# Show that the wedge product $dX \wedge dX = 0$ and $dY \wedge dY = 0$

So first I want to give you some background information:

begin of the background information

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

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.

Here:

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

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?

• That is by the defining properties of an anti-symmetric product. What is your understanding of the wedge or outer product? – LutzL May 24 at 16:23
• 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. – RukiaKuchiki May 24 at 16:43
• 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. – LutzL May 24 at 16:52

A defining property of the outer product is its anti-symmetry $$a∧b = -b∧a,$$ which implies that $$2\,a∧a=0.$$
• 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. – RukiaKuchiki May 25 at 17:39
• 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 $b\in\Lambda^lV$. – LutzL May 25 at 17:56