0
$\begingroup$

As is well known, there is a comparison principle for scalar differential equations $\dot x(t)=a(x(t))$ and $\dot y(t)=b(y(t))$ with $x(t_0)=y(t_0)$ and $a(\cdot)\leq b(\cdot)$, with the result being that $x(t)\leq y(t), \forall t\geq t_0$.

Is there a similar version of comparison principle for vector differential euqations $\dot x=F(x)$ and $\dot y=G(y)$? More specifically, if $\|F(x)\|\leq \|G(x)\|$ and $x(t_0)=y(t_0)$, what information can be obtained regarding the relationship of the two differential equations?

Improve this question
$\endgroup$
1
  • $\begingroup$ Your hypothesis $\|F(x)\|\leq \|G(x)\|$ is too weak to say something very useful, as illustrated in the answer by Kostya below. However, if you allow $x$ to take values in some space with a (sufficiently smooth) partial order relation $\le$ and $F(x) \le G(x)$ for all $x$, then you essentially have the same conclusions as in the scalar case. An example of such a space is $(\mathbb{R}^n,\le)$, where $(a_1,\ldots,a_n) \le (b_1,\ldots,b_n)$ if and only if $a_i \le b_i$ for all $i$. $\endgroup$
    – Igor Khavkine
    Commented Jun 21, 2017 at 12:49

1 Answer 1

Reset to default
1
$\begingroup$

Well, take $F(x)=x$ and $G(x)=2ix$, viewing $x$ as a complex number. Then, $x(t)=e^{t}x(t_0)$ is unbounded, but $y(t)=e^{2it}y(t_0)$ is bounded.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .