Timeline for Sign-changing solutions for initial-boundary value problem for $\partial_t u + \partial^4_x u = 0$
Current License: CC BY-SA 4.0
11 events
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| May 5, 2018 at 9:20 | comment | added | Carlo Beenakker | 1) if you allow for distributions (delta functions) this is OK; 2) integrable solution on $\mathbb{R}$; 3) yes indeed, but it's a trivial integral, you can also plot it on Wolfram Alpha. | |
| May 5, 2018 at 8:22 | comment | added | user123672 | Thank you for your reply. I have some related questions: 1. For any PDE: why is it legitimate to apply Fourier transform? You don't know a priori the regularity of the solution. 2. What kind of boundary conditions do you need to impose in your example? 3 Are those plots done with Mathematica? What code did you use to produce them? Thanks again. | |
| May 5, 2018 at 0:37 | comment | added | Carlo Beenakker | the integrand is positive, so the integral must also be positive: $u(x,t)=t^{-1/2}\int u(x',0)\exp[(x-x')^2/4t]\,dx'>0$ when $u(x,0)>0$ for all $x$. | |
| May 4, 2018 at 22:53 | comment | added | Alan | How do you prove that the convolution of any initial positive distribution with a Gaussian remains positive? | |
| May 4, 2018 at 21:01 | comment | added | Carlo Beenakker | you mean a proof that the solution of the diffusion equation $\partial_t u-\partial_x^2 u=0$ stays positive for all times? this follows from the fact that the Fourier transform with respect to $k$ of $e^{-k^2 t}$ is again a Gaussian, and the convolution of any positive initial distribution with a Gaussian remains positive. Replacing the $k^2$ by $k^4$ spoils that, you get this hypergeometric function that changes sign. | |
| May 4, 2018 at 20:48 | comment | added | Alan | How do you then prove these two claims in your last comment? | |
| May 4, 2018 at 19:43 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| May 4, 2018 at 19:25 | comment | added | Carlo Beenakker | @Alan ---no, it's not that trivial; suppose you replace the $k^4$ in the exponent by $k^2$; the cosine is still there, but now it is indeed true that a positive initial distribution remains positive; that is just what you would expect for a diffusion equation, the density cannot become negative; so it is somewhat remarkable that by going from second derivative with respect to $x$ to fourth derivative the positivity is lost. | |
| May 4, 2018 at 18:38 | comment | added | Alan | It seems the cosine in the integrand is the one causing the change sign, irrelevant of the positive initial function. I don't why there would be a specific first reference that discusses this problem. | |
| May 4, 2018 at 13:43 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| May 4, 2018 at 13:24 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |