completed Fourier_1 anyway. but further correction should be made.

Author: govin08

_posts/2024-10-05-Fourier_1.md

@@ -396,7 +396,7 @@ is a solution for the wave equation.
 
 $$u_m(x,t)=\sin mx\left(A_m\cos mt + B_m\sin mt\right)$$ -->
 
-<div class="notice--danger">
+<div class="notice--info" markdown="1">
 <b>Remark </b> <br>
 Letting, for example, $A_m=1$, $B_m=0$,
 $$u_1(x,t)=\cos t\sin x$$
@@ -503,10 +503,13 @@ $$
 
 By Euler identity, (we can choose $a_m=\frac{A_m'-A_mi}2$ and $a_{-m}=\frac{A_m'+A_mi}2$)
 
+<div class="notice--success" markdown="1">
 $$
 F(x) = \sum_{m=-\infty}^\infty a_me^{imx}.
 $$
 
+</div>
+
 Multiplying $e^{-inx}$ and integrating over $[-\pi, \pi]$,
 
 $$
@@ -568,7 +571,7 @@ Note that the function $f$ in the last expression is the extended version (on $\
 
 ### 2.1 Derivation of the heat equation
 
-<div class="notice--info">
+<div class="notice--info" markdown="1">
 책의 내용이 잘 이해가 가지 않아서, 따로 찾아보았고 별도로 유도해보았다.
 <a href="http://www-personal.umd.umich.edu/~adwiggin/TeachingFiles/FourierSeries/Resources/HeatEquationDerivation.pdf"> 미시건 대학교 자료</a>가 굉장히 이해가 잘 되게 설명되어 있어서 이 자료를 가장 많이 참고했다.
 이 자료에서 Fourier's law가 소개되고 사용되고 있는데, 사실 Stein의 책과 다른 자료들에서도 Fourier's law는 기본으로 깔고 시작하는 것 같다.
@@ -688,7 +691,7 @@ $$
 k\left[\frac{\partial^2u}{\partial x^2}(x,y,t)+\frac{\partial^2u}{\partial x^2}(x,y,t)\right].
 $$
 
-<div class="notice--info">
+<div class="notice--info" markdown="1">
 책의 내용을 이해했다.
 처음에 열에너지($H$)를 정의하는 식은 이해가 안간다.
 아마도 대학교 수준의 열역학을 이해해야 알 수 있을 듯하다.
@@ -796,6 +799,125 @@ $$
 
 This is called the **(time dependent) heat equation**.
 
+
+### 2.2 Steady-state heat equation in the disc
+
+As time passes by, the temperature at each point might converge to some equilibrium temperature.
+In this steady state, we can assume $\frac{\partial u}{\partial t}=0$.
+The heat equation then becomes,
+
+$$
+\frac{\partial^2u}{\partial x^2}
++
+\frac{\partial^2u}{\partial y^2}
+=0\tag{10}
+$$
+
+Let $\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$.
+It is called the *Laplacian* operator.
+(10) now becomes a simple equation $\Delta u=0$
+By 3.10, we can rewrite it as
+
+$$\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac{\partial}{\partial r}+\frac1{r^2}\frac{\partial^2}{\partial y^2}.$$
+
+<div class="notice--info" markdown="1">
+<b>Remark </b> <br>
+Such $u$ satisfying $\Delta u=0$ is called a *harmonic function*.
+</div>
+
+Let $D$ and $C$ be a unit circle and a unit circle which can be represented as
+
+$$
+\begin{aligned}
+D&=\{(x,y):x^2+y^2\lt1\}=\{(r,\theta):r\lt1\}\\
+C&=\{(x,y):x^2+y^2=1\}=\{(r,\theta):r=1\}
+\end{aligned}
+$$
+
+in the Cartesian and the Polar coordinates.
+Here is a famous and traditional problem in the field of PDE.
+
+<div class="notice--info" markdown="1">
+<b>Dirichlet Problem</b> <br>
+Given $u$ on $C$, find the steady-state solution $u$ on $D$
+</div>
+
+Let $f=u|C$ specify the temperature of the boundary.
+For convinience, we can assume that $f$ has its argument $\theta$ only.
+The boundary condition then becomes $f(\theta)=u(1,\theta)$.
+Using the polar form of Laplacian, the condition $\Delta u=0$ becomes
+
+$$
+\begin{gathered}
+\frac{\partial^2u}{\partial r^2}+\frac1r\frac{\partial u}{\partial r}+\frac1{r^2}\frac{\partial^2u}{\partial y^2}=0\\
+r^2\frac{\partial^2u}{\partial r^2}+r\frac{\partial u}{\partial r}=-\frac{\partial^2u}{\partial y^2}
+\end{gathered}
+$$
+
+Separating the variables so that $u(r,\theta)=F(r)G(\theta)$, then
+
+$$
+\begin{gathered}
+r^2F''(r)G(\theta)+rF'(r)G(\theta)=-F(r)G''(\theta)\\
+\frac{r^2F''(r)+rF'(r)}{F(r)}=-\frac{G''(\theta)}{G(\theta)}
+\end{gathered}
+$$
+
+Let both sides be $\lambda$ as before, we have a system
+
+$$
+\begin{aligned}
+G''(\theta)+\lambda G(\theta)&=0\\
+r^2F''(r)+rF'(r)-\lambda F(r)&=0
+\end{aligned}
+$$
+
+If $\lambda\lt 0$, then $G$ is not oscillatory.
+So, suppose $\lambda\ge0$ and let $m^2=\lambda$.
+The first differential equation then becomes (1) in the simple harmonic motion again.
+We conclude, using Euler identity,
+
+$$
+\begin{aligned}
+G(\theta)
+&=\tilde A\cos m\theta+\tilde B\sin m\theta\\
+&=Ae^{im\theta} + Be^{-im\theta}
+\end{aligned}
+$$
+
+for some constants $A$ and $B$.
+For the second differential equation, suppose further that $m$ is an integer and apply 3.11.
+If $m\neq0$, then $F(r)=\alpha r^m+\beta r^{-m}$.
+Since a negative power of $r$ diverges as $r\to0^+$, take $F(r)=r^{|m|}$.
+If $m=0$, then $F(r)=\alpha+\beta\log r$.
+Again, $\log r$ diverges as $r\to0^+$.
+So take $F(r)=1$.
+In general, we can write $F(r)=r^{|m|}$.
+
+Therefore,
+
+$$u(r,\theta)=r^{|m|}\left(A_m e^{im\theta}+B_me^{-im\theta}\right)$$
+
+is a solution to $\Delta u=0$ for each integer $m$.
+Superposing this class of solutions we can write
+
+<div class="notice--success" markdown="1">
+$$
+u(r,\theta)=\sum_{m=-\infty}^\infty r^ma_me^{im\theta}.
+$$
+
+</div>
+
+The boundary condition $f(\theta)=u(1,\theta)$ becomes
+
+$$
+f(\theta)=\sum_{m=-\infty}^\infty a_me^{im\theta}
+$$
+
+The solution and the boundary is just simlar to ones that we encountered in the wave equation.
+And we may wonder if such a sequence $\\{a_m\\}$ exists that equate the boundary condition.
+
+
 ## 3. Exercises
 
 <p class='text-size-12'> 3.1 </p>
@@ -1061,7 +1183,7 @@ $$
 
 Therefore, $f_n\to f$ uniformly.
 
-<div class="notice--info">
+<div class="notice--info" markdown="1">
 전체적인 증명은 
 <a href="https://en.wikipedia.org/wiki/Weierstrass_M-test">Wikipedia : Weierstrauss M-test</a>
 의 증명을 참고하였다.
@@ -1517,7 +1639,7 @@ $$
 \left|\frac{\partial u}{\partial\theta}\right|^2
 $$
 
-<div class="notice--info">
+<div class="notice--info" markdown="1">
 이 계산은 군대에서 전역하고 대학교 2학년으로 복학하던 시점에 했던 기억이 있다.
 지금 다시 해봐야지.
 </div>
@@ -1644,12 +1766,12 @@ $$
 \end{aligned}
 $$
 
-<p class='text-size-12'> 3.10 </p>
+<p class='text-size-12'> 3.11 </p>
 Let $n$ be an integer and let $F$ be a twice differentiable function satisfying
 
 $$r^2F''(r)+rF'(r)-n^2F(r)=0$$
 
-for $rgt0$.
+for $r\gt0$.
 The solution to the ODE is
 
 $$
@@ -1771,7 +1893,7 @@ $$F(r)=\alpha r^n+\beta r^{-n}$$
 
 taking $\alpha=\frac c{2n}$ and $\beta=c'''$.
 
-<div class="notice--info">
+<div class="notice--info" markdown="1">
 책에 나온 힌트를 따라가기는 했으나 마지막에 나오는 ODE를 풀지 못해 mathexchange에 
 <a href="https://math.stackexchange.com/questions/4986596/simple-ode-whose-solution-is-rn-or-r-n/4986639#4986639">질문</a>
 을 했었다.
@@ -1795,7 +1917,7 @@ taking $\alpha=\frac c{2n}$ and $\beta=c'''$.
 
 ### 4.1 Homogeneous second order linear ODE with constant coefficient
 
-<div class="notice--info">
+<div class="notice--info" markdown="1">
 책을 읽다보니 미분방정식을 풀어야 할 상황이 많이 나온다.
 학부 때 '미분방정식' 수업을 들었었고, 또 '미분방정식론', '편미분방정식' 등의 수업도 듣고 학점도 잘 나왔던 것으로 기억하지만, 정작 기본적인 '미분방정식' 수업에서 제대로 된 설명을 듣지 못했기에 기본이 부족하다.
 설명을 제대로 듣지 못했다면, 그때 따로 공부를 했으면 될 일이긴 하지만, 2학년이었던 당시에는 해석학과 선형대수, 집합론을 공부하는 데만 해도 시간이 부족했다.
@@ -1811,12 +1933,14 @@ $$y''(t)+c^2y(t)=0$$
 $$y''(t)+\lambda y(t)=0$$
 와 같은 식에서 $\lambda$가 음수이면 일반해가 어떻게 나오는지가 책의 내용만으로는 명확히 설명되지 않는다.
 특히 1.2절의 (3)번 식인 
+
 $$
 \begin{aligned}
 \psi''(t)-\lambda\psi(t)&=0\\
 \phi''(x)-\lambda\phi(x)&=0
 \end{aligned}\tag3
 $$
+
 의 경우에도 $\lambda$의 부호에 대해 두루뭉실하게 넘어가는 것 같다.
 마찬가지의 상황이 2.2절의
 $$G''(\theta)+\lambda G(\theta)=0$$
@@ -1882,25 +2006,19 @@ ar^2+br+c=0.\tag2
 $$
 
 Here is a theorem :
+<div class="notice--success" markdown="1">>
 
-<div class="notice--info">
 If the characteristic equation (2) has two distinct real roots $r_1$ and $r_2$, then the general solution to (1) is given by
-
 $$
 y=c_1e^{r_1x}+c_2e^{r_2x}.
 $$
-
 If (2) has a single (repeated) root  $r$, then the general solution to (1) is given by
-
 $$
 y=(c_1x+c_2)e^{rx}.
 $$
-
 If (3) has imaginary roots $\alpha\pm\beta i$, then the general solution to (1) is given by
-
 $$
 y=e^\alpha(c_1\cos\beta x+c_2\sin\beta x).
 $$
-
 In the above solutions, $c_1$ and $c_2$ are arbitrary constants.
 </div>