Download mw.mw

I don't understand your comment "The exports in the Maple interface do not produce command line executable code."  

In your Maple worksheet do File -> Export As and then select Maple input (.mpl).

That writes a plain text file containing the worksheet's commands.  You can then feed that file into the command-line Maple.  I just verified that it works in Maple 2024.  I expect it to work in earlier versions as well.

Compiler:-Compile translates the Maple code into a C++ program, which performs integer arithmetic in hardware, and therefore it is limited to the largest integer that the hardware can handle.  That is typically 2^32-1 or 2^64-1, depending on the hardware and compiler.  Apparently your JonesM1(4) exceeds that limit and hence the overflow error.

Here is the relevant paragraph from the Compiler:-Compile help page:

The set of procedures that can be successfully translated by the numeric code compiler is very limited. In general, procedures are limited to those that manipulate data types that have fairly direct representations as hardware integers, floats, and arrays containing these types of values. In part, procedures are limited to those that can be successfully translated to C by the CodeGeneration package.
 

Rolling of a ball on an uneven surface would be quite a challenge to model, but it appears that you are really asking for the motion of point mass, not an actual ball.

Even the motion of a point mass has its challenges.  What happens, for instance, when the point mass leaves the surface?  What happens at the impact when it falls back on the surface?

In the attached worksheet I model the motion of a point mass as it slides frictionlessly on a prescribed surface.  The surface can be anything actually -- change it as you will -- but we assume that the motion is slow enough so that the mass does not lift off the surface, or otherwise it is constrained to remain on the surface, say due to a magnetic attachment or something like that.

>  restart; >  with(plots): Kinetic energy >  T := m/2*( diff(x(t),t)^2 +  diff(y(t),t)^2 +  diff(z(t),t)^2 ); Potential energy >  V := m*g*z(t); Landscape >  f := (x,y) -> 1/2*sin(x+y) + 1/4*cos(x-y); Plot the landscape >  p1 := plot3d(f(x,y), x=-1..6, y=-7..1, color="Orange", style=surface, lightmodel=light4): The Lagrangian >  L := T - V + lambda*(f(x(t),y(t))-z(t)); The equations of motion >  VariationalCalculus:-EulerLagrange(L, t, [x(t),y(t),z(t)]): eqns := remove(type, %, equation); Eliminating z(t) and lambda leaves us with two equations in the unknowns x(t) and y(t).. >  subs(z(t)=f(x(t),y(t)), eqns): isolate(%[3], lambda): sys :=  simplify(subs(%, %%[1..2])); A numerical experiment: >  params := m=1, g=1; >  ic := { x(0)=0, y(0)=0, D(x)(0)=1/2, D(y)(0)=1/4 }; Solve the initial value problem >  subs(params, sys) union ic: dsol := dsolve(%, numeric, output=operator); Extract the x(t) and y(t) from the solution: >  X := subs(dsol, x); Y := subs(dsol, y); Plot the particle's orbit and superimpose it on the surface >  spacecurve([ X(t), Y(t), 1e-2 + f(X(t), Y(t))],         t=0..10, color="Black", thickness=3): p2 := display(%, p1): Animate the motion >  animate(pointplot3d, [ [ X(t), Y(t), f(X(t), Y(t)) ], symbol=solidsphere, symbolsize=20, color="Red"], t=0..10, frames=100, background=p2, scaling=constrained);

Download hills.mw

Plot your determinant to get a graphical representation of the roots.  Then specify the appropriate interval to calculate the root within that interval.  Here are the first four roots.

>  plot(det, beta=0..14, view=-1e-3..1e-3); >  fsolve(det, beta=4..5); >  fsolve(det, beta=8..8.5); >  fsolve(det, beta=8.5..9.5); >  fsolve(det, beta=12..13);

[Zero is also a root, if that's of interest.]

Height = infinity

Wrapping paper area = infinity

Volume of the boxes = Zeta(3/2) = 2.612375349

Volume enclosed by the wrapping paper = 2.885469966

It's not clear what you have in mind by "represent states".  It seems to me that an animation would be the natural thing to do.  I have modified your worksheet a little in order to produce the following, where t groes from 0 to 10 and a goes from 0 to 5, as you requested.

enfonction_p_2_cas1-animate.mw

Download mw.mw

Comment added later: Some or all of the calculations shown here may be wrong.  See my comment to Kitonum's reply.

Let's say you solve a pde in the unknown u(x,t), as in

pdsol := pdsolve(whatebver, numeric);

To plot the graph of u(0.5,t), we do

pdsol:-plot(x=0.5, t=0..1);

Aside: It will make it easier on others who respond to your question if you include a worksheet along with your question.

I assume that you know the difference between a membrane and a plate.  [A membrane does not support bending moments while a plate does.]

There is not much that Maple can do to help you derive the equations of membranes and plates.  There are various mathematical models that vary depending on the assumptions that you put in.

The simplest plate model is that of Kirchhoff and Love.  See the Wikipedia page for the details of how that model is derived.  Here is the resulting equation in Maple notation, where h is the plate's thickness and w(x,y,t) the the displacement at the point (x,y) at time t, and q(x,y,t) is the applied load:

Download mw.mw

The eigenvalues ω of that PDE are the vibration frequencies, and the eigenfunctions a(x,y) are the corresponding mode shapes.

As in vv's diagram, replace each of the three radii by a sine curve.

Download zz.mw

Maple correctly distinguishes between the n=4 and n≠4 cases because when n=4, the numerator and demominator of the summand in the Fourier series expansion of the solution are zero.

When calculating the Fourier coefficients by hand, one plugs in n=4 before carrying out the required integration.  Maple attempts to "cheat" by calculating the coefficients for general n, and then taking the limit of the general term by letting n→4. That's where things go wrong because n→4 makes no sense when n is an integer.

To see what happens, consider 

(-1 + (-1)^n) / (n-4);
limit(%, n=4);

Maple returns iπ.  The reason for that may be seen more transparently by looking at the stripped-down case

(-1 + (-1)^x)/x;
limit(%, x=0);

where Maple again returns iπ.  That makes sense because (-1)^x as x goes to zero makes no sense without bringing in complex numbers.

Here is the contents of the attached worksheet.  Ignore the artifacts

introduced through this website's interface.  The worksheet itself

is free of those artifacts.

The hiker draws on the map an isosceles triangle FAS where FA=SA and the angle at the vertex A is 2*alpha.  Of the two possible such triangles, he picks the one where the vertex A lies closer to him  He draws a circle centered at A and radius AF, and then discards the part of the circle that lies on the other side of the segment FS.  At any point M on the remaining arc, the angle FMS equals alpha.

He repeats the construction on the line segment SK and angle beta.  The intersection P of the two circular arcs constructed this way mark his location since the angles FPS and SPK equal alpha and beta, respectively

Then he reads off the coordinates of the point P on the map in the same way that he read the coordinates of F, S, and K.

if (l1-l2) . (l1-l2) = 0 then
    print("Equal Vectors");
end if;