@mmcdara You have correctly described the classical Poiseuille flow.  In that context, the radial velocity v is zero and the axial velocity u is parabolic.

The problem originally stated in this thread is not quite the Poiseuille flow—the pipe leaks along its lateral surface as indicated by the boundary condition v=p+a at r=1.  Consequently, the radial velocity is nonzero, and as far as I can see, there is no solution in terms of elementary functions.  A numerical solution should work.

@mmcdara I think the pbar in the original statement is meant to indicate the average pressure.  Thus, the average pressure at z=-1 is zero and at z=+1 is p_a.

I don't know whether these are sufficient boundary conditions to make a well-posed problem.

@Ronan the last command in my worksheet contains the construction
    seq(frame(t), t=0..100, 0.4)

This produces animation frames over the time t=0 to t=100, in steps of 0.4.  You may change those numbers as you wish.

As to the blog that you referred to, I find the interface non-intuitive and therefore am having difficulty navigating through it.  I searched there for your statement regarding the ratio of the angular momentum and kinetic energy but was unable to locate it.  Can you post a direct URL?  The statement should involve more than just the ratio, since that ratio is not a dimensionless quantity.

@vv You are right; it's not difficult to compute moments of inertia of the T-handle exactly through direct integration.  Following your idea, in the attached worksheet I calculate the moments of inertia while accounting for the overlap between the two cylinders.  As expected, the results are quite close to the approximate calculation where the overlap is ignored.

T-handle-flip-extra.mw

@Carl Love Your function Z is a prototypical infinitely differentiable function which is not analytic.  As you have noted, all derivatives of Z at x=0 are zero.  As a consequence, a naive attempt at expanding Z(x) in Maclaurin series leads to an infinite sum of zeros which certainly does not equal Z(x).  This is because Z is not analytic, and that ties in to your comment about complex numbers.

A function closely related to your Z is
    Y := x -> 2.252283621*piecewise(abs(x) < 1, exp(-1/(1-x^2)), 0);
which is also infinitely differentiable (but not analytic), and has compact support.  The purpose of the coefficient 2.252...  is to normalize the function so that the area under the graph is 1.

Define Y_ε(x) = Y(x/ε). The function Y_ε is called a mollifier. For any function F in L²(R), let's write F_ε ≡ Y_ε o F for the convolution of Y_ε and F.  It can be shown that:

• F_ε is infinitely differentiable;
• F_ε converges to F in L² as ε goes to zero.

Note that although a function F in L² is generally not differentiable, the mollified version, F_ε, always is, and therefore the derivative F ^'_ε is well-defined.  Additionally, if F is such that F ^'_ε converges to something in L², let's say G, then G is called the weak derivative of F.

Sobolev in Russia and Friedrichs in the US capitalized on this and developed what is now the standard theory of weak solutions and regularity of PDEs.

The Wikipedia page https://en.wikipedia.org/wiki/Mollifier has some details and historical remarks on mollifiers.

Saving Maple's graphics to PDF produces a useless result.  I have reported this to the Tech Support multiple times over the last five years or so, and each time I have received acknowledgment of the existence of the problem.  I am still waiting for a fix.

To save a graphics produced by Maple's plot(), I right-click on the plot, select Export and then one of the several choices of graphics formats.  The options PNG, GIF, JPEG, EPS work fine—they produce graphics files whose bounding boxes correspond to the extents of the image.  Saving to PDF misbehaves—it produces the equivalent of an 8.5'' x 11'' paper and inserts the graphics somewhere near the upper left corner.

I am absolutely at a loss to see the utility of that.  What in the world is the use of an  8.5'' x 11'' export? Shouldn't exporting to PDF produce a natural bounding box as exporting to other image formats do? 

@Kitonum No need for pointplot().  With the output=operator option to dsolve(), we may use spacecurve() as you did in your first example:

restart;
sol:=dsolve({
    diff(x(t),t)=-sin(t),
    diff(y(t),t)=cos(t),
    diff(z(t),t)=0.3, x(0)=1,y(0)=0,z(0)=0},
    numeric, output=operator):
plots:-spacecurve(eval([x(t),y(t),z(t)], sol),
    t=0..4*Pi, colorscheme = [blue, green]);

M(something) denotes a function M applied to "something".  To express M multiplied by something, leave a space after M.

See:

https://www.mapleprimes.com/questions/225627-Subsection-In-The-Insert-Menu

@tomleslie I agree with you that it would probably be faster to write Maple code from scratch.

But the obvious question is, why bother?  The equation may be solved immediately with Maple's dsolve():

restart;
de := diff(u(x),x,x) + u(x)*diff(u(x),x) - u(x) = exp(2*x);
bc := u(0)=exp(0), u(1)=exp(1);
dsol := dsolve({de,bc}, numeric);
plots:-odeplot(dsol, view=0..3);

@torabi I have shown that the trivial solution is the only possible one, and I did not use Maple.  It's just simple mathematics, independent of Maple.  If some papers propose a nontrivial solution, then either they are misinformed or they are attempting a different problem than the one that you have described.

@herclau In the Stackexchange link that you have cited we read: To make the distribution even, I rotated the lines for a specific angle ranging from 0 < h < 2 π.

In other words, the angle between the radii was taken to be constant.

But it could have equally well said: To make the distribution even, I picked equally spaced points around the ellipse.

Those two choices lead to different results. That's what everyone has been trying to tell you.

You will have to think about what the purpose of your calculation is, and then apply the proper procedure to obtain an answer that makes sense in its context.

The idea of "the arithmetic mean" in this setting is ill-defined.  Let t be the angle that the radius makes with respect to the x axis, and let s be the arc length measured along the ellipse.

The length L of the radius may be expressed as a function of t or s.  If L is viewed as a function of t, for example, then the arithmetic mean would be 1/(t2-t1)*int(L(t), t=t1..t2).  On the other hand, if L is viewed as a function or s, then the arithmetic mean would be 1/(s2-s1)*int(L(s), s=s1..s2), which is something quite different.

L also may be viewed as a function of x, or y, where (x,y) is the coordinates of the tip of the radius.  You can think of other variables that can serve to parametrize L.  There is a different "arithmetic mean" corresponding to each choice.

Although you don't state this explicitly in your worksheet, it is clear that the intention is to solve the partial differential equation of a vibrating Euler's beam through separation of variables.  It is also clear that ω² is the separation constant that arises in the process.

Things go wrong when you assign a value to ω², that is, where you set ω²=a² c. Where does that come from?

To fix, go over the process of separating the variables carefully.  Additionally, examine the physical assumptions that lead to the mathematical model.  Specifically, the motion of the beam certainly depends on how you get the motion started.  At no place in your worksheet you have said how the motion gets started.  Look up the term "initial condition" in your reference book.

@jefryyhalim I don't know how to compute the next eigenvalue but I doubt that it would be useful.  Generalizing from the 1-D model, the higher eigenvalues and their associated buckled shapes are likely to correspond to unstable equilibria and therefore you would not see them in experiments.

If your actual experiments show wrinkled shapes, then perhaps your mathematical model is at fault.  As I have said earlier, a plate is inherently a 2-D object and the proper model for its deformation is a partial differential equation (PDE).  I have no idea how you go from there to your system of ordinary differential equations (ODEs).  If I were to investigate the source of the mismatch between computations and experiment, I would start with a close examination of the derivation of the mathematical model.