ece205: catch up on last week

docs/2a/ece205.md

@@ -308,6 +308,145 @@ Substituting it into the IBVP results in a **separation constant** $-\lambda$.
 
 $$\boxed{\frac{T'(t)}{a^2T(t)}=\frac{X''(x)}{X(x)}=-\lambda}$$
 
+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:
+
+$$
+u_{tt}=a^2u_{xx},0<x<L,t>0 \\
+u(0,t)=u(L,t)=0,t>0 \\
+u(x,0)=f(x), 0\leq x\leq L \\
+u_t(x,0)=g(x), 0\leq x\leq L
+$$
+
+The following conditions must be met:
+
+$$
+u(x,t)=\sum^\infty_{n=1}\sin(\frac{n\pi x}{L})\left(\alpha_n\cos(\frac{n\pi a}{L}t)+\beta_n\sin(\frac{n\pi a}{L}t)\right) \\
+\boxed{f(x)=\sum^\infty_{n=1}\alpha_n\sin(\frac{n\pi x}{L}),0\leq x\leq L} \\
+\boxed{g(x)=\sum^\infty_{n=1}\frac{n\pi a}{L}\beta_n\sin(\frac{n\pi x}{L}), 0\leq x\leq L}
+$$
+
+### Fourier symmetry
+
+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).
+
+!!! example
+    For $f(x)=\begin{cases}\frac\pi 2 & [0,\frac\pi 2] \\ x-\frac\pi 2 & (\frac\pi2,\pi]\end{cases}$:
+    
+    
+    \begin{align*}
+    a_n&=\frac 2 L\int^L_0f(x)\cos(\frac{n\pi x}{L})dx \\
+    &=\frac 2\pi \int^{\pi/2}_0\frac\pi 2\cos(\frac{n\pi x}{\pi})dx + \frac 2 \pi\int^\pi_{\pi/2}(x-\frac\pi2)\cos(\frac{n\pi x}{\pi})dx \\
+    &=\frac{2}{n^2\pi}[(-1)^n-\cos(\frac{n\pi}{2})+\frac{n\pi}{2}\sin(\frac{n\pi}{2}) \\
+    \\
+    a_0&=\frac2\pi\int^\pi_0f(x)\cos(0)dx \\
+    &=\frac{3\pi}{4} \\
+    \\
+    \therefore f(x)&=\frac{3\pi}{8}+\sum^\infty_{n=1}\frac{2}{n^2\pi^2}[(-1)^n-\cos(\frac{n\pi}{2})+\frac{n\pi}{2}\sin(\frac{n\pi}{2})]\cos(nx),x\in[0,\pi]
+    \end{align*}
+    
+!!! example
+    For:
+    
+    $$
+    u_t=2u_{xx},0<x<\pi,t:0 \\
+    u(0,t)=u(\pi,t)=0,t>0 \\
+    u(x,0)=\begin{cases}
+    	\frac\pi 2 & [0,\frac\pi 2] \\
+    	x-\frac\pi 2 & (\frac\pi 2,\pi]
+    \end{cases}
+    $$
+    
+    We have $L=\pi,a=\sqrt 2$.
+    
+    \begin{align*}
+    u(x,t)&=\sum^\infty_{n=1}\alpha_ne^{-left(\frac{n\pi\sqrt 2}{\pi}\right)^2t}\sin(\frac{n\pi x}{\pi}) 
+    &=\sum^\infty_{n=1}\apha_ne^{-2n^2t}\sin(nx) \\
+    \alpha_n&=\frac 2 L\int^L_0f(x)\sin(\frac{n\pi x}{L})dx \\
+    &=\frac2\pi\int^{\pi/2}_0\frac\pi 2\sin(nx)dx+\frac2\pi\int^\pi_{\pi/2}(x-\frac\pi2\sin(nx)dx \\
+    &=\frac 1 n[1+(-1)^{n+1}-\cos(\frac{n\pi}{2})-\frac{2}{n\pi}\sin(\frac{n\pi}{2}]
+    \end{align*}
+
+### Convergence of Fourier series
+
+!!! definition
+    - $f(x^+)=\lim_{h\to0^+}f(x+h)$
+    - $f(x^-=\lim_{h\to0^-}f(x+h)$
+
+If $f$ and $f'$ are piecewise continuous on $[-L, L]$ for $x\in(-L,L)$, where $a_n$ and $b_n$ are from the Euler-Fourier formulae:
+
+$$\frac{a_0}{2}+\sum^\infty_{n=1}a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})=\boxed{\frac 1 2[f(x^+)+f(x^-)]}$$
+
+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 $|f(x)-[\frac{a_0}{2}+\sum^N_{n=1}a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})]|<\epsilon$ for all $N\geq N_0$ and all $x\in(-\infty,\infty)$.
+
+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 $[-L,L]$.
+
+Then, for any $x\in[-L,L]$:
+
+$$\int^x_{-L}f(t)dt=\int^x_{-L}\frac{a_0}{2}dt+\sum^\infty_{n=1}\int^x_{-L}(a_n\cos(\frac{n\pi t}{L})+b_n\sin(\frac{n\pi t}{L}))dt$$
+
+
 ## Resources
 
 - [Laplace Table](/resources/ece/laplace.pdf)