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How is the change of variables really performed in Lotka-Volterra?

I'm trying to understand what happens in the following document:

https://www.maths.dur.ac.uk/~ktch24/term1Notes(10).pdf, p. 17

I tried plugging the variables in, but I don't see how it produces the result.

mavavilj
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2 Answers2

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Start with the first equation

\begin{align} \frac{{\rm d}x}{\rm {d}t} &= ax - bxy \\ \frac{{\rm d}(dx/c)}{{\rm d}t} &= a(dx/c) - b(dx/c)y & \text{multiply by } (d/c) \\ \frac{{\rm d}\hat{x}}{{\rm d}t} &= a\hat{x} - b\hat{x}y & \text{define } \hat{x}= dx/c \\ \frac{{\rm d}\hat{x}}{{\rm d}(at)} &= a\hat{x}/a - \hat{x}(by/a) & \text{divide by } a \\ \frac{{\rm d}\hat{x}}{{\rm d}\hat{t}} &= \hat{x} - \hat{x}\hat{y} & \text{define } \hat{t}=at \text{ and } \hat{y}=by/a\\ \end{align}

you can use the same definitions of $\hat{x}$, $\hat{y}$ and $\hat{t}$ for the second equation, will leave that one to you

caverac
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  • @LutzLehmann Thanks, just updated the answer – caverac Feb 04 '21 at 10:15
  • Why/how does one decide to multiply by d/c? It seems easy if one reads backwards in the notes, but if one starts from scratch, then "multiply by d/c" seems random. – mavavilj Feb 04 '21 at 10:37
  • Does one expect to make $\hat{x}-\hat{x}\hat{y}$? And then sees how to produce it? Progress variable by variable eliminating the leading coefficients? – mavavilj Feb 04 '21 at 10:50
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Let's try it from a slightly different angle: Decompose every variable into a constant scale factor and a "normalized" variable, $x=x_*\hat x$ etc. Then insert into the equations$\newcommand{\dd}{{\rm d}}$ \begin{align} \frac{\dd(x_*\hat x)}{\dd(t_*\hat t)} &= a\,x_*\hat x-b\,x_*\hat x\,y_*\hat y\\ \frac{\dd(y_*\hat y)}{\dd(t_*\hat t)} &= c\,x_*\hat x\,y_*\hat y-d\,y_*\hat y \end{align} Now remove the scale factors from the left side and collect the scale factors \begin{align} \frac{\dd\hat x}{\dd\hat t} &= (at_*)\,\hat x-(bt_*y_*)\,\hat x\hat y\\ \frac{\dd\hat y}{\dd\hat t} &= (ct_*x_*)\,\hat x\hat y-(dt_*)\,\hat y \end{align} Now set the combined coefficients to $1$ as long as that is possible: $$ at_*=1\implies t_*=\frac1a,\\ bt_*y_*=1\implies y_*=\frac{a}{b},\\ ct_*x_*=1\implies x_*=\frac{a}{c},\\ dt_*=\frac{d}{a}=\hat d\text{ has no free parameter},\\ $$ This gives the normalized system \begin{align} \frac{\dd\hat x}{\dd\hat t} &= \hat x-\hat x\hat y\\ \frac{\dd\hat y}{\dd\hat t} &= \hat x\hat y-\hat d\,\hat y \end{align}

The other variant to simplify the second equation is to have both coefficients equal, so that the resulting factorization and thus the stationary points is/are normalized. $$ ct_*x_*=dt_*\implies x_*=\frac{d}{c}. $$ The normalized system is then \begin{align} \frac{\dd\hat x}{\dd\hat t} &= \hat x-\hat x\hat y\\ \frac{\dd\hat y}{\dd\hat t} &= \hat d\,(\hat x\hat y-\hat y) \end{align}

Lutz Lehmann
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