Skip to main content

differential equations - Why should the spatial derivative order of the ODE *not* exceed two?


Following this question I came across this strange behaviour.


Let me define a 1 D interval implicitely



a = 1; b = 2;
Ω = ImplicitRegion[a <= r <= b, {r}]

and let me try to solve PDEs on this interval.


This works as expected (second order ODE)


w1 = NDSolveValue[{w''[r] == 1/2,
DirichletCondition[w[r] == 0, r == a],
DirichletCondition[w[r] == 0, r == b]}, w, {r, a, b}]

Plot[w1[x], x ∈ Ω]


Mathematica graphics


and so does this


w2 = NDSolveValue[{w''[r] == 1/2,
DirichletCondition[w[r] == 0, r == a],
DirichletCondition[w[r] == 0, r == b]}, w, r ∈ Ω]

This works again as expected (this time 4th order ODE)


w1 = NDSolveValue[{w''''[r] == 1/2,
DirichletCondition[w[r] == 0, r == a],

DirichletCondition[w'[r] == 0, r == a],
DirichletCondition[w[r] == 0, r == b],
DirichletCondition[w'[r] == 0, r == b]}, w, {r, a, b}]


Plot[w1[x], x ∈ Ω]

Mathematica graphics


Question




But why does this fail?



w2 = NDSolveValue[{w''''[r] == 1/2,
DirichletCondition[w[r] == 0, r == a],
DirichletCondition[w'[r] == 0, r == a],
DirichletCondition[w[r] == 0, r == b],
DirichletCondition[w'[r] == 0, r == b]}, w, r ∈ Ω]

with the error message The spatial derivative order of the PDE may not exceed two. >>


The only difference being using r ∈ Ω as a domain instead of {r,a,b}.




Answer



Right now (V11.3) NDSolve uses FEM for elliptic PDEs and that code does up to 2nd order spatial derivatives. The fact that this PDE can be viewed as a time dependent ODE is a coincidence in 1D as pointed out by @MichaelE2 and time dependent ODE are specified via explicit bounds. Another, harder, issue is that it is not trivial to write a general test to see if an ImplicitRegion is "Simple". Also, this test needs time to execute. This needs to be run for every ImplicitRegionalso ParametricRegions basically everything that is RegionQ and the number of cases where it could help is, I feel, very limited compared to the time it is going to take. Having to specify explicit boundaries is not that bad I think. What possibly could work is to add an indication to the message that this could be solved as a transient ODE or have an example in the >> link. I'll make that a suggestion.


Update:


You can find the message documentation for this now in the help system by directing it to FEMDocumentation/ref/message/InitializePDECoefficients/femcmsd


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...