Maple 17 Questions and Posts

These are Posts and Questions associated with the product, Maple 17

In the below paper 

ODEs are solved by using Matlab Software

Is it possible to solve this in Matlab software?  [I think you mean Maple!--Carl Love]

I have attached the Article

 

j.physa.2019.123959.pdf

Hi 
I need to represent the contourplot of sum a function as

contourplot(sum(BesselJ(n,r),n=0..5),r=-10..10).
can anyone know what the problem is ?

Thanks


I m Stucked with this step

restart; with(plots)

eq1 := 2*n*4^n*eta^((n+1)*(1/2))*(diff(f(eta), `$`(eta, 2)))^(n-1)*(diff(f(eta), `$`(eta, 3)))+4^n*(n+1)*eta^((n-1)*(1/2))*(diff(f(eta), `$`(eta, 2)))^n+4*f(eta)*(diff(f(eta), `$`(eta, 2)))-4*m*(diff(f(eta), eta))^2+m-2*M*(diff(f(eta), eta)) = 0;

2*n*4^n*eta^((1/2)*n+1/2)*(diff(diff(f(eta), eta), eta))^(n-1)*(diff(diff(diff(f(eta), eta), eta), eta))+4^n*(n+1)*eta^((1/2)*n-1/2)*(diff(diff(f(eta), eta), eta))^n+4*f(eta)*(diff(diff(f(eta), eta), eta))-4*m*(diff(f(eta), eta))^2+m-2*M*(diff(f(eta), eta)) = 0

(1)

eq2 := 2*eta*(diff(theta(eta), `$`(eta, 2)))+2*(diff(theta(eta), eta))+Pr*(f(eta)*(diff(theta(eta), eta))-s*(diff(f(eta), eta))*theta(eta))+Nb*(diff(theta(eta), eta))*(diff(phi(eta), eta))+Nt*(diff(theta(eta), `$`(eta, 2)))^2 = 0;

2*eta*(diff(diff(theta(eta), eta), eta))+2*(diff(theta(eta), eta))+Pr*(f(eta)*(diff(theta(eta), eta))-s*(diff(f(eta), eta))*theta(eta))+Nb*(diff(theta(eta), eta))*(diff(phi(eta), eta))+Nt*(diff(diff(theta(eta), eta), eta))^2 = 0

(2)

eq3 := 2*eta*(diff(phi(eta), `$`(eta, 2)))+2*(diff(phi(eta), eta))+Sc*(f(eta)*(diff(phi(eta), eta))-s*(diff(f(eta), eta))*phi(eta))+Nb*(2*eta*(diff(theta(eta), `$`(eta, 2)))+2*(diff(theta(eta), eta)))/Nt = 0;
``

2*eta*(diff(diff(phi(eta), eta), eta))+2*(diff(phi(eta), eta))+Sc*(f(eta)*(diff(phi(eta), eta))-s*(diff(f(eta), eta))*phi(eta))+Nb*(2*eta*(diff(diff(theta(eta), eta), eta))+2*(diff(theta(eta), eta)))/Nt = 0

(3)

eq4 := 2*eta*(diff(chi(eta), `$`(eta, 2)))+2*(diff(chi(eta), eta))+Lb*(f(eta)*(diff(chi(eta), eta))-s*(diff(f(eta), eta))*chi(eta))-Pe*(2*eta*chi(eta)*(diff(phi(eta), `$`(eta, 2)))+2*chi(eta)*(diff(phi(eta), eta))+2*eta*(diff(chi(eta), `$`(eta, 2)))*(diff(phi(eta), `$`(eta, 2)))) = 0;

2*eta*(diff(diff(chi(eta), eta), eta))+2*(diff(chi(eta), eta))+Lb*(f(eta)*(diff(chi(eta), eta))-s*(diff(f(eta), eta))*chi(eta))-Pe*(2*eta*chi(eta)*(diff(diff(phi(eta), eta), eta))+2*chi(eta)*(diff(phi(eta), eta))+2*eta*(diff(diff(chi(eta), eta), eta))*(diff(diff(phi(eta), eta), eta))) = 0

(4)

bcs := (D(f))(a) = 0, f(a) = 2*s*a*(D(phi))(a)/Sc, theta(a) = 1, phi(a) = 1, chi(a) = 1, (D(f))(10) = 1/2, theta(10) = 0, phi(10) = 0, chi(10) = 0;

(D(f))(a) = 0, f(a) = 2*s*a*(D(phi))(a)/Sc, theta(a) = 1, phi(a) = 1, chi(a) = 1, (D(f))(10) = 1/2, theta(10) = 0, phi(10) = 0, chi(10) = 0

(5)

params := {Lb = .1, M = .1, Nb = .6, Nt = .2, Pe = 5, Pr = 6.2, Sc = .1, a = 0.1e-1, m = 1/3, n = 1, s = .1};

{Lb = .1, M = .1, Nb = .6, Nt = .2, Pe = 5, Pr = 6.2, Sc = .1, a = 0.1e-1, m = 1/3, n = 1, s = .1}

(6)

sol := dsolve(eval([eq1, eq2, eq3, eq4, bcs], params), numeric, output = listprocedure, maxmesh = 1024)

Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

 

odeplot(sol, [[eta, f(eta)][eta, phi(eta)], [eta, theta(eta)], [eta, chi(eta)]], eta = eval(c, params) .. 10)

Error, invalid subscript selector

 

NULL


 

Download Pblm2.mw

When I try to solve this problem above error displays

Here is the file
 

Download Main.mw


When i try to solve the problem i got error like this 

Please help me get the solution

restart

with(plots):

eq1 := (2*(eta*(diff(f(eta), `$`(eta, 3)))+diff(f(eta), `$`(eta, 2))))/((1-phi[1])^2.5*(1-phi[2])^2.5)+((1-phi[2])*(1-phi[1]+phi[1]*rho[s1]/rho[f])+phi[2]*rho[s2]/rho[f])*f(eta)*(diff(f(eta), `$`(eta, 2)))-sigma[hnf]*M*(diff(f(eta), eta))/(sigma[f]*(m^2+1)) = 0;

2*(eta*(diff(diff(diff(f(eta), eta), eta), eta))+diff(diff(f(eta), eta), eta))/((1-phi[1])^2.5*(1-phi[2])^2.5)+((1-phi[2])*(1-phi[1]+phi[1]*rho[s1]/rho[f])+phi[2]*rho[s2]/rho[f])*f(eta)*(diff(diff(f(eta), eta), eta))-sigma[hnf]*M*(diff(f(eta), eta))/(sigma[f]*(m^2+1)) = 0

(1)

eq2 := k[hnf]*(eta*(diff(theta(eta), `$`(eta, 2)))+diff(theta(eta), eta))/k[f]+(1/2)*Pr*((1-phi[2])*(1-phi[1]+phi[1]*`ρC`[p][s1]/`ρC`[p][f])+phi[2]*`ρC`[p][s2]/`ρC`[p][f])*f(eta)*(diff(theta(eta), eta)) = 0;

k[hnf]*(eta*(diff(diff(theta(eta), eta), eta))+diff(theta(eta), eta))/k[f]+(1/2)*Pr*((1-phi[2])*(1-phi[1]+phi[1]*`ρC`[p][s1]/`ρC`[p][f])+phi[2]*`ρC`[p][s2]/`ρC`[p][f])*f(eta)*(diff(theta(eta), eta)) = 0

(2)

eq3 := 2*(1-phi[1])^2.5*(1-phi[2])^2.5*(eta*(diff(chi(eta), `$`(eta, 2)))+diff(chi(eta), eta))+Sc*f(eta)*(diff(chi(eta), eta))-beta*Sc*(1+delta*theta(eta))^n*exp^(-E/(1+delta*theta(eta)))*chi(eta) = 0;

2*(1-phi[1])^2.5*(1-phi[2])^2.5*(eta*(diff(diff(chi(eta), eta), eta))+diff(chi(eta), eta))+Sc*f(eta)*(diff(chi(eta), eta))-beta*Sc*(1+delta*theta(eta))^n*exp^(-E/(1+delta*theta(eta)))*chi(eta) = 0

(3)

bcs := f(c) = (1/2)*lambda*c, (D(f))(c) = (1/2)*lambda, theta(c) = 1, chi(c) = 1, (D(f))(10) = (1-lambda)*(1/2), theta(10) = 0, chi(10) = 0;

f(c) = (1/2)*lambda*c, (D(f))(c) = (1/2)*lambda, theta(c) = 1, chi(c) = 1, (D(f))(10) = 1/2-(1/2)*lambda, theta(10) = 0, chi(10) = 0

(4)

params := {E = .5, M = .1, Pr = 6.2, Sc = .5, beta = .5, c = .1, delta = .5, lambda = -1, m = .5, n = .1, k[f] = .6130, k[hnf] = 1.039626683, phi[1] = .1, phi[2] = .1, rho[f] = 997.1, rho[s1] = 5180, rho[s2] = 1800, sigma[f] = 0.5e-1, sigma[hnf] = 0.8888888337e-1, `ρC`[p][f] = 4.1668809*10^6, `ρC`[p][s1] = 3470600, `ρC`[p][s2] = 1290600};

{E = .5, M = .1, Pr = 6.2, Sc = .5, beta = .5, c = .1, delta = .5, lambda = -1, m = .5, n = .1, k[f] = .6130, k[hnf] = 1.039626683, phi[1] = .1, phi[2] = .1, rho[f] = 997.1, rho[s1] = 5180, rho[s2] = 1800, sigma[f] = 0.5e-1, sigma[hnf] = 0.8888888337e-1, `ρC`[p][f] = 4166880.900, `ρC`[p][s1] = 3470600, `ρC`[p][s2] = 1290600}

(5)

ODEs := [eq1, eq2, eq3]

[2*(eta*(diff(diff(diff(f(eta), eta), eta), eta))+diff(diff(f(eta), eta), eta))/((1-phi[1])^2.5*(1-phi[2])^2.5)+((1-phi[2])*(1-phi[1]+phi[1]*rho[s1]/rho[f])+phi[2]*rho[s2]/rho[f])*f(eta)*(diff(diff(f(eta), eta), eta))-sigma[hnf]*M*(diff(f(eta), eta))/(sigma[f]*(m^2+1)) = 0, k[hnf]*(eta*(diff(diff(theta(eta), eta), eta))+diff(theta(eta), eta))/k[f]+(1/2)*Pr*((1-phi[2])*(1-phi[1]+phi[1]*`ρC`[p][s1]/`ρC`[p][f])+phi[2]*`ρC`[p][s2]/`ρC`[p][f])*f(eta)*(diff(theta(eta), eta)) = 0, 2*(1-phi[1])^2.5*(1-phi[2])^2.5*(eta*(diff(diff(chi(eta), eta), eta))+diff(chi(eta), eta))+Sc*f(eta)*(diff(chi(eta), eta))-beta*Sc*(1+delta*theta(eta))^n*exp^(-E/(1+delta*theta(eta)))*chi(eta) = 0]

(6)

``

sol := dsolve(eval([ODES, bcs], params), numeric, output = listprocedure)

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

 

``


 

Download 1.mw
 

 

 

Hi

cant get the answer, any comments plz

f := (x, y) ->(1/2)*a*(sinh(y-x^2)+tanh(x-y^3));

evalf(int(int(f(x, y), x = -6 .. 5), y = -5 .. 5)):


One forum had a topic related to such a platform. You can download a video of the movement of this platform from the picture at this link. The manufacturer calls the three-degrees platform, that is, having three degrees of freedom. Three cranks rotate, and the platform is connected to them by connecting rods through ball joints. The movable beam (rocker arm) has torsion springs.  I counted 4 degrees of freedom, because when all three cranks are locked, the platform remains mobile, which is camouflaged by the springs of the rocker arm. Actually, the topic on the forum arose due to problems with the work of this platform. Neither the designers nor those who operate the platform take into account this additional fourth, so-called parasitic degree of freedom. Obviously, if we will to move the rocker with the locked  cranks , the platform will move.
Based on this parasitic movement and a similar platform design, a very simple device is proposed that has one degree of freedom and is, in fact, a spatial linkage mechanism. We remove 3 cranks, keep the connecting rods, convert the rocker arm into a crank and get such movements that will not be worse (will not yield) to the movements of the platform with 6 degrees of freedom. And by changing the length of the crank, the plane of its rotation, etc., we can create simple structures with the required design trajectories of movement and one degree of freedom.
Two examples (two pictures for each example). The crank rotates in the vertical plane (side view and top view)
PLAT_1.mw


and the crank rotates in the horizontal plane (side view and top view).

The program consists of three parts. 1 choice of starting position, 2 calculation of the trajectory, 3 design of the picture.  Similar to the programm  in this topic.

 

 

Controlled platform with 6 degrees of freedom. It has three rotary-inclined racks of variable length:

and an example of movement parallel to the base:

Perhaps the Stewart platform may not reproduce such trajectories, but that is not the point. There is a way to select a design for those specific functions that our platform will perform. That is, first we consider the required trajectories of the platform movement, and only then we select a driving device that can reproduce them. For example, we can fix the extreme positions of the actuators during the movement of the platform and compare them with the capabilities of existing designs, or simulate your own devices.
In this case, the program consists of three parts. (The text of the program directly for the first figure : PLATFORM_6.mw) In the first part, we select the starting point for the movement of a rigid body with six degrees of freedom. Here three equations f6, f7, f8 are responsible for the six degrees of freedom. The equations f1, f2, f3, f4, f5 define a trajectory of motion of a rigid body. The coordinates of the starting point are transmitted via disk E for the second part of the program. In the second part of the program, the trajectory of a rigid body is calculated using the Draghilev method. Then the trajectory data is transferred via the disk E for the third part of the program.
In the third part of the program, the visualization is executed and the platform motion drive device is modeled.
It is like a sketch of a possible way to create controlled platforms with six degrees of freedom. Any device that can provide the desired trajectory can be inserted into the third part. At the same time, it is obvious that the geometric parameters of the movement of this device with the control of possible emergency positions and the solution of the inverse kinematics problem can be obtained automatically if we add the appropriate code to the program text.
Equations can be of any kind and can be combined with each other, and they must be continuously differentiable. But first, the equations must be reduced to uniform variables in order to apply the Draghilev method.
(These examples use implicit equations for the coordinates of the vertices of the triangle.)

Hi friends,

I have the next list of strings and I want to convert it to a list of vectors.

 

with(StringTools)

Generate(3, "012");
["000", "001", "002", "010", "011", "012", "020", "021", "022",

  "100", "101", "102", "110", "111", "112", "120", "121", "122",

  "200", "201", "202", "210", "211", "212", "220", "221", "222"]

 

Or maybe there's an easiest way to generate all the n-ary vectors of a given lenght.

Thank's in advance.

As an addition to the post.
Non-orientable surface in the sequence of orientable surfaces. In the picture we see the equations corresponding to the current surface plot.
Just entertainment.
surfaces.mw

 

Hi, I have a problem related to the cancellation of the square root with a square in its argument, see the image below.

Thanks in advance,

Santiago

Why in geom3d[FindAngle] we cannot get the value of the angle of a triangle greater than Pi / 2?
For example, I build a chord of a circle of unit radius along the sides of the triangle and calculate the center angle that corresponds to the given angle of the triangle. But it's not very convenient.
TR_ANGLE.mw

One way to find the equation of an ellipse circumscribed around a triangle. In this case, we solve a linear system of equations, which is obtained after fixing the values of two variables ( t1 and t2). These are five equations: three equations of the second-order curve at three vertices of the triangle and two equations of a linear combination of the coordinates of the gradient of the curve equation.
The solving of system takes place in the ELS procedure. When solving, hyperboles appear, so the program has a filter. The filter passes the equations of ellipses based on by checking the values of the invariants of the second-order curves.
FOR_ELL_ТR_OUT_PROCE_F.mw  ( Fixed comments in the text  01, 08, 2020)

An attempt to find the equation of an ellipse inscribed in a given triangle. 
The program works on the basis of the ELS procedure.  After the procedure works, the  solutions are filtered.
ELS procedure solves the system of equations f1, f2, f3, f4, f5 for the coefficients of the second-order curve.
The equation f1 corresponds to the condition that the side of the triangle intersects t a curve of the second order at one point.
The equation f2 corresponds to the condition that the point x1,x2  belongs to a curve of the second order.
Equation f3 corresponds to the condition that the side of the triangle is tangent to the second order curve at the point x1,x2.
The equation f4 is similar to the equation f2, and the equation f5 is similar to the equation f3.
FOR_ELL_ТR_PROCE.mw
For example

Hi there,

Can someone explain to me why is it happening? If smaller equal than coloured well, if smaller then just the equal line appears. Why?

with(plots);
dm1 := .23; pm12 := .7; dm12 := .23;
inequal(pm1*(dm12+am21)-pm12*(dm1+am21) <= 0, pm1 = 0 .. 1, am21 = 0 .. 1, color = "Nautical 1");


 

inequal(pm1*(dm12+am21)-pm12*(dm1+am21) < 0, pm1 = 0 .. 1, am21 = 0 .. 1, color = "Nautical 1")

Thank you!

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