## How I can pdsolve this partial fractional equatio...

How I can pdsolve this partial fractional  equation?

###The Riemann–Liouville fractional integral   is written in a convolution-form as:

#where (...) represents the Gamma function

 (1)

 (2)

## Bifurcation Procedure for the Rossler Attractor...

Hi,

I'm currently studying the Rossler Attractor, which is the following system:

.

I have found the points of equilibrium of the system.

But now I need to draw bifurcation diagrams: one for varying a, one for varying b, and one for varying c.

## How find a point over this curve?...

Hi!

Consider, fixed an integer m>1, the mapping given by the following procedure:

G := proc (t) local k, C; C := NULL; C := t; for k from 2 to d do C := C, 1/2-(1/2)*cos(Pi*m^(k-1)*t) end do; return [C] end proc

Then, it can be proved that given x in the cube [0,1]^{d} there is t in [0,1] such that the norm of x-G(t) is less, or equal, than sqrt(d-1)/m. Indeed, dividing the cube [0,1]^{d} into m^{d-1} subcubes of side-length 1/m x ... x 1/m x 1, the point x belongs to some of these subcubes, say J. As, by the properties of the cosines function, the curve G(t) lies in J whenever t in certain subinterval of [0,1], the result follows.

In other words, computing all the solutions of the equation

1/2*(1-cos(Pi*m^(d-1)*t)) = x[d], (j-1)/m <= t and t <= j/m

for some of these solutions the desired t is obtained, where j is such that x1 in [(j-1)/m,j/m] (x1 is the first coordinate of the point x). However, for large values of m and d, the above equation have many solutions, I have tried find all of them and the process is extremely slow....Other way to find such a t can be the following: find a t satisfying the following system of inequalities

EQ := abs(t-x[1]) <= 1/m; for k from 2 to d do EQ := EQ, abs(1/2*(1-cos(Pi*m^(k-1)*t))-x[k]) <= 1/m end do

and then, a solution of this system is a such t. I do not know how to find, efficiently, a t such that of x-G(t) is less, or equal, than sqrt(d-1)/m   :(

Some idea?

## Solving pell's equation...

Hi,
I am new to using maple and was wondering if i could get some help with this problem...

I'm having problems with 2b)

pell := proc (d, K)

local y;

y := sqrt(d*x^2+1);

d := 2;

if y <= K then print*(x, y) end if

end proc

this is what i have come up with, feel like it is completely wrong though.

Any help would be much appreciated.

## can Maple solve the Burger's PDE for viscous fluid...

I am having trouble getting Maple 2017.3 with latest Physics update to give solution to Burger's PDE for viscous fluid flow with the following initial condition. May be I am not doing something right. I tried different HINTS, but no luck.

Maple can solve the PDE without the initial conditions.

May be a Maple expert can find work around or show what I might be doing wrong.

restart;
pde := diff(u(x, t), t) + u(x, t)*diff(u(x, t), x) = mu*diff(u(x,t),x$2); ic := u(x,0) = PIECEWISE([0,x>=0],[1,x<0]); sol := pdsolve({pde,ic}, u(x, t)) assuming mu>0;  Maple returns () as solution. This PDE can be solved analytically. Here is Mathematica' solution ClearAll[u,x,y,mu] pde = D[u[x,t],{t}]+u[x,t]*D[u[x,t],{x}]==mu*D[u[x,t],{x,2}]; ic = u[x,0]==Piecewise[{{1,x<0},{0,x>=1}}]; sol = DSolve[{pde,ic},u[x,t],{x,t},Assumptions->mu>0] ## why Maple mserver.exe hangs on this convert() call... Asked by: This is on Maple 2017.3 under windows After obtaining solution from pdsolve(), I tried to see if Maple can convert it to hyperbolic trig functions by calling convert(sol,trigh). I waited and waited and nothing happened. Then clicked on the interrupt current operation button at top of menu. But I found that mserver.exe has hanged in a loop. Taking 100% CPU and still running. So Had to terminate it from task manager. Question is: It is ok if Maple can't do the conversion, but why does it hang? Maybe if I want for one hr it will finish, I do not know. But the important part, why does interrupt current operation does not work, in the sense that the mserver.exe is still running? Is this common thing to happen? Should this be fixed? Here is example restart; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; f:=x-> piecewise(x>0 and x<1/2, 2*x, x>1/2 and x<1, 2-2*x); bc:=u(0,y)=0,u(1,y)=0,u(x,0)=f(x),u(x,2)=f(x); sol:=pdsolve([pde,bc],u(x,y)) assuming x>0,y>0;  The above works OK and generates a solution. Now the next call hangs mserver.exe convert(sol,trigh); In case your Maple version can't solve the above PDE. Here is the solution obtained, so you can try this below without having to solve the PDE sol := u(x, y) = Sum(8*sin((1/2)*n*Pi)*sin(Pi*x*n)*(exp(Pi*n*(3*y-2))- exp(Pi*n*(3*y-4))+exp(Pi*y*n)-exp(Pi*n*(y-2)))*exp(-2*Pi*n*(y-2))/(Pi^2*n^2*(exp(4*n*Pi)-1)), n = 0 .. infinity); convert(sol,trigh);  I also tried convert(rhs(sol),trigh); but it made no difference. ## Why won't it append? Any help would be humbly an... Asked by: This is my code. It's not appending as I need it to. Any help would be humbly and gratefully appreciated!  >  >  (1)  >  (2)  >  (3)  > Download Sort4_320.mw ## what is the integral of this function ?... Asked by: f(x)=sqrt(sin(x)); x over the intervall [0,T] can anyone help me to find this integral ? ## How to make Maple solve the beam PDE?... Asked by: I might be doing something wrong since I expected Maple to be able to solve this. Could some Maple manage to make Maple solve the following beam PDE problem taken from a textbook? This is what I tried. restart; pde:=diff(u(x,t),t$2)+diff(u(x,t),x$4)=0; bc:=u(0,t)=-12*t^2,u(1,t)=1-12*t^2,D[1,1](u)(0,t)=0,D[1,1](u)(1,t)=12; ic:=u(x,0)=x^4,D[2](u)(x,0)=0; sol:=pdsolve({pde,ic,bc},u(x,t)); But Maple returns no solution. I am using Maple 2017.3 on windows. ## Mixed type Levi-Civita's... Asked by: I have two tensors, E_ and F_ below, that I believe should be equal. But they are not, and I cannot understand why. The problem does not appear in, say, Schwarschild spacetime, but it appears in Boyer-Lindquist spacetime, metric [5,29,1]; perhaps it appears only if the vierbein is nondiagonal? The code: Loading the packages and the metric: restart: with(Physics): with(Tetrads): g_[[5,29,1]]; # The Boyer-Lindquist metric Set up galilean and nongalilean Levi-Civita's, respectively, following the recipe given elsewhere: Define(varepsilon[a,b,c,d] = Array((1..4)$4,rhs(LeviCivita[nonzero])),quiet):
Setup(levicivita = nongalilean):
# Checking that the Levi-Civita's are indeed different
varepsilon[1,2,3,4];   # The galilean case
LeviCivita[1,2,3,4];   # The nongalilean case

Define the two tensors E_ and F_, using mixed type Levi-Civita's for the latter:

Define(
E_[~a,mu] = varepsilon[~a,~b,~c,~d]*LeviCivita[mu,nu,rho,sigma]*e_[b,~nu]*e_[c,~rho]*e_[d,~sigma],
F_[~a,mu] = varepsilon[~a,b,c,d]*LeviCivita[mu,~nu,~rho,~sigma]*e_[~b,nu]*e_[~c,rho]*e_[~d,sigma]
,quiet):
E_[definition];
F_[definition];

Compare the two expressions, which should be equal, I believe.

expr := simplify(TensorArray(E_[~a,mu] - F_[~a,mu])) assuming a::real,theta > 0,theta < Pi;
eval(expr,{a = 1,m = 1,r = 2,theta = Pi/4});   # Just to make the difference completely obvious

[I have trouble copy-pasting the output from these two lines, so you will have to execute the worksheet provided below to see it.]

However, they are not equal. Why not?

## Can Maple write out the first n terms of a summati...

Suppose I am given the following summation:

Can Maple write the first few terms of the summation? For example, if I want to see first three terms of the summation, I'd like Maple to output the following:

a + 2 a + 3 a

My use case is that if I am given a complicated summation, it can be useful to look at the first few terms to see if there are any patterns.

The things I've tried are:

 (1)

 (2)

 (3)

 (4)

In the realm of tetrads where both world indices and Lorentz indices are present, contractions, say, using simultaneously the Minkowskian (galilean) Levi-Civita symbol,

and the curvilinear Levi-Civita (pseudo-)tensor,

can be considered. Although each of the two types of Levi-Civitas can easily be obtained separately by specifying Setup(levicivita = galilean) or Setup(levicivita = nongalilean), I cannot figure out how to have them both available at the same time. Any suggestions?

PS: I am, of course, aware of the fact that the two Levi-Civitas are related by some appropiate square-root of the determinant of the metric, but I have no desire to fiddle around with explicit such determinants if they can be avoided.

This is probably a question to Edgardo: In another thread, the following quantity is considered [Eqs. (5) and (6) combined]:

expand(gamma_[definition]);

This is all very well, but it seems to depend on the metric loaded: if the Schwarzschild metric g_[sc] is loaded, then the above output results, but if the Minkowski metric g_[minkowski] is loaded, then the output of the above expansion is identically zero. Does that make sense? Is that intentional? The explicitly evaluated Ricci rotation coefficients vanish identically for the Minkowski metric, of course, but if evaluation is performed for that case, then why not also for the Schwarschild metric [not meaning to say that I want evaluation]?

## Why does Maple do this?...

Does anyone know why Maple doesn't simplify the following expression?

Thanks!

## solving Laplace inside disk problem. pdsolve...

I am trying to see if Maple can solve Laplace PDE inside the disk in polar coordinates. Standard textbook problem. Radius of disk is a. The boundary conditions on the disk is f(theta). One of the conditions needed also is that the solution is finite in the center of the disk.

I do not know how to tell Maple that the solution should be finite in the center of the disk. If I do not give this conditions, Maple gives me strange looking solution, which does not look like anything close to the standard series solution one gets from hand solution. There is not even a series solution.

This is what I tried

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta);
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0


Now, how to tell it that u(0,theta) is bounded? So that the ln(r) solution do not show up? Adding u(0,theta)<infinity to the boundary conditions, gives error

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta),u(0,theta)<infinity;
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0


The standard solution to this PDE is

Where c0 and cn and kn above are found from boundary conditions at $u(a,\theta)$.

How can one get Maple to give the above solution? How to tell it that $u$ is bounded at $r=0$?

 1 2 3 4 5 6 7 Last Page 1 of 23
﻿