Timeline for Why ODE's naive finite difference matrix works well for different boundary conditions
Current License: CC BY-SA 3.0
13 events
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| Mar 18, 2018 at 9:08 | history | edited | xzczd♦ | CC BY-SA 3.0 |
The formal solution is wrong, now (I think) it's corrected.
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| Jan 13, 2018 at 10:06 | comment | added | xiaohuamao | And I compared naive FDM and yours in another problem of coupled ODEs. Please see here if you're interested. | |
| Jan 13, 2018 at 9:04 | comment | added | xiaohuamao | Further, changing the right b.c. value does not matter, either. Such insignificance of left/right b.c. occurs once mesh is dense enough (actually 20 is the critical No. of points here). So 2nd part of my understanding is b.c. effects are rendered more and more unimportant in some, if not all, eigenvalue problem as we densify the mesh. | |
| Jan 13, 2018 at 9:03 | comment | added | xiaohuamao |
Thanks for the link. I found that in your code, replacing the left b.c. $y'(0)=0$ by $y(0)=\textbf{any value including 0}$ also works well if only we set eps nonzero but vanishingly small to avoid singularity, which reassures the 1st point I noted in the post. This crazily reminds me of the $\epsilon$-$\delta$ language of limit, which is actually the 1st part of my tentative understanding of why the wrong $y(0)=0$ in the naive FDM works.
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| Jan 12, 2018 at 12:42 | comment | added | xzczd♦ |
This is the first time I used pdetoae for eigenvalue problem (you see this type of problem is relatively infrequent in this site, and to be honest I'm not quite familiar with it), but I think the solving process won't be that different from solving ODE/PDE. (The only extra step seems to be transforming the equation(s) to $Lu=\lambda u$ form and take the left part. ) (Update: Oh I forgot I used to write this answer: mathematica.stackexchange.com/a/128275/1871 This method is more efficient than the one above, but less flexible and a bit harder to understand. )
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| Jan 12, 2018 at 10:58 | comment | added | xiaohuamao | Thanks a lot for the extra solution. Have you used this to solve some eigenvalue problem of several coupled ode or pde? (Just would like to learn some code example if you've ever shown somewhere.) Will it be very different for pde? $\qquad$ What I actually want to study is a few coupled 1D ode and a few coupled 2D pde with periodic b.c. in one direction. | |
| Jan 12, 2018 at 10:07 | history | edited | xzczd♦ | CC BY-SA 3.0 |
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| Jan 12, 2018 at 9:53 | comment | added | xzczd♦ |
Sadly I can't think out an explanation for the first question at the moment. As to the last question, the answer is no. Just repeat our analysis above: y'[x] + l/x y[x] + f[x] y[x] == \[Lambda] y[x] /. l -> 0/.x->0 gives f[0] y[0] + Derivative[1][y][0] == \[Lambda] y[0] so there's no implicit b.c. in this case. A simple counter example: test = y'[x] + l/x y[x] + f[x] y[x] == \[Lambda] y[x] /. l -> 0 /. f -> (0 &);D[DSolve[test, y[x], x], x] /. x -> 0
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| Jan 12, 2018 at 9:50 | history | edited | xzczd♦ | CC BY-SA 3.0 |
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| Jan 12, 2018 at 8:55 | history | edited | xzczd♦ | CC BY-SA 3.0 |
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| Jan 12, 2018 at 8:28 | comment | added | xiaohuamao | This sounds to make sense. My question is: for what reason implicit B.C. prevails over the wrong B.C. imposed by the matrix form? Why is this contradiction just insignificant? BTW, to analyse behavior at $x\rightarrow 0$, is it correct to say $y'=0$ for something like $y'+\frac{l}{x}y+f(x)y=\lambda y$ when $l=0$ and $f(0)$ is finite? | |
| Jan 12, 2018 at 7:13 | history | edited | xzczd♦ | CC BY-SA 3.0 |
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| Jan 12, 2018 at 7:04 | history | answered | xzczd♦ | CC BY-SA 3.0 |