Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hello!

 

With Maple 2015, license Student, is it possible calculate distance between point and plane in R³ ? Package geom3d is recommended for this or another? How can I calculate this?

 

Thank you so much!

Hello

anyone plz help me to find adomian decomposition method (ADM) maple code, also how to find the adomian polynomials separately using maple, i search it for a long time but did'nt succeeded,


The PDE & BC project , a very nice and challenging one, also one where Maple is pioneer in all computer algebra systems, has restarted, including now also the collaboration of Katherina von Bülow.

Recapping, the PDE & BC project started 5 years ago implementing some of the basic methods found in textbooks to match arbitrary functions and constants to given PDE boundary conditions of different kinds. At this point we aim to fill gaps, and the first one we tackled is the case of 1st order PDE that can be solved without boundary conditions in terms of an arbitrary function, and where a single boundary condition (BC) is given for the PDE unknown function, and this BC does not depend on the independent variables of the problem. It looks simple ... It can be rather tricky though. The method we implemented is a simple however ingenious use of differential invariants  to match the boundary condition.


The resulting new code, the portion already tested, is available for download in the Maplesoft R&D webpage for Differential Equations and Mathematical Functions (the development itself is bundled within the library that contains the new developments for the Physics package, in turn within the zip linked in the webpage).


The examples that can now be handled, although restricted in generality to "only one 1st order linear or nonlinear PDE and only one boundary condition for the unknown function itself", illustrate well how powerful it can be to use more advanced methods to tackle these tricky situations where we need to match an arbitrary function to a boundary condition.


To illustrate the idea, consider first a linear example, among the simplest one could imagine:

PDEtools:-declare(f(x, y, z))

f(x, y, z)*`will now be displayed as`*f

(1)

pde := diff(f(x, y, z), x)+diff(f(x, y, z), y)+diff(f(x, y, z), z) = f(x, y, z)

diff(f(x, y, z), x)+diff(f(x, y, z), y)+diff(f(x, y, z), z) = f(x, y, z)

(2)

Input now a boundary condition (bc) for the unknownf(x, y, z) such that this bc does not depend on the independent variables {x, y, z}; this bc can however depend on arbitrary symbolic parameters, for instance

bc := f(alpha+beta, alpha-beta, 1) = alpha*beta

f(alpha+beta, alpha-beta, 1) = alpha*beta

(3)

With the recent development, this kind of problem can now be solved in one go:

sol := pdsolve([pde, bc])

f(x, y, z) = (1/4)*(x-2*z+2+y)*(x-y)*exp(z-1)

(4)

Nice! And how do you verify this result for correctness? With pdetest , which actually also tests the solution against the boundary conditions:

pdetest(sol, [pde, bc])

[0, 0]

(5)

And what has been done to obtain the solution (4)? First the PDE was solved regardless of the boundary condition, so in general, obtaining:

pdsolve(pde)

f(x, y, z) = _F1(-x+y, -x+z)*exp(x)

(6)

In a second step, the arbitrary function _F1(-x+y, -x+z) got determined such that the boundary condition f(alpha+beta, alpha-beta, 1) = alpha*beta is matched. Concretely, the mapping _F1 is what got determined. You can see this mapping reversing the solving process in two steps. Start taking the difference between the general solution (6) and the solution (4) that matches the boundary condition

(f(x, y, z) = _F1(-x+y, -x+z)*exp(x))-(f(x, y, z) = (1/4)*(x-2*z+2+y)*(x-y)*exp(z-1))

0 = _F1(-x+y, -x+z)*exp(x)-(1/4)*(x-2*z+2+y)*(x-y)*exp(z-1)

(7)

and isolate here _F1(-x+y, -x+z)

PDEtools:-Solve(0 = _F1(-x+y, -x+z)*exp(x)-(1/4)*(x-2*z+2+y)*(x-y)*exp(z-1), _F1(-x+y, -x+z))

_F1(-x+y, -x+z) = (1/4)*exp(-x+z-1)*(x^2-2*x*z-y^2+2*y*z+2*x-2*y)

(8)

So this is the value _F1(-x+y, -x+z) that got determined. To see now the actual solving mapping _F1, that takes for arguments -x+y and -x+z and returns the right-hand side of (8), one can perform a change of variables introducing the two parameters `τ__1` and `τ__2` of the _F1 mapping:

{tau__1 = -x+y, tau__2 = -x+z, tau__3 = z}

{tau__1 = -x+y, tau__2 = -x+z, tau__3 = z}

(9)

solve({tau__1 = -x+y, tau__2 = -x+z, tau__3 = z}, {x, y, z})

{x = -tau__2+tau__3, y = -tau__2+tau__1+tau__3, z = tau__3}

(10)

PDEtools:-dchange({x = -tau__2+tau__3, y = -tau__2+tau__1+tau__3, z = tau__3}, _F1(-x+y, -x+z) = (1/4)*exp(-x+z-1)*(x^2-2*x*z-y^2+2*y*z+2*x-2*y), proc (u) options operator, arrow; simplify(u, size) end proc)

_F1(tau__1, tau__2) = -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2)

(11)

So the solving mapping _F1 is

_F1 = unapply(rhs(_F1(tau__1, tau__2) = -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2)), tau__1, tau__2)

_F1 = (proc (tau__1, tau__2) options operator, arrow; -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2) end proc)

(12)

Wow! Although this pde & bc problem really look very simple, this solution (12) is highly non-obvious, as is the way to get it just from the boundary condition f(alpha+beta, alpha-beta, 1) = alpha*beta and the solution (6) too. Let's first verify that this mapping is correct (even when we know, by construction, that it is correct). For that, apply (12) to the arguments of the arbitrary function and we should obtain (8)

(_F1 = (proc (tau__1, tau__2) options operator, arrow; -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2) end proc))(-x+y, -x+z)

_F1(-x+y, -x+z) = -(1/4)*exp(-x+z-1)*(-x+y)*(x-2*z+2+y)

(13)

Indeed this is equal to (8)

normal((_F1(-x+y, -x+z) = -(1/4)*exp(-x+z-1)*(-x+y)*(x-2*z+2+y))-(_F1(-x+y, -x+z) = (1/4)*exp(-x+z-1)*(x^2-2*x*z-y^2+2*y*z+2*x-2*y)))

0 = 0

(14)

Skipping the technical details, the key observation to compute a solving mapping is that, given a 1st order PDE where the unknown depends on k independent variables, if the boundary condition depends on k-1 arbitrary symbolic parameters alpha, beta, one can always seek a "relationship between these k-1parameters and the k-1differential invariants that enter as arguments in the arbitrary function _F1 of the solution", and get the form of the mapping _F1 from this relationship and the bc. The method works in general. Change for instance the bc (3) making its right-hand side be a sum instead of a product

bc := f(alpha+beta, alpha-beta, 1) = alpha+beta

f(alpha+beta, alpha-beta, 1) = alpha+beta

(15)

sol := pdsolve([pde, bc])

f(x, y, z) = (x-z+1)*exp(z-1)

(16)

pdetest(sol, [pde, bc])

[0, 0]

(17)

An interesting case happens when the boundary condition depends on less than k-1 parameters, for instance:

bc__1 := subs(beta = alpha, bc)

f(2*alpha, 0, 1) = 2*alpha

(18)

sol__1 := pdsolve([pde, bc__1])

f(x, y, z) = ((x-z+1)*_C1+x-y)*exp(((z-1)*_C1+y)/(1+_C1))/(1+_C1)

(19)

As we see in this result, the additional difficulty represented by having few parameters got tackled by introducing an arbitrary constant _C1 (this is likely to evolve into something more general...)

pdetest(sol__1, [pde, bc__1])

[0, 0]

(20)

Finally, consider a nonlinear example

PDEtools:-declare(u(x, y))

u(x, y)*`will now be displayed as`*u

(21)

pde := 3*(u(x, y)-y)^2*(diff(u(x, y), x))-(diff(u(x, y), y)) = 0

3*(u(x, y)-y)^2*(diff(u(x, y), x))-(diff(u(x, y), y)) = 0

(22)

Here we have 2 independent variables, so for illustration purposes use a boundary condition that depends on only one arbitrary parameter

bc := u(0, alpha) = alpha

u(0, alpha) = alpha

(23)

All looks OK, but we still have another problem: check the arbitrary function _F1 entering the general solution of pde when tackled without any boundary condition:

pdsolve(pde)

u(x, y) = RootOf(-y^3+3*y^2*_Z-3*y*_Z^2+_Z^3-_F1(_Z)-x)

(24)

Remove this RootOf to see the underlying algebraic expression

DEtools[remove_RootOf](u(x, y) = RootOf(-y^3+3*y^2*_Z-3*y*_Z^2+_Z^3-_F1(_Z)-x))

-y^3+3*y^2*u(x, y)-3*y*u(x, y)^2+u(x, y)^3-_F1(u(x, y))-x = 0

(25)

So this is a pde where the general solution is implicit, actually depending on an arbitrary function of the unknown u(x, y) The code handles this problem in the same way, just that in cases like this there may be more than one solution. For this very particular bc (23) there are actually three solutions:

pdsolve([pde, bc])

u(x, y) = x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)-((1/2)*I)*3^(1/2)*x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)+((1/2)*I)*3^(1/2)*x^(1/3)+y

(26)

Verify these three solutions against the pde and the boundary condition

map(pdetest, [u(x, y) = x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)-((1/2)*I)*3^(1/2)*x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)+((1/2)*I)*3^(1/2)*x^(1/3)+y], [pde, bc])

[[0, 0], [0, 0], [0, 0]]

(27)

:)


Download PDEs_and_Boundary_Conditions.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

I have an expression of derrivative, lets say (D[2, 2](G))(x, y).

How to get this numbers - [2,2] programmicaly from it?

hi maple is new to me.

 

i am trying to make a sinus graph f(x)=0,84*sin(0,017*x-1,3)+12,13

 

But its not possible to get the graph, 

anywhere i can see how to do this so the graph is made?.

I made reference to the show the enemy below and included part of the script an interesting part in the show. 

I did a mapleprimes search - only two returns none of which were what I was looking for.  I also did a google search which turned up nothing. 

Was my post deleted?

Hello. Using a symbol ":", you can hide the derivation of the expression. However, it lost the label of expression. Can I hide the expression of retaining the label for further use? It would be handy when working with large expressions.

Hello,

the attached code return error for  this values [0, 0.2, 0.4, 0.6] but work perfectly for this [0.2, 0.4, 0.6, 0.8]. I want it to start from 0 value. How do i correct the error? See the worksheet here com.mw

Thanks.

The first instalment in the Hollywood Math webinar series will be returning live this September 17th! It will be presented by Daniel Skoog, our Maple Product Manager.

Over its storied and intriguing history, Hollywood has entertained us with many mathematical moments in film. John Nash in “A Beautiful Mind,” the brilliant janitor in “Good Will Hunting,” the number theory genius in “Pi,” and even Abbott and Costello are just a few of the Hollywood “mathematicians” that come to mind.

During this webinar we will present a number of examples of mathematics in film, including those done capably, as well as questionable and downright “creative” treatments. See relevant, exciting examples that you can use to engage your students, or attend this webinar simply for its entertainment value. Have you ever wondered if the bus could really have jumped the gap in “Speed?” We’ve got the answer! Anyone with an interest in mathematics, especially high school and early college math educators, will be both entertained and informed by attending this webinar.

To join us for the live presentation, click here to register.

Is there a maple function that'll allow me to generate the matrix M(5 by 5) below given that matrix A and B are 3 by 3 M:=LinearAlgebra:-ZeroMatrix(5) A:=Matrix([[a11,a12,a13],[a21,a22,a33],[a31,a32,a33]]) B:=Matrix([[b11,b12,b13],[b21,b22,b33],[b31,b32,b33]]) Matrix(5, 5, {(1, 1) = a[11], (1, 2) = a[12], (1, 3) = a[13], (1, 4) = 0, (1, 5) = 0, (2, 1) = a[12], (2, 2) = a[22], (2, 3) = a[23], (2, 4) = 0, (2, 5) = 0, (3, 1) = a[13], (3, 2) = a[23], (3, 3) = a[33]+b[11], (3, 4) = b[12], (3, 5) = b[13], (4, 1) = 0, (4, 2) = 0, (4, 3) = b[12], (4, 4) = b[22], (4, 5) = b[23], (5, 1) = 0, (5, 2) = 0, (5, 3) = b[13], (5, 4) = b[23], (5, 5) = b[33]});

Good day everyone,

 

I want to construct groebner bases over rings of differential Operators.

Thus I used the following:

 

with(Ore_algebra);

with(Groebner); N := 3;

A := skew_algebra(diff = [D[1], x[1]], diff = [D[2], x[2]], diff = [D[3], x[3]], comm = i, alg_relations = i^2+1); T := MonomialOrder(A, tdeg(D[1], D[2], D[3]));

A["polynomial_indets"];  [returns    {D[1], D[2], D[3]}]

A["rational_indets"];       [returns    {i, x[1], x[2], x[3]}]

 

So far everything seems as it should be: The ring i wanted to define here is the third Ring of  Differential operators over the field of complex 'rational' functions and maple returns that indeed it will handle the D[i] as monomials and the rest as coefficients for them.

 

Hover, when i use the Skew product, the following happens:

skew_product(x[1],D[1],A)                    [returns D[1]x[1]+1}]

skew_product(x[1],D[1],A)                    [returns D[1]x[1]]

Both is wrong, but maple seems to 'know' this. It used the correct relation to in the first product, the only thing it didn't do was switch D[1] and x[1]. I think maple handles the result of the skew product as if it was a commutative product and always places the D[i] at the left but still 'knows' what the actual result is.

The actual results should have been x[1]D[1]+1 for the first and x[1]*D[1] for the second product.

 

In the second product, it seems like maple treated x[1]*D[1] as if the Elements were switched already.

 

What i want though is for maple to correctly display the skew products and return the products so that in every summand the D[i] are at the right side and their coefficients are at the left side (and if possible (i do not know how to do that yet) sort the result of a skew product in a way that displays every different Power product of the D[i] with their coefficients, so that i get (x[1]+x[2])*D[2] and not x[1]*D[2]+x[2]*D[2]).

 

Can anyone help me here?

 


                              
                     
                   
                 

Hi all,

I want use a timer to let Maple program run automatically.

Timer will make Maple run by a time step. Maple will plot photo base on a datadase, the database is a SQL database.

The timer is from Jave or somewhere we can used.

Please help me.

Thanks

Hi dear friends 

Is there an interactive package management utility or a way for solving following problem?

for m>=4 It dosent work!

restart:
Digits :=30: m := 3: g :=0.3: nu := 0.2: a := 1:
w := sum(b[n]*cos(n*r), n = 1 .. m):
W := simplify( subs( solve( { subs(r = 1, diff(w, r$2)+nu*diff(w,r))},{ b[1] }),w)):
d1:=diff(W,r):
d2:=diff(d1,r):
F:= int( ((d2+d1/r)^2-(2*(1-nu))*d2*d1/r)*r*(1+g*r/a)^3,r = 0 .. a) /int( d1^2*r,r = 0 .. a):
FW := simplify( subs(solve( {seq(subs(r=n/m,diff(F,b[n])),n=2..m)},{ seq(b[j], j = 2 .. m) }),F));

 

Hello.

 

Bellow I have an example of my current problem:

I want to put a sequence of numbers in the place of Tau in the same command, so that it returns the answers in the same line divided by commas. On the picture I've done the closest I could to what I want, but if you know how to solve this, I think it would be cleaner to use an alternative.

Screenshot of the problem

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