Goal: to make a 3D parametric plot of a complex function u(x,t) obtained by pdsolve.  I tried to follow one example of a plot of a complex expression/function.  I was unable to get any plot.  I got some warning and error.  Please let me know what have I done wrong.

I follow this example so that I can get a 3D parametric plot of a complex function u(x,t) for a fix x:

with(plots);

complexplot(exp(t*I), t = 0 .. 2*Pi, scaling = constrained);

plot([cos(t), sin(t), t = 0 .. 2*Pi], scaling = constrained);

z := t -> exp(t*I);

spacecurve([Re(z(t)), Im(z(t)), t], t = 0 .. 4*Pi, axes = normal, labels = ["Re", "Im", "t"]);

It was successful.  With the following pde, I obtain an analytical solution using pdsolve.

a := 3;
L := 2*Pi;
d := 0.5;
T0 := 2*Pi;
PDE := diff(u(x, t), t, t) = a^2*diff(u(x, t), x, x) - d*diff(u(x, t), t);
IBC := u(x, 0) = cos(Pi*x/L), D[2](u)(x, 0) = 0, D[1](u)(0, t) = 0, D[1](u)(L, t) = 0;
pds := pdsolve({IBC, PDE}, u(x, t));

It fail 2D plotting using the following:

with(plots);
complexplot(u(0, t), t = 0 .. 6);

So I break down the pds with the following:

RP := Re(pds);
IP := Im(pds);
RP0 := eval(RP, x = 0);
RP0t := unapply(RP0, t);
IP0 := eval(IP, x = 0);
IP0t := unapply(IP0, t);

And I was hopeful as the following give me real numbers when I approximate them:

RP0t(1);
                             "(->)"

                     Re(u(0, 1)) = 0.20248

RP0t(2);
                             "(->)"

                     Re(u(0, 2)) = -0.57765

But it fails in both 2D and 3D plot:

plot([RP0t(t), t = 0 .. 6])

spacecurve([RP0t(t), IP0t(t), t], t = 0 .. 6, axes = normal)

Dear sir,

i want to draw velocity and temperature plots for fluid and dust phases. but getting error in that showing 12 BC need only 8 
i have tried with 8 Bc also but showing different error how to rectify it

dust_phase_work_error.mw 

I have two equations developed in different ways and I want to compare them.
The first is a development done using maple, while the second is an algebraic analysis.

I need to find out if both are identical and, therefore, if the subtraction would be equal to zero.

How should I proceed?

I had thought about simplifying the first and comparing it with the second; is that a good approach?

Download TESTE_MAPLE.mw

my question is a little bit long but is not complicated, i want find thus  unkown but  realy i am don't know how apply on it by maple, i have two best paper which explain very well i just want to find thus dimensional Lie algebra which is be invariant or not satisfy condition or not which i have to used or which i have not to use it also the importan part how find them in paper 1 first , for equation fisher 
How find eq(29) which is i think is two -dimensional Lie algebra of equation, also the best part is reduction which by apply this we can change PDE to ode but i don't know how apply eq(31) or even find it yet  is related to eq(27-28)  and by replacing equation eq(34) we can get our ode i am just loking for the ode, for the eq(76) and eq(85) have same procedures,  i will mention the paper link too 

Lie.mw

paper-1

paper-2

given

ode:=x^2*diff(y(x),x)+cos(2*y(x)) = 1; 
ic:=y(infinity) = 5/4*Pi; 
mysol:=cot(y(x))=2/x+1;

the above solution is verified against the ode, but Maple do not give zero for the IC part. It gives 

                                 [0, -Pi*_Z6 + Pi]

But we can see the IC are verified also

limit(eval(mysol,y(x)=5/4*Pi),x=infinity)

Gives

                                    1 = 1

my  code checks if the entry in the result of odetest is zero or not. Hence for -Pi*_Z6 + Pi it thinks now the solution is wrong because it is not zero.   Ofcourse looking at it on the screen, we see that for _Z6=1 then it becomes zero.  

The question, why Maple odetest do not return zero for the IC part?

eq:=cot(y)=-A;
solve(eq,y)

eq:=cot(y)=A;
solve(eq,y)

I would expect solution to first to be arccot(-A) and for second to be arccot(A) but Maple likes to write the solution for the first one as Pi-arccot(A).

Of course Maple solution is correct. But why make it so complicated? Why not just give arccot(-A) as solution?  A is just a symbol.

I tried 4 different cas systems and they all give arccot(-A) for first one, except Maple gives Pi-arccot(A)

Is there some subtle reason why Maple gives solution like this instead of the simpler one? 

And is there a way to tell Maple not to do this? This seems to be something hardcoded internally in its automatic trig simplifications? I just do not see the point of writing it this way. May be someone knows why.

How can I display numerical values alongside a plot as shown in the sample below? What is the syntax?

Q_legend.mw

Hi,

I am trying to reproduce a polar curve, as shown in the attached picture.
I have tested several polar expressions and different plotting options in Maple, but I can’t manage to obtain an exact match, 

If you have any ideas about a possible polar equation, a combination of functions, or plotting parameters (sampling, domain, polarplot options, etc.), I would be very grateful.

Thanks in advance for your help!

Q_Polar.mw

Dear Power Users, I do hope someone is willing to help me out. I want to do the analysis of 3 cycles of an extension-load measurement. I would like to calculate:

- the slope of the upgoing part of the cycle between 1 and 4 N (loading stiffness)

- the slope of the unloading part of the cycle between 4 and 1 N

- the surface of each cycle.

- the extension at the minimum and maximum value of each cycle

I did this before using a special addin of mathcad 15 but as this is no longer available I hope to do this in maple. 

AnalysisCyclicData2.maple

Hi

I’m currently trying to animate a graph with about 50 vertices of this type, and I was thinking of using the GraphTheory package for that purpose.

If you have any suggestions or possible directions to explore, I’d be very happy to hear them.

Thanks again for all the helpful tips and support — they’re truly appreciated
 

I want to present my regional plot similar to the sample shown, using appropriate legends and labels. What syntax should I use to achieve this? Also, the axis labels need to be clear and bold, as they currently appear faded. What modifications should I make in the syntax?

Q1.mw

SAMPLE:

according to help under Explore, it says

I wanted to add title at top of the Explore window. But title do not show. 

Explore(plot(a*sin(b*t),t=0..10),
   parameters=[              
        [a=1..20],
        [b=1..20]
        ],
  initialvalues=[a=6,b=5],placement='right',width=300,'title'="My Explore");

I was expecting My Explore to show at top of frame above. What Am I doing wrong?

Maple 2025.2

Is this a bug?  Or one can not mix slider for one variable and textarea for second variable in Explore?

Or do I need to initialize the textarea is special way? Nothing in help shows an example with textarea and slider together.

Download explor_feb_19_2026.mw

I am not able to find an example or how to do this in help.

In Explore, when making a slider, using say 

parameters=[  [ a=1..1000 ], .....]

Is there is a way to tell it to increment "a" by say 10 for each movement of the slider using the mouse? it seems to default in this case to 1 each time.

Actually it is very hard now to make the slider go to the exact value I want. Is there a way to better control what value "a" should be?

In Mathematica, slider supports increment. Also it is possible to open each slider and then click on + and -  in the small window that opens below the slider, where now each click on + now moves the slider by the increment one decides on.  But in Maple I see no place to tell it what increment to use?

If I try this (similar to when using seq command, as in seq(i, i=1..1000,10), it does not work

[ a=1 .. 100, 10 ]

Hoping it will increment/decrement by 10 each time, it gives error.

Imagine a slider that goes from 1 to 10,000. Then how will one control the slider to be say exactly 5,550 ?  It is impossible now. Here is an example

Explore(plot(a*sin(x),x=-3*Pi..3*Pi,'gridlines'),
   parameters=[[a=1 .. 10000]],
   initialvalues=[a=1000],
   width=500,placement='right');

Now try using the slider above to make it go to a=5550. 

very hard to control the slider as it is very sensetive to slider motion, a small touch makes it move by large amount.   

I was hoping that if I can make it move by say 10 at time, so it will be at least easier to control.

Any suggestions?

I see a web page called Slider Component   but have no idea if this is related to Explor or not and how to use it there.

I know I can change the slider to "controller = textarea", and this way I can type in the value I want instead of using a slider. But want to see first if standard slider has a way to allow changing the increment amout. If not, then may be will use textarea instead of slider.

I define coordinates, a metric, and a tensor. Works good.

I calculate a covariant derivative, and I get an answer in terms of an unevaluated sum of Christoffel symbols. Calculating a Christoffel symbol works well. How to evaluate the result of the covariant derivative?

Thanks in advance.