Timeline for Construction of elliptic equation with Neumann boundary condition from a minimization problem
Current License: CC BY-SA 4.0
16 events
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| Jun 3, 2019 at 8:29 | comment | added | mnmn1993 | I start from a functional. The functional is called Ginzburg Landau functional. | |
| Jun 2, 2019 at 14:03 | history | edited | DCM | CC BY-SA 4.0 |
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| Jun 2, 2019 at 12:58 | comment | added | DCM | @mnmn1993: just out of curiosity... where does your $E$ come from? In particular, did you start with the functional or the PDE? | |
| Jun 2, 2019 at 12:53 | history | edited | DCM | CC BY-SA 4.0 |
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| May 30, 2019 at 18:51 | history | edited | DCM | CC BY-SA 4.0 |
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| May 30, 2019 at 18:45 | history | edited | DCM | CC BY-SA 4.0 |
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| May 30, 2019 at 18:37 | history | edited | DCM | CC BY-SA 4.0 |
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| May 29, 2019 at 21:39 | history | edited | DCM | CC BY-SA 4.0 |
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| May 29, 2019 at 5:55 | comment | added | mnmn1993 | Sorry ,finally I see what you concern. Suppose we have that $u \in H^2$, then we can integrate by part to obtain$$\int_B -\Delta u \, \xi + 4(u^2-1)u \xi dx + \int_{\partial B } \dfrac{\partial u}{\partial n}\xi- Q'(u)\xi d\mathcal{H}^2 = 0. $$ The PDE makes the first part vanish, so we have $$\int_{\partial B } \dfrac{\partial u}{\partial n}\xi- Q'(u)\xi d\mathcal{H}^2 = 0. $$ for any $\xi \in H^1(B_1)$ . So I think I also get the boundary condition in the pointwise sense. | |
| May 28, 2019 at 21:33 | history | edited | DCM | CC BY-SA 4.0 |
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| May 28, 2019 at 9:18 | comment | added | DCM | I nice analogy here is 'plotting the graph of a function'. $E'(u)$ is a linear function(al) on $H^1(B)$; we learn about its `shape' behaviour by plugging in values and seeing what we get. The behaviour of $E'(u)$ then tells us things about $u$. | |
| May 28, 2019 at 9:15 | comment | added | DCM | I think the slickest way to get the interior equation and the boundary equation alone is by choosing test functions $v$ which satisfy a suitable PDE of their own (a PDE designed to make one of the two integrals vanish). | |
| May 28, 2019 at 8:47 | comment | added | DCM | In particular, I'd be inclined to see what you get by using Green's identity (thought of in a distributional sense, at least initially) in the equation in your third display, and seeing what drops out of that for different choices of $v$. | |
| May 28, 2019 at 8:46 | comment | added | DCM | I think the general pattern you follow is sound. Assuming $E:H^1(B)\to \mathbb{R}$ is differentiable, a necessary condition for $u\in H^1(B)$ to be a minimiser is that $E'(u)v = 0$ for all $v\in H^1(B)$. Choosing $v\in H^1_0(B)$ gives you your "EL" equation, for example. The key thing is that $u$ has to satisfy the $E'(u)v = 0$ whatever $v\in H^1(B)$ you choose, which is the main mechanism through which you're able to deduce things about its behaviour. | |
| May 28, 2019 at 5:16 | comment | added | mnmn1993 | Yes, you are right. For a general $H^1(B)$ function, I think it may be have a well-defined derivative on the boundary and also the $\Delta u$ (ever in weak or distributional sense?). However my main goal is as the title, I do not know if I construct it in the correct way. I also look for another construction if mine is not good enough. | |
| May 28, 2019 at 0:19 | history | answered | DCM | CC BY-SA 4.0 |