critical points, differential equation
I have two differential equations and my assignment is to prove that this system have a unique stationary point. $$\begin{align} \frac{dx}{dt}&=a-(b+1)x+x^2 y\\ \frac{dy}{dt}&=bx-x^2y\\ \end{align}$$
I have tried to solve it, by putting them equal to zero and then solve x and y. But i get strange answers. Like x depends on y to make it zero. Is that a criteria that make it a non statonary point?
In the second equation you can tell if x is zero it will become zero.
Can someone help me solve this problem? How do I now that it is a unique point? And what is this point?
$\endgroup$2 Answers
$\begingroup$What are the assumptions on the parameters $a,b$?.
You have the correct idea, by definition you have to equate both lines to $0$.
The equilibrium point is unique if, well, you can find it and show that there is no other. Equivalently, the simultaneous system $\left\{a-b(x+1)x+x^2y=0;\quad bx-x^2y=0\right\}$ has a unique solution.
From the second equation you have $x(b-xy)=0$. The last equation has 2 solutions.
- $x=0$ implies in the first equation: $a=0$ and $y$ arbitrary.
- $b=xy$ implies in the first equation: $a-(xy+1)x+x^2y=0$ and therefor $a=x$, $y=b/a$.
In conclusion if $a\neq 0$ there is a unique equilibrium point (for each fixed value of $a$ and $b$) at $(a,b/a)$.
$\endgroup$ 4 $\begingroup$At a stationary point $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=0$
$a-(b+1)x+x^2 y=0$ and $bx-x^2y=0$
$x=0$ is solution of the second, but not of the first condition. So, the second condition is $b-xy=0$ leading to $xy=b$
Bringing it back into the first condition leads to $a-(b+1)x+b x=0$ which gives $x=a$
So, the condiotions are $x=a$ and $y=\frac{b}{a}$.
As a consequence this point is stationary. We saw that the system of two equations has no other solution, hense the poind found is unique.
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