## How do show all roots in the equation with Bairsto...

I'm trying show the all roots of equations using Bairstow's methodt, but only shows the roots of Quadratic Factor and don't show the others roots of the other equation. Thanks

This the code:

 >
 1   -0.6441699    0.1381090       0.3558        1.138   2   -0.5111131    0.4697336       0.1331       0.3316   3   -0.4996865    0.5002023      0.01143      0.03047   4   -0.5000001    0.5000000   -0.0003136   -0.0002023   5   -0.5000000    0.5000000    6.413e-08    9.268e-09   6   -0.5000000    0.5000000            0            0 Q(x)=(1)x^3 + (-4)x^2 + (5.25)x^1 + (-2.5) Remainder: 0(x-(-0.5))+(0) Quadratic Factor: x^2-(-0.5)x-(0.5) Zeros: 0.4999999993, -0.9999999997

Only show two roots: 0.4999999993, -0.9999999997, but the other roots are missing: 2, -1, 1+0.5i, 1-0.5i approximately, any solution?

I think it's in this part, but I can't think of how to implement it to get the missing roots.

## Coupled PDEs:Error, (in pdsolve/numeric/animate) u...

Analysis of the semiclassical (SC) momentum rate equations

Plotting the ICs and BCs and examining sensitivity to the Re and Im forces

MRB: 24/2/2020, 27/2/2020, 2/3/2020.

We examine solution of the SC version of the momentum rate equations, in which O terms for  are removed. A high level of sensitivity to ICs and BCs makes solution finding difficult.

 > restart;
 > with(PDETools): with(CodeTools):with(plots):

We set up the initial conditions:

 > ICu := {u(x, 0) = .1*sin(2*Pi*x)}; ICv := {v(x, 0) = .2*sin(Pi*x)};
 (1)
 > plot([0.1*sin(2*Pi*x),0.2*sin(Pi*x)],x = 0..2, title="ICs:\n u(x,0) (red), v(x,0) (blue)",color=[red,blue],gridlines=true);

The above initial conditions represent a positive velocity field  (blue) and a colliding momentum field (red).

Here are the BCs

 > BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t))};
 (2)
 > BCv := {v(0,t) = 0.5*sin(2*Pi*t),v(2,t)=-0.5*sin(2*Pi*t)};
 (3)
 > plot([0.5*(1-cos(2*Pi*t)),0.5*sin(2*Pi*t),-0.5*sin(2*Pi*t)],t=0..1,color=[red,blue,blue],linestyle=[dash,dash,dot],title="BCs:\n u(0,t) (red-dash),\n v(0,t) (blue-dash), v(1,t) (blue-dot)",gridlines=true);
 >

We can now set up the PDEs for the semiclassical case.

 > hBar:= 1:m:= 1:Fu:= 0.2:Fv:= 0.1:#1.0,0.2
 > pdeu := diff(u(x,t),t)+u(x,t)/m*(diff(u(x,t),x)) = Fu;
 (4)
 > 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;
 (5)
 > ICu:={u(x,0) = 0.1*sin(2*Pi*x)};
 (6)
 > ICv:={v(x,0) = 0.2*sin(Pi*x/2)};
 (7)
 > IC := ICu union ICv;
 (8)
 > BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t)), D[1](u)(2,t) = 0.1*cos(2*Pi*t)};
 (9)
 > BCv := {v(0,t) = 0.2*(1-cos(2*Pi*t))};
 (10)
 > BC := BCu union BCv;
 (11)

We now set up the PDE solver:

 > pds := pdsolve({pdeu,pdev},{BC[],IC[]},time = t,range = 0..2,numeric);#'numeric' solution
 (12)
 > Cp:=pds:-animate({[u, color = red, linestyle = dash],[v,color = blue,linestyle = dash]},t = 30,frames = 400,numpoints = 400,title="Semiclassical momentum equations solution for Re and Im momenta u(x,t) (red) and v(x,t) (blue) \n under respective constant positive forces [0.2, 0.1] \n with sinusoidal boundary conditions at x = 0, 1 and sinusoidal initial conditions: \n time = %f ", gridlines = true,linestyle=solid):Cp;
 (13)

Observations on the quantum case:

The classical equation for  is independent of the equation for .   (red) is a solution of the classical Burgers equation subject to a force 0.2, but  is NOT influenced by .  On the otherhand,  (blue) is a solution of the quantum dynamics equation subject to force 0.1 and is influenced by .   This one way causality (u )  is a feature of the semiclassical case, and it emphasises the controlling influence of the classical , which modulates the quantum solution for .  Causally, we have u.

The initial conditions are of low momentum amplitude:0.1 for the classical  (red) field and.2 for  (blue)  but their influence is soon washed out by the boundary conditions  and  that drive the momentum dynamics.

The temporal frequency of the boundary condition on the -field is twice that of the classical -field. This is evident in the above blue transient plot. Moreover, the boundary condition on the classical -momentum (red), drives that field in the positive direction, initially overtaking the quantum  field, as consistent with the applied forces [0.2, 0.1]. Although initially of greater amplitude than the classical field, the  momentum field is asymptotically of the same amplitude as the  field, but has greater spatial and temporal frequency, owing to the boundary conditions.

Referring to the semiclassical momentum rate equations, we note that the classical field  (red) modulates the quantum momentum rate equation for .

 >

I am having difficulty getting solutions to a pair of PDEs.  Would anyone like to cast an eye over the attached file, incase I am missing something.

Thanks

Melvin

## Maple 2018.2 crashs while using plot3d()...

Hello,

For a few days Maple crashs everytime i try to use the command "plot3d()".

I had'nt this problem befor and I have no idea what the reason could be. It ist irrelevant what Funktion I try to visualize,  the window just get closed evertime.

I hope someone can help me.

Thank you!

Tom

## How do I simplify KdV equation in Maple by using t...

How do I simplify KdV equation in Maple by using =fxt))xx)?)

 >

I am by using =2*difffxtxx)
My aim is to get the form

 >
 >
 >
 >
 >
 (1)
 >
 (2)
 >
 (3)
 >
 (4)
 >
 (5)
 >
 >
 >
 >

## How do I find Lax pair for this equation?...

Hello
In this example, we have the KdV equation
t] - 6 uux] + xxx] = 0
I would like to find the Lax pair for the KdV equation, which are
Lψ=λψ
ψ[t] = Mψ

Lt+ML-LM = 0  called a compatibility condition
So, I will start from this purpose
Then we will assume M in the form
will assume M in the form
M := a3*Dx^3+a^2+a1*Dx+a0
thenb using M and L in the for L[tL-LM = 0can find
Dx^5+( ) Dx^4+( ) Dx^3+( ) Dx^2+( ) Dx+( )=0
then wean find a_i =0,1,2,3
In the following maple code to do that
my question is
.How I canoue the soluo get a_i2,3 usinmaple code
any maple packge to find Lax pair for PDE -

 >

in this exampile we have KdV equation

I would likeind the Lax pair for the KdV equation, which are

Lψ=λψ

where

+ML-LM = 0    called  apatibility  condition

So, I  will start this purpose

L:=-Dx^2+u;

then we will assume M the m

Ma3*Dx^3+a2*Dx^2+Dx+a0

then busing in the form +ML-LM = 0 can find

( ) Dx^5+( ) Dx^4+( ) Dx^3+( ) Dx^2+( ) Dx+( )=0

then we can find a_i ;i=,2,3

the fllowile code to

my queion is ;

1) How I can continue the solution  to get a_i ;i=0,1,2,3 using maple code  ?

2) isir any maple packge to find  Lax pair for PDE ?

 >
 (1)
 >
 (2)
 >
 (3)
 >
 (4)
 >
 >
 (5)
 >
 (6)
 >
 (7)
 >
 (8)
 >
 (9)

 >
 >
 >
 >
 >

## How to define differential forms as function of so...

I want to define the co-ordianates (phi, PI, ....)  as functions of some variable eg:- x,y.

## Numeric solve of a system with coupled pde/ode...

I am studying the motion of a beam coupled to piezoelectric strips. This continuous system is modelled by two DE:

`YI*diff(w(x,t), x\$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x\$2)+c*diff(w(x,t), t)+`&rho;A`*diff(w(x,t), t\$2)+C[em,m]*v(t) = 0;`

And:

`C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));`

where "w(x,t)" stands for the beam's vibration and "v(t)" means the electric voltage, which is constant throught the beam. I would like to numerically solve both DE simultaneosly, but maple will not let me do it. I would like to know why. I am getting the following error:

`Error, (in pdsolve/numeric/process_PDEs) number of dependent variables and number of PDE must be the same`

I suppose it is because "w(x,t)" depends on "x" and "t", while "v(t)" depends solely on time, but I am not sure. Could someone help me out? Here is my current code:

```restart:
with(PDEtools):
declare(w(x,t), v(t)):

YI*diff(w(x,t), x\$4)-N[0]*cos(2*omega*t)*diff(w(x,t), x\$2)+c*diff(w(x,t), t)+`&rho;A`*diff(w(x,t), t\$2)+C[em,m]*v(t) = 0;
pde1:= subs([YI = 1e4, N[0] = 5e3, c = 300, omega = 3.2233993, C[em,m] = 1], %):
ibc1:= w(0,t) = 0, D[1,1](w)(0,t) = 0, w(ell,t) = 0, D[1,1](w)(ell,t) = 0, D[2](w)(x,0) = 0, w(x,0) = sin(Pi*x/ell):

C[p]*diff(v(t), t)+1/R[l]*v(t) = C[em,e]*(D[1,2](w)(0,t)-D[1,2](w)(ell,t));
pde2:= subs([C[p] = 10, R[l] = 1000, C[em,e] = 1, ell = 5], %):
ibc2:= v(0) = 0:

pdsolve({pde1, pde2}, {ibc1, ibc2}, numeric);```

Thanks.

## Firewall blocking Maple 2018 after upgrading comp...

I have just upgraded my laptop from Windows 7 to Windows 10.  On starting up Maple 2018,  I receive the attached message on screen.  This is after previously loading the worksheet successfully.   Today,  I am not able to do so.  I need to permanently register my firewall to allow Maple to run; can anyone help?

Melvin

## Maple 2018.2 is not compatible with macOS Catalina...

Hi everybody

l I tried to install Maple 2018.2 on the golden master of Catalina and it didn't work : the installation process ended after the entering of the password to authorize the installation. In fact Maple 2018.2 still contains 32 bit elements. Is there a solution ? Thank you

David

## Chebyshev series in different intervals...

Hello,

My question is mathematical in nature, so it might be a little out of place but I though I would give it a shot.

You have a series of chebyshev coefficients in two connecting subdomains lets say S1 = [0,0.5] and S2=[0.5,1]. So far you are still in the spectral space. If you want to compute the solution in real space you can sum the coefficients with the Chebyshev polynomials.

Now imagine you change the interval to S1 = [0,0.6] and S2 = [0.6,1]. Is there a way to manipulate the Chebyshev coefficients from both initial subdomains to create a new set of Chebyshev coefficients that fit the solution in the new subdomains.

The brute force method would be to create the real solution of Chebyshev polynomials and then use that to form a new set of Chebyshev coefficients. Or you can use Clenshaw to compute the solution at several points, and then use the points to create new Chebyshev coefficients.

But what if we can stay in spectral space and create the new chebyshev coefficients. Is that possible? If so, how?

## I Have a problem in Do loop...

Dears, greeting for all

I have a problem, I try to explain it by a figure

This formula does not work.

I need to substitute n=0 to give G_n+1 as a function of the parameter s, then find the limit.

.where G_n is a function in s.

this is the result

## Wrong values for Eigenvalues, depending on Digits ...

Hello!

I want to calculate Eigenvalues. Depending on values for digits and which datatype I choose Maple sometimes returns zero as Eigenvalues. Maybe there is a problem with the used routines: CLAPACK sw_dgeevx_, CLAPACK sw_zgeevx_.

 >

Problems LinearAlgebra:-Eigenvalues, Digits, ':-datatype' = ':-sfloat', ':-datatype' = ':-complex'( ':-sfloat' )

 > restart;
 > interface( ':-displayprecision' = 5 ):
 > infolevel['LinearAlgebra'] := 5; myPlatform := kernelopts( ':-platform' ); myVersion := kernelopts( ':-version' );
 (1.1)

Example 1

 > A1 := Matrix( 5, 5, [[0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0], [0, 0, 0, 0, 1], [-10201/1000, 30199/10000, -5049/250, 97/50, -48/5]] );
 (1.1.1)
 > LinearAlgebra:-Eigenvalues( A1 );
 CharacteristicPolynomial: working on determinant of minor 2 CharacteristicPolynomial: working on determinant of minor 3 CharacteristicPolynomial: working on determinant of minor 4 CharacteristicPolynomial: working on determinant of minor 5
 (1.1.2)
 > A11 := Matrix( op( 1, A1 ),( i,j ) -> evalf( A1[i,j] ), ':-datatype' = ':-sfloat' );
 (1.1.3)
 > Digits := 89; LinearAlgebra:-Eigenvalues( A11 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.1.4)
 > Digits := 90; LinearAlgebra:-Eigenvalues( A11 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.1.5)
 > A12 := Matrix( op( 1, A1 ),( i,j ) -> evalf( A1[i,j] ), ':-datatype' = ':-complex'( ':-sfloat' ) );
 (1.1.6)
 > Digits := 100; LinearAlgebra:-Eigenvalues( A12 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.1.7)
 > Digits := 250; LinearAlgebra:-Eigenvalues( A12 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.1.8)
 >
 >

Example 2

 > A2 := Matrix(3, 3, [[0, 1, 0], [0, 0, 1], [3375, -675, 45]]);
 (1.2.1)
 > LinearAlgebra:-Eigenvalues( A2 );
 IntegerCharacteristicPolynomial: Computing characteristic polynomial for a 3 x 3 matrix IntegerCharacteristicPolynomial: Using prime 33554393 IntegerCharacteristicPolynomial: Using prime 33554383 IntegerCharacteristicPolynomial: Used total of  2  prime(s)
 (1.2.2)
 > A21 := Matrix( op( 1, A2 ),( i,j ) -> evalf( A2[i,j] ), ':-datatype' = ':-sfloat' );
 (1.2.3)
 > Digits := 77; LinearAlgebra:-Eigenvalues( A21 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.2.4)
 > Digits := 78; LinearAlgebra:-Eigenvalues( A21 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.2.5)
 > A22 := Matrix( op( 1, A2 ),( i,j ) -> evalf( A2[i,j] ), ':-datatype' = ':-complex'( ':-sfloat' ) );
 (1.2.6)
 > Digits := 58; LinearAlgebra:-Eigenvalues( A22 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.2.7)
 > Digits := 59; LinearAlgebra:-Eigenvalues( A22 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.2.8)
 >
 >

Example 3

 > A3 := Matrix(4, 4, [[0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [-48841, 8840, -842, 40]]);
 (1.3.1)
 > LinearAlgebra:-Eigenvalues( A3 );
 IntegerCharacteristicPolynomial: Computing characteristic polynomial for a 4 x 4 matrix IntegerCharacteristicPolynomial: Using prime 33554393 IntegerCharacteristicPolynomial: Using prime 33554383 IntegerCharacteristicPolynomial: Used total of  2  prime(s)
 (1.3.2)
 > A31 := Matrix( op( 1, A3 ),( i,j ) -> evalf( A3[i,j] ), ':-datatype' = ':-sfloat' );
 (1.3.3)
 > Digits := 75; LinearAlgebra:-Eigenvalues( A31 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.3.4)
 > Digits := 76; LinearAlgebra:-Eigenvalues( A31 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.3.5)
 > A32 := Matrix( op( 1, A3 ),( i,j ) -> evalf( A3[i,j] ), ':-datatype' = ':-complex'( ':-sfloat' ) );
 (1.3.6)
 > Digits := 100; LinearAlgebra:-Eigenvalues( A32 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.3.7)
 > Digits := 250; LinearAlgebra:-Eigenvalues( A32 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.3.8)
 >
 >

Example 4

 > A4 := Matrix(8, 8, [[0, 1, 0, 0, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1], [-1050625/20736, 529925/1296, -15417673/10368, 3622249/1296, -55468465/20736, 93265/108, -1345/8, 52/3]]);
 (1.4.1)
 > LinearAlgebra:-Eigenvalues( A4 );
 CharacteristicPolynomial: working on determinant of minor 2 CharacteristicPolynomial: working on determinant of minor 3 CharacteristicPolynomial: working on determinant of minor 4 CharacteristicPolynomial: working on determinant of minor 5 CharacteristicPolynomial: working on determinant of minor 6 CharacteristicPolynomial: working on determinant of minor 7 CharacteristicPolynomial: working on determinant of minor 8
 (1.4.2)
 > A41 := Matrix( op( 1, A4 ),( i,j ) -> evalf( A4[i,j] ), ':-datatype' = ':-sfloat' );
 (1.4.3)
 > Digits := 74; LinearAlgebra:-Eigenvalues( A41 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.4.4)
 > Digits := 75; LinearAlgebra:-Eigenvalues( A41 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_dgeevx_
 (1.4.5)
 > A42 := Matrix( op( 1, A4 ),( i,j ) -> evalf( A4[i,j] ), ':-datatype' = ':-complex'( ':-sfloat' ) );
 (1.4.6)
 > Digits := 100; LinearAlgebra:-Eigenvalues( A42 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.4.7)
 > Digits := 250; LinearAlgebra:-Eigenvalues( A42 );
 Eigenvalues: calling external function Eigenvalues: initializing the output object Eigenvalues: using software external library Eigenvalues: CLAPACK sw_zgeevx_
 (1.4.8)
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >

## Visualising solutions to sums of polynomials...

I'm working towards creating a way to visualise real polynomial ideals! (or at least the solutions of the polynomials in the ideals) this code creates a plot showing the solutions to all the polynomials in the ideal generated by P1 and P2 (these are specified in the code)

with(plots);
P1 := x^2+2*y^2-3;
solve(P1, y);
Plot1 := plot([%], x = -2 .. 2);

P2 := -2*x^2+2*x*y+3*y^2+x-4;
solve(%, y);
Plot2 := plot([%], x = -4 .. 2);

P2*a+P1;
solve(%, y);
seq(plot([%], x = -4 .. 2), a = 0 .. 10, .1);
display(%, Plot1, Plot2)

This is because when you multiply two polynomials their set of solution curves is just the union of the sets of curves associated with the previous polynomials.

For the next step I'd like to create a graph of the solutions associated with an ideal with three generators. To stop this from being excessively messy I'd like to do it with the RGB value of the colour of a curve is determined by  a and b where the formula for a generic polynomial that we are solving and graphing is given by:

P1+a*P2+b*P3;

where P3 is given by

P3 := x*y-3

I've tried various ways to use cury to make this work (my intuition is cury is the right function to use here)  but got no where. Any ideas how to procede?

## pdsolve 2D equations with unknown parameter...

possible to solve following equation with unknown parameter omega.

parameter constant.

I see before for one dimension ode this type equation was solved.

Now for 2d equation is possible?

can consider or I can send again.

Best

2d-2