@sand15 

...for Your detailed and helpful advice and the literature reference.

However, I'm not currently concerned with solving a specific practical problem. Rather, for other reasons, I want to construct a "pathological" example. It should demonstrate what theory has long known: Solving the variational problem as a minimum problem is equivalent to solving the resulting Euler differential equation as a necessary condition. My goal was to show, in an intuitive way, that the solution of Euler's differential equation corresponds, at least approximately, to the calculation of the functional minimum and to represent both in a single plot. I am particularly interested in calculating the minima of functionals, especially convex functions (see Rockafellar, Hiriart-Urruty/Lemarechal).

Since I've only recently begun to explore the world of Maple, there are several inaccuracies and errors in the command sequence of my file. Hence my plea for help.

Later I would like to try out so-called quadratic functionals (Lit.: Michlin "Variation Problems of Mathematical Physics") in Maple.

Addition:

Thank you again for your advice. The frequency analysis is particularly helpful. I still need to thoroughly understand the commands for it. Fifty years ago, this practical application (I know the theory) would have been very welcome.

@nm 

... is y(x)=const.=pi. Is this particular solution allowed?

@nm 

...So, this is about the famous pendulum equation. Then choosing the initial value y(∞)=π is physically meaningless, because this deflection angle describes an indifferent equilibrium state, which, according to the derivation of the equation, does not capture this state. In contrast, the pendulum equation for deflections smaller than π is derived from equilibrium conditions or conservation laws. Overall, the textbook problem therefore appears to me to be in a confusing form. It would be better to ask: what limiting value allows us to estimate the solution – and that is π.

edited:

From my point of view, only the reversal of the problem statement according to the textbook makes sense: With an initial displacement of "close" pi, the mathematical pendulum will swing infinitely long.

Apart from the fact that calculating the solution of a second-degree ODE also requires two initial values, solutions exist, for theoretical reasons, only for finite abscissas of the initial values. This becomes "experimentally" obvious when the second-degree ODE is treated as a system of two first-degree equations with the same initial value.

p. s.

test.mw

@acer 

I wasn't familiar with "convert". Last question: How can the lhs be converted to the rhs in (4) of the attached file test2, instead of using the command "is" in the argument of "is"?

Download test2.mw

@dharr 

The goal of my question was not to solve the specific problem, it's easy to solve and an exact solution function is known. The background of my question is: How can an integral equation with an unknown function at the integration limit be solved in Maple if the integrand doesn't have a conventional/closed-form antiderivative? How can the coefficients of a linear combination of appropriately chosen functions be determined to obtain an approximate function?

Using the example chosen here, whose solution is known, I tried to test this using the `minimize` function. The result not only showed the coefficients but also a special value of t corresponding to the minimum. What options need to be added to `minimize` to obtain coefficients and then an approximate function over a t-interval?

The problem I remember with the unknown function at the integration limit dates back to the 1990s. The integrand had a terrifying structure; it was nonlinear and had no antiderivative. Back then, my only option was to use the Ritz method in Mathcad. Since then, hardware and software options have improved. Therefore, I'm attempting to solve this old problem again using these newer methods.

@Rouben Rostamian

...

is interesting:

It is identical to every special solution to the initial value of the form y(u)=u, where the additional conditions for u and y(u) must be met.

@Carl Love 

I would never have thought of that. I need to take a closer look at the 1D and 2D input methods and their differences. Where can I find more information about that?

I have got it. :-)

@Carl Love 

Your method is, of course, much easier to apply than my experiments with parameterized real and imaginary parts. It's interesting to observe, using examples, how the zeros of polynomials "arise" and paths in the complex plane of polynomials are mapped into this plane.

@Christian Wolinski 

As a Maple beginner, I'm not yet familiar with all the commands You use and the logical structure of nested code. Therefore, I'd appreciate your advice regarding the error message in the attached file. It doesn't seem to be the parentheses...

test.mw

@acer @nm

Learned something new again. Now, playing with polynomials in the complex plane is enjoyable. I am currently interested in how polynomials map given paths in the complex plane to this plane.

@nm 

...how can the polynomial be factored using radicals instead of decimals?

Download test.mw

@Kitonum 

...are easily understood with pen and paper. It's enough to apply the rules for powers and logarithms. The original equation only looks dangerous, but after simplification it is not. I managed to do that without Maple before. Maple is, of course, much faster. A plot for variants of C also looks quite nice.

Currently, the "surd" command is very interesting to me, as I want to explore the complex plane with Maple, so thank You for that.

@dharr 

Your very first sentence, with the remark about "default behavior," has already solved the problem. I will begin with the "residue counter" (residue theorem) and refresh my memory of the good-old book by Hurwitz and Courant.

@dharr 

...gives in Maple2024.2: [o,undefined] also - strange. 

Assumption:

ic:=y(infinity)=5*pi/4 is during the insertion test interpreted in Maple as the limit x-->infinity. This limit is not uniquely determinable, analogous to lim(sin(x), x-->infinity). Therefore, y(infinity) can be anything according to pi/4+k*pi. In any case, ic can be varied mod pi without changing the solution.

BTW:

I forgot to mention that while a general solution exists in the domain for x>0 according to the Picard-Lindelöf theorem, the proof presupposes the finiteness of the abscissa for the initial value. And y(infinity), however, lies, in a sense, on the boundary, which is not covered by the existence and uniqueness theorems. It is therefore not surprising that Maple leads classical solution methods to strange results.