Timeline for first-order linear differential equation with boundary conditions
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 26, 2012 at 11:15 | vote | accept | Mathman | ||
| Apr 24, 2012 at 0:18 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
more informative title
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| Apr 23, 2012 at 15:56 | answer | added | Thierry Combot | timeline score: 1 | |
| Apr 9, 2012 at 8:35 | comment | added | Robert Israel | $f(x) = f(p^k x)$ means $f(p^{-k} t) = f(t)$ where $t = p^k x$, so you haven't changed anything. | |
| Apr 6, 2012 at 10:12 | comment | added | Mathman | umm then perhaps my condition is too restrictive , how about only invariance under $ f(x)= fp^{k}x) $ here 'p' means all the primes $ p =2 ,3,5,7,..... $ and k=1,2,3,4,5,.. $ is still $ f=0$ ?? thanks :) | |
| Apr 6, 2012 at 0:39 | comment | added | Robert Israel | Actually no: if $f$ is a distribution and $g$ is a test function, $\int f(rx) g(x)\ dx = r^{-1} \int f(t) g(t/r) dt$ is a continuous function of $r \ne 0$, so the sum can't converge in the sense of distributions unless all the terms are $0$. | |
| Apr 5, 2012 at 21:39 | comment | added | Mathman | aja.. however if we assume that $ f$ is a DISTRIBUTION instead of a funciton is there a possibility to get a different result to $ f=0 $ ?? thanks for your answers | |
| Apr 5, 2012 at 21:09 | comment | added | Robert Israel | If $f$ is continuous, such a sum won't converge unless $f=0$ everywhere. | |
| Apr 5, 2012 at 21:00 | comment | added | Mathman | wouldn't the function $ F(x)= \sum_{q}f(qx) $ with a sum taken over all the positive rational would satisfy the boundary conditions ?? with $ F(0)=0 $ and $ \int_{0}^{\infty} F(x)dx =0 $ | |
| Apr 5, 2012 at 17:37 | comment | added | Robert Israel | As for the differential equation, its general solution is $f(x) = c x^{iE_n - 1/2}$ where $c$ is an arbitrary constant. | |
| Apr 5, 2012 at 17:34 | comment | added | Robert Israel | Your "boundary condition" does not appear to be a boundary condition in the usual sense. Is this supposed to be for all $x$? If $f(x) = f(p^kx)$ for all $x$ and all primes $p$ and integers $k$, then $f(x) = f(rx)$ for all positive rationals $r$, and then if $f$ is continuous it is constant. | |
| Apr 5, 2012 at 16:15 | comment | added | Mathman | here D is the derivative operator with respect to 'x' and the boundary conditions apply to ALL the primes $ p=2,3,5,7,.... $ | |
| Apr 5, 2012 at 15:41 | history | asked | Mathman | CC BY-SA 3.0 |