@KatePirs
I don't get that error message, but after supplying values for a and m I get complaints about initial/boundary conditions.
You have boundary conditions, but no initial conditions!

The help page for pdsolve,numeric says:

"The pdsolve(PDEsys,conditions,numeric,vars,options) command returns a module that can be used to compute numerical solutions for time-based PDE systems over a fixed finite 1-space interval."


That must mean that initial conditions must be given also.

@KatePirs
I don't get that error message, but after supplying values for a and m I get complaints about initial/boundary conditions.
You have boundary conditions, but no initial conditions!

The help page for pdsolve,numeric says:

"The pdsolve(PDEsys,conditions,numeric,vars,options) command returns a module that can be used to compute numerical solutions for time-based PDE systems over a fixed finite 1-space interval."


That must mean that initial conditions must be given also.

@Axel Vogt Actually the worksheet was saved with the data. If you execute the assignment to stiff you will see that it is a symmetric tridiagonal matrix with diagonal elements = 2 and the other -1, as would have been produced by:
LinearAlgebra:-BandMatrix([-1,2,-1], 1, 21);

It would be more interesting and useful if you gave us the 21x21 matrix. Maybe upload a worksheet.

Works for me too in 16.02.

Works for me too in 16.02.

Why repost?

http://www.mapleprimes.com/questions/145196-How-To-Compare-Or-Perform-Phase-Portrait--Diffthetat

For help with the syntax you could look at my answer to your question about the damped driven pendulum.

http://www.mapleprimes.com/questions/145198-The-Damped-Driven-Pendulum

I don't get anything I can see.

Could you give us the equations in Maple syntax in text form we can copy and paste or as an uploaded worksheet?

I don't see your equations. Could you give us them in text form or as an uploaded worksheet?

"but throughout mathematics teachings I've always learned cos(45) as 1/sqrt(2) as I'm sure

the rest of you all have as well"

Not this one!

I don't know anything about CUDA, but from what I can read in the help pages this technology accelerates floating point matrix computations only.

One of the eigenvalues is zero, since you make sure that det(K.M) = det(K)*det(M) = 0. The corresponding eigenvectors are computed fast:

restart;
M := Matrix(3, symbol = m, shape = symmetric);
det := LinearAlgebra:-Determinant(M);
M[1, 2] := solve(det, m[1, 2])[1];
K := Matrix(3, symbol = k, shape = symmetric);

LinearAlgebra:-LinearSolve(K.M, <0,0,0> ,free=t);

@dhonkabulo I have posted my answer to this as an 'answer' and not a 'comment'.  The system happens to be so simple that no numerical computation is necessary. That makes the solution simpler. I have done that in my newly posted answer.

@dhonkabulo I have posted my answer to this as an 'answer' and not a 'comment'.  The system happens to be so simple that no numerical computation is necessary. That makes the solution simpler. I have done that in my newly posted answer.

However, with that approach also

w();

and

w(77,bb,pp); #or whatever

would result in 99.