1.8 Closer look at orbits and tangent vectors

This section takes a closer look at orbits and tangent vectors $\xi ,\eta$ which are the core of Lie symmetry method. By definition

$$\begin{array}{rl}\text{(1)}& \xi (x,y)& ={\frac{d\overline{x}}{dϵ}|}_{ϵ=0}\eta (x,y)& ={\frac{d\overline{y}}{dϵ}|}_{ϵ=0}\end{array}$$

Hence $\xi (x,y)$ shows how $\overline{x}$ changes as function of $(x,y)$. And $\eta (x,y)$ shows how $\overline{y}$ changes as function of $(x,y)$. This is because

$$\begin{array}{rl}\text{(2)}& \overline{x}& =x+\xi ϵ\overline{y}& =y+\eta ϵ\end{array}$$

Comparing (2) to equation of motion where $\overline{x}$ represents final position and $x$ is initial position, then $\xi$ is the speed and $ϵ$ is the time. When time is zero, initial and final position is the same. As time increases final position changes depending on the speed as time (here represented as $ϵ$) increases. So it helps to think of $\xi ,\eta$ as the rate at which $\overline{x},\overline{y}$ change location depending on the value $ϵ$. $\xi ,\eta$ are calculated when $ϵ$ is very small in the limit as it reaches zero.

As $ϵ$ increases the point $(x,y)$ moves closer to the final destination point $(\overline{x},\overline{y})$. So these quantities $\xi ,\eta$ specify the orbit shape. The orbit is the path taken by point transformation from $(x,y)$ to $(\overline{x},\overline{y})$ and depends on $ϵ$ such that the ode remain invariant in $\overline{x},\overline{y}$ and points on solution curves are mapped to points on other solution curves.

Different $\xi ,\eta$ give different orbits between two solution curves. The following example shows this. Given the ode

$$y^{\prime }=\frac{x-y}{x+y}$$

This is Abel type ode. Also Homogeneous class A$.$

It has two solutions. One solution is given by Mathematica as $y=-x-\sqrt{c_1+2x^2}$. A small program was now written that plots the orbit for 4 solutions $\xi ,\eta$ found for the similarity conditions. The similarity solution were found by Maple’s symgen command.

Figure 5: Command used to find $\xi ,\eta$

The program starts from the same $(x,y)$ point from one solution curve and determines $(\overline{x},\overline{y})$ location on anther solution curve using each pair of $\xi ,\eta$ found. The same solution curves are used in order to compare the orbits. The following plot was generated showing the result

Figure 6: Different orbits using different $\xi ,\eta$

The source code used to generate the above plot is

<<MaTeX` 
ode=y'[x]==(x-y[x])/(x+y[x]); 
ysol=DSolve[ode,y[x],x] 
ysol=-x-Sqrt[C[1]+2 x^2]; 
 
x1 = 1.5; 
y1 = ysol /. {C[1] -> 1, x -> x1}; 
 
ysol2=ysol/.C[1]->1.1 
 
getSolutions[inf_List, titles_List, x_Symbol, ysol1_, ysol2_, x1_, 
  y1_, from_, to_] := 
 Module[{xbar, ybar, eps, eq, soleps, p, data, n, xi, eta, texStyle}, 
  data = Table[0, {n, Length@inf}]; 
  texStyle = {FontFamily -> "Latin Modern Roman", FontSize -> 12}; 
 
  Do[ 
   xi = First[inf[[n]]]; 
   eta = Last[inf[[n]]]; 
   xbar = x1 + eps*xi ; 
   ybar = y1 + eps*eta; 
   eq = ybar == ysol2 /. x -> xbar; 
   soleps = SolveValues[eq, eps]; 
   soleps = First@SortBy[soleps, Abs]; 
   ybar = ybar /. eps -> soleps; 
   xbar = xbar /. eps -> soleps; 
   p = Plot[{ysol1, ysol2}, {x, from, to}, 
     PlotLabel -> MaTeX[titles[[n]], Magnification -> 1.5], 
     BaseStyle -> texStyle, 
     Epilog -> {{Arrowheads[.02], Arrow[{{x1, y1}, {xbar, ybar}}]}, 
       Text[MaTeX["\\left( x,y \\right)"], {x1, y1}, {-1, -1}], 
       Text[ 
        MaTeX["\\left( \\bar{x},\\bar{y}\\right)"], {xbar, ybar}, {1, 
         1}]}, 
     ImageSize -> 400]; 
   data[[n]] = p 
   , 
   {n, 1, Length@inf} 
   ]; 
 
  data 
 
  ]; 
 
inf = {{1/x1, -1/x1}, 
   {0, 1/(x1 + y1)}, 
   {-(x1^2 - 2*x1*y1 - y1^2)/(x1 - y1), 0}, 
   {2*x1 + y1, x1} 
   }; 
titles = {"\\xi=\\frac{1}{x},\\eta=-\\frac{1}{x}", 
   "\\xi=0,\\eta=\\frac{1}{x+y}", 
   "\\xi=\\frac{-(x^2-2 x y-y^2)}{x=y},\\eta=0", "\\xi=2 x+y,\\eta=x"}; 
data = getSolutions[inf, titles, x, ysol /. C[1] -> 1, ysol2, x1, y1, 
   1.45, 1.51]; 
p = Grid[Partition[data, 2], Frame -> All, Spacings -> {1, 1}]