很高兴能在网上和别人一起探讨问题,我的小专栏又有了一点更新的动力!
前面的文章讲了已知坐标系变换的情况下如何求无穷小生成元,例如
\left.\begin{array}{l} \tilde{x}=e^{\varepsilon} x \\ \tilde{y}=e^{2 \varepsilon} y \end{array}\right\} \Longrightarrow \left.\begin{array}{l} \xi_{x}=\left.\frac{\partial \tilde{x}(x,
\varepsilon)}{\partial \varepsilon}\right|_{\varepsilon=0}=\left.e^{\varepsilon} x\right|_{\varepsilon=0}=x \\ \xi_{y}=\left.\frac{\partial \tilde{y}(x, \varepsilon)}{\partial
\varepsilon}\right|_{\varepsilon=0}=\left.2 e^{\varepsilon} y\right|_{\varepsilon=0}=2 y \end{array}\right\}
可知它的 X=x \frac{\partial}{\partial x}+2 y \frac{\partial}{\partial y} ,但是我们在求解PDE的时候就是要找到一个合适的LTG,以求解方程,也就是说我们是不知道坐标系变换的,那么在这种情况下如何求解 X 呢?
延拓
对于一个已知的李变换群,我们对无穷小生成元的定义是
\tilde{\boldsymbol{x}}=\Phi(\boldsymbol{x} ; \varepsilon) \Longleftrightarrow X=\sum_{i=1}^{n} \xi_{i}(\boldsymbol{x}) \frac{\partial}{\partial x_{i}}
对于一个对称函数,我们是这样定义
F(\boldsymbol{x})=0 \quad \Longleftrightarrow \quad F(\tilde{\boldsymbol{x}})=0 \left.\Longleftrightarrow \quad X F(\boldsymbol{x})\right|_{F=0}=0
我们需要将这一概念延拓到同样对于微分方程 F\left(\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{y}^{(1)}, \ldots\right)=0 适用.
LTG : \tilde{\boldsymbol{x}}=(\tilde{x}, \tilde{y})=\boldsymbol{\Phi}(\boldsymbol{x} ; \varepsilon)=(\Phi(x, y ; \varepsilon), \Psi(x, y ; \varepsilon)) =\boldsymbol\Phi(x, y ;
\varepsilon)
求导 : y_{k}=y^{(k)}=\frac{\mathrm{d}^{k} y}{\mathrm{d} x^{k}}
把LTG延拓到 \left(x, y, y^{(1)}, \dots\right) 中,有求导法则有
\mathrm{d} y_{k}=y_{k+1} \mathrm{d} x \quad \longrightarrow \quad y_{k+1}=\frac{\mathrm{d} y_{k}}{\mathrm{d} x} \\ \mathrm{d} \tilde{y}_{k}=\tilde{y}_{k+1} \mathrm{d} \tilde{x} \quad
\longrightarrow \quad \tilde{y}_{k+1}=\frac{\mathrm{d} \tilde{y}_{k}}{\mathrm{d} \tilde{x}}
由LTG可知求导条件可以写成
\begin{aligned} &\mathrm{d} \tilde{y}=\mathrm{d} \Psi(\boldsymbol{x} ; \varepsilon)=\frac{\partial \Psi}{\partial x}(\boldsymbol{x} ; \varepsilon) \mathrm{d} x+\frac{\partial
\Psi}{\partial y}(\boldsymbol{x} ; \varepsilon) \mathrm{d} y\\ &\mathrm{d} \tilde{x}=\mathrm{d} \Phi(\boldsymbol{x} ; \varepsilon)=\frac{\partial \Phi}{\partial x}(\boldsymbol{x} ;
\varepsilon) \mathrm{d} x+\frac{\partial \Phi}{\partial y}(\boldsymbol{x} ; \varepsilon) \mathrm{d} y \end{aligned}
再将这两个式子带入到求导法则中 \mathrm{d} \tilde{y}=\tilde{y}_{1} \mathrm{d} \tilde{x} 可以得到 \tilde{y}_{1} 了
\frac{\partial \Psi}{\partial x}(\boldsymbol{x} ; \varepsilon) \mathrm{d} x+\frac{\partial \Psi}{\partial y}(\boldsymbol{x} ; \varepsilon) \mathrm{d} y=\tilde{y}_{1}\left[\frac{\partial
\Phi}{\partial x}(\boldsymbol{x} ; \varepsilon) \mathrm{d} x+\frac{\partial \Phi}{\partial y}(\boldsymbol{x} ; \varepsilon) \mathrm{d} y\right]
\Rightarrow \tilde{y}_{1}=\frac{\frac{\partial \Psi}{\partial x}(\boldsymbol{x} ; \varepsilon)+y_{1} \frac{\partial \Psi}{\partial y}(\boldsymbol{x} ; \varepsilon)}{\frac{\partial
\Phi}{\partial x}(\boldsymbol{x} ; \varepsilon)+y_{1} \frac{\partial \Phi}{\partial y}(\boldsymbol{x} ; \varepsilon)}
这意味着 \tilde{y}_{1}=\Psi_{1}\left(x, y, y_{1} ; \varepsilon\right) 我们就由最开始的LTG算出了 \tilde{y}_{1},也就是说我们把变量空间从 (x,y) 延拓到了 (x,y,y_1) .
这时候我们的LTG : \tilde{x}=\Phi(x, y ; \varepsilon) ; \tilde{y}=\Psi(x, y ; \varepsilon) ; \tilde{y}_{1}=\Psi_{1}\left(x, y, y_{1} ; \varepsilon\right)
后面我们把所有的 y_k 看成是相互独立的变量,而这样的变量空间就是Jet-Raum.
同时如果 \tilde{x}=\Phi(x, y, \varepsilon), \tilde{y}=\Psi(x, y, \varepsilon) 是一个李变换群,那么 \tilde{x}=\Phi(x, y ; \varepsilon) ; \tilde{y}=\Psi(x, y ; \varepsilon) ;
\tilde{y}_{1}=\Psi_{1}\left(x, y, y_{1} ; \varepsilon\right) 同样也是李变换群.
对于任意的 \tilde{y}_{k}=\Psi_{k}\left(x, y, y_{1}, \ldots, y_{k} ; \varepsilon\right)=\frac{\frac{\partial \Psi_{k-1}}{\partial x}+y_{1} \frac{\partial \Psi_{k-1}}{\partial y}+\cdots+y_{k}
\frac{\partial \Psi_{k-1}}{\partial y_{k-1}}}{\frac{\partial \Phi}{\partial x}+y_{1} \frac{\partial \Phi}{\partial y}}
这个不难算的.
接下来延拓无穷小生成元了.
对于二元变量空间的李变换群
\begin{aligned} &\tilde{x}=\Phi(x, y ; \varepsilon)=x+\varepsilon \xi(x, y)+O\left(\varepsilon^{2}\right)\\ &\tilde{y}=\Psi(x, y ; \varepsilon)=y+\varepsilon \eta(x,
y)+O\left(\varepsilon^{2}\right) \end{aligned} 与之相对应的无穷小生成元就是
X=\xi(x, y) \frac{\partial}{\partial x}+\eta(x, y) \frac{\partial}{\partial y}
那么在Jet-Raum里面延拓之后的变量空间 \begin{aligned} &\tilde{x}=\Phi(x, y ; \varepsilon)=x+\varepsilon \xi(x, y)+O\left(\varepsilon^{2}\right)\\ &\tilde{y}=\Psi(x, y ;
\varepsilon)=y+\varepsilon \eta(x, y)+O\left(\varepsilon^{2}\right)\\ &\tilde{y}_{1}=\Psi_{1}\left(x, y, y_{1} ; \varepsilon\right)=y_{1}+\varepsilon \eta^{(1)}\left(x, y,
y_{1}\right)+O\left(\varepsilon^{2}\right)\\ &\tilde{y}_{k}=\Psi_{k}\left(x, y, y_{1}, \ldots, y_{k} ; \varepsilon\right)=y_{k}+\varepsilon \eta^{(k)}\left(x, y, y_{1}, \ldots,
y_{k}\right)+O\left(\varepsilon^{2}\right) \end{aligned}
延拓之后的无穷小生成元依据最开始的定义 X= \boldsymbol{\xi}(\boldsymbol{x})\cdot \nabla=\sum_{i=1}^{n}{\xi_{i}(x)\frac{\partial}{\partial x_{i}}} 在Jet-Raum(y_k 是相互独立的变量)里面就是
\begin{aligned} \mathrm{pr}^{(k)} X=X^{(k)}=& \xi(x, y) \frac{\partial}{\partial x}+\eta(x, y) \frac{\partial}{\partial y}+\eta^{(1)}\left(x, y, y_{1}\right) \frac{\partial}{\partial
y_{1}} +\cdots+\eta^{(k)}\left(x, y, y_{1}, \ldots, y_{k}\right) \frac{\partial}{\partial y_{k}} \end{aligned}
到此就成功地把 X 延拓到微分方程 F\left(\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{y}^{(1)}, \ldots\right)=0 了.
全微分算子 (Jet-Raum): \frac{\mathrm{D}}{\mathrm{D} x}=\frac{\partial}{\partial x}+y_{1} \frac{\partial}{\partial y}+y_{2} \frac{\partial}{\partial y_{1}}+\cdots+y_{n+1}
\frac{\partial}{\partial y_{n}}
接下来的问题是怎么求 \eta^{(1)}, \ldots, \eta^{(k)} ?
按照定义来算就好了 \tilde{y}_{k}=\Psi_{k}\left(x, y, y_{1}, \ldots, y_{k} ; \varepsilon\right)=\frac{\frac{\partial \Psi_{k-1}}{\partial x}+y_{1} \frac{\partial \Psi_{k-1}}{\partial
y}+\cdots+y_{k} \frac{\partial \Psi_{k-1}}{\partial y_{k-1}}}{\frac{\partial \Phi}{\partial x}+y_{1} \frac{\partial \Phi}{\partial y}}
=\frac{\frac{D\Psi_{k-1}}{Dx}}{\frac{D\Phi}{Dx}}=\frac{{D(y_{k-1}+\varepsilon\eta^{k-1})}}{Dx}/\frac{D(x+\varepsilon\xi)}{Dx}
=(\frac{Dy_{k-1}}{Dx}+\varepsilon\frac{D\eta^{k-1}}{Dx})/(1+\varepsilon\frac{D\xi}{Dx})
=\frac{y_{k}+\varepsilon\frac{D\eta^{k-1}}{Dx}}{(1+\varepsilon\frac{D\xi}{Dx})}=(y_{k}+\varepsilon\frac{D\eta^{k-1}}{Dx})(1-\varepsilon\frac{D\xi}{Dx}) =y_{k}-\varepsilon y_{k}
\frac{D\xi}{Dx}+\varepsilon\frac{D\eta^{k-1}}{Dx}-\varepsilon^2\frac{D\eta^{k-1}}{Dx}\frac{D\xi}{Dx}
=y_{k}+\varepsilon (\frac{D\eta^{k-1}}{Dx}-y_{k} \frac{D\xi}{Dx})=y_{k}+\varepsilon\eta^{k} (计算过程有级数展开和略去高阶无穷小量)
至此就得出了 \eta^{k}=\frac{D\eta^{k-1}}{Dx}-y_{k} \frac{D\xi}{Dx}
约定 \frac{\partial y_{i}}{\partial x_{j}}=y_{i, j} , \frac{\partial^{2} y_{i}}{\partial x_{j} \partial x_{k}}=y_{i, jk} ...
那么 y_{m, i}=\frac{\mathrm{D} y_{m}}{\mathrm{D} x_{i}}=\frac{\partial y_{m}}{\partial x_{i}}+y_{k, i} \frac{\partial y_{m}}{\partial y_{k}}+y_{k, ij} \frac{\partial y_{m}}{\partial y_{k,
j}}+\ldots =y_{k, i} \delta_{m k}=y_{m, i} 这说明链式法则跟全微分算子是一致的.
我们将它用在LTG上 \frac{\mathrm{D} \Psi_{k}}{\mathrm{D} x_{i}}=\frac{\mathrm{D} \Phi_{m}}{\mathrm{D} x_{i}} \cdot \underbrace{\frac{\mathrm{D} \tilde{y}_{k}}{\mathrm{D}
\tilde{x}_{m}}}_{\tilde{y}_{k, m}}\Rightarrow \frac{\mathrm{D} \Psi_{k}}{\mathrm{D} x_{i}}=\tilde{y}_{k, m} \cdot \frac{\mathrm{D} \Phi_{m}}{\mathrm{D} x_{i}}
可得 \eta_{k ; i_{1} i_{2} \ldots i_{s}}=\frac{\mathrm{D} \eta_{k;i_{1} i_{2}\ldots i_{s-1}}}{\mathrm{D} x_{i_{s}}}-y_{k ; i_{1} \ldots i_{s-1} m} \frac{\mathrm{D} \xi_{m}}{\mathrm{D}
x_{i_{s}}}
延拓算子为 X^{(s)}=\xi_{i} \frac{\partial}{\partial x_{i}}+\eta_{k} \frac{\partial}{\partial y_{k}}+\eta_{k ; i_{1}} \frac{\partial}{\partial y_{k, i_{1}}}+\eta_{k ; i_{1} i_{2}}
\frac{\partial}{\partial y_{k, i_{1} i_{2}}}+\ldots
准备工作就全部做完了,刀磨好了,开始杀羊试试.
试试ODE吧,这个计算比较简单,哈哈哈!!!
还是由对称函数的定义,我们用在ODE上就是 \left.X^{(s)} F\right|_{F=0}=0,F\left(\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{y}^{(1)}, \ldots, \boldsymbol{y}^{(s)}\right)=0
举例
y^{\prime \prime}+\frac{y^{\prime}}{x}-e^{y}=0 显然这不是线性的.
第一步写出延拓算子
X^2=\xi(x,y)\frac{\partial}{\partial x}+\eta(x,y)\frac{\partial}{\partial y}+\eta^{(1)}(x,y,y_{1})\frac{\partial}{\partial y_{1}}+\eta^{(2)}(x,y,y_{1},y_{2})\frac{\partial}{\partial
y_{2}}
\eta^{k}=\frac{D\eta^{k-1}}{Dx}-y_{k} \frac{D\xi}{Dx}
可得 \eta^{(1)}=\frac{\partial \eta}{\partial x}+y_{1}(\frac{\partial \eta}{\partial y}-\frac{\partial \xi}{\partial x})-y_{1}^{2}\frac{\partial \xi}{\partial y} \\ \eta^{(2)}=\frac{D
\eta^{(1)}}{Dx}-y_{2}\frac{D \xi}{Dx}=\frac{\partial ^{2} \eta}{\partial x^{2}}+y_{1}(2\frac{\partial^{2} \eta}{\partial x \partial y}-\frac{\partial^{2} \xi}{\partial
x^{2}})+y_{1}^{2}(\frac{\partial^{2} \eta}{\partial y^{2}}-2\frac{\partial^{2} \xi}{\partial x \partial y}) \\-y_{1}^{3}\frac{\partial^{2} \xi}{\partial y^{2}}+y_{2}(\frac{\partial
\eta}{\partial y}-2\frac{\partial \xi}{\partial x})-3y_{2}y_{1}\frac{\partial \xi}{\partial y}
第二步 \left.X^{(s)} F\right|_{F=0}=0
\begin{aligned} X^{(2)} F &=\left[\xi \frac{\partial}{\partial x}+\eta \frac{\partial}{\partial y}+\eta^{(1)} \frac{\partial}{\partial y_{1}}+\eta^{(2)} \frac{\partial}{\partial
y_{2}}\right]\left(y_{2}+\frac{y_{1}}{x}-e^{y}\right) \\ =&-\xi \frac{y_{1}}{x^{2}}-\eta e^{y}+\eta^{(1)} \frac{1}{x}+\eta^{(2)} \\ =&-\xi \frac{y_{1}}{x^{2}}-\eta e^{y} \\
&+\left[\frac{\partial \eta}{\partial x}+y_{1}\left(\frac{\partial \eta}{\partial y}-\frac{\partial \xi}{\partial x}\right)-y_{1}^{2} \frac{\partial \xi}{\partial y}\right]
\frac{1}{x} \\ &+\frac{\partial^{2} \eta}{\partial x^{2}}+y_{1}\left(2 \frac{\partial^{2} \eta}{\partial x \partial y}-\frac{\partial^{2} \xi}{\partial
x^{2}}\right)+y_{1}^{2}\left(\frac{\partial^{2} \eta}{\partial y^{2}}-2 \frac{\partial^{2} \xi}{\partial x \partial y}\right) \\ &-y_{1}^{3} \frac{\partial^{2} \xi}{\partial
y^{2}}+y_{2}\left(\frac{\partial \eta}{\partial y}-2 \frac{\partial \xi}{\partial x}\right)-3 y_{2} y_{1} \frac{\partial \xi}{\partial y}=0 \end{aligned}
第三步替换 F=0
\begin{array}{c} y_{2}=-\frac{y_{1}}{x}+e^{y} \Longrightarrow \quad-\xi \frac{y_{1}}{x^{2}}+\cdots+\left(-\frac{y_{1}}{x}+e^{y}\right)\left(\frac{\partial \eta}{\partial y}-2
\frac{\partial \xi}{\partial x}\right)-3\left(-\frac{y_{1}}{x}+e^{y}\right) y_{1} \frac{\partial \xi}{\partial y}=0 \end{array}
这是很重要的一步,不然缺少条件最后求不出来的.
第四步 依据导数的次数整理所有的式子
\begin{aligned} &-y_{1}^{3} \frac{\partial^{2} \xi}{\partial y^{2}}+\left[\frac{\partial^{2} \eta}{\partial y^{2}}-2 \frac{\partial^{2} \xi}{\partial x \partial y}+\frac{2}{x}
\frac{\partial \xi}{\partial y}\right] y_{1}^{2} +\left[2 \frac{\partial^{2} \eta}{\partial x \partial y}-\frac{\partial^{2} \xi}{\partial x^{2}}+\frac{\partial \xi}{\partial x}
\frac{1}{x}-\frac{\xi}{x^{2}}-3 e^{y} \frac{\partial \xi}{\partial y}\right] y_{1}\\ &+\frac{\partial^{2} \eta}{\partial x^{2}}+\frac{\partial \eta}{\partial x}
\frac{1}{x}+\left(\frac{\partial \eta}{\partial y}-2 \frac{\partial \xi}{\partial x}-\eta\right) e^{y}=0 \end{aligned}
第五步 求解 \eta,\xi
因为对于任意的 y 均要成立,所以系数表达式只能为0.
y_{1}^{3}: \quad \frac{\partial^{2} \xi}{\partial y^{2}}=0 \quad \Longrightarrow \quad \xi=A_{1}(x) y+A_{2}(x)
\begin{array}{} y_{1}^{2}: \frac{\partial^{2} \eta}{\partial y^{2}}=2 \frac{\partial^{2} \xi}{\partial x \partial y}-\frac{2}{x} \frac{\partial \xi}{\partial y}=2
A_{1}^{\prime}-\frac{2}{x} A_{1} \\ \Longrightarrow \quad \frac{\partial \eta}{\partial y}=\left(2 A_{1}^{\prime}-2 \frac{A_{1}}{x}\right) y+A_{3}(x) \\ \Longrightarrow
\eta=\left(A_{1}^{\prime}(x)-\frac{A_{1}}{x}\right) y^{2}+A_{3}(x) y+A_{4}(x) \end{array}
\begin{aligned} &y_{1}: \quad 2 \frac{\partial^{2} \eta}{\partial x \partial y}-\frac{\partial^{2} \xi}{\partial x^{2}}+\frac{\partial \xi}{\partial x} \frac{1}{x}-\frac{\xi}{x^{2}}-3
e^{y} \frac{\partial \xi}{\partial y}=0\\ &\begin{aligned} \Longrightarrow & 4\left(A_{1}^{\prime \prime}(x)-\frac{A_{1}^{\prime}(x)}{x}+\frac{A_{1}(x)}{x^{2}}\right) y+2
A_{3}^{\prime}(x) \\ &-A_{1}^{\prime \prime}(x) y-A_{2}^{\prime \prime}(x)+\frac{A_{1}^{\prime}(x)}{x} y+\frac{A_{2}^{\prime}}{x}-\frac{A_{1}(x)}{x^{2}} y-\frac{A_{2}(x)}{x^{2}} \\ &
\underbrace{-3 e^{y} A_{1}(x)}_{=0} =0 \end{aligned} \end{aligned}
显然 A_{1}(x)=0 剩余的 2 A_{3}^{\prime}(x)-A_{2}^{\prime \prime}(x)+\frac{A_{2}^{\prime}}{x}-\frac{A_{2}}{x^{2}}=0
\begin{aligned} y_{1}^{0}: & \frac{\partial^{2} \eta}{\partial x^{2}}+\frac{\partial \eta}{\partial x} \frac{1}{x}+\left(\frac{\partial \eta}{\partial y}-2 \frac{\partial \xi}{\partial
x}-\eta\right) e^{y}=0 \\ \Longrightarrow & A_{3}^{\prime \prime}(x) y+A_{4}^{\prime \prime}(x)+\frac{A_{3}^{\prime}(x)}{x} y \\ &+\frac{A_{4}^{\prime}(x)}{x}+\left(A_{3}(x)-2
A_{2}^{\prime}(x)-\underbrace{A_{3}(x) y}_{=0} -A_{4}(x)\right)e^{y}=0 \end{aligned}
显然 A_{3}(x)=0
从 y_{1} 可知 \begin{aligned} &-A_{2}^{\prime \prime}(x)+\frac{A_{2}^{\prime}(x)}{x}-\frac{A_{2}(x)}{x^{2}}=0\Rightarrow &\left.\begin{array}{l} x^{2} A_{2}^{\prime \prime}(x)-x
A_{2}^{\prime}(x)+A_{2}(x)=0 \\ A_{2}(x)=x^{n} \end{array}\right\} \end{aligned}
解得 A_{2}(x)=C_{1}xlnx+C_{2}x
剩余的 y_{1}^{0} : \underbrace{A_{4}^{\prime \prime}(x)+\frac{A_{4}^{\prime}(x)}{x}}_{=0}+\underbrace{\left(-2 A_{2}^{\prime}(x)-A_{4}(x)\right)}_{=0} e^{y}=0
解得 A_{4}(x)=-2(C_{1}(1+ln(x))+C_{2})
至此可知所求 \xi= C_{1}xln(x)+C_{2}x \\ \eta=-2(C_{1}(1+ln(x))+C_{2})
我们要求的坐标变换的LTG就是 X=( C_{1}xln(x)+C_{2}x )\frac{\partial}{\partial x}-2(C_{1}(1+ln(x))+C_{2})\frac{\partial}{\partial y} 知道如何变换坐标系之后只需将方程改写就可以进行求解了,后面我有时间再补充.
眼睛快瞎了,大家同情一下读到点个赞,笑嘻嘻

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查看更多优质解析为什么我会想直接求二阶导数.然后证明为凸函数就行了.囧.第二个化为m（lnx+x）=x^2 /2 有且有一个跟令H（x）=x^2/2-m(lnx+x) 让H（x）的零点为1个就行了.不过我还是挺纠结.凸函数要怎么证明呢.忘了,因为感觉像是公式一样.不需要证明的样子.囧解析看不懂？免费查看同类题视频解析查看解答
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