So first I want to give you some background information:

begin of the background information

I'm currently reading an abstact about the Lotka Volterra differential equations:

$$ x^{'} = x -xy $$ $$ y^{'} = -y +xy $$

We know that the most numerical methods give us spiraling solutions instead of cyclic. So I want to try to show that a modified forward Euler Method leads to cyclic solutions.


$$ \frac{x_{n+1}-x_n}{\Delta t} = x_n -x_ny_n $$ $$\frac{y_{n+1}-y_n}{\Delta t} = -y_n -x_{n+1}y_n $$

There is a proof to show that this modification does not spiral.

end of background information

Now I have some questions:

So let us say to simplify notitation $$ X = \Delta tx +x - \Delta txy$$ and $$ Y = -\Delta ty+ y + \Delta tXy $$ where we set $X:=x_{n+1}, Y:=y_{n+1}$, $x:= x_n$ and $ y := y_n$ solved for the unknown $X$ and $Y$.

Taking derivatives: $$ dX = \Delta tdx +dx - \Delta tdxy - \Delta txdy $$ and $$ dY = -\Delta tdy+ dy + \Delta tdXy + \Delta tXdy $$

Finally we arrived to one point where I stucked.

I need that $ dX \wedge dX = 0 $ and $ dY \wedge dY = 0 $. Can you help me out here?

  • 1
    $\begingroup$ That is by the defining properties of an anti-symmetric product. What is your understanding of the wedge or outer product? $\endgroup$ – LutzL May 24 at 16:23
  • $\begingroup$ That sounds reasonable. Do you know what confuses me? In the abstract they say that this follows after some "manipulations". So I thought that I have to do something more. $\endgroup$ – RukiaKuchiki May 24 at 16:43
  • $\begingroup$ Usually, the symplectic Euler method only works for Hamiltonian systems $\dot x=\partial_yH(x,y)$, $\dot y=-\partial_xH(x,y)$. While the LV system has a first integral $F(x,y)=x-\ln x+y-\ln y$, it is not a Hamiltonian function of the system. $\endgroup$ – LutzL May 24 at 16:52

A defining property of the outer product is its anti-symmetry $$ a∧b = -b∧a, $$ which implies that $$ 2\,a∧a=0. $$

  • $\begingroup$ Dear LutzL, I found the property $ a \wedge b = (-1)^{kl} b \wedge a $. Sorry for the stupid question, but why do you have only $ -b \wedge a$ . Like I said I completely new at this topic. $\endgroup$ – RukiaKuchiki May 25 at 17:39
  • $\begingroup$ This elementary formula is for $a,b\in V=\Lambda^1V$. The extended formula you cite is for the case that $a,b$ are already higher order outer products, $a\in\Lambda^kV$ and $b\in\Lambda^lV$. $\endgroup$ – LutzL May 25 at 17:56

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