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      Method for solving underdetermined systems of nonlinear equations. The idea of the method is to find a connected subset of a set of solutions of the system by moving along this subset from one point in different directions. The direction of movement can be changed in each point.

      Very simple example of  single equation with three variables:

                                   (x1 ^ 4 + x2 ^ 4 - 2) ^ 2 + x3 ^ 4 - 1 = 0;

      From the point (0, -1.31607, 0) or (0, 1., 0) or any point if it is a solution, we first move for a variety of solutions along a curve parallel to the axis Ox3, and then from each point of this curve is moving in a direction parallel to x1Ox2 or vice versa. So we get all the solutions.
      This works for any space of any number of the equations when the number of equations is less than the number of variables.
underdetermined_system.mw

 

 

 

Hi guys, 

       After my calculations, I got the Lagrange function having three gereralized co-ordinates namely u,v and w. Hence, according to theory it should give three equations of motion. 

      I have trouble finding equations of motion from following Lagrange function(L = T -U), Can anyone guide me with this?

      Moreover, kinetic energy and strain energy equations are in the form of double integral!! 

 

 

hi,

I have bought maple18 student edition. I want to learn GPU programming through Maple. Please suggest how to do this. I have a notebook with i7 processor and NVIDIA geforce 750m graphics. I want to solve system of algebraic equations, integral equations etc in parallel using GPUs.

thanks

The equations of motion for a rigid body can be obtained from the principles governing the motion of a particle system. Now we will solve with Maple.

 

Dinamica_plana_de_cuerpos_rigidos.mw

(in spanish)

Atte.

Lenin Araujo Castillo

Corrección ejercico 4

 

4.- Cada una de las barras mostradas tiene una longitud de 1 m y una masa de 2 kg. Ambas giran en el plano horizontal. La barra AB gira con una velocidad angular constante de 4 rad/s en sentido contrario al de las manecillas del reloj. En el instante mostrado, la barra BC gira a 6 rad/s en sentido contrario al de las manecillas del reloj. ¿Cuál es la aceleración angular de la barra BC?

Solución:

restart; with(VectorCalculus)

NULL

NULL

m := 2

L := 1

theta := (1/4)*Pi

a[G] = x*alpha[BC]*r[G/B]-omega[BC]^2*r[G/B]+a[B]NULL

NULL

a[B] = x*alpha[AB]*r[B/A]-omega[AB]^2*r[B/A]+a[A]

NULL

aA := `<,>`(0, 0, 0)

`&alpha;AB` := `<,>`(0, 0, 0)

rBrA := `<,>`(1, 0, 0)

`&omega;AB` := `<,>`(0, 0, 4)

aB := aA+`&x`(`&alpha;AB`, rBrA)-4^2*rBrA

Vector[column](%id = 4411990810)

(1)

`&alpha;BC` := `<,>`(0, 0, `&alpha;bc`)

rGrB := `<,>`(.5*cos((1/4)*Pi), -.5*sin((1/4)*Pi), 0)

aG := evalf(aB+`&x`(`&alpha;BC`, rGrB)-6^2*rGrB, 5)

Vector[column](%id = 4412052178)

(2)

usando "(&sum;)M[G]=r[BC] x F[xy]"

rBC := `<,>`(.5*cos((1/4)*Pi), -.5*sin((1/4)*Pi), 0)

Fxy := `<,>`(Fx, -Fy, 0)

NULL

`&x`(rBC, Fxy) = (1/12*2)*1^2*`&alpha;bc`

(.2500000000*sqrt(2)*(-.70710*`&alpha;bc`-25.456)+(.2500000000*(57.456-.70710*`&alpha;bc`))*sqrt(2))*e[z] = (1/6)*`&alpha;bc`

(3)

 

"(&sum;)Fx:-Fx=m*ax"           y             "(&sum;)Fy:Fy=m*ay"

ax := -28.728+.35355*`&alpha;bc`

-28.728+.35355*`&alpha;bc`

(4)

ay := .35355*`&alpha;bc`+12.728

.35355*`&alpha;bc`+12.728

(5)

Fx := -2*ax

57.456-.70710*`&alpha;bc`

(6)

Fy := 2*ay

.70710*`&alpha;bc`+25.456

(7)

`&x`(rBC, Fxy) = (1/12*2)*1^2*`&alpha;bc`

(.2500000000*sqrt(2)*(-.70710*`&alpha;bc`-25.456)+(.2500000000*(57.456-.70710*`&alpha;bc`))*sqrt(2))*e[z] = (1/6)*`&alpha;bc`

(8)

.2500000000*sqrt(2)*(-.70710*`&alpha;bc`-25.456)+(.2500000000*(57.456-.70710*`&alpha;bc`))*sqrt(2) = (1/6)*`&alpha;bc`

.2500000000*2^(1/2)*(-.70710*`&alpha;bc`-25.456)+(14.36400000-.1767750000*`&alpha;bc`)*2^(1/2) = (1/6)*`&alpha;bc`

(9)

"(->)"

[[`&alpha;bc` = 16.97068481]]

(10)

NULL

 

Download ejercicio4.mw

Hi,

I'm pretty new into MAPLE andI'm trying to get into it.

I have these four equations:


eq1:=1.6*10^(-7)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)-1.6*10^(-14)*R^2*cos(t)^2+4.2*10^(-14)*R^2-1.3+2.1*10^(-9)*R*cos(t)=0; eq2 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.9*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0; eq3 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.2*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0; eq4 :=2.1*10^(-9)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)+1.6*10^(-14)*R^2*cos(t)^2+2.6*10^(-14)*R^2-1.3+1.6*10^(-7)*R*cos(t)=0;
 

Hello friends;

I want to solve 3 - second orger equations with  6 unkowns. 

When i say just dsolve without boundry conditions it is solving but so complicated. 

_C6Ci  it is giving like that which i dont know the meaning of Ci.

But with boundry conditions it is not solving .

May you help me please where i am doing wrong. I downloaded the file.

Thanks.

Curvedbeam_static_variablethickness_analitik4.mw

At the internet site of The Heun Project, a strong declaration is made that only Maple incorporates Heun functions, which arise in the solution of differential equations that are extremely important in physics, such as the solution of Schroedinger's equation for the hydrogen atom.  Indeed solutions appear in Heun functions, which are highly obscure and complicated to use because of their five or six arguments, but when one tries to convert them to another function, nothing seems to work.  For instance, if one inquires of FunctionAdvisor(display, HeunG), the resulting list contains

"The location of the "branch cuts" for HeunG are [sic, is] unknown ..." followed by several other "unknown" and an "unable". Such a solution of a differential equation is hollow.

Incidentally, Maple's treatment of integral equations is very weak -- only linear equations with simple solutions, although procedures by David Stoutemyer from 40 years ago are available to enhance this capability.

When can we expect these aspects of Maple to work properly, for applications in physics?

I've got the following equations :

x^3-4x=y, y^3-3y=x that I want to plot together over an appropriate range for x and y

I've tried plots[multiple]([eq1,x=-2..2,y=-2..2],[eq2,x=-2..2,y=-2..2]) 

but that doesn't seem to work. How can I rectify this error?

I've got a pair of equations :

x^3-4x=y and y^3-4y=x

 which I've defined as eqns:={,}

and 9 solutions as solns1:={,,..}

and being stored as s1,s2,..s9

when I run a command such as testeq(subs(s1,eqns[1])=subs(s1,eqns[2])

I get an error of passing invalid arguments into testeq. What I essentially need to show is that on substituting for x,y from each s1,..s9; both equations get the same result. What am I doing incorrectly?

I've also noticed that just subs(s1,eqns[1]) returns an equality; I don't quite understand why

I'm given the following two equations:

x^3-4x=y, y^3-4y=x

to solve the system, I've just used

eqns:={x^3-4x=y,y^3-4y=x};

vars:={x,y};

solns:=solve(eqns,vars);

and have obtained only four solutions when I should instead get 9. Is there a mistake in my approach?


eqn1 := (3*y/(y^2+1)^(5/2)+(3*(x+y))/(1+(x+y)^2)^(5/2)+(3*(y+z))/(1+(y+z)^2)^(5/2)+(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*x^2/(x^2+1)^2+1/(x^2+1)-2*x*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*x*(x+z)/(1+(x+z)^2)^2+1/(1+(x+z)^2)-2*x*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2))+(-3*x/(x^2+1)^(5/2)-(3*(x+y))/(1+(x+y)^2)^(5/2)-(3*(x+z))/(1+(x+z)^2)^(5/2)-(3*(x+y+z))/(1+(x+y+z)^2)^(5/2))*(-2*y^2/(y^2+1)^2+1/(y^2+1)-2*y*(x+y)/(1+(x+y)^2)^2+1/(1+(x+y)^2)-2*y*(y+z)/(1+(y+z)^2)^2+1/(1+(y+z)^2)-2*y*(x+y+z)/(1+(x+y+z)^2)^2+1/(1+(x+y+z)^2)):

eqn2 := x/(x^2+1)+x/(1+(x+y)^2)+x/(1+(x+z)^2)+x/(1+(x+y+z)^2)-y/(y^2+1)-y/(1+(x+y)^2)-y/(1+(y+z)^2)-y/(1+(x+y+z)^2):

eqn3 := subs({x = (tan(alpha)-tan(beta)+tan(gamma))*(1/2), y = (tan(alpha)+tan(beta)-tan(gamma))*(1/2), z = (-tan(alpha)+tan(beta)+tan(gamma))*(1/2)}, eqn1):

eqn4 := subs({x = (tan(alpha)-tan(beta)+tan(gamma))*(1/2), y = (tan(alpha)+tan(beta)-tan(gamma))*(1/2), z = (-tan(alpha)+tan(beta)+tan(gamma))*(1/2)}, eqn2):

My question is how to solve eqn3 and eqn4 of tan(alpha)&&tan(beta).

I want to solve the equations eqn3 and eqn4 to solve the  tan(alpha)  and tan(beta),  give me a help .thanks a lot 

Download 0112.mw

Good day everyone, could you please help use Gauss Elimination method for these system of equations. See the worksheet here F1.mw

Thanks.

Dear All,

I am solving 6 ODE equations with boundary conditions using Runge kutta Felbergh 45 (Maple 12). then, i got this problem.. any suggestion??

Thank you :)

ISPC3.mw

``

restart; with(plots); M := 3; k = .2; blt := 6; r := 2; l := .1; Pr := 6.8; Ec := 2; N := .5; rho := .5; Tv := .5; Tt := .5; c := 1; cm := .1; cp := .1

Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0;

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-3*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0

(1)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0;

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(2)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0;

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(3)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0;

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(4)

Eq5 := diff(theta(eta), eta, eta)+Pr*(f(eta)*(diff(theta(eta), eta))-2*(diff(f(eta), eta))*theta(eta))+N*Pr*(theta1(eta)-theta(eta))/(rho*c*Tt)+N*Pr*Ec*(F(eta)-(diff(f(eta), eta)))^2/(rho*Tv) = 0;

diff(diff(theta(eta), eta), eta)+6.8*f(eta)*(diff(theta(eta), eta))-13.6*(diff(f(eta), eta))*theta(eta)+13.60000000*theta1(eta)-13.60000000*theta(eta)+27.20000000*(F(eta)-(diff(f(eta), eta)))^2 = 0

(5)

Eq6 := 2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+cp*(theta1(eta)-theta(eta))/(c*cm*Tt) = 0;

2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+2.000000000*theta1(eta)-2.000000000*theta(eta) = 0

(6)

bcs1 := f(0) = r, (D(f))(0) = -1, (D(f))(blt) = 0, F(blt) = 0, G(blt) = -f(blt), H(blt) = k, theta(0) = 1, theta(blt) = 0, theta1(blt) = 0;

f(0) = 2, (D(f))(0) = -1, (D(f))(6) = 0, F(6) = 0, G(6) = -f(6), H(6) = k, theta(0) = 1, theta(6) = 0, theta1(6) = 0

(7)

L := [0.1e-2];

[0.1e-2]

(8)

for k to 1 do R := dsolve(eval({Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, bcs1}, B = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta1(eta)], numeric, output = listprocedure); Y || k := rhs(R[2]); YP || k := rhs(R[3]); YR || k := rhs(R[4]); YQ || k := rhs(R[5]) end do

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

 

R

R

(9)

print([(YP || (1 .. 1))(0)]);

[YP1(0)]

(10)

``

P1 := plot([YP || (1 .. 1)], 0 .. 14, labels = [eta, (D(f))(eta)]):

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

 

plots:-display([P1]);

 

``

``


Download ISPC3.mw

D_Method.mw

The classical Draghilev’s method.  Example of solving the system of two transcendental equations. For a single the initial approximation are searched 9 approximate solutions of the system.
(4*(x1^2+x2-11))*x1+2*x1+2*x2^2-14+cos(x1)=0;
2*x1^2+2*x2-22+(4*(x1+x2^2-7))*x2-sin(x2)=0; 
x01 := -1.; x02 := 1.;

Hi there,

I would like to compute and display the nullclines of a set of ordinary differential equations.

AFAIK, I can compute the nullclines in Maple by defining the equations and solving the system

e.g.:

# Define the equations
eq1 := u(t)*(1-u(t)/kappa)-u(t)*v(t) = 0;
eq2 := g*(u(t)-1)*v(t) = 0;

# Solve the system (i.e. compute the nullclines)
sol := solve({eq1, eq2}, {u(t), v(t)});

However, I am not quite able to imagine how to display them over a dfieldplot or a phaseportrait.

Attached is an example with some differential equations, and their vector field and trajectories: MaplePrimes_Predator_prey_model_nullclines.mw.

It can be use to illustrate how to (compute and) display the nullclines.

 

Thank you,

jon

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