Timeline for Is it possible to prove unboundedness of 3rd order ODE?
Current License: CC BY-SA 4.0
9 events
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| Jan 27, 2019 at 11:13 | comment | added | Robert Bryant | @user35202: You're welcome. By the way, I realized, after I wrote my previous comment (which I can't edit now), that, since $a<0$, we can never have $a^3 = (n{+}1)/n^2$ anyway, so $g_a(\tau)$ is always real-analytic. Meanwhile, the expression above would work to describe unstable solutions associated to the positive roots (if any) of $a^3{+}Aa{+}1=0$ in the regime $t<<0$, but only as long as $a^3 \not= (n{+}1)/n^2$ for some $n\ge 2$. | |
| Jan 26, 2019 at 18:41 | comment | added | user35202 | Thank you for explaining it so clearly. Very interesting. | |
| Jan 26, 2019 at 14:47 | comment | added | Robert Bryant | @user35202: Here's a little more about this, in case you are interested: If we let $a<0$ be the (unique) negative root of $a^3 + Aa^2+1=0$, i.e., $A = -(a+1/a^2)$, then as long as $a^3\not=(n+1)/n^2$ for any integer $n\ge 2$, there is an analytic function $g_a(\tau)$ on an open neighborhood of $\tau=0$ such that $x(t) = g(e^{at})$ solves the equation when $t>0$ is sufficiently large, where $$g_a(\tau) = \tau + \frac{a^2\,\tau^2}{(3{-}4a^3)}+\frac{2a^4\,\tau^3}{(3{-}4a^3)(4{-}9a^3)}+ \frac{4a^6(13{-}21a^3)\,\tau^4}{3(3{-}4a^3)^2(4{-}9a^3)(5{-}16a^3)} +\cdots .$$ This $x(t)$ converges to $0$. | |
| Jan 25, 2019 at 21:16 | comment | added | Robert Bryant | @user35202: As far as I know, there's not an explicit parametrization of the (1-dimensional) stable manifold in this case, though there are certainly numerical techniques that will describe it approximately. In cases such as this, there is typically a countable number of values of the parameter $A$ for which the stable manifold is only $C^k$ at the origin (for some finite $k$). Of course, it will be real-analytic everywhere else. That may possibly have some effect on the stability of numerical schemes for describing the stable manifold. | |
| Jan 25, 2019 at 16:54 | comment | added | user35202 | To find the stable manifold/stable paths I assume would be possible only numerically? | |
| Jan 25, 2019 at 1:36 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed an error in the sign of the root, which simplified the argument considerably.
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| Jan 25, 2019 at 1:20 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added a remark to deal with the case that $A$ is negative.
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| Jan 25, 2019 at 0:19 | vote | accept | user2175783 | ||
| Jan 24, 2019 at 22:59 | history | answered | Robert Bryant | CC BY-SA 4.0 |