I attempted to understand what it is that you aim to do but couldn't.  So far as I can see, you have a differential equation dg(x)/dt = f(x, g(x)), where f is given. You have boundary conditions g'(a)=0 and g'(b)=0, where a and b are unspecified.  What do you want to do then?

I have simplified your notation here, and I hope I have not changed the meaning.

You say "Im working on an assignment" and "Im supposed to implement", which is fine.  What do you have so far, and where do you need help?

Earl, I put together a tutorial on barycentric coordinates which may answer some of your questions.  Ask if other questions remain.  Here it is:  barycentric-coords-tutorial.mw

As to %T, that's Maple's notation for the transpose of a matrix or a vector.  Thus

u := < a, b >;

u^%T;

u^%T . u;

Carl: Ah yes, I was inattentive to the "otherwise" clause of the defintion.  Now the double-summation makes sense to me. 

I can't make sense of the math as it is presented.

To explain my confusion, let's forget about your problem for a moment.  Let's consider a real-valued function f defined on the interval [0,1] and another real-valued function g defined on the interval [5,6].  Question: What does the sum f(t) + g(t) mean?  Answer: The sum is meaningless—it makes no sense to add functions defined on different domains.

Now, going back to your problem, your functions b[n,m](t) are defined on generally non-overlapping subintervals, which leads to the question: What do you mean by that double-sum formula in your post?

To help others to help you, upload your worksheet.  Use the big green up-arrow that shows up in the window in which you type your response.

Since you have the result for one of the v values, what is preventing you from doing it for the other v values?

Consider uploading your worksheet so that people can see what you have done and what remains to be done.  To upload a worksheet, use the big green up-arrow that shows up in the window in which you type your response.

@I_Mariusz Your reaction is somewhat like saying a BMW coupe is not a powerful automobile because it can't do the job of a tank in a battlefiled.  Well, it can't, but that's not a good way to judge its usefulness.

@I_Mariusz Your equations are correct now, but unfortunately Maple is unable to solve them.

I haven't used Maple's PDE solvers at all, so I dug into the doucmentation to understand the cause.  So far as I was able to learn, Maple's numeric PDE solvers are not designed to solve elliptic PDEs.  Looking into MaplePrimes archives I found a few other requests for elliptic PDE solvers, and the respondants said that no such thing exists.

So I don't have a useful suggestion for you.  If I had to solve your problem, I would try one of the following:

1. Your domain is a rectangle, therefore implementing a finite difference scheme for a numerical solution should not be hard.
2. If I had more time and motivation, I would write a finite element scheme to solve the problem.
3. If I needed a solution immediately, I would use Comsol Multiphysics, but this would not be a useful option for you if you don't have access to Comsol.

@Ken_obi The title area is for the title of a message.  What you say should go in the body of the message, not in the title.

In particular, putting a link in the title doesn't do what you expect.

I don't understand the part that begins with "I differentiate the above equation ...".  You differentiate with respect to what?  And what does that have to do with a table?

@Markiyan Hirnyk Thanks for the link.  O'Rourke's theorem (page 126 in your referenced PDF) says that we need no more than floor( (n + 2h))/3 ) cameras, where n is the number of the vertices (counting vertices on the outer walls as well as on the holes) and h is the number of holes.

In the current problem we have 8 vertices on the outer walls and 4 vertices on the pollar, therefore n = 12.  We have one hole, therefore h = 1.  Then according to O'Rourke's theorem, the upper bound on the number of cameras is floor(14/3) = 4.  The solution that I posted earler, shows that 3 cameras will do, that is, the theorem's upper bound is an overestimate in this case.

@Markiyan Hirnyk Yes, I have.  I didn't see anything there pertaining to non-simple polygons.  Did you?

@vv The art gallery problem and the associated theorem by Chvatal/Fisk pertain to simple polygons, that is, polygons with no holes.  The "gallery" in the question asked here is not a simple polygon due to the pillar in the center.  There may exist extentions of the Chvatal/Fisk theorem to non-simple polygons but that is beyond my realm of knowledge.

@Markiyan Hirnyk That's quite curious and makes me wonder about the nature of this bug.  Your workaround is excellent when it works, but has its limitations.  The following still causes problems:

plots:-shadebetween(1/sqrt(1-x^2), 1/sqrt(1-x^3), x = 0 .. (1/2)*sqrt(3));

Two of the methods suggested on this page work and produce the correct result in this case:

1. The method suggested by acer
2. The method of "faking it" suggested by me, that is
   plots:-shadebetween(1/sqrt(1.0001-x^2), 1/sqrt(1.0001-x^3), x = 0 .. (1/2)*sqrt(3));