connections

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courses.maths.ox.ac.uk/pluginfile.php/21926/mod_folder/content/0 ... 0-%20Connections.pdf
Connections – Part Ⅰ

Consider the heat equation

Find the steady state solution when ${∂T\over∂t}=0$, so we have $\nabla^{2} T=\frac{\partial^{2} T}{\partial x^{2}}+\frac{\partial^{2} T}{\partial y^{2}}=0$.
Find first separable solutions. Say $T(x,y)=X(x)Y(y)$. Putting this into Laplace's equation we get $X^{\prime \prime}(x) Y(y)+X(x) Y^{\prime \prime}(y)=0$ and find $\frac{X^{\prime \prime}}{X}=\frac{-Y^{\prime \prime}}{Y}=\alpha\text { (constant) }$
we know $X(0)=X(\pi)=Y(0)=0$
Then $X(x)=\left\{\begin{array}{ll}c_{1} e^{\sqrt{\alpha} x}+c_{2} e^{-\sqrt{\alpha} x} & \alpha>0 \\ c_{1} x+c_{2} & \alpha=0 \\ c_{1} \sin \sqrt{-\alpha }x+c_{2} \cos \sqrt{-\alpha }x & \alpha<0\end{array}\right.$
For non-trivial solutions we need $α<0$
$\begin{array}{l}X(0)=0 \Rightarrow c_{2}=0 \\ X(\pi)=0 \Rightarrow c_{1} \sin \sqrt{-\alpha} \pi=0\end{array}$
For $c_1≠0$, we need $\array{\sqrt{-α}=n≥1\\⇒α=-n^2}$
So $X(x)=c_1\sin nx$
and $y''=-αy=n^2y$
Hence $Y(y)=c_3e^{ny}+c_4e^{-ny}$
But $Y(0)=0$ means $c_3+c_4=0$
and $Y$ is proportional to $\sinh ny$
So separable solutions are $T_n(x,y)=\sin nx\sinh ny$
which satisfy $0=T_n(0,y)=T_n(π,y)=T_n(x,0)$

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Connections – Part Ⅱ

Consider the heat equation

We found separable solutions $T_n(x,y)=\sin nx\sinh ny\quad n≥1$
Consider $T(x, y)=\sum_{n=1}^{\infty} \alpha_n \sin n x \sinh n y$, we wish $1=T(x, \pi)=\sum_{n=1}^{\infty} \alpha_n\sin nx\sinh n\pi$
Recall that $\int_{0}^{\pi} \sin m x \sin n x{\rm d}x=\left\{\begin{array}{ll}0 & m \neq n \\ \frac{\pi}{2} & m=n\end{array}\right.$
as $\int_{0}^{\pi} \sin ^{2}mx\ {\rm d} x=\int_{0}^{\pi} \frac{1}{2}(1-\cancel{\cos2mx})\ {\rm d}x=\frac{\pi}{2}$
Then
$\begin{aligned}\underset{\lower3ex\Large∥}{\langle 1, \sin nx\rangle}&=\frac{\pi}{2}α_n \operatorname{sinh}{n\pi } \\ \int_{0}^{\pi} \sin nx{\rm d}x &=\left[-\frac{\cos nx}{n}\right]_{0}^{\pi} \\ &=-\frac{1}n\left((-1)^n-1\right)\\ \frac{\pi}{2} \alpha_n\operatorname{sinh}nπ &=\left\{\begin{array}{cc}0 & n\text{ even}\\ \frac{2}n &n \text { odd}\end{array}\right.\end{aligned}$
$\begin{aligned} T(x, y) &=\sum_{n=1}^{\infty} \alpha_{n} \sin nx \operatorname{sinh}ny \\ &=\frac{4}{\pi} \sum_{n \text { odd }} \frac{\sin nx \sinh ny}{n\sinh n\pi} \end{aligned}$
we will refer to this $T_s(x,y)$

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Connections – Part Ⅲ

In 11.1 and 11.2 we found a solution $T_s(x,y)$ to the Dirichlet problem

Say $T_1$ and $T_2$ are two solutions and set $u=T_1-T_2$. Now

Recall Green's theorem states$$\iint_{D}\left(\frac{\partial p}{\partial x}+\frac{\partial q}{\partial y}\right){\rm d}A=\int_{\partial D} p{\rm d}y-q{\rm d}x$$
Choose $\array{p=uu_x\\q=uu_y}$
then in $D$
\begin{aligned}p_x+q_y&=u_x^2+u_y^2+u\!\!\cancelto0{(u_{xx}+u_{yy})}\\&\ge0\end{aligned}
Green's theorem shows $$\iint_{D}\left(u_{x}^{2}+u_{y}^{2}\right){\rm d} A=0$$
As $u_{x}^{2}+u_{y}^{2} \geqslant 0$ then $u_{x}=0=u_{y}$ in $D$
Then $u$ is constant. As $u=0$ on $\partial D$
then $u\equiv0$ by continuity.

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Connections – Part Ⅳ

Consider the heat equation

we found a steady state solution $T_s(x, y)$
If we set $T(x, y, t)=u(x, y, t)+T_{s}(x, y)$
Then $u$ satisfies

Then $W=\left\{u(x, y, t)  \middle| \begin{array}l&\frac{\partial u}{\partial t}=κ \nabla^{2} u \\ &u(x, 0)=u(x, \pi) \\ &=u(0, y)=u(\pi, y)=0\end{array}\right\}$
is a vector space. If we consider separable solutions $u(x,y,t)=A(x)B(y)C(t)$
Arguing as in 11.1 we can find $\array{A(x)\propto\sin mx\quad m\geqslant1\\B(y) \propto \sin n y \quad n \geqslant 1}$
Putting $u$ into $u_t=κ\nabla^2u$ we get
$C^{\prime} / C=κ(\underbrace{A^{\prime \prime} / A}_{-m^{2}}+\underbrace{B^{\prime \prime} / B}_{-n^{2}})$
and hence
$C(t)=\exp \left(-κ\left(m^{2}+n^{2}\right) t\right)$
and
$u_{m,n}(x, y, t)=\exp \left(-κ\left(m^{2}+n^{2}\right) t\right) \sin m x\sin ny\quad m,n≥1$
$\big[$Given $u(x,y,0)$ can solve using Fourier analysis to get $\sum_{m, n \geq 1} \alpha_{m, n} u_{m, n}(x, y, t)\big]$
Write $u_{m,n}(x, y, t)=e^{-κ\left(m^{2}+n^{2}\right) t} \underbrace{F_{m, n}(x, y)}_{\sin m x \sin ny}$
we have continuous function $f,g$ on $[0,π]^2$ and we can introduce an inner product
$$\left\langle f, g\right\rangle=\iint_{\text {square }} f(x, y) g(x, y){\rm d}A$$
The functions $F_{m,n}(x, y)=\sin m x\sin ny$ are orthogonal with respect to this inner product.
Further note that $F_{m,n}(x,y)$ are eigenvectors of the Laplacian
$\nabla^{2}=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}$
and so
$\begin{aligned} \nabla^{2} F_{m,n} &=\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)(\sin mx \sin ny) \\ &=-\left(m^{2}+n^{2}\right) F_{m,n} . \end{aligned}$
In what way is the Laplacian "symmetric" at all?
For a matrix symmetry means ${\rm A}={\rm A}^{\sf T}$
This is equivalent
${\rm A}{\bf x}·{\bf y}=({\rm A}{\bf x})^{\sf T}{\bf y}={\bf x}^{\sf T}{\rm A}^{\sf T}{\bf y}={\bf x}^{\sf T}{\rm A}{\bf y}={\bf x}·{\rm A}{\bf y}=\RHS$
Does the Laplacian have a similar property? We'd want
$$\left\langle\nabla^{2} u, v\right\rangle=\left\langle u, \nabla^{2} v\right\rangle$$
We'd need
$$\iint_{\text {square }}\left(\nabla^{2}u\right)v-u\left(\nabla^2v\right){\rm d}A=0$$
Remember Gauss' Theorem states$$\iint_{D}\left(\frac{\partial p}{\partial x}+\frac{\partial q}{\partial y}\right){\rm d}A=\int_{\partial D} p{\rm d}y-q{\rm d}x$$
In our integrand we have$$v u_{x x}+vu_{y y}-uv_{x x}-u v_{y y}$$
If we choose $\begin{array}{l}p=u_{x} v-uv_{x} \\ q=-uv_{y}+vu_{y}\end{array}$
Then $\begin{aligned} p_{x}+q_{y}=& u_{x x} v+\cancel{u_x v_x}-\cancel{u_x v_x}-u v_{x x} \\ &-u v_{y y}+vu_{yy} \end{aligned}$
By Gauss' Theorem
\begin{aligned} &\iint_{\text {square }}\left(\nabla^{2} u\right)v-u\left(\nabla^{2} v\right){\rm d}A \\ &=\int_{\text {square }}(p_{x}+q_{y}) {\rm d}A \\ &=\int_{\rm boundary} p{\rm d}y-q{\rm d}x \\ &=\int_{\rm boundary}(u_{x} \cancelto{=0}v-\cancelto{=0}u v_{x}){\rm d}y \\ &\quad\quad\quad\quad\quad\quad\quad-(-\underset{\substack{\small∥\\0}}uv_{y}+\underset{\substack{\small∥\\0}}{v} u_{y}){\rm d}x \\ &=0 . \end{aligned}
So $\left\langle\nabla^{2} u, v\right\rangle=\left\langle u, \nabla^{2} v\right\rangle$