@Athar Shahabinejad This discussion gives the impression that you are just shooting in the dark, hoping that you will hit a target by accident.  That's not a very productive way of doing science.

You say that ph(t)+pc(t)+pa(t)+pgd(t)+prd(t)+prj(t)+pf(t)=1  for all t. That is not consistent with your system of differential equations, as Preben has demonstrated.  This shows that there is an error in the logic that has lead you to those equations.   Instead of asking "Should I change parameters?", your question should be "What is wrong with my logic?".

See if you can review the derivation of the equations, and find out what it is that leads you to the inconsistency.  Only after you have fixed that issue it makes sense to worry about evaluating the integral that you had asked in the beginning.

There is something called the Routh-Hurwitz stability criterion which may be relevant to your quesiton.  After you have produced the correct set of equations, I will see if the Routh-Hurwitz criterion provides useful information about it.

Just yesterday you asked for the solution of a very closely related problem, to which Carl Love gave a complete solution.  Can't you modify his answer to solve your current problem?  The purpose of these homework assignments is to exercise your brain. What's the point in asking others to do the thinking for you?

@javaneh0 Your copy/paste method is not a good way of communicating.  As I suggested in an eralier reply, upload your worksheet. 

In the fragments that you have shown, you have:

pf(t) = ...
int(1-pf(t), t = 0 .. infinity);

That's definitely incorrect, so it's not surprising that you are not getting an answer.  If you upload your worksheet, people will be able to point out where things are wrong, and how to fix it.

@Markiyan Hirnyk A more challenging problem is to produce a picture of the cut-out sphere, like the gray 3D picuture shown, in Maple.  I tried, and then had to give up.

I conjecture that the error is due to missing multiplication signs in your worksheet, however It is impossible to be sure based what you have posted.

Consider uploading your worksheet.  Use the large green up-arrow in the reply screen for that.

@tellap I can't tell whether you are asking for help with the math or with Maple.  Do you know how to determine the equation of the shock in terms of a and b?  Once you have that equation, you can ask Maple to plot it for you.  Maple can't do that on its own.

Your reference/textbook should tell you how to determine the shock speed.  Look it up.

@tellap Forget about the cases (i) through (vi) for now.  Take a simple case.  Suppose that the inital data is a when x < 0, and b when x > 0.   There are two cases:

Case (1): a < b    [example: a = -1, b = 2]

Case (2): a > b    [example: a = 3, b = -2]

Can you do the plots for cases (1) and (2)?  Unless you can do these, there is no point in attempting the more complex cases (i) through (vi).

@tellap After you remove the Magenta and Cyan plots, don't forget to remove the comma after the Blue plot.

@tellap Yes, they are.

Matlab?  What is Matlab?

Do you have a question about Maple?

@laporte bernard Copy/paste to mapleprimes is not a good idea at all.  If you do that, the contents of your worksheet is shown as a set of little pictures.  If I want to test your code in Maple, then I will have to retype what you have typed by looking at those pictures, and try very hard not to make mistakes.  Most people will just ignore your post because that is too much trouble.

Instead of copy/paste, you should upload your worksheet with the big GREEN ARROW which I noted earlier.  See, for example, the link labeled de.mw in:

http://www.mapleprimes.com/questions/204117-Test-Solution-In-A-Complicated-Differential

This is a little simpler:

x := 3;

x$5;

      3, 3, 3, 3, 3

Hi laporte bernard,

I can copy the code that you have posted into Maple, execute it, and see the results of the computations, and the plots of the circles.

If your own worksheet contains more information than what you have already posted, then you may attach it to your message with the help of the big GREEN ARROW which appears when you edit your message.  But if it is the same as what you have already shown, then that wouldn't be necessary.

You should make an effort, however, to clearly state your question.  As it it stated now, I have no idea what you are asking.

Sure.  Jusy say

A;

F;

If you meant to ask something else, it is not clear from your wording.

I did not expect my remark regarding the root x ≈ 1/2015! to generate so many followups.  Had I known that, I would have provided more details to begin with which, in particular,  would have shown that

1/2015! - h*(1/2015!)^2  <  x  <  1/2015!

where

h = 1 + 1/2 + 1/3 + ... + 1/2015 ≈ 8.2,

and would have spared Markiyan some confusion.  The explanation is a bit long.  However, it may be of some interest due to the novel(?) idea of a using a differential equation to solve an algebraic equation.  I have used this method on a few occasions in the past, and I like to think that it's my own invention, although I dare not claim it, since it wouldn't be surprising that others have stumbled upon it as well. I will explain the idea in the context of Markiyan's problem, but the method is general and is not tied to that specific equation.

Let

p(x) = x(x+a1)(x+a2)...(x+an),


where the ai's are distinct positive constants. We want to solve the equation p(x) = t for some x > 0 and  t > 0.  Ultimately we are interested in t = 1, but we let t be an arbitrary parameter for now.

We regard p(x) = t  as an implicit definition of x as a function of t.  We write x(t) for that function.  It follows that p(x(t)) = t, and in particular p(x(1)) = 1.  Therefore, x(1) is the solution of p(x) = 1, and solving Markiyan's problem is equivalent to computing x(1).  For future reference let us note that x(0) = 0.

Upon differentiating p(x(t)) = t with respect to t we obtain p'(x(t)) * x'(t) = 1, therefore x(t) satisfies the differential equation

dx/dt = 1/p'(x),    x(0) = 0.       (*)

We solve the initial value problem (*) by any possible means, and then evaluate x(1) which, as explained above, would be the solution of p(x) = 1.

We choose to solve (*) by the method of power series here.  Experimenting in Maple with a few smallish values of n (recall that p(x) is defined in terms of n) we see that the solution of (*) is an alternating series which begins with

x(t) = (1/A)t - (h/A2) t2 + O(t3)

where

A = a1 a2 ... an,

and

h = 1 + 1/2 + 1/3 + ... + 1/n.

It then follows that x(1), which is the solution of p(x) = 1, is given by the alternating series

x(1) = 1/A - h/A2 + ...


We conclude that

1/A - h/A2  < x(1) < 1/A.


QED