I see now you flipped the sign of c in your substitution, to make every thing consistently x+y-c*t.So the authors consistently used x+y+c*t. To get the plot orientation right you need x+y+c*t or use your convention and a negative c value in producing the plot.

Here it is altogether - you can change the line with xtc to change the convention.

derivation.mw

@salim-barzani Excellent. You used xi = x+y-c*t. I figured out that for the plots you need xi = x+y+c*t to get the correct orientation, but then pde2 is not satisfied. So the authors must have used xi = x+y-c*t in deriving the substitution in 3.4 but xi = x+y+c*t elsewhere. It is in any case just a matter of changing the sign of c.

@salim-barzani You have u and v in pde1 and pde2 in a way that I didn't understand. The paper says xi = x+y+c*t but you have x+y-c*t (which seems better to me). Either way pde1 is satisfied but pde2 is not. Perhaps something is not right about the derivation in Eqs 3.1-3.5?

P5-pde.mw

Here's further back to psi and fig 1. I had v=u^3/3 instead of v=u^2/3 earlier; now everything looks correct.

Download fp3.mw

@salim-barzani Here's how to transform back to Eq 3.5, which was the original starting point of this question

fp3.mw

I think the subcases in the paper are not all different when e1=0, so it is simpler to go through them without the classification scheme. Knowing when the quartic is a perfect square is useful to simplify the integral, but it is not clear to me that knowing when the various roots are real or not has much to do with evaluating the integral. Probably the solutions can be interconverted. In the end, Maple can just solve the ode directly, so I'm not sure I can see the point.

roots.mw

@Alfred_F That's a nice observation. e0 is lost, but there are now two integration constants. But the square root is gone, which I think is more important.

Here's a simpler way to find two of the solutions, and of course the general solution is easy to find.

roots2.mw

@dharr I found a slightly different tan solution which only works for a special relationship between e0 and e2, so probably that is what is meant. 

fp3.mw

@salim-barzani OK, lets take epsilon=1 then I agree with Eq 3.7. I needed W because you can't integrate wrt a function in Maple, i.e. int(..., w(xi__1)) doesn't work.

But as I showed there is no way to transform F to F2, so the integral that gives the arctan result is invalid, and so Eq 3.13 is incorrect, so it will not satisfy the ode. You can perfom the integral with e2 and e0 in (just change the Int to int). The problem is then that you get xi__1 = f(w) but here and in general f(w) has multiple w in, so you cannot easily invert it to get w as a function of xi__1.

The solution to this is to solve the ode in Eq 3.7. dsolve will do this directly, or you could take the square root and integrate. Since you get a general solution it will work for arbitrary e2 and e0. If you want to then find cases that are real for various choices of e2 and e0 you can make assumptions that make the things under the square roots positive.

@salim-barzani Almost there.

@salim-barzani Well if it were changed I might look at it. I have several times pointed out that it makes it harder to help you, and I have ignored several of your recent questions for this reason. In the present case using w instead of r will be especially difficult since the paper uses w later for a different purpose.

I notice that as usual you changed the notation in your worksheet to make it harder to follow the paper. Perhaps you wish to correct that.

@sand15 Sorry; I must have google searched it instead of clicking the link

@sand15 Your second link does not work (not a web address).