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Tetrads and Weyl scalars in canonical form

 

The material below is about a new development that didn't arrive in time for the launch of Maple 2016 (March) and that complements in a relevant way the ones introduced in Physics in Maple 2016. It is at topic in general relativity, the computation of a canonical form of a tetrad, so that, generally speaking (skipping a technical description) the Weyl scalars are fixed as much as possible (either equal to 0 or to 1) regarding transformations that leave invariant the tetrad metric in a tetrad system of references. Bringing a tetrad in canonical form is a relevant step in the tackling of the equivalence problem between two spacetime metrics (Mapleprimes post), and it is relevant in connection with the digitizing in Maple 2016 of the database of solutions to Einstein's equations of the book Exact Solutions to Einstein Field Equations.

The reference for this development is the book "General Relativity, an Einstein century survey", edited by S.W. Hawking (Cambridge) and W. Israel (U. Alberta, Canada), specifically Chapter 7 written by S. Chandrasekhar, and more specifically exploring what is said in page 388 about the Petrov classification.


A canonical form for the tetrad and Weyl scalars admits alternate forms; the implementation is as implicit in page 388:

 

`Ψ__0`

`Ψ__1`

`Ψ__2`

`Ψ__3`

`Ψ__4`

Residual invariance

Petrov type I

0

"<>0"

"<>0"

1

0

none

Petrov type II

0

0

"<>0"

1

0

none

Petrov type III

0

0

0

1

0

none

Petrov type D

0

0

"<>0"

0

0

`&Psi;__2`  remains invariant under rotations of Class III

Petrov type N

0

0

0

0

1

`&Psi;__4` remains invariant under rotations of Class II

 

The transformations (rotations of the tetrad system of references) used are of Class I, II and III as defined in Chandrasekar's chapter - equations (7.79) in page 384, (7.83) and (7.84) in page 385. Transformations of Class I can be performed with the command Physics:-Tetrads:-TransformTetrad using the optional argument nullrotationwithfixedl_, of Class II using nullrotationwithfixedn_ and of Class III by calling TransformTetrad(spatialrotationsm_mb_plan, boostsn_l_plane), so with the two optional arguments simultaneously.

 

In this development, a new optional argument, canonicalform got implemented to TransformTetrad so that the whole sequence of three transformations of Classes I, II and III is performed automatically, in one go. Regarding the canonical form of the tetrad, the main idea is that from the change in the Weyl scalars one can derive the parameters entering tetrad transformations that result in a canonical form of the tetrad. 

 

with(Physics); with(Tetrads)

`Setting lowercaselatin letters to represent tetrad indices `

 

0, "%1 is not a command in the %2 package", Tetrads, Physics

 

0, "%1 is not a command in the %2 package", Tetrads, Physics

 

[IsTetrad, NullTetrad, OrthonormalTetrad, PetrovType, SimplifyTetrad, TransformTetrad, e_, eta_, gamma_, l_, lambda_, m_, mb_, n_]

(1)

(Note the Tetrads:-PetrovType command, unfinished in the first release of Maple 2016.) To run the following computations you need to update your Physics library to the latest version from the Maplesoft R&D Physics webpage, so with this datestamp or newer:

Physics:-Version()

"/Users/ecterrab/Maple/lib/Physics2016.mla", `2016, April 20, 12:56 hours`

(2)

An Example of Petrov type I

There are six Petrov types: I, II, III, D, N and O. Start with a spacetime metric of Petrov type "I"  (the numbers always refer to the equation number in the "Exact solutions to Einstein's field equations" textbook)

g_[[12, 21, 1]]

`Systems of spacetime Coordinates are: `*{X = (t, x, y, phi)}

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (t, x, y, phi)}

 

`The McLenaghan, Tariq (1975), Tupper (1976) metric in coordinates `[t, x, y, phi]

 

`Parameters: `[a, k, kappa0]

 

"`Comments: `_k parametrizes the most general electromagnetic invariant with respect to the last 3 Killing vectors"

 

`Resetting the signature of spacetime from "+ - - -" to \`- + + +\` in order to match the signature in the database of metrics:`

 

g[mu, nu] = (Matrix(4, 4, {(1, 1) = -1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 2*y, (2, 1) = 0, (2, 2) = a^2/x^2, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = a^2/x^2, (3, 4) = 0, (4, 1) = 2*y, (4, 2) = 0, (4, 3) = 0, (4, 4) = x^2-4*y^2}))

(3)

The Weyl scalars

Weyl[scalars]

psi__0 = (1/4)*((4*I)*x^3*abs(x)^3-abs(x)^6+abs(x)^4*x^2+abs(x)^2*x^4-x^6)/(a^2*abs(x)^4*x^2), psi__1 = 0, psi__2 = -(1/4)*(x^2+abs(x)^2)*(x^4+abs(x)^4)/(a^2*abs(x)^4*x^2), psi__3 = 0, psi__4 = (1/4)*((4*I)*x^3*abs(x)^3-abs(x)^6+abs(x)^4*x^2+abs(x)^2*x^4-x^6)/(a^2*abs(x)^4*x^2)

(4)

... there is abs around. Let's assume everything is positive to simplify formulas, use Capital Physics:-Assume  (the lower case assume  command redefines the assumed variables, so it is not compatible with Physics, DifferentialGeometry and VectorCalculus among others).

Assume(x > 0, y > 0, a > 0)

{a::(RealRange(Open(0), infinity))}, {x::(RealRange(Open(0), infinity))}, {y::(RealRange(Open(0), infinity))}

(5)

The scalars are now simpler, although still not in "canonical form" because `&Psi;__4` <> 0 and `&Psi;__3` <> 1.

Weyl[scalars]

psi__0 = I/a^2, psi__1 = 0, psi__2 = -1/a^2, psi__3 = 0, psi__4 = I/a^2

(6)

The Petrov type

PetrovType()

"I"

(7)

The  call to Tetrads:-TransformTetrad two lines below transforms the current tetrad ,

e_[]

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078512745638)

(8)

into another tetrad such that the Weyl scalars are in canonical form, which for Petrov "I" type happens when `&Psi;__0` = 0, `&Psi;__4` = 0 and `&Psi;__3` = 1.

TransformTetrad(canonicalform)

Matrix(%id = 18446744078500192254)

(9)

Despite the fact that the result is a much more complicated tetrad, this is an amazing result in that the resulting Weyl scalars are all fixed (see below).  Let's first verify that this is indeed a tetrad, and that now the Weyl scalars are in canonical form

"IsTetrad(?)"

`Type of tetrad: null `

 

true

(10)

Set (9) to be the tetrad in use and recompute the Weyl scalars

"Setup(tetrad = ?):"

Inded we now have `&Psi;__0` = 0, `&Psi;__4` = 0 and `&Psi;__3` = 1 

simplify([Weyl[scalars]])

[psi__0 = 0, psi__1 = (-1/2-(3/2)*I)/a^4, psi__2 = (-1+I)/a^2, psi__3 = 1, psi__4 = 0]

(11)

So Weyl scalars computed after setting the canonical tetrad (9) to be the tetrad in use are in canonical form. Great! NOTE: computing the canonicalWeyl scalars is not really the difficult part, and within the code, these scalars (11) are computed before arriving at the tetrad (9). What is really difficult (from the point of view of computational complexity and simplifications) is to compute the actual canonical form of the tetrad (9).

 

An Example of Petrov type II

Consider this other solution to Einstein's equation (again, the numbers in g_[[24,37,7]] always refer to the equation number in the "Exact solutions to Einstein's field equations" textbook)

g_[[24, 37, 7]]

`Systems of spacetime Coordinates are: `*{X = (u, v, x, y)}

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (u, v, x, y)}

 

`The Stephani metric in coordinates `[u, v, x, y]

 

`Parameters: `[f(x), a, Psi1(u, x, y)]

 

"`Comments: `Case 6 from Table 24.1:_Psi1(u,x,y): diff(_Psi1(u,x,y),x,x)+diff(_Psi1(u,x,y),y,y)=0, diff(x*diff(_M(u,x,y),x),x)+x*diff(_M(u,x,y),y,y)=_kappa0*(diff(_Psi(u,x,y),x)^2+diff(_Psi(u,x,y),y)^2)"

 

g[mu, nu] = (Matrix(4, 4, {(1, 1) = -2*x*(f(x)+y*a), (1, 2) = -x, (1, 3) = 0, (1, 4) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 3) = 1/x^(1/2), (3, 4) = 0, (4, 4) = 1/x^(1/2)}, storage = triangular[upper], shape = [symmetric]))

(12)

Check the Petrov type

PetrovType()

"II"

(13)

The starting tetrad

e_[]

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078835577550)

(14)

results in Weyl scalars not in canonical form:

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = (1/8)/x^(3/2), psi__3 = 0, psi__4 = -((3*I)*a-2*x*(diff(diff(f(x), x), x))-3*(diff(f(x), x)))/(x^(1/2)*(4*y*a+4*f(x)))

(15)

For Petrov type "II", the canonical form is as for type "I" but in addition `&Psi;__1` = 0. Again let's assume positive, not necessary, but to get simpler formulas around

Assume(f(x) > 0, x > 0, y > 0, a > 0)

{a::(RealRange(Open(0), infinity))}, {x::(RealRange(Open(0), infinity)), (-f(x))::(RealRange(-infinity, Open(0))), (f(x))::(RealRange(Open(0), infinity))}, {y::(RealRange(Open(0), infinity))}

(16)

Compute now a canonical form for the tetrad, to be used instead of (14)

TransformTetrad(canonicalform)

Matrix(%id = 18446744078835949430)

(17)

Set this tetrad and check the Weyl scalars again

"Setup(tetrad = ?):"

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = (1/8)/x^(3/2), psi__3 = 1, psi__4 = 0

(18)

This result (18) is fantastic. Compare these Weyl scalars with the ones (15) before transforming the tetrad.

 

An Example of Petrov type III

g_[[12, 35, 1]]

`Systems of spacetime Coordinates are: `*{X = (u, x, y, z)}

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (u, x, y, z)}

 

`The Kaigorodov (1962), Cahen (1964), Siklos (1981), Ozsvath (1987) metric in coordinates `[u, x, y, z]

 

`Parameters: `[Lambda]

 

g[mu, nu] = (Matrix(4, 4, {(1, 1) = 0, (1, 2) = exp(-2*z), (1, 3) = 0, (1, 4) = 0, (2, 2) = exp(4*z), (2, 3) = 2*exp(z), (2, 4) = 0, (3, 3) = 2*exp(-2*z), (3, 4) = 0, (4, 4) = 3/abs(Lambda)}, storage = triangular[upper], shape = [symmetric]))

(19)

Assume(z > 0, Lambda > 0)

{Lambda::(RealRange(Open(0), infinity))}, {z::(RealRange(Open(0), infinity))}

(20)

The Petrov type and the original tetrad

PetrovType()

"III"

(21)

e_[]

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078349449926)

(22)

This tetrad results in the following scalars

Weyl[scalars]

psi__0 = -2*Lambda*2^(1/2)+(11/4)*Lambda, psi__1 = -(1/2)*Lambda*2^(1/2)+(3/4)*Lambda, psi__2 = (1/4)*Lambda, psi__3 = -(1/2)*Lambda*2^(1/2)-(3/4)*Lambda, psi__4 = 2*Lambda*2^(1/2)+(11/4)*Lambda

(23)

that are not in canonical form, which for Petrov type III is as in Petrov type II but in addition we should have `&Psi;__2` = 0.

Compute now a canonical form for the tetrad

TransformTetrad(canonicalform)

Matrix(%id = 18446744078500057566)

(24)

Set this one to be the tetrad in use and recompute the Weyl scalars

"Setup(tetrad = ?):"

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = 0, psi__3 = 1, psi__4 = 0

(25)

Great!``

An Example of Petrov type N

g_[[12, 6, 1]]

`Systems of spacetime Coordinates are: `*{X = (u, v, y, z)}

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (u, v, y, z)}

 

`The Defrise (1969) metric in coordinates `[u, v, y, z]

 

`Parameters: `[Lambda, kappa0]

 

"`Comments: `_Lambda < 0 required for a pure radiation solution"

 

g[mu, nu] = (Matrix(4, 4, {(1, 1) = 0, (1, 2) = -(3/2)/(y^2*Lambda), (1, 3) = 0, (1, 4) = 0, (2, 2) = -3/(y^4*Lambda), (2, 3) = 0, (2, 4) = 0, (3, 3) = 3/(y^2*Lambda), (3, 4) = 0, (4, 4) = 3/(y^2*Lambda)}, storage = triangular[upper], shape = [symmetric]))

(26)

Assume(y > 0, Lambda > 0)

{Lambda::(RealRange(Open(0), infinity))}, {y::(RealRange(Open(0), infinity))}

(27)

PetrovType()

"N"

(28)

The original tetrad and related Weyl scalars are not in canonical form:

e_[]

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078404437406)

(29)

Weyl[scalars]

psi__0 = -(1/4)*Lambda, psi__1 = -((1/4)*I)*Lambda, psi__2 = (1/4)*Lambda, psi__3 = ((1/4)*I)*Lambda, psi__4 = -(1/4)*Lambda

(30)

For Petrov type "N", the canonical form has `&Psi;__4` <> 0 and all the other `&Psi;__n` = 0.

Compute a canonical form, set it to be the tetrad in use and recompute the Weyl scalars

TransformTetrad(canonicalform)

Matrix(%id = 18446744078518486190)

(31)

"Setup(tetrad = ?):"

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = 0, psi__3 = 0, psi__4 = 1

(32)

All as expected.

An Example of Petrov type D

 

g_[[12, 8, 4]]

`Systems of spacetime Coordinates are: `*{X = (t, x, y, z)}

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (t, x, y, z)}

 

`The  metric in coordinates `[t, x, y, z]

 

`Parameters: `[A, B]

 

"`Comments: `k = 0, kprime = 1, not an Einstein metric"

 

g[mu, nu] = (Matrix(4, 4, {(1, 1) = -B^2*sin(z)^2, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 2) = A^2, (2, 3) = 0, (2, 4) = 0, (3, 3) = A^2*x^2, (3, 4) = 0, (4, 4) = B^2}, storage = triangular[upper], shape = [symmetric]))

(33)

Assume(A > 0, B > 0, x > 0, 0 <= z and z <= (1/4)*Pi)

{A::(RealRange(Open(0), infinity))}, {B::(RealRange(Open(0), infinity))}, {x::(RealRange(Open(0), infinity))}, {z::(RealRange(0, (1/4)*Pi))}

(34)

PetrovType()

"D"

(35)

The default tetrad and related Weyl scalars are not in canonical form, which for Petrov type "D" is with `&Psi;__2` <> 0 and all the other `&Psi;__n` = 0

e_[]

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078503920694)

(36)

Weyl[scalars]

psi__0 = (1/4)/B^2, psi__1 = 0, psi__2 = (1/12)/B^2, psi__3 = 0, psi__4 = (1/4)/B^2

(37)

Transform the  tetrad, set it and recompute the Weyl scalars

TransformTetrad(canonicalform)

Matrix(%id = 18446744078814996830)

(38)

"Setup(tetrad=?):"

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = -(1/6)/B^2, psi__3 = 0, psi__4 = 0

(39)

Again the expected canonical form of the Weyl scalars, and `&Psi;__2` <> 0 remains invariant under transformations of Class III.

 

An Example of Petrov type O

 

Finally an example of type "O". This corresponds to a conformally flat spacetime, for which the Weyl tensor (and with it all the Weyl scalars) vanishes. So the code just interrupts with "not implemented for conformally flat spactimes of Petrov type O"

g_[[8, 33, 1]]

`Systems of spacetime Coordinates are: `*{X = (t, x, y, z)}

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (t, x, y, z)}

 

`The  metric in coordinates `[t, x, y, z]

 

`Parameters: `[K]

 

"`Comments: `_K=3*_Lambda, _K>0 de Sitter, _K<0 anti-de Sitte"

 

g[mu, nu] = z

(40)

PetrovType()

"O"

(41)

The Weyl tensor and its scalars all vanish:

Weyl[nonzero]

Physics:-Weyl[mu, nu, alpha, beta] = {}

(42)

simplify(evala([Weyl[scalars]]))

[psi__0 = 0, psi__1 = 0, psi__2 = 0, psi__3 = 0, psi__4 = 0]

(43)

TransformTetrad(canonicalform)

Error, (in Tetrads:-CanonicalForm) canonical form is not implemented for flat or conformally flat spacetimes of Petrov type "O"

 

NULL

 

Download TetradsAndWeylScalarsInCanonicalForm.mw

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

Suppose I have a function like this: f=cos(2t/m)+cos(2(t+5)/m).

 

Now for each fixed m, we get the maximum value of f. Then I want to build a plot where x-axis is m and y-axis is f, how could I do that? Please help!

 

I have written the following coade in Maple:
r := 50;
l1 := 0.2742e-10;
s := I*w;
l := (-1.342110665*10^22*c^2*(Pi^4)-4.225000000*10^25*c^2*(Pi^2)+2.316990000*10^11*c1*(Pi^2)-1)/(-1.342110665*10^22*c^2*c1*(Pi^4)-7.140250000*10^43*c^2*c1*r^2*(Pi^4)+1.957856550*10^33*c^2*(Pi^4)+9.789282750*10^32*c*c1*(Pi^4)-1.690000*10^22*c*(Pi^2)-4.22500*10^21*c1*(Pi^2));
z1 := (c*l1*s^2+1)/(c*s);
z2 := l*s/(c1*l*s^2+1);
h := (z1+2*z2)*((z1+2*r)*(z1+3*z2)/(2*r)-2*z2)/z2-(1/2)*z2*(z1+2*r)*r;
f := h*(z1+3*z2)/z2-(z1+2*r)(2*r)*(z1+3*z2)+2*z2;
gain := 2*z2/f;
a := abs(gain);
d := diff(a, w);
s := subs(w = 2*pi*0.325e11, d)
Now, I have a function named "s" which I want to set to zero, and calculate the relationship between variables c & c1 in order to achieve this. How should it be done?
Thanks.


The year 2015 has been one with interesting and relevant developments in the MathematicalFunctions  and FunctionAdvisor projects.

• 

Gaps were filled regarding mathematical formulas, with more identities for all of BesselI, BesselK, BesselY, ChebyshevT, ChebyshevU, Chi, Ci, FresnelC, FresnelS, GAMMA(z), HankelH1, HankelH2, InverseJacobiAM, the twelve InverseJacobiPQ for P, Q in [C,D,N,S], KelvinBei, KelvinBer, KelvinKei, KelvinKer, LerchPhi, arcsin, arcsinh, arctan, ln;

• 

Developments happened in the Mathematical function package, to both compute with symbolic sequences and symbolic nth order derivatives of algebraic expressions and functions;

• 

The input FunctionAdvisor(differentiate_rule, mathematical_function) now returns both the first derivative (old behavior) and the nth symbolic derivative (new behavior) of a mathematical function;

• 

A new topic, plot, used as FunctionAdvisor(plot, mathematical_function), now returns 2D and 3D plots for each mathematical function, following the NIST Digital Library of Mathematical Functions;

• 

The previously existing FunctionAdvisor(display, mathematical_function) got redesigned, so that it now displays more information about any mathematical function, and organized into a Section with subsections for each of the different topics, making it simpler to find the information one needs without getting distracted by a myriad of formulas that are not related to what one is looking for.

More mathematics

 

More mathematical knowledge is in place, more identities, differentiation rules of special functions with respect to their parameters, differentiation of functions whose arguments involve symbolic sequences with an indeterminate number of operands, and sum representations for special functions under different conditions on the functions' parameters.

Examples

   

More powerful symbolic differentiation (nth order derivative)

 

Significative developments happened in the computation of the nth order derivative of mathematical functions and algebraic expressions involving them.

Examples

   

Mathematical handling of symbolic sequences

 

Symbolic sequences enter various formulations in mathematics. Their computerized mathematical handling, however, was never implemented - only a representation for them existed in the Maple system. In connection with this, a new subpackage, Sequences , within the MathematicalFunctions package, has been developed.

Examples

   

Visualization of mathematical functions

 

When working with mathematical functions, it is frequently desired to have a rapid glimpse of the shape of the function for some sampled values of their parameters. Following the NIST Digital Library of Mathematical Functions, a new option, plot, has now been implemented.

Examples

   

Section and subsections displaying properties of mathematical functions

 

Until recently, the display of a whole set of mathematical information regarding a function was somehow cumbersome, appearing all together on the screen. That display was and is still available via entering, for instance for the sin function, FunctionAdvisor(sin) . That returns a table of information that can be used programmatically.

With time however, the FunctionAdvisor evolved into a consultation tool, where a better organization of the information being displayed is required, making it simpler to find the information we need without being distracted by a screen full of complicated formulas.

To address this requirement, the FunctionAdvisor now returns the information organized into a Section with subsections, built using the DocumentTools package. This enhances the presentation significantly.

Examples

   

These developments can be installed in Maple 2015 as usual, by downloading the updates (bundled with the Physics and Differential Equations updates) from the Maplesoft R&D webpage for Mathematical Functions and Differential Equations


Download MathematicalFunctionsAndFunctionAdvisor.mw

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

The well known William Lowell Putnam Mathematical Competition (76th edition)  took place this month.
Here is a Maple approach for two of the problems.

1. For each real number x, 0 <= x < 1, let f(x) be the sum of  1/2^n  where n runs through all positive integers for which floor(n*x) is even.
Find the infimum of  f.
(Putnam 2015, A4 problem)

f:=proc(x,N:=100)
local n, s:=0;
for n to N do
  if type(floor(n*x),even) then s:=s+2^(-n) fi;
  #if floor(n*x) mod 2 = 0  then s:=s+2^(-n) fi;
od;
evalf(s);
#s
end;

plot(f, 0..0.9999);

 

min([seq(f(t), t=0.. 0.998,0.0001)]);

        0.5714285714

identify(%);

So, the infimum is 4/7.
Of course, this is not a rigorous solution, even if the result is correct. But it is a valuable hint.
I am not sure if in the near future, a CAS will be able to provide acceptable solutions for such problems.

2. If the function f  is three times differentiable and  has at least five distinct real zeros,
then f + 6f' + 12f'' + 8f''' has at least two distinct real zeros.
(Putnam 2015, B1 problem)

restart;
F := f + 6*D(f) + 12*(D@@2)(f) + 8*(D@@3)(f);

dsolve(F(x)=u(x),f(x));

We are sugested to consider

g:=f(x)*exp(x/2):
g3:=diff(g, x$3);

simplify(g3*8*exp(-x/2));

So, F(x) = k(x) * g3 = k(x) * g'''
g  has 5 distinct zeros implies g''' and hence F have 5-3=2 distinct zeros, q.e.d.

 

How do I construct the seuqence 1/16 , 1/32 , 1/64 , 1/128 , 1/256 in maple?

 

What's the syntax?

I looked at the examples in here:

http://www.maplesoft.com/support/help/maple/view.aspx?path=seq

 

But didn't find something similar.

 

 

 

Consider a taper steel plate of uniform thickness t := 25mm as shown in the Fig. In addition to its self weight, the plate is subjected to a point load P := 100N at its mid point. Find the global force vector [F] , global stiffness matrix [K] , displacement in each element (1 and 2) , stresses in each element  (1 and 2) and reaction force at the support.Take E := 2*10^5N/mm2; rho := 8.2*10^(-5)kg/m3;

restart

t__1 := 150:

t__3 := 75:

w := 25:

l := 600:

t__2 := (t__1-t__3)/l*((1/2)*l)+t__3 = 225/2

A__1 := t__1*w = 3750``

A__2 := t__2*w = 5625/2``

A__3 := t__3*w = 1875``

Revised areas:

A__1e := (A__1+A__2)*(1/2) = 13125/4``

A__2e := (A__2+A__3)*(1/2) = 9375/4``

  E := 2*10^11:m2; F__1 := R__1:is support reaction N; F__2 := 100:N;``

rho__1 := 82*10^(-6) = 41/500000  N/mm2

rho__2 := 82*10^(-6) = 41/500000 N/mm2

l := 600:``

Number of elements,

n__e := 2:

l__e := 300 = 300````

q__0 := 100:N/m ; l := 1: m; n__e := 4:  elementsl  l__e := l/n__e: m;

We shall consider a two element system as shown in the Fig.
For element 1 Stiffness matrix K is

                                           Vector[row](2, {(1) = 1, (2) = 2})
K__1 := A__1e*E/l__e.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[2625000000000000, -2625000000000000], [-2625000000000000, 2625000000000000]])  Vector(2, {(1) = 1, (2) = 2})

For element 2 Stiffness matrix K is

                                         Vector[row](2, {(1) = 2, (2) = 3})
K__2 := A__2e*E/l__e.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[1875000000000000, -1875000000000000], [-1875000000000000, 1875000000000000]])  Vector(2, {(1) = 2, (2) = 3})

Global stiffness matrix obtained by adding all the elemental stiffness matrices and given b

           Vector[row](3, {(1) = 0, (2) = 0, (3) = 0})

K__g := Matrix(3, 3, {(1, 1) = K__1[1, 1], (1, 2) = K__1[1, 2], (1, 3) = 0, (2, 1) = K__1[2, 1], (2, 2) = K__1[1, 2]+K__2[1, 1], (2, 3) = K__2[1, 2], (3, 1) = 0, (3, 2) = K__2[2, 1], (3, 3) = K__2[2, 2]}) = Matrix([[K__1[1, 1], K__1[1, 2], 0], [K__1[2, 1], K__1[1, 2]+K__2[1, 1], K__2[1, 2]], [0, K__2[2, 1], K__2[2, 2]]])  Vector(3, {(1) = 0, (2) = 0, (3) = 0})

For element 1 Load matrix F is

  F__1e := (1/2)*`&rho;__1`*A__1e*l__e*(Vector(2, {(1) = 1, (2) = 1})) = Vector[column]([[861/25600], [861/25600]]) Vector(2, {(1) = 1, (2) = 2})

``

For element 2 Load matrix F isNULL

F__2e := (1/2)*A__2e*l__e*`&rho;__2`*(Vector(2, {(1) = 1, (2) = 1})) = Vector[column]([[123/5120], [123/5120]]) 

``

 

Download wrong_answers.mwwrong_answers.mwwrong_answers.mw

Ramakrishnan V

rukmini_ramki@hotmail.com

 

``

 

I would appreciate if anyone lets me know how to write circular references  (say 1 inside a circle to refer element 1. At present i do a drawing insert text and using.

 

Also i do not know how to remove the boundary of the overall drawing.

NULL

 

Download A_DOUBT_to_be_sent_to_prime_community.mw

Ramakrishnan V

rukmini_ramki@hotmail.com

There have come unwanted lines and marks . I donot know how to remove them. Using doc.block, remove block seems to be little tough to incorporate! Please enlighten me. Modified doc. is most welcome. Thanks. Ramakrishnan V 

Gaussian Elimination Method

 

 

Given*the*equations

  restartreset:

with(Student[LinearAlgebra])``

(1)
Coefficient Tanle

Equation 1

Equation 2

Equation 3

Equations

`m__1,1` := 3:
`` 

`m__2,1` := 2:
``

`m__3,1` := 1:
``

`m__1,1`*x__1+`m__1,2`*y+`m__1,3`*z = `m__1,4`; = 3*x__1+y-z = 3

`m__2,1`*x__1+`m__2,2`*y+`m__2,3`*z = `m__2,4`; = 2*x__1-8*y+z = -5

```m__3,1`*x__1+`m__3,2`*y+`m__3,3`*z = `m__3,4`; = x__1-2*y+9*z = 8

The equations in matrix form is given by

Matrix([[3, 1, -1, 3], [2, -8, 1, -5], [1, -2, 9, 8]])

(2)

The Gaussian Elimination gives the simplified natrix equation as given below:

Matrix([[3, 1, -1, 3], [0, -26/3, 5/3, -7], [0, 0, 231/26, 231/26]])

(3)

``The equations in simplified form are:

3*x+y-z = 3

(4)

-(26/3)*y+(5/3)*z = -7

(5)

(231/26)*z = 231/26

(6)

``

The aolution ia obtained by solving the above equations in reverse order

{x = 1, y = 1, z = 1}

(7)

 

``

 

Download GausianFinal15Nov2015.mwGausianFinal15Nov2015.mw

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Maple Product Management

Are there any commands in maple that will help me find a suitable function that approximates the numerical solution of:



  restart;
  PDE := diff(v(x, t), t) = diff(v(x, t), x, x);
  JACOBIINTEGRAL := int(JacobiTheta3(0, exp(-Pi^2*s))*v(1, t-s)^4, s = 0 .. t);
  IBC:= D[1](v)(0,t)=0,
        D[1](v)(1,t)=-0.000065*v(1, t)^4,
        v(x,0)=1;
#
# For x=0..1, t=0..1, the solution varies only very slowly
# so I have increased the timestep/spacestep, just to speed
# up results generation for diagnostic purposes
#
  pds := pdsolve( PDE, [IBC], numeric, time = t, range = 0 .. 1,
                  spacestep = 0.1e-1, timestep = 0.1e-1,
                  errorest=true
                )

diff(v(x, t), t) = diff(diff(v(x, t), x), x)

 

int(JacobiTheta3(0, exp(-Pi^2*s))*v(1, t-s)^4, s = 0 .. t)

 

(D[1](v))(0, t) = 0, (D[1](v))(1, t) = -0.65e-4*v(1, t)^4, v(x, 0) = 1

 

_m649569600

(1)

#
# Plot the solution over the ranges x=0..1,
# time=0..1. Not a lot happens!
#
  pds:-plot(x=1, t=0..1);

 

#
# Plot the estimated error over the ranges x=0..1,
# time=0..1
#
  pds:-plot( err(v(x,t)), x=1,t=0..1);

 

#
# Get some numerical solution values
#
  pVal:=pds:-value(v(x,t), output=procedurelist):
  for k from 0 by 0.1 to 1 do
      pVal(1, k)[2], pVal(1, k)[3];
  od;

 

t = 0., v(x, t) = Float(undefined)

 

t = .1, v(x, t) = .999977377613528229

 

t = .2, v(x, t) = .999967577518313666

 

t = .3, v(x, t) = .999959874331053822

 

t = .4, v(x, t) = .999952927885405241

 

t = .5, v(x, t) = .999946262964885979

 

t = .6, v(x, t) = .999939702966688881

 

t = .7, v(x, t) = .999933182128311282

 

t = .8, v(x, t) = .999926675964661227

 

t = .9, v(x, t) = .999920175361791563

 

t = 1.0, v(x, t) = .999913676928735229

(2)

 

 

 

Download PDEprob2_(2).mw

 

I am refering to the first graph, is there a way in maple to find an explicit suitable approximating function?

I.e, I want the function to have the same first graph obviously, it seems like addition of exponent and a line function, I tried plotting exp(-t)-0.3*t, it doesn't look like it approximates it very well. Any suggestion on how to implement this task in maple?

Thanks.

 

Dear collagues

Hi,

I write a code to solve a system of ODE. It solve the ODES in a wide range of parameters but as I decrease NBT below 0.5, it doesnt converge. I do my best but I couldn't find the answer. Would you please help me? Thank you

Here is my code and it should be run for 0.1<NBT<10. the value of NBT is input directly in res1.

restart:
EPSILONE:=1000:
Digits:=15:

a[mu1]:=5.45:
b[mu1]:=108.2:
a[k1]:=1.292:
b[k1]:=-11.99:




rhop:=4175:
rhobf:=998.2:
mu1[bf]:=9.93/10000:
k1[bf]:=0.597:

rhost(eta):=1-phi(eta)+phi(eta)*rhop/rhobf;
k:=unapply(k1[bf]*(1+a[k1]*phi(eta)+b[k1]*phi(eta)^2),eta);


eq1:=(diff(u(eta), eta))*a[mu1]*(diff(phi(eta), eta))+2*(diff(u(eta), eta))*b[mu1]*phi(eta)*(diff(phi(eta), eta))+((diff(u(eta), eta))+(diff(u(eta), eta))*a[mu1]*phi(eta)+(diff(u(eta), eta))*b[mu1]*phi(eta)^2)/(eta+EPSILONE)+diff(u(eta), eta, eta)+(diff(u(eta), eta, eta))*a[mu1]*phi(eta)+(diff(u(eta), eta, eta))*b[mu1]*phi(eta)^2+1-phi(eta)+phi(eta)*rhop/rhobf:
eq2:=(1+a[k1]*phi(eta)+b[k1]*phi(eta)^2)*(diff(T(eta), eta))/(eta+EPSILONE) + (a[k1]*(diff(phi(eta), eta))+2*b[k1]*phi(eta)*(diff(phi(eta), eta)))*(diff(T(eta), eta))+(1+a[k1]*phi(eta)+b[k1]*phi(eta)^2)*(diff(T(eta), eta, eta));
eq3:=diff(phi(eta),eta)+phi(eta)/(N_bt*(1+gama1*T(eta))^2)*diff(T(eta),eta):

eq2:=subs(phi(0)=phi0,eq2):
eq3:=subs(phi(0)=phi0,eq3):


eval(dsolve({eq3,phi(1)=phiv},phi(eta)),(T)(1)=1):
Phi:=normal(combine(%)):
Teq:=isolate(eval(eq2,Phi),diff(T(eta),eta)):
ueq1:=eval(eq1,Phi)=0:
ueq2:=subs(Teq,ueq1):


lambda:=0;
Ha:=0;
N_bt:=cc*NBT+(1-cc)*0.8;
kratio:=k1[p]/k1[bf]:






GUESS:=[T(eta) =0.0001*eta, u(eta) =0.1*eta, phi(eta) = 0.3*(eta-1)^4];
res1 := dsolve(subs(NBT=0.48,gama1=0.2,phiv=0.06,{eq1,eq2,eq3,u(0)=lambda*D(u)(0),D(u)(1)=0,T(0)=0,phi(1)=phiv,T(1)=1}), numeric,method=bvp[midrich],maxmesh=4000,approxsoln=GUESS, output=listprocedure,continuation=cc):
G0,G1,G2:=op(subs(subs(res1),[phi(eta),u(eta),diff(T(eta),eta)])):

masst:=evalf(int((1-G0(eta)+G0(eta)*rhop/rhobf)*G1(eta), eta = 0..1));
heatt:=(1+a[k1]*G0(0)+b[k1]*G0(0)^2)*G2(0);

plots:-odeplot(res1,[[eta,T(eta)]],0..1,legend=[T],color=["Black"],linestyle=Solid,axes=boxed,thickness=3);
plots:-odeplot(res1,[[eta,u(eta)]],0..1,legend=[u],color=["Black"],linestyle=Solid,axes=boxed,thickness=3);
plots:-odeplot(res1,[[eta,phi(eta)]],0..1,legend=[phi],color=["Black"],linestyle=Solid,axes=boxed,thickness=3);

>
>

 

Thank you

 

Amir

Hello all

I am trying to write  a tutorial about systems of linear equations, and I want to demonstrate the idea that when you have a system of 3 euqtions with 3 unknowns, the solution is the intersection point between these planes. Plotting 3 planes in Maple 2015 is fairly easy (you plot one and just drag the others in), but I don't know how to plot the intersection point. Can you help please ?

 

My equations are:

x-2y+z=0

2y-8z=8

-4x+5y+9z=-9

The intersection point is (29,16,3)

 

Thank you !

Dear Collegues

I have a system of odes as follows

restart:
#gama1:=0.2:

#rhop:=5180:
#rhobf:=998.2:
#a[mu1]:=39.11:
#b[mu1]:=533.9:
#a[k1]:=7.47:
#b[k1]:=0:

Teq := N_bt*T(eta)^2*(exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta)))))^2*(diff(T(eta), eta, eta))*gama1^2*b[k1]+N_bt*T(eta)^2*exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta))))*(diff(T(eta), eta, eta))*gama1^2*a[k1]+2*N_bt*T(eta)*(exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta)))))^2*(diff(T(eta), eta, eta))*gama1*b[k1]+N_bt*T(eta)^2*(diff(T(eta), eta, eta))*gama1^2;


UEQ:=(a[mu1]*(-(diff(T(eta), eta))/(N_bt*(1+gama1)*(1+gama1*T(eta)))+(T(eta)-1)*gama1*(diff(T(eta), eta))/(N_bt*(1+gama1)*(1+gama1*T(eta))^2))*exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta))))+2*b[mu1]*(exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta)))))^2*(-(diff(T(eta), eta))/(N_bt*(1+gama1)*(1+gama1*T(eta)))+(T(eta)-1)*gama1*(diff(T(eta), eta))/(N_bt*(1+gama1)*(1+gama1*T(eta))^2)))*(diff(u(eta), eta))+(1+a[mu1]*exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta))))+b[mu1]*(exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta)))))^2)*(diff(u(eta), eta, eta))+1-exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta))))+exp(-(T(eta)-1)/(N_bt*(1+gama1)*(1+gama1*T(eta))))*rhop/rhobf;

I want to solve them with the following boundary conditions

T(0)=0, T(1)=1

u(0)=L*D(u)(0), D(u)(1)=0

However I tried, I cannot find the solution in a closed form. I want to know that is there a closed form solution for the above odes?

Thank you

Amir

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