Definition: let \(f: (0,\infty) \to \mathbb{R}\) be a piecewise continuous function that complies to the demand: \(\exists s_0 \geq 0, \mu > 0: |f(t)| \leq \mu e^{s_0 t}\), then the Laplace transform \(\mathcal{L}[f]\) is defined by
\[
\mathcal{L}[f](s) := \int_0^\infty e^{-st} f(t)dt = F(s),
\]
where \(F(s)\) exists for all \(s > s_0\).
Basic properties
Linearity: if \(f,g: (0,\infty) \to \mathbb{R}\) both have Laplace transforms, then \(f + g\) also has a Laplace transform, and
\[
\mathcal{L}[f + g] = \mathcal{f} + \mathcal{g},
\]
on the interval where both are defined.
Proof:
Will be added later.
If \(c \in \mathbb{R}\) then \(cf\) also has a Laplace transform and,
\[
\mathcal{L}[cf] = c \mathcal{L}[f].
\]
Shifting: if \(f\) has a Laplace transform \(F\) on \((s_0,\infty)\) and \(a \in \mathbb{R}\) then the function \(g\) given by
\[
g(t) = e^{at} f(t)
\]
has a Laplace transform \(G\) on \((\mathrm{max}(s_0 + a),0),\infty\), and
\[
G(s) = F(s-a)
\]
on this interval
Proof:
Will be added later.
More shifting: let \(a>0\), if \(f\) has a Laplace transform \(F\) on \(s_0, \infty\) then the function \(g\) given by
\[
g(t) = \begin{cases} f(t-a) \qquad &\text{if } t \geq a, \\ 0 \qquad &\text{if } t < a \end{cases}
\]
has a Laplace transform G on \((s_0,\infty)\), and
\[
G(s) = e^{-as}F(s)
\]
on this interval.
Proof:
Will be added later.
Scaling: let \(a > 0\). If \(f\) has a Laplace transform \(F\) on \((s_0, \infty)\) then the function \(g\) given by
\[
g(t) = f(at)
\]
has a Laplace transform G on \((as_0, \infty)\), and
\[
G(s) = \frac{1}{a} F\Big(\frac{s}{a}\Big)
\]
on this interval.
Proof:
Will be added later.
Derivatives: if \(f\) has a derivative \(g\) having a Laplace transform \(G\) on the interval \((s_0,\infty)\) then \(f\) has a Laplace transform on the same interval, and
\[
G(s) = sF(s) - f(0).
\]
More generally, for higher derivatives we have (under analogous assumptions)
\[
\mathcal{L}[f^{(n)}](s) = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n-1-k)}(0)
\]
Proof:
For large enough \(s\), the case \(n=1\) follows by integration by parts
\[
\begin{align*}
\mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\
&= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\
&= sF(s) - f(0),
\end{align*}
\]
suppose \(\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\) is true for \(k \in \mathbb{N}\). Then by assumption
\[
\begin{align*}
\mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt, \\
&= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\
&= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\
&= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0).
\end{align*}
\]
Examples
Solving a second order linear ODE: with \(y: \mathbb{K} \to \mathbb{R}\) given by
\[
\ddot y + 4 \dot y + 4y = t \qquad \text{with } y(0) = 1 \text{ and } \dot y(0) = 0
\]
using the Laplace transform
\[
\begin{align*}
\mathcal{L}[\ddot y + 4 \dot y + 4y](s) &= \frac{1}{s^2} \qquad \text{let } \mathcal{L}[y](s) = Y(s), \\
s^2 Y(s) -s + 4(sY(s) - 1) + 4 Y(s) &= \frac{1}{s^2}, \\
(s^2 + 4s + 4)Y(s) &= \frac{1}{s^2} + s + 4, \\ Y(s) &= \frac{s^3 + 4s^2 +1}{s^2(s+2)^2},
\end{align*}
\]
then it may be solved with partial fraction decomposition and the inverse transform.
Solving a linear system of ODEs: with \(\mathbf{y}: \mathbb{K} \to \mathbb{R}^2\) given by
\[
\mathbf{\dot y}(t) = \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathbf{y}(t) \qquad \text{with } \mathbf{y}(0) = \begin{pmatrix} -3 \\ 7 \end{pmatrix}
\]
using the Laplace transform
\[
\begin{align*}
\mathcal{L}[\mathbf{\dot y}](s) &= \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathcal{L}[\mathbf{y}](s) \qquad \text{let } \mathcal{L}[\mathbf{y}](s) = \mathbf{Y}(s), \\
s \mathbf{Y}(s) - \mathbf{y}(0) &= \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathbf{Y}(s), \\
s \mathbf{Y}(s) + \begin{pmatrix} 3 \\ -7 \end{pmatrix} &= \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathbf{Y}(s), \\
\begin{pmatrix} 5 - s & 1 \\ 1 & 5 - s \end{pmatrix} \mathbf{Y}(s) &= \begin{pmatrix} 3 \\ -7 \end{pmatrix},
\end{align*}
\]
using Cramer's rule
\[
\begin{align*}
&Y_1(s) = \frac{\mathrm{det}\begin{pmatrix} 3 & 1 \\ -7 & 5 - s \end{pmatrix}}{(5-s^2)-1}, \\
\\
&Y_2(s) = \frac{\mathrm{det}\begin{pmatrix} 5 - s & 3 \\ 1 & -7\end{pmatrix}}{(5-s^2)-1},
\end{align*}
\]
both can be solved with partial fraction decomposition and the inverse transform.