Finished ODEs
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\subfile{sections/solution_methods.tex}
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\subfile{sections/picard_lindelof.tex}
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\subfile{sections/linear_odes.tex}
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\end{document}
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chapters/sections/linear_odes.tex
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chapters/sections/linear_odes.tex
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% !TeX root = ../../script.tex
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\documentclass[../../script.tex]{subfiles}
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\begin{document}
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\section{Linear Differential Equation Systems}
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\begin{defi}
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Let $I$ be an open interval, and $A: I \rightarrow \realn^{n \times n}$, $b: I \rightarrow \realn^n$.
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Then the ODES
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\[
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y' = A(x)y + b(x)
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\]
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is said to be a linear differential equation system. If $b$ is the zero function, then the system is homogeneous (otherwise it's inhomogeneous).
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If $A(x) = \const.$, then the system is said to have constant coefficients.
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\end{defi}
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\begin{rem}
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\begin{enumerate}[(i)]
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\item By using substitution we can transform the equation
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\[
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y^{(n)} = a_{n-1}(x) y^{(n-1)} + a_{n-2}(x) y^{(n-2)} + \cdots + a_0 y + b(x)
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\]
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into the system
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\begin{align*}
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y_{n-1}' &= a_{n-1}(x) y_{n-2} + a_{n-2}(x) y_{n-3} + \cdots + a_0 y + b(x) \\
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y_1 &= y' \\
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y_2 &= y_1' \\
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&~~\vdots \\
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y_{n-1} &= y_{n-2}'
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\end{align*}
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\item Let $y, z$ be solutions of $y' = A(x) y + b(x)$, then $y - z$ is the solution of the related homogeneous equation $y' = A(x)y$.
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This follows from
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\begin{align*}
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(y - z)'(x) &= A(x)y(x) + b(x) - (A(x)z(x) + b(x)) \\
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&= A(x) (y - z)(x)
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\end{align*}
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\end{enumerate}
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\end{rem}
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\begin{lem}[Grönwall's Lemma]
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Let $I$ be an open interval, $x_0 \in I$, $y: I \rightarrow [0, \infty)$ continuous, $a, b \ge 0$ and
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\[
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y(x) \le a + b \abs{\int_{x_0}^x y(t) \dd{t}}
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\]
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Then
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\[
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y(x) \le a e^{b\abs{x - x_0}}
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\]
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\end{lem}
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\begin{proof}
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Here we only prove $x > x_0$, but the proof for $x \le x_0$ works analogously.
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Let $\epsilon > 0$ be arbitrary and choose
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\begin{equation}
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z(x) := a + \epsilon + b \int_{x_0}^x y(t) \dd{t}
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\end{equation}
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Then
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\begin{equation}
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z'(x) = by(x) \le bz(x) ~~\forall x \in I
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\end{equation}
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And since
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\begin{equation}
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z(t) \ge a + \epsilon > 0
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\end{equation}
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we get
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\begin{align}
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\int_{x_0}^x \frac{z'(t)}{z(t)} \dd{t} &\le b(x - x_0) \\
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\int_{x_0}^x \frac{z'(t)}{z(t)} \dd{t} &= \ln(x) - \ln(z)
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\end{align}
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Due to the monotony of the exponential function
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\begin{equation}
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z(x) \le z(x_0) e^{b(x - x_0)} = (a + \epsilon) e^{b(x - x_0)}
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\end{equation}
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So
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\begin{equation}
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y(x) \le z(x) \le (a + \epsilon) e^{b(x - x_0)} \le a e^{b(x - x_0)} ~~\forall x \in I
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\end{equation}
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\end{proof}
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From now on $I$ will always be an open interval, and
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\begin{align*}
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A: I &\rightarrow \realn^{n \times n} \\
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b: I &\rightarrow \realn^n
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\end{align*}
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are continuous, $x_0 \in I$ and $y_0 \in \realn$.
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\begin{cor}
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The IVP
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\begin{align*}
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y' = A(x)y + b(x) && y(x_0) = y_0
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\end{align*}
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has a unique maximal solution that is defined on all of $I$.
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\end{cor}
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\begin{proof}
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\begin{equation}
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\begin{split}
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f: I \times \realn^n &\longrightarrow \realn^n \\
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(x, y) &\longmapsto A(x) y + b(x)
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\end{split}
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\end{equation}
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We need to show that $f$ fulfils a local Lipschitz condition in $y$.
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Let $(x_1, y_1) \in I \times \realn^n$. Choose a compact $I_1$ such that $x_1 \in I_1 \subset I$.
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Then $A(x)$ is bounded on $I_1$, i.e.
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\begin{equation}
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\exists L > 0: ~~\norm{A(x)} \le L ~~\forall x \in I_1
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\end{equation}
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And then $\forall (x, y), (x, z) \in I_1 \times \realn^n$
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\begin{equation}
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\norm{f(x, y) - f(x, z)} = \norm{A(x)(y - z)} \le \norm{A(x)}\norm{y - z} \le L \norm{y - z}
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\end{equation}
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So $f$ fulfils a local Lipschitz condition, and thus there exists a unique maximal solution.
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Let $a, b \in \realn \cup \set{-\infty, \infty}$ such that $y: (a, b) \rightarrow \realn^n$ is the maximal solution.
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Assume $b \in I$ (so $y$ isn't defined on all of $I$). Then there exists $M, K > 0$ such that $\norm{A(x)} \le M$ and $\norm{b(x)} \le K$ and $[x_0, b]$ and
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\begin{equation}
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\begin{split}
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\norm{y(x)} = \norm{y_0 + \int_{x_0}^x y'(t) \dd{t}} &= \norm{y_0 + \int_{x_0}^x A(t)y(t) + b(t) \dd{t}} \\
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&\le \norm{y_0} + \int_{x_0}^x \norm{A(t)}\norm{y(t)} \dd{t} + \int_{x_0}^x \norm{b(t)} \dd{t} \\
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&\le \norm{y_0} + K(b - x_0) + M\int_{x_0}^x \norm{y(t)} \dd{t}
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\end{split}
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\end{equation}
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Applying Grönwall's Lemma onto $\norm{y(t)}$ yields
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\begin{equation}
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\norm{y(x)} \le \left(\norm{y_0} + K(b - x_0)\right) e^{M\abs{x - x_0}} \le \left(\norm{y_0} + K(b - x_0)\right) e^{M(b - x_0)}
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\end{equation}
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and thus $y$ is bounded on $[x_0, b)$.
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So none of the conditions from \Cref{thm:837} are satisfied, and therefore $b \notin I$.
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This mean that $y$ is defined up to the right boundary of $I$.
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\end{proof}
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\begin{rem}
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One can show that for linear systems, the Picard iteration leads to a solution that converges on all of $I$. This would lead to an alternative proof.
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\end{rem}
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\begin{cor}
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Let $y, z: I \rightarrow \realn^n$ be solutions of the ODES
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\[
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y' = A(x)y + b(x)
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\]
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Then the following are equivalent
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\begin{enumerate}[(i)]
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\item $y(x) = z(x) ~~\forall x \in I$
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\item $y(x_0) = z(x_0)$
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\item $y(x) = z(x) ~~\text{ for some } x \in I$
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\end{enumerate}
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\end{cor}
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\begin{proof}
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$(i) \implies (ii)$, $(ii) \implies (iii)$ is trivial.
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To prove $(iii) \implies (i)$, let $x_1 \in I$ such that $y_1 = y(x_1) = z(x_1)$.
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Then $y, z$ are solutions to the IVP
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\begin{align}
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y' = A(x)y + b(c) && y(x_1) = y_1
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\end{align}
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Since this problem has unique solutions
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\begin{equation}
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y = z
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\end{equation}
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must hold.
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\end{proof}
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\begin{thm}
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The solution set of the homogeneous ODES
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\[
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y' = A(x)y
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\]
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so
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\[
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V := \set[y'(x) = A(x) y(x) ~~\forall x \in I]{y: I \rightarrow \realn^n}
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\]
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is an $n$-dimensional linear subspace of $C^1(I, \realn^n)$.
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\end{thm}
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\begin{proof}
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Proving that $V$ is a vector space is trivial.
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So let $e_1, \cdots, e_n$ be a basis of $\realn^n$ and let $y_i$ be the unique solutions of the initial value problem
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\begin{align*}
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y' = A(x) y && y(x_0) = e_i ~~i \in \set{1, \cdots, n}
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\end{align*}
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Then $y_1, \cdots, y_n$ is a basis of $V$.
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To prove their linear independence, let $\alpha_1, \cdots, \alpha_n \in \realn$ such that
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\begin{equation}
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\alpha_1 y_1 + \cdots + \alpha_n y_n = 0
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\end{equation}
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then
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\begin{equation}
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\alpha_1 y_1(x_0) + \cdots + \alpha_n y_n(x_0) = \alpha_1 e_1 + \cdots + \alpha_n e_n = 0
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\end{equation}
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Since the $e_1, \cdots, e_n$ are linear independent
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\begin{equation}
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\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0
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\end{equation}
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To prove that the $y_1, \cdots, y_n$ span $V$, set $z \in V$ and choose $\alpha_1, \dots, \alpha_n \in \realn$ such that
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\begin{equation}
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\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n = z(x_0)
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\end{equation}
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Then the $z$ and $\alpha_1 y_1 + \cdots + \alpha_n y_n$ are maximal solutions of the ODES that are equal in $x_0$. Thus
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\begin{equation}
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z = \alpha_1y_1 + \cdots + \alpha_n y_n
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\end{equation}
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\end{proof}
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\begin{defi}
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A basis $y_1, \cdots, y_n$ of $V$ is said to be a fundamental system of the ODES
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\[
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y' = A(x) y
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\]
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Analogously, $n$ linearly independent solutions of the equation
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\[
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y^{(n)} = a_{n-1}(x) y^{(n-1)} + a_{n-2}(x) y^{(n-2)} + \cdots + a_0 y
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\]
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are said to be a fundamental system.
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\end{defi}
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\begin{eg}
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Consider the inhomogeneous equation
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\[
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y' = \sin(x)y + \sin(x)\cos(x)
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\]
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First, find the solutions to the homogeneous equation
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\[
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\frac{y'}{y} = \sin(x)
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\]
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This can be done via integration
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\begin{align*}
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\int \frac{y'(t)}{y(t)} \dd{t} &= -\cos(x) + c \\
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\ln y + c &= -\cos(x) + c
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\end{align*}
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Then the solution is
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\[
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y = K e^{-\cos(x)}
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\]
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The fundamental system in this case is $e^{-\cos x}$.
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We can use a technique called "variation of the constant" to find a solution of the inhomogeneous equation. Define
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\[
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y(x) = C(x) e^{-\cos(x)}
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\]
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Deriving this gives
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\[
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y'(x) = C'(x) e^{-\cos(x)} - C(x) \sin(x) e^{-\cos(x)}
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\]
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Resubstituting this into the initial equation yields
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\begin{align*}
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C'(x) e^{-\cos(x)} + \cancel{C(x)\sin(x)e^{-\cos(x)}} &= \cancel{C(x)\sin(x)e^{-\cos(x)}} + \sin(x)\cos(x) \\
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C'(x) e^{-\cos(x)} &= \sin(x)\cos(x) \\
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C'(x) &= \sin(x)\cos(x)e^{\cos(x)} \\
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C(x) &= (1 - \cos(x))e^{\cos(x)}
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\end{align*}
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So the general solution to the ODE is
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\[
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y(x) = 1 - \cos(x) + K e^{-\cos(x)}
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\]
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\end{eg}
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\begin{thm}
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Let $y_1, \cdots, y_n$ be a fundamental system for $y' = A(x) y$.
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Define an $n \times n$-matrix
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\[
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W(x) := (y_1(x), y_2(x), \dots, y_n(x))
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\]
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Then $W(x)$ is invertible $\forall x \in I$ and
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\begin{align*}
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z: I &\longrightarrow \realn^n \\
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x &\longmapsto W(x) \int_{x_0}^x \inv{W(t)}b(t) \dd{t}
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\end{align*}
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is a solution to the inhomogeneous system
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\[
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y' = A(x)y + b(x)
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\]
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\end{thm}
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\begin{proof}
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According to the prerequisites the $y_1, \cdots, y_n$ are linearly independent, so the $y_1(x), \dots, y_n(x)$ are also linearly independent in $\realn^n$.
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Thus
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\begin{equation}
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\det W(x) \ne 0 \implies W(x) \text{ invertible}
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\end{equation}
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Deriving this yields
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\begin{equation}
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W'(x) = A(x)W(x)
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\end{equation}
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which means the $i$-th column of this equation is $y_i'(x) = A(x)y_i(x)$.
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Deriving $z$ gives us
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\begin{equation}
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\begin{split}
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z'(x) &= W'(x) \int_{x_0}^x \inv{W(t)} b(t) \dd{t} + W(x)\inv{W(x)} b(x) \\
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&= A(x) z(x) + b(x)
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\end{split}
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\end{equation}
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To apply the fundamental theorem, $W(t)b(t)$ should be continuous.
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The mapping $A \mapsto \inv{A}$ is continuous on $Gl(n)$ (space of invertible matrices).
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\end{proof}
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\begin{eg}
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Consider the system
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\begin{align*}
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u' = v + \sin(x) && v' = -u + \cos(x)
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\end{align*}
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The homogeneous system in this case is
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\[
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\begin{pmatrix}
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u \\ v
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\end{pmatrix}'
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=
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\begin{pmatrix}
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0 & 1 \\ -1 & 0
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\end{pmatrix}
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\begin{pmatrix}
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u \\ v
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\end{pmatrix}
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\]
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The fundamental system is
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\begin{align*}
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y_1(x) = \begin{pmatrix}
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\sin \\ \cos
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\end{pmatrix}(x)
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&&
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y_2 = \begin{pmatrix}
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\cos \\ -\sin
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\end{pmatrix}(x)
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\end{align*}
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Then define
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\begin{align*}
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z(x) &= C_1(x)y_1(x) + C_2(x)y_2(x) \\
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&= \underbrace{\begin{pmatrix}
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\sin(x) & \cos(x) \\
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\cos(x) & -\sin(x)
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\end{pmatrix}}_{W(x)}
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\begin{pmatrix}
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C_1(x) \\ C_2(x)
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\end{pmatrix}
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\end{align*}
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Deriving this yields
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\begin{align*}
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z'(x) &= C_1'(x)y_1(x) + \cancel{C_1(x)y_1'(x)} + C_2'(x)y_2(x) + \cancel{C_2(x)y_2'(x)} \\
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&= \cancel{C_1(x)Ay_1(x)} + \cancel{C_2(x)Ay_2(x)} + b(x) \\
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&= b(x)
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\end{align*}
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This can be explicitly solved
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\begin{align*}
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C_1'(x)\sin(x) + C_2'(x)\cos(x) &= \sin(x) \\
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C_1'(x)\cos(x) - C_2'(x)\sin(x) &= \cos(x)
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\end{align*}
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Leading to
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\begin{align*}
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C_1'(x) &= C_1'(x)(\sin^2(x) + \cos^2(x)) = \sin^2(x) + \cos^2(x) = 1 \\
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C_2'(x) &= C_2'(x)(\cos^2(x) - \sin^2(x)) = 0
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\end{align*}
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Thus
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\begin{align*}
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C_1(x) &= x \\
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C_2(x) &= 0
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\end{align*}
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So the general solution of the homogeneous equation is
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\[
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y_h = \begin{pmatrix}
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x \sin(x) \\ x \cos(x)
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\end{pmatrix}
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\]
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Our next goal is to find a solution of $y' = Ay$ with $A \in \realn^{n \times n}$ constant.
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In one dimension the solution would be
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\[
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y = Ce^{Ax}
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\]
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Does this also hold for $n > 1$?
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\end{eg}
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\begin{rem}
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Let $A \in \realn^{n \times n}$
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\[
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e^{Ax} = \sum_{k=0}^{\infty} \rec{k!}(Ax)^k = \sum_{k=0}^{\infty} \rec{k!} A^k x^k
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\]
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We have
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\[
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\sum_{k=0}^{\infty} \rec{k!} \norm{A^k x^k} \le \sum_{k=0}^{\infty} \frac{\abs{x}^k}{k!} \norm{A}^k = e^{\norm{x}\norm{A}} < \infty
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\]
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Thus, $e^{Ax}$ is defined $\forall A \in \realn^{n \times n}, ~\forall x \in \realn$. Deriving this yields
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\[
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\dv{x}e^{Ax} = \sum_{k=1}^{\infty} \rec{k!} A^k x^{k-1} = A \sum_{k=1}^{\infty} \rec{(k-1)!} A^{k-1} x^{k-1} = Ae^{Ax}
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\]
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\end{rem}
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\begin{thm}
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Let $A \in \realn^{n \times n}$. The IVP
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\begin{align*}
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y' = Ay && y(x_0) = y_0
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\end{align*}
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is solved exactly by
|
||||
\[
|
||||
y(x) = e^{A(x - x_0)}y_0
|
||||
\]
|
||||
\end{thm}
|
||||
\begin{proof}
|
||||
Without proof.
|
||||
\end{proof}
|
||||
|
||||
\begin{rem}
|
||||
\begin{enumerate}[(i)]
|
||||
\item The problem of solving IVPs can be reduced to a problem of calculating a matrix exponential.
|
||||
\item The following does NOT generall hold
|
||||
\begin{align*}
|
||||
\dv{t}e^{A(x)} &= A'(x) e^{A(x)} \\
|
||||
e^{A + B} &= e^A e^B
|
||||
\end{align*}
|
||||
\item Let $v$ be an eigenvector of $A$ to the eigenvalue $\lambda$. Then
|
||||
\begin{align*}
|
||||
e^{Ax} v = \left(\sum_{k=0}^{\infty} \rec{k!} A^k x^k\right) v &= \sum_{k=0}^{\infty} \frac{x^k}{k!} A^k v \\
|
||||
&= \left(\sum_{k=0}^{\infty} \frac{x^k}{k!} \lambda^k\right) v = e^{\lambda x} v
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
\end{rem}
|
||||
|
||||
\begin{eg}
|
||||
Consider the IVP
|
||||
\begin{align*}
|
||||
\begin{pmatrix}
|
||||
y \\ z
|
||||
\end{pmatrix}'
|
||||
= \underbrace{\begin{pmatrix}
|
||||
0 & 1 \\ 1 & 0
|
||||
\end{pmatrix}}_A
|
||||
\begin{pmatrix}
|
||||
y \\ z
|
||||
\end{pmatrix}
|
||||
&& y_0 = \begin{pmatrix}
|
||||
1 \\ 0
|
||||
\end{pmatrix}
|
||||
\end{align*}
|
||||
This $A$ is diagonalizable and has the eigenvalues
|
||||
\begin{align*}
|
||||
\lambda_1 = -1 && \lambda_2 = 1
|
||||
\end{align*}
|
||||
and the eigenvectors
|
||||
\begin{align*}
|
||||
v_1 = \begin{pmatrix}
|
||||
1 \\ 1
|
||||
\end{pmatrix}
|
||||
&&
|
||||
v_2 = \begin{pmatrix}
|
||||
1 \\ -1
|
||||
\end{pmatrix}
|
||||
\end{align*}
|
||||
So we can solve this ODES by calculating
|
||||
\begin{align*}
|
||||
e^{Ax} y_0 = e^{Ax} \cdot \frac{1}{2}(v_1 + v_2) &= \frac{1}{2}\left(e^{\lambda_1 x} v_1 + e^{\lambda_2 x} v_2\right) \\
|
||||
&= \frac{1}{2} \left(e^x v_1 + e^{-x} v_2\right)
|
||||
\end{align*}
|
||||
And thus
|
||||
\begin{align*}
|
||||
y(x) = \frac{1}{2} \left(e^x + e^{-x}\right) && z(x) = \frac{1}{2} \left(e^x - e^{-x}\right)
|
||||
\end{align*}
|
||||
\end{eg}
|
||||
|
||||
\begin{rem}
|
||||
Often the process above is formulated as follows:
|
||||
Start by defining
|
||||
\[
|
||||
y(x) = c \cdot e^{\lambda x} v ~~c, \lambda \in \field \text{ and } v \in \realn
|
||||
\]
|
||||
Insert this into the ODE
|
||||
\[
|
||||
c\lambda e^{\lambda x} = c e^{\lambda x} Av
|
||||
\]
|
||||
So $\lambda$ is an eigenvalue of $A$ to the eigenvector $v$.
|
||||
\end{rem}
|
||||
|
||||
\begin{thm}
|
||||
Let $A \in \realn^{n \times n}$ be diagonalizable, and $v_1, \cdots, v_n$ is a basis of eigenvectors to the eigenvalues $\lambda_1, \cdots, \lambda_n$.
|
||||
Then the functions
|
||||
\[
|
||||
y_i(x) = e^{\lambda_i x} v_i ~~i \in \set{1, \cdots, n}
|
||||
\]
|
||||
are a fundamental system to the ODES
|
||||
\[
|
||||
y' = Ay
|
||||
\]
|
||||
\end{thm}
|
||||
\begin{proof}
|
||||
We have
|
||||
\begin{equation}
|
||||
e^{Ax} v_i = e^{\lambda_i x} v_i
|
||||
\end{equation}
|
||||
In $x = 0$ the
|
||||
\begin{equation}
|
||||
y_1(0) = v_1, ~y_2(0) = v_2, ~\cdots, ~y_n(0) = v_n
|
||||
\end{equation}
|
||||
are linearly independent, so the $y_1, \cdots, y_n$ are also linearly independent.
|
||||
\end{proof}
|
||||
|
||||
\begin{rem}
|
||||
\begin{enumerate}[(i)]
|
||||
\item There is a special case, where $A \in \realn^{n \times n}$ is not diagonalizable in the real number space, but in the complex number space.
|
||||
Let $\lambda = \lambda_r + \lambda_i$ be the eigenvalue to the eigenvector $v = v_r + v_i$. Then
|
||||
\begin{gather*}
|
||||
e^{\lambda_r x} (v_r \sin(\lambda_i x) + v_i \cos(\lambda_i x)) \\
|
||||
e^{\lambda_r x} (v_r \cos(\lambda_i x) + v_i \sin(\lambda_i x))
|
||||
\end{gather*}
|
||||
be linearly independent, real-valued solutions. To solve the IVP
|
||||
\begin{align*}
|
||||
y(x) = C e^{\lambda x} v && y(0) = y_0
|
||||
\end{align*}
|
||||
we want to transform it into an eigenvalue problem and find a solution to that. Doing that gives us
|
||||
\[
|
||||
y(x) = C_1 e^{\lambda_1 x} v_1 + \cdots + C_n e^{\lambda_n x} v_n
|
||||
\]
|
||||
By inserting the initial condition we can find
|
||||
\[
|
||||
C_1 v_1 + C_2 v_2 + \dots + C_n v_n = y_0
|
||||
\]
|
||||
Finding the $C_1, \cdots, C_n$ shows us that the solution is automatically real.
|
||||
|
||||
\item If $A$ is not diagonalizable one can try and bring $A$ into Jordan normal form.
|
||||
\end{enumerate}
|
||||
\end{rem}
|
||||
|
||||
\begin{eg}
|
||||
Consider the IVP
|
||||
\begin{align*}
|
||||
\begin{pmatrix}
|
||||
y \\ z
|
||||
\end{pmatrix}'
|
||||
=
|
||||
\begin{pmatrix}
|
||||
0 & 1 \\ -1 & 0
|
||||
\end{pmatrix}
|
||||
\begin{pmatrix}
|
||||
y \\ z
|
||||
\end{pmatrix}
|
||||
&&
|
||||
y_0 = \begin{pmatrix}
|
||||
1 \\ 0
|
||||
\end{pmatrix}
|
||||
\end{align*}
|
||||
The eigenvalues and eigenvectors are
|
||||
\begin{align*}
|
||||
\lambda_1 &= i & \lambda_2 &= -i \\
|
||||
v_1 &= \begin{pmatrix}
|
||||
1 + i \\ -1 + i
|
||||
\end{pmatrix}
|
||||
&
|
||||
v_2 &= \begin{pmatrix}
|
||||
1 - i \\ -1 - i
|
||||
\end{pmatrix}
|
||||
\end{align*}
|
||||
Thus we have the general solution
|
||||
\[
|
||||
C_1 e^{ix} v_1 + C_2 e^{-ix} v_2
|
||||
\]
|
||||
which expands to
|
||||
\begin{align*}
|
||||
(i + 1) C_1 \cancel{e^{i0}} + (1 - i)C_2 \cancel{e^{-i0}} &= 1 \\
|
||||
(i - 1) C_1 \cancel{e^{i0}} + (-1 - i)C_2 \cancel{e^{-i0}} &= 0
|
||||
\end{align*}
|
||||
and solves to
|
||||
\begin{align*}
|
||||
C_1 = \frac{1}{4} (1 - i) && C_2 = \frac{1}{4} (1 + i)
|
||||
\end{align*}
|
||||
So the solution to the IVP is
|
||||
\begin{align*}
|
||||
y(x) = \cos(x) && z(x) = -\sin(x)
|
||||
\end{align*}
|
||||
\end{eg}
|
||||
|
||||
\begin{thm}
|
||||
Let $a_1, \cdots, a_{n-1} \in \cmpln$. Let $\lambda_1, \cdots, \lambda_k$ be the roots of the polynomial
|
||||
\[
|
||||
a_0 + a_1 \lambda + \cdots + a_{n-1} \lambda^{n-1} + \lambda^n
|
||||
\]
|
||||
and $\nu_1, \cdots, \nu_k$ their multiples. Then the functions
|
||||
\[
|
||||
x \longmapsto x^l e^{\lambda_i x} ~~i \in \set{1, \cdots, k}, l \in \set{0, \cdots, \nu_{i_1}}
|
||||
\]
|
||||
form a fundamental system for
|
||||
\[
|
||||
a_0 y + a_1 y' + \cdots + a_{n-1} y^{(n-1)} + y^{(n)}
|
||||
\]
|
||||
\end{thm}
|
||||
\end{document}
|
||||
|
|
@ -326,7 +326,7 @@ We define $\metric$ to be a metric space, $x \in X$ and $A \subset X$. Then
|
|||
d(x, A) = \inf\set[y \in A]{d(x, y)}
|
||||
\]
|
||||
|
||||
\begin{thm}
|
||||
\begin{thm}\label{thm:837}
|
||||
Let $D \subset \realn \times \realn^n$ be open, $(x_0, y_0) \subset D$ and $f: D \rightarrow \realn^n$ continuous and satisfying the local Lipschitz condition in terms of $y$.
|
||||
Let $a, b \in \realn \cup \set{-\infty, \infty}$ such that
|
||||
\[
|
||||
|
|
|
|||
BIN
script.pdf
BIN
script.pdf
Binary file not shown.
|
|
@ -35,6 +35,7 @@
|
|||
%\DeclareMathOperator{\dim}{dim}
|
||||
\DeclareMathOperator{\sgn}{sgn}
|
||||
\DeclareMathOperator{\diag}{diag}
|
||||
\DeclareMathOperator{\const}{const}
|
||||
|
||||
\newcommand{\natn}{\mathbb{N}}
|
||||
\newcommand{\intn}{\mathbb{Z}}
|
||||
|
|
|
|||
Loading…
Reference in a new issue