@Markiyan Hirnyk I'm not try to prove anything about the expression F in the mathematical sense of proof. I'm trying to prove, if you could call it that, or verify, that plot was unable to evaluate F at any of the values that it chose, except for k = 0. It may well be that there are other values of k at which F can be evaluated.

Now I am convinced that this situation---where plot can only evaluate the expression at one value---should be considered a bug, because it should give the same warning message that one gets if it can't evaluate the expression at any values. Displaying empty axes with no warning is not helpful.

Compare plot(1+x*I, x= 0..1) with plot(1+x*I, x= 1..2).

PDEtools:-Solve just passes the problem on to dsolve if it detects that they're ODEs. If dsolve is used for this problem without the numeric option, it will spend a huge amount of time and memory trying to find a symbolic solution. I didn't let it finish, as I was getting close to my memory limit.

PDEtools:-Solve just passes the problem on to dsolve if it detects that they're ODEs. If dsolve is used for this problem without the numeric option, it will spend a huge amount of time and memory trying to find a symbolic solution. I didn't let it finish, as I was getting close to my memory limit.

@spradlig

Plain text is easiest to work with. Use one of the three clipboard icons on the toolbar in MaplePrimes to cut-and-paste directly into your post. Maple understands plaintext, but not Word AFAIK. Or upload a worksheet. Why take the extra step of going through Word?

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@spradlig

Plain text is easiest to work with. Use one of the three clipboard icons on the toolbar in MaplePrimes to cut-and-paste directly into your post. Maple understands plaintext, but not Word AFAIK. Or upload a worksheet. Why take the extra step of going through Word?

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That definition of T(x) should be enough. But your initial conditions are ill-posed:

R(x) = 0, X(x) = 10;

Shall I assume that that should be
R(0) = 0, X(0) = 10
?

That definition of T(x) should be enough. But your initial conditions are ill-posed:

R(x) = 0, X(x) = 10;

Shall I assume that that should be
R(0) = 0, X(0) = 10
?

Another great algorithm, Kitonum. As a point of reference, these structures are formally called (in English) "polyominoes" (rather than "polygons of the match"), and there is much literature on them.

Could you please edit the first paragraph of your post?  The right sides of the lines are cut off; they did not word wrap.

And how did you export an array of plots from Maple? Perhaps by using an external plotter? When I try (in Standard GUI), it only lets me select one cell.

@Markiyan Hirnyk Well, of course, that's just -ln(2). But that is the only point in the plot range that plot was able to evaluate. You can see that in the plot structure:

P:= plot(F, k= 0..2):  # F cut-and-paste from OP's warning msg.
op(P);

CURVES(Matrix(2, 2, {(1, 2) = -.693147180559945, (2, 1) = HFloat(HFloat(undefined)), (2, 2) = HFloat(HFloat(undefined))}, datatype = float[8], storage = rectangular, order = Fortran_order, shape = []), COLOUR(RGB, .47058824, 0., 0.54901961e-1, _ATTRIBUTE("source" = "mathdefault"))), AXESLABELS(k, ""), VIEW(0. .. 2., DEFAULT, _ATTRIBUTE("source" = "mathdefault"))

For every other value of k that plot tried, it got the error about abs not being differentiable, but it traps those errors in a try ... end try block. It won't draw a curve with only one point, so it just displays empty axes.

@Markiyan Hirnyk Well, of course, that's just -ln(2). But that is the only point in the plot range that plot was able to evaluate. You can see that in the plot structure:

P:= plot(F, k= 0..2):  # F cut-and-paste from OP's warning msg.
op(P);

CURVES(Matrix(2, 2, {(1, 2) = -.693147180559945, (2, 1) = HFloat(HFloat(undefined)), (2, 2) = HFloat(HFloat(undefined))}, datatype = float[8], storage = rectangular, order = Fortran_order, shape = []), COLOUR(RGB, .47058824, 0., 0.54901961e-1, _ATTRIBUTE("source" = "mathdefault"))), AXESLABELS(k, ""), VIEW(0. .. 2., DEFAULT, _ATTRIBUTE("source" = "mathdefault"))

For every other value of k that plot tried, it got the error about abs not being differentiable, but it traps those errors in a try ... end try block. It won't draw a curve with only one point, so it just displays empty axes.

@Markiyan Hirnyk I realize that my plot is not what the OP was trying to plot. I was trying to illustrate that the warning message was unrelated to the failure to get a plot. Rather, the issue is as Preben suggested: the inability to evaluate the derivative of abs that appears in the OP's function.

@Markiyan Hirnyk I realize that my plot is not what the OP was trying to plot. I was trying to illustrate that the warning message was unrelated to the failure to get a plot. Rather, the issue is as Preben suggested: the inability to evaluate the derivative of abs that appears in the OP's function.

@Markiyan Hirnyk If you were asking the question to be Socratic with the OP, then I am sorry for interfering here. But if you really were asking because you didn't understand the OP's problem, then C = Limit(z(t), t= infinity).

@Joe Riel I'm glad you switched to using has. Your prior procedure for backtracking for the previously used vertices was taking the vast majority of your cpu time. I will run a comparison.

Edit: I ran the comparison, and your most recent linked-list version does use less memory. The run times are about the same, more-or-less depending on when the garbage collection kicks in. I'll have to run with some bigger examples to even out the gc()s.

One potential drawback to the linked lists is the need to flatten them for most practical uses.

@Joe Riel I'm glad you switched to using has. Your prior procedure for backtracking for the previously used vertices was taking the vast majority of your cpu time. I will run a comparison.

Edit: I ran the comparison, and your most recent linked-list version does use less memory. The run times are about the same, more-or-less depending on when the garbage collection kicks in. I'll have to run with some bigger examples to even out the gc()s.

One potential drawback to the linked lists is the need to flatten them for most practical uses.