Your ode needs to be written with phi(x), not just phi. Note that diff(phi,x) is zero, and so your ODE was not actually an ODE. Then Maple needs to solve for that derivative, and there are two solutions, so it gives up, You can help it, as shown here (changes in red), though I might not have picked the right solution. The ode gets solved, but the plots are not what you expected - maybe the initial condition I chose is incorrect, or the parameter values need to be changed.

Download rk4.mw

Your minus in the expression for Ghat has some extra backticks. If these are removed, it works as expected.

asyapprox.mw

Note: There appears to be an error in the paper. In the expression for omega[i] in Eq 11, I find q[i] =0 (see eq 15 in the worksheet)

LongWave1solition.mw

Here's how to do the limiting parts shaded in green. I do not yet know how to arrive at the limiting form for fN.

Download LongWaveLimit.mw

Yes, they are related.

Download proc.mw

It's hard to guess what you mean from your description. So this may be completely wrong. I'm guessing you want to find the ode that is satisfied by psi(xi) = diff(phi(xi),xi). firint in the DEtools package does this, but your case is too difficult unless you know what the form of the integating factor is.

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map[3](op@FileTools:-ListDirectory,currentdir(),'all','select'=~["*.mw","*.eps"]);

You odes and initial conditions need to be together in a set (or list). You also have three functions A(t), Ab(t), R(t) and only two odes; I'm assuming Ab(t) was meant to be A(t). Then it works. (I made c3 smaller to see something on the plot.)

dsolve_problem.mw

Your error disappears if you run the worksheet after a restart. Your graphs are all the same because although you change S[t] when you define your substitutions p1, p2 etc., there is no S[t] in the expression you are substituting into (which I've renamed ff since you are using f(x) for other purposes in your worksheet). You should be careful about using both S[t] and S as variables; if you set a value for S then S[t] will not be defined -- to solve this use S__t instead of S[t]; they look almost the same on output (the first has the t in italics).

To change the view for the plot, use view=[0..1,0..1].

sstf.mw

After transforming to xi, evalc(Re()) and evalc(Im()) work as you would expect, and you get Eq. (1.5). (I find the pdes easier to manipulate if you leave off the =0, which is usually assumed anyway for many Maple commands, such as pdsolve.)

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You can subtract a common term from each side of an equation or inequality using elementwise subtraction, assuming you know what term you want to subtract. If you don't, it is simpler to subtract it all, which is essentially @nm's answer. (I would typically set up an equation or inequality in the ineq3 form originally, then you do not have this issue.) Here it turns out the simplification makes the inequality harder to solve. 

For the assumptions you gave in the worksheet, solve gives two cases.

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The other question about phase shifts has now been deleted. .pdf of it is here: phase_shift.pdf

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To get Maple to work this out, turns out to be surprisingly difficult, if you want to avoid inputting things like the cosine rule that you expect Maple to know. Here's a solution. Really, it should be easier.

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If I trust isolve (?), then there is only one solution, with the tangents 1, 2 and 3. In the figure below tan(A) = 3, tan(B) = 2, tan(C) = 1.

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The exponentials are gone in eq22, so it has worked. I assume that after the substitution, the expression with V simplifies so that there is no longer a V.

If fact you can verify this by doing simplify on eq2.

p.s. use indets(eq22,name) to only see the variable names and not more complicated expressions.

@salim-barzani This iterates over all possibilities of mu[1]=0,1;mu[2]=0,1; etc, and is the only way I can make sense of it.

Download formula.mw