Lotka-Volterra model for two competing species in Boyce's Elementary Differential Equations and Boundary Value problems' is given by: \begin{align*} \frac{dx_{1}(t)}{dt}&=r_{1}x_{1}-a_{11}x_{1}x_{1}-a_{12}x_{1}x_{2} \\ &=r_{1}x_{1}(1-\frac{a_{11}}{r_{1}}x_{1})-a_{12}x_{1}x_{2} \\ &=r_{1}x_{1}(1-\frac{x_{1}}{K_{1}})-a_{12}x_{1}x_{2},\quad x_{1}(0)>0 \\ \frac{dx_{2}(t)}{dt}&=r_{2}x_{2}-a_{22}x_{2}x_{2}-a_{21}x_{1}x_{2} \\ &=r_{2}x_{2}(1-\frac{a_{22}}{r_{2}}x_{2})-a_{21}x_{1}x_{2} \\ &=r_{2}x_{2}(1-\frac{x_{2}}{K_{2}})-a_{21}x_{1}x_{2},\quad x_{2}(0)>0 \\ K_{1}= \frac{r_{1}}{a_{11}}\\ K_{2}= \frac{r_{2}}{a_{22}} \end{align*} $ r_{i}, i=1, 2 $:is the intrinsic growth rate for prey species.

$ a_{ii} $ :are intra-species interference coefficient of the two prey species.

$ K_{i}, i=1, 2 $ : the environmental carrying capacity for ith prey species.

$ a_{ij} $: are inter-species interference coefficient of the two prey species.

While in the book by Murray 'Mathematical Biology: I. An Introduction' is given by: \begin{align*} \frac{dx_{1}(t)}{dt}&=r_{1}x_{1}\left(1-\frac{x_{1}}{K_{1}}-\frac{a_{12}}{K_{1}}x_{2}\right),\quad x_{1}(0)>0\\ \frac{dx_{2}(t)}{dt}&=r_{2}x_{2}\left(1-\frac{x_{2}}{K_{2}}-\frac{a_{21}}{K_{2}}x_{1}\right),\quad x_{2}(0)>0\\ \end{align*} Trying to match them I get \begin{align*} \frac{dx_{1}(t)}{dt}&=r_{1}x_{1}\left(1-\frac{x_{1}}{K_{1}}-\frac{a_{12}}{K_{1}}x_{2}\right) \\ &=r_{1}x_{1}- \frac{r_{1}}{K_{1}}x_{1}x_{1}- \frac{r_{1}}{K_{1}}a_{12}x_{1}x_{2} \\ &=r_{1}x_{1}- a_{11}x_{1}x_{1}- a_{11}a_{12}x_{1}x_{2},\quad x_{1}(0)>0\\ \frac{dx_{2}(t)}{dt}&=r_{2}x_{2}\left(1-\frac{x_{2}}{K_{2}}-\frac{a_{21}}{K_{2}}x_{1}\right) \\ &=r_{2}x_{2}- \frac{r_{2}}{K_{2}}x_{2}x_{2}- \frac{r_{2}}{K_{2}}a_{21}x_{2}x_{1} \\ &=r_{2}x_{2}- a_{22}x_{2}x_{2}- a_{22}a_{21}x_{2}x_{1},\quad x_{2}(0)>0\\ K_{1}= \frac{r_{1}}{a_{11}}\\ K_{2}= \frac{r_{2}}{a_{22}} \end{align*} so here we have an extra factor in the inter-species terms. How they are equivalent?

  • 1
    $\begingroup$ Do you use upper and lower case ‘k’ by purpose for Murray’s version? $\endgroup$ – fileunderwater Nov 10 '19 at 21:44
  • $\begingroup$ Sorry for the mistanke it is upper case K in all , the environmental capacity. $\endgroup$ – F.O Nov 11 '19 at 5:59

This is probably not a very satisfactory answer. The very next step in Murray's book (3rd Ed) there is a non-dimensionalised version of the model (eq 3.32) which looks like the one given in Boyce's book (10th Ed) except $\epsilon_i$ and $\sigma_i$ (eq 2, Ch 9). If you plug in $\epsilon_i = \sigma_i = 1$ in Boyce's book you get the same fixed point expression (eq 36) as Murray's book.

The idea is: in the RHS of a $\frac{dx_i}{dt}$ term the $x_i x_i$ represents intra-specific interaction and the $x_i x_j$ represents inter-specific interaction between species $i$ and $j$. You can put constants wherever you like to scale these terms with an interpretation. But the qualitative behavior of the system doesn't change and the coefficients persist in algebra, based on how you plugged them.

But to literally interpret the extra factor in the second set of equations: you can think of them like if $a_{ii}$ is the scaling of intra-specific interference, then $a_{ij}$ is a scaling of $a_{ii}$ for inter-species interaction, i.e. the species $j$ interfere with species $i$ with $a_{ij} a_{ii}$.

  • $\begingroup$ Thanks for the contribution. The problem is that both books define the parameters biologically in similar way which doesn’t make sense for me. I think $a_{ij} $ is scaled with $a_{ii} $ and not the opposite without any explanation given. $\endgroup$ – F.O Nov 18 '19 at 15:26
  • $\begingroup$ I agree with @Cheesebread I think that one of the books misdefines the parameters because rescaling is super common and often times mathematically-oriented authors more easily forget about the underlying biological interpretation. This is especially true, as pointed out in this answer, when the qualitative behavior is the same. $\endgroup$ – Chris Moore Dec 6 '19 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.