常微分方程

利用首次积分法(First Integral)求解对称形式的常微分方程组：
$$\frac{{\rm\,d}x}{-x+y+z}=\frac{{\rm\,d}y}{x-y+z}=\frac{{\rm\,d}z}{x+y-z}$$

$$\frac{{\rm\,d}x}{-x^2+y^2+z^2}=\frac{{\rm\,d}y}{x^2-y^2+z^2}=\frac{{\rm\,d}z}{x^2+y^2-z^2}$$

$$\begin{align*} &&\frac{{\rm\,d}x}{-x+y+z}&=\frac{{\rm\,d}y}{x-y+z}\\ &\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{-2\left(x-y\right)}&=\frac{{\rm\,d}\left(x+y+z\right)}{x+y+z}\\ &\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{\left(x-y\right)}&=-\frac{2{\rm\,d}\left(x+y+z\right)}{\left(x+y+z\right)}\\ &\Rightarrow&\ln\left|x-y\right|&=-2\ln\left|x+y+z\right|+\ln|C_1|\\ &\Rightarrow&\left(x-y\right)\left(x+y+z\right)^2&=C_1\\ \end{align*}$$

同理
In like manner
$$\begin{align*} &&\frac{{\rm\,d}y}{x-y+z}&=\frac{{\rm\,d}z}{x+y-z}\\ &\Rightarrow&\frac{{\rm\,d}\left(y-z\right)}{-2\left(y-z\right)}&=\frac{{\rm\,d}\left(x+y+z\right)}{x+y+z}\\ &\Rightarrow&\frac{{\rm\,d}\left(y-z\right)}{\left(y-z\right)}&=-\frac{2{\rm\,d}\left(x+y+z\right)}{\left(x+y+z\right)}\\ &\Rightarrow&\ln\left|y-z\right|&=-2\ln\left|x+y+z\right|+\ln|C_2|\\ &\Rightarrow&\left(y-z\right)\left(x+y+z\right)^2&=C_2\\ \end{align*}$$

$$\begin{align*} \left\{ \begin{aligned} \frac{\text{d}x}{-x+y+z}&=\frac{\text{d}y}{x-y+z}\\ \frac{\text{d}y}{x-y+z}&=\frac{\text{d}z}{x+y-z}\\ \end{aligned} \right. \Rightarrow \left\{ \begin{aligned} \left( x-y \right) \left( x+y+z \right) ^2&=C_1\\ \left( y-z \right) \left( x+y+z \right) ^2&=C_2\\ \end{aligned} \right. \end{align*}$$

$$\begin{align*} \left\{ \begin{aligned} \frac{\text{d}x}{-x+y+a}&=\frac{\text{d}y}{x-y+a}\\ \frac{\text{d}y}{b-y+z}&=\frac{\text{d}z}{b+y-z}\\ \end{aligned} \right. \Rightarrow \left\{ \begin{aligned} x-y&=C_1e^{-\frac{x+y}{z}}\\ y-z&=C_2e^{-\frac{y+z}{x}}\\ \end{aligned} \right. \end{align*}$$

 

$$\begin{align*} \left\{ \begin{aligned} \frac{\text{d}x}{-x^2+y^2+a^2}&=\frac{\text{d}y}{x^2-y^2+a^2}\\ \frac{\text{d}y}{b^2-y^2+z^2}&=\frac{\text{d}z}{b^2+y^2-z^2}\\ \end{aligned} \right. \Rightarrow \left\{ \begin{aligned} x-y&=C_1e^{-\frac{\left( x+y \right) ^2}{2a^2}}\\ y-z&=C_2e^{-\frac{\left( y+z \right) ^2}{2b^2}}\\ \end{aligned} \right. \end{align*}$$
另外：
$$\begin{align*} &&\frac{{\rm\,d}x}{-x^3+y^3+a^3}&=\frac{{\rm\,d}y}{x^3-y^3+a^3}\\ &\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{-2\left(x^3-y^3\right)}&=\frac{{\rm\,d}\left(x+y\right)}{2a^3}\\ &\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{{\rm\,d}\left(x+y\right)}&=\frac{-2\left(x^3-y^3\right)}{2a^3}\\ &\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{{\rm\,d}\left(x+y\right)}&=\frac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{a^2}\\ &\Rightarrow&\frac{\left(x-y\right){\rm\,d}\left(x-y\right)}{{\rm\,d}\left(x+y\right)}&=\frac{-\left(x-y\right)^2\left(3\left(x+y\right)^2+\left(x-y\right)^2\right)}{a^2}\\ &\Rightarrow&\frac{{\rm\,d}\left(\left(x-y\right)^2\right)}{{\rm\,d}\left(x+y\right)}&=-\frac{3\left(x+y\right)^2}{a^2}\left(x-y\right)^2-\frac{1}{a^2}\left(x-y\right)^4\\ &\Rightarrow&\frac{{\rm\,d}u}{{\rm\,d}v}&=-\frac{3v^2}{a^2}u-\frac{1}{a^2}u^2\\ &\Rightarrow&\\ &\Rightarrow&\\ \end{align*}$$

%http://kuing.orzweb.net/viewthread.php?tid=5752&tdsourcetag=s_pctim_aiomsg

$$y'''=\frac{3\left(y''\right)^2+x\!\cdot\!\left(y'\right)^5}{y'}$$
$$\begin{align*} \frac{{\rm\,d}x}{{\rm\,d}y}&=\frac{1}{y'}\\ \frac{{\rm\,d}^2x}{{\rm\,d}y^2}&=\frac{{\rm\,d}}{{\rm\,d}y}\left(\frac{{\rm\,d}x}{{\rm\,d}y}\right) =\frac{{\rm\,d}}{{\rm\,d}y}\left(\frac{1}{y'}\right) =\frac{{\rm\,d}}{{\rm\,d}x}\left(\frac{1}{y'}\right)\frac{{\rm\,d}x}{{\rm\,d}y}\\ &=-\frac{y''}{\left(y'\right)^2}\frac{{\rm\,d}x}{{\rm\,d}y}=-\frac{y''}{\left(y'\right)^3}\\ \frac{{\rm\,d}^3x}{{\rm\,d}y^3}&=\frac{{\rm\,d}}{{\rm\,d}y}\left(\frac{{\rm\,d}^2x}{{\rm\,d}y^2}\right) =\frac{{\rm\,d}}{{\rm\,d}y}\left(-\frac{y''}{\left(y'\right)^3}\right) =\frac{{\rm\,d}}{{\rm\,d}x}\left(-\frac{y''}{\left(y'\right)^3}\right)\frac{{\rm\,d}x}{{\rm\,d}y}\\ &=-\frac{y'''\!\cdot\!\left(y'\right)^3-3\left(y'\right)^2y''\!\cdot\!y''}{\left(y'\right)^6}\frac{{\rm\,d}x}{{\rm\,d}y}=-\frac{y'''\!\cdot\!\left(y'\right)^3-3\left(y'\right)^2\left(y''\right)^2}{\left(y'\right)^7}\\ &=\frac{3\left(y'\right)^2\left(y''\right)^2-y'''\!\cdot\!\left(y'\right)^3}{\left(y'\right)^7}=\frac{3\left(y''\right)^2-y'y'''}{\left(y'\right)^5} \end{align*}$$

$$\begin{align*} &&\frac{{\rm\,d}^3x}{{\rm\,d}y^3}&=\frac{\color{red}{3\left(y''\right)^2-y'y'''}}{\left(y'\right)^5}=\frac{\color{red}{-x\!\cdot\!\left(y'\right)^5}}{\left(y'\right)^5}=-x\\ &\Rightarrow&\frac{{\rm\,d}^3x}{{\rm\,d}y^3}+x&=0\\ &\Rightarrow&x'''+x&=0\\ \end{align*}$$

$$\begin{align*} x'''+x&=0\\ x&=C_1e^{-y}+C_2e^{\frac{y}{2}}\cos\left(\frac{\sqrt{3}}{2}y\right)+C_3e^{\frac{y}{2}}\sin\left(\frac{\sqrt{3}}{2}y\right) \end{align*}$$

更一般：

$$y ^ { \prime } y ^ {\prime \prime \prime } - 3 \left( y ^ { \prime \prime } \right) ^ { 2 } + a \left( y ^ { \prime } \right) ^ { 2 } y ^ { \prime \prime } - b \left( y ^ { \prime } \right) ^ { 4 } - c x \cdot \left( y ^ { \prime } \right) ^ { 5 } = 0$$

$$\xrightarrow[\frac{{\rm\,d}^3x}{{\rm\,d}y^3}=\frac{3\left(y''\right)^2-y'y'''}{\left(y'\right)^5}] {\frac{{\rm\,d}x}{{\rm\,d}y}=\frac{1}{y'}\quad\frac{{\rm\,d}^2x}{{\rm\,d}y^2}=-\frac{y''}{\left(y'\right)^3}}$$

$$f ^ { \prime } ( x ) y ^ { \prime } y ^ { \prime \prime \prime } - 3 f ^ { \prime } ( x ) \left( y ^ { \prime \prime } \right) ^ { 2 } + 3 f ^ { \prime \prime } ( x ) y ^ { \prime } y ^ { \prime \prime } + a f ^ { \prime } ( x ) \left( y ^ { \prime } \right) ^ { 2 } y ^ { \prime \prime } = a f ^ { \prime \prime } ( x ) \left( y ^ { \prime } \right) ^ { 3 } + b f ^ { \prime } ( x ) \left( y ^ { \prime } \right) ^ { 4 } + f ^ { \prime \prime \prime } ( x ) \left( y ^ { \prime } \right) ^ { 2 } + c f ( x ) \left( y ^ { \prime } \right) ^ { 5 }$$

令$u(y)=f(x)$
$$\frac { \mathrm { d } ^ { 3 } u ( y ) } { \mathrm { d } y ^ { 3 } } + a \frac { \mathrm { d } ^ { 2 } u ( y ) } { \mathrm { d } y ^ { 2 } } + b \frac { \mathrm { d } u ( y ) } { \mathrm { d } y } + c u ( y ) = 0.$$
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$$\begin{align*} \min_x\quad f(x)=\sum_{i=1}^{n} f_i(x),\\ s.t.\quad g_i(x)\leq 0, A_ix=b_i,x\in\cap_{i=1}^n\Omega_i. \end{align*}$$

$$\begin{align*} \frac{d}{dt}\left( \begin{array}{c} y_i\\ \lambda _i\\ \mu _i\\ x_i\\ \end{array} \right) &=\left( \begin{array}{c} -y_i+x_i-\nabla f_i\left( x_i \right) -\left( \nabla g_i\left( x_i \right) \right) ^T\left( \lambda _i+g_i\left( x_i \right) \right) ^+-A_{i}^{T}\left( \mu _i+A_ix_i-b_i \right) +u_i\\ -\lambda _i+\left( \lambda _i+g_i\left( x_i \right) \right) ^+\\ A_ix_i-b_i\\ x=P_{\Omega _i}\left( y_i \right)\\ \end{array} \right) \\ u_i &=\mathcal{k}_P\sum_{j=1}^n{a_{ij}\left( x_j-x_i \right)}+\mathcal{k}_I\int_0^t{\sum_{j=1}^n{a_{ij}\left( x_j\left( s \right) -x_i\left( s \right) \right) ds}} \end{align*}$$