Maple 2024 Questions and Posts

These are Posts and Questions associated with the product, Maple 2024

I want to ask a question, but I cannot submit it using the link below. I tried several times, but it didn’t work. I wrote it like this, but I didn’t get the desired answer.

http://www.mapleprimes.com/questions/231499-Changing-The-Variables-

NULL

restart:
with(PDEtools):
tr1:={x=xi- lambda*t - delta,u(x,t)=Theta(xi) };


PDE := diff(u(x,t),t) + diff(u(x,t),x) + alpha*u(x,t)^2*diff(u(x,t),x)
      + beta*diff(u(x,t), t, x$2);

ode := dchange(tr1,PDE,[xi,Theta(xi)]);

{x = -lambda*t-delta+xi, u(x, t) = Theta(xi)}

 

diff(u(x, t), t)+diff(u(x, t), x)+alpha*u(x, t)^2*(diff(u(x, t), x))+beta*(diff(diff(diff(u(x, t), t), x), x))

 

diff(Theta(xi), xi)+alpha*Theta(xi)^2*(diff(Theta(xi), xi))

(1)

NULL NULL

Download 963.mw

Hi,

I want to solve the equation shown in the image, along with its given conditions, using Maple and obtain the same results as in the image, but it does not work. Could you please help me?

NULL

restart:
interface(showassumed = 0):


df := diff(w(xi), xi) = rho + eta*w(xi)^2:

 

# Condition (1): rho * eta>0,
assume(rho * eta>0);

w1 := dsolve([df, w(0)=0]);

w(xi) = tan(xi*(rho*eta)^(1/2))*(rho*eta)^(1/2)/eta

(1)
 

 

Download 852.mw

It's possible to carve a hole through a unit cube, without splitting it into pieces, so that another unit cube can pass through that hole.  This is know as the Prince Rupert problem and was first analyzed by John Wallis, a contemporary of Isaac Newton.  Here's what the result looks like:

If your computer can play audio, have a look at  Ruperts Cube with music!

Here is the worksheet that produced the cube and the animation: ruperts-cube.mw

i have out come with different coieficent  i can do by hand one by one seperate them and make them equal to zero but how i can make  list of equation  with leading term is apear  too not just the equation i have to know which related the equation and leading term too, then find unkown coiefficient?

system.mw

i don't know how replace the function which remove the sign of integral even i know the author replaced by sin(t) and t and combined them like the picture but i don't know why my pdes not give me zero 

h-pde-M.mw

Here is a 3D plot of a hemisphere punctured by a cylinder.  I had to think a bit about how to produce this graphics in Maple.  I am posting it here for your amusement, and a challenge for your spare time.

in this pde I have to find R, and already in the paper finded but  in my openion is not correct even when i tested is not give me solution, in some paper like that i finded R variable by finding some condition in pde but i want a coding for arrange that condition for example in pde we have 4 function of t like C(t),B(t),N(t),and E(t) i have to change them for finding R  like exchange C(t)=C(t), B(t)=C(t),N(t)=-C(t),E(t)=E(t) also we can equal them to one too, this is one of the condition how many condition we can have for substitute each on like that and tested for finding R, one of this condition will make help to find that R  like  this paper

but my equation is different is this equation  in below 

i tested the author outcome and not make my pde to zero, also dipersion relation without changing any any function of t is so different and i think he just use the penlevy test variable 6/beta*diff(phi,x) as a tranformation but i think is mistake, so where is problem and how we can find all the condition 

pde-condition.mw

in this function contain another function which contain integral but when i replace is just take some of the function and other function is not acceptable i want ask  there is any way for plotting the function when i substitute all other function, is work for csc,sec,cosh, and for other not work and  infinity sign appear likesome other  like t,sin,cos,tan,tanh,sinh, so on 

 t-plot.mw

I would like to experiment with error estimation in the symbolic solution of ordinary differential equations. I've written a simple example in the attached file. I would now like to plot both the left side of the ODE and the solution y(x) together in the same coordinate system. I can't do this, and I'm asking for help. How can the cumbersome numerical terms in the solution y(x) be converted to floating-point numbers?

As I said, this is a recreational experiment ;-) .DGL_test.mw

restart

ode := x^2 = sum(x*(diff(y(x), [`$`(x, i)]))/factorial(i), i = 0 .. 3)

x^2 = x*y(x)+x*(diff(y(x), x))+(1/2)*x*(diff(diff(y(x), x), x))+(1/6)*x*(diff(diff(diff(y(x), x), x), x))

(1)

``

``

ics := y(0) = 0, (D(y))(0) = 0, ((D@@2)(y))(0) = 0

y(0) = 0, (D(y))(0) = 0, ((D@@2)(y))(0) = 0

(2)

dsolve({ics, ode})

y(x) = x-1+((1/12)*(2-2^(1/2))*(1+2^(1/2))^(2/3)+(1/12)*2^(1/2)*(1+2^(1/2))^(1/3)+1/3)*exp(((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)-1)*x)+((1/12)*(2^(1/2)-2)*(1+2^(1/2))^(2/3)-(1/12)*2^(1/2)*(1+2^(1/2))^(1/3)+2/3)*exp(-(1/2)*((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)+2)*x)*cos((1/2)*3^(1/2)*(1+2^(1/2))^(1/3)*(2^(1/2)*(1+2^(1/2))^(1/3)-(1+2^(1/2))^(1/3)+1)*x)-3^(1/2)*(-1+(2^(1/2)-1)*(1+2^(1/2))^(1/3))*2^(1/2)*exp(-(1/2)*((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)+2)*x)*sin((1/2)*3^(1/2)*(1+2^(1/2))^(1/3)*(2^(1/2)*(1+2^(1/2))^(1/3)-(1+2^(1/2))^(1/3)+1)*x)/((1+2^(1/2))^(2/3)*(12*2^(1/2)-12))

(3)

NULL``

Download DGL_test.mw

I have a lot of maple code for analysing electronic design uncertainties and would like to place a clip from the schematic before outputing the analysis results.  I have been working for hours on this with no success but I'm sure it must be simple.

I see that ImageTools,Preview kind of works but it outputs a very small, blurry image.  Is there just a simple command that I can read in my .png image and output it to the display?

undesired_textplot_format.mw

When using textplot() AND display(), how do I ensure that the expression I want in the figure is not modified by display()? See my attached spreadsheet where I demostrate this undesired behaviour. 

i did seperating but but is seems is not correct what is mistake in here?

issue1.mw

in here  for jacobieliptic i got two lambda[0] same as paper did but in second one is different and is true make my ode be zero, but is so different from paper 

test-Dr.D.mw

...but it's easily solvable with the help of a trick. My main concern is the path to the solution for the attached Diophantine equation. I was able to solve the problem both with pen and paper and then in Maple. There's certainly a more elegant way. I'm particularly interested in a special Maple command. However, I don't want to ask for it here yet, as it would give a hint of the trick and spoil the fun of solving the puzzle.

 

419*x^2+116*x*y-426*x*z+78*y^2-142*y*z+133*z^2-1604*x-682*y+1086*z+2306 = 0

NULL

Download Diophant.mw

in here i want the system which when i found the system it contain a singularity but by some changing of system we can remove it, i know how he did that but i can't do it by coding, also after removing which i did by hand , there is another problem which equalibruiom points is to long how i can make it be shorter becuase contain two function i don't know how do that can i change the coiefecient of any function by another letter? also for phase portrait i need `conserved quantity` which last code is not work for finding this kind of system ? and how decide about jacobian if the `conserved quantity` is know by that after determine the points it will be clear on plots, i write the code untill finding the `conserved quantity` becuase code not run i didn't find jacobian and phase portrait but i will put the code of jacobian and phase portrait too! 

bi-1.mw

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