The Laplace transform is a wonderful operation to convert a function of $t$ into a function of $s$. Where $s$ is an unknown variable independent of $t$:
$$
\mathcal L\{f(t)\}=F(s)=\int^\infty_0e^{-st}f(t)dt, s > 0
Clearly $f(t)$ is piecewise ocontinuous on $[0,\infty)$ and has an exponential order of -1 when $t\geq 1$ and 0 when $0\leq t<1$.Thus$\mathcalL\{f(t)\}$isdefinedfor$s>0$.
The Heaviside and unit step functions are identical:
$$
H(t-c)=u(t-c)=u_c(t)=\begin{cases}
0 & t <c \\
1 & t \geq c
\end{cases}
$$
Piecewise continuous functions can be manipulated into a single equation via the Heaviside function.
For a Heaviside transform $\mathcal L\{u_c(t)g(t)\}$, if $g$ is defined on $[0,\infty)$, $c\geq 0$, and $\mathcal L\{g(t+c)\}$ exists for some $s>s_0$:
The **period** of a function is an increment that always returns the same value: $f(x+T)=f(x)$, and its **fundamental period** of a function is the smallest possible period.
!!! example
The fundamental period of $\sin x$ is $2\pi$, but any $2\pi K,k\in\mathbb N$ is a period.
The fundamental periods of $\sin \omega x$ and $\cos\omega x$ are both $\frac{2\pi}{\omega}$.
If functions $f$ and $g$ have a period $T$, then both $af+bg$ and $fg$ also must have period $T$.
Functions are **orthogonal** on an interval when the integral of their product is zero, and a set of functions is **mutually orthogonal** if all functions in the set are orthogonal to each other.
Possible values for the separation constant are known as **eigenvalues**, and their corresponding **eigenfunctions** contain the unknown constant $a_n$:
$$
\lambda_n=\left(\frac{n\pi}{L}\right)^2 \\
X_n(x)=a_n\sin(\frac{n\pi x}{L})
$$
### Wave equation
A string stretched between two secured points at $x=0$ and $x=L$ can be represented by the following IBVP:
To find a Fourier series for functions defined only on $[0, L]$ instead of $[-L, L]$, a **periodic extension** can be used.
A **half-range sine expansion (HRS)** is used for odd functions:
$$
f_o(x)=\begin{cases}
f(x) & x\in(0, L) \\
-f(-x) & x\in(-L, 0)
\end{cases}
$$
A **half-range cosine expansion (HRC)** is used for even functions:
$$
f_e(x)=\begin{cases}
f(x) & x\in(0, L) \\
f(-x) & x\in(-L, 0)
\end{cases}
$$
Thus if a Fourier series on $(0,L)$ exists, it can be expressed as either a **Fourier sine series** (via HRS) or a **Fourier cosine series** (via HRC).
At $x=\pm L$, the series converges to $\frac 1 2[f(-L^+)+f(L^-)]$. This implies:
- A continuous $f$ converges to $f(x)$
- A discontinuous $f$ has the Fourier series converge to the average of the left and right limits
- Extending $f$ to infinity using periodicity allows it to hold for all $x$
!!! example
The square wave function $f(x)=\begin{cases}-1 & -\pi<x<0 \\ 1&0<x<\pi\end{cases},f(x+2\pi)=f(x)$:
$f$ and $f'$ are piecewise continuous, but the function is discontinuous at $k\pi,k\in\mathbb Z$. Thus at $x=\pm\pi$, the series converges to $\frac 1 2(-1+1)=0$. At $x=0$, the series converges to $\frac 1 2(1+(-1))=0$.
If $f$ is 2L-periodic and continuous on $-\infty,\infty$, and $f'$ is piecewise continuous on $[-L,L]$, the Fourier series converges **uniformly** to $f$ on $[-L,L]$ and thus any interval.
More formally, for every $\epsilon>0$, there exists an integer $N_0$ depending on $\epsilon$ such that $|f(x)-[\frac{a_0}{2}+\sum^N_{n=1}a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})]|<\epsilon$forall$N\geqN_0$andall$x\in(-\infty,\infty)$.
More intuitively, for a high enough summation of the Fourier series, the value must lie in an **$\epsilon$-corridor** of $f(x)$ such that $f(x)$ is always between $f(x)\pm\epsilon$.
!!! example
- The Fourier series for the triangle wave function **is** uniformly convergent.
- The Fourier series for the square wave function **is not** uniformly convergent, which means that Gibbs overshoots would not fit in an arbitrarily small $\epsilon$-corridor.
The **Weierstrass M-test** states that if $|a_n(x)|\leq M_n$ for all $x\in[a,b]$ and if $\sum^\infty_{n=1}M_n$ converges, then $\sum^\infty_{n=1}a_n(x)$ converges uniformly to $f(x)$ on $[a,b]$.
!!! example
$\sum^\infty_{n=1}\frac{1}{n^2}\cos(nx)$ converges uniformly on any finite closed interval $[a,b]$.
$|\frac{\cos(nx)}{n^2}|\leq\frac{1}{n^2}$ for all $x$, and $\sum^\infty_{n=1}\frac{1}{n^2}$ also converges. Thus the result follows from the M-test.
### Differentiating Fourier series
You can termwise differentiate the Fourier series of $f(x)$ only if:
- $f(x)$ is continuous on $(-\infty,\infty)$ and 2L-periodic
- $f'(x),f''(x)$ are both piecewise continuous on $[-L,L]$
You can termwise integrate the Fourier series of $f(x)$ only if $f(x)$ is piecewise continuous on $[-L,L]$.
The Fourier coefficients $c_n$ map to the amplitude spectrum $|c_n|$. **Parseval's theorem** maps the frequency domain ($\{c_n\}$) to and from the time domain ($f(t)$):
If a 2L-periodic function $f(t)$ has a complex Fourier series $f(t)=\sum^\infty_{n=-\infty}c_ne^{\frac{in\pi x}{L}}$:
To convert a Fourier series back to the original function, the following conditions must hold:
- there must not be any infinite discontinuities: $\int^\infty_{-\infty}|f(x)|dx<\infty$
- in any finite interval, there must be a finite number of extrema and discontinuities
Then, the **Fourier integral** / **inverse Fourier transform** converges to $f(x)$ wherever continuous and $\frac 1 2[f(x^+)+f(x^-)]$ at discontinuities.
