Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

It would seem that if a package is loaded at the "top-level" via with() and the worksheet also has a procedure definition which contains a 'uses' statement for the same package (OK, admittedly a bit redundant), the commands from the relevant package actually "go missing".

See the output from test4() in the attached.

I can't believe that this is deliberate, because when trying to 'compartmentalise' code, then a 'uses' statement in a procedure seems like a good idea. To have this fail because the "top-level" worksheet contains a with() statement loading the same package seems perverse.

Please don't post workarounds - I already know several ways to do achieve it. I'm trying to find out if this behaviour is "deliberate" or a "bug". If the latter, it has been around for a long time because I have checked all the way back to Maple 18: every version exhibits the same behaviour

  restart:

  kernelopts(version);
  test1:= proc(M::Matrix)
               uses LinearAlgebra:
               return MatrixInverse(M):
          end proc:
  test2:= proc(M::Matrix)
               return LinearAlgebra:-MatrixInverse(M):
          end proc:
  with(LinearAlgebra):
  test3:= proc(M::Matrix)
               return MatrixInverse(M):
          end proc:
  test4:= proc(M::Matrix)
               uses LinearAlgebra:
               return MatrixInverse(M):
          end proc:
  test1( Matrix( [[1,2],[3,4]]));
  test2( Matrix( [[1,2],[3,4]]));
  test3( Matrix( [[1,2],[3,4]]));
  test4( Matrix( [[1,2],[3,4]]));

`Maple 2020.0, X86 64 WINDOWS, Mar 4 2020, Build ID 1455132`

 

Matrix(2, 2, {(1, 1) = -2, (1, 2) = 1, (2, 1) = 3/2, (2, 2) = -1/2})

 

Matrix(2, 2, {(1, 1) = -2, (1, 2) = 1, (2, 1) = 3/2, (2, 2) = -1/2})

 

Matrix(2, 2, {(1, 1) = -2, (1, 2) = 1, (2, 1) = 3/2, (2, 2) = -1/2})

 

MatrixInverse(Matrix(%id = 18446744074373391174))

(1)

 

Download usewith.mw

Please everybody, help me to make phaseportrait 3D

Maple's pdsolve() is quite capable of solving the PDE that describes the motion of a single-span Euler beam.  As far as I have been able to ascertain, there is no obvious way of applying pdsolve() to solve multi-span beams.  The worksheet attached to this post provides tools for solving multi-span Euler beams.  Shown below are a few demos.  The worksheet contains more demos.

 

A module for solving the Euler beam with the method of lines

beamsolve

 

The beamsolve proc solves a (possibly multi-span) Euler beam equation:``

"rho ((∂)^2u)/((∂)^( )t^2)+ ((∂)^2)/((∂)^( )x^2)(EI ((∂)^(2)u)/((∂)^( )x^(2)))=f"

subject to initial and boundary conditions.  The solution u = u(x, t) is the

transverse deflection of the beam at point x at time t, subject to the load
density (i.e., load per unit length) given by f = f(x, t). The coefficient rho 

is the beam's mass density (mass per unit length), E is the Young's modulus of

the beam's material, and I is the beam's cross-sectional moment of inertia

about the neutral axis.  The figure below illustrates a 3-span beam (drawn in green)
supported on four supports, and loaded by a variable density load (drawn in gray)
which may vary in time.  The objective is to determine the deformed shape of the
beam as a function of time.


The number of spans, their lengths, and the nature of the supports may be specified

as arguments to beamsolve.

 

In this worksheet we assume that rho, E, I are constants for simplicity. Since only
the product of the coefficients E and I enters the calculations, we lump the two

together into a single variable which we indicate with the two-letter symbol EI.

Commonly, EI is referred to as the beam's rigidity.

 

The PDE needs to be supplied with boundary conditions, two at each end, each

condition prescribing a value (possibly time-dependent) for one of u, u__x, u__xx, u__xxx 
(that's 36 possible combinations!) where I have used subscripts to indicate

derivatives.  Thus, for a single-span beam of length L, the following is an admissible

set of boundary conditions:
u(0, t) = 0, u__xx(0, t) = 0, u__xx(L, t) = 0, u__xxx(t) = sin*t.   (Oops, coorection, that last
condition was meant to be uxxx(L,t) = sin t.)

Additionally, the PDE needs to be supplied with initial conditions which express

the initial displacement and the initial velocity:
"u(x,0)=phi(x),   `u__t`(x,0)=psi(x)."

 

The PDE is solved through the Method of Lines.  Thus, each span is subdivided into

subintervals and the PDE's spatial derivatives are approximated through finite differences.

The time derivatives, however, are not discretized.  This reduces the PDE into a set of

ODEs which are solved with Maple's dsolve().  

 

Calling sequence:

        beamsolve(L, n, options)

 

Parameters:

        L:  List of span lengths, in order from left to right, as in [L__1, L__2 .. (), `L__ν`].

        n The number of subintervals in the shortest span (for the finite difference approximation)

 

Notes:

• 

It is assumed that the spans are laid back-to-back along the x axis, with the left end
of the overall beam at x = 0.

• 

The interior supports, that is, those supports where any two spans meet, are assumed
to be of the so-called simple type.  A simple support is immobile and it doesn't exert
a bending moment on the beam.  Supports at the far left and far right of the beam can
be of general type; see the BC_left and BC_right options below.

• 

If the beam consists of a single span, then the argument L may be entered as a number
rather than as a list. That is, L__1 is equivalent to [L__1].

 

Options:

        All options are of the form option_name=value, and have built-in default values.

        Only options that are other than the defaults need be specified.

 

        rho: the beam's (constant) mass density per unit length (default: rho = 1)

        EI: the beam's (constant) rigidity (default: EI = 1)

        T: solve the PDE over the time interval 0 < t and t < T (default: T = 1)

        F: an expression in x and t that describes the applied force f(x, t)  (default: F = 0)
        IC: the list [u(x, 0), u__t(x, 0)]of the initial displacement and velocity,  as
                expressions in x (default: IC = [0,0])

        BC_left: a list consisting of a pair of boundary conditions at the left end of
                the overall (possibly multi-span beam.  These can be any two of
                u = alpha(t), u_x = beta(t), u_xx = gamma(t), u_xxx = delta(t). The right-hand sides of these equations

                can be any expression in t.  The left-hand sides should be entered literally as indicated.

                If a right-hand side is omitted, it is taken to be zero.   (default: BC_left = [u, u_xx] which

                corresponds to a simple support).

        BC_right: like BC_left, but for the right end of the overall beam (default: BC_right = "[u,u_xx])"

 

The returned module:

        A call to beamsolve returns a module which presents three methods.  The methods are:

 

        plot (t, refine=val, options)

                plots the solution u(x, t) at time t.  If the discretization in the x direction

                is too coarse and the graph looks non-smooth, the refine option

                (default: refine=1) may be specified to smooth out the graph by introducing

                val number of intermediate points within each discretized subinterval.

                All other options are assumed to be plot options and are passed to plots:-display.

 

        plot3d (m=val, options)

                plots the surface u(x, t).  The optional m = val specification requests

                a grid consisting of val subintervals in the time direction (default: "m=25)"

                Note that this grid is for plotting purposes only; the solution is computed

                as a continuous (not discrete) function of time. All other options are assumed

                to be plot3d options and are passed to plots:-display.

 

        animate (frames=val, refine=val, options)

                produces an animation of the beam's motion.  The frames option (default = 50)

                specifies the number of animation frames.  The refine option is passed to plot
                (see the description above. All other options are assumed to be plot options and
                are passed to plots:-display.

Note:

        In specifying the boundary conditions, the following reminder can be helpful.  If the beam

        is considered to be horizontal, then u is the vertical displacement, `u__x ` is the slope,  EI*u__xx

        is the bending moment, and EI*u__xxx is the transverse shear force.

 

A single-span simply-supported beam with initial velocity

 

The function u(x, t) = sin(Pi*x)*sin(Pi^2*t) is an exact solution of a simply supported beam with

"u(x,0)=0,   `u__t`(x,0)=Pi^(2)sin(Pi x)."  The solution is periodic in time with period 2/Pi.

sol := beamsolve(1, 25, 'T'=2/Pi, 'IC'=[0, Pi^2*sin(Pi*x)]):
sol:-animate(size=[600,250]);

The initial condition u(x, 0) = 0, u__t(x, 0) = 1  does not lead to a separable form, and

therefore the motion is more complex.

sol := beamsolve(1, 25, 'T'=2/Pi, 'IC'=[0, 1]):
sol:-animate(frames=200, size=[600,250]);


 

A single-span cantilever beam

 

A cantilever beam with initial condition "u(x,0)=g(x),  `u__t`(x,0)=0," where g(x) is the
first eigenmode of its free vibration (calculated in another spreadsheet).  The motion is
periodic in time, with period "1.787018777."

g := 0.5*cos(1.875104069*x) - 0.5*cosh(1.875104069*x) - 0.3670477570*sin(1.875104069*x) + 0.3670477570*sinh(1.875104069*x):
sol := beamsolve(1, 25, 'T'=1.787018777, 'BC_left'=[u,u_x], 'BC_right'=[u_xx,u_xxx], 'IC'=[g, 0]):
sol:-animate(size=[600,250]);

If the initial condition is not an eigenmode, then the solution is rather chaotic.

sol := beamsolve(1, 25, 'T'=3.57, 'BC_left'=[u,u_x], 'BC_right'=[u_xx,u_xxx], 'IC'=[-x^2, 0]):
sol:-animate(size=[600,250], frames=100);


 

A single-span cantilever beam with a weight hanging from its free end

 

sol := beamsolve(1, 25, 'T'=3.57, 'BC_left'=[u,u_x], 'BC_right'=[u_xx,u_xxx=1]):
sol:-animate(size=[600,250], frames=100);


 

A single-span cantilever beam with oscillating support

 

sol := beamsolve(1, 25, 'T'=Pi, 'BC_left'=[u=0.1*sin(10*t),u_x], 'BC_right'=[u_xx,u_xxx]):
sol:-animate(size=[600,250], frames=100);


 

A dual-span simply-supported beam with moving load

 

Load moves across a dual-span beam.

The beam continues oscillating after the load leaves.

d := 0.4:  T := 4:  nframes := 100:
myload := - max(0, -6*(x - t)*(d + x - t)/d^3):
sol := beamsolve([1,1], 20, 'T'=T, 'F'=myload):
sol:-animate(frames=nframes):
plots:-animate(plot, [2e-3*myload(x,t), x=0..2, thickness=1, filled=[color="Green"]], t=0..T, frames=nframes):
plots:-display([%%,%], size=[600,250]);


 

A triple-span simply-supported beam with moving load

 

Load moves across a triple-span beam.

The beam continues oscillating after the load leaves.

d := 0.4:  T := 6: nframes := 100:
myload := - max(0, -6*(x - t)*(d + x - t)/d^3):
sol := beamsolve([1,1,1], 20, 'T'=T, 'F'=myload):
sol:-plot3d(m=50);
sol:-animate(frames=nframes):
plots:-animate(plot, [2e-3*myload(x,t), x=0..3, thickness=1, filled=[color="Green"]], t=0..T, frames=nframes):
plots:-display([%%,%], size=[600,250]);

z3d;


 

A triple-span beam, moving load falling off the cantilever end

 

In this demo the load move across a multi-span beam with a cantilever section at the right.

As it skips past the cantilever end, the beam snaps back violently.

d := 0.4:  T := 8: nframes := 200:
myload := - max(0, -6*(x - t/2)*(d + x - t/2)/d^3):
sol := beamsolve([1,1,1/2], 10, 'T'=T, 'F'=myload, BC_right=[u_xx, u_xxx]):
sol:-animate(frames=nframes):
plots:-animate(plot, [1e-2*myload(x,t), x=0..3, thickness=1, filled=[color="Green"]], t=0..T, frames=nframes):
plots:-display([%%,%], size=[600,250]);


 


Download worksheet: euler-beam-with-method-of-lines.mw

 

f:= (ln(4x-4)/4)-(ln(4x+4)/4)+(arctan(x)/2)

When I do plots([f(x),f(x)-2,f(x)+2]), the graphs don't extend infinitely like I want. They stop just themselves. How get I the plots to extended more? I attached a picture. 

it really confused, with generating different sample some times the command fsolve is working and some times don't . is there any other way to solve this system of equations. because I am going to generate 1000 samples and for each sample, I supposed to solve this system.
 

``

with(LinearAlgebra):
NULL

 

 

``

f[1] := VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`+`(n_x, n_y), 1/theta1), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`+`(theta2, -1), x[t]), 1/VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(theta1, theta2), x[t]), 1)), t = 1 .. n_x))), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`+`(2, R_x[t]), x[t]), 1/VectorCalculus:-`*`(VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(theta1, theta2), x[t]), 1)^VectorCalculus:-`+`(1, VectorCalculus:-`-`(1/theta2)), VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(theta1, theta2), x[t]), 1)^(1/theta2), -1), theta3))), t = 1 .. n_x))), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`+`(theta2, -1), y[T]), 1/VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(theta1, theta2), y[T]), 1)), T = 1 .. n_y))), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`+`(2, R_y[T]), y[T]), 1/VectorCalculus:-`*`(VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(theta1, theta2), y[T]), 1)^VectorCalculus:-`+`(1, VectorCalculus:-`-`(1/theta2)), VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(theta1, theta2), y[T]), 1)^(1/theta2), -1), theta4))), T = 1 .. n_y))):

N_x := 30:

n_x := 30:

NULL

R_x := [seq(0, nl = 1 .. 30)];

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

 

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

(1)

`&lambda;&lambda;` := 5; -1; QQ := 1; -1; `&alpha;&alpha;1` := 2; -1; `&alpha;&alpha;2` := 4

4

(2)

W := GenerateUniform(n_x, 0, 1):
``

fsolve({f[1], f[2], f[3], f[4]}, {theta1 = 0 .. infinity, theta2 = 0 .. infinity, theta3 = .5 .. infinity, theta4 = .5 .. infinity})

 

(3)

``

``


 

Download ask_maple.mw

I just need someone to explain to me why the following occurs:

 

 

 

showstat(rand)


rand := proc(r::{posint, numeric .. numeric}, $)
   1   if nargs = 0 then
   2     RandomTools:-MersenneTwister:-GenerateInteger()
       else
   3     try
   4       if type(r,('numeric') .. ('numeric')) then
   5         if rhs(r) < lhs(r) then
   6           error "invalid range, end points are out of order"
             elif not type(r,('integer') .. ('integer')) then
   7           return () -> RandomTools:-Generate(float(('range') = r,('method') = ('uniform')))
             end if
           end if;
   8       return RandomTools:-MersenneTwister:-NewGenerator(('range') = r)
         catch :
   9       error
         end try
       end if
end proc

 

NULL

showstat(RandomTools:-MersenneTwister:-GenerateInteger)


RandomTools:-MersenneTwister:-GenerateInteger := proc({range::{integer, integer .. integer} := 1000000000000}, $)
local bits, divisor, offset;
   1   if nargs = 0 then
   2     return MTKernelInterface(4)
       end if;
   3   if type(range,'integer') then
   4     divisor := range;
   5     offset := 0
       else
   6     if rhs(range) < lhs(range) then
   7       error "illegal range, end points out of order"
         end if;
   8     if lhs(range) = rhs(range) then
   9       return lhs(range)
         end if;
  10     offset := lhs(range);
  11     divisor := rhs(range)-offset+1
       end if;
  12   if divisor = 0 then
  13     return offset
       end if;
  14   bits := MTKernelInterface(5,divisor);
  15   MTKernelInterface(6,divisor,bits)+offset
end proc

 

NULL

showstat(MTKernelInterface)

Error, (in showstat) procedure name expected

 

``

``

 

 

Download rand_showstat.mw

Does maple provide any method for finding the saddle point of a function? Thank you

I am trying to write maple program for Gauss Siedel method. I needed two for loop. One for running the iterative method and the other for checking where (the number of iteration required) the results converge.

Please, I need help. 

 

Thanks. 

Hi guys! I'm having trouble using int with a few expressions, for example:

is returning

however simpler expressions such as

int(x^2 + x, x)

or int(x/2 + 1/2, x)

are working correctly and returning 

respectively. I'm using Maple 2020 in worksheet mode using Math 2D Input/Output.

Here's my screen, any help would be much appreciated thanks!

Hi,

I'm trying to solve for Double Inequality. (1.1) and (1.3) combined.

Thanks in Advance
 

Example: Solving a Double Inequality

 

-3 <= 6*x-1

0 <= 6*x+2

(1.1)

"(->)"

[[-1/3 <= x]]

(1.2)

6*x-1 < 3

6*x < 4

(1.3)

"(->)"[[x < 2/3]]

"ex1(x):=6 x-1"

proc (x) options operator, arrow, function_assign; 6*x-1 end proc

(1.4)

"->"

 

"->"

"ex2(y):=3"

proc (y) options operator, arrow, function_assign; 3 end proc

(1.5)

"ex3(y):=-3"

proc (y) options operator, arrow, function_assign; -3 end proc

(1.6)

NULL


 

Download Double_Inequality.mw

Hi,

I was wondering is there an easier way for me to solve an Equation Involving an Absolute Value?


 

Example: Solving an Equation Involving an Absolute Value

 

x^2-3*x = -4*x+6

x^2-3*x = -4*x+6

(1.1)

"(->)"

[[x = 2], [x = -3]]

(1.2)

-x^2+3*x = -4*x+6

-x^2+3*x = -4*x+6

(1.3)

"(->)"

[[x = 1], [x = 6]]

(1.4)

"ex7(x):=|x^(2)-3 x|+4 x-6"

proc (x) options operator, arrow, function_assign; abs(x^2-3*x)+4*x-6 end proc

(1.5)

"->"

 

``

NULL


 

Download Absolute_Value.mw

Mapple doesn' work at document mode.How can i solve the problem.(My Os is Win 10 )

Hi!
I am trying to solve ODE with bvp[midrich], and I need to prove the validity of this method. 
Where can i read about realization of bvp[midrich]. What scheme does it use? Аre there any articles on this subject?
 

 

In Maple 2019.2, I'm trying to compute series expansions. For the input
series(1/z^4/(1-z), z=0, 1);
Maple gives me a series of the expression around z=0, truncated to 1 term, i.e.
z^(-4) + O(z^(-3))

But with the input
series(1/z^4/(1-z), z=0, 2);
Maple gives me the result
z^(-4) + z^(-3) + z^(-2) + z^(-1) + 1 + z + O(z^2)

Apparently Maple has decided not to give a series with 2 terms, but a series up to O(z^2). I'm surprised that the third argument of "series" is not treated in a consistent manner. I've restarted the Maple server between the two computations, so it's not a memory effect.

In the above example, this is just a minor annoyance, but I'm actually dealing with huge expressions, and for performance reasons it's important to truncate the series expansion at the desired order and stop further computations. Has anyone encountered this problem?

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