@acer Thanks for your thorough explanation, as usual. That unexpected behavior is not a big deal, and if it is done that way for the sake of efficiency, I am all for it.

@rlopez Hello Robert, thanks for pointing out the TNBFrame().  I didn't know about it.  It's good to know.

Happy new year!

g@sheriph05 You ask: 'Can disease-free equilibrium be obtained in a similar way as the endemic equilibrium point?",

I don't quite understand what that is asking, but I have several comments which may be relevant.

• We have seen that there are two equilibria for your system of equations.  One of them has I=0, which is a good thing—if the solution go there, then the disease dies out.  In your previous messages you seemed to want to dismiss that solution.  I don't quite see why.
• Referring to the two solutions noted above, the one with nonzero I does not necessarily indicate an acceptable solution because if I is negative, then it is not meaningful in the epidemics context.

  I suppose that your interest in Descartes's Rule of Signs stems from a desire to determine whether that I is negative or positive.

• Having established that there is an equilibrium with a positive I value does not necessarily mean that the disease will take hold and stay forever. If the equilibrium is unstable, then its existence is pretty much irrelevant because solutions will not go toward it.

• Similarly, if the equilibrium with I=0 is unstable, then its existence is pretty much irrelevant.  On the other hand, it it is stable, that's great news because it says that the disease eventually dies out.

In conclusion, you have two tasks ahead of you. (a) determine which equilibria are of interest, that is, identify those whose S, M, V, I values are either zero or positive, and (b) analyze the stability of those equilibria.  Only then you may be able to make interesting statements about them.

A primary method for analyzing the stability of  equilibria the Routh-Hurwitz criterion which you may be familiar with already.  If not, then Wikipedia's article on Routh-Hurwitz may be a good starting point.

@Kitonum That last line should be
plot3d(F, x=0..5, y=0..5);  # Visualization

@wswain 

Bill,

Maple's user interface is quite customizable.  It offers a Document Mode and a Worksheet Mode.  It offers a 1-D Input Mode and 2-D Input Mode.  Their combinations already yield four possible configurations.  There are many more options.

Maple's default configuration is the Document Mode and 2-D Input Mode.  I, along with many long-time Maple users, favor the configuration consisting of the 1-D Input Mode and the Worksheet Mode.  You may want to give that a try.  See https://userpages.umbc.edu/~rostamia/math481/config/maple.html for instructions.

Please note that the new configuration will not convert any of your previous documents. It will apply only to new worksheets that you make after this configuration.

Having done that, and with the help of a good guidebook to show the way, the basic operations in Maple are not too difficult to learn.  You shouldn't be groping in the dark.  I suggest Ian Thompson's Understanding Maple.  I must admit that I haven't read the book myself.  My recommendation is based (A) on having viewed a few sample pages on amazon's website; and (B) customer reviews, especially the negative ones.

That last comment may seem puzzling, so let me explain. The negative comments say that Thompson focuses on the 1-D Input Mode and the Worksheet Mode.  I say that's exactly the way it should be.  I wouldn't recommend a book that does otherwise.

I suggest that you begin reading the book from the beginning, and wend your way forward.  Should you run into difficulties, ask here and I am sure there will be many who will offer help.

@sheriph05 The attached worksheet shows that the equilibrium I is expressible in terms of the roots of a cubic polynomial.  You may apply Decartea's Rule of Signs to determine whether there are positive roots.  That requires determining the signs of the cubic's coefficients.  I have isolated those coefficients within the worksheet but I have no idea about the meanings of the numerous parameters of your equations, and therefore cannot make any statement about the signs of the coefficients.

calculations.mw

@wswain Bill, I can see the point of doing things your way if your purpose is to practice with Maple's commands. If, however, your goal is merely to arrive at expressions for the moment and shear, there are better ways.

The attached worksheet does the calculations twice: first, the short way through Maple's dsolve(), and second, the long way, manually.  Of course the first method is the normal way.

manual-calculation.mw

@wswain 

Bill, you wrote: "just trying to understand the Maple capability, approach and language/function effort the computer can/will do to resolve this dependencies".

That, it seems to me, is giving Maple too much undue credit.  Euler figured out the differential equation of the deflection of a beam.  That was the hard part. The equation turned out to be quite elementary.  It doesn't take any special powers to solve it.  A freshman can do it with paper and pencil.  There is nothing magical in what Maple does in this case.

The magic lies in the beam's differential equation.  Understanding and formulating the boundary conditions requires some thought.  Solving it is quite trivial.

Regarding a torsional spring joint, it is a matter of replacing the boundary condition (D@@2)(u)(1)=0 in the worksheet with (D@@2)(u)(1)=-K*D(u)(1), where K the the desired stiffness of the spring.  Experiment with various values of K.  Try K=0, 1, 5, 10.

Bill, you have quite a few questions there.  I will try to answer as many as I can.

• As to: Where can I find this referenced?

  The Euler beam equation is a standard subject in elementary structural mechanics textbooks.  Since you refer to the EI stiffness, I suppose you must be familiar with some of that literature.  For a quick review you may begin with Euler–Bernoulli beam theory in Wikipedia.

• As to: Did you select the Dirac intuitively?

  The beam's static equilibrium is described by the fourth order ordinary differential equation [EI u''(x)]'' = f(x), where u(x) is the beam's deflection at point x due to the application of the load f(x).  The load is measured as force per unit length.

  A point load, that is, a load that is concentrated at some point x=a, is expressed as f(x) = Dirac(x-a).  Dirac's delta function was invented specifically for such a purpose; see Dirac delta function in Wikipedia.

  You will see in that article that Dirac's function is, in some sense, the derivative of the Heaviside function.  Therefore it is not surprising that integrating a differential equation that involves the Dirac function, we obtain a solution that involves the Heaviside function.

• As to: My main objective is to impart onto the hinge a K-theta compliance spring stiffness and solve for the moment and x,y component reactions.

  The bending moment at any point of the beam is given by EI u''(x).  Therefore if the endpoint at x=1 is supported by a springy hinge, then the boundary condition there would be EI u''(1) = K*u'(1).

  In particular, if there is no spring, then we have K = 0, and therefore the boundary condition is u''(1) = 0.  In Maple that's expressed as (D@@2)(u)(1) = 0, and that's what I took in my worksheet.

A further comment

You should be aware that the calculation in animate-beam.mw does not account for dynamic effects. The animation is produced by merely displaying a sequence of pictures of static equilibria.  That's correct as far it goes, however that's not the same as a load moving across the beam as a function of time.  The proper model for that is a partial differential equation which you will find in the section titled "Dynamic beam equation" in the Wikipedia article which I referenced earlier.  Maple can solve that equation as well, but perhaps that should wait until you clarify your understanding of the static case.

@sheriph05 Unlike the previous case, your equations are now nonlinear, and therefore you may expect more than one solution. Indeed, we have two. One solution has I=0. the other solution doesn't.  From what you have said, it's the second solutions that you are interested in.

Here is an overview of what to do.  We solve the equations as usual:
endemic := solve({ fS, fM, fV, fI },  { S, M, V, I } ):
Then the command
nops([endemic]);
returns 2, indicating that there are two solutions.  To extract the value of I from the first solution, we do
eval(I, endemic[1]);
and we see that I=0, so we ignore that solution.  To extract the value of I from the second solution, we do
eval(I, endemic[2]);
and we see that this I is nonzero, so this is the solution that we keep.

Aside:
Your differential equations are:
dS/dt = fS,  dM/dt = fM,  dV/dt = fV,  dI/dt = fI.
Adding these up we get:
d(S+M+V+I) = fS + fM + fV + fI
The right-hand side simplifies to
    (-mu+gamma__V)*S + (-kappa-mu-tau)*I
        -Lambda*rho__M+Lambda*rho__m-M*mu-V*mu+Lambda
This looks somewhat suspicious to me.  Usually in epidemics models the sum of the various subsets of the population equals the total population which is generally a constant.  In your case the sum is not a constant.  This may not be wrong but it's worth verifying that the model does what it is supposed to do.

@tsunamiBTP Transcendental equations don't have solutions in terms of elementary function in general.  Why are you expecting this from your equation?

You haven't said anything about the parameter N that appears in your equation.  If we assume that N is a positive integer, then it is possible to show—see the attached worksheet—that solving your equation reduces to solving an (N−1)-degree polynomial.   Therefore for N=1,2,3,4,5 you get explicit solutions; see the worksheet.  You probably know that there is no solution in terms of elementary functions for polynomials of degree greater than four. That's as far as you can take this.

allsolutions.mw

@Jjjones98 OK, here is how.

@AndreaAlp It seems that in your my_quadric, you have written z where you should have y.  If we make that change, then graph that we obtain agrees with the photo of the bone that you have shown.

But this raises the question of what you mean by "the center of the ellipse".  If you rotate the graph in the worksheet (given in the link below) you will see that I have calculated and marked the true center of the ellipse.  (That mark is not visible in the picture shown above).   I am afraid, however, that's not what you are really looking for.  I suspect that you are looking for some sort of a "center" on the red surface.  If so, the meaning of such a "center" is unclear.  You may want to rethink it.

mw3.mw

@AndreaAlp Youwrite: "Still, those surfaces should intersect and create an ellipse":

Which surfaces?  We seem to have agreed that g(x,z)=0 and g(x,z)=y are different.

@AndreaAlp You seem to be confusing f(x,y)=0 with f(x,y)=z.  They are certainly not the same things!  For instance, x^2 + y^2 - 1 = 0 is a cylinder, while x^2 + y^2 - 1 = z is a paraboloid.

Similarly, g(x,z)=0 and g(x,z)=y are quite different things.   For instance, x^2 + z^2 - 1 = 0 is a cylinder, while x^2 + z^2 - 1 = y is a paraboloid.