@sarra

1. Suppose n=5.  Where is e(3)*e(1) defined in your worksheet?

2. Consider two vectors u = <1,2,3,4> and v = <1,-1,1,-1>.  How would you calculate u*v?

@sarra I don't understand what you mean by " "*" denotes product of two vectors ".  Suppose n=5.  What do the products e(3)*e(1) or e(1)*e(3) mean?

1. You have:

       # The vectors e(i) satify the folowing conditions
       e(0)*e(1)=e(n-1) assuming  1<n;

   Does the "*" denote some sort of multiplication?  How does one multiply two vectors?
   
   
2.  You have:
   
       for i from 4 to n do
   What do you expect this to do?  You haven't said what n is.
   

@Doug Meade I think what you have shown may be made more transparent if you first extract the solutions x(t) and y(t) out of sol, as in

sol := dsolve({de1, de2, ic}, numeric, output=listprocedure);
X := eval(x(t), sol);
Y := eval(y(t), sol);

Then [X(1), Y(1)] would be the coordinates of a point on the solution curve at time t=1.  The attached worksheet has all the details.

mw.mw

@vv Your calculation shows great insight.  There is magic in the definition of H.  I was looking for something like that but was unable to derive it.  I have to admit that even after seeing it I still don't quite understand why it works, but that's OK since it gives me something to think about.

@Axel Vogt I don't see where you get the 0..10 range in z1.  That surely can't be correct.  Did you calculate that by hand or did you use a Maple library or something?

This is not a solution but only a sketch of a strategy which after a lot of effort will reduce the problem to a triviality.

The germ of the idea is observing that the integrand is constant on any hyperplane x₁ + x₂ + … + x₁₀ = r where r is a constant.  Therefore, changing the coordinates so that one of them points in the direction of that hyperplane's normal, the ten-fold integral reduces to an integral of a single variable in that direction.  The price we pay is having to deal with  geometric complications that arise from that change of coordinates.

I will demonstrate the method by solving the case of the triple integral.  The ten-fold integral is left as an exercise for the interested reader ;-)

Thus, let us look at the problem

where g is any function of one variable.  Change from the (x,y,z) variables to a new (u,v,w) variables defined through

x = u,
y = v - u,
z = w - v.

We see that x + y + z = w, therefore the integrand g(x+y+z) equals g(w), and the problem reducdes to that of the integration of a function of a single variable.

For the purpose of explanation, it helps to look at the more general problem

where f(x,y,z) is not necessarily of the form g(x+y+z).  To perform the change of variables, we examine the intersection of the cube [0,1]x[0,1]x[0,1] with the plane x+y+z=w and we see that there are three distinct types of intersections, depending on the value w.  When w is between 0 and 1, the intersection is an equilaterial triangle.  When w is between 1 and 2, the intersection is a hexagon.  When w is between 2 and 3, the intersection is again an equilateral triangle.  Thus, we slice the cube with the plane x+y+z=w as w varies from 0 to 3, integrate over each cross-section, and add up the results, which yields the vallue of the desired integral.

After a bit of geometric scrutinizing, we see that i₀ = i₁ + i₂ + i₃ + i₄, where

In partucular, if f(x,y,z) = g(x+y+z), each of these reduces to an integral of single variable g(w).  As an example:

which is approximately equal to 0.1589542911.

The attached worksheet has some extra details.

funky-integral.mw

@mskalsi dsubs is indeed the right tool for this.  I had forgotten about its existence.  Thanks for your note.

@mskalsi Ah, now I see what it is that you are asking.  I don't have a neat solution for you, but perhaps the attached worksheet will do.

zero-derivatives.mw

@Markiyan Hirnyk Yes, of course.  It is within the context of the solution of that system of PDEs that diff(u,x)=0.  Outside of that context u is just an undefined symbol.  That's why I prefaced my original comment by saying "I assume that you are loading the PDEtools package because you intend to solve a PDE".  I thought that my intention was clear.

@Markiyan Hirnyk A PDE is defined over a domain—that would be an open subset of the (x,y,t) space in this case.  One specifies initial and/or boundary conditions on the domain's boundary to arrive at a well-posed problem.

@Markiyan Hirnyk By "throughout" I meant "throughout the domain".

@Carl Love Your suggestion is a good one for a conventional eigenvalue problem, that is, finding ω so that det(A - ω I) = 0, where I is the identity matrix, or the more general det(A - ω C) = 0, where C is any matrix. The problem posted in the first worksheet in this thread, that is, eign.mw,  is neither. It is a nonlinear eigenvalue problem which seeks the zeros of det(A - C(ω)) = 0, where C(ω) is a nonlinear function of ω.

The problem in this thread's the second worksheet, that is, 1.mw, is of the form det(A - ω^2 I) = 0, which is amenable to your suggested method, as vv has observed.

@Carl Love Your suggestion is a good one for a conventional eigenvalue problem, that is, finding ω so that det(A - ω I) = 0, where I is the identity matrix, or the more general det(A - ω C) = 0, where C is any matrix. The problem posted in this thread is neither. It is a nonlinear eigenvalue problem which seeks the zeros of det(A - C(ω)) = 0, where C(ω) is a nonlinear function of ω.

This may be operating system dependent.  I tried your code on Maple 2015.2 on 64-bit Linux.  The memory usage stayed at 20.18 MB throughout the computation with n=4, 5, and 6.  I did not even "restart"  between the tests.