5
$\begingroup$

Given the following system of differential equations \begin{equation} \begin{cases} \frac{dx}{dt}=-x+xy \\\frac{dy}{dt}=-2y-x^2 \end{cases} \end{equation} I want to prove that the maximal solutions of the system are defined on all $\mathbb{R}$ and that $$\lim_{t\rightarrow +\infty } {(x(t),y(t))}=0$$

As shown here I tried proceeding same way but I don't have any information about $t=0$. $$x′x+y′y=-x^2-2y^2\leq 0$$ $$x′x+y′y= x'+y'\cdot x+y =\frac{1}{2}\frac{d}{dt}|| x+y ||^2$$ $$\frac{1}{2}\frac{d}{dt}|| x+y ||^2\leq 0$$ Hence I have information about the monotony, but I don't know how to show that, for example, the solution is limited on a subset of $\mathbb{R}$ (I could use this hypothesis to prolong the solution on all $\mathbb{R}$).

$\endgroup$
  • 1
    $\begingroup$ I don't know what you mean by the norm of $x+y$. But your first inequality shows that $V(x,y)=x^2+y^2$ is a strong Lyapunov function, so for $t \to +\infty$ things follow immediately from Lyapunov's theorem. To show that the solution is defined for all $t < 0$ would require some other argument. $\endgroup$ – Hans Lundmark Nov 29 '17 at 11:26
  • $\begingroup$ Possible duplicate of Differential Problem $\endgroup$ – Hans Lundmark Dec 2 '17 at 9:46
2
$\begingroup$

Note that you already have $$x′x+y′y=-x^2-2y^2$$ which implies $$ \frac{1}{2}(x^2+y^2)'=-x^2-2y^2\le-x^2-y^2=-(x^2+y^2). $$ So $$ (x^2+y^2)'+2(x^2+y^2)\le0$$ or $$ e^{2t}(x^2+y^2)'+2e^{2t}(x^2+y^2)\le0$$ or $$ [e^{2x}(x^2+y^2)]'\le0. $$ Integrating from $0$ to $t$ gives $$ e^{2t}(x^2(t)+y^2(t))\le x^2(0)+y^2(0) $$ or $$ x^2(t)+y^2(t)\le e^{-2t}[x^2(0)+y^2(0)]. $$ So $$ \lim_{t\to\infty}(x^2(t)+y^2(t))=0. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @C.Rose, see the update. $\endgroup$ – xpaul Nov 29 '17 at 21:42
5
$\begingroup$

There are many ways to approach such things, which can be studied on a Dynamical Systems course.

(1) With some brute force, you can prove by solving each equation separately :

$$x(t) = c_1e^{\int [y(t)-1]dt}$$

$$y(t) = \frac{-\int e^{2t}x^2(t)dt+c_2}{e^{2t}}$$

and then make conclusions.

(2) We have the system :

$$\begin{equation} \begin{cases} x'=-x+xy \\y'=-2y-x^2 \end{cases} \end{equation}$$

We observe that $O(0,0)$ is a stationary point for the system.

The Jacobian of this system, is :

$$J(x,y) = \begin{bmatrix} -1+y & x \\ -2x & -2 \end{bmatrix}$$

and thus, the Jacobian of regarding the stationary point $O(0,0)$ is :

$$J(0,0)=\begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}$$

Calculating the eigenvalues of the $J(0,0)$ :

$$\det(J(0,0)-λI)=0\Rightarrow(-1-λ)(-2-λ)=0 \Leftrightarrow \begin{cases} λ_1=-1 \\ λ_2 = -2 \end{cases}$$

Thus, it is : $λ_2 < λ_1 < 0$ and $λ_1\cdot λ_2 > 0$, which means $O(0,0)$ is an asymptotically stable node.

From the image of the phase plane that corresponds to an asymptotically stable node (check here), you can conclude that the solutions "collapse" to $O(0,0)$, thus : $\lim_{t\rightarrow +\infty } {(x(t),y(t))}=0$

(3) As mentioned in one of the comments, the first expression you have written (the inequality), shows that the functional $V(x,y) = x^2 + y^2$ is a strong Lyapunov one, so for $t \to + \infty$ you can make a conlcusion from Lyapunov's theorem. Though, as mentioned again, that needs an extra argument for $t<0$.

(4) Via Linearization with polar coordinates, you can transform your system as :

$$x=r\cosθ$$

$$y=r\sinθ$$

$$x^2 + y^2 = r^2$$

Differentiating the last equation, yields :

$$rr' = xx' + yy'$$

To find the angle $θ$, we take :

$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$

Using the quotient rule, we get :

$$\theta' = \dfrac{x y' - y x'}{r^2}$$

Making substitutions you can thus yield results for a new system :

$$\begin{cases} r' = \dots \\ θ' = \dots \end{cases}$$

and make conclusions.

Regarding polar coordinates, you can check an answer of mine, here.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Could you explain "From the image of the phase plane that corresponds to an asymptotically stable node (check here), you can conclude that the solutions "collapse" to ...". This would be true for a linear system. But the given system is nonlinear. We could have an asymptotically stable origin in combination with an unstable or semistable limit cycle, which we did not rule out. By linearization, we only know that the origin is locally asymptotically stable in an environment around the origin. $\endgroup$ – MrYouMath Nov 29 '17 at 15:08
  • $\begingroup$ @MrYouMath All the types of stationary points hold their topological properties from the lineariased to the non-linear. The only type of stationary point that does not hold its topological property is case of a center, which is not the case here. $\endgroup$ – Rebellos Nov 29 '17 at 15:30
  • $\begingroup$ I am not talking about the local behaviour but about the global behaviour. You said $\lim_{t\to \infty}(x(t),y(t))=0$, but I think that this claim is not necessarily true for any initial condition and a given asymptotic equilibrium of a nonlinear system. You first need to show global asymptotic stability to make this conclusion. $\endgroup$ – MrYouMath Nov 29 '17 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.