Timeline for Growth at infinity of a solution to a parabolic PDE
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2017 at 20:15 | comment | added | Mateusz Kwaśnicki | I did not follow your argument closely, but I fail to see any obvious reason why this could be wrong. | |
| Oct 15, 2017 at 11:12 | comment | added | Kore-N | I have tried to formalize the suggestion of following the transport equations. But I come to the surprising result that the growth of $b$ does not really matter any more. Do you think this is completely wrong? I couldn't come up with a counterexample, since the transport with $b =|x|^{ \epsilon}$ shows the same behaviour. | |
| Sep 21, 2017 at 16:22 | comment | added | Mateusz Kwaśnicki | If $|b(t,x)|\leqslant C(1+|x|)^{1-\epsilon}$, the polynomial bound for the solution is reasonable. If we ignore the Laplacian, we get a transport equation with characteristics $x(t)$ growing at most polynomially (for $\epsilon=0$ we would have exponential growth, while for $\epsilon<0$ – blowup in finite time), so the solution $u(t,x)$ would also have polynomial growth. Adding the Laplace operator should not change this scenario in an essential way. | |
| Sep 21, 2017 at 12:58 | comment | added | Kore-N | But the equation does have some algebraic structure (forgetting about regularity and stuff, so in the following the functions may well be considered smooth). Indeed $b(t,x) = h(t,x) + g(t,x) $ and $f(t,x) = 2h^2(t,x) + g^2(t,x),$ all with polynomial growth as small as desired. So in particular it is a positive forcing. And $u_0 = 0.$ | |
| Sep 21, 2017 at 12:55 | comment | added | Kore-N | I do have a particular case in mind. But it's much more horrible than this. I would like to take something similar to $b (t,x)= \partial_x Y(t,x)$ where $Y$ is the solution the the stochastic heat equation: $(\partial_t - \Delta)Y = \xi$ with $\xi$ white noise. So in particular $b$ will be only a distribution in some H\"older space. | |
| Sep 21, 2017 at 12:30 | comment | added | Mateusz Kwaśnicki | Do you have a particular case in mind? I believe that a polynomial bound for the solution is true under appropriate additional assumptions. My guess is that this is true if, for example, $x b(t, x) < 0$ when $|x|$ is large enough. (It is clearly the linear term that causes problems). | |
| Sep 21, 2017 at 12:25 | comment | added | Kore-N | Yes, you responded to a comment I cancelled, because I figured as well that $u_0 = 0$ is not sufficient (just take $v = u-x$ above and it will solve the equation with a polynomial forcing). But your comment is interesting nonetheless. | |
| Sep 21, 2017 at 12:14 | comment | added | Mateusz Kwaśnicki | (Apparently I responded to a comment that is no longer there...) | |
| Sep 21, 2017 at 12:13 | comment | added | Mateusz Kwaśnicki | I believe you can use the maximum principle to show that if $u_0$ and $f$ are bounded and if you impose an extra growth condition on the solution, such as $|u(t,x)| \le A \exp(B|x|^2)$, then the solution is indeed bounded. (The extra growth condition is needed to prevent exotic solutions of the heat equation that accumulate energy "at infinity"). However, this will not work without a bound on $f$: if $u_0(x)=0$, $b(t,x)=0$ and $f(t,x)=x$ for $t\in[0,1]$, then $u(1,x)=x$, and so after time $1$ you are back to the unbounded initial value scenario (I mean, $u(1+t,x)$ has unbounded initial data). | |
| Sep 21, 2017 at 12:01 | vote | accept | Kore-N | ||
| Sep 21, 2017 at 10:35 | comment | added | Kore-N | I guess that solves the question quite simply :) | |
| Sep 21, 2017 at 10:35 | vote | accept | Kore-N | ||
| Sep 21, 2017 at 12:01 | |||||
| Sep 21, 2017 at 10:13 | history | answered | Mateusz Kwaśnicki | CC BY-SA 3.0 |