Timeline for $L^\infty-L^1$ smoothing effect for the heat equation
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2016 at 1:45 | comment | added | Nate Eldredge | If $\alpha > 1$ or $\alpha < 1$, we can send $b \to 0$ or $b \to +\infty$ to conclude $\|u(t)\|_{L^\infty} = 0$ which is absurd. | |
| Jan 10, 2016 at 1:45 | comment | added | Nate Eldredge | @Deer: Suppose the inequality holds as you stated it. Let $u_0$ be any nonzero function and $u(t)$ the corresponding solution, which does not vanish identically at any time $t$. For any constant $b$, since the heat equation is linear, the scaled function $bu$ is a solution with initial condition $bu_0$. Then the claimed inequality says $\|b u(t)\|_{L^\infty} \le C t^{-\gamma} \|b u_0\|_{L^1}^\alpha$. Rearranging, $\|u(t)\|_{L^\infty} \le C b^{\alpha - 1} t^{-\gamma} \|u_0\|^\alpha_{L^1}$. ... | |
| Jan 9, 2016 at 22:22 | comment | added | Deer | @NateEldredge I agree it should be one but could you tell me what you mean with your scaling argument? | |
| Jan 9, 2016 at 1:03 | comment | added | Nate Eldredge | Where did the $\alpha$ exponent come from? It seems to me that just by scaling, you can't expect anything except $\alpha = 1$. | |
| Jan 8, 2016 at 23:48 | comment | added | Piero D'Ancona | Take a look at Chapter 2 in the book by Davies, Heat kernels and spectral theory | |
| Jan 8, 2016 at 18:39 | review | First posts | |||
| Jan 8, 2016 at 18:52 | |||||
| Jan 8, 2016 at 18:39 | history | asked | Deer | CC BY-SA 3.0 |