Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

This must be so complicated process to describe the exact mechanism in human or animals ... and there are many reserchs on web for this subject ! Not expected Large answer !

By this way i am just curious to know by support of signal processing methods available in maple we can model a very basic vibrator as larynx and variable duct (what kind of filter we can call ?) or cavity that play the role of tongue or lips for changing of main stream to made various phonemes or frequencies

Here is a movie that show the section of human vocal system :

Anatomy

I found that while using maple gui cases of corruption, stuck while typing etc. Hope this problem will get solved in coming version

Need the code of the following:

 

h is the step size could be 0.1 or 0.01 etc

Dear experts

I am trying to study linear version of the Navier-Stokes (NS) equation. I define NS equations in a way suggested in this page

restart;

with(PDEtools): with(Student[VectorCalculus]):

SetCoordinates(cartesian[x,y,z]):

V:= VectorField(< v[1](x,z,t),0,v[3](x,z,t)>);

NavierStokes:= diff(V,t) - nu*Laplacian(V) +1/rho* Gradient(p(x,z,t)) - VectorField(< 0,0,-g>)=0;

the velocity components are intriduced as following

v[1]:=(x,z,t)->diff(f(z),z)*exp(I*k*x+(-1)*I*omega*t);

v[2]:=(x,z,t)->0;

v[3]:=(x,z,t)->(-1)*I*k*f(z)*exp(I*k*x+(-1)*I*omega*t);

So the problem is 2D model so that v[1] and v[3] satisfy incompresibility condition. It is clear that substituting velocity components into NS leads to differential equations for f(z). The general form of f(z) is

f(z) = a1*cosh(k*z)+a2*sinh(k*z)+a3*cosh(kappa*z)+a4*sinh(kappa*z).

my problems are

1) apply v[1], v[3] and find f(z) with Maple. then apply boundary conditions to find coeffs a[i] with Maple

2) In addition, Substituting the f(z) into BCs leads to a linear homogeneous system of equations for the coefficients. The vanishing of the determinant of this system which is the condition for the existence of a nontrivial solution is interested. How can I do that?

3) f(z) is for the case that the bottom is z=0 and surface is z=h. in my case bottom is z=-h and surface is z=0. how to do this in the solution. I dont know how to apply it.

4) calculate pressure

this file is what I did.

NSE.mw

 Mathematical model for the temperature profiles of steel pipes quenched by water cooling rings

Hi

 Can anyone solve the given equations along with the boundary conditions analytically with Maple and draw the graphs ???????????

maple save variables is greyed out  would anyone happen to know why this could be the case?

I have 4 ode equations. i just want to know can i use any option or simplification to have a analytical solution or NOT? Thanks in Advance

 

``

restart:

ode1 := -2*diff(lambda(t),t)*y1(t) - lambda(t)*diff((y1)(t),t)-0*diff(eta(t),t) - diff((y1)(t),t$3) + diff((y1)(t),t)*(y1(t)^2 + y2(t)^2) +4*y1(t)*sqrt(y1(t)^2 + y2(t)^2)*diff(sqrt(y1(t)^2 + y2(t)^2),t)+diff((y1)(t),t)/r^2
+ y1(t)^2*diff(y1(t),t) + y1(t)*y2(t)*diff(y2(t),t) - 2*diff(y1(t),t)/r^2 ;

 

-2*(diff(lambda(t), t))*y1(t)-lambda(t)*(diff(y1(t), t))-(diff(diff(diff(y1(t), t), t), t))+(diff(y1(t), t))*(y1(t)^2+y2(t)^2)+2*y1(t)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))-(diff(y1(t), t))/r^2+y1(t)^2*(diff(y1(t), t))+y1(t)*y2(t)*(diff(y2(t), t))

(1)

ode2 := diff((lambda)(t),t$2) + lambda(t)*(y1(t)^2 + y2(t)^2) - 2*y1(t)*diff((y1)(t),t$2) - y1(t)^2*(y1(t)^2 + y2(t)^2) - y1(t)^2/r^2 - diff((y1)(t),t)^2 - 2*diff(sqrt(y1(t)^2 + y2(t)^2),t)^2 - 2*sqrt(y1(t)^2 + y2(t)^2)*diff(sqrt(y1(t)^2 + y2(t)^2),t$2) - diff((y2)(t),t)^2 - 2*y2(t)*diff((y2)(t),t$2) - y2(t)^2*(y1(t)^2 + y2(t)^2)

diff(diff(lambda(t), t), t)+lambda(t)*(y1(t)^2+y2(t)^2)-2*y1(t)*(diff(diff(y1(t), t), t))-y1(t)^2*(y1(t)^2+y2(t)^2)-y1(t)^2/r^2-(diff(y1(t), t))^2-(1/2)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))^2/(y1(t)^2+y2(t)^2)-2*(y1(t)^2+y2(t)^2)^(1/2)*(-(1/4)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))^2/(y1(t)^2+y2(t)^2)^(3/2)+(1/2)*(2*(diff(y1(t), t))^2+2*y1(t)*(diff(diff(y1(t), t), t))+2*(diff(y2(t), t))^2+2*y2(t)*(diff(diff(y2(t), t), t)))/(y1(t)^2+y2(t)^2)^(1/2))-(diff(y2(t), t))^2-2*y2(t)*(diff(diff(y2(t), t), t))-y2(t)^2*(y1(t)^2+y2(t)^2)

(2)

ode3 := 2*diff((lambda)(t),t)*y2(t) + lambda(t)*diff((y2)(t),t) - y1(t)*y2(t)*diff((y1)(t),t) - 4*y2(t)*sqrt(y1(t)^2 + y2(t)^2)*diff((sqrt(y1(t)^2 + y2(t)^2)),t) - y2(t)^2*diff((y2)(t),t) - (y1(t)^2 + y2(t)^2)*diff((y2)(t),t) - diff((y2)(t),t$3) ;

2*(diff(lambda(t), t))*y2(t)+lambda(t)*(diff(y2(t), t))-y1(t)*y2(t)*(diff(y1(t), t))-2*y2(t)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))-y2(t)^2*(diff(y2(t), t))-(y1(t)^2+y2(t)^2)*(diff(y2(t), t))-(diff(diff(diff(y2(t), t), t), t))

(3)

ode4 := lambda(t)*y1(t)/r + mu(t)*r - diff((y1)(t),t$2)/r -1/r*y1(t)*(y1(t)^2 + y2(t)^2) - y1(t)/r^3-2/r*diff(y1(t),t$2)

lambda(t)*y1(t)/r+mu(t)*r-3*(diff(diff(y1(t), t), t))/r-y1(t)*(y1(t)^2+y2(t)^2)/r-y1(t)/r^3

(4)

sys := [ode1, ode2, ode3, ode4]:

dsolve(sys,[y1(t),y2(t),lambda(t),mu(t)],'implicit')

``

``


 

Download 1.1.mw

In graph theory, the crossing number cr(G) of a graph G is the lowest number of edge crossings of a plane drawing of the graph G. I'd like to use maple to compute crossing number of some graphs (any graphs is ok,  for example: cr(K5)=1 ). Someone can help me ? thanks. 

I know the problem of computing the crossing number is NP, but I guess we have some good  ways to compute in some small graphs.

Sorry if this is stupid but How do I use the interactive debugger?   the button is greyed out for me and 
 

stopat(EAdd(x^3+2x+1,x, 13, 1, 2, 1, 2), 1);

Error, missing operator or `;`

 

stopat(EAdd, 1);

                             [EAdd]
 

debug button is still greyed out IDK what to do now


p.s. is there a way to set breakpoints in your code by clicking like in other programming?
 

I want to factor large numbers, for this purpose I want to utilize multi cores on my computer.

Anyone who can help me in solving this problem?

I am utilizing Lenovo Thinkpad T440s with specifications core i7-4600 CPU @ 2.19 GHz.

In addition I would like to ask if anyone can help me in integrating GPU's with Maplesoft 2019 for factorization.

 

I am having a cluster of 64 GPU's i.e. AMD RX-460.

 

Any help in this regard will be appreciated.

 

I am using a maple workbook to develop, test, and install a package I am developing.  Several days ago, I noticed that the workbook was not updating the package to reflect the most recent versions of my source code (.mpl). I initially assumed the problem caused by some unintentional change I made to the file, but in the end I found there is a general problem with the URIs in the maple workbook. Everything worked in 2019.1 and I hadn't made any changes after I updated to 2019.2.1.

To confirm the problem, I created a new directory(in fully updated Windows 10 ) and copied my source file into the directory. Then I created a worksheet in maple and saved it into the directory as xx.mw. Using the Navigator panel I then saved the worksheet as a workbook creating xx.maple in the directory.  I then tried to attach the mpl file to the workbook.  In the Navigator menu, the item "attach a file" is unavailable( greyed out).  This is also the case in any of the other earlier workbooks I have created.  I have confirmed the problem is not affecting worksheets because I was able to read my mpl file into a worksheet and load the module using a with(modulename) command.

Please advise.

Dear experts

I have a complex equation (eq=0) which is a function of K, Q, p and s. K and Q can be complex variables and pa and s are parameters.

equation is

eq:= (K*( Q*sinh(K)*cosh(Q)-K*cosh(K)*sinh(Q))*(1+s*K^2)
   +p*(-4*K^2*Q*(K^2+Q^2)
       +Q*(Q^4+2*K^2*Q^2+5*K^4)*cosh(K)*cosh(Q)
       -K*(Q^4+6*K^2*Q^2+K^4)*sinh(K)*sinh(Q)))/(K^2*Q*cosh(Q)) =0 :

and we know that Q:= sqrt(K^2-I*omega); where omega is complex variable

i) How can I study and plot K vs p if Q is 1) real and 2) pure imaginary. 3) complex variable and vice versa

ii) How can I study and plot p vs s

 

Hey there

I have noticed that after some time, Maple freezes and says in the bottom left corner that it is "evaluting" when I  copy something. This is becoming unbearable, as it makes every assignment take a lot longer. Why is this happening, and how can I prevent it?

I  can't seem to get a solution to the following problem.  Can anyone see where I am going wrong I thought I had correct IBC s but they may be wrong/ill-posed

Melvin
 

Two 1-D coupled Burgers equations - semiclassical case: remove O( `&hbar;`^2) terms for u(x,t) but retain O( `&hbar;`) terms for v(x,t):

In the quantum case, there are two coupled quantum Burgers equations, which each include the quantum potential terms.  As in the classical case above, we apply constant external forces Fuand Fv.  Our aim is to display the profiles of u(x, t) and v(x, t) as strings on x, t space.

#hBar := 'hBar': m := 'm':Fu := 'Fu': Fv := 'Fv': # define constants

hBar := 1:m := 1:Fu := 0.2:Fv := 0.1: # set constant values - same as above ...consider reducing

At O( `&hbar;`^2) the real quantum potential term is zero, leaving the classical expression:

pdeu := diff(u(x,t),t)+u(x,t)/m*(diff(u(x,t),x)) = Fu;

diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = .2

(1)

On the otherhand, the imaginary quantum potential equation for v(x,t) has only O( `&hbar;`) terms and so is retained as semiclassical

pdev := diff(v(x,t),t)+u(x,t)/m*(diff(v(x,t),x))-hBar*(diff(u(x,t),x$2))/(2*m)+v(x,t)*(diff(u(x,t),x))/m = Fv;

diff(v(x, t), t)+u(x, t)*(diff(v(x, t), x))-(1/2)*(diff(diff(u(x, t), x), x))+v(x, t)*(diff(u(x, t), x)) = .1

(2)

By inspection of the derivatives in above equations we now set up the ICs and BCs for u(x, t) and "v(x,t)."

The quantum initial and boundary conditions are similar to the classical case, but also comprise additional boundary condition terms for v and for u, notably a 1st derivative BC term for u.

IBCu := {u(x,0) = 0.1*sin(2*Pi*x),u(0,t) = 0.5-0.5*cos(2*Pi*t),D[1](u)(0,t) = 2*Pi*0.1*cos(2*Pi*t)};# IBC for u

{u(0, t) = .5-.5*cos(2*Pi*t), u(x, 0) = .1*sin(2*Pi*x), (D[1](u))(0, t) = .6283185308*cos(2*Pi*t)}

(3)

IBCv := {v(x,0) = 0.2*sin((1/2)*Pi*x),v(0,t)=0.2-0.2*cos(2*Pi*t)};# IBC for v

{v(0, t) = .2-.2*cos(2*Pi*t), v(x, 0) = .2*sin((1/2)*Pi*x)}

(4)

IBC := IBCu union IBCv;

{u(0, t) = .5-.5*cos(2*Pi*t), u(x, 0) = .1*sin(2*Pi*x), v(0, t) = .2-.2*cos(2*Pi*t), v(x, 0) = .2*sin((1/2)*Pi*x), (D[1](u))(0, t) = .6283185308*cos(2*Pi*t)}

(5)

pds:=pdsolve({pdeu,pdev},IBC, time = t, range = 0..0.2,numeric);# 'numeric' solution

_m2606922675232

(6)

The following quantum animation is in contrast with the classical case, and illustrates the delocalisation of the wave form caused by the quantum diffusion and advection terms:

T:=2; p1:=pds:-animate({[u, color = green, linestyle = dash], [v, color = red, linestyle = dash]},t = 0..T, gridlines = true, numpoints = 2000,x = 0..0.2):p1;

2

 

Error, (in pdsolve/numeric/animate) unable to compute solution for t>HFloat(0.0):
matrix is singular

 

p1

(7)

Note that this plot also shows that there are regions in which pIm < 0,  pRe > 0.  Below, the 3D plot of u(x,t),v(x,t) also illustrates the quantum delocalisation of features:

T := 3; p1 := pds:-plot3d({[u, shading = zhue], [v, color = red]}, t = 0 .. T, x = -0.1e-2 .. 2,transparency = 0.0, orientation = [-146, 54, 0], title = print("Coupled quantum solution \n u(x, t) zhue, v(x,t) red", numeric),scaling=unconstrained):p1;

3

 

"Coupled quantum solution 
 u(x, t) zhue, v(x,t) red", numeric

 

Error, (in pdsolve/numeric/plot3d) unable to compute solution for t>HFloat(0.0):
matrix is singular

 

``


 

Download BurgersEqns.mw

Here it is:

#hBar := 'hBar': m := 'm':Fu := 'Fu': Fv := 'Fv': # define constants
hBar := 1:m := 1:Fu := 0.2:Fv := 0.1: # set constant values - same as above ...consider reducing
At O( 
`&hbar;`^2;
) the real quantum potential term is zero, leaving the classical expression:
pdeu := diff(u(x,t),t)+u(x,t)/m*(diff(u(x,t),x)) = Fu;
              / d         \           / d         \      
      pdeu := |--- u(x, t)| + u(x, t) |--- u(x, t)| = 0.2
              \ dt        /           \ dx        /      
On the otherhand, the imaginary quantum potential equation for v(x,t) has only O( 
`&hbar;`;
) terms and so is retained as semiclassical
pdev := diff(v(x,t),t)+u(x,t)/m*(diff(v(x,t),x))-hBar*(diff(u(x,t),x$2))/(2*m)+v(x,t)*(diff(u(x,t),x))/m = Fv;
                                                     2         
         / d         \           / d         \   1  d          
 pdev := |--- v(x, t)| + u(x, t) |--- v(x, t)| - - ---- u(x, t)
         \ dt        /           \ dx        /   2    2        
                                                    dx         

              / d         \      
    + v(x, t) |--- u(x, t)| = 0.1
              \ dx        /      
By inspection of the derivatives in above equations we now set up the ICs and BCs for 
u(x, t);
 and 
v(x,t).;
The quantum initial and boundary conditions are similar to the classical case, but also comprise additional boundary condition terms for 
v;
 and for 
u;
, notably a 1st derivative BC term for 
u;
.
IBCu := {u(x,0) = 0.1*sin(2*Pi*x),u(0,t) = 0.5-0.5*cos(2*Pi*t),D[1](u)(0,t) = 2*Pi*0.1*cos(2*Pi*t)};# IBC for u
          IBCu := {u(0, t) = 0.5 - 0.5 cos(2 Pi t), 

            u(x, 0) = 0.1 sin(2 Pi x), 

            D[1](u)(0, t) = 0.6283185308 cos(2 Pi t)}
IBCv := {v(x,0) = 0.2*sin((1/2)*Pi*x),v(0,t)=0.2-0.2*cos(2*Pi*t)};# IBC for v
                   /                                 
          IBCv := { v(0, t) = 0.2 - 0.2 cos(2 Pi t), 
                   \                                 

                             /1     \\ 
            v(x, 0) = 0.2 sin|- Pi x| }
                             \2     // 
IBC := IBCu union IBCv;
         /                                 
 IBC := { u(0, t) = 0.5 - 0.5 cos(2 Pi t), 
         \                                 

   u(x, 0) = 0.1 sin(2 Pi x), v(0, t) = 0.2 - 0.2 cos(2 Pi t), 

                    /1     \  
   v(x, 0) = 0.2 sin|- Pi x|, 
                    \2     /  

                                           \ 
   D[1](u)(0, t) = 0.6283185308 cos(2 Pi t) }
                                           / 
pds:=pdsolve({pdeu,pdev},IBC, time = t, range = 0..0.2,numeric);# 'numeric' solution
                     pds := _m2606922675232
The following quantum animation is in contrast with the classical case, and illustrates the delocalisation of the wave form caused by the quantum diffusion and advection terms:
T:=2; p1:=pds:-animate({[u, color = green, linestyle = dash], [v, color = red, linestyle = dash]},t = 0..T, gridlines = true, numpoints = 2000,x = 0..0.2):p1;
                             T := 2
Error, (in pdsolve/numeric/animate) unable to compute solution for t>HFloat(0.0):
matrix is singular
                               p1
 

At first, I appology my poor English and I am a non-proessional on mathematics and programing. I am an old retired engineer/researcher.

I found strange phenomena with maple.
Could someone please verify this result?
This is sound like a stupid joke.
This queation is related to factorization of composite number N=p*q
p, q are large prime number respectively.

Degits:=200;
1.    Let p is large prime such as 50 degits
2.    Let R is small integer such as 3
3.    Let q is nearest prime R*p
4.    Let N=p*q
5.    Let f=frac(N/(q+s))

j:=1;
m:=2^b*2^c;
r:=m/2^b;
b:=8; c:=0;
F:=[ ];
for s from -m to m by r do
f:= frac(N/(q+s));
      if f=0 OR 1.0 then f:=100000; # zero dev exeption;
      elif f <= 0.5 then f:=log10(f);
      elif f > 0.5 AND f < 1.0 then f:=-log10(1.0-f);
      else print(“error”) break;
      end if:
F:=[F[ ], [s, f]];  # maple plot format
j::=j+1;
end do:

6.    Draw a Graph s vs F

You wil find a oscillating graph with step at s=0;

7. b:=8 and c:=75; and try same caluclation.

You will find a graph with dip at s=0;
Caution! Resolution and Scanwidth can not go together.

8.    Let qx=q+10^25;
9.    Nx:=p*qx;
10.    Let ps:=sqrt(N/3.0) and let qs:=3.0*sqrt(N/3.0);
11.    fs:=frac(Nx/(qx+s));
Caution!  Nx not equal Ns:=p*qs
12.     Draw a graph s vs Fs with Nx, p, qs
You will find a similar graph dip point shifted from s=0
13. Verify qx=ps-dip point s

My questions are
1.    Is it a kaind of factorization of Nx?
2.    This phenomena hava posibirity to make a vulneravility for RSA crypto?
3.    Power spectrum of F( c=0) indicates R=q/p; with an peak
4.    Is it an rediscovery of “Wheel”?  Are thea any papers similar to this phenomena

I appreciate to read this questions.

Environment: Windows 10 for workstation and Maple 2019(64bit) CPU: Intel Xeon.

15 Dec 2019  some miss spelling corrected

                       A graph added   frac(N(q+s)) vs s b=8 c=75 

 

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