1. Ok so i'm not sure if factorisation is the crux of the problem. The non-zero steady states of a population model are given by the roots of the following polynomial, with 'r' and 'q' being parameters whose value is greater than zero. using the conditions of a "double root" i am to show that the curve in r,q space that partitions the space into regions where there are either 1 or 3 positive steady states is given parametrically by,  .

As i understand it, since negative or complex population numbers make no physical sense and as there are three roots to the cubic above it follows that in regions where there are three unique roots only one of these will be real. Hence it is only in the region where a double root occurs that three solutions (two unique) will occur.

So my first thought was to use the discriminant of a cubic, which i pulled from wiki, however that just led to pages of scribbling and crossing out. (i hate that) So i instead decided that i'd utilise the fact that a multiple root of any polynomial is also a stationary point of said polynomial. ie in the case of a cubic then for .

So with some quick differentiation i find that when u = b, and thanks to a formula they made me learn in grade 9-10ish it follows that, .

now the only remaining thing i can think of doing is to recall that we require the solution to be real, thus . But this has not gotten me the answer i so desperately seek. If someone could point out a mistake or suggest how to continue this it would be much appreciated.

... never mind  2.

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