Consider the system $$\dot{x} = y \\ \dot{y} = -4x $$ ($\dot{x}$ means $\displaystyle \frac{dx}{dt}$ and $\dot{y}$ means $\displaystyle \frac{dy}{dt})$

I need to prove that the fixed point $\mathbf{x^{*} = 0}$ is Liapunov stable.

For reference, according to my textbook (Strogatz), a fixed point $\mathbf{x^{*}}$ is said to be Liapunov stable if $\forall \epsilon > 0$, $\exists \delta >0$ such that $|\mathbf{x(t)} - \mathbf{x^{*}}| < \epsilon$ whenever $t \geq 0$ and $|\mathbf{x(0)}- \mathbf{x^{*}}|< \delta$

Now, if we let $x(0)=x_{0}$, $y(0)=y_{0}$, I was able to solve this system using eigenvalues and eigenvectors to have the following general solution: $$ x(t) = x_{0} \cos (2t) + \frac{y_{0}}{2}\sin(2t) \\ y(t) = -2x_{0}\sin(2t) + y_{0}\cos(2t)$$

According to my solutions manual, "it can be observed from the equations in the solution that $\delta < |x_{0}|$. This implies that $|x(t),y(t)|<2|x_{0}|$, where $2|x_{0}|$ is the radius $\epsilon$. Thus, it can be concluded that $|x(t),y(t)|<\epsilon$."

However, I don't understand how they were able to show this. We never really learned how to do these kinds of $\epsilon-\delta$ proofs for systems in my class, and this solution doesn't give a very good (meaning "detailed") explanation as to how they went about doing it.

Could someone please explain to me step-by-step how they were able to surmise all of this? I would like to be able to apply this to other problems going forward, but unless I see one done out in detail, I am afraid I will not be able to.

Thank you for your time and patience.

  • $\begingroup$ What is the book? $\endgroup$ – CroCo May 18 '18 at 3:58

The solution to this problem is an ellipse parameterized by the initial conditions $x_0,y_0$ that has a center at $\{0,0\}$. Therefore, for a given set of initial conditions the maximum distance a solution could have at any time from the fixed point at the origin will be a fixed finite number. Intuitively then your solution is Liapunov stable.

For a rigorous proof of this fact you will use the $\epsilon-\delta$ definition you gave. The idea is this: $\epsilon$ is assumed to be some given small quantity and it is your task to find the $\delta$ that makes the definition true.

Your $\delta$ will be parameterized by the initial conditions and $\epsilon$. If you knew the maximum distance from the origin a solution could attain(semimajor axis), then you could use that knowledge to define $\delta$.

Since $x^*=0$ you will not have the usual algebraic difficulties associated with this kind of proof. Start with the statement $|x(t)-x^*|$ and develop a chain of inequalities leading to the required $\delta$.

Reference to the solution manual: In general, $\delta$ will depend on $x_0$ and $y_0$. However, i think the trick here is to observe that if you find a $\delta<\text{some distance}: d(x_0,y_0)$ that works then choosing the magnitude of the $x-$component as a new $\delta$ must also work because it could not possibly be bigger than the original.


The norm of the solution is $$ \|{\bf x}(t)\|=\sqrt{x^2(t)+y^2(t)}=\sqrt{\left(x_0\cos 2t+\frac{y_0}2\sin 2t\right)^2+\left(-2x_0\sin 2t+y_0\cos 2t\right)^2}. $$ One can use the triangle inequality: $$ \sqrt{(a_1+b_1)^2+(a_2+b_2)^2}\le \sqrt{a_1^2+a_2^2}+\sqrt{b_1^2+b_2^2} $$ in order to obtain $$ \|{\bf x}(t)\|\le\sqrt{x_0^2\cos^22t+4x_0^2\sin^22t}+\sqrt{\frac{y_0^2}4\sin^22t+y_0^2\cos^22t} $$ $$ =2|x_0|\sqrt{\frac14\cos^22t+\sin^22t}+|y_0| \sqrt{\frac14\sin^22t+\cos^22t}. $$ Notice that $$ \sqrt{\frac14\cos^22t+\sin^22t}\le\sqrt{\cos^22t+\sin^22t}=1, $$ $$ \sqrt{\frac14\sin^22t+\cos^22t}\le\sqrt{\sin^22t+\cos^22t}=1, $$ hence, $\|{\bf x}(t)\|\le2|x_0|+|y_0|$ ($\|{\bf x}(t)\|\le2|x_0|$ in the book is possibly a typo).

We have $|x_0|\le \|{\bf x}(0)\|$ and $|y_0|\le \|{\bf x}(0)\|$, thus, $\|{\bf x}(t)\|\le3\|{\bf x}(0)\|$. It means that for any $\epsilon>0$ one can take $\delta=\frac13\epsilon$ to prove the stability: $$ \|{\bf x}(0)\|<\delta=\frac13\epsilon\;\Rightarrow\;\|{\bf x}(t)\|\le 3\|{\bf x}(0)\|<3\delta=\epsilon. $$


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