Differential Problem

Question

I can't solve this exercise my professor gave me.. is there anyone who is able to?

Study:

(P) $\begin{cases} \frac{dx}{dt}=-x+xy\\ \frac{dy}{dt}=-2y-x^{2} \end{cases}$

a) Demonstrate that the maximal solutions of P are defined on all $\mathbb{R}$ and that

$\underset{t\rightarrow+\infty}{lim}(x(t),y(t))=0$

b) Demonstrate that if $(x(0),y(0))\neq(0,0)$ ,

$\forall t\epsilon\mathbb{R}$ :

$(x(t),y(t))\neq(0,0)$

Thanks a lot!

EDIT:

Thanks for the answer to everyone of you, but the problem is that we haven't studied the Lyapunov Functions yet.. is there any other way to solve it?

Answer 1

Let's try out the Lyapunov Function : $V(x,y) = x^2 + y^2$, so it is :

$$\dot{V}(x,y)=2x(-x+xy)+2y(-2y-x^2)=-2(x^2+2y^2) <0\forall(x,y)\in\mathbb R^2-(0,0)$$

It's also obvious to see that $O(0,0)$ is a stationary point for the system and given the Lyapunov considerations, it's asymptotically stable.

Considering that : $V(x,y) = x^2+y^2 > 0 \forall (x,y)\in \mathbb R^2-(0,0)$ and that it's continuously differentiable, can you figure out if that and the one before leads you somewhere for part $(a)$ and part $(b)$ ?

Answer 2

Read up on Lyapunov functions, the standard one $V=x^2+y^2$ should work. $$ (\dot Vf)(x,y)=\nabla V\cdot(\dot x,\dot y)^T=2x(-x+xy)+2y(-2y-x^2)=-2(x^2+2y^2)\in[-4V,-2V] $$

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Using $v(t)=V(x(t),y(t))$ we thus got $$ -4v\le \dot v\le -2 v $$ which can be solved via separation or integrating factor to find $$ v(0)e^{-4t}\le v(t)\le v(0)e^{-2t} $$ for $t>0$. The second inequality answers the first question, a bounded solution exists for all times, and the first inequality gives a quantitative answer to the second question.

Answer 3

In a sense this is just a rewriting of the previous answers, but it avoids using explicitly the Lyapunov formalism.

For part a), consider $V(t)=x(t)^2+y(t)^2$. Then $$ V'(t)= 2x(t)x'(t)+ 2y(t)y'(t) = -2x(t)^2 -4y(t)^2 \leq -2x(t)^2 -2y(t)^2 = -2V(t)$$ In particular, by Gronwall's lemma, we have $V(t)\leq f(t)$ for all $t\geq 0$, where $f$ is the solution of $$ f(0)=V(0), \quad f'(t)=-2f(t)$$ i.e. $$ V(t)\leq V(0)e^{-2t} \quad \forall t\geq 0$$ In particular this implies that, for any initial data, the solutions exist for all $t\geq 0$ since the norm $\Vert x(t),y(t)\Vert$ stays bounded (there is no explosion time). Moreover, since $V(t)\geq 0$ and $V(0)e^{-2t}\to 0$ as $t\to\infty$, we also obtain that for any initial data $$V(t)\to 0 \text{ as } t\to\infty$$

For part b), first of all observe that every term in your ODE is smooth, therefore local existence and uniqueness of the solutons is granted. Moreover, $(0,0)$ is a fixed point, therefore $x(t)=0$, $y(t)=0$ is a solution for the initial data $x(0)=0$, $y(0)=0$ (and by uniqueness it is the only solution).

Now assume that, for some $t$, you have $x(t)=y(t)=0$, i.e. the solution $(x(t),y(t))$ and the solution constantly $0$ "touch" at $t$, then by uniqueness you must have that the two solutions coincide on the domain in which they are both defined, thus $x(0)=y(0)=0$ as well. Therefore, if $(x(0),y(0))\neq(0,0)$, you must have $(x(t),y(t))\ne(0,0)$ for all $t$.