Download hanging-rope.mw

We know that the derivative of a constant is zero.  As far as Maple knows, u[i] does not depend on x and therefore diff(u[i],x)=0.  You need to tell Maple that u[i] is actually a function of x, like this:

n := 0;

for k from 0 to n do
#A[k] := sum((Diff(u[i](x), x))*(Diff(u[k-i](x), x)), i = 0 .. k);
A[k] := sum((diff(u[i](x), x))*(diff(u[k-i](x), x)), i = 0 .. k)
end do;

Here's an animation to go with vv's illustration.

Worksheet: classic-puzzle.mw

In this variant the pieces move simultaneously:

Worksheet: classic-puzzle-alt.mw

This will get you [−1,0]. There is no particularly useful way to get (−1,0) since Maple will strip away the parentheses.

a := solve({-1 < x, 0 < 2*x/(x^2 - 1), x < 1}, {x});
[(op~(a) minus {x})[]];

Earl, have a look at the (lengthy!) discussion in 
https://www.mapleprimes.com/questions/125273-Dsolve-Events-How-To-Control-For-A-Sign-Change
regarding events.

In particular this response from Allan Wittkopf
https://www.mapleprimes.com/questions/125273-Dsolve-Events-How-To-Control-For-A-Sign-Change#comment125426
shows how to use the evenfired option.  I think he is the author of Maple's events handler, but I can be wrong.

with(Student[Calculus1], SurfaceOfRevolution):
plots:-display([
  SurfaceOfRevolution(r, r=0..5, axis=vertical, output=plot, transparency=0.5),
  SurfaceOfRevolution(12-r^2, r=0..3, axis=vertical, output=plot)
], scaling=constrained, color=["Gold", "DarkRed"], caption="");

Download mw.mw

This custom-made graphics imposes a color grid over the domain in order to visualize which parts get mapped where.

Worksheet: mw.mw

The original domain is on the left.  The mapped domain is on the right.

You are attempting to solve a system of 5 linear differential equations symbolically.  The symbolic solution would be expressed in terms of the roots of the characteristic polynomial which is of the fifth degree.  But such a polynomial cannot be solved symbolically in general. That's a theorem!.

You should solve the system numerically.  I have edited your worksheet by commenting out some lines (marked with ###) and adding a few lines at the end which solve the system numerically and produce a couple of representative plots.

Download mw.mw

The short answer to your question is that the ODE won't be any different from the usual one which is m*x'' + k*x = 0 where m is the mass and k is Hooke's modulus.

The question to be answered is what is the meaning of Hooke's modulus for a spring whose properties vary along its length.

The answer requires a mini-lesson in elasticity.  Here I will sketch some of the essential points.

The usual characterization of a uniform spring is through the formula F=k*x where F is the applied force and x is the corresponding stretch.  That is not the best characterization of a uniform spring since k depends not only on the material of the spring (rubber, steel, etc.) but also on its length.  It's better to write that as F=c/L*x, where L is the spring's length.  This improved formulation has the advantage that the coefficient c is independent of the spring's length.

In a nonuniform spring the coefficient c is a locally defined quantity (since it is independent of length).  That is, there is a c value at any point along the spring.  Let's write that as c(s), where s indicates the location in the spring.  In places where the spring is stiff, c(s) is large.  Where the spring is soft, c(s) is small.  By the way, c(s) is called the spring's elasticity at the point s.

If you pull the end of such a nonuniform spring, you will not be able to detect the nonuniformities. You will feel the spring's averaged elasticity, that is, the spring will feel as if it obeys Hooke's law F=k*x for some k. The value of k may be calculated through the formula k = 1 / int(1/c(s), s=0..L) which is related to the harmonic mean of c(s).

Additional remarks:

1. I have assumed that we are dealing with small deformations.  This is the usual assumption that leads to Hooke's linear displacement–force formula.  Under large deformations the force versus displacement is nonlinear even for a uniform spring.

2. I have assumed that the spring's mass is negligible.  That's what one normally does to arrive at the harmonic oscillator equation.  If the spring's mass is significant (e.g. as in a Slinky), then the equation of motion will be a partial differential equation.

I don't have Ralph's book and without it I cannot tell what his code attempts to do. It would have helped if you have included a statement of the mathematical problem.  In the absence of that, I have tried to guess what the problem is, and have solved it in Maple. See if that is useful to you.

But if you want to learn the details of the code in your book, I suggest that you try to apply it first to the simple problem that I stated earlier.

Download mw.mw

As Polya once wrote, for any problem that you cannot solve, there is an easier problem that you cannot solve.  Try that one first!

So I suggest that you try applying your method to solving the much simpler problem

diff(u(x,t),t) = alpha*diff(u(x,t),x,x) + 1;

with zero initial condition and zero boundary conditions.

Can you do that?

Replace your 3rd for-loop with

for i from 1 to k do
  C[i] := B[i] -~ Am[i];
end do;

The -~ combnation says that we want to subtract the scalar Am[i] from each entry of the vector B[i].

Solving x.A=b is equivalent to solving A'.x' = b', where a prime indicates the transpose.  Here is an example.

Download mw.mw

Expr := (D@@2)(T)(0) + :-O(1) + 1/2*sin(1/2*Pi*T(x))^2 + 1/12*sin(1/2*Pi*T(x))*Pi*diff(T(x), x)*cos(1/2*Pi*T(x)) + :-O(2/3*(3/2*Pi^2*diff(T(x), x)*cos(1/2*Pi*T(x))^2*diff(T(x), x, x) - Pi^3*diff(T(x), x)^3*cos(1/2*Pi*T(x))*sin(1/2*Pi*T(x)) + sin(1/2*Pi*T(x))*Pi*diff(T(x), x, x, x)*cos(1/2*Pi*T(x)) - 3/2*sin(1/2*Pi*T(x))^2*Pi^2*diff(T(x), x, x)*diff(T(x), x))/Pi^2);
convert(Expr, polynom);