# ( Proof Explanation ) Show that a certain system preserves the weighted area $(dx \wedge dy)/xy$

I already told few questions ago that I'm currently reading an abstract about the Lotka Volterra differential equations. But now I have a proof, where I need explanations. Consider: $$\dot{x} = -xy\frac{\delta H}{ \delta y} , x(0) = \hat{x}$$ $$\dot{y} = xy\frac{\delta H}{ \delta x} , y(0) = \hat{y}$$where $$H(x,y) = x + y - ln(x) -ln(y)$$.

I have to show that this System preserve the weighted area $$(dx \wedge dy)/xy$$. I marked my Questions in the proof below.

Proof:

Let $$\Omega_0$$ be a subset of $$\mathbb{R}^2$$ at time $$t_0$$ and $$\Omega_1$$ the set into which $$\Omega_0$$ is mapped by the system above at time $$t_1$$. Preservation of $$(dx \wedge dy)xy$$ is equivalent to $$\int_{\Omega_0} \frac{1}{xy}dxdy = \int_{\Omega_1} \frac{1}{xy} dxdy$$ first Question: why is this equivalent? We now look at the Domain $$D$$ in x,y,t space with bondary $$\delta D$$ given by $$\Omega_0$$ at $$t_0$$, $$\Omega_1$$ at $$t_1$$ and the set of trajectories emerging from the boundary of $$\Omega_0$$ and ending on the boudnary of $$\Omega_1$$. Consider the vector field $$v := \frac{1}{xy}(\dot{x},\dot{y},1)^T$$ in $$x,y,t$$ space. Integrating this vector field over the boundary $$\delta D$$ of $$D$$, we obtain $$\int_{\delta D} v \cdot n = \int_{\Omega_0} v \cdot n_0 + \int_{\Omega_1} v \cdot n_1 = \int_{\Omega_0} \frac{1}{xy} dxdy - \int_{\Omega_1} \frac{1}{xy}dxdy$$ where $$n_0 =(0,0,-1)^T$$ denote the unit outward normal of $$\Omega_0$$ and $$\Omega_1$$.Second question & Third question: Can you explain why we integrate $$v \cdot n$$ ? I thought we integrate $$v$$ and can you explain the first equation above? There is no other contribution to the surface integral, because the vector field $$v$$ is by contruction parallel to the trajectories, which form the rest of the bondary $$\delta D$$. Forth question: Can you explain why vector field is parallel to the trajectories? Applying the divergence theorem to the left hand side of the same equation, we get $$\int_{\delta D} v \cdot n = \int_D \nabla v = \int_D - \frac{\delta H^2}{\delta x \delta y} + \frac{ \delta H^2}{\delta x \delta y} + 0 = 0$$ which concludes the proof.

I hope that my questions are not to easy, but I'm a beginner.

I believe the proof is taken from Mickens, Applications of Nonstandard Finite Difference Schemes. Preservation of the area weighted by the factor $$\rho$$ by definition means that $$\int_{\Omega(t_0)} \rho \, dS = \int_{\Omega(t)} \rho \, dS$$ if the set $$\Omega(t_0)$$ is mapped to $$\Omega(t)$$ by the system (the omegas are sets of $$(x, y)$$ points). We want to show that this holds for the given system and for $$\rho(x, y) = 1/(x y)$$.
Suppose we parametrize $$\partial \Omega(t_0)$$ by $$\phi$$. A point $$(x, y, t)$$ on the surface $$\mathcal S$$ is given by specifying $$\phi$$ and $$t$$: $$x$$ and $$y$$ are the solution of the system at time $$t$$ with the initial conditions given by $$\phi$$. If we fix $$\phi$$ and vary $$t$$, we'll get a curve $$(x, y, t)$$, which lies on $$\mathcal S$$ by construction. Since $$\boldsymbol v = (\dot x, \dot y, 1)$$ is tangent to the curve, it is also tangent to $$\mathcal S$$.
Then we take $$\partial D = \mathcal S \cup \Omega(t_0) \cup \Omega(t)$$ and take $$\hat {\boldsymbol n}$$ to be the outward unit normal to $$\partial D$$. Since $$\boldsymbol v \cdot \hat {\boldsymbol n} = 0$$ on $$\mathcal S$$ and $$\dot \rho = 0$$,
$$\int_{\partial D} \rho \hspace {1px} \boldsymbol v \cdot \hat {\boldsymbol n} \, dS = -\int_{\Omega(t_0)} \rho \,dS + \int_{\Omega(t)} \rho \, dS, \\ \int_{\partial D} \rho \hspace {1px} \boldsymbol v \cdot \hat {\boldsymbol n} \, dS = \int_D \nabla \cdot (\rho \hspace {1px} \boldsymbol v) \, dV = \int_D \nabla \cdot \left( -\frac {\partial H} {\partial y}, \frac {\partial H} {\partial x}, \rho \right) dV = 0.$$