## 65 Reputation

6 years, 288 days

## Numerical solution of the "fractal-like"...

Maple 17

Hello.

I would like to find the numerical solution of the Ginzburg-Landau equation for the1D, written in the form

For d=1 and with the boundary conditions z(-infinity)=0 and z(infinity)=1 the solution is well-known, this is tanh(x) function.

My goal is to undertand the evolution of the numerical solution for the interval of the dimensionality 0<d<1. However, I faced with the problem that even for d=1 (the classical case) I get the error "Error, (in dsolve/numeric/bvp) Newton iteration is not converging".

Here is my simple code

```restart;

with(plots):

R0 := (diff(z(x), x, x))*kappa^2-(1-d)*kappa^2*(diff(z(x), x))/x+z(x)-z(x)^3;

kappa := 2; d := 1;

p := dsolve({R0, z(-15) = 0, z(15) = 1}, type = numeric, range = -15 .. 15, maxmesh = 50000):

odeplot(p);```

## The issue with mtaylor for a function of...

Maple 17

Hello.

I want to check my analytical calculations and to compare my results with the series expansion for x=0 and y=0 of the function

R0 := -(1-tanh((1/2)*l*sqrt(k*y^2+x^2)*d)^2)*x/(k*y^2+x^2)+2*tanh((1/2)*l*sqrt(k*y^2+x^2)*d)*x/(l*(k*y^2+x^2)^(3/2)*d)

by means of the command

mtaylor(R0, [x, y], 5);

However, I have a message from Maple 17

Error, (in mtaylor) does not have a taylor expansion, try series()

I read here a little bit about some bugs of mtaylor function, but how to resolve this issue?

## Discrepancy between numerical and analyt...

Maple

As usual, I have a tricky question. There is an integral that Maple can take numerically

R0 := 1/(a-sqrt(b+c*cos(x)));

Now let's put the coefficients, e.g.

a := 0.9; b := 4.5; c :=0.1

and take the integral from 0 to 2*Pi

R1 := evalf(int(R0, x = 0 .. 2*Pi));

Also, there is an exact analytical result that Maple gives (I give it after simplifying it to avoid division by zero for the limit x=0 and x=2*Pi)

R2:=-4*((a^2-b+c)*EllipticK(sqrt(-2*c/(b-c)))-a^2*EllipticPi(2*c/(a^2-b+c), sqrt(-2*c/(b-c))))/((a^2-b+c)*sqrt(b-c));

As it turns out, the results are completely different. In the first case -5.145818656, while for the second case -3.612771378+0.I

Moreover, If we change the coefficients to a := 0.9; b := 4.5; c := -4 then I obtain Float(undefined)+3.662506136*I and -2.362349457+3.662506117*I , respectively.

My question: how to avoid this descepancy?

## How to solve an integro-functional equat...

Maple

Hello everybody.

My goal is to solve the following integro-functional equation:

```(int(p^2/(f(p)-f(p+q)+omega), p = a .. b))/omega^2 = ln(omega^2+q^2)
```

(int(p^2/(f(p)-f(p+q)+omega), p = a .. b))/omega^2 = ln(omega^2+q^2)

where where is the unknown function and a, b are some numerical values as well as ω and q are real positive variables.

I would be grateful for any ideas.

## Divergence or convergence?...

Maple

Hello everybody.

I have a short question. I have an expression related to the calculation of the Gaussian curvature

`-24.*y^6*(y-1.)^7*(y+1.)^7*(4.+6.*y^8-22.*y^6+30.*y^4-18.*y^2)/(4.*y^18-28.*y^16+84.*y^14+4.-136.*y^12+116.*y^10-24.*y^8-52.*y^6+56.*y^4-24.*y^2)^2`

that is diverged for y=1 as (y-1)^(-2).

At the same time substitution of something like y:=1.00001 gives the value almost zero.

What is the trick?

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