4.7 KiB
The Laplace transform
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 := \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^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' &= \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} &= \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 &= \frac{1}{s^2} \qquad \text{let } \mathcal{L}y = 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} &= \begin{pmatrix} 5 & 1 \ 1 & 5 \end{pmatrix} \mathcal{L}\mathbf{y} \qquad \text{let } \mathcal{L}\mathbf{y} = \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.