Skip to main content

programming - NDSolve and {C, K, Slot} and other built-ins as a variable name


The following problem is an exploration of what causes the error "Input is not an ordinary differential equation" in Mathematica as it seems to have changed from version 8 to version 9.


Specifically I have found that in Mathematica 9 I have come across the issue that


NDSolve[{x'[t] == 5, C'[t] == 5, x[0] == 1, C[0] == 1}, {x, C}, {t, 0, 100}]


gives the error, whereas in version 8 it does not. So something is up, but my question has more to do with if this is a bug in v8 or a bug in v9, that is, I get that C is a built in symbol: so is this error a correct catching of an input mistake? If so then


NDSolve[{C'[t] == 5, C[0] == 1}, C, {t, 0, 100}]

Should also give an error and it does not.


I really want to understand what is possibly going on, since as an ecologist I have to deal with equations that have C in them for "consumer" and I am forced to troubleshoot any issues that might come up with people not understanding the shadowing issue.


Clarification


It has been pointed out that it is bad coding style to use user defined symbols that start with a capital letter, as this will avoid clashes with Mathematica built-in definitions. That is fine, but does that matter?


What I am trying to ask more generally is whether the issue is that Mathematica is specifically checking for C and related symbols in NDSolve? Because I have no problems using symbols that have existing UpValues/DownValues as variable names (upper or lowercase) and even Protected in the exact same context.


What I want to try and understand is are there semantic reasons that this fails, ie could I cause the same error with user generated code? Or is this just a hard coded one off that Wolfram is doing for this case?


Further Clarification



If this is just an issue of using built-ins why can I use C in Solve and even DSolve! if it is an error in NDSolve? Again seems more and more like a specific check in NDSolve. For example


 DSolve[{x'[t] == 5, C'[t] == 5, x[0] == 1, C[0] == 1}, {x, C}, t]

Gives back an answer ... so suddenly this same input string is valid ODE again?


Now there are other built-ins that have this behavior such as K and Slot. As a challenge to all that would just tell me to code differently (which I do ... this is a question of why Mathematica works this way) can you use any of the reasons that you given to make a user defined symbol that will trigger the same error? If not then I will conclude that Mathematica is specifically checking for such symbols and move on. But I want to be totally clear that this is not just a question about coding style but if there are differences between system symbols and user defined symbols in this context. What if I make a package that has defined symbols in it, will I ever be able to cause the same problem if a naive user shadows some of my imported symbols into their global context?


One interesting suggestion is that somehow NDSolve is introducing C or K or Slot when it processes the input and that is causing errors. If so then I should never be able to create this problem with user defined symbols, but I do not yet know if this is the correct idea.



Answer



From @Szabolcs comment we see that we can recreate this issue using an arbitrary System symbol so it is not an issue of some kind of functionality of the built in symbol, though he mentions that he can find models that it will still work for so it seems to be a bug, not a check.


Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more