Timeline for Solving a differential system
Current License: CC BY-SA 3.0
26 events
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| Oct 11, 2016 at 10:56 | comment | added | Pietro Majer | I edited and modified, taking into accounts the fact that $\Phi(t)$ goes to $+\infty$ at $0$ and $1$, which semplifies the discussion. | |
| Oct 11, 2016 at 10:55 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 11, 2016 at 10:27 | comment | added | Pietro Majer | Let us continue this discussion in chat. | |
| Oct 11, 2016 at 10:26 | comment | added | CodeGolf | @ Pietro Majer $\Phi(0)=\int_b^0F^{-1}(s)ds<0$ as $F^{-1}$ is positive, where $F^{-1}$ denotes the inverse of $F$? | |
| Oct 11, 2016 at 10:26 | comment | added | Pietro Majer | But in fact i think $\Phi(0)=\Phi(1)=+\infty$, which simplify things. Just write $\Phi(0)$ $\Phi(1)$ in terms of the density $\mu(x)$ as a double integral, and use Fubini. | |
| Oct 11, 2016 at 10:03 | comment | added | Pietro Majer | Actually no, $\Phi(0)$ is positive (the integrand is negative but $\int_b^0=-\int_0^b$). $\Phi$ is convex (the derivative is $\Phi'(t) = F^{-1}(t)$, increasing). Note also that $|\Phi'(t)|\to\infty$ for $t\to0$ and $t\to1$, which implies $\Psi_\pm'(x)\to0$ resp. for $x\to\Phi(0)$ and for $x\to\Phi(1)$. | |
| Oct 11, 2016 at 9:55 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 11, 2016 at 9:52 | comment | added | CodeGolf | @ Pietro Majer Thanks again for the reply. In your answer, should it be $\|\Phi\|_{\infty}=\max\{-\Phi(0), \Phi(1)\}$ and $\Psi_-:=(-\Phi_{|(0,b]})^{-1}: [0,-\Phi(0))\to (0,b]$? As $\Phi(0)$ is negative. | |
| Oct 11, 2016 at 9:37 | comment | added | Pietro Majer | To be precise, it seems to me that the solutions $\phi_\pm$ have domains $[0,\tau_\pm)$, and $\phi_\pm(t)\to\pm\infty$ for $t\to\tau_\pm$, where respectively $\tau_-=2\int_0^{\Phi(0)} \frac{dy}{\Psi(y)}$ and $\tau_+=2\int_0^{\Phi(1)} \frac{dy}{\Psi(y)}$. This implies that at least one among $\tau_-$ and $\tau_+$ is infinite, and both are infinite if and only if $\Phi(0)=\Phi(1)$. | |
| Oct 10, 2016 at 16:49 | comment | added | Pietro Majer | Sorry, I mean $\Psi(t)=|\{\Phi\ge t\}|$ (fixed). The m.d.r. of $\Phi$ is a decreasing function on $(0,1)$ whose super-level sets have the same measure of the corresponding super-level sets of $\Phi$. Since the function $\Phi$ is convex on $(0,1)$ and the sub-level set $\{\Phi< t\}$ is the interval $(\Psi_-(t),\Psi_+(t) )$ its complement has length $\Psi(t)=|\{\Phi\ge t\}|$. | |
| Oct 10, 2016 at 16:37 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 10, 2016 at 16:33 | comment | added | CodeGolf | @ Pietro Majer Could you please specify a bit more about the definition of $\Psi$, i.e. ``$\Psi(t):=1-\Psi_+(t)+\Psi_-(t)=|\{\Psi\ge t\}|$ is the inverse function of the monotone decreasing rearrangement of $\Psi$''. Thanks a lot! | |
| Oct 10, 2016 at 16:08 | comment | added | CodeGolf | @ Pietro Majer Very nice solution. Thanks a lot! | |
| Oct 10, 2016 at 12:53 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 10, 2016 at 12:29 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 10, 2016 at 12:12 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 9, 2016 at 23:01 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 9, 2016 at 22:36 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 9, 2016 at 22:30 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 9, 2016 at 22:00 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 9, 2016 at 20:47 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 9, 2016 at 20:46 | comment | added | Nawaf Bou-Rabee | That's an interesting transformation. | |
| Oct 9, 2016 at 20:34 | comment | added | Pietro Majer | sorry, it was a typo, thank you, fixed. | |
| Oct 9, 2016 at 20:34 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Oct 9, 2016 at 20:31 | comment | added | Nawaf Bou-Rabee | Why is $\mu(-\infty,0] = \mu[0,+\infty)$? | |
| Oct 9, 2016 at 20:13 | history | answered | Pietro Majer | CC BY-SA 3.0 |