@awass You misunderstand what Re is intended to do. It doesn't extract the real part of an expression, except in the trivial case that the expression is in the form a + b*I with a and b real (* footnote). Rather, Re is a placeholder that represents the real part of an expression, and that real part may be extracted by some later command, very likely evalc. So since Re((2-I*x)^4) returns unevaluated (even under assuming x::real), that expression is what is passed to expand. Then the procedure `expand/Re` tells expand that the argument to Re can be expanded. Once expanded, the expression is in the form a + b*I, the Re is evaluated again, and this time it works.

*footnote: If a procedure `Re/f` exists, then the call Re(f(x)) calls that procedure if f(x) returns unevaluated.

@magiblot Each of your problems is specifed over a region in the plane, i.e., a 2-dimensional space. Each constraint, when converted to an equality, is a line, a 1-dimensional space. Each pair of these equalities when solved together give a finite set of points (or no solution at all), which is a 0-dimensional space.

If the original problem were specifed in a 3D region, you'd need to check the interior of the region for critical points, then check each 2D surface that forms the boundaries, then each 1D curve that forms the boundaries of those, then finally the 0D corner points.

Do you understand?

@magiblot The commands in the Optimization package only find local optima for nonlinear problems. You want the global optima, and you should use the Lagrange strategy that I outlined above. I don't have time for a full Answer at the moment, but I'll probably write one this evening.

Since your objective function is quadratic, and your constraints linear, you can also try Optimization:-QPSolve.

@tomleslie Tom: For the symbolic solution of global optimization problems with inequality constraints (typical of Calculus III problems), you find the critical points in the interior of the region as the zeros of the gradient. Then you change the inequality constraints to equalities and use Lagrange on each boundary. Then you find the pairwise intersections of the boundaries, steps down one dimension, and use Lagrange again. When you're down to 0 dimensions, the corner points are your final critical/candidate points. Then you filter the candidate points, discarding any that don't satisfy the orginal constraints. Then you evaluate the objective function at the remaining points and pick the min and/or max.

Of course, all of the above assumes a differentiable objective and piecewise-smooth boundaries with a finite number of pieces that form a compact set.

@9009134 This is not a fractional DE; it's an ODE with a derivative raised to a fractional power. Because of the particularly simple form, that power can be moved to the other side of the equation. Because the dependent variable doesn't appear on the right side of the equation, it's simply an integration problem. Solving your problem is equivalent to integrating

(A*cosh(y) + B*sinh(y) + C*y + D)^E

for noninteger E.

@marians Sorry, I was using the wrong circle. I was using the circle of radius 1 centered at sqrt(11/2). It's supposed to be the circle of radius sqrt(11/2) centered at 1.

@9009134 I'm not sure, but I doubt it.

The exp at the end of your integrand is missing its argument.

@9009134 I searched extensively for a real value of A3 that worked, and I couldn't find one. The graphical evidence that I collected leads me to strongly believe that no real A3 exists. For real A3, the minimal value of u(1) is approximately 1.4 and is attained at A3 approximately 2.4.

@azwa saleh From the article's abstract, it seems interesting. However, I can't read any more than the abstract because the article itself is behind a pay wall.

Could you provide some reference materials, some web pages perhaps, that define what you're talking about? I don't understand "stability region" or "diagonally implicit".

@Al86 I don't have enough knowledge of the underlying science to answer your question directly without further reference material. If you get a Maple error, I can very likely figure out the cause. If you supply an algorithm and want to ascertain whether some Maple code correctly implements it, I can very likely do that. I presume that the algorithm is in the article that you linked in your OP, but I can't access that article because it is behind a pay wall.

@macmp91 You can't effectively make an assignment to both B and B[b] in the same code. One assignment will overwrite the other. If you want to have B subscripted by b for stylistic reasons (i.e., because you like the way that it appears in the prettyprinted code), then use B__b (double underscore).

@Preben Alsholm I'm a bit skeptical about the sum of the squares of the residuals (i.e., the vertical distance between point and curve) being the objective function. Kitonum's function is clearly a better fit of the amplitude of the data. DirectSeach:-DataFit comes with 9 stock objective functions, including sum of squares of residuals. I haven't tried any of these yet.

Just eyeballing your graph, it looks like A*x^2*sin(B*x) where A and B are positive constants to be determined from the data. A is approximately 0.015 and B is approximately 280.