Problem :
Given the non-linear system: $$ x'=(a-b\cdot y)\cdot x $$ $$y'=(-c+d\cdot x)\cdot y $$ $$ a,b,c,d>0$$ also known as the predator–prey equation.
Also, Lyapunov's function $$V=(y^a\cdot e^{-b\cdot y})\cdot (x^c\cdot e^{-d\cdot y})$$ is given
- Prove that the non-linear system is stable to the stationary point $$A\bigg(\dfrac{c}{d},\dfrac{a}{b}\bigg)$$
Firstly, I solved the system take four different cases:
For $$x=0$$ $$y=0$$ i take the stationary point $$O(0,0)$$
For $$x\neq 0$$ $$y=0$$ which is not possible
For $$x=0$$ $$y\neq 0$$ which is not possible
and finally. For $$x\neq 0$$ $$y\neq 0$$ we take the given stationary point $$A(\dfrac{c}{d},\dfrac{a}{b})$$
I differentiate the liapunov's function and i take: $$V'=(y^a\cdot e^{-b\cdot y}\cdot c\cdot x^{-d\cdot x}-d\cdot y^a\cdot e^{-b\cdot y}\cdot x^c\cdot e^{-d\cdot x})\cdot x' + (a\cdot y^{a-1}\cdot e^{-b\cdot y}\cdot x^c \cdot e^{-d\cdot x}-b\cdot y^a\cdot e^{-b\cdot y}\cdot x^c\cdot e^{-d\cdot x})\cdot y'$$
Ι have to find that $$V'\leq 0$$ so to prove that A is a stable point but i have some difficulties to prove this.Can anyone help with this?
I would really appreciate a thorough solution and explanation.
Thanks in advance!
By differentiating the two components of $V$ separately, you can see that $g(y) = y^a\cdot e^{-b\cdot y}$ has a global maximum at $y = \frac{a}{b}$ and that $h(x)=x^c\cdot e^{-d\cdot x}$ has a global maximum at $x = \frac{c}{d}$. Thus $V$ has a global maximum at $\left(\frac{c}{d},\,\frac{a}{b}\right)$.