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Section 4 - Laplace Transformations
Section 4.1 - Definition
This section is from Paul's Online Math Notes.
Definition. The Laplace transform of a function is given by the following:
\mathcal{L} \{f(t)\}(s) = F(s) = \int_0^{\infty} e^{-st}f(t) dt
Section 4.2 - Properties
The Laplace Transformation is a linear transformation over functions in \mathbb{R}[t]. That is, given a, b \in \mathbb{R}, f(t), g(t) \in \mathbb{R}[t], we know that
\mathcal{L} \{a f(t)\ + b g(t) \}(s) = a F(s) + b G(s)
Section 4.3 - Inverse Laplace Transformation
Given F(s), we define the Inverse Laplace Transformation as the following;
f(t) = \mathcal{L}^{-1} \{F(s)\}
Section 4.4 - Step Function
The step/Heaviside function u_c(t) is defined as 0 if t < c, and 1 if t > c.
Alternatively, u(t - c) = H(t - c) is 0 if t < c, and 1 if t > c.
Applying this to the Laplace transform,
\begin{align}
\mathcal{L} \{ u_c(t) f(t-c) \} &= \int_0^{\infty} e^{-st}u_c(t)f(t) dt \\
&= \int_c^{\infty} e^{-st}f(t) dt
\end{align}
If we let u = t - c,
\begin{align}
\mathcal{L} \{ u_c(t) f(t-c) \} &= \int_0^{\infty} e^{-s(u+c)}f(u) du \\
&= \int_0^{\infty} e^{-su}e^{-cs}f(u) du \\
&= e^{-cs} \int_0^{\infty} e^{-su}f(u) du \\
&= e^{-cs} F(s)
\end{align}
Section 4.5 - Laplace Transformation applied to IVPs
Theorem. Given a function f(t) with C^n continuity, then
\mathcal{L} \{ f^{(n)} (t) \} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \ldots - s f^{(n-2)} (0) - f^{(n-1)} (0)
For n=1, 2 we see that
\begin{align}
\mathcal{L} \{ y' \} &= sY(s) - y(0) \\
\mathcal{L} \{ y'' \} &= s^2 Y(s) - s y(0) - y'(0)
\end{align}
We can take the Laplace transformation of an IVP, solve for Y(s), then take the inverse to find the solution.
Section 4.6 - Non-constant Coefficient IVPs
If f(t) is piecewise continuous on [0, \infty), then \lim_{s \rightarrow \infty} F(s) = 0.
Definition. A function f(t) is said to be of exponential order \alpha if there exists positive constants T, M such that for all t \geq T, |f(t)| \leq Me^{\alpha t}.
To check this, simply compute \lim_{t \rightarrow \infty} \frac{|f(t)|}{e^{\alpha t}}. If this is finite for some \alpha, then the function is of exponential order \alpha.
Section 4.7 - IVPs with Step Functions
Recall that \mathcal{L} \{u_c(t)f(t-c)\} = e^{-cs}F(s). Then, we can solve IVPs involving step functions.
Section 4.8 - Dirac Delta Function
The Dirac Delta function has several properties. First, \delta(t - a) = 0 when t \neq a. Notably, though,
\int_{\mathbb{R}} f(t) \delta(t - a) dt = f(a)
Note that this is not an actual function, buy instead a generalized function or distribution, as several functions can express this property using infinite limits.
Then, we can see that \mathcal{L} \{\delta(t-a)\} = \int_0^\infty e^{-st} \delta(t-a) dt by definition. Then, applying the properties of the Delta function, \mathcal{L} \{\delta(t-a)\} = e^{-as}, given a > 0.
Section 4.9 - Convolution Integrals
Consider two functions F(s) and G(s) such that F(s) G(s) = H(s), of which we want to find an inverse Laplace transform.
We define a convolution integral (f*g)(t) as
(f*g)(t) = \int_0^t f(t - \tau)(g - \tau) d\tau
A unique property of this integral is that (f*g) = (g*f).
With this, we see that \mathcal{L} \{f * g\} = F(s)G(s), or that \mathcal{L}^{-1} \{F(s)G(s)\} = (f * g)(t).