Alfred_F

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These are replies submitted by Alfred_F

@Rouben Rostamian  

That is great :-)

@vv 

One solution would be to calculate it in detail ;-) . Question: How long is a coastline that is being measured in ever greater detail? What happens with polygon in lim ---> oo ?

@Carl Love 

No rings - or topologically equivalent figures of higher genus, otherwise everything is allowed.

@janhardo 

Maple and Mathematica cannot find a solution in your source either. Without special transcendental functions, it will generally not work. For this:
https://www.12000.org/my_notes/solving_ODE/current_version/chapters/Differential_Gleichungen_E_Kamke_3rd_ed_Chelsea_Pub_NY_1948/indexsubsection1585.htm
7.11, 1601, #10565

@janhardo 

The "wheel" for this topic was already described in 1902 and 1926 (Painleve, Ince). A reference to a general overview of the treatment of ordinary differential equations is of little help. The fact is that the ODE in question can only be solved in closed form in special cases. More precise information on the coefficients is required.

If I understand your question correctly, you are investigating a nonlinear ordinary differential equation of the structure
a*y´´+b*y^3+c*y=0.
Equations of this type are discussed in E. Kamke's book "Differential equations, solution methods and solutions", volume 1, from page 544. The references to further literature are interesting. According to this, so-called closed solutions only exist in special cases. For practical cases, the "numerical club" is usually required. And this is easily possible if a system of two first-order equations is generated from the original equation.

@one man 

"I really like moving pictures." I like them too. Unfortunately, as a Maple newbie, I am not yet able to create a worksheet with animated graphics myself. Do you know Holditch's theorem? An animation of it could help learners, for example, to understand the thinking behind Guldin and Cavalieri's rules as an introduction to the concepts of "integral" and "topological product".
Would you give me Maple help getting started with an animation?

Thank you for this article. It reminds me of the beauty of projective geometry, which is now easier to calculate with the help of computers. Such interesting theorems, which make the duality "... lie on ..." and "... go through ..." visible, are illustrated in a way that is easy for learners to understand. Last but not least, I am also thinking of Desargue's theorem ;-) and a "difficult" task posed here.

@shashi598 

First of all, the question:
What is the aim of your work? Is it to solve a numerically specific problem or is it to design a solution algorithm for a category of equations?
In contrast to ordinary differential equations, there is no general existence or uniqueness statement for solutions to partial differential equations. There are solution methods for the classical equations of theoretical physics, but the general theory of PDEs is currently treated individually for categories of equations. And there is a mountain of literature on this, which quickly leads one to sink into the depths of functional analysis.
According to this, two methods have prevailed in practice: the FEM method, according to which the area of ​​the definition domain is discretized/networked and solution ordinates are calculated over the grid points using clever methods from the difference calculation (there is free software for this on the internet) and the minimum method already described. For the latter, you can find information in Michlin's book "Differential equations in mathematical physics", page 81, under "energy functional". In any case, the spectral method you have chosen currently appears to have "only" theoretical value. But the literature describes that in this method the Chebyshev polynomials are numerically more favorable than Legendre polynomials.
As a Maple newbie, I cannot say anything about Maple's capabilities in pde matters.

@Rouben Rostamian  

Yes, this is a well-known graphical solution "for the road" of Pothenot's problem.

As a Maple beginner, I played around with the ODE for practice and chose coefficients. I have difficulty entering initial values ​​(x must logically be different from zero). And I am also unable to numerically approximate the tapeworm terms in the solution. I am asking for help with this in the attached file.

restart

c1 := 1

1

(1)

``

c1*(diff(f(x), `$`(x, 4)))+c2*(diff(f(x), `$`(x, 3)))/x+c3*(diff(f(x), `$`(x, 2)))/x^2+c4*(diff(f(x), x))/x^3+c5*f(x) = 0

diff(diff(diff(diff(f(x), x), x), x), x)+c2*(diff(diff(diff(f(x), x), x), x))/x+c3*(diff(diff(f(x), x), x))/x^2+c4*(diff(f(x), x))/x^3+c5*f(x) = 0

(2)

ode := c1*(diff(f(x), `$`(x, 4)))+c2*(diff(f(x), `$`(x, 3)))/x+c3*(diff(f(x), `$`(x, 2)))/x^2+c4*(diff(f(x), x))/x^3+c5*f(x) = 0

diff(diff(diff(diff(f(x), x), x), x), x)+c2*(diff(diff(diff(f(x), x), x), x))/x+c3*(diff(diff(f(x), x), x))/x^2+c4*(diff(f(x), x))/x^3+c5*f(x) = 0

(3)

c3 := -1

-1

(4)

c2 := -1

-1

(5)

c4 := 1

1

(6)

c5 := 1

1

(7)

dsolve(ode)

f(x) = c__1*hypergeom([], [1/2, 3/8-(1/8)*13^(1/2), 3/8+(1/8)*13^(1/2)], -(1/256)*x^4)+c__2*x^2*hypergeom([], [3/2, 7/8+(1/8)*13^(1/2), 7/8-(1/8)*13^(1/2)], -(1/256)*x^4)+c__3*x^(5/2-(1/2)*13^(1/2))*hypergeom([], [1-(1/4)*13^(1/2), 9/8-(1/8)*13^(1/2), 13/8-(1/8)*13^(1/2)], -(1/256)*x^4)+c__4*x^(5/2+(1/2)*13^(1/2))*hypergeom([], [1+(1/4)*13^(1/2), 9/8+(1/8)*13^(1/2), 13/8+(1/8)*13^(1/2)], -(1/256)*x^4)

(8)

NULL

Download test.mwtest.mw

@nm 

It's a must to have fun - but now to the point:
The ODE is a homogeneous linear ODE with non-constant coefficients. Equations of this type are covered in "Kamke, Differential Equations, Solution Methods and Solutions, Volume 1". A similar example is given in this book under 4.37 on page 534 with a solution. And as is to be expected with linear ODEs, the solution is formed from a linear combination with coefficients that are to be calculated from initial values. My first answer referred to these coefficients. In the present case, it is therefore to be expected that the summands contain Bessel functions, for example. And from this point on, the fun of deriving a closed solution as a term stops for me. Therefore, the equation should be converted into a first-order system and treated numerically. This requires the specific specification of the coefficients C_i and the initial values.

Calculating the coefficients requires initial values. Otherwise, only a general solution to the ODE can be determined. First, however, the equation should be divided by c1. Then four initial values ​​are sufficient.

How do you want to calculate the general solution?

@vv 

For me as a beginner, it is impressive what is possible. Yes, Maple seems to be weak when it comes to Diophantine equations. And since it has been proven that there is no general solution algorithm (Hilbert No. 10), categories of equation types should be implemented with their solution methods.

There are methods for tasks of the current type, e.g. https://www.alpertron.com.ar/QUAD.HTM. It even provides recursions that are based on the known basic solutions (3;1) and (5;1).

Unfortunately, I have no idea about IT, I am just a theoretician and that is why I am asking questions here. At the moment I am trying in vain to insert text from another file into a Maple worksheet using copy and paste between Maple lines. I would like some advice.

@Mariusz Iwaniuk 

Thank you, even the Binet form of the formation law is shown. Only the recursions are still missing.

Similar to how sqrt(5) plays a role in the Fibonacci-Numbers, the importance of sqrt(30) becomes clear here. The "30" can be explained if the original equation is transformed to
(x-6*y)^2 + (4*x +12*y - 3) = 30*y^2.
:-)

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