@tomleslie The equation shown is slight modification of a standard difference scheme for a parabolic PDE.  There is no "step j" and "step j+1".  The OP's equation expresses a transformation from the array a[] to the array A[], where a[] is the known solution at a certain time step, and A[] is the solution at the next time step.  One needs boundary conditions (not an initial condition) to complete the definition of that transformation  The initial condition only enters at a later stage when that transformation is applied recursively to generate the solution of the PDE.

I can't make sense of the phrase "compute A[j] for a[j]".  It will help if you reword.

Furthermore, what are the boundary conditions?  What is the range of the j index?

@Carl Love I haven't used Maple's PDE solver in any substantive way and I know very little about it.  I made a naive attempt, just for learning purposes, to solve a much simpler version of the OP's equations.  Specifically, let's look at Burgers equation

@Oloyemike I stopped responding because I just don't trust those equations, and therefore I don't want to waste my time.

If you clarify by explaining the physics behind the problem, the geomertry of the domain, and the meanings of the various variables, then I may consider taking another look.

@MDD This begins to smell like a homework assignment, so here is a hint.  Write a proc which will do the equivalent of ArrayTools:-IsZero(). Such a proc is quite short and straightforward.

The elements of a basis set should be linearly independent, by definition.  So if A, B, C, D form a basis, then you cannot express C as a linear combination of A and B.  Reformulate your question.

@Markiyan Hirnyk I have not thought at all about the existence of solutions of such an equation, or the convergence of Euler's scheme.

@Preben Alsholm I suspect that Maple's algorithm is not equipped to deal with this kind of equation, which is actually not a delay differential equation in the conventional sense, since the delay itself is unknown.  It appears that it applies an algorithm to it blindly and just by luck picks up some parts of the correct solution.  As you have noted, its "solution" extends to x = 1.0058, which is probably wrong, since all evidence suggests that the solution should break down before reaching x = 0.85.

As to the incorrect plot in the x<0 region, that's not something particular to that specific equation.   This happens even in the simplest case:

eq:=diff(y(x),x)= -y(x-1);
res:=dsolve({eq,y(0)=1},numeric);
plots:-odeplot(res,[x,y(x)],-1..10);

The graph in the x>0 region is correct, and actually corresponds to y(x)≡1 in x<0, as intended.  The graph in the x<0 region is incorrect, however, in the same way as the graph for the more complex equation that you have shown.  I have no idea how Maple comes up with that graph.

@NickB Alejandro's and my results are equivalent.  You will get his result by applying the asympt() function to mine.  Choose whichever form best suits your needs.

@Preben Alsholm The equation's variable delay is tau = x-y(x)+2, as you have noted. If we plot the delay, as in:

plots:-odeplot(res,[x,x-y(x)+2],0..1);

we see that it becomes negative after around x=0.85.  Is the solution meaningful after that point?

@Oloyemike You will have to examine your equations more closely.  As things stand right now, they don't make sense:

1. If, as you say, Q in the first pde is Q(r,tau), then U in the first PDE will depend on r, but you have said U is a function of x and tau only.
2. Similarly, if U in the BC is U(x, tau), then Q will depend on x.

@acer It goes to show that it's good to pay attention to existing code before embarking on its modifications.

Acer and Carl, thank you very much for your answers and discussion.  In the application which brought this up, the efficiency of pefromance is not an issue at all because the table is small and the number of lookups are few.  I will go with the assigned() test since it is simple to use and expresses the intent well.

A few minutes ago I posted a worksheet in response to a thread to which both of you had contributed:

http://www.mapleprimes.com/questions/204335-Can-Rotate-3d-Text-Like-This-Be-Done-In-Maple

Within the worksheet I use the assigned() test to tell whether a given entry is in the table.

@acer That's a very clever idea. I knew nothing about the this STL business at all.

I have used your idea to produce a couple of procs that automate some of the constructions.  In particular, the procs account for varying widths of the characters.  For instance, the charcater "M" is subtantially wider than the character "I", so we don't want to place them in uniform intervals when processing a string such as "MIAMI".

The procs extract information about the character widths, then arrange them so that the gap between the characters is prescribed.  For instance, with the gap set to zero we get:

while with the gap set to 15 we get

We may wrap the text around a cylinder:

or produce an animation:

Here is the worksheet that produced these:  3d-letters.mw

@Markiyan Hirnyk That's the hard way of proving the symmetry.  The obvious way is to observe that y = (1-x^5)^(1/5) is equivalent to x^5 + y^5 = 1, and therefore the graph is invariant under the interchange of x and y.

It is based on that symmetry that I grouped the diagram's six subregions into three pairs of areas of a, b, c.

It is also clear that the power 5 is nothing special.  The argument works for x^p + y^p = 1 for any positive p.