常微分方程图解

%我的 tex 模版
\documentclass[UTF8,a1paper,landscape]{ctexart}%UTF8 中文支持,a1paper 纸张大小,landscape 横向版面,ctexart 中文文章
\usepackage{tikz}%图包
\usetikzlibrary{trees}%树包

\usepackage{amsmath}

\usepackage{geometry}%页边距设置
\geometry{top=5cm,bottom=5cm,left=5cm,right=5cm}

\usepackage{fancyhdr}%页头页尾页码设置
\pagestyle{fancy}
\begin{document}
    \title{常微分方程图解}
    \author{dengchaohai}
    \maketitle
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    \newpage%另起一页
    \part{一阶常微分方程}
    \section*{一阶常微分方程逻辑关系图解}
    \begin{center}%居中
        \begin{tikzpicture}
        [
        grow=right,
        r/.style={rectangle,draw,fill=red!20,align=center,rounded corners=.8ex},
        g/.style={rectangle,draw,fill=green!20,align=center,rounded corners=.8ex},
        b/.style={rectangle,draw,fill=blue!20,align=center,rounded corners=.8ex},
        grow via three points={one child at (4,-4) and two children at (4,-4) and (4,-8)},
        edge from parent path={(\tikzparentnode.south)|-(\tikzchildnode.west)},
        ]%属性定义
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \node(-1)at(0,12)[r]{解的存在唯一性\\$/x-x_0/\leq h$}
        child{node(g)[g]{解的延拓}}
        child{node(h)[g]{解对初值的连续可微性}}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{};
        \node(-2)at(10,8)[b]{解的存在空间(局部)\\$|x-x_0|\leq h$};
        \node(-3)at(15,8)[b]{通解\\$y=y(c,x)$};
        \node(-4)at(20,8)[b]{定解\\$y=y(x_0,y_0,x)$};
        \node(-5)at(25,8)[b]{解的存在空间(饱和)\\$(c,d)$};
        \node(-6)at(15,0)[b]{包络};
        \node(-7)at(20,4)[b]{解对初值的连续可微性\\$y=y(x_0,y_0,x,\lambda)$};
        \node(-8)at(40,0)[b]{奇解}
        child{node[b]{$c-$判别曲线\\$\Phi(c,x,y)=0,\Phi‘_c=0$}}
        child{node[b]{$p-$判别曲线\\$F(x,y,p)=0,F‘_p=0$}};

        \draw[->](-2)--(-3);
        \draw[->](-3)--(-6);
        \draw[->](-4)--(-7);
        \draw[->](-3)to node[above]{初值$(x_0,y_0)$}(-4);
        \draw[->](-4)--(-5);
        \draw[->](-2)--(10,9)to node[above]{延拓}(25,9)--(-5);
        \draw[->](-6)--(-8);

        \node(0)at(0,0)[r]{一阶常微分方程\\$F(x,y,y‘)=0$}
        child{node[g]{显式\\$y‘=f(x,y)$}
            child{node(a)[b]{3分式微分方程\\$\frac{dy}{dx}=\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}$}}
            child{node(b)[b]{7伯努利微分方程\\$\frac{dy}{dx}=P(x)y+Q(x)y^n$}}}
        child[missing]{}
        child[missing]{}
        child{node[g]{隐式\\$F(x,y,y‘)=0$}
            child{node(c)[b]{8显解$x$\\$x=f(y,y‘)$}}
            child{node(d)[b]{9显解$y$\\$y=f(x,y‘)$}}
            child{node(e)[b]{10不含$x$\\$F(y,y‘)=0$}}
            child{node(f)[b]{11不含$y$\\$F(x,y‘)=0$}}};

        \node(1)at(15,-8)[b]{2齐次微分方程\\$\frac{dy}{dx}=f(\frac{y}{x})$};
        \node(2)at(20,-8)[b]{1变量分离方程\\$\frac{dy}{dx}=\psi(x)\varphi(y)$};
        \node(3)at(25,-8)[b]{4恰当微分方程\\$M(x,y)dx+N(x,y)dy=0$};
        \node(4)at(15,-12)[b]{6非齐次线性微分方程\\$\frac{dy}{dx}=P(x)y+Q(x)$};
        \node(5)at(20,-12)[b]{5齐次线性微分方程\\$\frac{dy}{dx}=P(x)y$};

        \draw[->](a)--(1);
        \draw[->](1)--(2);
        \draw[->](2)--(3);
        \draw[->](b)--(4);
        \draw[->](4)--(5);
        \draw[->](5)--(2);
        \draw(-1)--(0);
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \end{tikzpicture}
    \end{center}
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    \newpage
    \section*{一阶常微分方程图解对应解法}
    \begin{itemize}
        \item 1.
        \[
        \begin{split}
        given:\quad\frac{dy}{dx}=\psi(x)\varphi(y).\        if\quad\varphi(y)=0\        &\Rightarrow \varphi(y)=0\        &\Rightarrow y=y_0\        if\quad\varphi(y)\ne0\        &\Rightarrow \frac{1}{\varphi(y)}dy=\psi(x)dx\        &\Rightarrow \int\frac{1}{\varphi(y)}dy=\int\psi(x)dx+c\        &\Rightarrow \Phi(c,x,y)=0\        &\Rightarrow y=y(c,x)\        \end{split}
        \]
        \item 2.
        \[
        \begin{split}
        given:\quad\frac{dy}{dx}=f(\frac{y}{x}).\        let\quad u=\frac{y}{x}
        &\Rightarrow y=ux\        &\Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u\        &\Rightarrow \frac{du}{dx}x+u=f(u)\        &\Rightarrow \frac{du}{dx}=\frac{f(u)-u}{x}\        &\Rightarrow \frac{du}{dx}=\psi(u)\varphi(x)\        \end{split}
        \]
        \item 3.
        \[
        \begin{split}
        given:\quad\frac{dy}{dx}=\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}.\        if\quad c_1=c_2=0\        &\Rightarrow \frac{dy}{dx}=\frac{a_1+b_1\frac{y}{x}}{a_2+b_2\frac{y}{x}}\        &\Rightarrow \frac{dy}{dx}=f(\frac{y}{x})\        let\quad u=\frac{y}{x}\        &\Rightarrow \frac{dy}{dx}=\psi(u)\        if\quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k\\
        &\Rightarrow \frac{dy}{dx}=k\        if\quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=k\ne\frac{c_1}{c_2}\        let\quad u=a_1x+b_1y\        &\Rightarrow \frac{du}{dx}=a_1+b_1\frac{dy}{dx}=a_1+b_1\frac{ku+c_1}{u+c_2}\        &\Rightarrow \frac{du}{dx}=\psi(u)\        if\quad \frac{a_1}{a_2}\ne\frac{b_1}{b_2}\        &\Rightarrow a_1x+b_1y+c_1=0,a_2x+b_2y+c_2=0\        &\Rightarrow x=x_0,y=y_0\        let\quad x=X+x_0,y=Y+y_0\        &\Rightarrow \frac{dy}{dx}=\frac{dY}{dX}=\frac{a_1X+b_1Y}{a_2X+b_2X}=\frac{a_1+b_1\frac{Y}{X}}{a_2+b_2\frac{Y}{X}}\        let\quad u=\frac{Y}{X}\        &\Rightarrow \frac{du}{dX}=\psi(u)\varphi(X)\\
        \end{split}
        \]
        \item 4.
        \[
        \begin{split}
        given:\quad M(x,y)dx+N(x,y)dy=0.\        if\quad \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{-M}=\varphi(y)\        &\Rightarrow \mu=e^{\int\varphi(y)dy}\        if\quad \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\psi(x)\        &\Rightarrow \mu=e^{\int\varphi(x)dx}\        let\quad \mu Mdx+\mu Ndy=0\        &\Rightarrow u=\int\mu Mdx+\varphi(y)\        &\Rightarrow \frac{du}{dy}=\mu N\        &\Rightarrow \varphi(y)\        &\Rightarrow u=\Phi(x,y)\        &\Rightarrow \Phi(x,y)=c\        \end{split}
        \]
        \item 5.
        \[
        \begin{split}
        given:\quad\frac{dy}{dx}=P(x)y.\        &\Rightarrow y=ce^{\int P(x)dx}\        \end{split}
        \]
        \item 6.
        \[
        \begin{split}
        given:\quad\frac{dy}{dx}=P(x)y+Q(x).\        &\Rightarrow y=e^{\int P(x)dx}(\int Q(x)e^{-\int P(x)dx}dx+c)\        \end{split}
        \]
        \item 7.
        \[
        \begin{split}
        given:\quad\frac{dy}{dx}=P(x)y+Q(x)y^n.\        y^{-n}\        &\Rightarrow y^{-n}\frac{dy}{dx}=y^{-n}P(x)y+y^{-n}Q(x)y^n\        &\Rightarrow y^{-n}\frac{dy}{dx}=P(x)y^{1-n}+Q(x)\        let\quad z=y^{1-n}\        &\Rightarrow \frac{dz}{dx}=(1-n)y^{-n}\frac{dy}{dx}\        &\Rightarrow \frac{dz}{dx}=(1-n)P(x)z+(1-n)Q(x)\        &\Rightarrow \frac{dz}{dx}=\psi (x)z+\varphi(x)\        \end{split}
        \]
        \item 8.
        \item 9.
        \[
        \begin{split}
        given:\quad y=f(x,y‘).\\
        let\quad p=y‘\\
        &\Rightarrow y=f(x,p)\        &\Rightarrow \frac{dy}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial p}\frac{dp}{dx}\        &\Rightarrow p=\varphi(c,x)\        &\Rightarrow x=\psi(c,p)\        &\Rightarrow y=f(\psi(c,p),p)\        \end{split}
        \]
        \item 10.
        \item 11.
        \[
        \begin{split}
        given:\quad F(x,y‘)=0.\\
        let\quad p=y‘=p(t,x)\\
        &\Rightarrow x=\psi(t)\        &\Rightarrow p=\varphi(t)\        &\Rightarrow dy=pdx=\varphi(t)\psi‘(t)dt\\
        &\Rightarrow y=\int\varphi(t)\psi‘(t)dt+c\\
        \end{split}
        \]
    \end{itemize}

    \newpage
    \part{高阶常微分方程}
    \section*{高阶常微分方程逻辑关系图解}
    \begin{center}%居中
        \begin{tikzpicture}
        [
        grow=right,
        r/.style={rectangle,draw,fill=red!20,align=center,rounded corners=.8ex},
        g/.style={rectangle,draw,fill=green!20,align=center,rounded corners=.8ex},
        b/.style={rectangle,draw,fill=blue!20,align=center,rounded corners=.8ex},
        grow via three points={one child at (4,-2) and two children at (4,-2) and (4,-4)},
        edge from parent path={(\tikzparentnode.south)|-(\tikzchildnode.west)},
        ]%属性定义
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \node(0)at(0,0)[r]{高阶常微分方程}
        child[missing]{}
        child[missing]{}
        child{node(a)[g]{齐次微分方程}
            child{node[b]{常系数线性齐次微分方程}
                child{node[b]{特征根法求通解}
                    child{node[b]{单根}
                        child{node[b]{实单根\\$e^{\lambda_1t}$}}
                        child{node[b]{$e^{\lambda t}=e^{\alpha t}(cos(\beta t)+sin(\beta t))$\\复单根\\(复$=2$,单$=1\Rightarrow 2\times1$的矩阵)\\$e^{\lambda_1t}$\\$e^{\overline{\lambda_1}t}$}}}
                    child[missing]{}
                    child[missing]{}
                    child{node[b]{重根}
                        child{node[b]{实$k$重根\\$\underbrace{e^{\lambda_1t},te^{\lambda_1t},\dots,t^{k-1}e^{\lambda_1t}}_k$}}
                        child[missing]{}
                        child{node[b]{$e^{\lambda t}=e^{\alpha t}(cos(\beta t)+sin(\beta t))$\\复$k$重根\\(复$=2$,单$=k\Rightarrow 2\times k$的矩阵)\\$\underbrace{e^{\lambda_1t},te^{\lambda_1t},\dots,t^{k-1}e^{\lambda_1t}}_k$\\$\underbrace{e^{\overline{\lambda_1}t},te^{\overline{\lambda_1}t},\dots,t^{k-1}e^{\overline{\lambda_1}t}}_k$}}}}}}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child[missing]{}
        child{node(b)[g]{非齐次微分方程}
            child{node[b]{常系数非线性齐次微分方程}
                child{node[b]{待定系数法求特解}
                    child{node[b]{$f(t)=(b_0t^m+b_1t^{m-1}+\dots+b_m)e^{\lambda t}$}
                        child{node[b]{$k$是特征值$\lambda$的重数,若$\lambda$不是特征值,那么$k=0$\\\~{x}$=t^k((b_0t^m+b_1t^{m-1}+\dots+b_m)e^{\lambda t})$}}}
                    child[missing]{}
                    child[missing]{}
                    child{node[b]{$f(t)=((P(t)cos(\beta t)+Q(t)sin(\beta t))e^{\alpha t}$}
                        child{node[b]{$e^{\lambda t}=e^{\alpha t}(cos(\beta t)+sin(\beta t))$\\$k$是特征值$\lambda$的重数,若$\lambda$不是特征值,那么$k=0$\\\~{x}$=t^k(((P(t)cos(\beta t)+Q(t)sin(\beta t))e^{\alpha t})$}}}}}};

        \node(1)at(4,0)[g]{降阶法\\$x=x(t)$}
        child{node[b]{不显含$x$,降$k$阶\\$F(t,x^{(k)},x^{(k+1)},\dots,x^{(n)})=0$}}
        child{node[b]{不显含$t$,降$1$阶\\$F(x,x‘,x‘‘,\dots,x^{(n)})=0$}};
        \draw[->](0)--(1);
        \draw[->,green](a)to node[right]{常数变易法$c\Rightarrow c(x)$}(b);
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \end{tikzpicture}
    \end{center}
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    \newpage
    \section*{高阶常微分方程图解对应解法}
    \begin{itemize}
        \item 1.降$1$阶
        \[
        \begin{split}
        given\quad F(x,x‘,x‘‘,\dots,x^{(n)})=0.\\
        let\quad y=x‘\\
        &\Rightarrow x‘‘=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=y‘\frac{dy}{dx}\\
        &\Rightarrow F(x,y,y‘\frac{dy}{dx},\dots,\cdots\frac{dy}{dx})=0\\
        &\Rightarrow y=y(t,c_1,c_2,\dots,c_{n-1})\        &\Rightarrow \int ydx=\int x‘dx\\
        &\Rightarrow x\\
        \end{split}
        \]
        \item 2.降$k$阶
        \[
        \begin{split}
        given\quad F(t,x^{(k)},x^{(k+1)},\dots,x^{(n)})=0.\        let\quad y=x^{(k)}\        &\Rightarrow F(t,y,y‘,\dots,y^{(n-k)})=0\\
        &\Rightarrow y=y(t,c_1,c_2,\dots,c_{n-k})\        &\Rightarrow \int ydx=\int x^{(k)}dx\        &\Rightarrow x=\underbrace{\idotsint}_k x^{(k)}dx\\
        \end{split}
        \]
    \end{itemize}
    \newpage
    \part{常微分方程组}
    \section*{常微分方程组逻辑关系图解}
    \begin{center}%居中
        \begin{tikzpicture}
        [
        grow=right,
        r/.style={rectangle,draw,fill=red!20,align=center,rounded corners=.8ex},
        g/.style={rectangle,draw,fill=green!20,align=center,rounded corners=.8ex},
        b/.style={rectangle,draw,fill=blue!20,align=center,rounded corners=.8ex},
        grow via three points={one child at (4,-2) and two children at (4,-2) and (4,-4)},
        edge from parent path={(\tikzparentnode.south)|-(\tikzchildnode.west)},
        ]%属性定义
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        123

        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \end{tikzpicture}
    \end{center}
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}
时间: 2024-11-09 17:12:04

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