Timeline for Solution of ODE with discontinuity
Current License: CC BY-SA 4.0
14 events
when toggle format | what | by | license | comment | |
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Apr 27, 2019 at 16:47 | answer | added | Bazin | timeline score: 1 | |
Apr 26, 2019 at 15:10 | review | Close votes | |||
May 2, 2019 at 3:05 | |||||
Apr 25, 2019 at 12:24 | answer | added | Carlo Beenakker | timeline score: 1 | |
Apr 25, 2019 at 12:12 | comment | added | Riku | @CarloBeenakker Yes. And the real problem is computing $\Phi_2$. | |
Apr 25, 2019 at 12:11 | comment | added | Carlo Beenakker | your notation is very confusing, but let me make an effort to parse it: if I only look at the first component of each of your equations, then I see that $G_1=0$, so $\partial_t\Phi_1=0$, and since $\Phi_1=x_1$ at $t=0$, I conclude that $\Phi_1=x_1$ at all $t$. Is that what you want? | |
Apr 25, 2019 at 11:55 | history | edited | Riku | CC BY-SA 4.0 |
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Apr 25, 2019 at 11:55 | comment | added | Riku | @CarloBeenakker That is a typo. Thank you. I'll edit it. | |
Apr 25, 2019 at 11:54 | comment | added | Carlo Beenakker | this still makes no sense; if $\Phi=(\Phi_1,\Phi_2)\in\mathbb{R}^2$, then $\Phi_2\in\mathbb{R}$; but in the definition of $G(x,y)$ you use $F(y)$ and if as you say $y$ should be replaced by $\Phi_2$, then you wish to evaluate the function $F$ at a point in $\mathbb{R}$, however, you have defined $F:\mathbb{R}^2\rightarrow\mathbb{R}$. | |
Apr 25, 2019 at 11:48 | history | edited | Riku | CC BY-SA 4.0 |
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Apr 25, 2019 at 11:48 | comment | added | Riku | @user64494 $\Phi:\mathbb R^2 \to \mathbb R^2$. We use $\Phi_1$ as $x$ and $\Phi_2$ as $y$. | |
Apr 25, 2019 at 8:05 | review | Close votes | |||
Apr 25, 2019 at 18:51 | |||||
Apr 25, 2019 at 4:20 | comment | added | user64494 | Also $G$ has two arguments, but you wrote $G(\Phi)$. | |
Apr 25, 2019 at 4:06 | comment | added | user64494 | BTW, this is a PDE, not an ODE. Please fix it in the question.. | |
Apr 25, 2019 at 0:59 | history | asked | Riku | CC BY-SA 4.0 |