robertocooper

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These are replies submitted by robertocooper

thank you @vv,

when I run the following code:

restart:
local gamma:
PDE:=I*diff(u(x,t),t)+diff(u(x,t),x$2)+alpha*(abs(u(x,t))^2)*u(x,t)+ I*( gamma[1]*diff(u(x,t),x$3) + gamma[2]*(abs(u(x,t))^2)*u(x,t) + gamma[3]*diff((abs(u(x,t))^2),x)*u(x,t) )=0;

I get the following result:

what does abs(1,u(x,t))  mean?

or should I use the evalc as folows? Which one is correct defining the PDE?
 

PDE:=I*diff(u(x,t),t)+diff(u(x,t),x$2)+alpha*(evalc(abs(u(x,t))^2))*u(x,t)+ I*( gamma[1]*diff(u(x,t),x$3) + gamma[2]*(evalc(abs(u(x,t))^2))*u(x,t) + gamma[3]*diff((evalc(abs(u(x,t))^2)),x)*u(x,t) )=0;

Thank you very much Dear @Kitonum. The procedure is GREAT.

Thank you very much dear @dharr,

Yes, the Pde is complex. So, we can plot the absolute

I don't know what is the exact solution for the PDE. So, I don't guess initial or boundary conditions.

Is there no chance to find them? :)

It's great. Thank you very much @dharr,

Can we draw the 3D graphics of the solutions of the PDE for the appropriate parameters?

 

For example;

sigma:=1;
k:=-1;
c:=1;
varphi:=0;

 

Thanks for such a good procedure @Kitonum 

1. I want the rectangle should be opaque and white. Because if there is no empty space within the plot, the rectangle will overlap the lines. I don't know whether the opaque rectangular is possible or not. If it is possible, we can overcome this situation as follows:  

matlab legends ile ilgili görsel sonucu

  • In fact, I wanted to add an option to change the location of the legends like 'northeast' or 'southwest' etc. But I am not capable of achieving this.

2. Yes, you are right. If the procedure works for one function or 3 functions, etc, it will be nice. For doing this, can you help?

3. By defining G list, using the codes of Maple's plots package is a really good idea. I have not thought that or seen that before. Thanks also for this.

Thank you dear @acer, I edited the question.

I look forward to your valuable ideas for solving the question.

Best regards.

Thanks, @Rouben Rostamian , you are right.

After you said, I exported the same graphic in Maple 2016, I got the better result than Maple 2020.

Do you think which version has better EPS exporting capability in Maple 2016-2019?

Is it 2018?

Thanks for your interest Dear @Kitonum,

If it is possible, could you share an example code?

Thanks, dear @vv,

Your figure can be acceptable but the texts of axis x and y and also texts of -10,-5,...5,10 are tiny. Yes, it can be increased by Maple. 

But I will add the 3d graphics produced by Maple to the scientific articles. When I submitted the article which includes png figures like yours, the editor/reviewer wanted to correct the texts as vectorial texts. 

What would you suggest to make those texts vectorial? to use another program?

Dear @vv,

I export the above figure to a high resolution .png. But the quality of figure is still very low.

Could you share your all Maple code in order to accomplish the example above?

Thanks for your interest dear @nm,

It's a nice idea. But, I don't understand the following step:

epspdf filename.ps
pdfcrop --margins 10  filename.pdf  filename.pdf

I think you run the code in latex. But I don't see the steps above in your Latex code?

\documentclass[11pt]{article}
\usepackage{graphicx}
\begin{document}
This is my graphics generated in Maple 2020.2 

\includegraphics[width=0.9\textwidth]{file_name_here}
\end{document}

 

Thanks, @tomleslie,

My goal was to find symbolically b[0],b[1], and U(x,t,y) and then was to test it for some numerical values of beta, alpha such as beta=2,alpha=2,... and then was to plot U(x,t,y) .

On the other hand, your code is almost what I want. 

Just changing tanA:= to  tanA:=xi-> 

and changing

tanA*(sqrt(4*alpha*sigma - beta^2)/2*xi):

to

tanA(sqrt(4*alpha*sigma - beta^2)/2*xi):

I get the same results as Carl's love.

I want to ask one more question:

Suppose that b[0] has a minus and plus value such as b[0]:=±4+x+y, and suppose that we will calculate b[0]^3 and plot it (or more difficult calculations). My question is that what do you prefer in these cases? What is the trick for you?

Thanks, @Carl Love.

Your code is what I want.

 

In fact,  even if I don' t change (x,y,t)-> test to unapply(test, [x,y,t]), I get same results as yours just changing your step 1 and 2.

  1. In order to define functions, in which case do you prefer " ->" , in which case do you prefer "unapply" ?
  2. On the other hand, suppose that b[0] has a minus and plus value such as b[0]:=±4+x+y, and suppose that we will calculate b[0]^3 and plot it (or more difficult calculations). My question is that what do you prefer in these cases? What is the trick for you?

Many thanks, dear @Carl Love and @Kitonum.

I understand much better now.

 

Dear @Kitonum @Carl Love,

What is the purpose of the left quote ` and also ~ operators? 

What do you suggest to me for understanding the operators? (Examples, books, weblinks, etc.)

Thank you.

 

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