I'm taking an Hons ODE class and the teacher just assumes everyone is comfortable with analysis (which I never took before) and I'm having trouble understanding how it is applied.
We're studying the stability of ODE solutions, particularly we want to show stability/unstability using the epsilon-delta definition, i.e,
$$ \text{If for any } \ \epsilon > 0 \ ,\exists \ \delta > 0 \ \text{such that } $$
$$ \text{If } \ ||y(0) – \widetilde{y}||\le \delta \ \text{then} \ ||y(t) – \widetilde{y}|| \le \epsilon , \ \forall t \ge 0 \quad ,\text{then $\widetilde{y}$ is stable.} $$
where $y(t)$ is a solution to the initial ODE, and $\widetilde{y}$ is an equilibrium point.
The particular problem on which we're trying to use this definition is
\begin{equation}
y' = y^{2} -(a+b)y +ab \qquad \text{where} \ 0<a<b. \tag{1}
\end{equation}
After long hours of trying to solve this mess (since it is separable), I gave up, you have to solve the ODE then truncate the implicit expression in three functions and so on and so on , the math just gets messier. I did find the roots $y=a , \ y=b$ which correspond to the equilibrium points.
The teaching assistant hinted my for this problem ; $(1)$'s right-hand side is actually a parabola. We can graph this parabola where the y-axis is $y'$ and the x-axis is $y$. From this graph it is then evident which are stable/unstable by looking at the slope's sign on the graph around each point $a$ and $b$.
From here I'm stuck I don't understand how to prove this using the epsilon-delta proof, because we don't have explicitly the solution $y(t)$, but to use the definition this quantity is required. Does anyone have any idea how to proceed ?
Best Answer
This is a Riccati equation with the known solutions $y\equiv a$ and $y\equiv b$, which allows to transform it to a Bernoulli equation and so on.
The other trick is to set $y=-\frac{u'}{u}$ which results in $$ u''+(a+b)u'+abu=0 $$ which can be handily solved using the methods of linear DE with constant coefficients.
What you can see from the graph is that the solutions have to be monotonous in time, either moving to the left forever or moving to the right for ever and since ever. Thus if you start in some small interval around $a$, you will never leave that interval. If you start in a small interval around $b$, you will always leave that interval.
This situation becomes only complicated in higher dimensions where the vector field on the spheres around the equilibrium point does not point inward everywhere or outward everywhere. You have to allow that the solution (temporarily) leaves the ball $B(\tilde y;δ)$, but can demand that it does not go too far away, staying inside some larger $B(\tilde y;ϵ)$.