@nm chini_dgl_a.mw

Happy Easter!

@Alfred_F 

This "unsolvable" equation has a numerical solution for the initial value (1;-1) in the interval [1; 2.700592...]. After a long calculation, I was unable to find a symbolic solution.

@Ronan 

How can point coordinates of the curve be directly retrieved in the plot of the function curve without tabulating (getdata)?

@Ronan 

I didn't know the command "rhs" yet.

@nm 

...exists, but not for every arbitrary initial value. A special solution exists for (sqrt(3);1). There is also another initial value (an incredibly long term).

Quote:

"...For this Abel ode   y'=x+y^3, which is known not to be solvable, Maple hangs on ..."

So the existence and uniqueness theorems of Peano, Picard, and Lindelöf do not apply to this equation if the right-hand side is continuous? But, contrary to the statement in the quote, these very theorems ensure the existence of at least a local solution.

@nm 

Reply to quotes:

"But without using the limit, how else will one find the constants of integration? Are you saying the series must be assumed to have  uniform convergence to use the limit as above?"

Yes.

"I assume Maple also used the limit internally to solve for C2 in this example and give the answer it did. How else could it have found the solution it did otherwise?"

That's probably the case. In such cases, a hint from the software would be helpful.

@nm 

...it should also be noted that uniform convergence of the three summand series must be assumed in the open interval x > 0. Only then can y(t) be assumed to converge uniformly in this interval. And only then is it permissible to calculate y(0) as the sum of the limits of the three summand series.

@nm 

...but that doesn't change the fact that when applying the explicit equation form y = f(t,y), as used in proofs of the existence and uniqueness of solutions, the right-hand side is not continuous/Lipschtz continuous at x = 0. Then Maple's error message should look different, or perhaps a note should appear indicating that a solution was only possible after continuing the solution to the edge of the domain of f(t,y).

Regarding your maple_sol solution, I'd like to know what Maple does with the natural logarithm it contains for x = 0.
The general solution may be correct; I haven't checked it. But classically, no special solution is possible for the initial value at x = 0.

@dharr 

You're right. 2^m is correct instead of 2^(m+1). Thanks for this hint and the tips with the table.

@dharr 

How can coordinates be retrieved from the plot of my "test" file, and how is a table of values ​​created, e.g., from {(m; y) } with y from term (1) from my "test" file for, e.g., m=2...17?

...to a problem that shouldn't be forgotten. There are solutions in the literature that are about 100 years old, and they're quite challenging in theory. However, with today's computational capabilities, it's now possible to come up with new proof ideas and practice mathematical reasoning. Perhaps I'll report on that later ;-) .
BTW: I chose this problem because I want to learn how to use Maple commands from the example solutions so that I can work on specific problems in a targeted manner later. I would be very happy to receive more examples with strongly asymmetrical contours, which do not have to be convex.

...it helped a lot.

@mmcdara 

...I expressed myself in a confusing way, for which I apologize. I only posed the original problem to learn how to handle constraints (here, m not equal to n) in Maple. In this problem, the inner sum over n and the outer sum over m must be calculated uniquely.

If the series elements are written as a matrix in the usual way, then, depending on the transposition, they must be added first column by column and then row by row, omitting the "forbidden" elements on the main diagonal. Rearranging the summation order and commuting the elements leads to different limit values, depending on the desired result. This is based, for example, on the so-called "Great Rearrangement Theorem" (Cauchy). It is one of the most powerful theorems in series theory. Accordingly, commutating is only permitted if absolute convergence exists. Your cited example of problem 21 is proof that reversing the order produces different results.

Therefore, the only possible limit for the original problem is pi^2/8 in terms of absolute value, which, depending on the transpose of the matrix, is either pi^2/8 or -pi^2/8 (the latter is my solution).
I hope our misunderstanding has been cleared up.

@acer 

The "remainder term" (3/4)/m^2 can be calculated using the inner sum (over n), which is initially finite up to N. It is calculated using the transformation sum(1/(m^2-n^2)) = -1/(2*m)*sum(1/(n-m)-1/(n+m)) = 1/(2*m)*(sum (k=1 to N+m)1/k - sum(k=1 to N-m)1/k) - 3/(4*m^2). The two sums in the parentheses are harmonic sums (sum of consecutive reciprocals). These are replaced in the usual way by the natural logarithm, the Euler constant, and O(1/n). After the limit N-->00, the difference in the sums in the parentheses becomes zero. The remainder term remains. A new limit is created.
Unfortunately, I can't yet represent this in Maple. I only have an old derive file. If you wish, I can put them into a text.