Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

i want to plot

u(x,y,z,t)=tanh(x+y+z-wt),  (w is a constant)

could you help me please? 

I'm using Win10, maple 2016.

I'm using high contrast mode (dark mode, white on black). Maple looks broken, the text is black on black, and in the side panel, some of the buttons are shiney white.

How do I set maple to support high contrast?

I am trying to calculate the following integra   
r*rr*g1^2*h1^2*f1^2*fh1^2*exp(-2*t)/t

here g1 is a kummerM functin in s, and also h1 is another kummerM function  in ss, and f1 and fh1 are the HeunB functions with complex arguments in r and rr. and t=sqrt((r-rr)^2+(s-ss)^2).I would like to integte over dr drr ds dss

   Dear friends,on the left I created two Tasks named a and aa.However,I do not know how to delete these two Tasks.Then I refer to the help pages below.

The text I marked says I should enter full filename.I do not know the fullfile name of customized Tasks.So does anyone know how to delete them?

 

Using mode with axis we can choose linear or log.  But what if we want some other custom scaled axis?

Seems if we want a y^3 y-axis, we'll have to cube root the data then use plottools to form our own scale.  Unless there's some other way I don't know. 

In the help file the permitted declartions for the second argument of the Escape command from the StringTools package are html,regexp and xml, to not specify one, which as far as I can tell instructs maple to take all characters as plain text with no escape characters.

Is this the complete list of options for the form when using the escape command?

Or is it possible to custom build escape rules?

 

I thought the easiest way to show the world map, a projected flat map into 3d was to use the builtin one and just transform it.  You can zoom into it and rotate it no problem but unforunately it's not as clean as I thought.  Is it possible to have cleaner shading manipulating the Builtin map to 3d?

with(plots):
with(plottools):
with(DataSets):
with(Builtin):
m := WorldMap():
m1 := Display(m)
                                

to3d := transform((x, y) -> [x, y, 0]):
m2 := to3d(m1)
                               

display(m2)

 

 

If a maple command or function are not available on the target language  of the code generation of maple, is it possible to set myself the expected output for such cases so that the Csharp(...)  recognizes the cases and generates the expect code?

for example 

h := proc(x::Array(1 .. 3, 1 .. 3), y::Array(1 .. 3, 1 .. 3)) local z; z := evalm(x &* y); return z[1, 1] + z[2, 2]; end proc;
CSharp(h);

The function names {`&*`, evalm} can not be recognized in the target language

but for the &* it shoud be easy to add a template with the desired C# output. 

Is it possible to add templates in existing languages but not new language definitions?


 

I'm using variable names that have subscripts, not as a table index but literal i.e. R__1 as a unique variable name.  It seems whenever I make assumptions on variables that have subscripts, when I use them the variables that have subscripts are printed twice:

 

Can anyone explain why this happens and how to get around it?

 

Thanks in advance.

Is it italic when copied and pasted?  Is it bold when copied from maple 8?  I just ahve not been able to work it out.

The only way that I can think of doing it is by multiplying by a tetrad.  Even then it does not work well see my worksheet:  The Dirac Equation in Robertson-Walker spacetime.


 

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`ℏ`^2 terms for u(x, t) 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)};

{u(x, 0) = .1*sin(2*Pi*x)}

 

{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 v(x, 0) (blue) and a colliding momentum field u(x, t)(red).

 

Here are the BCs

BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t))};

{u(0, t) = .5-.5*cos(2*Pi*t)}

(2)

BCv := {v(0,t) = 0.5*sin(2*Pi*t),v(2,t)=-0.5*sin(2*Pi*t)};  

{v(0, t) = .5*sin(2*Pi*t), v(2, t) = -.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;

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

(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;  

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

(5)

ICu:={u(x,0) = 0.1*sin(2*Pi*x)};  

{u(x, 0) = .1*sin(2*Pi*x)}

(6)

ICv:={v(x,0) = 0.2*sin(Pi*x/2)};  

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

(7)

IC := ICu union ICv;  

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

(8)

BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t)), D[1](u)(2,t) = 0.1*cos(2*Pi*t)};

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

(9)

BCv := {v(0,t) = 0.2*(1-cos(2*Pi*t))};  

{v(0, t) = .2-.2*cos(2*Pi*t)}

(10)

BC := BCu union BCv;  

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

(11)

We now set up the PDE solver:

pds := pdsolve({pdeu,pdev},{BC[],IC[]},time = t,range = 0..2,numeric);#'numeric' solution

_m2592591229440

(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;

Error, (in pdsolve/numeric/animate) unable to compute solution for t>HFloat(0.0):
Newton iteration is not converging

 

Cp

(13)

Observations on the quantum case:

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

 

The initial conditions are of low momentum amplitude:"+/-"0.1 for the classical u(x, 0) (red) field and`&+-`(0).2 for v(x, 0) (blue)  but their influence is soon washed out by the boundary conditions "u(0,t) ~1, v(0,t)~0.5" and "v(1,t)~0.5" that drive the momentum dynamics.

 

The temporal frequency of the boundary condition on the v-field is twice that of the classical u-field. This is evident in the above blue transient plot. Moreover, the">=0" boundary condition on the classical u-momentum (red), drives that field in the positive direction, initially overtaking the quantum v(x, t) field, as consistent with the applied forces [0.2, 0.1]. NULLAlthough initially of greater amplitude than the classical u(x, t)field, the v(x, t) momentum field is asymptotically of the same amplitude as the u(x, t) 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 u(x, t) (red) modulates the quantum momentum rate equation for v(x, t).

``

 

 

 


 

Download SC-plots.mw

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

Please I found out that the MatrixInverse on the assignment statement P3 does not run for about three days now. Please kindly help to simplify the code. Thank you and kind regards.

restart; omega := v/h;
r := a[0]+a[1]*x+a[2]*sinh(omega*x)+a[3]*cosh(omega*x)+a[4]*cos(omega*x)+a[5]*sin(omega*x);
b := diff(r, x);

c := eval(b, x = q) = f[n];
d := eval(r, x = q+3*h) = y[n+3]; e := eval(b, x = q+3*h) = f[n+3];
g := eval(b, x = q+2*h) = f[n+2];
i := eval(b, x = q+h) = f[n+1];
j := eval(b, x = q+4*h) = f[n+4];
k := solve({c, d, e, g, i, j}, {a[0], a[1], a[2], a[3], a[4], a[5]});
Warning,  computation interrupted
assign(k);
cf := r;
s4 := y[n+4] = simplify(eval(cf, x = q+4*h));
s3 := y[n+2] = simplify(eval(cf, x = q+2*h));
s2 := y[n+1] = simplify(eval(cf, x = q+h));
s1 := y[n] = simplify(eval(cf, x = q));

with(LinearAlgebra);
with(plots);
h := 1;
YN_1 := seq(y[n+k], k = 1 .. 4);
A1, a0 := GenerateMatrix([s1, s2, s3, s4], [YN_1]);
eval(A1);
YN := seq(y[n-k], k = 3 .. 0, -1);
A0, b1 := GenerateMatrix([s1, s2, s3, s4], [YN]);
eval(A0);
FN_1 := seq(f[n+k], k = 1 .. 4);
B1, b2 := GenerateMatrix([s1, s2, s3, s4], [FN_1]);
eval(B1);
FN := seq(f[n-k], k = 3 .. 0, -1);
B0, b3 := GenerateMatrix([s1, s2, s3, s4], [FN]);
eval(B0);
ScalarMultiply(R, A1)-A0;
det := Determinant(ScalarMultiply(R, A1)-A0);
P1 := A1-ScalarMultiply(B1, z);
P2 := combine(simplify(P1, size), trig);
P3 := MatrixInverse(P2);
P4 := A0-ScalarMultiply(B0, z);
P5 := MatrixMatrixMultiply(P3, P4);
P6 := Eigenvalues(P5);
f := P6[4];
T := unapply(f, z);
implicitplot(f, z = -5 .. 5, v = -5 .. 5, filled = true, grid = [5, 5], gridrefine = 8, labels = [z, v], coloring = [blue, white]);

 

Why the performance of Maple GUI is bad. It is really hard to type anything in thew gui. 

HI

I am using a simple piece of code to generate Lie derivatives, and I am interested in adapting it so that I can use it with vector Fields where that include RootOf expressions. 

LieDer_with_rootof.mw

The above includes an example where it works, and an example where RootOf appears in the vector field and it does not work. 

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