Timeline for Method of characteristics
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 14, 2015 at 13:01 | comment | added | GregVoit | @DeaneYang I did ask it there some time ago but received absolutely no feedback | |
| Jul 13, 2015 at 16:20 | comment | added | Deane Yang | This question really should be migrated to math.stackexchange.com | |
| Jul 13, 2015 at 13:49 | vote | accept | GregVoit | ||
| Jul 13, 2015 at 13:48 | comment | added | Joonas Ilmavirta | @GregVoit, you need to specify the value of $f$ at one point on each $\gamma$, but there are many curves. In your example you specify the values of a function of three variables ($f$ restricted to the surface). It might help to consider every $\gamma$ separately first and then remember that you need to have many of them to cover all of $\mathbb R^4$. | |
| Jul 13, 2015 at 13:37 | comment | added | GregVoit | thank you very much! Just last clarification: you keep saying valueS, and I just wrote down one expression for the initial value $f(x_{1},0,x_{3},x_{4})$. That would be just one function, where do the others come from? @Joonas Ilmavirta | |
| Jul 13, 2015 at 13:35 | comment | added | Joonas Ilmavirta | @GregVoit, if all components of $X$ are non-vanishing (everywhere), then you can pick any $i=1,2,3,4$ and $c\in\mathbb R$ and fix the values on the hyperplane where $x_i=c$. Your example with $i=2$ and $c=0$ works. Then you have to prescribe the values of $f$ for all choices of $x_1,x_3,x_4$ to cover all integral curves. | |
| Jul 13, 2015 at 13:30 | comment | added | GregVoit | oh, ok now I see that my equations were wrong. So about the initial data: my vector field $X$ has non-vanishing components, i.e. $X_{i}\neq 0$, $i=1,2,3,4$. Does it mean then that I can choose myself on which hyper surface to pick the values? For example, if I take $x_{2}=0$, then I have to prescribe value $f(x_{1},0,x_{3},x_{4})$ right? But that's just one value… @Joonas Ilmavirta | |
| Jul 13, 2015 at 13:27 | comment | added | Joonas Ilmavirta | @GregVoit, I added some details. Your curve had one dimension too many. If you write $F=f\circ\gamma$ and $U=u\circ\gamma$, the ODE I wrote becomes $F'=UF$, which you should be able to solve. | |
| Jul 13, 2015 at 13:25 | history | edited | Joonas Ilmavirta | CC BY-SA 3.0 |
added 712 characters in body
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| Jul 13, 2015 at 13:18 | comment | added | GregVoit | and second question: are my equations correct then? And how does the final solution $f=…$ then look like (in terms of my $\gamma$)? | |
| Jul 13, 2015 at 13:17 | comment | added | GregVoit | thank you for the response. Could you please specify/give an example of what you mean by "fix the values on a three dimensional surface that meets every integral curve exactly once"? I have seen such statement before in the textbooks, but i don't see what it means in my concrete example? Could you write down what you mean? @Joonas Ilmavirta | |
| Jul 13, 2015 at 13:11 | history | answered | Joonas Ilmavirta | CC BY-SA 3.0 |