Consider the two species competition model given by
$$ \frac{da}{dt }= \frac {λ_1 a} {a+K_1} - r_{ab}\cdot ab - da, \ \ \ \ \ \ \ \ \ \ (1)$$
$$\frac{db}{dt }= λ_2 b (1-\frac{b}{K_2}) - r_{ba}\cdot ab , \ \ \ \ \ \ \ t>0,\ \ \ \ \ \ \ \ (2)$$
for two interacting species denoted $a=a(t)$ and $b=b(t)$, with initial conditions $a=a_0$ and $b=b_0$ at $t=0$. Here $λ_1, λ_2, K_1,K_2$, $r_{ab}$,$r_{ba}$ and $d$ are all positive parameters.,
NOW it is decided that the model is to non-dimensionalised such that
$a= Xu$, $b=K_2 v$, and $t= \frac{\tau}{Y}$,
where $X$ and $Y$ are parameter scalings to be determined. Here $u,v$ and $\tau$ are all non-dimensional variables.
The application of this non-dimensionalisation leads to the equation for $a(t)$ becoming \begin{equation} \frac {du}{d\tau}=\frac{\overline{\lambda}_{1} u}{u+1} -\overline{r}_{ab}uv -\overline{d}u \end{equation} where $\overline{\lambda}_{1}= \frac {\lambda_{1}}{\lambda_{2} K_{1}}$ Determine X and Y, all of the remaining non-dimensional parameters and their relationship to the dimensional parameters and the non-dimensional equation describing b(t).
=>i have started by finding b(t), but sure whether is right or wrong,
\begin{equation}\implies \frac {dv}{d\tau}={\overline{\lambda}_{2}}(1-v) -\overline{r}_{ba}uv \end{equation}
where I think $ \overline{\lambda_{2}}=\lambda_{2}K_{2}$ I need to determine X and Y and also all the remaining non-dimensional parameters like $\overline{r}_{ba},\overline{\lambda}_{2}, \overline{\lambda}_{1},\overline{r}_{ab},\overline{d}$ and their relationship to the dimensional parameters, by using the two equations of a(t) and b(t).
Please help me, I am really confused. Thanks!
Assume that $\frac{\mathrm da}{\mathrm dt}=F(a(t),b(t))$ and that $(a,b,t)=(\alpha/A,\beta/B,\tau/T)$ for some positive parameters $(A,B,T)$, then $$\frac{\mathrm d\alpha}{\mathrm d\tau}=A\frac{\mathrm da}{\mathrm d\tau}=\frac{A}T\frac{\mathrm da}{\mathrm dt}=\bar F(\alpha(t),\beta(t))$$ if the function $\bar F$ is such that $\bar F(Aa,Bb)=\frac{A}TF(a,b)$, that is, $$ \bar F(\alpha,\beta)=\frac{A}TF\left(\frac{\alpha}A,\frac{\beta}B\right). $$ The function $F$ in the RHS of (1) is $$ F(a,b)=\frac {λ_1 a} {a+K_1} - r_{ab}\cdot ab - da $$ hence one gets $$ \bar F(\alpha,\beta)=\frac{A}T\left(\frac {λ_1 \frac{\alpha}A} {\frac{\alpha}A+K_1} - r_{ab}\cdot \frac{\alpha}A\frac{\beta}B - d\frac{\alpha}A\right). $$ The ratio in the first part of the RHS must yield a multiple of $\frac{\alpha}{\alpha+1}$ and this happens if $$AK_1=1. $$ When this condition is met $$ \bar F(\alpha,\beta)=\frac {\bar λ_1 \alpha} {\alpha+1} - \bar r_1\cdot \alpha\beta - \bar d\alpha, $$ with $$ \bar\lambda_1=\frac{A}T\lambda_1=\frac{\lambda_1}{K_1T},\qquad\bar d=\frac{d}T,\qquad\bar r_1=\frac{r_{ab}}{BT}. $$ One can apply the same technique to (2) with the function $G$ defined by $$ G(a,b)=λ_2 b \left(1-\frac{b}{K_2}\right) - r_{ba}\cdot ab, $$ to rewrite it as $$ \frac{\mathrm d\beta}{\mathrm d\tau}=\bar G(\alpha,\beta), $$ for some function $\bar G$ defined by $$ \bar G(\alpha,\beta)=\bar λ_2 \beta (1-\beta) - \bar r_2\cdot \alpha\beta. $$ The factor $\beta(1-\beta)$ arises if $$ BK_2=1. $$ When this condition is met, $$ \bar\lambda_2=\frac{\lambda_2}T,\qquad\bar r_2=\frac{r_{ba}}{AT}=\frac{r_{ba}K_1}T,\qquad\bar r_1=\frac{r_{ab}K_2}{T}. $$ To sum up, the parameters $(A,B)$ are determined and the parameter $T$ is free but if one further imposes the (rather mysterious) condition $$ \bar\lambda_1=\frac{\lambda_1}{\lambda_2K_1}, $$ then the parameter $T$ is determined as well and must be $$ T=\lambda_2. $$