From 076bc5efc60458951a46298d3ea75c9ee930cda4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Petr=20Veli=C4=8Dka?= Date: Tue, 22 Apr 2025 22:12:05 +0200 Subject: [PATCH] theory - q1,q2,q3 --- theory/Makefile | 7 +++++ theory/question1.tex | 63 +++++++++++++++++++++++++++++++++++++++++++ theory/question2.tex | 46 +++++++++++++++++++++++++++++++ theory/question3.tex | 49 +++++++++++++++++++++++++++++++++ theory/theory.pdf | Bin 0 -> 81510 bytes theory/theory.tex | 39 +++++++++++++++++++++++++++ 6 files changed, 204 insertions(+) create mode 100644 theory/Makefile create mode 100644 theory/question1.tex create mode 100644 theory/question2.tex create mode 100644 theory/question3.tex create mode 100644 theory/theory.pdf create mode 100644 theory/theory.tex diff --git a/theory/Makefile b/theory/Makefile new file mode 100644 index 0000000..ab2fac7 --- /dev/null +++ b/theory/Makefile @@ -0,0 +1,7 @@ +.PHONY: clean + +theory.pdf: $(wildcard *.tex) + tectonic theory.tex + +clean: + rm theory.pdf diff --git a/theory/question1.tex b/theory/question1.tex new file mode 100644 index 0000000..5fdb1ff --- /dev/null +++ b/theory/question1.tex @@ -0,0 +1,63 @@ +\section{Question 1} + +\subsection{Part A} + +Linear 1st-order PDE is the equation +$$ \sum_{i=1}^n a_i(x) \pdv{u}{x_i}(x) = f(x) $$ +which we consider for $x \in \Omega \subset \R^n$. +By its solution we mean a continuously differentiable function $u(x)$ such that the previous equation holds pointwise for all $x \in \Omega$. + +Characteristic system for this equation is a system of ODEs $\odv{x_i}{t}(t) = a_i(x(t))$ for $i = 1, \dots, N$ and $x \in \Omega$. + +\begin{theorem}[characterization of solutions of 1st order linear PDE] + Function $\psi(x_1, \dots, x_N)$ is a solution for a linear homogeneous PDE if and only if it is constant along each of the solutions of the characteristic system. +\end{theorem} + +\begin{proof} + $\Rightarrow$: Let $\psi$ be such that $\sum_{i=1}^N a_i(x) \odv{\psi(x)}{x_i} = 0$. + Let $\{\varphi_1(t), \dots, \varphi_N(t)\}$ be one of its characteristics. Then + $$ \odv*{\psi(\varphi_1(t),\dots,\varphi_N(t))}{t} = \sum_{i=1}^N \pdv{\psi}{x_i}(\varphi_1(t), \dots, \varphi_N(t)) \odv{\varphi_i(t)}{t} = $$ + $$ \sum_{i=1}^N \pdv{\psi}{x_i}(\varphi_1(t), \dots, \varphi_N(t)) a_i(\varphi_1(t), \dots, \varphi_N(t)) = 0. $$ + + $\Leftarrow$: Let $\psi$ be constant along every characteristic. Choose $\xi \in \Omega$. Since $\{a_i\}_{i=1}^N$ are continuous, at least one characteristic goes through the point $\xi$ which corresponds to some interval $(t_1, t_2)$. Fix such characteristic $(\varphi_1(t), \dots, \varphi_N(t))$ and a point $\tau$ such that $(\varphi_1(\tau), \dots, \varphi_N(\tau)) = (\xi_1, \dots, \xi_N)$. + + Since $\psi(\varphi_1(t), \dots, \varphi_N(t))$ is constant on $(t_1, t_2)$, we have that + $$ 0 = \odv{\psi}{t}(\varphi_1(t), \dots, \varphi_N(t)) = \sum_{i=1}^N \pdv{\psi}{x_i}(\varphi_1(t),\dots, \varphi_N(t))a_i(\varphi_1(t), \dots, \varphi_N(t)). $$ + Now it is sufficient to choose $t = \tau$ and we get that $\psi$ is a solution at point $\xi$. Since this point was chosen arbitrarily, we conclude that it is a solution on the entire $\Omega$. +\end{proof} + +\subsection{Part B} + +Fundamental solution of the Poisson equation is the function +$$ \mathcal{E}(x) := \begin{cases} - \frac{1}{2\pi} \log |x|, N = 2;\\\frac{1}{(N - 2}\kappa_N|x|^{N - 2}, N \geq 3,\end{cases} $$ +where $\kappa_N$ is the measure of the $n$-dimensional unit ball. + +\begin{theorem}[Three Potentials Theorem] + Let $\omega \subset \R^N$ be a bounded region with at least Lipschitz boundary, $u \in C^2(\bar \Omega)$ and $\mathcal{E}$ is the fundamental solution of the Poisson equation. Then for all $x \in \Omega$, + $$ u(x) = -\int_\omega \mathcal{E}(x - y) \Delta u(y) dy + \int_{\delta\Omega} (\mathcal{E}(x - y) \pdv{u}{n}(y) - u(y) \pdv{\mathcal{E}}{n_y}(x - y)) dS(y), $$ + where + $$ \pdv{u}{n}(y) := \nabla u(y) n(y), \pdv{\mathcal{E}}{n_y}(x - y) := \nabla_y \mathcal{E}(x - y) n(y) $$ + and $n$ is the outer normal to $\delta\Omega$. +\end{theorem} + +\begin{proof} + Let $x \in \Omega$ and let $B_\varepsilon(x) \subset \Omega$. Let $V_\varepsilon := \Omega \setminus \overline{B_\varepsilon(x)}$. Using Green's theorem, we obtain + $$ \int_{V_\varepsilon} \mathcal{E}(x - y)\Delta u(y) dy = \int_{V_\varepsilon} \delta_y \mathcal{E}(x - y) u(y) dy + \int_{\delta V_\varepsilon} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y) - $$ + $$ \int_{\delta V_\varepsilon} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y). $$ + Denote the individual integrals $I_1 = I_2 + I_3 - I_4$ and send $\varepsilon$ to zero. + + First we can write (the functions in question are ``nice" enough) + $$I_1 \to \int_{\Omega} \mathcal{E}(x - y)\Delta u(y) dy$$ + + $I_2$ is obviously zero (fundamental solution). + + $I_3$ we can rewrite in the following manner: + $$ I_3 = \int_{\delta \Omega} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y) - \int_{\delta B_\varepsilon(x)} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y) \to $$ + $$ \int_{\delta \Omega} \mathcal{E}(x - y)\pdv{u}{n}(y) dS(y). $$ + + For the last integral, we can write + $$ I_4 = \int_{\delta \Omega} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y) - \int_{\delta B_\varepsilon(x)} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y) \to $$ + $$ \int_{\delta \Omega} u(y) \pdv{\mathcal{E}}{n_y}(x - y)dS(y) + u(x). $$ + + From this follows the equality in question. +\end{proof} diff --git a/theory/question2.tex b/theory/question2.tex new file mode 100644 index 0000000..0269329 --- /dev/null +++ b/theory/question2.tex @@ -0,0 +1,46 @@ +\section{Question 2} + +\subsection{Part A} + +Functions $u_1(x_1, \dots, x_n), u_2(x_1, \dots, x_N), u_n(x_1, \dots, x_N)$ are dependent on the set $\bar O$ ($O$ is open and bounded), if there exists a continuously differentiable function $F(u_1, \dots, u_n)$ such that +\begin{enumerate} + \item $F$ is not identically zero on any region $G \subset \R^n$. + \item for all $x = (x_1, \dots, x_N) \in \bar O$ we have that $F(u_1(x), \dots, u_n(x)) = 0$. +\end{enumerate} + +\begin{theorem}[Jacobi's criterion for function dependence] + Let $\Omega \subset \R^N$ be open, $u_i \in C^1(\Omega)$ for $i = 1,\dots,N$. Then $u_1(x), \dots, u_n(x)$ are dependent in $\Omega$ if and only if + $$ J_u(x) = \det + \begin{pmatrix} + \pdv{u_1(x)}{x_1} & \cdots & \pdv{u_1(x)}{x_N} \\ + \vdots & \ddots & \vdots \\ + \pdv{u_N(x)}{x_1} & \cdots & \pdv{u_N(x)}{x_N} + \end{pmatrix} = 0. + $$ +\end{theorem} + +\begin{theorem}[Dependence of $N$ solutions] + Let $\psi_1, \dots, \psi_N$ be solutions of a non-trivial linear homogeneous 1st-order PDE in $\Omega \subset \R^N$. Then they are dependent in $\Omega$. +\end{theorem} + +\begin{proof} + Assume that $\psi_1, \dots, \psi_N$ are not dependent in $\Omega$. Then there exists a point $x_0 \in \Omega$ such that $J_\psi(x_0) \neq 0$. From continuity of solutions we have that the determinant in question is non-zero on some neighborhood $U(x_0)$. This implies that the system + $ \begin{pmatrix} + \pdv{u_1(x)}{x_1} & \cdots & \pdv{u_1(x)}{x_N} \\ + \vdots & \ddots & \vdots \\ + \pdv{u_N(x)}{x_1} & \cdots & \pdv{u_N(x)}{x_N} + \end{pmatrix} y = 0 $ + only has a trivial solution, which is in contradiction with the assumption that our original PDE was non-trivial. +\end{proof} + +\subsection{Part B} + +Consider the following problem: $\pdv[order={2}]{v(t, x; s)}{t} - c^2\pdv[order={2}]{v(t, x; s)}{x} = 0$, $v(s, x; s) = 0$ and $\pdv{f(t, x; s)}{t}_{t=s} = f(t, x)$. + +Then $u(t, x) := \int_0^t v(t, x; s)ds$ solves general wave equation with RHS $f$. + +For $N = 1$ we have $v(t, x; s) = \frac{1}{2c}\int_{B_{c(t-s)}(x)} f(s, y)dy$ and thus +$$ u(t, x) = \frac{1}{2c} \int_0^t \int_{x-c(t-s)}^{x+c(t-s)}f(s, y) dy ds = \frac{1}{2c} \int_0^t \int_{x-c\tau}^{x+c\tau} f(t - \tau, y)dy d\tau. $$ + +For $N = 2$ we have $v(t, x; s) = \frac{1}{2\pi c} \int_{B_{c(t-s)}(x)} \frac{f(s, y)}{\sqrt{c^2(t-s)^2}-|y-x|^2} dy$. +Thus $$u(t, x) = \frac{1}{2\pi c} \int_{B_{c(t-s)}(x)} \frac{f(s, y)}{\sqrt{c^2(t-s)^2}-|y-x|^2} dy ds.$$ diff --git a/theory/question3.tex b/theory/question3.tex new file mode 100644 index 0000000..dd611e8 --- /dev/null +++ b/theory/question3.tex @@ -0,0 +1,49 @@ +\section{Question 3} + +\subsection{Part A} + +Linear 1st-order PDE is the equation +$$ \sum_{i=1}^n a_i(x) \pdv{u}{x_i}(x) = f(x) $$ +which we consider for $x \in \Omega \subset \R^n$. +By its solution we mean a continuously differentiable function $u(x)$ such that the previous equation holds pointwise for all $x \in \Omega$. + +Functions $u_1(x_1, \dots, x_n), u_2(x_1, \dots, x_N), u_n(x_1, \dots, x_N)$ are dependent on the set $\bar O$ ($O$ is open and bounded), if there exists a continuously differentiable function $F(u_1, \dots, u_n)$ such that +\begin{enumerate} + \item $F$ is not identically zero on any region $G \subset \R^n$. + \item for all $x = (x_1, \dots, x_N) \in \bar O$ we have that $F(u_1(x), \dots, u_n(x)) = 0$. +\end{enumerate} + +Let $\psi_1, \dots, \psi_{N-1}$ are solutions of the 1st-order PDE. For every subregion $\Omega' \subset \Omega$ let there be a point $x_0 \in \Omega'$ such that the matrix +$$ \begin{pmatrix}\pdv{\psi_1}{x_1} & \cdots & \pdv{\psi_1}{x_N}\\ +\vdots & \ddots & \vdots \\ +\pdv{\psi_{N-1}}{x_1} & \cdots & \pdv{\psi_{N-1}}{x_N} \end{pmatrix}$$ +has rank $N - 1$. Then $\psi_1, \psi_{N-1}$ make the fundamental system of the linear PDE. + +\begin{theorem}[Maximum Number of Independent Solutions] + Let there exist a point $x \in \Omega'$ for every subregion $\Omega' \subset \Omega$ such that $\sum |a_i(x)| > 0$. Let $\psi_1, \dots, \psi_{N-1}$ be the fundamental system of said PDE. Then $\theta \in C^1(\Omega)$ solves the PDE iff $\psi_1, \dots, \psi_{N - 1}, \theta$ are dependent in $\Omega$. +\end{theorem} + +\begin{proof} + $\implies$ was already proven before + $\impliedby$: We assume that $J_{\vec\psi, \theta} = 0$. Therefore the system + $y_1\pdv{\psi_1}{x_i} + \dots + y_{N-1}\pdv{\psi_{N-1}}{x_i} + y_N\pdv{\theta}{x_i} = 0$ for $i=1, \dots, N$ has a non-trivial solution for every $x \in \Omega$. Multiplying this equation by $a_i(x)$ and summing all of the together gives us + $$ \sum_j y_j(x)\sum_i a_i(x)\pdv{\psi_j}{x_i} + y_N(x) \sum_i a_i(x) \pdv{\theta}{x_i} = 0. $$ + Since $\psi_i$ solve the equation, we get that + $$ y_N(x) \sum_i a_i(x) \pdv{\theta}{x_i} = 0. $$ + Assume that $\theta$ does not solve are equation. Therefore we have an $x_0$ such that $\sum_{i=1}^N a_i(x_0)\pdv{\theta(x_0)}{x_i} \neq 0$. + By the usual notion of continuity we get that there exists a neighborhood of $x_0$ where this formula is non-zero. Therefore $y_N = 0$ on some $U(x_0)$. Therefore $y_1\pdv{\psi_1}{x_i} + \dots + y_{N-1}\pdv{\psi_{N-1}}{x_i} = 0$. Since all the original functions are independent, there is a point $x_1$ where the system only has a trivial solution and we get that $J_{\vec\phi, \theta}(x_1) \neq 0$, which contradicts our assumption of independence. +\end{proof} + +\subsection{Part B} + +Consider the following system: $-\Delta u = f$ on some $\Omega$, $u = g$ on $\delta\Omega$. If $\Omega$ is bounded, we talk about the inner Dirichlet problem, if $\R^n \setminus \bar \Omega$ is bounded, we talk about the outer Dirichlet problem (in this case we only care about functions $u(x) = O\left(\frac{1}{|x|^{N - 2}}\right)$ for $|x| \to \infty$. + +For the uniqueness of the solution for the inner problem, we only need to prove that the zero problem for $f = g = 0$ only has the trivial solution $u = 0$, which in fact follows from the weak maximum principle ($0 = \min_{\delta\Omega} u = \min_\Omega u \leq \max_\Omega u = \max_{\delta\Omega} u = 0$). + +Now we shall prove the uniqueness for the outer problem. Assume that $u_1$ and $u_2$ are two solutions. Then $u_1 - u_2 =: w$ solves +$\Delta w = 0$ on $\Omega$, $w = 0$ on the boundary and, $w(x) = O\left(\frac{1}{|x|^{N-2}}\right)$. + +First assume that $N \geq 3$, therefore $w(x) \to 0$ as $|x|$ goes to infinity. Let $x_0 \in \Omega$ and $\varepsilon > 0$ be given. Then there exists $R > 0$ such that $x_0, G \subset B_R$, where $G := \R^N \setminus \bar\Omega$ and $|w| \leq \varepsilon$ on $\delta B_R$. Now apply the maximum principle for the set $\Omega_R := \Omega \cup B_R$. We have that $w \leq \varepsilon$ on $\Omega_R$. The same can be said for $-w$, which means that we are done. + +Let now $N = 2$. WLOG let the origin be in $G := \R^N \setminus \bar\Omega$. Fix an open ball $B_R \subset G$. Define a mapping $F: \R^N \setminus \{0\} \to \R^N \setminus \{0\}$ defined as $F(x) = x' := \frac{xR^2}{|x|^2}$. This means that $x$ and $F(x)$ lie on the same half-line originating in the origin. Then $F$ maps $\Omega$ to some region $\Omega^* \subset B_R$, in fact, $F(\Omega) = \Omega^* \setminus \{0\}$. Define $w'(x') = w(F^{-1}(x')) = w(x)$. Obviously $w' = 0$ on $\delta\Omega^*$. We will show that $w'$ is harmonic on $\Omega^* \setminus \{0\}$. Set $r = |x|$ ad $\rho = |x'|$. Then set +$\tilde w(r, \phi) = w(x_1, x_2)$ and $\tilde w'(\rho, \phi') = w'(x_1', x_2')$ and moving to the polar coordinates we get the desired equality. Now the function is obviously bounded, therefore we can extend it harmonically. 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