ecterrab

Setup(tensorsimplifier)...

Answered: ecterrab 14540

March 30 2019

3 0

Hi Escorpsy,

Using Maple 2019 I get a result - no hangs, as shown below. Perhaps more important, (take a look at the help page for Physics:-Setup ) have in mind that simplification matters are tricky, in some sense more art than exact science. The default tensor simplifier aims at doing a good job in a wide range of algebraic situations, but there is no single alternative for all cases. For that reason, the design is that you can set it to anything else more suitable for your problem if you prefer; you see below how to play around with that.

>

Physics:-g_[mu, nu] = Matrix(%id = 18446744078650807950)

(1)

First, just let it run. In my computer I get a result in ~90 sec.

>

(2)

Check the tensor simplifier being used: it is a wide-range cocktail of simplifications called Physics:-TensorSimplifier.

>

(3)

Set it to normal (ie, just the very basic stuff)

>

(4)

>

(5)

:) :) :)

But the lengths are

>

(6)

:( :(

Anyway you can simplify after having computed, to get

>

(7)

>

(8)

Playing around with the simplifier can get you the best result in different situations. Here is another option, apparently the best one for your problem that has no radicals and only trig functions

>

(9)

>

(10)

>

Download Kerr_-_tensorsimplifier_(reviewed).mw

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

PDEtools:-charstrip...

Answered: ecterrab 14540

March 27 2019

2 0

The command that computes the characteristic strip for 1st order PDEs is PDEtools:-charstrip. Take a look at its help page. The command for plotting the PDE solution following the characteristcs is PDEtools:-PDEplot.

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

dchange and PDE symmetries...

Answered: ecterrab 14540

March 27 2019

4 0

Hi Carl, Alaloush,

>

(1)

>

(2)

The first substitution, , is not a change of variables but only a substitution

>

(3)

>

(4)

The next one, , is a change of variables. Unless you tell how you are transforming the dependent variable (you didn't), I'd assume you are transforming . So this is how you use dchange to perform such a transformation

>

(5)

>

(6)

This is the call to dchange

>

(7)

Note however that the resulting PDE has derivatives with respect to and therefore the transformation you show (including the equation I guessed for the dependent variable) does not reduce the number of independent variables.

What Carl did, that I reproduce here, implicitly assumes which amounts to just removing by hand one variable from the problem. In this example works for no evident reason (the reason is shown further below)

So this is Carl's input/output

>	$PDE1 := diff(m(t, x), t)+diff((u(t, x)^2-(diff(u(t, x), x))^2)m(t, x), x) = 0; PDE2 := eval(PDE1, m(t, x) = u(t, x)-(diff(u(t, x), x, x))); ODE := convert(simplify(eval(PDE2, u(t, x) = U(-at+x)), {-a*t+x = y}), diff)$

(8)

>

U(y) = _C1, Intat(1/(_a^2-a-(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = U(y))-y-_C3 = 0, Intat(1/(_a^2-a+(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = U(y))-y-_C3 = 0, Intat(-1/(_a^2-a-(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = U(y))-y-_C3 = 0, Intat(-1/(_a^2-a+(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = U(y))-y-_C3 = 0

(9)

You now reverse the substitution and you get

>

Intat(1/(_a^2-a-(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = u(t, x))+a*t-x-_C3 = 0

(10)

>

(11)

But note however that what you did, Carl, amounts to actually (and only) removing one variable by hand. For example: forget about

>

(12)

The above is also an ODE, there is no t around, and this happens regardless of , no need for it. Also, if the idea is to use instead of , then the answer to your question on how to do it using dchange (instead of simplify/siderels as you did, Carl) is

>	$tr__2 := {t = tau, x = a*tau+y, u(t, x) = upsilon(y)}$

(13)

>

(14)

And there you are with your ODE

>

upsilon(y) = _C1, Intat(1/(_a^2-a-(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = upsilon(y))-y-_C3 = 0, Intat(1/(_a^2-a+(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = upsilon(y))-y-_C3 = 0, Intat(-1/(_a^2-a-(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = upsilon(y))-y-_C3 = 0, Intat(-1/(_a^2-a+(-4*_C1*_a+a^2+4*_C2)^(1/2))^(1/2), _a = upsilon(y))-y-_C3 = 0

(15)

You see ths solution is the same I computed in (10) for your ODE.

All OK, but how is that by simply removing one variable by hand you reduce the number of independent variables? The interesting question for me is, actually, how do you unveil the possible transformations that reduce the number of independent variables.

The answer is in InvariantSolutions . It is one of the most powerful and advanced commands in the Maple library. In brief, InvariantSolutions returns a sequence of lists of the form $[{inverse_transformation}, {transformation}]$ , where transformation involves equations for independent and dependent variables (in this case there is only one PDE but the command works for PDE systems as well). The independent variables that do not change are not mentioned.

So these are the transformations you can actually use to reduce the number of independent variables (remember that is being displayed as u)

>

$[{_t1 = t, _u1(_t1) = u(t, x)}, {t = _t1, u(t, x) = _u1(_t1)}], [{_t1 = x, _u1(_t1) = u(t, x)}, {x = _t1, u(t, x) = _u1(_t1)}], [{_t1 = x, _t2 = t, _u1(_t1) = u(t, x)*t^(1/2)}, {t = _t2, x = _t1, u(t, x) = _u1(_t1)/_t2^(1/2)}]$

(16)

The first and second one are the same as removing by hand one of the variables, which happens to work for this particular PDE (and not in general, at all).

These transformations are derived from the symmetries of PDE2. To see the infinitesimals of the corresponding symmetry transformations use Infinitesimals

>

(17)

In the above you also see an arbitrary function _F1, which means there are more than just the transformations shown in (16), obtained by (automatically) specializing these arbitrary functions.

So let's see how you use dchange to transform the variables using these transformations TR to actually diminish the number of independent variables of PDE2 in a way that is systematic and with understandable origin for the transformation.

>

(18)

>

(19)

The second transformation in TR is equivalent to what you did, Carl

>

(20)

>

(21)

>

${t = tau, x = y, u(t, x) = upsilon(y)/tau^(1/2)}$

(22)

>	$dchange({t = tau, x = y, u(t, x) = upsilon(y)/tau^(1/2)}, PDE2, simplify)$

(1/2)*(6*upsilon(y)^2*(diff(upsilon(y), y))-2*upsilon(y)^2*(diff(diff(diff(upsilon(y), y), y), y))-8*upsilon(y)*(diff(upsilon(y), y))*(diff(diff(upsilon(y), y), y))-2*(diff(upsilon(y), y))^3+2*(diff(upsilon(y), y))^2*(diff(diff(diff(upsilon(y), y), y), y))+4*(diff(upsilon(y), y))*(diff(diff(upsilon(y), y), y))^2-upsilon(y)+diff(diff(upsilon(y), y), y))/tau^(3/2) = 0

(23)

In the three cases you go an ODE, each case leads to a different solution to PDE2. Equation (10) seems to depend on but it doesn't: it only occurs in the denominator - the numerator is free of .

>

Download changes_of_variables_and_reduction_of_their_number.mw

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

PDEtools:-dchange...

Answered: ecterrab 14540

March 27 2019

2 2

See PDEtools:-dchange, it is the command for changing variables in PDEs and algebraic expressions in general.

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

Yes it is possible....

Answered: ecterrab 14540

March 26 2019

4 2

Set the coordinates, type of letter to represent space indices and define as space (3D) tensors

>

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

$[coordinatesystems = {X}, spaceindices = lowercaselatin, tensors = {Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], Physics:-gamma_[a, b], h[i, j], omega[j], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}]$

(1)

This is the expression you posted

>

f*(-omega[i]*Physics:-d_(Physics:-SpaceTimeVector[`~i`](X))+Physics:-d_(t))*(-omega[a]*Physics:-d_(Physics:-SpaceTimeVector[`~a`](X))+Physics:-d_(t))-h[i, j]*Physics:-d_(Physics:-SpaceTimeVector[`~i`](X))*Physics:-d_(Physics:-SpaceTimeVector[`~j`](X))/f

(2)

For the line element to be in the way Setup understands you need to sum over the repeated indices

>

Define now the metric:

>

$[metric = {(1, 1) = (f^2*omega[1]^2-h[1, 1])/f, (1, 2) = (1/2)*(2*f^2*omega[1]*omega[2]-h[1, 2]-h[2, 1])/f, (1, 3) = (1/2)*(2*f^2*omega[1]*omega[3]-h[1, 3]-h[3, 1])/f, (1, 4) = -f*omega[1], (2, 2) = (f^2*omega[2]^2-h[2, 2])/f, (2, 3) = (1/2)*(2*f^2*omega[2]*omega[3]-h[2, 3]-h[3, 2])/f, (2, 4) = -f*omega[2], (3, 3) = (f^2*omega[3]^2-h[3, 3])/f, (3, 4) = -f*omega[3], (4, 4) = f}]$

(3)

Check it out

>

Physics:-g_[mu, nu] = Matrix(%id = 18446744078687098150)

(4)

>

Download metric_in_index_form.mw

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

Not correctly formulated...

Answered: ecterrab 14540

March 24 2019

3 2

Hi guru kido

Some things are not properly formulated, and if I execute the worksheet you posted, I do not see what you show as output. First, in (1) you input things like u11, u22 before having defined them (you define them in (2) -> (6)). But the output of (1) shows u11, u22, ... already defined as in (2) -> (6), so you are doing something else that is not shown here, the order of execution of your worksheet is not from top to bottom. Then, think about: if you input W[i, j, k, l, nonzero], what do you see? You do not show it, but you'd see an equation where the right-hand side is a set of equations. You cannot do arithmetics with such an object, so your Weyl[~i,j,k,l,nonzer] + W[~i,j,k,l,nonzero] has no meaning, it is a nonsensical operation. Your input "halfflatmetrics in 4 D" has no natural sense too. Guru kido, these commands are powerful but you need to read their help pages to understand how to use it. An approach merely looking for a result is of no help; some understanding is.

So answering your question skipping these problems I see in your worksheet: suppose you have two tensors J[a,b,c,d] and K[a,b,c,d] and you want to show that their addition or subtraction (cannot be both, of course) is equal to 0. How do you do that? It is easy: take one minus the other one and send the expression to TensorArray, as in TensorArray(J[a,b,c,d] - K[a,b,c,d], simplifier=simplify). Now you see these are tensors of 4 indices so you cannot see a "matrix" for output. But you can use ArrayElems (take a look at the help pages of these commands before proceeding), so your input will be ArrayElems(TensorArray (J[a,b,c,d] - K[a,b,c,d], simplifier=simplify)), and if J and K are equal then the output will be an empty set; and if the output is not an empty set then they are not equal. I hope this helps you.

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

A reduction to one equation of order two...

Answered: ecterrab 14540

March 23 2019

4 0

>

Hi Jjjones98,

The following is the worksheet you posted, with adjustments and comments.

First, n the DEs you have and . Maple is case sensitive. I suppose you know that, and it is a typographical mistake, so I am rewriting as . Also, don't take this wrongly but it is good practice to always look at the help of a command before using it. In the help page of dsolve it tells that the first argument is either a differential equation, or a set or list of them. In your worksheet it is neither: you wrote the two equations in sequence, that is: not a set nor a list of equations. Fixing that too, your system is

>

sys := [((2*d^2*F(y)/dy^2*F(y))*G(y)-(d*F(y)/dy*(d*G(y)/dy))*F(y)+(n-3)*(d*F(y)/dy)^2*G(y))/(4*F(y)*G(y)^2) = A, -(3*(2*d^2*F(y)/dy^2-(d*F(y)/dy)^2*G(y)-(d*F(y)/dy*(d*G(y)/dy))*F(y)))/(F(y)^2*G(y)) = B]

Yet, the syntax for derivatifes is wrong. Look at what is behind the input aboe, copied from your worksheet:

>

[1/4*(d^2*F(y)^2/dy^2*G(y)+(n-3)*d^2*F(y)^2/dy^2*G(y))/F(y)/G(y)^2 = A, -3*(2*d^2*F(y)/dy^2-2*d^2*F(y)^2/dy^2*G(y))/F(y)^2/G(y) = B]

There are no derivatives here. In Maple, you input derivatives using the diff command. So the input for your system should have been

>

sys := [2*(diff(F(y), y, y))*F(y)*G(y)-(diff(F(y), y))*(diff(G(y), y))*F(y)+(n-3)*(diff(F(y), y))^2 = A, (3*(2*(diff(F(y), y, y))-(diff(F(y), y))^2*G(y)-(diff(F(y), y))*(diff(G(y), y))*F(y)))/(F(y)^2*G(y)) = B]

[2*(diff(diff(F(y), y), y))*F(y)*G(y)-(diff(F(y), y))*(diff(G(y), y))*F(y)+(n-3)*(diff(F(y), y))^2 = A, 3*(2*(diff(diff(F(y), y), y))-(diff(F(y), y))^2*G(y)-(diff(F(y), y))*(diff(G(y), y))*F(y))/(F(y)^2*G(y)) = B]

(1)

So this is a system with two unknowns, nonlinear and coupled. Let's first use a more compact display for it

>

(2)

Now it looks like

>

[2*(diff(diff(F(y), y), y))*F(y)*G(y)-(diff(F(y), y))*(diff(G(y), y))*F(y)+(n-3)*(diff(F(y), y))^2 = A, 3*(2*(diff(diff(F(y), y), y))-(diff(F(y), y))^2*G(y)-(diff(F(y), y))*(diff(G(y), y))*F(y))/(F(y)^2*G(y)) = B]

(3)

Decoupling this system is straightfoward

>

(4)

The above means: you need to solve only one third order equation, nonlinear, for . If you can get that solution, substituting for that solution in the first equation gives you .

So let's concentrate on the equation for F

>

(5)

This equation admits one easy symmetry with infinitesimals of degree 1

>

(6)

Using that symmetry you can reduce the differential order by 1, obtaining a 2nd order equation as follows

>

$F(y) = ODESolStruc(_a, [{diff(diff(_b(_a), _a), _a) = -12*_a*(diff(_b(_a), _a))^3/(_b(_a)^2*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*((6*B*_a^3-54*A*_a^2)*_b(_a)^3+(18*_a^2*n-54*_a^2-54*_a)*_b(_a))*(diff(_b(_a), _a))^2/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*(-6*A*B*_b(_a)^6*_a^3+(12*B*_a^3*n-36*B*_a^3+12*B*_a^2+18*A*_a)*_b(_a)^4)*(diff(_b(_a), _a))/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*(A*_a^4*B^2*_b(_a)^9+(-B^2*_a^4*n+3*B^2*_a^4)*_b(_a)^7+9*A*_b(_a)^5+(-9*n+27)*_b(_a)^3)/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))}, {_a = F(y), _b(_a) = 1/(diff(F(y), y))}, {y = Int(_b(_a), _a)+_C1, F(y) = _a}])$

(7)

So you'd now need to solve the equation for _b(_a)

>

$ode_b := op([2, 2, 1, 1], F(y) = ODESolStruc(_a, [{diff(diff(_b(_a), _a), _a) = -12*_a*(diff(_b(_a), _a))^3/(_b(_a)^2*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*((6*B*_a^3-54*A*_a^2)*_b(_a)^3+(18*_a^2*n-54*_a^2-54*_a)*_b(_a))*(diff(_b(_a), _a))^2/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*(-6*A*B*_b(_a)^6*_a^3+(12*B*_a^3*n-36*B*_a^3+12*B*_a^2+18*A*_a)*_b(_a)^4)*(diff(_b(_a), _a))/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*(A*_a^4*B^2*_b(_a)^9+(-B^2*_a^4*n+3*B^2*_a^4)*_b(_a)^7+9*A*_b(_a)^5+(-9*n+27)*_b(_a)^3)/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))}, {_a = F(y), _b(_a) = 1/(diff(F(y), y))}, {y = Int(_b(_a), _a)+_C1, F(y) = _a}]))$

(8)

There is a constant solution that is not 0

>

${_b(_a) = 0}, {_b(_a) = RootOf(A*_Z^2-n+3)}$

(9)

With it you can construct a solution for ode_F

>

$DEtools[buildsol](F(y) = ODESolStruc(_a, [{diff(diff(_b(_a), _a), _a) = -12*_a*(diff(_b(_a), _a))^3/(_b(_a)^2*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*((6*B*_a^3-54*A*_a^2)*_b(_a)^3+(18*_a^2*n-54*_a^2-54*_a)*_b(_a))*(diff(_b(_a), _a))^2/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*(-6*A*B*_b(_a)^6*_a^3+(12*B*_a^3*n-36*B*_a^3+12*B*_a^2+18*A*_a)*_b(_a)^4)*(diff(_b(_a), _a))/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))-(1/6)*(A*_a^4*B^2*_b(_a)^9+(-B^2*_a^4*n+3*B^2*_a^4)*_b(_a)^7+9*A*_b(_a)^5+(-9*n+27)*_b(_a)^3)/(_b(_a)^2*_a*((-B*_a^2+3*A*_a)*_b(_a)^2-3*_a*n+9*_a-3))}, {_a = F(y), _b(_a) = 1/(diff(F(y), y))}, {y = Int(_b(_a), _a)+_C1, F(y) = _a}]), op(({_b(_a) = 0}, {_b(_a) = RootOf(A*_Z^2-n+3)})[2]))$

F(y) = -(-y+_C1)/((n-3)/A)^(1/2)

(10)

However, this is solution involves only one integration constant, while ode_F is of third order - a general solution for it would involve three integration constants. (10) is just a particular solution, and as we see next it is of no use (leads to ). For that purpose get the equation for as a function of

>

G(y) = (-3*(diff(F(y), y))^2*n+9*(diff(F(y), y))^2+3*A+6*(diff(diff(F(y), y), y)))/(B*F(y)^2+6*(diff(diff(F(y), y), y))*F(y)+3*(diff(F(y), y))^2)

(11)

>

(12)

So what you need to do is to search for a more general solution to (8), the equation for . The dsolve command along cannot find a solution for it. The way to go is to try different things using the commands in PDEtools and DEtools for symmetry and integrating factor methods, plus any idea you may have regarding changes of variables that may simplify the problem (PDEtools:-dchange is the command with which you perform those changes).

The first question is whether ode_b admits more point symmetries. Below I do not go further but show you what would be the first steps of such a human-guided search for a possible soution. First rewrite the equation using more readable variables

>

(13)

>

(14)

The system of equations for for point symmetry infinitesimals is

>

(15)

>

(16)

If this sytsem admits solutions other than then you can perform further reductions of order (from 2 to 1, the restart and from 1 to 0, ie the solution). To see if a solution of that sort exists the input is

>

But the above took more than 2 minutes in my computer so I interrupted, I do not have more time for this. What I verified beyond the above is that there are no polynomial solutions to this system.

The commands that could help you, from PDEtools: Infinitesimals and InvariantSolutions with their various options - they require some understanding of the symmetry methods though, also ConservedCurrents (has to do with integrating factors).

From DEtools: the commands odepde and gensys, to generate systems of equations whose solution would help you solve ode_H, and symgen and intfactor to tackle either ode_H or any 1-equation intermediate step.

Hope this helps - concretely the above only reduces your system of two equations, two unknowns, differential order 2 to one equation in one unknown of differential order 2.

>

Download Ric_Equations_(reviewed).mw

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

Fixed, within the Physics Updates v.335...

Answered: ecterrab 14540

March 19 2019

4 0

Hi nm
Good catch, the issue is fixed and the fix available to everybody within the Maplesoft Physics Updates v.335 or higher, that only works with Maple 2019.

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

It was a problem in int, not pdsolve. Fi...

Answered: ecterrab 14540

March 17 2019

4 4

Hi nm,

The problem is fixed and the fix uploaded for everybody within the Maplesoft Physics Updates v.333 (or higher).

Regarding the problem itself, it was in int, not pdsolve, and unrelated to installing or not the Maplesoft Physics Updates.

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

The Physics Updates for Maple 2018.2...

Answered: ecterrab 14540

March 15 2019

4 3

Hi

The Physics package is developed by Maplesoft and is an integral part of the Maple system. It is not sold separately.

On the other hand, for Maple 2018.2, there is the Physics Updates, also developed by Maplesoft, that is just the advanced version of the Physics package (the one to appear in the next release), that you can install from the MapleCloud (click that icon on the upper-right corner of your Maple worksheet) and includes a relevant part of the new functionality for tensors that appeared in Maple 2019.

These Physics Updates are produced along the year, with uploads several times per week, including bug fixes and new developments frequently following feedback by users (here in Mapleprimes, or sent directly to physics@maplesoft.com). That is important. If you have a computational request that has some generality and it is clearly and properly conveyed, it is considered as a candidate for extending the functionality already installed. You feedback is welcome, and you may have it working already in a day or a week within one of these Updates.

The current version of the Physics Updates for Maple 2018 is 329, so along the last 12 month the Updates where updated 329 times. Version 329 is also the last one for Maple 2018. I plan to upload on Monday the first version for Maple 2019 with new developments for the computation of exact solutions to partial differential equations with boundary conditions, as well as some further enhancements in tensor simplification, things that were not ready for Maple 2019.0. By the way tensor simplification is an area that will see more improvements along this year.

Regarding (2) and (3), I suggest you take a look at A Complete Guide for Tensor computations using Physics. At the end of that post there are two links, to a pdf version with all the sections open, and to a mw Maple worksheet with the same contents, that you can use to experiment yourself. If what you need is not there, then you can post here a worksheet showing what is what you intended to compute and couldn't.

Update March/17: The first Physics Updates for Maple 2019 is already available for installation via MapleCloud. In this moment, that also means that you cannot use the MapleCloud to install the last version for Maple 2018. So to install the Updates in Maple 2018 open Maple and enter PackageTools:-Install("5137472255164416", version = 329, overwrite)

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

The Physics Updates is now available for...

Answered: ecterrab 14540

March 15 2019

4 0

Hi

I'm busy with a presentation tomorrow in India. The Physics Updates v.329 is the last one for Maple 2018. Note anyway that Maple 2019 already contains all the developments you see in the Physics Updates for Maple 2018.

Hopefully, on Monday, we will have the first Physics Updates for Maple 2019. There are already developments post Maple 2019 for PDE & BC (more equations solved) and in the simplification of tensorial expressions. I will write in this thread again.

Update March/17: The first Physics Updates for Maple 2019 is already available for installation through the MapleCloud. Also, at this moment you cannot use the MapleCloud to install the Physics Updates for Maple 2018. So, to install the last version of the Updates for Maple 2018, open Maple 2018 and enter PackageTools:-Install("5137472255164416", version = 329, overwrite)

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

A = B = C is not correct...

Answered: ecterrab 14540

March 13 2019

0 1

In (7) you have a malformed equation, with two equal signs `=`. Try correcting that then if you could explain as bit better what do you mean by elimination qx,qy,qz in favor of T?

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

Generic differential operator (Physics)...

Answered: ecterrab 14540

March 09 2019

3 0

An alternative to having an operator that is a combination of operators (as you show in your post) but it is also not a combination of operators, is to set a generic differential operator.

For that purpose, in the Physics package you can set differential operators (see ?Physics:-Setup, you indicate the name of the operator and the differentiation variables related to it). The key observation is that you can set them without saying how these operators are defined, and then compute using then within algebraic expressions, with multiplication `*` representing also their application, as we do with paper and pencil. These differential operators are non-commutative, so no problem here: after you load Physics, the product operator understands noncommutative and anticommutative operands. When you want to have these generic operators applied you can use Physics:-Library:-ApplyProductsOfDifferentialOperators (see ?Physics,Library). And then, when it is convenient in your formulation, you can specify the form of the differential operators you set using Physics:-Setup, so that your definition is taken into account.

For examples of all the above, see ?updates,Maple2016,Physics (the section on "Operatorial Algebraic Expressions Involving the Differential Operators of the Physics package") and ?updates,Maple2018,Physics (the section "User-defined algebraic differential operators" - this is kinda new stuff introduced during 2017).

Some comments to the side: in Physics you also have ToFieldComponents and ToSuperfields, both relevant for working with superfunctions. Also, during 2018 significant improvements happened in Physics:-Gtaylor. This command, for computing series of superfunctions, got rewritten and now handles rather general cases.

I imagine that you may have more questions - if so, could you please post the question with a worksheet containing the formulation of what you want to do up to what you know how to do it. That helps to understand and so giving you feedback that is more specific.

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

Try using Physics...

Answered: ecterrab 14540

March 09 2019

5 8

DualOfTheWeylTensor.mw

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

This is using the Ph...

Answered: ecterrab 14540

March 07 2019

2 0

This is using the Physics package, I find it simpler

>

$`Default differentiation variables for d_, D_ and dAlembertian are:`*{X = (x1, x2, x3, x4)}$

(1)

From what you posted, this is your line element:

>

(2)

To have everything GR related you only need to set the metric

>

(3)

Check it out just to be sure this is what you wanted - and if it is not, change the line element (2)

>

Physics:-g_[mu, nu] = Matrix(%id = 18446744078538592734)

(4)

Now, you realize this is a metric in which all the components are constants, there is no curvature, the Christoffel symbols (proportional to derivatives o the metric with respect to the coordinates) are equal to zero. Are you sure that is what you want? ... I will assume here that the three ujk are actually functions of X

>

(5)

>

(6)

>

(7)

>

Physics:-g_[mu, nu] = Matrix(%id = 18446744078538571038)

(8)

Now you have, for example,

>

Physics:-Christoffel[1, mu, nu] = Matrix(%id = 18446744078538560926)

(9)

For a more compact notation avoiding the redundant display of the functionality (X) of the ujk, and displaying derivatives indexed, use

>

(10)

>

Physics:-Christoffel[1, mu, nu] = Matrix(%id = 18446744078538560926)

(11)

For the symbols of the second kind, when indicating the indices, write the first one contravariant

>

Physics:-Christoffel[`~1`, mu, nu] = Matrix(%id = 18446744078369100254)

(12)

All the nonzero at once

>

(13)

>

(14)

The same applies to all the general relativity tensors (also to tensors you define), for example:

>

Physics:-Christoffel[alpha, mu, nu] = (1/2)*Physics:-d_[nu](Physics:-g_[alpha, mu], [X])+(1/2)*Physics:-d_[mu](Physics:-g_[alpha, nu], [X])-(1/2)*Physics:-d_[alpha](Physics:-g_[mu, nu], [X])