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I have to solve the following simple ODE using symmetry methods: $y''=0$

I have done the following: Symmetry Generator is given by: $G = \xi(x,y)\frac{\partial}{\partial x} + \eta(x,y)\frac{\partial}{\partial y} \label{eq:gen} \tag{2}$ then we need what is called the first order generator (G^1) is given as:

$G^1=G+\zeta^1\frac{\partial}{\partial \dot{y}} \label{eq:firstGen} \tag{3}$ The quantity $\zeta^1$ is called the first order prolongation of the generator $G$. Then there's the second order generator $G^2$. That is, \begin{equation} G^2 = G^1 + \zeta^2 \frac{\partial}{\partial \ddot{y}}. \end{equation}

The Prolongations

Before we move further, we need to understand the concept of the Total Derivative Operator: $D=\frac{\partial}{\partial x}+\dot{y}\frac{\partial}{\partial y} \label{eq:5} \tag{5}$ NB:$DF(x,y)=F_x+y'F_y$

The $k^{th}$ prolongation is defined by:

$\zeta^{(k)}(x,y)=D\zeta^{(k-1)}(x,y)-y^kD\xi \label{eq:Prol} \tag{6}$ where $\zeta^{(0)}=\eta(x,y)$

Using the above definition we get the following:

First Prolongation \begin{align} \zeta^{(1)} & =D\eta(x,y) -\dot{y}D(\xi(x,y)) \nonumber \\ & = \eta_x +\dot{y}\eta_y-\dot{y} \left( \xi_x+\dot{y}\xi_y\right) \nonumber \\ & = \eta_x +\dot{y}\eta_y -\dot{y}\xi_x - \dot{y}^2\xi_y \nonumber\\ \label{eq:ProlOne} \tag{7} \end{align} Second Prolongation \begin{align} \zeta^{(2)} & = D\zeta^{(1)}-\ddot{y}D\xi(x,y) \hspace{10mm} \text{$ \ddot{y}=0$ so the second term falls away} \nonumber \\ & = \eta_{xx}+\dot{y}\eta_{xy}-\dot{y}\xi_{xx}-\dot{y}^2\xi{xy}+\dot{y}\left(\eta_{xy}+\dot{y}\eta_{yy}-\dot{y}\xi_{xy}-\dot{y}^2\xi_{yy}\right) \nonumber\\ &=\eta_{xx} +\left(2\eta_{xy}-\xi_{xx}\right)\dot{y} +\left(\eta_{yy}-2\xi_{xy}\right)\dot{y}^2-\xi_{yy}\dot{y}^3 \nonumber\\ \label{ProlTwo} \tag{8} \end{align}

The Determining Equation: The \textbf{invariance criterion } is: $\zeta^{(2)}=0 \hspace{10mm} \text{when} \hspace{10mm} \ddot{y}=0 \label{eq:InvCrit} \tag{9}$ We expand \ref{eq:InvCrit} to get: $\eta_{xx} +\left(2\eta_{xy}-\xi_{xx}\right)\dot{y} + \left(\eta_{yy}-2\xi_{xy}\right)\dot{y}^2-\xi_{yy}\dot{y}^3=0 \label{eq:Det} \tag{10}$ From \ref{eq:Det} we get the following system of equations which Bluman \& Anco refer to as the system of determining equations: $\eta_{xx}=0\label{eq:a} \tag{a}$

$2\eta_{xy}-\xi_{xx}=0 \label{eq:b} \tag{b}$

$\eta_{yy}-2\xi_{xy}=0 \label{eq:c} \tag{c}$

$\xi_{yy}=0 \label{eq:d} \tag{d}$

The monomials From \ref{eq:d} we get: $\xi(x,y)=A(x)y+B(x) \label{eq:11} \tag{11}$ where $A$ and $ B$ are arbitrary functions of $x$

we substitute \ref{eq:11} into \ref{eq:c} and to get: \begin{align*} \eta_{yy}-2A'(x) & = 0\\ \eta_{yy} & = 2A'(x) \\ \end{align*} therefore we have: $\eta(x,y)=A'(x)y^2+C(x)y+D(x) \label{eq:12} \tag{12}$ where $A, B, C, \text{ and } D$ are arbitrary functions of $x$

we substitute \ref{eq:11} and \ref{eq:12} into equations \ref{eq:a} and \ref{eq:b} to get: $3A''(x)y+2C'(x)-B''(x)=0 \label{eq:13} \tag{13}$

$A^{(3)}(x)y^2+C''(x)y+D''(x)=0 \label{eq:14} \tag{14}$

We equate the powers of $y$ from equations \ref{eq:13} and \ref{eq:14} to get the following system of equations:

$A''(x)=0 \label{eq:15.1} \tag{15.1}$

$C''(x)=0 \label{eq:15.2} \tag{15.2}$

$D''(x)=0 \label{eq:15.3} \tag{15.3}$

$2C'(x)=B''(x) \label{eq:15.4} \tag{15.4}$

From \ref{eq:15.1} \ref{eq:15.2} and \ref{eq:15.3} we have $A(x)=c_1x+c_2 \label{eq:16.1} \tag{16.1}$

$C(x)=c_3x+c_4 \label{eq:16.2} \tag{16.2}$

$D(x)=c_5x+c_6 \label{eq:16.3} \tag{16.3}$

we substitute \ref{eq:16.2} into \ref{eq:15.4} : \begin{align} B''(x) & = 2c_3 \\ \nonumber B(x) & = c_3x^2 +c_7x +c_8 \\ \label{eq:16.4} \tag{16.4} \end{align}

We use \ref{eq:16.1} - \ref{eq:16.4} to get the following equations for $\eta$ and $\xi$

$\xi(x,y) = c_1xy +c_2y+c_3x^2+c_7x+c_8 \label{eq:17} \tag{17}$

$\eta(x,y) = c_1y^2+c_3xy+c_4y+c_5x+c_6 \label{eq:18} \tag{18}$

The Generators

The infinitesimal generator is defined as: $X=\sum_{i=1}^{8}c_iX_i$ where

$X_1=xy\partial x+y^2\partial y$

$X_2=y \partial x$

$X_3=x^2 \partial x +xy \partial y $

$X_4=y \partial y$

$X_5=x \partial y$

$X_6=\partial y$

$X_7=x \partial x$

$X_8=\partial x$

From here I am unsure what the next steps are. I think I need to find the commutators for each pair of $X_i$

The Commutators

\begin{align*} \left[ X_1 X_2 \right] = & \left( xy \partial x + y^2 \partial y \right) \left( y \partial x \right) - \left( y \partial x \right) \left( xy \partial x + y^2 \partial y \right)\\ = & xy \partial x y \partial x + y^2 \partial y y \partial x - \left( y \partial x (xy) \partial x + y \partial x y^2 \partial y \right)\\ =& 0 + y^2\partial x - (y^2 \partial x + 0 ) \\ =&0 \end{align*}

Through similar calculations we determine each commutator. The results are in the table below:

\begin{array} {c| c c c c c c c c} & X_1 & X_2 & X_3 & X_4 & X_5 & X_6 & X_7 & X_8 \\ \hline X_1 & 0& 0& 0 & -X_1 & -X_3 & -X_3 & 0 & -X_2\\ X_2 & 0 & 0 & X_1 & -X_2 & X_4-X_7 &-X_8 & X_2 & 0 \\ X_3 & 0 & -X_1 & 0 & 0 & 0 & -X_5 & -X_3 & -(2X_7+X_4) \\ X_4 & X_3 & X_2 & 0 & 0 & -X_5 & -X_6 & 0 & 0 \\ X_5 & X_3 & X_7-X_4 & 0 & X_5 & 0 & 0 & -X_5 & -X_6\\ X_6 & X_3 & X_8 & X_5 & X_6 & 0 & 0 & 0 & 0 \\ X_7 & 0 & X_2 & X_3 & 0 & X_5 & 0 & 0 & -X_8\\ X_8 & X_2 & 0& 2X_7+X_4 & 0 & X_6 & 0 & X_8 & 0 \\ \end{array}

I have tried solving by finding canonical coordinates: $$ X_1 = xy\partial x + y^2 \partial y $$

We have determined, in the previous section that: $$ \begin{align*} r_1 = \frac{y}{x} &&\text{and} & & s_1 = \frac{-1}{y}\\ \end{align*} $$

We now find $\frac{ds}{dr}$ using the formula: $$\frac{ds}{dr} = \frac{s_x + s_y y'}{r_x + r_y y'} \label{eq:1.22} \tag{1.22}$$ Substituting our values of $r_1 \text{ and } s_1$ we have:

$$ \begin{align*} \frac{ds}{dr} = & \frac{0 + \frac{1}{y^2}y'}{\frac{-y}{x^2}+\frac{1}{x}y'}\\ \\ = & \frac{y'}{y^2}\left(\frac{x}{y'-r}\right)\\ \\ = & \frac{sy'}{r}\left(\frac{1}{r -y'} \right)\\ \\ =& \frac{sy'}{r^2-ry'}\\ \\ ds\left(r^2-ry'\right) = & sy' dr\\ \\ \frac{1}{sy'}ds = &\frac{1}{r^2-ry'}dr\\ \\ \frac{1}{y'}\ln (s) = &\frac{1}{y'} \ln \left(\frac{y'}{r}-1\right)\\ \\ \ln s = & \ln \left(\frac{y'}{r}-1\right) \\ \\ s = & \frac{y'}{r}-1\\ \\ \end{align*} $$

We now substitute our values for $s$ and $r$ in terms of $x$ and $y$

$$ \begin{align*} y' = rs + r'\\ \\ = & \frac{-1}{x}+ \frac{y}{x}\\ \\ = & \left( \frac{y - 1 }{x} \right )\\ \\ \frac{dy}{dx} = & \frac{y-1}{x}\\ \\ \frac{1}{x}dx = \frac{1}{y-1}dy \\ \\ \ln \left( y - 1 \right) = & \ln x \\ \\ y -1 = & x\\ \\ y = & x+1 \end{align*} $$

However when I try to do this with $X_4$ it all goes pear shaped:(

$$X_4 = y \partial y$$

$$ \begin{align*} r_4 =x & & \text{and} & & s_4 = \ln y\\ \end{align*} $$

Applying \ref{eq:1.22} to solve for $\frac{ds}{dr}$ : $$ \begin{align*} \frac{ds}{dr} = & \frac{\frac{1}{y}y'}{1}\\ \\ = & \frac{1}{y}y' \\ \\ \end{align*} $$ But this was getting me some crazy solutions when I tried to apply the same method I had used for $X_1$

So I did this instead: $$ y' =y\frac{ds}{dr} $$ We now determine $\frac{d^2s}{dr^2}$

$$ \begin{align*} \frac{d^2s}{dr^2} = &\frac{d}{dr}\left(\frac{y'}{y}\right)\\ \\ =& \frac{d}{dx}\left( \frac{y'}{y} \right)\\ \\ = & \frac{y'' \cdot y - y' \cdot y'}{y^2} \\ \\ = & \frac{- \left(y'\right)^2}{y^2}\\ \\ = & -\left( \frac{ds}{dr}\right)^2 \end{align*} $$

Now we let $\frac{ds}{dr}=z$

$$ \begin{align*} \frac{dz}{dr} = & -z^2 \\ \\ \frac{-1}{z^2}dz = & dr \\ \\ \frac{1}{z} = & r + \alpha_1\\ \\ z = \frac{1}{r + \alpha_1}\\ \end{align*} $$

Now this gives us: $$ \begin{align*} \frac{ds}{dr} = &\frac{1}{r+ \alpha_1}\\ \\ s = & \ln \left(r + \alpha_1 \right) + \alpha_2\\ \\ \ln y = & \ln \left(x + \alpha_1 \right) + \alpha_2\\ \\ \frac{y}{x + \alpha_1} = & e ^ {\alpha_2}\\ \\ y = & \alpha_3 x + \alpha_3 \alpha_1\\ \\ y = Ax + B\\ \end{align*} $$

However if I try this method with $X_1$ I get stuck.

What do I do???

Please help.

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  • $\begingroup$ @MrYouMath can you show me how? $\endgroup$ Feb 9, 2018 at 17:53
  • $\begingroup$ Added an answer, hope it is useful. $\endgroup$
    – MrYouMath
    Feb 9, 2018 at 22:02
  • $\begingroup$ @MrYouMath I added my attempted solution. I was solving based on each generator. Is that incorrect? $\endgroup$ Feb 10, 2018 at 9:20

1 Answer 1

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The easiest way to get the general solution to your problem is by combining the two symmetries $X_8$ and $X_6$. Such that we have the infinitesimal generator

$$X = \partial_x+c_1\partial y.$$

Now, we use the invariant surface condition

$$X\left[y-y(x)\right]\bigr|_{y=y(x)}=0.$$

Note, that $y$ is treated as an independent variable and $y(x)$ is a dependent variable. Hence we obtain:

$$ \left[\partial_x+c_1\partial y\right]y-\left[\partial_x+c_1\partial y\right]y(x)=0$$ $$\implies c_1-\dfrac{\partial y(x)}{\partial x}=0 \implies \dfrac{dy(x)}{dx}=c_1.$$

Integration will yield

$$y(x) = c_1x+c_2,$$

which is the general solution to the initial ODE. It is important to always verify that the candidate solution is really a solution to the original ODE.

Note, that the evaluation of $y=y(x)$ was not important here, but it will be important in the general setting.

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  • $\begingroup$ Thanks, you're choosing vectors $X_i$ and $X_j$ for which $[X_i X_j ] = 0$ right? I have added my attempted solutions. However I still don't think what I'm doing is correct. Please have a look at it for me. :) $\endgroup$ Feb 10, 2018 at 7:32
  • $\begingroup$ Can I use this method with the same vector? With canonical coordinates I will get solutions per vector. So can I pair X1 with X1 and get a solution? Or does this method only work with pairing distinct vectors? $\endgroup$ Feb 13, 2018 at 6:38

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