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Problem with HPM...

Asked by: mojtaba75 15
nonlinear coefficients perturbation
May 28 2016
1  8

Hi everyone.

I'm going to solve a problem with HPM in Maple. I wrote some initial codes but now I'm confused becouse of P^0 coefficients in A1 and B1. I mean I can't reach to f0 and g0.

I upload that file. these are codes that i typed. could you please help me how can I reach to them(f0 & g0)?

http://www.filehosting.org/file/details/573095/Maple%20Project+.mw

Bra-Ket Simplification Question...

Asked by: Kevin Wheelock 15
May 28 2016
0  1

Hi all,

Using the Physics package, I have defined a density operator, shown in the image below:

Definition of Density Operator

(I did assume(s ≠i), assume(v≠h).

I then operate on it with the Bracket as shown in the next image:

Bracket on Density Operator

 

I think this should evaluate to (|Y,s,h><Y,s,h|)/2; or at least that's the result I want. However, Maple evaluates this to zero.

Is there a where to convince Maple to evaluate this to what I want/expect?

Thanks in advance,

Kevin

How to solve polynomial system?...

Asked by: farahnaz 15
polynomial solve
May 27 2016
0  1

how i can solve equations and gain{p1,p2,q1,q2}?

thanks

solve.mw

Q1 := -mu1*p1-upsilon1*q1+gamma11*q1*(p1^2+q1^2)+gamma12*q1*(p2^2+q2^2)-delta1*(2*p1*q1*p2-q2*(p1^2-q1^2))-s11*F*q1+s12*F*q2 = 0

-mu1*p1-upsilon1*q1+gamma11*q1*(p1^2+q1^2)+gamma12*q1*(p2^2+q2^2)-delta1*(2*p1*q1*p2-q2*(p1^2-q1^2))-s11*F*q1+s12*F*q2 = 0

 (1)

Q2 := -mu1*q1+upsilon1*p1-gamma11*p1*(p1^2+q1^2)-gamma12*p1*(p2^2+q2^2)-delta1*(2*p1*q1*q2+p2*(p1^2-q1^2))-s11*F*p1-s12*F*p2 = 0

-mu1*q1+upsilon1*p1-gamma11*p1*(p1^2+q1^2)-gamma12*p1*(p2^2+q2^2)-delta1*(2*p1*q1*q2+p2*(p1^2-q1^2))-s11*F*p1-s12*F*p2 = 0

 (2)

Q3 := -mu2*p2-upsilon2*q2+gamma21*q2*(p1^2+q1^2)+gamma22*q2*(p2^2+q2^2)+delta2*(3*p1^2*q1-q1^3)+s21*F*q1 = 0

-mu2*p2-upsilon2*q2+gamma21*q2*(p1^2+q1^2)+gamma22*q2*(p2^2+q2^2)+delta2*(3*p1^2*q1-q1^3)+s21*F*q1 = 0

 (3)

Q4 := -mu2*q2+upsilon2*p2-gamma21*p2*(p1^2+q1^2)-gamma22*p2*(p2^2+q2^2)+delta2*(-p1^3+3*p1*q1^2)-s21*F*p1 = 0

-mu2*q2+upsilon2*p2-gamma21*p2*(p1^2+q1^2)-gamma22*p2*(p2^2+q2^2)+delta2*(-p1^3+3*p1*q1^2)-s21*F*p1 = 0

 (4)

SOLL := solve({Q1, Q2, Q3, Q4}, real)

Error, (in PolynomialSystemSolvers:-PseudoResultant) numeric exception: division by zero

  

SOLL := solve({Q1, Q2, Q3, Q4}, {p1, p2, q1, q2})

Warning,  computation interrupted

  

``

 

Download solve.mw

 

How to differentiate with respect to y?...

Asked by: iman 35
differentiation
May 27 2016
1  1

how i can differential with respect to ''y'' variable

thanks...help_(2).mw

T(r, theta) := .5*alpha+sum([gamma*n*(r^n+R^(2*n)+r^(-n))*sin(n*theta)], n = 1 .. infinity)

.5*alpha+[-gamma*r*(r^2-1)*sin(theta)/(4*cos(theta)^2*r^2-4*cos(theta)*r^3+r^4-4*cos(theta)*r+2*r^2+1)-((1/2)*I)*R^2*gamma*(exp(I*theta)*R^4-exp((3*I)*theta)*R^4-exp(I*theta)+exp((3*I)*theta))/((R^2*exp(I*theta)-1)^2*(-R^2+exp(I*theta))^2)+((1/2)*I)*r*gamma*(exp(I*theta)*r^2-exp((3*I)*theta)*r^2-exp(I*theta)+exp((3*I)*theta))/((exp(I*theta)-r)^2*(exp(I*theta)*r-1)^2)]

 (1)

``

y := r, proc (theta) options operator, arrow; r*sin(theta) end proc

r, proc (theta) options operator, arrow; r*sin(theta) end proc

 (2)

 

 

diff(T, y)

Error, invalid input: diff received proc (theta) options operator, arrow; r*sin(theta) end proc, which is not valid for its 3rd argument

  

``

 

Download help_(2).mw

About export ...

Asked by: PaulMSheldon 5
animation export
May 27 2016
0  1

I don't know the use to me of having PDF exports having animations requiring huge emails since I don't own website. Yet, I've been fooling with Apple textbook maker app that has movie imort to PDFs. Someday, I'll use google drive links...

So, action item, maybe Maple would like to have a worksheet drop in to this apple textbook app Like movies. I hope it's easy from Apple taunting html5.

Parametric linear system ...

Asked by: logiaco 5
linear matrix
May 27 2016
1  10

Hi,

 

I am trying to solve a simple system of the form AX=0, where A is a N*N matrix, X is an N*1 vector (and the right-hand side of the equation is an N*1 vector of zeros, I apologize for the inexact notation). The difficulty comes from the fact that the values of A are parameterized by 2*N parameters (that I will write as the 2*N vector P), and I would like to get a solution in the form X=f(P).

 

One solution is to try to use LinearAlgebra[LinearSolve], but it only returns the trivial solution X=0, which I am not interested in.

Another solution is to compute analytically the Moore-Penrose pseudoinverse Ag of A, as the general solution is of the form

(I - Ag A)f ;

where f is a vector of free parameters. However, even for a small matrix size (N=4), Maple is still computing after 3 hours on my (fairly powerful) machine, and it is taking more and more memory over time. As the results are polynomial/rational equations in the parameters P, I was actually expecting Maple to be more powerful than other softwares, but for this particular problem, Matlab's symbolic toolbox (muPAD) gives quick solutions until N=6. I need, in the end, to solve additional polynomial/rational equations that are derived from the solutions X=f(P), where Matlab fails. This is why I would really like to be able to solve the above-mentioned problem AX=0 with Maple in order to try to solve the subsequent step of the problem (polynomial system) with Maple.

 

Any suggestions on how to do this would be highly appreciated! Thank you very much for your time and help.

 

Laureline

Expansion of all polynomials in workbook...

Asked by: jlgross 10
polynomial expand
May 27 2016
1  1

I would prefer that all the polynomials generated in my workbook by MAPLE were in expanded form.  For instance, it the elements of a matrix are polynomials, I want to see the expanded form for all of them.  What do I type into a workbook to make this happen.  (I am a new user of MAPLE.) 

Font size in workbook...

Asked by: jlgross 10
font size
May 27 2016
1  1

12pt type is difficult for me to read.  I would like to fix a workbook so that everything yet to appear will be in 18pt type, without my having to enlarge things one at a time.   (I am a new MAPLE user.)  

How to sequentially print two matrix?...

Asked by: 2vdba2 5
matrix print
May 27 2016
1  1

When I print 2 matrix it always try to multiply them. How to force just to show?

Pairing of elements of two lists ...

Asked by: sikjw1305 45
May 27 2016
1  9

Hi

Given two lists I and J of the same size, how can we find all perfect mathings of two lists? 

In other way, how can we find all pairing of elements of I and J? 

For exapmle, given I = [1,2,3] and J = [4,5,6], i would like to get 

[[1,4],[2,5],[3,6]], [[1,4], [2,6],[3,5]], [[1,5],[2,4],[3,6]], [[1,5],[2,6],[3,4]], [[1,6],[2,4],[3,5]], [[1,6],[2,5],[3,4]]

I can see how I can use the permutation of one list and match component-wise in order will do it, but would there be more efficient way to complete  the task? 

Ultimately, I would like do this for a list of lists, that is, for example,

I = [[3,5],[6,7,9,12]] and J = [[8,10], [1,2,4,11]] then I would like to get 

[[3,8],[5,10],[6,1],[7,2],[9,4],[12,11]] and by the above example, get 24 of product of disjoint cycles like such.

Thanks

How to do composite differentiaition? ...

Asked by: mskalsi 290
differentiation
May 26 2016
1  7

Please help me to differentiate function "u" wrt t in following manner:

 

> 

with(PDEtools):

-t*c[2]*k[2]+x*k[2]

 (1)
> 

DepVars := [F(xi), G(xi)]

[F(xi), G(xi)]

 (2)
> 

alias(F = F(xi), G = G(eta))

F, G

 (3)
> 

declare(F, G(xi))

F(xi)*`will now be displayed as`*F

  

G(xi)*`will now be displayed as`*G

 (4)
> 

u := a[0]+(F*a[1]+G*a[2]+kappa[1])/(mu[0]+mu[1]*(diff(F, xi))+mu[2]*(diff(G, eta)))+(F^2*a[3]+F*G*a[4]+G^2*a[5]+kappa[2])/(mu[0]+mu[1]*(diff(F, xi))+mu[2]*(diff(G, eta)))^2

a[0]+(a[1]*F+a[2]*G+kappa[1])/(mu[0]+mu[1]*(diff(F, xi))+mu[2]*(diff(G, eta)))+(a[3]*F^2+a[4]*F*G+a[5]*G^2+kappa[2])/(mu[0]+mu[1]*(diff(F, xi))+mu[2]*(diff(G, eta)))^2

 (5)
> 

xi := -t*c[1]*k[1]+x*k[1]; 1; eta := -t*c[2]*k[2]+x*k[2]

-t*c[1]*k[1]+x*k[1]

  

-t*c[2]*k[2]+x*k[2]

 (6)
> 

diff(u, t)

Error, invalid input: diff received -t*c[1]*k[1]+x*k[1], which is not valid for its 2nd argument

  
> 

``

 

Download [1063]_Sub-equation_Method.mw

Regards

Long Commands in Maple...

Asked by: Parham2016 70
syntax
May 26 2016
0  6

 

hi all,

How could I write long commands in Maple?

For instance: a long vector in Matlab could be written in this way:

A=[ 3, 4, 5, 6, 6, 45, 37...

5, 4, 67, 39, -967 ];

But what is that in Maple??? 

How to export undirected graph to C#?...

Asked by: longqnh 10
May 26 2016
2  2

Hi all,

I drew a undirected graph using Maple

restart;
with(GraphTheory);
with(SpecialGraphs);
with(RandomGraphs);
G := Graph(undirected, {[{1, 2}, 4], {1, 4}, {2, 3}, {2, 5}, {3, 4}});
DrawGraph(G, style = circle);

I want to import this graph to picturebox or panel on C#, please help me!

Problem with ODE system...

Asked by: haani 5
ode dsolve numeric
May 26 2016
1  2

Hi, i am trying to solve my PDEs with HPM method ,but i get strange errors.

first one is :Error, (in trig/reduce/reduce) Maple was unable to allocate enough memory to complete this computation.  Please see ?alloc,

but when i run my last function again,the error chages,let me show you.


restart;
lambda:=0.5;K[r]:=0.5;Sc:=0.5;Nb:=0.1;Nt:=0.1;Pr:=10;
                              0.5
                              0.5
                              0.5
                              0.1
                              0.1
                               10
> EQUATIONS;


equ1:=diff(f(eta),eta$4)-R*(diff(f(eta),eta)*diff(f(eta),eta$2)-f(eta)*diff(f(eta),eta$2))-2*K[r]*diff(g(eta),eta)=0;

equ2:=diff(g(eta),eta$2)-R*(diff(f(eta),eta)*g(eta)-f(eta)*diff(g(eta),eta))+2*K[r]*diff(f(eta),eta)=0;

equ3:=diff(theta(eta),eta$2)+Pr*R*f(eta)*diff(theta(eta),eta)+Nb*diff(phi(eta),eta)*diff(theta(eta),eta)+Nt*diff(theta(eta),eta)^2=0;

equ4:=diff(phi(eta),eta$2)+R*Sc*f(eta)*diff(phi(eta),eta)+diff(theta(eta),eta$2)*(Nt/Nb)=0;
/  d   /  d   /  d   /  d         \\\\     //  d         \ /  d  
|----- |----- |----- |----- f(eta)|||| - R ||----- f(eta)| |-----
\ deta \ deta \ deta \ deta       ////     \\ deta       / \ deta

   /  d         \\          /  d   /  d         \\\
   |----- f(eta)|| - f(eta) |----- |----- f(eta)|||
   \ deta       //          \ deta \ deta       ///

         /  d         \    
   - 1.0 |----- g(eta)| = 0
         \ deta       /    
     /  d   /  d         \\
     |----- |----- g(eta)||
     \ deta \ deta       //

            //  d         \                 /  d         \\
        - R ||----- f(eta)| g(eta) - f(eta) |----- g(eta)||
            \\ deta       /                 \ deta       //

              /  d         \    
        + 1.0 |----- f(eta)| = 0
              \ deta       /    
  /  d   /  d             \\               /  d             \
  |----- |----- theta(eta)|| + 10 R f(eta) |----- theta(eta)|
  \ deta \ deta           //               \ deta           /

           /  d           \ /  d             \
     + 0.1 |----- phi(eta)| |----- theta(eta)|
           \ deta         / \ deta           /

                             2    
           /  d             \     
     + 0.1 |----- theta(eta)|  = 0
           \ deta           /     
    /  d   /  d           \\                /  d           \
    |----- |----- phi(eta)|| + 0.5 R f(eta) |----- phi(eta)|
    \ deta \ deta         //                \ deta         /

                     /  d   /  d             \\    
       + 1.000000000 |----- |----- theta(eta)|| = 0
                     \ deta \ deta           //    
> BOUNDARY*CONDITIONS;


ics:=
f(0)=0,D(f)(0)=1,g(0)=0,theta(0)=1,phi(0)=1;
f(1)=lambda,D(f)(1)=0,g(1)=0,theta(1)=0,phi(1)=0;
   f(0) = 0, D(f)(0) = 1, g(0) = 0, theta(0) = 1, phi(0) = 1
  f(1) = 0.5, D(f)(1) = 0, g(1) = 0, theta(1) = 0, phi(1) = 0
> HPMs;


hpm1:=(1-p)*(diff(f(eta),eta$4)-2*K[r]*diff(g(eta),eta))+p*(diff(f(eta),eta$4)-R*(diff(f(eta),eta)*diff(f(eta),eta$2)-f(eta)*diff(f(eta),eta$2))-2*K[r]*diff(g(eta),eta))=0;

hpm2:=(1-p)*(diff(g(eta),eta$2)+2*K[r]*diff(f(eta),eta))+p*(diff(g(eta),eta$2)-R*(diff(f(eta),eta)*g(eta)-f(eta)*diff(g(eta),eta))+2*K[r]*diff(f(eta),eta))=0;

hpm3:=(1-p)*(diff(theta(eta),eta$2))+p*(diff(theta(eta),eta$2)+Pr*R*f(eta)*diff(theta(eta),eta)+Nb*diff(phi(eta),eta)*diff(theta(eta),eta)+Nt*diff(theta(eta),eta)^2)=0;

hpm4:=(1-p)*(diff(phi(eta),eta$2)+diff(theta(eta),eta$2)*(Nt/Nb))+p*(diff(phi(eta),eta$2)+R*Sc*f(eta)*diff(phi(eta),eta)+diff(theta(eta),eta$2)*(Nt/Nb))=0;

        //  d   /  d   /  d   /  d         \\\\
(1 - p) ||----- |----- |----- |----- f(eta)||||
        \\ deta \ deta \ deta \ deta       ////

         /  d         \\     //  d   /  d   /  d   /  d         \
   - 1.0 |----- g(eta)|| + p ||----- |----- |----- |----- f(eta)|
         \ deta       //     \\ deta \ deta \ deta \ deta       /

  \\\     //  d         \ /  d   /  d         \\
  ||| - R ||----- f(eta)| |----- |----- f(eta)||
  ///     \\ deta       / \ deta \ deta       //

            /  d   /  d         \\\       /  d         \\    
   - f(eta) |----- |----- f(eta)||| - 1.0 |----- g(eta)|| = 0
            \ deta \ deta       ///       \ deta       //    
        //  d   /  d         \\       /  d         \\     //  d  
(1 - p) ||----- |----- g(eta)|| + 1.0 |----- f(eta)|| + p ||-----
        \\ deta \ deta       //       \ deta       //     \\ deta

   /  d         \\
   |----- g(eta)||
   \ deta       //

       //  d         \                 /  d         \\
   - R ||----- f(eta)| g(eta) - f(eta) |----- g(eta)||
       \\ deta       /                 \ deta       //

         /  d         \\    
   + 1.0 |----- f(eta)|| = 0
         \ deta       //    
                                       /                         
        /  d   /  d             \\     |/  d   /  d             \
(1 - p) |----- |----- theta(eta)|| + p ||----- |----- theta(eta)|
        \ deta \ deta           //     \\ deta \ deta           /

  \               /  d             \
  | + 10 R f(eta) |----- theta(eta)|
  /               \ deta           /

         /  d           \ /  d             \
   + 0.1 |----- phi(eta)| |----- theta(eta)|
         \ deta         / \ deta           /

                           2\    
         /  d             \ |    
   + 0.1 |----- theta(eta)| | = 0
         \ deta           / /    
        //  d   /  d           \\
(1 - p) ||----- |----- phi(eta)||
        \\ deta \ deta         //

                 /  d   /  d             \\\     //  d   /  d   
   + 1.000000000 |----- |----- theta(eta)||| + p ||----- |-----
                 \ deta \ deta           ///     \\ deta \ deta

          \\                /  d           \
  phi(eta)|| + 0.5 R f(eta) |----- phi(eta)|
          //                \ deta         /

                 /  d   /  d             \\\    
   + 1.000000000 |----- |----- theta(eta)||| = 0
                 \ deta \ deta           ///    
f(eta)=sum(f[i](eta)*p^i,i=0..1);
                f(eta) = f[0](eta) + f[1](eta) p
g(eta)=sum(g[i](eta)*p^i,i=0..1);
                g(eta) = g[0](eta) + g[1](eta) p
theta(eta)=sum(theta[i](eta)*p^i,i=0..1);
          theta(eta) = theta[0](eta) + theta[1](eta) p
phi(eta)=sum(phi[i](eta)*p^i,i=0..1);
             phi(eta) = phi[0](eta) + phi[1](eta) p
> FORequ1;


A:=collect(expand(subs(f(eta)=f[0](eta)+f[1](eta)*p,g(eta)=g[0](eta)+g[1](eta)*p,hpm1)),p);
/      /  d            \ /  d   /  d            \\
|-1. R |----- f[1](eta)| |----- |----- f[1](eta)||
\      \ deta          / \ deta \ deta          //

                 /  d   /  d            \\\  3   /
   + R f[1](eta) |----- |----- f[1](eta)||| p  + |
                 \ deta \ deta          ///      \
      /  d            \ /  d   /  d            \\
-1. R |----- f[0](eta)| |----- |----- f[1](eta)||
      \ deta          / \ deta \ deta          //

          /  d            \ /  d   /  d            \\
   - 1. R |----- f[1](eta)| |----- |----- f[0](eta)||
          \ deta          / \ deta \ deta          //

                 /  d   /  d            \\
   + R f[0](eta) |----- |----- f[1](eta)||
                 \ deta \ deta          //

                 /  d   /  d            \\\  2   //  d   /  d   /
   + R f[1](eta) |----- |----- f[0](eta)||| p  + ||----- |----- |
                 \ deta \ deta          ///      \\ deta \ deta \

    d   /  d            \\\\       /  d            \
  ----- |----- f[1](eta)|||| - 1.0 |----- g[1](eta)|
   deta \ deta          ////       \ deta          /

          /  d            \ /  d   /  d            \\
   - 1. R |----- f[0](eta)| |----- |----- f[0](eta)||
          \ deta          / \ deta \ deta          //

                 /  d   /  d            \\\  
   + R f[0](eta) |----- |----- f[0](eta)||| p
                 \ deta \ deta          ///  

     /  d   /  d   /  d   /  d            \\\\
   + |----- |----- |----- |----- f[0](eta)||||
     \ deta \ deta \ deta \ deta          ////

         /  d            \    
   - 1.0 |----- g[0](eta)| = 0
         \ deta          /    
A1:=diff(f[0](eta),eta$4)-2*K[r]*(diff(g[0](eta),eta))=0;
A2:=diff(f[1](eta),eta$4)-2*K[r]*(diff(g[1](eta),eta))-R*(diff(f[0](eta),eta))*(diff(f[0](eta),eta$2))+R*f[0](eta)*(diff(f[0](eta),eta$2))=0;
/  d   /  d   /  d   /  d            \\\\       /  d            \   
|----- |----- |----- |----- f[0](eta)|||| - 1.0 |----- g[0](eta)| =
\ deta \ deta \ deta \ deta          ////       \ deta          /   

  0
/  d   /  d   /  d   /  d            \\\\       /  d            \
|----- |----- |----- |----- f[1](eta)|||| - 1.0 |----- g[1](eta)|
\ deta \ deta \ deta \ deta          ////       \ deta          /

       /  d            \ /  d   /  d            \\
   - R |----- f[0](eta)| |----- |----- f[0](eta)||
       \ deta          / \ deta \ deta          //

                 /  d   /  d            \\    
   + R f[0](eta) |----- |----- f[0](eta)|| = 0
                 \ deta \ deta          //    
icsA1:=f[0](0)=0,D(f[0])(0)=1,g[0](0)=0,f[0](1)=lambda,D(f[0])(1)=0,g[0](1)=0;
icsA2:=f[1](0)=0,D(f[1])(0)=0,g[1](0)=0,f[1](1)=0,D(f[1])(1)=0,g[1](1)=0;
   f[0](0) = 0, D(f[0])(0) = 1, g[0](0) = 0, f[0](1) = 0.5,

     D(f[0])(1) = 0, g[0](1) = 0
    f[1](0) = 0, D(f[1])(0) = 0, g[1](0) = 0, f[1](1) = 0,

      D(f[1])(1) = 0, g[1](1) = 0
>
FORequ2;


B:=collect(expand(subs(f(eta)=f[0](eta)+f[1](eta)*p,g(eta)=g[0](eta)+g[1](eta)*p,hpm2)),p);
/      /  d            \          
|-1. R |----- f[1](eta)| g[1](eta)
\      \ deta          /          

                 /  d            \\  3   /
   + R f[1](eta) |----- g[1](eta)|| p  + |
                 \ deta          //      \
      /  d            \          
-1. R |----- f[0](eta)| g[1](eta)
      \ deta          /          

          /  d            \          
   - 1. R |----- f[1](eta)| g[0](eta)
          \ deta          /          

                 /  d            \
   + R f[0](eta) |----- g[1](eta)|
                 \ deta          /

                 /  d            \\  2   //  d   /  d            
   + R f[1](eta) |----- g[0](eta)|| p  + ||----- |----- g[1](eta)
                 \ deta          //      \\ deta \ deta          

  \\       /  d            \        /  d            \          
  || + 1.0 |----- f[1](eta)| - 1. R |----- f[0](eta)| g[0](eta)
  //       \ deta          /        \ deta          /          

                 /  d            \\     /  d   /  d            \\
   + R f[0](eta) |----- g[0](eta)|| p + |----- |----- g[0](eta)||
                 \ deta          //     \ deta \ deta          //

         /  d            \    
   + 1.0 |----- f[0](eta)| = 0
         \ deta          /    
B1:=diff(g[0](eta),eta$2)+2*K[r]*(diff(f[0](eta),eta))=0;
B2:=diff(g[1](eta),eta$2)+2*K[r]*(diff(f[1](eta),eta))-R*(diff(f[0](eta),eta))*g[0](eta)+R*f[0](eta)*(diff(g[0](eta),eta))=0;
     /  d   /  d            \\       /  d            \    
     |----- |----- g[0](eta)|| + 1.0 |----- f[0](eta)| = 0
     \ deta \ deta          //       \ deta          /    
       /  d   /  d            \\       /  d            \
       |----- |----- g[1](eta)|| + 1.0 |----- f[1](eta)|
       \ deta \ deta          //       \ deta          /

              /  d            \          
          - R |----- f[0](eta)| g[0](eta)
              \ deta          /          

                        /  d            \    
          + R f[0](eta) |----- g[0](eta)| = 0
                        \ deta          /    
icsB1:=f[0](0)=0,D(f[0])(0)=1,g[0](0)=0,f[0](1)=lambda,D(f[0])(1)=0,g[0](1)=0;
icsB2:=f[1](0)=0,D(f[1])(0)=0,g[1](0)=0,f[1](1)=0,D(f[1])(1)=0,g[1](1)=0;
   f[0](0) = 0, D(f[0])(0) = 1, g[0](0) = 0, f[0](1) = 0.5,

     D(f[0])(1) = 0, g[0](1) = 0
    f[1](0) = 0, D(f[1])(0) = 0, g[1](0) = 0, f[1](1) = 0,

      D(f[1])(1) = 0, g[1](1) = 0
> FORequ3;


C:=collect(expand(subs(theta(eta)=theta[0](eta)+theta[1](eta)*p,phi(eta)=phi[0](eta)+phi[1](eta)*p,f(eta)=f[0](eta)+f[1](eta)*p,hpm3)),p);
 /                                     
 |                /  d                \
 |10. R f[1](eta) |----- theta[1](eta)|
 \                \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[1](eta)| |----- theta[1](eta)|
          \ deta            / \ deta              /

                               2\                              
          /  d                \ |  3   /                /  d   
    + 0.1 |----- theta[1](eta)| | p  + |10. R f[0](eta) |-----
          \ deta              / /      \                \ deta

                \                   /  d                \
   theta[1](eta)| + 10. R f[1](eta) |----- theta[0](eta)|
                /                   \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[0](eta)| |----- theta[1](eta)|
          \ deta            / \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[1](eta)| |----- theta[0](eta)|
          \ deta            / \ deta              /

                                                            /
          /  d                \ /  d                \\  2   |/
    + 0.2 |----- theta[0](eta)| |----- theta[1](eta)|| p  + ||
          \ deta              / \ deta              //      \\

     d   /  d                \\
   ----- |----- theta[1](eta)||
    deta \ deta              //

                      /  d                \
    + 10. R f[0](eta) |----- theta[0](eta)|
                      \ deta              /

          /  d              \ /  d                \
    + 0.1 |----- phi[0](eta)| |----- theta[0](eta)|
          \ deta            / \ deta              /

                               2\  
          /  d                \ |  
    + 0.1 |----- theta[0](eta)| | p
          \ deta              / /  

      /  d   /  d                \\    
    + |----- |----- theta[0](eta)|| = 0
      \ deta \ deta              //    
C1:=diff(theta[0](eta),eta$2)=0;
C2:=diff(theta[1](eta), eta, eta)+Pr*R*f[0](eta)*(diff(theta[0](eta), eta))+Nb*(diff(phi[0](eta), eta))*(diff(theta[0](eta), eta))+Nt*(diff(theta[0](eta), eta))^2=0;
                  d   /  d                \    
                ----- |----- theta[0](eta)| = 0
                 deta \ deta              /    
       /  d   /  d                \\
       |----- |----- theta[1](eta)||
       \ deta \ deta              //

                           /  d                \
          + 10 R f[0](eta) |----- theta[0](eta)|
                           \ deta              /

                /  d              \ /  d                \
          + 0.1 |----- phi[0](eta)| |----- theta[0](eta)|
                \ deta            / \ deta              /

                                     2    
                /  d                \     
          + 0.1 |----- theta[0](eta)|  = 0
                \ deta              /     
icsC1:=theta[0](0)=1,theta[0](1)=0;
icsC2:=theta[1](0)=0,theta[1](1)=0,phi[0](0)=0,phi[0](1)=0;
                theta[0](0) = 1, theta[0](1) = 0
 theta[1](0) = 0, theta[1](1) = 0, phi[0](0) = 0, phi[0](1) = 0
> FORequ4;


E:=collect(expand(subs(theta(eta)=theta[0](eta)+theta[1](eta)*p,phi(eta)=phi[0](eta)+phi[1](eta)*p,f(eta)=f[0](eta)+f[1](eta)*p,hpm4)),p);
                 3 /  d              \   /                /  d   
0.5 R f[1](eta) p  |----- phi[1](eta)| + |0.5 R f[0](eta) |-----
                   \ deta            /   \                \ deta

             \                   /  d              \\  2   //
  phi[1](eta)| + 0.5 R f[1](eta) |----- phi[0](eta)|| p  + ||
             /                   \ deta            //      \\

    d   /  d              \\
  ----- |----- phi[1](eta)||
   deta \ deta            //

                 /  d   /  d                \\
   + 1.000000000 |----- |----- theta[1](eta)||
                 \ deta \ deta              //

                     /  d              \\  
   + 0.5 R f[0](eta) |----- phi[0](eta)|| p
                     \ deta            //  

     /  d   /  d              \\
   + |----- |----- phi[0](eta)||
     \ deta \ deta            //

                 /  d   /  d                \\    
   + 1.000000000 |----- |----- theta[0](eta)|| = 0
                 \ deta \ deta              //    
E1:=diff(phi[0](eta),eta$2)+Nt*(diff(theta[0](eta),eta$2))/Nb=0;
E2:=diff(phi[1](eta),eta$2)+Nt*(diff(theta[1](eta),eta$2))/Nb+R*Sc*f[0](eta)*(diff(phi[0](eta),eta))=0;
       /  d   /  d              \\
       |----- |----- phi[0](eta)||
       \ deta \ deta            //

                        /  d   /  d                \\    
          + 1.000000000 |----- |----- theta[0](eta)|| = 0
                        \ deta \ deta              //    
         /  d   /  d              \\
         |----- |----- phi[1](eta)||
         \ deta \ deta            //

                          /  d   /  d                \\
            + 1.000000000 |----- |----- theta[1](eta)||
                          \ deta \ deta              //

                              /  d              \    
            + 0.5 R f[0](eta) |----- phi[0](eta)| = 0
                              \ deta            /    
icsE1:=theta[0](0)=1,theta[0](1)=0,phi[0](0)=1,phi[0](1)=0;
icsE2:=theta[1](0)=0,theta[1](1)=0,phi[1](0)=0,phi[1](1)=0;
 theta[0](0) = 1, theta[0](1) = 0, phi[0](0) = 1, phi[0](1) = 0
 theta[1](0) = 0, theta[1](1) = 0, phi[1](0) = 0, phi[1](1) = 0
       
theta[0](eta) = -(152675527/100000000)*eta+1;
                                152675527        
              theta[0](eta) = - --------- eta + 1
                                100000000        
U:=f[1](eta)=0;
                         f[1](eta) = 0
Dsolve(A1,B1,icsA1,icsB1);
                  Dsolve(A1, B1, icsA1, icsB1)


sys:={ diff(g[0](eta), eta, eta)+1.0*(diff(f[0](eta), eta)) = 0, diff(f[0](eta), eta, eta, eta, eta)-1.0*(diff(g[0](eta), eta)) = 0};
    //  d   /  d   /  d   /  d            \\\\
   { |----- |----- |----- |----- f[0](eta)||||
    \\ deta \ deta \ deta \ deta          ////

            /  d            \      
      - 1.0 |----- g[0](eta)| = 0,
            \ deta          /      

     /  d   /  d            \\       /  d            \    \
     |----- |----- g[0](eta)|| + 1.0 |----- f[0](eta)| = 0 }
     \ deta \ deta          //       \ deta          /    /
IC_1:={ f[0](0) = 0, (D(f[0]))(0) = 1, g[0](0) = 0, f[0](1) = .5, (D(f[0]))(1) = 0, g[0](1) = 0,f[0](0) = 0, (D(f[0]))(0) = 1, g[0](0) = 0, f[0](1) = .5, (D(f[0]))(1) = 0, g[0](1) = 0};
    {f[0](0) = 0, f[0](1) = 0.5, g[0](0) = 0, g[0](1) = 0,

      D(f[0])(0) = 1, D(f[0])(1) = 0}
ans1 := combine(dsolve(sys union IC_1,{f[0](eta),g[0](eta)}),trig);
Error, (in dsolve) expecting an ODE or a set or list of ODEs. Received `union`(IC_1, sys)
>
 

A New "Section" Style with Programmatic Table...

by: Samir Khan 2206 Maple 2016
May 25 2016
7  3

A few people have asked me how I created the sections in the Maple application in this video: https://youtu.be/voohdmfTRn0?t=572

Here's the worksheet (Maple 2016 only). As you can see, the “sections” look different what you would normally expect (I often like to experiment with small changes in presentation!)

These aren't, however, sections in the traditional Maple sense; they're a demonstration of Maple 2016's new tools for programmatically changing the properties of a table (including the visibility of its rows and columns). @dskoog gets the credit for showing me the technique.

Each "section" consists of a table with two rows.

  • The table has a name, specified in its properties.
  • The first row (colored blue) contains (1) a toggle button and (2) the title of each section (with the text in white)
  • The second row (colored white) is visible or invisible based upon the state of the toggle button, and contains the content of my section.

Each toggle button has

  • a name, specified in its properties
  • + and - images associated with its on and off states (with the image background color matching the color of the first table row)
  • Click action code that enables or disables the visibility of the second row

The Click action code for the toggle button in the "Pure Fluid Properties" section is, for example,

tableName:="PureFluidProperties_tb":
buttonName:="PureFluidProperties_tbt":
if DocumentTools:-GetProperty(buttonName, 'value') = "false" then   
     DocumentTools:-SetProperty([tableName, 'visible[2..]', true]);
else
     DocumentTools:-SetProperty([tableName, 'visible[2..]', false]);
end if;

As I said at the start, I often try to make worksheets look different to the out-of-the-box defaults. Programmatic table properties have simply given me one more option to play about with.

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