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.

However, with that approach also

w();

and

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

would result in 99.

Could you give us text in MaplePrimes or upload a worksheet instead of a picture?

@Markiyan Hirnyk Nonlinearity has nothing to do with. The jacobian is the result of linearizing about the equilibrium point. There would be no point in linearizing if the system was linear.

I resent the tone of your comment, as I have done with some of your comments in the past to me and to other people.

@Markiyan Hirnyk Nonlinearity has nothing to do with. The jacobian is the result of linearizing about the equilibrium point. There would be no point in linearizing if the system was linear.

I resent the tone of your comment, as I have done with some of your comments in the past to me and to other people.

@Markiyan Hirnyk Stability questions are sometimes difficult to settle, yes. But very often it is a very simple matter as is the case with the two points found by fsolve (still assuming the order given in my answer):

subs(sol1,[v1,v2,u1]);
J(op(%));
LinearAlgebra:-Eigenvalues(%);
#So that point is asymptotically stable.
subs(sol2,[v1,v2,u1]);
J(op(%));
LinearAlgebra:-Eigenvalues(%);
#So that point is unstable.

@Markiyan Hirnyk Stability questions are sometimes difficult to settle, yes. But very often it is a very simple matter as is the case with the two points found by fsolve (still assuming the order given in my answer):

subs(sol1,[v1,v2,u1]);
J(op(%));
LinearAlgebra:-Eigenvalues(%);
#So that point is asymptotically stable.
subs(sol2,[v1,v2,u1]);
J(op(%));
LinearAlgebra:-Eigenvalues(%);
#So that point is unstable.

@STHence The worksheet is your worksheet with a few lines commented out and with my additions, which are in red Maple notation.

MaplePrimes13-03-19D.mw

The computation of the integral with the given relative tolerance (epsilon) takes 1-2 minutes (I didn't time it), so if you want a graph of G(t) (or sqrt(G(t))) you may want to use the plot options adaptive = false and numpoints = whatever you have the patience for. I didn't try that either.

Update: An idea for plotting G(t) or sqrt(G(t)) which avoids repetitive calculation of the same integral is to begin by deciding which t-values you want to use in plotting. Say 0, t1, t2, ..., tN.
Then find the integrals int( g ,t0..t1), int( g ,t1..t2),  ....
Below I have done that with an equidistant sequence of t's.

t0:=time():
N:=15: #Modest
for i from 1 to N do
   t1:=2*(i-1)/N;
   t2:=2*i/N;
   val[i]:=evalf(Int(g,t1..t2,epsilon=1e-3));
end do;
time()-t0; #10-11 min on my not very fast machine
plot([[0,0],seq([2*i/N,add(val[j],j=1..i)],i=1..N)]);
plot([[0,0],seq([2*i/N,sqrt(add(val[j],j=1..i))],i=1..N)]); #sqrt(G(t))

@STHence The worksheet is your worksheet with a few lines commented out and with my additions, which are in red Maple notation.

MaplePrimes13-03-19D.mw

The computation of the integral with the given relative tolerance (epsilon) takes 1-2 minutes (I didn't time it), so if you want a graph of G(t) (or sqrt(G(t))) you may want to use the plot options adaptive = false and numpoints = whatever you have the patience for. I didn't try that either.

Update: An idea for plotting G(t) or sqrt(G(t)) which avoids repetitive calculation of the same integral is to begin by deciding which t-values you want to use in plotting. Say 0, t1, t2, ..., tN.
Then find the integrals int( g ,t0..t1), int( g ,t1..t2),  ....
Below I have done that with an equidistant sequence of t's.

t0:=time():
N:=15: #Modest
for i from 1 to N do
   t1:=2*(i-1)/N;
   t2:=2*i/N;
   val[i]:=evalf(Int(g,t1..t2,epsilon=1e-3));
end do;
time()-t0; #10-11 min on my not very fast machine
plot([[0,0],seq([2*i/N,add(val[j],j=1..i)],i=1..N)]);
plot([[0,0],seq([2*i/N,sqrt(add(val[j],j=1..i))],i=1..N)]); #sqrt(G(t))

Thanks!
After I posted the answer I got to think (my usual order of doing things). Why 10 when 1*3 + 3*2 = 9?