ATRACTOR

Now that we know that the solution curves are periodic (of period \(T\)) we can determine the mean values (over a period) of \(x\) and \(y\). By analogy to the definition of the arithmetic mean of a finite sample of real numbers, and taking into account that \(x\) and \(y\) are continuous variables, these mean values are computed by the expressions \[\begin{array}{ccc} \bar{x}=\frac{\intop_{0}^{T}x(t)dt}{T} & & \bar{y}=\frac{\intop_{0}^{T}y(t)dt}{T}\end{array}\] where the sum of the concept of arithmetic mean is replaced here by the integral. It should be remembered, however, that we do not know \(x(t)\) and \(y(t)\) explicitly. But, as we have previously concluded, \[x'=Ax-Bxy=(A-By)x\Longleftrightarrow\frac{1}{x}x'=A-By\] and the integration of \(x^{'}\) from \(0\) to \(T\) leads to \[\ln x(T)-\ln x(0)=\intop_{0}^{T}A-By(t)dt\] Since \(x\) is periodic of period \(T,\) \(x(T)=x(0)\) and, consequently, \(\ln x(T)-\ln x(0)=0;\) then \[\intop_{0}^{T}A-By(t)dt=0,\] that is, \[AT-B\intop_{0}^{T}y(t)dt=0,\] that is \[\intop_{0}^{T}y(t)dt=\frac{AT}{B}\]

Then \(\bar{y}=\frac{\intop_{0}^{T}y(t)dt}{T}=\frac{A}{B}.\)

In an analogous way we can show that \(\bar{x}=\frac{C}{D}.\)