This is the third time that you've posted this problem since March 7---only 1-1/2 months. Three of the top seven responders on this site have already given you input. Understand that no one here can help you further. It's not that this problem is uninteresting or that you've inadequately described it. I find it interesting. It's just too hard. Perhaps it's time to accept that this problem can only be solved numerically, not symbolically. There're a great many such problems. Do you have some reason to believe that this problem can be solved symbolically?

If you make progress on this problem, then post a followup to one of your previous threads. Don't make a new Question unless it's truly new. If you encounter any Maple-related issues in your investigations, then please do post them as new Questions. For example, your Question "D meaning in Maple" was a fine Question even though it involved this same problem.

Try posting in Stack Exchange's Math Overflow forum. That's their forum for research-level problems.

If I've misinterpretted your intent, and you actually do want to make a numerical comparison of these five functions, then ignore my comments above and please accept my apology: I was mistakenly inferring your intent based on your previous posts about this problem. There're numerous ways that we can use Maple to make a numerical comparison.

@c4meleon 

If you're truly sorry for what you did, then you will put back the original Question with the code that was in it, or some close approximation of that, and change the title to something appropriate. And if you're not sorry, then I'll never respond to your posts again.

And, no, I don't respect you, yet. But I could come quickly around to it if you come correct.

@testht06 

You wrote:

Of course I known R^4 = M and I checked by simply comnand Expand~(R^4) mod p;. But the problem is that if I excuted Your proc with M:= [ [x^2, x,1,1], [x^3,x^2+x,x^2+1,x^2+x],[x^3+x+1,x^3+x^2+x, x^3+x,x^2+x],[x,x,x^3+x^2+x+1, x^3+x+1], but not receive R= [[0,1,0,0],[0,0,1,0],[0,0,0,1],[1,1,x,x^2].

But using the values for R and M that you give in the above paragraph R^4 is NOT equal to M!! That is clearly shown by the easy computation Expand~(R^4) mod p---a computation that doesn't involve my program at all. You say "Of course" and "I checked ... Expand~(R^4) mod p." Well, then attach a worksheet showing that you checked. Because if you're getting that R^4 = M, then there's a serious bug in Maple---a bug in basic matrix multiplication that's nearly beyond belief. I even did the matrix multiplication by hand, and I did not get M.

And with the following matrix

M:= < x^3+x^2+x, x+1, x^3+x+1, x^5+x+1; 
 x, x, 1, x+1; 
 x^3+x^2+1, 1, x^3+x+1, x^5+x^3; 
 x^4+x^3+x, x^2+x, x^4+x^2+1, x^6+x^2+x+1 >:

also not receive R=[[0,1,0,x],[1,0,1,0],[x,0,1,3],[1,0,1,x^2]], while R^3 = M

In this case it is indeed true that R^3 = M. But nth roots are of course not generally unique. In the cases that we've dealt with so far, the fourth roots of the eigenvalues were unique. That means that R was uniquely determined. In the case at hand, each eigenvalue has three distinct cube roots. That means that M has 3^4 = 81 different cube roots. Now, if you want, I could search through all 81 cube roots for those that can be represented in the original field. But that can be a lengthy search---which quickly grows much longer as ROOT and the order of the matrix increases.

Moreover, the input of this proc is any matrix, need to find R (R is completely unknown)

Uh-uh. First we work out all the bugs, kinks, and issues in the worksheet version. Then I'll make it into an actual proc procedure that will work for an arbitrary finite-field matrix M and positive integer ROOT.

@rbd66 

After or under makes no difference. Please post the worksheet showing it not working. Here's my copy of the worksheet showing the command working.

Earthquake_2_eval.mw

@Markiyan Hirnyk Yes, I forgot about the Explore command, and I agree with you.

DON'T EVER DELETE OR EDIT OUT THE BODY OF OR CHANGE THE TITLE OF YOUR ORIGINAL QUESTION AFTER IT HAS RESPONSES!!!! YOU HAVE JUST MADE THIS WHOLE THREAD WORTHLESS FOR OTHER/FUTURE READERS!!! IT'S NOT ALL ABOUT YOU AND YOUR QUESTION! HAVE SOME RESPECT FOR THE OTHER READERS AND FOR THE WRITERS WHO TOOK THE TIME TO RESPOND TO YOU.

Hello, and welcome to MaplePrimes.

1. Your boundary conditions use a parameter q which your post doesn't give the numeric value of. I can't debug this without that q.

2. I note that your f and g are not coupled; they can be solved separately. So my first step to debugging this would be to try to solve them separately.

3. pdsolve(..., numeric) is often very sensitive to its optional arguments spacestep, timestep, etc. I've often gotten results that were complete garbage (as you're getting) on the first try that could be corrected by adjusting the options.

4. What is the significance of your boundary positions -100 and 100? Are they stand-ins for infinity? If yes, then try making those -10 and 10 or -5 and 5. Get it working for those values first, and then, if need be, gradually spread them out.

5. In the future, please post your code as plaintext and/or as an attached worksheet. (Plaintext is the format that you used for the parameters r0 and theta0.) This is so that no-one has to retype it. To attach a worksheet, use the green uparrow tool, which is the last item on the second row of the toolbar in the MaplePrimes editor.

@matmxhu 

Your "catch string" should not include the first part, "Error, (in fsolve/polyill) "; it should be simply "time expired". So try this:

g:= proc()  [solve(eq)] end proc:

try
     sols:= timelimit(300, g())
catch "time expired":
     sols:= SolveEquations(eq)
end try;

@testht06 

No, it doesn't check out. You can see for yourself simply by computing R^4 using the proposed fourth root as the R. 

R:= <0,1,0,0;
          0,0,1,0;
          0,0,0,1;
          1,1,x,x^2>:

Expand~(R^4) mod p;

@testht06 

You wrote:

Please help me convert the entries of R (and R^4) from GF(2^16) back to GF(2^8).

Okay, it's done, amazingly enough. And R has a very simple representation in the original field. This conversion back to the original field will not be possible in the general case. I had to construct an explicit field isomorphism element by element to do this. I could not come up with an algebraic trick for it. I hope that someone here can help me with that.

I think, GF(2^8) is the sub set of GF(2^16).

Yes, but that subset is not a subfield, in particular it is not the subfield of GF(2^16) which is isomorphic to our original GF(2^8). To see that such a situation is not even possible, consider this: y^7 is a member of that subset, but its square, y^14, is not. So the subset is not closed under multiplication and is thus not a subfield.

And while convert R and R^4 back to GF(2^8), the entries ≤ 255 will be not changed

No, they must be changed, for the reason in the preceding paragraph. The only elements that can stay the same are the members of the prime subfield---in this case just 0 and 1.

Furthermore, for the calculation of the roots of the factor (t² +231t +30), can You choose an irreducible polynomial ext2 over GF(2^8)? (ie ext2 = t² + at + b, where a, b in GF(2^8). GF((2⁸)² is the composition field)

Well t^2+231T+30 itself is irreducible, and that is what I chose, essentially. But the only way that Maple will work with a composite field is by "flattening" the representation from (2^8)^2 to 2^16 via Primfield.

Here is the new worksheet, which is mostly the same as the previous worksheet, but it has isomorphism at the bottom and the conversion of R back to the original field.

Download Matrix_powers_finite_field.mw

@matmxhu 

This is a new question completely unrelated to the theme of this thread: limiting the time of processes! Please post as a new Question, a new thread!

@tomleslie 

Yes, I find it very surprising that frem has no symbolic capability and that it cannot be evaluated by evalhf. My solution seems, so far, to correct all problems.

You wrote:

[A]ll "solutions" I have seen are based on the premise: please don't check the arguments to frem() until floats are actually presented.

My (second) solution does not rely on that.

[I]f frem(x,y) when presented with one or two symbolic arguments. simply returned frem(x,y), then the issue would be avoided.

It is trivial to write an frem variant that does that, but it wouldn't solve all problems. For example, you wouldn't be able to symbolically integrate or differentiate it. My solution relies on the fact that floor is fully symbolic.

@patient For what it's worth (which I don't know at this point), the FindSingularity procedure that I gave will work for backwards integration, say using 0 as the initial point and integrating towards -1. No modification is needed---simply call FindSingularity(sol, -1).

@Preben Alsholm 

Try a time comparison of your procedural dsolve with my symbolic version of frem:

Frem:= (x,y)-> x - y*floor(x/y+1/2);

Ironically, Frem can be evaluated by evalhf, but frem can't be. For that reason, I think that it'll be faster.

@Axel Vogt 

What you called "acer's trick" was my trick. It seems that the setting of Digits determines whether declaring a method is needed. At Digits = 10, no method is needed. My guess is that the native Maple methods can't guarantee the accuracy at the higher setting.

Regarding the symbolic integration, see my new Answer about a symbolic equivalent to frem. It integrates easily.