Catchup 1

docs/math/diffeq/4-laplace.md

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+# Section 4 - Laplace Transformations
+## Section 4.1 - Definition
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx).
+
+**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
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/LaplaceTransforms.aspx).
+
+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
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/InverseTransforms.aspx).
+
+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
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx).
+
+**Theorum**. Given a function $f(t)$ with $C^n$ continuity, then
+
+$$
+\mathcal{L} \{ f^{(n)} (t) \} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - ... - s f^{(n-2)} (0) - f^{(n-1)} (0)
+$$
+
+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 - Nonconstant Coefficient IVPs
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx).