Fourier 1 : finished section 1 (i.e. 1.2 and 1.3 today)

Author: govin08

_posts/2024-10-05-Fourier_1.md

@@ -6,9 +6,7 @@ tags: [analysis, Fourier analysis, Fourier Series]
 use_math: true
 publish: false
 author_profile: false
-toc:
-  enabled: True
-  level: 2
+toc: true
 ---
 
 약 일년 반 전, 확률 및 통계 스터디를 하는 과정에서 Fourier analysis에 대해 잠시 공부한 적이 있었다. 
@@ -143,15 +141,6 @@ $$
 
 We analyze and solve the standard one and simply apply change of variables for the general cases.
 
-### 1.2 Solution to the wave equation
-
-<!-- There are two kinds of solution to the wave equation ; -->
-We can get two kinds of solution to the wave equation ;
-(1) using traveling waves
-(2) using the superposition of standing waves
-
-#### A solution using traveling waves
-
 Consider a physical problem of a string where $0\le x\le \pi$ :
 
 $$
@@ -168,6 +157,15 @@ u(0,t)=u(\pi,t)=0
 \tag a
 $$
 
+### 1.2 Solution to the wave equation
+
+<!-- There are two kinds of solution to the wave equation ; -->
+We can get two kinds of solution to the wave equation ;
+(1) using traveling waves
+(2) using the superposition of standing waves
+
+#### A solution using traveling waves
+
 We claim that (a) has the general solution of the form
 
 $$
@@ -211,11 +209,8 @@ Then
 
 $$
 \begin{align*}
-u(0,t)
-&=\frac12\left[f(t)+f(-t)\right]+\frac12\int_{-t}^tg(y)\,dy\\
-&=0\\
-u(\pi,t)&=\frac12\left[f(\pi+t)+f(\pi-t)\right]+\frac12\int_{\pi-t}^{\pi+t}g(y)\,dy\\
-&=0.
+u(0,t)&=\frac12\left[f(t)+f(-t)\right]+\frac12\int_{-t}^tg(y)\,dy=0\\
+u(\pi,t)&=\frac12\left[f(\pi+t)+f(\pi-t)\right]+\frac12\int_{\pi-t}^{\pi+t}g(y)\,dy=0.
 \end{align*}
 $$
 
@@ -320,15 +315,255 @@ Note that $C_1'+C_2'=0$ since $f(x)=F(x)+G(x)$.
 Therefore,
 
 $$
-\begin{align*}
+\begin{aligned}
 u(x,t)
 &=F(x+t)+G(x-t)\\
 &=\frac12f(x+t)+\frac12\int_0^{x+t}g(y)\,dy+C_1'
 +\frac12f(x-t)-\frac12\int_0^{x-t}g(y)\,dy+C_2'\\
 &=\frac12\left[f(x+t)+f(x-t)\right]+\frac12\int_{x-t}^{x+t}g(y)\,dy.
-\end{align*}
+\end{aligned}
+$$
+
+#### A solution using superposition of standing waves
+
+We claim that (a) has the general solution of the form
+
+$$u(x,t)=\sum_{m=1}^\infty\sin mx\left(A_m\cos mt+B_m\sin mt\right)\tag c$$
+
+where
+
+$$
+\begin{aligned}
+A_n&=\frac2\pi\int_0^\pi f(x)\sin nx\,dx\\
+B_n&=\frac2{n\pi}\int_0^\pi g(x)\sin nx\,dx\\
+\end{aligned}
+$$
+
+
+Separating variables $x$ and $t$, let
+
+$$u(x,t)=\phi(x)\psi(t).$$
+
+Then, the wave equation becomes
+
+$$\phi(x)\psi''(t)=\phi''(x)\psi(t).$$
+
+This implies, if $\phi(x)\ne0$ and $\psi(t)\ne0$,
+
+$$\frac{\psi''(t)}{\psi(t)}=\frac{\phi''(x)}{\phi(x)}.$$
+
+Since LHS is independent of $x$ and RHS is independent of $t$, it is independent of both of them.
+Let this constant be $\lambda$.
+Then
+
+$$
+\begin{aligned}
+\psi''(t)-\lambda\psi(t)&=0\\
+\phi''(x)-\lambda\phi(x)&=0
+\end{aligned}\tag3
+$$
+
+As in the simple harmonic motion in (1), these ODEs has general solutions
+
+$$
+\begin{aligned}
+\psi(t)&=A\cos mt + B\sin mt\\
+\phi(x)&=\tilde A\cos mx + \tilde B\sin mx
+\end{aligned}
 $$
 
+by 3.6, where $m^2=-\lambda$.
+Apply the fourth condition of (a).
+Since the string has zero displacement at $x=0$,
+
+$$0=\tilde A\left(A\cos mt + B\sin mt\right)$$
+
+for every $t$.
+Thus, $\tilde A=0$.
+At $x=\pi$,
+
+$$0=\tilde B\sin m\pi\left(A\cos mt + B\sin mt\right)$$
+
+for every $t$.
+Avoiding the trivial case when $\tilde B=0$, $m$ is an integer.
+For every integer $m$,
+
+$$u_m(x,t)=\sin mx\left(A\cos mt + B\sin mt\right)$$
+
+is a solution for the wave equation.
+
+<!-- Let
+
+$$u_m(x,t)=\sin mx\left(A_m\cos mt + B_m\sin mt\right)$$ -->
+
+<div class="notice--danger">
+<b>Remark </b> <br>
+Letting, for example, $A_m=1$, $B_m=0$,
+$$u_1(x,t)=\cos t\sin x$$
+is called the fundamental tone or the first harmonic.
+$$u_2(x,t)=\cos2t\sin2x$$
+is called the first overtone or the second harmonic.
+$$u_3(x,t)=\cos3t\sin3x$$
+is called second overtone or the third harmonic, and so on.
+
+For fixed $t$, $u_m(x,t)=0$ if $x=\frac\pi mk$ for all $k\in\mathbb Z$.
+For each $k$, $x=\frac\pi mk$ is called a node, while $x=\frac\pi m\left(k+\frac12\right)$ is called an antinode.
+</div>
+
+The wave equation is linear in that any linear combination of solutions for the wave equation solves the equation too.
+Thus,
+
+$$u(x,t)=\sum_{m=-\infty}^\infty\sin mx\left(A_m\cos mt+B_m\sin mt\right)$$
+
+also satisfy the wave equation.
+Suppose that this solution is a general one (we omit the proof here.)
+Since $u_m=0$ for $m=0$ and
+
+$$
+\begin{aligned}
+u_{-m}(x,t)
+&=\sin(-mx)\left(A_m\cos(-mt)+B_m\sin(-mt)\right)\\
+&=\sin mx\left((-A_m)\cos mt+B_m\sin mt\right),
+\end{aligned}
+$$
+
+we can write, alternatively as
+
+$$u(x,t)=\sum_{m=1}^\infty\sin mx\left(A_m\cos mt+B_m\sin mt\right)\tag4$$
+
+for different $A_m$ and $B_m$.
+By the initial displacement condition $u(x,0)=f(x)$ and by the initial velocity condition $u_t(x,0)=g(x)$
+
+$$
+\begin{aligned}
+f(x)&=\sum_{m=1}^\infty A_m\sin mx\\
+g(x)&=\sum_{m=1}^\infty mB_m\sin mx.
+\end{aligned}
+$$
+
+Multiplying $\sin nx$ and integrating over $[0,\pi]$ for $n\ge1$,
+
+$$
+\begin{aligned}
+\int_0^\pi f(x)\sin nx\,dx
+&=\sum_{m=1}^\infty A_m\int_0^\pi\sin mx\sin nx\,dx\\
+&=\sum_{m=1}^\infty\frac{A_m}2\int_0^\pi-\cos(m+n)x+\cos(m-n)x\,dx\\
+&=\frac{A_n}2\int_0^\pi\,dx\\
+&=A_n\times\frac\pi2\\
+\int_0^\pi g(x)\sin nx\,dx
+&=\sum_{m=1}^\infty mB_m\int_0^\pi\sin mx\sin nx\,dx\\
+&=\sum_{m=1}^\infty\frac{mB_m}2\int_0^\pi-\cos(m+n)x+\cos(m-n)x\,dx\\
+&=\frac{nB_n}2\int_0^\pi\,dx\\
+&=nB_n\times\frac\pi2.
+\end{aligned}
+$$
+
+Thus, (Fourier sine coeffcient of $f$)
+
+$$
+\begin{aligned}
+A_n&=\frac2\pi\int_0^\pi f(x)\sin nx\,dx\\
+B_n&=\frac2{n\pi}\int_0^\pi g(x)\sin nx\,dx.
+\end{aligned}
+$$
+
+And the claim is now proved.
+
+Note that this argument assumes that a *reasonable* function $f$ on $[0,\pi]$ can be expressed as a linear combination of sine functions ; 
+
+$$
+f(x)=\sum_{m=1}^\infty A_m\sin mx\tag{$\ast$}
+$$
+
+Suppose further that this class of function can also be expressed as cosine series too ;
+
+$$
+g(x)=\sum_{m=1}^\infty A'_m\cos mx
+$$
+
+If an odd function $f$ on $[-\pi,\pi]$ is given, the restriction on $[0,\pi]$ can be expressed as $(\ast)$ and $(\ast)$ is also valid for $f$ on $[-\pi,\pi]$ too;
+
+$$
+f(-x)=-f(x)=-\sum_{m=1}^\infty A_m\sin mx=\sum_{m=1}^\infty A_m\sin(-mx).
+$$
+
+If an even function $g$ on $[-\pi,\pi]$ is given, the restriction on $[0,\pi]$ can be expressed as cosine series.
+The equation is valid for $g$ on $[-\pi, \pi]$ too, by the similar fashion.
+
+Now suppose that a function $F$ on $[-\pi, \pi]$ is given.
+Since $F$ can be expressed as $f+g$,
+
+$$
+\begin{aligned}
+F(x)
+&=f(x)+g(x)\\
+&=\sum_{m=1}^\infty\left(A_m\sin mx + A'_m\cos mx\right).
+\end{aligned}
+$$
+
+By Euler identity, (we can choose $a_m=\frac{A_m'-A_mi}2$ and $a_{-m}=\frac{A_m'+A_mi}2$)
+
+$$
+F(x) = \sum_{m=-\infty}^\infty a_me^{imx}.
+$$
+
+Multiplying $e^{-inx}$ and integrating over $[-\pi, \pi]$,
+
+$$
+\begin{aligned}
+\int_{-\pi}^\pi F(x)e^{-inx}\,dx
+&=\sum_{m=-\infty}^\infty a_m\int_{-\pi}^\pi e^{i(m-n)x}\,dx\\
+&=a_n\int_{-\pi}^\pi\,dx\\
+&=a_n\times2\pi.
+\end{aligned}
+$$
+
+Thus,
+
+$$
+a_n = \frac1{2\pi}\int_{-\pi}^\pi F(x)e^{-inx}\,dx.
+$$
+
+Such $a_n$ is called the *Fourier coefficient* of $F$.
+
+### 1.3 Example : The plucked string
+
+Consider the case when the initial string is located as the two line segments
+
+$$
+f(x)
+=\begin{cases}
+\frac{hx}p&(0\le x\le p)\\
+\frac{h(\pi-x)}{\pi-p}&(p\le x\le\pi)
+\end{cases}
+$$
+
+and when there is no initial movement ; $g(x)=0$.
+Following (c),
+
+$$
+f(x)=\sum_m A_m\sin mx,\quad
+A_m=\frac{2h\sin mp}{m^2p(\pi-p)}
+$$
+
+by 3.9.
+Since $g(x)=0$, $B_m=0$ for all $m$.
+Thus (4) now reduces to 
+
+$$
+\begin{aligned}
+u(x,t)
+&=\sum_{m=1}^\infty A_m\sin mx\cos mt\\
+&=\frac12\sum_{n=1}^\infty A_m\left[\sin m(x+t)+\sin m(x-t)\right]\\
+&=\frac{\sum_{n=1}^\infty A_m\sin m(x+t)+\sum_{n=1}^\infty A_m\sin m(x-t)}2\\
+&=\frac{f(x+t)+f(x-t)}2
+\end{aligned}
+$$
+
+The above equation also corresponds to (b) since $g(x)=0$.
+Note that the function $f$ in the last expression is the extended version (on $\mathbb R$) of the original function $f$ on $[0,\pi]$.
+
+
 ## 2. The Heat Equation
 
 ## 3. Exercises
@@ -972,4 +1207,41 @@ and
 $$
 \lim_{h\to0}\frac{F(x+h)+F(x-h)-2F(x)}{h^2}
 =\lim_{h\to0}\left(F''(x)+\phi(h)+\phi(-h)\right)=F''(x)
-$$ 
+$$ 
+
+### 3.9
+
+Consider the case of plucked string ; 
+
+$$
+f(x)
+=\begin{cases}
+\frac{hx}p&(0\le x\le p)\\
+\frac{h(\pi-x)}{\pi-p}&(p\le x\le\pi)
+\end{cases}
+$$
+
+Calculate the Fourier sine coefficient sine coefficient $A_m$.
+For what position of $p$ are the second, fourth, $\cdots$ harmonics missing?
+For what position of $p$ are the third, sixth, $\cdots$ harmonics missing?
+
+(proof)
+
+$$
+\begin{aligned}
+A_m
+&=\frac2\pi\int_0^\pi f(x)\sin mx\,dx\\
+&=\frac{2h}{\pi p}\int_0^px\sin mx\,dx
++\frac{2h}{\pi(\pi-p)}\int_p^{\pi}(\pi-x)\sin mx\,dx\\
+&=\frac{2h}{\pi p}\left(-\frac1mp\cos np+\frac1{m^2}\sin mp\right)
++\frac{2h}{\pi(\pi-p)}\left(\frac1m(\pi-p)\cos mp+\frac1{m^2}\sin mp\right)\\
+&=\frac{2h\sin mp}{m^2p(\pi-p)}
+\end{aligned}
+$$
+
+$A_2=0$ iff $p=\frac\pi2k$ and $A_4=0$ iff $p=\frac\pi4k$ for an integer $k$.
+Note also that neither $p=0$ nor $p=\pi$ as long as $h\ne0$.
+So if $p=\frac\pi2$, then $A_2=A_4=\cdots=0$ and the even harmonics are missing.
+
+$A_3=0$ iff $p=\frac\pi3k$ and $A_6=0$ iff $p=\frac\pi6k$ for an integer $k$.
+If $p=\frac\pi3$ or $p=\frac23p$, then $A_3=A_6=\cdots=0$ and the harmonics of $3k$ are missing.