@Jean-Michel I'm sorry, I am totally confused and don't understand at all what you are asking. Hope someone else can help you.

@salim-barzani I think it is just too complicated for pdetest to finish. It works for some choices of the constants and functions that reduce the integrands to zero, but note that xi isn't real. Too many square roots of square roots in this case.

f-2s.mw

@Jean-Michel I think you are referring to the sequence on that page:

intseq := [1, 2, 7, 23, 85, 314, 1207, 4682, 18493, 73688, 296671, 1202849, 4910689, 20158436, 83169871, 344628527, 1433631973, 5984532728, 25060514887, 105240685511, 443102517025, 1870054761632, 7909539602647, 33521289826778, 142330494633985, 605375433105734, 2578988979186127, 11003364185437517];

That one can be solved - did you have a problem with that one? But that doesn't mean the other two sequences you chose can be solved.

@salim-barzani Note the comment about the magic of dsubs. Toward the end you had not substituted for P(y,t), a[1](y,t) etc. Now it works.

F7.mw

@salim-barzani I put it (F4.mw) instead of my earlier answer.

I don't see any problem here. These seem to be A057167 and A005132 for which there are no generating functions or recurrence formulas given, so the fact that it is too hard for Maple doesn't seem surprising to me.

@salim-barzani I went back and checked where I saw that the third derivative had disappeared and fixed that up. Not sure why the original method did not work. Now I get the correct a[0] and Q is almost the same as in the paper. I worked it up with the phi(xi) solution you chose and everything works. However if I add the term with Int(K__1(y),y) to Q as in the paper it does not work.

Yes, Maple's solution to the pde system means f__1(t) and f__3(y) can be any functions of t or y respectively.

@acer So Maple 2026 misses some, but Maple 2025 finds some that are not actually solutions.

Download s_ex.mw

@acer @Alfred_F In Maple 2026 (but not in Maple 2024) convert(simplify(convert(z, cosh)), cosh); works.

I guess my point was that I would not expect eval to work with infinity, and I always use limit with infinity (for the reason that @sand15 mentioned: they are different things). Elsewhere in Maple infinity is handled slightly differently to mathematics, e.g., a*infinity is left as is until simplify(a*infinity) assuming a>0 leads to infinity. But I think this is fairly well thought out and generally doesn't lead to inconsistencies. Another principle that is often found in Maple is that if something is going to be fairly compute intensive it may not be done by default but you can force it with an option or another command. Here I do not think you want to overload eval with special cases since then writing efficient code for long calculations in a loop degrades performance.

@C_R @nm I guess with infinity it is safer to routinely use limit.

A workaround is to use expand

eval(expand((A*x-1)/x),x=infinity);

But it is hard to see any reason to get zero here. Many other places Maple gives sensible results with infinity. For example, infinity/infinity gives undefined, which would be a better result here than zero.

The fractional powers are very hard to deal with. I think you have to transform your ode to get rid of them

simplify(PDEtools:-dchange(V(xi) = U(xi)^3, Fode, [U(xi)])) assuming U(xi) >0

Of course if you really want V(xi) negative there is a mathematical problem about which cube root you want, but I'm guessing the equation was produced with the same disregard for rigor.

I'm struggling to understand the meaning here. It seems you want x=1 at t=0 since the units of the upper and lower limits should be the same and the integral should go to zero at t=0 (from the rhs). But in your approximation for x(t) you have ln(t) which implies you want t=1 to be the initial time?

In your mimimization you minimized t. Probably you meant to mimimize the integral of the squared difference over some range of t? 

Under my guessed interpretation, there is an easy solution:

intsolve.mw

These are not files listed on the File format help page, so you are out of luck for an easy way. I guess if you know the details of the binary file you could write a routine to read them, but I don't think that would count as "easy".