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Is it possible to read YAML in Maple?...

Asked by: LouiseHoffman 25
hash import yaml
May 06 2016
1  2

Hello all,

I have a hash structure with arrays in a YAML file that I would like to read in Maple.

Does there exist a module, so I can read it?

 

Problem with QR algorithm...

Asked by: spreka 5
algorithm svd eigenvalues
May 05 2016
1  10

I'm trying to implement the QR algorithm to find the Eigenvalues of the input matrix which will be forwarded to another implementation (of the SVD alg.) to find the singular values. My implementation goes as follows:

1. feeding input: A::Matrix(datatype=float) # a bidiagonal matrix
2. construct input matrix for the QR alg. of matrix A and Z (zeros of size A): C := Matrix([[Z,Transpose(A)],[A,Z]], datatype=float); # therefore C should be symmetric
3. find the eigenvalues of matrix C with an implementation of the QR alg.:

for k from 1 to 400 do
Q, R := QRDecomposition(C);
C:=R.Q;
end do:

At this point, the eigenvalues of C should be placed in the diagonal of the matrix, but they're randomly placed around the diagonal, with only ~0 elements (like 2,xxx * 10^(-13)) in the diagonal.

If anyone knows how to resolve this, let the knowledge flow through. Any help will be appriciated, thanks in advance.

Location for extrema value...

Asked by: mskalsi 290
extrema
May 05 2016
0  3

Please check why Maple is not returning location of Minima in following case:

 

> 

-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x)

-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x)

 (1)
> 

plot(-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x), x = -3.2 .. 3.2)

 
> 

readlib(extrema):

{-0.6447467154e-1, 0.6447467152e-1}

 (2)
> 

Minima := op(1, {-0.6447467154e-1, 0.6447467152e-1}); 1; Maxima := op(2, {-0.6447467154e-1, 0.6447467152e-1})

-0.6447467154e-1

  

0.6447467152e-1

 (3)
> 

minimize(-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x), x = 0 .. 3.5, location)

minimize(-0.6159648936e-1*sin(.9960622471*x)+0.1077739351e-1*sin(1.992124494*x)-0.6872829504e-3*sin(2.988186741*x)+0.3984248988e-4*sin(3.984248988*x), x = 0 .. 3.5, location), {}

 (4)

Why Maple is not returning location of minima?

 

Download Location_for_Max_Min.mw

Regards

Analysis and Design of Machine Foundation...

Asked by: moses 119
2dmath programming assignment
May 03 2016
1  0

 

ANALYSIS AND DESIGN OF MACHINE FOUNDATION

 


restart

Loading Optimization

 

Loading LinearAlgebra  

 

Loading plots  

with(ScientificConstants):

Loading DynamicSystems  

with(Units:-Standard)

with(Units:-Natural)

with(StringTools)

FormatTime("%m-%d-%Y, %H:%M")

FormatTime("%m-%d-%Y, %H:%M")

 (1)

NULL

Introduction
 

This document deals with vibration analysis and design of machine foundations subjected to dynamic load.

NULL

NULL

NULL

NULL

GetConstant(g);

standard_acceleration_of_gravity, symbol = g, value = 9.80665, uncertainty = 0, units = m/s^2

 (2)

g__SI := evalf(Constant(g, system = SI, units))

9.80665*Units:-Unit(('m')/('s')^2)

 (3)

NULL

Richart and Lysmer's Model
 

Richart et al. (1970) idealised the foundation as a lumped mass supported on soil which is idealised as frequency independent springs which he described in term of soil parameter

dynamic shear modulus or shear wave velocity of the soil for circular footing when footings having equivalent circular radius. The Tables below shows the different values of spring and damping vlaues as per Richart and Lysmer.

NULL

In which, G = dynamic shar modulus of he soil and is given G = `ρ__s`*V__s^2 ; ν = Piosson's ratio of the soil; ρs = mass density of the soil; Vs = shear wave velocity of the soil obtained

from soil testing; g = acceleration due to gravity; m = mass of the machine and foundation; J = mass moment of inertia of the machine and foundation about the appropriate axes; K = equivalent spring stiffness of the soil; C = damping value of the soil; B = interia factor contributing to the damping factor; D = damping ratio of the soil; r = equivalent radius of a circular foundation; L = length of foundation, and B = width of the foundation.

NULL

NULL

NULL

Sketch
 

NULL

NULL

NULL

NULL

nu := .25

Table : Values of soil springs as per Richart and Lysmer (1970) model
 

NULL

NULL

SI No.

Direction

Spring value

Equivalent radius

Remarks

1

Vertical

K__z = 4*G*r__z/(1-nu)"(->)"

r__z = sqrt(L*B/Pi)"(->)"

This is in vertical Z direction

2

Horizontal

K__x = (32*(1-nu))*G*r__x/(7-8*nu)
"(->)"

r__x = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal X

2.1

Horizontal

K__y = (32*(1-nu))*G*r__y/(7-8*nu)
"(->)"

r__y = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal Y

3

Rocking

`K__φx` = 8*G*`r__φx`^3/(3*(1-nu))"(->)"

`r__φx` = (L*B^3/(3*Pi))^(1/4)"(->)"

This produces roxking about Y axis

3.1

Rocking

`K__φy` = 8*G*`r__φy`^3/(3*(1-nu))"(->)"

`r__φy` = (L^3*B/(3*Pi))^(1/4)"(->)"

This produces roxking about X axis

4

Twisting

`K__ψ` = 16*G*`r__ψ`^3*(1/3)"(->)"

`r__ψ` = ((B^3*L+B*L^3)/(6*Pi))^(1/4)
"(->)"

This produces twisting about vertical Z axis
 

NULL

NULL

NULL

NULL

NULL

Table : Values of soil damping as per Richart and Lysmer (1970) model
 

NULL

SI No.

Direction

Mass ratio (B)

Damping ratio and Damping values

Remarks

1

Vertical

B__z = .25*m__U*(1-nu)*g__SI/(`ρ__s`*r__z^3)
"(->)"

`ζ__z` = .425/sqrt(B__z)"(->)"C__z = 2*`ζ__z`*sqrt(K__z*m__U)"(->)"

This damping value is in vertical Z direction

2

Horizontal

B__x = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`ρ__s`*r__x^3)
"(->)"

`ζ__x` = .288/sqrt(B__x)"(->)"

C__x = 2*`ζ__x`*sqrt(K__x*m__U)"(->)"

This damping value is in lateral X direction

2.1

Horizontal

B__y = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`ρ__s`*r__y^3)
"(->)"

`ζ__y` = .288/sqrt(B__y)"(->)"

`ζ__y` = .288/((2.145204688-2.451662500*nu)*m__U*Units:-Unit(('m')/('s')^2)/((1-nu)*`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)

 (5.1)

NULLError, invalid left hand side in assignmentError, invalid left hand side in assignment

`ζ__ψ` = .5/(1+117.6798000*`J__ψ`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`ρ__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

 (5.2)

C__y = 2*`ζ__y`*sqrt(K__y*m__U)"(->)"

This damping value is in lateral Y direction

3

Rocking

`B__φx` = (.375*(1-nu))*`J__φx`*g__SI/(`ρ__s`*`r__φx`^5)
"(->)"

`ζ__φx` = .15/((1+`B__φx`)*sqrt(`B__φx`))
"(->)"

`ζ__φx` = .15/((1+11.03248125*(1-nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))^(1/2))

 (5.3)

Error, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.15/((1+11.03248125*(1-nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))^(1/2)) = .15/((1+11.03248125*(1-nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))^(1/2))

 (5.4)

`C__φx` = 2*`ζ__φx`*sqrt(`K__φx`*`J__φx`)"(->)"

NULL

This damping value is for rocking about Y direction

3.1

Rocking

`B__φy` = (.375*(1-nu))*`J__φy`*g__SI/(`ρ__s`*`r__φy`^5)
"(->)"

NULL

`ζ__φy` = .15/((1+`B__φy`)*sqrt(`B__φy`))
"(->)"

.15/((1+`B__φy`)*`B__φy`^(1/2)) = .15/((1+`B__φy`)*`B__φy`^(1/2))

 (5.5)

`C__φy` = 2*`ζ__φy`*sqrt(`K__φy`*`J__φy`)"(->)"

NULL

This damping value is for rocking about X direction

4

Twisting

`B__ψ` = `J__ψ`*g__SI/(`ρ__s`*`r__ψ`^5)"(->)"

`ζ__ψ` = .5/(1+2*`B__ψ`)"(->)"NULLError, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.5/(1+117.6798000*`J__ψ`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`ρ__s`*((B^3*L+B*L^3)/Pi)^(5/4))) = .5/(1+117.6798000*`J__ψ`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`ρ__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

 (5.6)

`C__ψ` = 2*`ζ__ψ`*sqrt(`K__ψ`*`J__ψ`)"(->)"NULL

``

NULL

This damping value is valid for twisting about vertical Z axis
 

NULL

NULL

NULLNULL

 (4)

``

NULL

Vertical Motion Considering damping of the Soil
 

For vertical direction the equation becomes that of a lumped mass having single degree of freedom when

deq := m__U*(diff(z(t), t, t))+'C__z'*(diff(z(t), t))+'K__z'*`#mi("z")` = P__0*sin(`ω__m`*t)

m__U*(diff(diff(z(t), t), t))+C__z*(diff(z(t), t))+K__z*`#mi("z")` = P__0*sin(`ω__m`*t)

 (6.1)

NULL

t1 := subs(P__0*sin(`ω__m`*t)/m__U = F, expand(deq/m__U));

diff(diff(z(t), t), t)+1.085721853*(G*(L*B/Pi)^(1/2)*m__U/(1-nu))^(1/2)*(diff(z(t), t))/(m__U*(m__U*Units:-Unit(('m')/('s')^2)/(`ρ__s`*(L*B/Pi)^(3/2))-m__U*Units:-Unit(('m')/('s')^2)*nu/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))+4*G*(L*B/Pi)^(1/2)*`#mi("z")`/(m__U*(1-nu)) = F

 (6.2)

NULL

 (6.3)

By algebraically manipulating the expression, the form traditionally used by engineers is derived:

t2 := algsubs('C__z'/m__U = 2*zeta*omega, t1)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

 (6.4)

NULL

t3 := algsubs('K__z'/m__U = omega^2, t2)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

 (6.5)

This form includes the damping ratio , the natural frequency , and the external forcing term .  Consider only free vibration by setting

gen3 := subs(F = 0, t3)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = 0

 (6.6)

NULL

sol1 := dsolve({gen3, z(0) = P, (D(z))(0) = V}, z(t))

z(t) = -(1000000000/1178791942081753609)*Pi*m__U*exp(-(1085721853/1000000000)*Units:-Unit(('s')/('m')^(1/2))*t/(Pi*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*m__U))*(1085721853*V*Units:-Unit(('s')/('m')^(1/2))*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*`ρ__s`*L^2*B^2-4000000000*(L*B/Pi)^(1/2)*Pi*nu*`#mi("z")`+4000000000*`#mi("z")`*Pi*(L*B/Pi)^(1/2))/(Units:-Unit(('s')/('m')^(1/2))^2*B^2*L^2*`ρ__s`)+(4000000000/1085721853)*`#mi("z")`*G*Pi*(L*B/Pi)^(1/2)*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*t/((nu-1)*Units:-Unit(('s')/('m')^(1/2)))+(1/1178791942081753609)*(1085721853000000000*V*Units:-Unit(('s')/('m')^(1/2))*Pi*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*m__U*`ρ__s`*L^2*B^2+1178791942081753609*Units:-Unit(('s')/('m')^(1/2))^2*P*`ρ__s`*L^2*B^2-4000000000000000000*(L*B/Pi)^(1/2)*Pi^2*m__U*nu*`#mi("z")`+4000000000000000000*`#mi("z")`*Pi^2*(L*B/Pi)^(1/2)*m__U)/(B^2*L^2*`ρ__s`*Units:-Unit(('s')/('m')^(1/2))^2)

 (6.7)

NULL

NULL

 

Download Analysis_and_Design_of_Machine_Foundations_1.mw

Good Morning Mapleprime Community,

Would anybody please help in the attached worksheet. I'm trying to use the new function in Maple that is the clicable method, but I was having problem in some of my output such as zeta_y and zeta_phi as this two equations are generating an error message.

 

Regards,

Moses

Why are args being changed to integers?...

Asked by: Carl Love 26488
procedure float trace
May 02 2016
0  7

The following code is part of my attempt to answer the recent Question about the bifurcation of the map f:= x-> exp(x^2*(a-x)). Two very weird things are happening. They can be seen by applying trace to f. The first is that the input argument to f seems to be changed to a very large integer. The second is that for some real values of a and x, I get imaginary results from this obviously real-valued function. Why are these things happening?

> 

restart:
> 

f:= x-> exp(x^2*(a-x)):
> 

trace(f):
> 

Iterate:= proc(a, x0:= 1., n:= 2000)
local A:= hfarray(1..n, [x0]), f:= subs(:-a= a, eval(:-f));          
     #evalhf(
          proc(f, A, n)
          local k;
               for k from 2 to n do A[k]:= f(A[k-1]) end do
          end proc
          (f, A, n);
     #);
     evalf[4]~(convert(A[1000..], set))
end proc:
> 

Iterate(1.05);

{--> enter f, args = 4607182418800017408

  

HFloat(1.0512710963760241)

  

<-- exit f (now in unknown) = 4607413323290551347}
{--> enter f, args = 4607413323290551347

  

HFloat(0.9985962074909431)

  

<-- exit f (now in unknown) = 4607169774561176020}
{--> enter f, args = 4607169774561176020

  

HFloat(1.0525960836530153)

  

Warning,  computation interrupted

  
> 

Iterate(.75);

{--> enter f, args = 4607182418800017408

  

.754589752755861+.192678397202388*I

  

<-- exit f (now in unknown) = HFloat(0.7545897527558614)+HFloat(0.19267839720238844)*I}

Error, (in unknown) unable to store 'HFloat(0.7545897527558614)+HFloat(0.19267839720238844)*I' when datatype=float[8]

  
> 

 

 

Download bifurcation.mw

How to solve it?...

Asked by: torabi 85
fsolve
April 30 2016
1  15

hi.i am a problem for solving this non linear algebric equation.

please help me...thanks

FSOLVE.mw

FSOLVE.mw

 

How to rearrange equations...

Asked by: rdpdo 15
ode differential-equations
April 29 2016
1  2

Hello,

 

I have a complex set of non linear diff eqns in the form :

y1'' = f(y1',y1,y2'',y2',y2,y3'',y3',y3,y4'',....,y6'',y6',y6,u1,u2,u3,u4) ;

y2'' = f(y1'',y1',y1,y2',y2,y3'',y3',y3,y4'',....,y6'',y6',y6,u1,u2,u3,u4)

and so on ... y6''=(...)

As I want to resolve this coupled systeme in matlab using @ODE45... I wanted the equations in the form : y1''=f(y1',y1,y2',y2,....) and so on ... => X'[] = f(X[],U[])

 

How can I force maple to rearrange a system of coupled eqns with only the variables i want ?

 

I know this is possible beacause it is a nonlinear state space model but maple do not work with nonlinear state space model... It give me error when I tried to create statespace model with my non linear diff eqns.

 

Thanks a lot !

Unreasonable permissions?...

Asked by: John Fredsted 2238
editing delete
April 28 2016
0  7

Wanting to edit a post of mine, I accidentally clicked the 'More'-button and chose 'Edit' on anothers post. I immediately saw that it was not my post, and did of course not alter it. But it left me somewhat shocked, for it would seem to imply that anyone can edit (and delete, as well, have just investigated that) anyones post(s). That I think is something that really needs to be remedied as soon as possible.

Plotting problem:unsmooth and stepwise curves ...

Asked by: rostam 5
plot smoothing
April 28 2016
1  15

Hey

I have a very simple function and I need to plot it. The plot is however not smooth and there are annoying cuts and steps which don't go away no matter how I set the options. This problem has been puzzling me for a while and I really appreciate any help.

The followings are all the possible options I could find, none seems to work. You can take a look at this file: ploting_problem.mw

Thanks

Problems with Solar System Model...

Asked by: nrebman1 15
dsolve numeric differential-equations
April 27 2016
1  5

Howdy all,

I am trying to create a solar system model by defining a force equation then using the sequence function to create a differential equation and then solving those differential equations using the initial conditions (in X,Y,Z coordinates). So far I have the code below.


> 

m[1] = 1.989*10^30; m[2] = 3.301*10^23; m[3] = 4.867*10^24; m[4] = 5.972*10^24+7.346*10^22; m[5] = 6.417*10^23; m[6] = 1.899*10^27; m[7] = 5.685*10^26; m[8] = 8.682*10^25; m[9] = 1.024*10^26; m[10] = 1.471*10^22; m[11] = 9.3*10^20; m[12] = 2.6*10^20; m[13] = 2*10^20; m[14] = 8.67*10^19; m[15] = 3.9*10^19; mass := Matrix(15, 1, [1.989*10^30, 3.301*10^23, 4.867*10^24, 5.972*10^24+7.346*10^22, 6.417*10^23, 1.899*10^27, 5.685*10^26, 8.682*10^25, 1.024*10^26, 1.471*10^22, 9.3*10^20, 2.6*10^20, 2*10^20, 8.67*10^19, 3.9*10^19]); G := 6.67408*10^(-11)

mass := Vector(4, {(1) = ` 15 x 1 `*Matrix, (2) = `Data Type: `*anything, (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})

  

0.6674080000e-10

 (1)
> 

sqrt(sum(x[i](t)^2, i = 1 .. 3));

(x[1](t)^2+x[2](t)^2+x[3](t)^2)^(1/2)

  

proc (i, j) options operator, arrow; sqrt(sum((x[i, k](t)-x[j, k](t))^2, k = 1 .. 3)) end proc

  

((x[1, 1](t)-x[3, 1](t))^2+(x[1, 2](t)-x[3, 2](t))^2+(x[1, 3](t)-x[3, 3](t))^2)^(1/2)

  

proc (i, j) options operator, arrow; [x[j, 1](t)-x[i, 1](t), x[j, 2](t)-x[i, 2](t), x[j, 3](t)-x[i, 3](t)] end proc

  

[x[3, 1](t)-x[1, 1](t), x[3, 2](t)-x[1, 2](t), x[3, 3](t)-x[1, 3](t)]

  

x[3, 1](t)-x[1, 1](t)

  

x[j, 1](t)-x[i, 1](t), x[j, 2](t)-x[i, 2](t), x[j, 3](t)-x[i, 3](t)

  

proc (i, j) options operator, arrow; [seq(x[j, k](t)-x[i, k](t), k = 1 .. 3)] end proc

  

[x[3, 1](t)-x[1, 1](t), x[3, 2](t)-x[1, 2](t), x[3, 3](t)-x[1, 3](t)]

  

x[3, 1](t)-x[1, 1](t)

 (2)
> 

diff(x[0, 1](t), t, t) = force(0)[1]:
> 

initialPositions := Matrix([[0, 0, 0], [-0.210e8, 0.426e8, 0.541e7], [0.106e9, -0.244e8, -0.644e7], [-0.139e9, -0.569e8, 0.316e4], [-0.177e9, -0.155e9, 0.111e7], [-0.802e9, 0.131e9, 0.174e8], [-0.480e9, -0.142e10, 0.438e8], [0.280e10, 0.103e10, -0.324e8], [0.420e10, -0.157e10, -0.645e8], [0.132e10, -0.477e10, 0.127e9], [0.431e9, -0.690e8, -0.816e8], [0.228e9, 0.305e9, -0.368e8], [0.300e9, -0.351e9, 0.217e9], [-0.434e9, -0.841e7, -0.284e8], [-0.115e9, -0.466e9, -0.612e8]])

initialPositions := Vector(4, {(1) = ` 15 x 3 `*Matrix, (2) = `Data Type: `*anything, (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})

 (3)
> 

initialVelocities := Matrix([[0, 0, 0], [-0.462e7, -0.170e7, 0.285e6], [0.666e6, 0.293e7, 0.183e4], [0.936e6, -0.239e7, 83.3], [0.145e7, -0.140e7, -0.650e5], [-0.196e6, -0.106e7, 0.879e4], [0.745e6, -0.271e6, -0.250e5], [-0.208e6, 0.524e6, 0.467e4], [0.161e6, 0.442e6, -0.129e5], [0.463e6, 0.294e5, -0.137e6], [0.193e6, 0.143e7, 0.923e4], [-0.119e7, 0.974e6, 0.116e6], [0.978e6, 0.562e6, -0.470e6], [0.166e6, -0.156e7, -0.131e5], [0.132e7, -0.170e6, 0.395e6]])

initialVelocities := Vector(4, {(1) = ` 15 x 3 `*Matrix, (2) = `Data Type: `*anything, (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})

 (4)
> 

ic1 := seq(seq(x[i, k](0) = initialPositions[i+1, k], k = 1 .. 3), i = 0 .. N-1); ic2 := seq(seq((D(x[i, k]))(0) = initialVelocities[i+1, k], k = 1 .. 3), i = 0 .. N-1); equations := {ic1, ic2, ode}; sol := dsolve(equations, numeric)

Error, (in dsolve/numeric/process_input) invalid specification of initial conditions, got {x[0, 1](0) = 0, x[0, 2](0) = 0, x[0, 3](0) = 0, x[1, 1](0) = -0.210e8, x[1, 2](0) = 0.426e8, x[1, 3](0) = 0.541e7, x[2, 1](0) = 0.106e9, x[2, 2](0) = -0.244e8, x[2, 3](0) = -0.644e7, x[3, 1](0) = -0.139e9, x[3, 2](0) = -0.569e8, x[3, 3](0) = 0.316e4, x[4, 1](0) = -0.177e9, x[4, 2](0) = -0.155e9, x[4, 3](0) = 0.111e7, x[5, 1](0) = -0.802e9, x[5, 2](0) = 0.131e9, x[5, 3](0) = 0.174e8, x[6, 1](0) = -0.480e9, x[6, 2](0) = -0.142e10, x[6, 3](0) = 0.438e8, x[7, 1](0) = 0.280e10, x[7, 2](0) = 0.103e10, x[7, 3](0) = -0.324e8, x[8, 1](0) = 0.420e10, x[8, 2](0) ...

  
> 

plots[odeplot](sol, [x[1, 1](t), x[1, 2](t), x[1, 3](t)], t = 0 .. 20, numpoints = 1000, axes = normal)

Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

  
> 

plots[odeplot](sol, [seq(x[1, k](t), k = 1 .. 3)], t = 0 .. 20, numpoints = 1000, axes = normal)

Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

  
> 

plots[odeplot](sol, [seq([seq(x[i, k](t), k = 1 .. 3)], i = 0 .. N-1)], t = 0 .. 20, numpoints = 1000, axes = normal)

Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

  
> 

``


Download ass4.mw

Everything works fine until I try to execute the last line. When I do that I get an error that says "Error, in dsolve/numeric/process_input. invalid specifications of initial conditions.

At this point I am not sure if the problem lies in how I have defined my initial conditions or the way i've defined the force equation but I am open to any suggestions or ideas on where I should go from here.

Thanks in advance!

 

Nick

How to simplify results of pdsolve?...

Asked by: zhuxian 60
simplify pdsolve
April 27 2016
1  3

i use the pdsolve to find the solutions of a system of partial differential equations,

but the result contains some indefinite integrals, how to simplify it further?

thank you

code:

eq1 := {6*(diff(_xi[t](x, t, u), u))-3*(diff(_xi[x](x, t, u), u)), 12*(diff(_xi[t](x, t, u), u, u))-6*(diff(_xi[x](x, t, u), u, u)), 2*(diff(_xi[t](x, t, u), u, u, u))-(diff(_xi[x](x, t, u), u, u, u)), diff(_eta[u](x, t, u), t)+diff(_eta[u](x, t, u), x, x, x)+(diff(_eta[u](x, t, u), x))*u, 18*(diff(_xi[t](x, t, u), x, u))+3*(diff(_eta[u](x, t, u), u, u))-9*(diff(_xi[x](x, t, u), x, u)), 6*(diff(_xi[t](x, t, u), x, x))+3*(diff(_eta[u](x, t, u), x, u))-3*(diff(_xi[x](x, t, u), x, x)), 6*(diff(_xi[t](x, t, u), x, u, u))+diff(_eta[u](x, t, u), u, u, u)-3*(diff(_xi[x](x, t, u), x, u, u)), 12*(diff(_xi[t](x, t, u), u))-6*(diff(_xi[x](x, t, u), u))+6*(diff(_xi[t](x, t, u), x, x, u))-6*(diff(_xi[t](x, t, u), u))*u+3*u*(diff(_xi[x](x, t, u), u))-3*(diff(_xi[x](x, t, u), x, x, u))+3*(diff(_eta[u](x, t, u), x, u, u)), 12*(diff(_xi[t](x, t, u), x))-6*(diff(_xi[x](x, t, u), x))+2*(diff(_xi[t](x, t, u), t))+2*(diff(_xi[t](x, t, u), x, x, x))-4*(diff(_xi[t](x, t, u), x))*u+2*(diff(_xi[x](x, t, u), x))*u+_eta[u](x, t, u)-(diff(_xi[x](x, t, u), t))+3*(diff(_eta[u](x, t, u), x, x, u))-(diff(_xi[x](x, t, u), x, x, x))};

simplify(pdsolve(eq1))

 

Default method used in pdsolve/numeric...

Asked by: Al86 145
numeric pdsolve
April 27 2016
2  0

I read in the net that the method used in pdsolve numeric is the theta method, my question: is it the most efficient with regard to rate of convergence of the numerical solution of the PDE?

If not then why is it used as the default method?

 

Thanks.

 

How to find eigenvectors in terms of mmm?...

Asked by: ComputerUser 535
eigenvectors
April 27 2016
0  2

nullspace or reducedform or Eigenvectors still can not find eigenvector in terms of  mmm , how to find this?

 

mmm is a variable

Question about Export As LaTeX...

Asked by: naderj 5
latex
April 27 2016
0  4

Dear All, 

I am using the comand " export as" form the file menu to obatain a latex version of my worksheet. The generated latex file use a package called amplestd2e.sty that should be loaded for latex compiler to function proper. Do somebody know where to find it. Thank you. N. Jand 

Problem with delta as parameter...

Asked by: farahamirah 10
constant parameter
April 26 2016
0  0

I want to ask., I put delta as my constant in maple program and I want the answer are in delta as well., but the thing is., when running., it let delta=0, delta=-1, and delta=delta.,
the condition is we cannot let delta=1 or delta=0 because it is just same for s5 and s7.,.(delta is refer to the s8). How can I get answer as delta? with the condition? here I attach my maple programme..

 

> derivation := proc (A, n)
local i, j, k, t, s5, s7, s8, m, D,
sols5, sols7, sols8, eqns5, eqns7, eqns8,
BChange5, BChange7, BChange8; eqns5 := {}; eqns7 := {}; eqns8 := {};
D := matrix(n, n);
BChange5 := matrix(n, n); BChange7 := matrix(n, n); BChange8 := matrix(n, n);
for i to n do for j to n do for m to n do
s5 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+A[i, k, m]*D[k, j], k = 1 .. n));
s7 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+0*A[i, k, m]*D[k, j], k = 1 .. n));
s8 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+delta*A[i, k, m]*D[k, j], k = 1 .. n));
eqns5 := `union`(eqns5, {s5}); eqns7 := `union`(eqns7, {s7}); eqns8 := `union`(eqns8, {s8})
end do end do end do;
sols5 := [solve(eqns5)]; sols7 := [solve(eqns7)]; sols8 := [solve(eqns8)];
t := nops(sols5); t := nops(sols7); t := nops(sols8);
for i to t do for j to n do for k to n do
BChange5[k, j] := subs(sols5[i], D[k, j]);
BChange7[k, j] := subs(sols7[i], D[k, j]);
BChange8[k, j] := subs(sols8[i], D[k, j])
end do end do;
print("eqns&Assign;", eqns5); print("sols:=", sols5); print("BChange5:=", BChange5);
print("eqns&Assign;", eqns7); print("sols:=", sols7); print("BChange8:=", BChange7);
print("eqns&Assign;", eqns8); print("sols:=", sols8); print("BChange8:=", BChange8)
end do end proc;

> AS1 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 2) = 1]);
> derivation(AS1, 2);

> AS2 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1]);
> derivation(AS2, 2);

> AS3 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (2, 1, 2) = 1]);
> derivation(AS3, 2);

> AS4 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (2, 2, 2) = 1]);
> derivation(AS4, 2);

> AS5 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1, (2, 1, 2) = 1]);
> derivation(AS5, 2);

> AS1 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (3, 1, 2) = 1]);
> derivation(AS1, 3);

> AS2 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (3, 1, 2) = alpha]);
> derivation(AS2, 3);

> AS3 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (1, 2, 3) = 1, (2, 1, 3) = 1]);
> derivation(AS3, 3);

> AS4 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (2, 3, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS4, 3);

> AS5 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 3, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS5, 3);

> AS6 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(3, 1, 2) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS6, 3);

> AS7 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 2, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS7, 3);

> AS8 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS8, 3);

> AS9 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS9, 3);

> AS10 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS10, 3);

> AS11 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (2, 3, 2) = 1, (3, 1, 2) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS11, 3);

> AS12 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS12, 3);

> AS13 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS13, 3);

> AS14 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS14, 3);

> AS15 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS15, 3);

> AS16 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS16, 3);

> AS17 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS17, 3);
>

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