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get element of list...

Asked by: goebeld 65
September 07 2024
0  1

Dear all, how can I get the 1st which is (2) or 2nd element which is (3) of xxx ?

Regular Expression and StringTools RegSplitz...

Asked by: goebeld 65
stringtools subs incomplete-question
September 07 2024
0  1

Dear all, I'm struggeling with the regex and special characters in maple. The StringTools:-Subs works but StringTools:-RegSplit not. Why? I do have the same problem wird the double quotes.

Difficulty how to plot Numerical simulation of the...

Asked by: Paras31 245
ode dsolve numeric
September 07 2024
1  18

I am trying to create the plot for the dynamics of the discrete Klein - the equation gives Gordon equation (DKG) with friction for t=0 and t=3000.

As you can see below, creating the plot for t=0 is pretty easy. However, for t=3000, I tried to create a procedure to produce the desired plots, but when I ran the code, nothing happened. The worksheets are evaluated all the time. There are no results.

I am studying  for my thesis  the dynamics of the discrete Klein - the equation gives Gordon equation (DKG) with friction:

 

diff(U(t), t, t)-K(`U__n+!`-2*U__n+`U__n-1`)+delta*U__n+(D(W))(U__n) = 0, beta > 0, delta > 0

In the above equation, W describes the potential function:

W(U__n) = -(1/2)*`ω__d`^2*U__n^2+(1/4)*beta*`ω__d`^2*U__n^4

 

to which every coupled unit U__nadheres. In Eq. (1), the variable U__n(t)is the unknown displacement of the oscillator occupying the
n-th position of the lattice, and is the discretization parameter. We denote by the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient, while β s the coefficient of the nonlinear cubic term.

 

For the DKG chain (1), we will consider the problem of initial-boundary values, with initial conditions

 

U__n(0) = `U__n,0` and U__n(0) = `U__n,1` and `in`(`U__n,1`, 2*real^(k+2))

and Dirichlet boundary conditions at the boundary points x__0 = -(1/2)*Land "`x__k+1`=L/(2),"that is,

U__0 = `U__k+1` and `U__k+1` = 0, t >= 0*3

 

Therefore, when necessary, we will use the short notation `Δ__d`or the one-dimensional discrete Laplacian

`U__Δ__d__U__n∈ℤ` = `U__Error loading this structure.`-2*U__n+4*`U__n-1`

 

We investigate numerically the dynamics of the system (1)-(2)-(3). Our first aim is to conduct a numerical study of the property of Dynamic Stability of the system, which directly depends on the existence and linear stability of the branches of equilibrium points. 

For the discussion of numerical results, it is also important to emphasize the role of the parameter `ω__d`By changing the time variable proc (t) options operator, arrow; 1/h end proc we rewrite Eq.(1) in the form

 

diff(U(t), t, t)-`Δ__d`*U(t)+(diff(delta(x), x))*U__n = `Ω__d`(-`βU__n`^3+U__n)^2, t > 0, `Ω__d`^2 = h^2*`ω__d`^2, diff(delta(x), x) = `hδ`

 

The change in the scaling of the lattice parameter of the problem makes it important to comment here on the nature of the continuous and anti-continuous limit. For the anti-continuous limit, we need to consider the case in Eq. (1) where proc (`ω__d`) options operator, arrow; infinity end procIn the case of nonlinearity present in the governing equations, the continuous limit needs to be taken as well. On the other hand, for small values of the parameter, the system becomes significant. However, because of the model we consider, we take the asymptotic linear limit as

proc (`ω__d`) options operator, arrow; 0 end proc

 

We consider spatially extended initial conditions of the form:

U__n(0) = `U__n,0` and `U__n,0` = a*sin(`jπhn`/L), j = 1, () .. K

 

where h = L/(K+1) is the distance of the grid and  a > 0 is the amplitude of the initial condition
We also assume zero initial velocity:

 

"(`U__n`(0)=`U__n,1`=0) "

 

• 

t=0

 

restart; with(plots); a := 2; j := 2; L := 200; K := 99; h := L/(K+1); n := Vector([`$`(-(1/2)*L+h*i, i = 0 .. K+1)]); U_n0 := a*map(proc (x) options operator, arrow; sin(j*Pi*x/L) end proc, h*n)

plot([seq([n[i], U_n0[i]], i = 1 .. K+2)], style = point, color = blue, axes = boxed, gridlines, grid = [true, true], labelfont = [TIMES, 12])

  

Now I am trying to create plot for different values of a.

a = {2, 1.82, 1.85, 1.9, 1.95}, K = 99, L = 200, beta = 1, delta = 0.5e-1, `ω__d`^2 = 1

restart; with(plots)

L := 200; K := 99; j := 2; omega_d := 1; beta := 1; delta := 0.5e-1

h := L/(K+1); n := Vector([`$`(-(1/2)*L+h*i, i = 0 .. K+1)])

dt := 0.5e-1; tmax := 3000; tspan := [seq(0+dt*i, i = 0 .. trunc(tmax/dt))]

NULL

a := 1; U0 := a*map(proc (x) options operator, arrow; sin(j*Pi*h*x/L) end proc, n); U0[1] := 0; U0[K+2] := 0; Udot0 := Vector(K+2, 0)

discrete_laplacian := proc (U) local i, Laplacian; Laplacian := Vector(K+2); for i from 2 to K+1 do Laplacian[i] := U[i+1]-2*U[i]+U[i-1] end do; Laplacian[1] := 0; Laplacian[K+2] := 0; return Laplacian end proc

NULL

U := U0; Udot := Udot0; for t to num_steps do Lap := discrete_laplacian(U); Ucube := map(proc (x) options operator, arrow; x^3 end proc, U); for i from 2 to K+1 do Udot[i] := Udot[i]+dt*(2*omega_d^2*U[i]-beta*Ucube[i]-delta*Udot[i]+Lap[i]) end do; for i from 2 to K+1 do U[i] := dt*Udot[i]+U[i] end do; U[1] := 0; U[K+2] := 0; if `mod`(t, 1000) = 0 then plotU := plot(n, U, style = line, color = blue, title = cat("Displacement at t=", t*dt), labels = ["n", "Displacement"], grid = [true, true]); display(plotU) end if end do

NULL


My target is the following plots

Download Numerical_Simulation_of_discrete_Klein-Gordon_Equation.mw

Using square brackets with upper and lower integra...

Asked by: Tomás 5
2dmath typesetting subscript
September 07 2024
1  4

This is an example of what I need to display in Maple:

Notation

Is it possible to display boundary-conditions in Maple like this?

I tried to experiment with maple but I cannot type the lower and upper boundaries on top of each other. Unfortunately, they are not vertically aligned if I try to use superscript and subscript at the end of the right square bracket.

I hope there is a hack or a way to deal with this problem.

I know that the notation is a bit weird but that is how our teachers want us to display the intermediate calculations when we have to deliver our assignments. I know I could skip this intermidiate process but it is a tradition in my country to show the intermidiary calculations like I showed in the link.

Best regards.

How to count the number of operations in an object...

Asked by: 130
September 07 2024
0  2

Hi everyone!

I am trying to count the number of operations (addition, subtraction, multiplication, division) in object "A".  I have donea lot of searching around, but haven't been able to find out how to do this.  Some help would be appreciated!

Count_the_number_of_operations_in_A.mw

How to use for loops to build a matrix?...

Asked by: 130
matrix loop
September 07 2024
1  3

Hi everyone!

I am trying to build a matrix using for loops.  I have been unable to construct a matrix using one "double" for loop using variables i and j.

Instead, I have 5 "single" for loops.  Could someone show me how to do this using a "double" loop over i and j (rows and colums)?

Thanks so much!

How_to_do_this_using_one_double_loop.mw

using expand in RHS of applyrule?...

Asked by: nm 11353
expand applyrule evaluation
September 06 2024
2  1

applyrule is useful but seems limited. I can't do operation such as expand in the RHS of the rule.

For example, I wanted to applyrule that says to take sin(x::anything) and change it to cos(expand(x)), but it does not expand x. 

It seems because in the RHS of the rule, at the instance applyrule sees expand, x remains a symbol and not evaluated. So there is nothing to expand. It gets evaluated at later time, but by then too late for expand to do anything, it is gone. Just a guess.

But is there a trick to allow one to do more things in RHS of applyrule, such as expand or simplify? This would make applyrule much more useful. Otherwise, as it is, applyrule is of limited use. 

I know I can use evalindets ofcourse for this. 

> 

#expand does not work in RHS of applyrule
e1:=sin(2*sqrt(a*b)*(t+mu));
applyrule( sin(x::anything) = cos(expand(x)), e1);

sin(2*(a*b)^(1/2)*(t+mu))

cos(2*(a*b)^(1/2)*(t+mu))

> 

#but other operations worksheetdir
e1:=sin(2*sqrt(a*b)*(t+mu));
applyrule( sin(x::anything) = cos(x^2), e1);

sin(2*(a*b)^(1/2)*(t+mu))

cos(4*a*b*(t+mu)^2)

> 

expand(2*sqrt(a*b)*(t+mu))

2*(a*b)^(1/2)*t+2*(a*b)^(1/2)*mu

> 

#I could do this ofcourse
evalindets(e1,'specfunc'(sin),X->sin(expand(op(1,X))));

sin(2*(a*b)^(1/2)*t+2*(a*b)^(1/2)*mu)

 

 

Download apply_rule.mw


Using other software, there is no such problem using other operations on RHS of a rule

You see, expand worked on RHS of rule. 

I'd like to do same thing using applyrule in Maple. Is there a trick to allow this?

Maple 2024.1

How to count math operations with maple?...

Asked by: 130
September 06 2024
0  1

Hi,

When I enter this command into maple and press enter, nothing is produced.  That's the "first" problem.  However, what I am trying to do is to figure out how many calculations (multipication, addition, subtraction, division) are performed for each variable that "solve" is designed to solve.  Basically, how many math operations are performed when calculating A?

Thanks for your help!

Count_operations.mw

Procedure to show all steps used by solve, dsolve ...

Asked by: Scot Gould 1039
solve pdsolve
September 06 2024
0  2

What is the name of the procedure that provides the information about which steps Maple is using to solve an equation? 

I ask because I think pdsolve is caught in a loop, but I can't see what it is trying.

(I can't think of this name at the moment. And Google Gemini returns nonsense. So, I am hoping that by answering this question, Google Gemini will be "trained.") 

What is the technique of getting conditionaly solu...

Asked by: salim-barzani 1560
ode substitution differential-equations
September 06 2024
1  5

What is the idea  of getting equation (48),(49), and (50)

Factoring in long ODE and find parameters in bette...

Asked by: salim-barzani 1560
solve ode parameters
September 06 2024
0  4

I find thus equation of course i use a little bit seperation by hand, i have 11 equation but i can't get results of parameters i don't know why my code is a little bit old i think and i don't know there is another technique for finding them or not here is the file of mp, thanks for any help

F_P_Correct.mw

for apear each part  remove (:)

Single pendulum sweep frequency analysis...

Asked by: GFY 65
September 06 2024
0  2

Change the length of the pendulum to achieve the analysis of the pendulum frequency change, but I scan the frequency of the two images do not correspond to each other, what is the problem?

sweep.mw

Extend/Truncate list of Colours to plot ...

Asked by: Ronan 1341
color list plotting
September 05 2024
1  4

I need to automatically extend or truncate a colour list to suit number of points in plottools:-point plot.
extending the list the colour sequence could be repeated or if the is too awkard just add black.
I can't figure out how to select the list of colours to put into a new list to modify.

This will be inside a plotting procedure.

> 

restart
> 

with(ListTools)

[BinaryPlace, BinarySearch, Categorize, Classify, Collect, Deal, DotProduct, Enumerate, FindMaximalElement, FindMinimalElement, FindRepetitions, Flatten, FlattenOnce, Group, Interleave, InversePermutation, Join, JoinSequence, LengthSplit, MakeUnique, Occurrences, Pad, PartialSums, Reverse, Rotate, Search, SearchAll, SelectFirst, SelectLast, Slice, Sorted, Split, Transpose, Unpermute]

 (1)
> 

L:=[[1,2],[3,4],[-5,6],[-3,-2]]

[[1, 2], [3, 4], [-5, 6], [-3, -2]]

 (2)
> 

qty:=nops(L)

4

 (3)
> 

styles:=[symbol=solidcircle,colour=[red,green,blue,orange],symbolsize=12]

[symbol = solidcircle, colour = [red, green, blue, orange], symbolsize = 12]

 (4)
> 

 

> 

plots:-display(plottools:-point(L,styles[]))

 
> 

#Need to extend or truncate the list of colours to match the number of points
#styles can be copied to styles 1,2 etc
#the extended list would be a repeatition of the colour sequence given
#would need to look for color or colour;

has(styles,colour); #how to select colour our into a seperate list?

true

 (5)
> 

#too few point for the colours;
L1:=[[1,2],[3,4],[-5,6]]

[[1, 2], [3, 4], [-5, 6]]

 (6)
> 

plots:-display(plottools:-point(L1,styles[]))

Error, (in Plot:-Structure:-Points) number of colors must match number of points

  
> 

#too many point for the colours;
L2:=[[1,2],[3,4],[-5,6],[-3,-2],[6-1],[0,1]]

[[1, 2], [3, 4], [-5, 6], [-3, -2], [5], [0, 1]]

 (7)
> 

plots:-display(plottools:-point(L2,styles[]))

Error, (in plottools:-point) incorrect arguments for creating points structure, try providing the dimension option

  
 

 

Download 2024-09-06_Q_Extend_or_Truncate_List_of_Colours.mw

Min is bigger than max???...

Asked by: Sotto 60
optimization minimize
September 05 2024
2  6

Hello

in this example Untitled.mw Maple 2022 on MacOs  'finds' a minimum bigger than the maximum.

It looks like something is wrong...

Best Nicola

How do I mute the action of an accent applied to a...

Asked by: nmacsai 260
symbol inert accent
September 05 2024
2  2

How do I mute the action of an accent applied to a symbol so that the accented symbol is itself an inert object? See the attached example.

How_do_I_make_an_accented_symbol_inert.mw
 

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