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Background:

I am working on the mathematical modeling of infectious diseases, namely HIV and TB. In the process of proving global asymptotic stability of the Disease-Free Equilibrium, I must construct a Lyapunov function. To that end, I have arrived at the following problem and would like some help.

Problem: How can I go about finding a natural number $k$ such that the expression

$$W\bigg(A-\frac{X}{W}\bigg)^{2k-1}\bigg[X-AW-\frac{AY}{N}(B+D)-\frac{AZ}{N}(C+D+E+F+G)\bigg]+\frac{X}{N}[Y(A+H)(B+D)+ZA(C+D+E+F+G)]$$

is always negative?

In this problem, the parameters $$A,B,C,D,E,F,G,H,N,W,X,Y,Z$$ are all positive. In addition, $$N > A+B+C+D+E+F+G+H.$$

I hope you all can either guide me mathematically or explain how I can use software to help me, as I am relatively new to Maple.

Thanks for any help!

1 Answers1

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You can evaluate everything in capital letters to make your expression $a^{2k-1}b+c$ As all the capital letters are positive, we have $c \gt 0$ If $a,b$ have the same sign, you are sunk as the first term will be positive, too. As we have $b=Wa-stuff$, we assume $a \gt 0, b \lt 0$ So we have to assume $AW \gt X$. Then $b \lt 0$ will be satisfied. Note that $H$ only appears in the last term. If it is huge you will need $A-\frac XW \gt 1$ and $k$ large. Similarly, $X$ multiplies the whole last term. I don't think you will find a $k$ independent of the parameters. But if you want $a^{2k-1}b+c \lt 0$ you need $$a^{2k-1} \lt \frac c{-b}\\(2k-1)\log a \lt \log \frac c{-b}\\k \lt \frac {\log \frac {ac}{-b}}{2 \log a} $$ if $a \gt 1$ In case $a \lt 1$, dividing by its log reverses the inequality because the log is negative, giving $$k \gt \frac {\log \frac {ac}{-b}}{2 \log a} $$

Ross Millikan
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