930. The function is defined as

$$
\begin{equation}
\delta(t-t_0) = \left\{
\begin{aligned}
\infty &\hskip2em t = t_0, \\
0 &\hskip2em t \neq t_0, \\
\end{aligned}
\right. \hskip2em \text{and} \\
\int_0^{\infty} \delta(t-t_0)\ dt = 1
\label{ref:lapl_dirac}
\end{equation}
$$

In fact, nothing actually behaves like this, but it serves well for solving problems. From purely mathematical viewpoint, the Dirac delta is even not strictly a function.

Laplace transform of Dirac delta can be found from definition. Let us evaluate Laplace transform of $f(t)\ \delta(t-t_0)$ and transform of Dirac delta itself is a case if $f(t) = 1$ then:

$$
\begin{align}
{\scr L}[f(t)\ \delta(t-t_0)] &= \int_0^{\infty} e^{-st}f(t)\ \delta(t-t_0)\ dt = e^{-st}f(t)\int_0^{\infty}\delta(t-t_0)\ dt \implies \nonumber \\
\implies \color{red}{{\scr L}[f(t)\ \delta(t-t_0)]} &= \color{red}{e^{-st}f(t)}
\end{align}
$$

The integral is laid from zero to infinity. However $\delta$ has a value of zero anywhere except its peak. That implies that the limits of integral can be reduced to the vicinity of $t_0$. And the interval is so small that the value of functions $e^{-st}f(t)$ are kept constant, therefore can be moved in front of integral. And what is left inside integral is known to be 1 (from $(\ref{ref:lapl_dirac})$). So the Laplace transform of $f(t)\ \delta(t-t_0)$ is $e^{-st}f(t)$.

A mass of 1 kg is attached to a spring with spring constant $k = 1$ N/m. The mass is at rest. At $t = 0$ a hammer blows an impulse 1 to the mass and the second blow at $T=2\pi$. What is the response of the system?

We have some idea how the system behaves from the chapter modeling. We know that the DE of mass on the spring is described as $M\ddot{u} + C\dot{u} + Ku = f$. There is no damping in this example considered, so let us use the equation in form

$$ my'' + ky =f(t), \hskip2em y(0)=0,\ y'(0)=0 $$After substituting $m=1$, $k=1$ and $f(t) = \delta (t-0) + \delta(t-2\pi)$ we have DE to solve.

$$ \begin{align} y''+y &= \delta(t) + \delta(t-2\pi) \nonumber \\ s^2Y + Y &= e^{-0} + e^{-2\pi s} \nonumber \\ Y(s^2+1) &= 1 + e^{-2\pi s} \nonumber \\ Y &= \frac{1+e^{-2 \pi s}}{s^2+1} = \frac{1}{s^2+1} + \frac{e^{-2\pi s}}{s^2+1} \nonumber \end{align} $$The first fraction can be identified as a product of $\sin t$. The second is also $\sin t$ and term $e^{-2\pi s}$ records time shift at $t=2\pi$.

$$ y(t) = \sin t + u_{2\pi}(t)\sin(t-2\pi) = \sin t + u_{2\pi}(t)\sin(t) $$The final answer has to be written in the case form:

$$ y(t) = \left\{ \begin{aligned} \sin t &\hskip2em 0\leq t\leq 2\pi \\ 2\sin t &\hskip2em t\gt 2\pi \end{aligned} \right. $$Nonhomogeneous member on the right side of DE is $f(t) = 2\delta(t-\pi) -\delta(t-2\pi)$. Find the transformed function $F(s)$.

We are going to transform impulse (Dirac delta function). There are two
impulses, the first one at $t=\pi$, the second one at $t=2\pi$. These impulses
last for infinitesimally small timeframe and they are of unit size (area is 1
when integrated). That means if we use definition of Laplace transform
($\int_0^{\infty}e^{-st}\delta(t_0)\ dt$), the member $e^{-st}$ anywhere where
delta function is *nonzero* can be considered as not changing and can be taken as
$e^{-st_0}$ in front of the integral. Then $e^{-st_0}$ has to be multiplied by
integral. Only delta function remains in integral and it is known from the
definition of Dirac function that the value of integral is one.

Nonhomogeneous member on the right side of DE is $f(t) = \delta(t-2\pi)\cos t$. Find the transformed function $F(s)$.

The task is to transform $\delta(t-2\pi)\cos t$. If we use the definition of
Laplace transform, we have $\int_0^{\infty}e^{-st}\sin t\ \delta(t_0)\ dt$.
Anywhere except at $t_0 = 2\pi$ the Dirac delta function has value of zero.
Whatever makes any sense to multiply happens at $t=2\pi$. Since it is at single
point, $e^{-st}\sin t$ is not changing and can be taken outside of
the integral. And the value of integral is known to be *one* from the definition of the
Dirac function.

Find the first and the second antiderivative of $\delta(t-1)$.

For the first antiderivative we have to solve

$$ y' = \delta(t-1), \hskip2em y(0) = 0 $$Let us transform each member and solve:

$$ \begin{align} sY - y(0) &= e^{-s} \nonumber \\ sY &= e^{-s} \nonumber \\ Y(s) &= \frac{e^{-s} }{s} \nonumber \\ \end{align} $$The solution is then

$$ \underline{\underline{y(t) = 1\cdot u_1(t) = u_1(t)}} $$For the second antiderivative we have to solve

$$ y'' = \delta(t-1), \hskip2em y(0) = 0,\ y'(0) = 0 $$Let us transform each member and solve:

$$ \begin{align} s^2Y - sy(0) - y'(0) &= e^{-s} \nonumber \\ s^2Y &= e^{-s} \nonumber \\ Y(s) &= \frac{e^{-s} }{s^2} \nonumber \\ \end{align} $$The solution is then

$$ \underline{\underline{y(t) = (t-1)\cdot u_1(t) }} $$