5
$\begingroup$

The question is (from Arnol'd 's book : Ordinary Differential Equations):

"Which of the differential equations $ \dot x = x^ n$ determine on an affine line a phase velocity field that can be extended without singular points to the projective line?"

The answers that are given are n = 0,1,2.

In the book from where I got this question, phase velocity field is when you plot the function $ v (x) = \dot x = x^n $ against $x$. $x$ is on the vertical axis and $v(x)$ on the horizontal one.

For $\dot x = x$ and $\dot x = 1 $, the phase velocity fields are a diagonal line and a vertical line, respectively. So it makes sense that these are solutions. On the other hand, $\dot x = x^2$ has the phase velocity field that looks like $ y = \sqrt x $, which is what the phase velocity fields look like for higher values of n (although they get flatter faster).

So what I don't understand is why is $ n = 2 $ a solution, but not $ n > 2 $

So where am I going wrong? Any help is appreciated. Thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ I guess this is a quite good question. $\endgroup$ – Leonhardt von M Apr 3 '15 at 22:59
5
+50
$\begingroup$

What Arnol'd is asking is about the behaviour of the solution on $\mathbb R \cup \{\infty\}$, which is obtained by "gluing together" the ends of the real line at a "point at infinity". You can also visualize this as an attempt to map $\mathbb R$ to a circle $S^1$ stereographically. Take a look at the image below. We draw a line from the North pole $N$ of the circle, through any point $P$ on the circle, and then we identify the point $P'$ on the real line with the point $P$ on the circle. You can try obtaining an explicit formula however it's pretty much clear that it's one-to-one and onto - except for the case when $P=N$ - that is, when $NP$ and the real line become parallel. If $P$ approaches $N$ in an anticlockwise direction, the corresponding $P'$ approaches $+\infty$. For the other direction, it approaches $-\infty$. The projective real line is simply defined as having that one extra point, making the stereographic projection a bijection.

enter image description here

Now we're coming to what Arnol'd is asking. If we stay on the real line, a solution to a differential equation can diverge in finite time. However on the projective real line, nothing "diverges", it just reaches the point at infinity in finite time. Or: it reaches the North pole in finite time. Now if we move $P$ to $N$ and then continue going, we will end up on the negative real line.

Arnol'd asks: what solutions are nonsingular on the projective line? Well, yes we want to avoid "singular points", but we can think of this as passing from the positive real axis to the negative real axis continuously, with just an added minus sign that indicates we are now on the negative real line.

The solution to the equation $\dot x = x^n,n>1$ is $$x(t)= c_n\left(t_{\infty}-t\right){}^{\frac{1}{1-n}},$$ where $c_n:= (n-1)^{\frac{1}{1-n}}$, and $t_{\infty}:=\frac{x(0)^{1-n}}{n-1}$, so that $x$ blows up at $t=t_{\infty}$.

This is already in a form of "series expansion" around $t_{\infty}$. For instance, we can see that for $n=2$, we have $$x(t)=c_2 \frac{1}{t_{\infty}-t},$$

so that $$\lim_{\epsilon\rightarrow 0} x(t_{\infty}-\epsilon)=+\infty$$ and $$\lim_{\epsilon\rightarrow 0} x(t_{\infty}+\epsilon)=-\infty.$$

This is good, this is what we were talking about. If you plot the solution, you can see that it blows up, reaching its vertical asymptote, and then reappearing to the right of the asymptote from below, from $x=-\infty$. However, for $n>2$, $$\lim_{\epsilon\rightarrow 0} x(t_{\infty}-\epsilon)=\lim_{\epsilon\rightarrow 0} c_n \frac{1}{\epsilon^{\frac {1}{n-1}}},$$

yet

$$\lim_{\epsilon\rightarrow 0} x(t_{\infty}+\epsilon)=\lim_{\epsilon\rightarrow 0} c_n \frac{1}{(- \epsilon)^{\frac {1}{n-1}}}.$$

If we extract the minus, we get a factor of $(-1)^\frac {1}{1-n}=e^{\frac{i \pi}{1-n}}$, which is not a real number! Clearly we have gotten out of the projective real line by trying to extend time beyond $t_{\infty}$, and ended up on some new branch of the appropriate complex function. I hope this helped! And I will leave the asker to re-examine the cases $n=0,1$ from this point of view!

$\endgroup$
  • $\begingroup$ My issues is that for $n-1$ odd, then $$ \lim_{\epsilon \to 0} c_n \frac{1}{(-\epsilon)^{\frac{1}{n-1}}} $$ has a real branch. The odd square root of negative one always has a branch which is equal to negative one. If you were solving this differential equation you would have to choose one of the complex branches to actually use as the solution. Choose the branch where $(-\epsilon)^{\frac{1}{n-1}}$ gives negative reals. This would give - according to your explanation - a solution extended w/o singular points projective plane. $\endgroup$ – William Vickery May 23 '16 at 17:26
  • $\begingroup$ Why $\lim_{\epsilon \to 0} x(t_{\infty}+\epsilon)=-\infty$ and $\lim_{\epsilon \to 0} x(t_{\infty}-\epsilon)=+\infty$? $\endgroup$ – Александр Jan 5 at 20:39

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.