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Posted: ecterrab 14540 Product: Maple

June 27 2016

4 0

Hi,
The latest update to the differential equations Maple libraries (this week, can be downloaded from the Maplesoft R&D webpage for Differential Equations and Mathematical functions) includes new functionality in pdsolve, regarding whether the solution for a PDE or PDE system is or not a general solution.

In brief, a general solution of a PDE in 1 unknown, that has differential order N, and where the unknown depends on M independent variables, involves N arbitrary functions of M-1 arguments. It is not entirely evident how to extend this definition in the case of a coupled, possibly nonlinear PDE system. However, using differential algebra techniques (automatically used by pdsolve when tackling a PDE system), that extension to define a general solution for a DE system is possible, and also when the system involves ODEs and PDEs, and/or algebraic (that is, non-differential) equations, and/or inequations of the form involving the unknowns, and all of this in the presence of mathematical functions (based on the use of Maple's PDEtools:-dpolyform). This is a very nice case were many different advanced developments come together to naturally solve a problem that otherwise would be rather difficult.

The issues at the center of this Maple development/post are then:

a) How do you know whether a PDE or PDE system solution returned is a general solution?

b) How could you indicate to pdsolve that you are only interested in a general PDE or PDE system solution?

The answer to a) is now always present in the last line of the userinfo. So input infolevel[pdsolve] := 3 before calling pdsolve, and check what the last line of the userinfo displayed tells.

The answer to b) is a new option, generalsolution, implemented in pdsolve so that it either returns a general solution or otherwise it returns NULL. If you do not use this new option, then pdsolve works as always: first it tries to compute a general solution and if it fails in doing that it tries to compute a particular solution by separating the variables in different ways, or computing a traveling wave solution or etc. (a number of other well known methods).

The examples that follow are from the help page pdsolve,system, and show both the new userinfo telling whether the solution returned is a general one and the option generalsolution at work.The examples are all of differential equation systems but the same userinfos and generalsolution option work as well in the case of a single PDE.

Example 1.

Solve the determining PDE system for the infinitesimals of the symmetry generator of example 11 from Kamke's book . Tell whether the solution computed is or not a general solution.

(1.1)

The PDE system satisfied by the symmetries of Kamke's ODE example number 11 is

sys__1 := [diff(xi(x, y), y, y) = 0, diff(eta(x, y), y, y)-2*(diff(xi(x, y), y, x)) = 0, 3*x^r*y^n*(diff(xi(x, y), y))*a+2*(diff(eta(x, y), y, x))-(diff(xi(x, y), x, x)) = 0, 2*(diff(xi(x, y), x))*x^r*y^n*a-x^r*y^n*(diff(eta(x, y), y))*a+eta(x, y)*a*x^r*y^n*n/y+xi(x, y)*a*x^r*r*y^n/x+diff(eta(x, y), x, x) = 0]

This is a second order linear PDE system, with two unknowns and four equations. Its general solution is given by the following, where we now can tell that the solution is a general one by reading the last line of the userinfo. Note that because the system is overdetermined, a general solution in this case does not involve any arbitrary function

-> Solving ordering for the dependent variables of the PDE system: [xi(x,y), eta(x,y)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
<- Returning a *general* solution

${eta(x, y) = -_C1*y*(r+2)/(n-1), xi(x, y) = _C1*x}$

(1.2)

Next we indicate to pdsolve that and are parameters of the problem, and that we want a solution for , making more difficult to identify by eye whether the solution returned is or not a general one. Again the last line of the userinfo tells that pdsolve's solution is indeed a general one

[diff(diff(xi(x, y), y), y) = 0, diff(diff(eta(x, y), y), y)-2*(diff(diff(xi(x, y), x), y)) = 0, 3*x^r*y^n*(diff(xi(x, y), y))*a+2*(diff(diff(eta(x, y), x), y))-(diff(diff(xi(x, y), x), x)) = 0, 2*(diff(xi(x, y), x))*x^r*y^n*a-x^r*y^n*(diff(eta(x, y), y))*a+eta(x, y)*a*x^r*y^n*n/y+xi(x, y)*a*x^r*r*y^n/x+diff(diff(eta(x, y), x), x) = 0, n <> 1]

(1.3)

>	$`sol__1.1` := pdsolve(`sys__1.1`, parameters = {n, r})$

-> Solving ordering for the dependent variables of the PDE system: [r, n, xi(x,y), eta(x,y)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
tackling triangularized subsystem with respect to r
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to r
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to r
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to n
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
tackling triangularized subsystem with respect to xi(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to eta(x,y)
<- Returning a *general* solution

${n = 2, r = -5, eta(x, y) = y*(_C1*x+3*_C2), xi(x, y) = x*(_C1*x+_C2)}, {n = 2, r = -20/7, eta(x, y) = -(2/343)*(-6*_C1*x^2-98*x^(8/7)*_C1*a*y-147*_C2*a*x*y)/(x*a), xi(x, y) = _C1*x^(8/7)+_C2*x}, {n = 2, r = -15/7, eta(x, y) = -(1/343)*(-49*_C2*a*x*y-147*x^(6/7)*_C1*a*y+12*_C1*x)/(x*a), xi(x, y) = _C1*x^(6/7)+_C2*x}, {n = 2, r = r, eta(x, y) = -_C1*y*(r+2), xi(x, y) = _C1*x}, {n = -r-3, r = r, eta(x, y) = ((_C1*x+_C2)*r+4*_C1*x+2*_C2)*y/(r+4), xi(x, y) = x*(_C1*x+_C2)}, {n = n, r = r, eta(x, y) = -_C1*y*(r+2)/(n-1), xi(x, y) = _C1*x}$

(1.4)

(1.5)

Example 2.

Compute the solution of the following (linear) overdetermined system involving two PDEs, three unknown functions, one of which depends on 2 variables and the other two depend on only 1 variable.

sys__2 := [-(diff(F(r, s), r, r))+diff(F(r, s), s, s)+diff(H(r), r)+diff(G(s), s)+s = 0, diff(F(r, s), r, r)+2*(diff(F(r, s), r, s))+diff(F(r, s), s, s)-(diff(H(r), r))+diff(G(s), s)-r = 0]

The solution for the unknowns G, H, is given by the following expression, were again determining whether this solution, that depends on 3 arbitrary functions, , is or not a general solution, is non-obvious.

-> Solving ordering for the dependent variables of the PDE system: [F(r,s), H(r), G(s)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [r, s]
tackling triangularized subsystem with respect to F(r,s)
First set of solution methods (general or quasi general solution)
Trying differential factorization for linear PDEs ...
differential factorization successful.
First set of solution methods successful
tackling triangularized subsystem with respect to H(r)
tackling triangularized subsystem with respect to G(s)
<- Returning a *general* solution

${F(r, s) = _F1(s)+_F2(r)+_F3(s-r)-(1/12)*r^2*(r-3*s), G(s) = -(diff(_F1(s), s))-(1/4)*s^2+_C2, H(r) = diff(_F2(r), r)-(1/4)*r^2+_C1}$

(1.6)

(1.7)

Example 3.

Compute the solution of the following nonlinear system, consisting of Burger's equation and a possible potential.

sys__3 := [diff(u(x, t), t)+2*u(x, t)*(diff(u(x, t), x))-(diff(u(x, t), x, x)) = 0, diff(v(x, t), t) = -v(x, t)*(diff(u(x, t), x))+v(x, t)*u(x, t)^2, diff(v(x, t), x) = -u(x, t)*v(x, t)]

We see that in this case the solution returned is not a general solution but two particular ones; again the information is in the last line of the userinfo displayed

-> Solving ordering for the dependent variables of the PDE system: [v(x,t), u(x,t)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, t]
tackling triangularized subsystem with respect to v(x,t)
tackling triangularized subsystem with respect to u(x,t)
First set of solution methods (general or quasi general solution)
Second set of solution methods (complete solutions)
Trying methods for second order PDEs
Third set of solution methods (simple HINTs for separating variables)
PDE linear in highest derivatives - trying a separation of variables by *
HINT = *
Fourth set of solution methods
Trying methods for second order linear PDEs
Preparing a solution HINT ...
Trying HINT = _F1(x)*_F2(t)
Fourth set of solution methods
Preparing a solution HINT ...
Trying HINT = _F1(x)+_F2(t)
Trying travelling wave solutions as power series in tanh ...
* Using tau = tanh(t*C[2]+x*C[1]+C[0])
* Equivalent ODE system: {C[1]^2*(tau^2-1)^2*diff(diff(u(tau),tau),tau)+(2*C[1]^2*(tau^2-1)*tau+2*u(tau)*C[1]*(tau^2-1)+C[2]*(tau^2-1))*diff(u(tau),tau)}
* Ordering for functions: [u(tau)]
* Cases for the upper bounds: [[n[1] = 1]]
* Power series solution [1]: {u(tau) = tau*A[1,1]+A[1,0]}
* Solution [1] for {A[i, j], C[k]}: [[A[1,1] = 0], [A[1,0] = -1/2*C[2]/C[1], A[1,1] = -C[1]]]
travelling wave solutions successful.
tackling triangularized subsystem with respect to v(x,t)
First set of solution methods (general or quasi general solution)
Trying differential factorization for linear PDEs ...
Trying methods for PDEs "missing the dependent variable" ...
Second set of solution methods (complete solutions)
Trying methods for second order PDEs
Third set of solution methods (simple HINTs for separating variables)
PDE linear in highest derivatives - trying a separation of variables by *
HINT = *
Fourth set of solution methods
Trying methods for second order linear PDEs
Preparing a solution HINT ...
Trying HINT = _F1(x)*_F2(t)
Third set of solution methods successful
tackling triangularized subsystem with respect to u(x,t)
<- Returning a solution that *is not the most general one*

${u(x, t) = -_C2*tanh(_C2*x+_C3*t+_C1)-(1/2)*_C3/_C2, v(x, t) = 0}, {u(x, t) = -_c[1]^(1/2)*((exp(_c[1]^(1/2)*x))^2*_C1-_C2)/((exp(_c[1]^(1/2)*x))^2*_C1+_C2), v(x, t) = _C3*exp(_c[1]*t)*_C1*exp(_c[1]^(1/2)*x)+_C3*exp(_c[1]*t)*_C2/exp(_c[1]^(1/2)*x)}$

(1.8)

(1.9)

This example is also good for illustrating the other related new feature: one can now request to pdsolve to only compute a general solution (it will return NULL if it cannot achieve that). Turn OFF userinfos and try with this example

This returns NULL:

Example 4.

Another where the solution returned is particular, this time for a linear system, conformed by 38 PDEs, also from differential equation symmetry analysis

There are 38 coupled equations

(1.10)

When requesting a general solution pdsolve returns NULL:

A solution that is not a general one, is however computed by default if calling pdsolve without the generalsolution option. In this case again the last line of the userinfo tells that the solution returned is not a general solution

(1.11)

-> Solving ordering for the dependent variables of the PDE system: [eta[1](x,y,z,t,u), xi[1](x,y,z,t,u), xi[2](x,y,z,t,u), xi[3](x,y,z,t,u), xi[4](x,y,z,t,u)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, y, z, u]
tackling triangularized subsystem with respect to eta[1](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F1(x,y,z,t), _F2(x,y,z,t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, y, z, u]
tackling triangularized subsystem with respect to _F1(x,y,z,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F3(x,y,z), _F4(x,y,z)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y, z, t]
tackling triangularized subsystem with respect to _F3(x,y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F4(x,y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F5(y,z), _F6(y,z)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [y, z, x]
tackling triangularized subsystem with respect to _F5(y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F6(y,z)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F7(z), _F8(z)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [z, y]
tackling triangularized subsystem with respect to _F7(z)
tackling triangularized subsystem with respect to _F8(z)
tackling triangularized subsystem with respect to _F2(x,y,z,t)
First set of solution methods (general or quasi general solution)
Trying differential factorization for linear PDEs ...
Trying methods for PDEs "missing the dependent variable" ...
Second set of solution methods (complete solutions)
Third set of solution methods (simple HINTs for separating variables)
PDE linear in highest derivatives - trying a separation of variables by *
HINT = *
Fourth set of solution methods
Preparing a solution HINT ...
Trying HINT = _F3(x)*_F4(y)*_F5(z)*_F6(t)
Third set of solution methods successful

tackling triangularized subsystem with respect to xi[1](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F1(x,z,t), _F2(x,z,t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, z, y]
tackling triangularized subsystem with respect to _F1(x,z,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F2(x,z,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful

-> Solving ordering for the dependent variables of the PDE system: [_F3(x,t), _F4(x,t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, z]
tackling triangularized subsystem with respect to _F3(x,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to _F4(x,t)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F5(x), _F6(x)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [x, t]
tackling triangularized subsystem with respect to _F5(x)
tackling triangularized subsystem with respect to _F6(x)
tackling triangularized subsystem with respect to xi[2](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
-> Solving ordering for the dependent variables of the PDE system: [_F1(t), _F2(t)]
-> Solving ordering for the independent variables (can be changed using the ivars option): [t, z]
tackling triangularized subsystem with respect to _F1(t)
tackling triangularized subsystem with respect to _F2(t)
tackling triangularized subsystem with respect to xi[3](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
tackling triangularized subsystem with respect to xi[4](x,y,z,t,u)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
First set of solution methods successful
<- Returning a solution that *is not the most general one*

${eta[1](x, y, z, t, u) = (_C13*(_C10*(exp(_c[3]^(1/2)*z))^2+_C11)*(_C8*(exp(_c[2]^(1/2)*y))^2+_C9)*(_C6*(exp(_c[1]^(1/2)*x))^2+_C7)*cos((-_c[1]-_c[2]-_c[3])^(1/2)*t)+_C12*(_C10*(exp(_c[3]^(1/2)*z))^2+_C11)*(_C8*(exp(_c[2]^(1/2)*y))^2+_C9)*(_C6*(exp(_c[1]^(1/2)*x))^2+_C7)*sin((-_c[1]-_c[2]-_c[3])^(1/2)*t)+u*exp(_c[1]^(1/2)*x)*exp(_c[2]^(1/2)*y)*exp(_c[3]^(1/2)*z)*(_C1*t+_C2*x+_C3*y+_C4*z+_C5))/(exp(_c[1]^(1/2)*x)*exp(_c[2]^(1/2)*y)*exp(_c[3]^(1/2)*z)), xi[1](x, y, z, t, u) = -(1/2)*_C2*x^2+(1/2)*(-2*_C1*t-2*_C3*y-2*_C4*z+2*_C17)*x+(1/2)*(-t^2+y^2+z^2)*_C2+_C16*t+_C15*z+_C14*y+_C18, xi[2](x, y, z, t, u) = -(1/2)*_C3*y^2+(1/2)*(-2*_C1*t-2*_C2*x-2*_C4*z+2*_C17)*y+(1/2)*(-t^2+x^2+z^2)*_C3+_C20*t+_C19*z-_C14*x+_C21, xi[3](x, y, z, t, u) = -(1/2)*_C4*z^2+(1/2)*(-2*_C1*t-2*_C2*x-2*_C3*y+2*_C17)*z+(1/2)*(-t^2+x^2+y^2)*_C4+_C22*t-_C19*y-_C15*x+_C23, xi[4](x, y, z, t, u) = -(1/2)*_C1*t^2+(1/2)*(-2*_C2*x-2*_C3*y-2*_C4*z+2*_C17)*t+(1/2)*(-x^2-y^2-z^2)*_C1+_C20*y+_C22*z+_C16*x+_C24}$

(1.12)

(1.13)

Example 5.

Finally, the new userinfos also tell whether a solution is or not a general solution when working with PDEs that involve anticommutative variables set using the Physics package

(1.14)

Set first and as suffixes for variables of type/anticommutative (see Setup )

$[anticommutativeprefix = {Q, _lambda, theta}]$

(1.15)

A PDE system example with two unknown anticommutative functions of four variables, two commutative and two anticommutative; to avoid redundant typing in the input that follows and redundant display of information on the screen let's use PDEtools:-diff_table PDEtools:-declare

(1.16)

(1.17)

Consider the system formed by these two PDEs (because of the q diff_table just defined, we can enter derivatives directly using the function's name indexed by the differentiation variables)

(1.18)

(1.19)

The solution returned for this system is indeed a general solution

-> Solving ordering for the dependent variables of the PDE system: [_F4(x,y), _F2(x,y), _F3(x,y)]

-> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
tackling triangularized subsystem with respect to _F4(x,y)
tackling triangularized subsystem with respect to _F2(x,y)
tackling triangularized subsystem with respect to _F3(x,y)
First set of solution methods (general or quasi general solution)
Trying simple case of a single derivative.
HINT = _F6(x)+_F5(y)
Trying HINT = _F6(x)+_F5(y)
HINT is successful
First set of solution methods successful
<- Returning a *general* solution

(1.20)

This solution involves an anticommutative constant , analogous to the commutative constants where n is an integer.

Download PDE_general_solutions.mw

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

Posted: ecterrab 14540 Product: Maple

June 27 2016

8 3

New developments (after the release of Maple 2016) happened in the project on exact solutions for "Partial Differential Equations & Boundary Conditions". This is work in collaboration with Katherina von Bulow and the improvements are of wide range, representing a noticeable step forward in the capabilities of the Maple system for this kind of problem. As usual, these improvements can be installed in current Maple 2016 by downloading the updated library from the Maplesoft R&D webpage for Differential Equations and Mathematical functions (the update is distributed merged with the updates of the Physics package)

The improvements cover:

•	PDE&BC in semi-infinite domains for which a bounded solution is sought

•	PDE & BC problems in bounded spatial domains via eigenfunction (Fourier) expansions

•	Implementation of another algebraic method for tackling linear PDE & BC

•	Improvements in solving PDE & BC solutions by first finding the PDE's general solution.

•	Improvements in solving PDE & BC problems by using a Fourier transform.

•	PDE & BC problems that used to require the option HINT = `+` are now solved automatically

What follows is a set of examples solved now with these new developments, organized in sections according to the kind of problem. Where relevant, the sections include a subsection on "How it works step by step".

PDE&BC in semi-infinite domains for which a bounded solution is sought can now also be solved via Laplace transforms

Maple is now able to solve more PDE&BC problems via Laplace transforms.

How it works: Laplace transforms act to change derivatives with respect to one of the independent variables of the domain into multiplication operations in the transformed domain. After applying a Laplace transform to the original problem, we can simplify the problem using the transformed BC, then solve the problem in the transformed domain, and finally apply the inverse Laplace transform to arrive at the final solution. It is important to remember to give pdsolve any necessary restrictions on the variables and constants of the problem, by means of the "assuming" command.

A new feature is that we can now tell pdsolve that the dependent variable is bounded, by means of the optional argument HINT = boundedseries.

Consider the problem of a falling cable lying on a table that is suddenly removed (cf. David J. Logan's Applied Partial Differential Equations p.115).

If we ask pdsolve to solve this problem without the condition of boundedness of the solution, we obtain:

u(x, t) = -invlaplace(exp(s*x/c)*_F1(s), s, t)+invlaplace(exp(-s*x/c)*_F1(s), s, t)-(1/2)*g*t^2+invlaplace(exp(s*x/c)/s^3, s, t)*g

(1.1)

New: If we now ask for a bounded solution, by means of the option HINT = boundedseries, pdsolve simplifies the problem accordingly.

u(x, t) = (1/2)*g*(Heaviside(t-x/c)*(c*t-x)^2-c^2*t^2)/c^2

(1.2)

And we can check this answer against the original problem, if desired:

(1.3)

How it works, step by step

Let us see the process this problem undergoes to be solved by pdsolve, step by step.

First, the Laplace transform is applied to the PDE:

s^2*laplace(u(x, t), t, s)-(D[2](u))(x, 0)-s*u(x, 0)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s

(1.1.1)

and the result is simplified using the initial conditions:

>	$simplified_transformed_PDE := eval(transformed_PDE, {iv[1]})$

s^2*laplace(u(x, t), t, s)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s

(1.1.2)

Next, we call the function "laplace(u(x,t),t,s)" by the new name U:

(1.1.3)

And this equation, which is really an ODE, is solved:

U(x, s) = exp(-s*x/c)*_F2(s)+exp(s*x/c)*_F1(s)-g/s^3

(1.1.4)

Now, since we want a BOUNDED solution, the term with the positive exponential must be zero, and we are left with:

U(x, s) = exp(-s*x/c)*_F2(s)-g/s^3

(1.1.5)

Now, the initial solution must also be satisfied. Here it is, in the transformed domain:

(1.1.6)

Or, in the new variable U,

(1.1.7)

And by applying it to bounded_solution_U, we find the relationship

(_F2(s)*s^3-g)/s^3 = 0

(1.1.8)

(1.1.9)

so that our solution now becomes

U(x, s) = exp(-s*x/c)*g/s^3-g/s^3

(1.1.10)

to which we now apply the inverse Laplace transform to obtain the solution to the problem:

u(x, t) = (1/2)*g*(-t^2+Heaviside(t-x/c)*(c*t-x)^2/c^2)

(1.1.11)

Four other related examples

A few other examples:

pde[2] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))

(1.2.1)

(1.2.2)

(1.2.3)

(1.2.4)

(1.2.5)

(1.2.6)

The following is an example from page 76 in Logan's book:

pde[5] := diff(u(x, t), t) = diff(u(x, t), x, x)

u(x, t) = (1/2)*x*(int(f(_U1)*exp(-x^2/(4*t-4*_U1))/(t-_U1)^(3/2), _U1 = 0 .. t))/Pi^(1/2)

(1.2.7)

More PDE&BC problems in bounded spatial domains can now be solved via eigenfunction (Fourier) expansions

The code for solving PDE&BC problems in bounded spatial domains has been expanded. The method works by separating the variables by product, so that the problem is transformed into an ODE system (with initial and/or boundary conditions) problem, one of which is a Sturm-Liouville problem (a type of eigenvalue problem) which has infinitely many solutions - hence the infinite series representation of the solutions.

Here is a simple example for the heat equation:

pde__6 := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv__6 := u(0, t) = 0, u(l, t) = 0

u(x, t) = Sum(_C1*sin(_Z1*Pi*x/l)*exp(-k*Pi^2*_Z1^2*t/l^2), _Z1 = 1 .. infinity)

(2.1)

(2.2)

Now, consider the displacements of a string governed by the wave equation, where c is a constant (cf. Logan p.28).

pde__7 := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))

u(x, t) = Sum(sin(_Z2*Pi*x/l)*(sin(c*_Z2*Pi*t/l)*_C1+cos(c*_Z2*Pi*t/l)*_C5), _Z2 = 1 .. infinity)

(2.3)

(2.4)

Another wave equation problem (cf. Logan p.130):

pde__8 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = 0; iv__8 := u(0, t) = 0, (D[2](u))(x, 0) = 0, (D[1](u))(l, t) = 0, u(x, 0) = f(x)

u(x, t) = Sum(2*(Int(f(x)*sin((1/2)*Pi*(2*_Z3+1)*x/l), x = 0 .. l))*sin((1/2)*Pi*(2*_Z3+1)*x/l)*cos((1/2)*c*Pi*(2*_Z3+1)*t/l)/l, _Z3 = 1 .. infinity)

(2.5)

(2.6)

Here is a problem with periodic boundary conditions (cf. Logan p.131). The function stands for the concentration of a chemical dissolved in water within a tubular ring of circumference . The initial concentration is given by , and the variable is the arc-length parameter that varies from 0 to .

u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*sin(_Z4*Pi*x/l), x = 0 .. 2*l))*sin(_Z4*Pi*x/l)+(Int(f(x)*cos(_Z4*Pi*x/l), x = 0 .. 2*l))*cos(_Z4*Pi*x/l))*exp(-M*Pi^2*_Z4^2*t/l^2)/l, _Z4 = 1 .. infinity)

(2.7)

(2.8)

The following problem is for heat flow with both boundaries insulated (cf. Logan p.166, 3rd edition)

u(x, t) = Sum(2*(Int(f(x)*cos(_Z6*Pi*x/l), x = 0 .. l))*cos(_Z6*Pi*x/l)*exp(-k*Pi^2*_Z6^2*t/l^2)/l, _Z6 = 1 .. infinity)

(2.9)

(2.10)

This is a problem in a bounded domain with the presence of a source. A source term represents an outside influence in the system and leads to an inhomogeneous PDE (cf. Logan p.149):

u(x, t) = Int(Sum(2*(Int(p(x, tau1)*sin(_Z7*x), x = 0 .. Pi))*sin(_Z7*x)*sin(c*_Z7*(t-tau1))/(Pi*_Z7*c), _Z7 = 1 .. infinity), tau1 = 0 .. t)

(2.11)

Current pdetest is unable to verify that this solution cancells the mainly because it currently fails in identifying that there is a fourier expansion in it, but its subroutines for testing the boundary conditions work well with this problem

>	$pdetest_BC({ans__11}, [iv__11], [u(x, t)])$

(2.12)

Consider a heat absorption-radiation problem in the bounded domain :

pde__12 := diff(u(x, t), t) = diff(u(x, t), x, x)

u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*cos((1/2)*_Z8*Pi*x), x = 0 .. 2))*cos((1/2)*_Z8*Pi*x)+(Int(f(x)*sin((1/2)*_Z8*Pi*x), x = 0 .. 2))*sin((1/2)*_Z8*Pi*x))*exp(-(1/4)*Pi^2*_Z8^2*t), _Z8 = 1 .. infinity)

(2.13)

(2.14)

Consider the nonhomogeneous wave equation problem (cf. Logan p.213, 3rd edition):

pde__13 := diff(u(x, t), t, t) = A*x+diff(u(x, t), x, x)

u(x, t) = Int(Sum(2*A*(Int(x*sin(Pi*_Z9*x), x = 0 .. 1))*sin(Pi*_Z9*x)*sin(Pi*_Z9*(t-tau1))/(Pi*_Z9), _Z9 = 1 .. infinity), tau1 = 0 .. t)

(2.15)

>	$pdetest_BC({ans__13}, [iv__13], [u(x, t)])$

(2.16)

Consider the following Schrödinger equation with zero potential energy (cf. Logan p.30):

pde__14 := I*h*(diff(f(x, t), t)) = -h^2*(diff(f(x, t), x, x))/(2*m)

f(x, t) = Sum(_C1*sin(_Z10*Pi*x/d)*exp(-((1/2)*I)*h*Pi^2*_Z10^2*t/(d^2*m)), _Z10 = 1 .. infinity)

(2.17)

(2.18)

Another method has been implemented for linear PDE&BC

This method is for problems of the form

", w(`x__i`,0) = f(`x__i`), (&PartialD;w)/(&PartialD;t)() ? ()|() ? (t=0) =g(`x__i`)"

where M is an arbitrary linear differential operator of any order which only depends on the spatial variables .

Here are some examples:

pde__15 := diff(w(x1, x2, x3, t), t)-(diff(w(x1, x2, x3, t), x2, x1))-(diff(w(x1, x2, x3, t), x3, x1))-(diff(w(x1, x2, x3, t), x3, x3))+diff(w(x1, x2, x3, t), x3, x2) = 0

(3.1)

(3.2)

Here are two examples for which the derivative with respect to t is of the second order, and two initial conditions are given:

pde__16 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))

(3.3)

(3.4)

pde__17 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))

w(x1, x2, x3, t) = (1/2)*t^4*x1+t^2*x1^3+3*t^2*x1^2*x3+x1^3*x3^2+(1/6)*t^3-t*x2*x3+cos(x1)*t+sin(x1)

(3.5)

(3.6)

More PDE&BC problems are now solved via first finding the PDE's general solution.

The following are examples of PDE&BC problems for which pdsolve is successful in first calculating the PDE's general solution, and then fitting the initial or boundary condition to it.

pde__18 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

If we ask pdsolve to solve the problem, we get:

(4.1)

and we can check this answer by using pdetest:

(4.2)

How it works, step by step:

The general solution for just the PDE is:

(4.1.1)

Substituting in the condition , we get:

(4.1.2)

(4.1.3)

We then isolate one of the functions above (we can choose either one, in this case), convert it into a function operator, and then apply it to gensol

(4.1.4)

(4.1.5)

Three other related examples

pde__19 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0

u(x, y) = (2*sin(-y+((1/2)*I)*2^(1/2)*x)+(-I*2^(1/2)*x+2*y)*_F2(y-((1/2)*I)*2^(1/2)*x)+(I*2^(1/2)*x-2*y)*_F2(y+((1/2)*I)*2^(1/2)*x))/(I*2^(1/2)*x-2*y)

(4.2.1)

(4.2.2)

pde__20 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0

u(x, y) = (sinh((1/2)*(I*2^(1/2)*x-2*y)*2^(1/2))*2^(1/2)-(I*2^(1/2)*x-2*y)*(_F2(-y+((1/2)*I)*2^(1/2)*x)-_F2(y+((1/2)*I)*2^(1/2)*x)))/(I*2^(1/2)*x-2*y)

(4.2.3)

(4.2.4)

pde__21 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0

(4.2.5)

(4.2.6)

More PDE&BC problems are now solved by using a Fourier transform.

Consider the following problem with an initial condition:

pdsolve can solve this problem directly:

(5.1)

And we can check this answer against the original problem, if desired:

(5.2)

How it works, step by step

Similarly to the Laplace transform method, we start the solution process by first applying the Fourier transform to the PDE:

(5.1.1)

Next, we call the function "fourier(u(x,t),x,s1)" by the new name U:

(5.1.2)

And this equation, which is really an ODE, is solved:

(5.1.3)

Now, we apply the Fourier transform to the initial condition :

(5.1.4)

(5.1.5)

Or, in the new variable U,

(5.1.6)

Now, we evaluate solution_U at t = 0:

(5.1.7)

and substitute the transformed initial condition into it:

>	$eval(solution_U_at_IC, {trasnformed_IC_U})$

(5.1.8)

Putting this into our solution_U, we get

>	$eval(solution_U, {(rhs = lhs)(IPi(Dirac(s+1)-Dirac(s-1)) = _F1(s))})$

(5.1.9)

Finally, we apply the inverse Fourier transformation to this,

(5.1.10)

PDE&BC problems that used to require the option HINT = `+` to be solved are now solved automatically

The following two PDE&BC problems used to require the option HINT = `+` in order to be solved. This is now done automatically within pdsolve.

pde__23 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0

(6.1)

(6.2)

pde__24 := diff(u(x, y), y, y)+diff(u(x, y), x, x) = 6*x-6*y

iv__24 := u(x, 0) = x^3+11*x+1, u(x, 2) = x^3+11*x-7, u(0, y) = -y^3+1, u(4, y) = -y^3+109

(6.3)

(6.4)

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

Posted: ecterrab 14540 Product: Maple

April 29 2016

21 0

Below is the worksheet with the whole material presented yesterday in the webinar, “Applying the power of computer algebra to theoretical physics”, broadcasted by the “Institute of Physics” (IOP, England). The material was very well received, rated 4.5 out of 5 (around 30 voters among the more than 300 attendants), and generated a lot of feedback. The webinar was recorded so that it is possible to watch it (for free, of course, click the link above, it will ask you for registration, though, that’s how IOP works).

Anyway, you can reproduce the presentation with the worksheet below (mw file linked at the end, or the corresponding pdf also linked with all the input lines executed). As usual, to reproduce the input/output you need to have installed the latest version of Physics, available in the Maplesoft R&D Physics webpage.

Why computer algebra?

... and why computer algebra?

We can concentrate more on the ideas instead of on the algebraic manipulations

We can extend results with ease

We can explore the mathematics surrounding a problem

We can share results in a reproducible way

Representation issues that were preventing the use of computer algebra in Physics

Notation and related mathematical methods that were missing:

coordinate free representations for vectors and vectorial differential operators,

covariant tensors distinguished from contravariant tensors,

functional differentiation, relativity differential operators and sum rule for tensor contracted (repeated) indices

Bras, Kets, projectors and all related to Dirac's notation in Quantum Mechanics

Inert representations of operations, mathematical functions, and related typesetting were missing:

inert versus active representations for mathematical operations

ability to move from inert to active representations of computations and viceversa as necessary

hand-like style for entering computations and textbook-like notation for displaying results

Key elements of the computational domain of theoretical physics were missing:

ability to handle products and derivatives involving commutative, anticommutative and noncommutative variables and functions

ability to perform computations taking into account custom-defined algebra rules of different kinds

(commutator, anticommutator and bracket rules, etc.)

Examples

	The Maple computer algebra environment

Classical Mechanics

	Inertia tensor for a triatomic molecule

Classical Field Theory

	*The field equations for the model

	*Maxwell equations departing from the 4-dimensional Action for Electrodynamics

	*The Gross-Pitaevskii field equations for a quantum system of identical particles

Quantum mechanics

	*The quantum operator components of satisfy

	Quantization of the energy of a particle in a magnetic field

Unitary Operators in Quantum Mechanics

	*Eigenvalues of an unitary operator and exponential of Hermitian operators

Properties of unitary operators

Consider two set of kets and , each of them constituting a complete orthonormal basis of the same space.

One can always build an unitary operator that maps one basis to the other, i.e.:

	*Verify that implies on

	*Show that is unitary

	*Show that the matrix elements of in the and basis are equal

	Show that and have the same spectrum

Schrödinger equation and unitary transform

Consider a ket that solves the time-dependant Schrödinger equation:

and consider

where is a unitary operator.

Does evolves according a Schrödinger equation

and if yes, which is the expression of ?

	Solution

Translation operators using Dirac notation

In this section, we focus on the operator

	Settings

	The Action (translation) of the operator on a ket

	Action of on an operator

General Relativity

	*Exact Solutions to Einstein's Equations

*"Physical Review D" 87, 044053 (2013)

Given the spacetime metric,

$g[mu, nu] = (Matrix(4, 4, {(1, 1) = -exp(lambda(r)), (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = -r^2, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = -r^2*sin(theta)^2, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = exp(nu(r))}))$

a) Compute the Ricci and Weyl scalars

b) Compute the trace of

where is some function of the radial coordinate, is the Ricci tensor, is the covariant derivative operator and is the stress-energy tensor

$T[alpha, beta] = (Matrix(4, 4, {(1, 1) = 8*exp(lambda(r))*Pi, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 8*r^2*Pi, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 8*r^2*sin(theta)^2*Pi, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 8*exp(nu(r))*Pi*epsilon}))$

c) Compute the components of the traceless part of of item b)

d) Compute an exact solution to the nonlinear system of differential equations conformed by the components of obtained in c)

Background: paper from February/2013, "Withholding Potentials, Absence of Ghosts and Relationship between Minimal Dilatonic Gravity and f(R) Theories", by P. Fiziev.

	a) The Ricci and Weyl scalars

	b) The trace of

	b) The components of the traceless part of

	c) An exact solution for the nonlinear system of differential equations conformed by the components of

*The Equivalence problem between two metrics

From the "What is new in Physics in Maple 2016" page:

In the Maple PDEtools package, you have the mathematical tools - including a complete symmetry approach - to work with the underlying [Einstein’s] partial differential equations. [By combining that functionality with the one in the Physics and Physics:-Tetrads package] you can also formulate and, depending on the metrics also resolve, the equivalence problem; that is: to answer whether or not, given two metrics, they can be obtained from each other by a transformation of coordinates, as well as compute the transformation.

	Example from: A. Karlhede, "A Review of the Geometrical Equivalence of Metrics in General Relativity", General Relativity and Gravitation, Vol. 12, No. 9, 1980

	*Equivalence for Schwarzschild metric (spherical and Krustal coordinates)

Tetrads and Weyl scalars in canonical form

Generally speaking a canonical form is obtained using transformations that leave invariant the tetrad metric in a tetrad system of references, so that theWeyl scalars are fixed as much as possible (conventionally, either equal to 0 or to 1).

Bringing a tetrad in canonical form is a relevant step in the tackling of the equivalence problem between two spacetime metrics.

The implementation is as in "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, page 388:

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.

The determination of appropriate transformation parameters to be used in these rotations, as well as the sequence of transformations happens all automatically by using the optional argument, canonicalform of TransformTetrad .

restart; with(Physics); with(Tetrads)

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

(7.4.1)

	Petrov type I

	Petrov type II

	Petrov type III

	Petrov type N

	Petrov type D

Physics_2016_IOP_webinar.mw Physics_2016_IOP_webinar.pdf

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

Posted: ecterrab 14540 Product: Maple

April 20 2016

4 0

Formulating and solving the equivalence problem for Schwarzschild metric in a simple case

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. I was recently asked about a statement found in the "What is new in Physics in Maple 2016" page:

This question posed is a reasonable one: "could you please provide one example?" This post provides that example.

First of all the existing science behind: in my opinion, the main reference regarding the equivalence problem is at the paper "A Review of the Geometrical Equivalence of Metrics in General Relativity", General Relativity and Gravitation, Vol. 12, No. 9, 1980, by A. Karlhede (University of Stockholm). This approach got refined later by others and, generally speaking, it is currently know as the Cartan-Karlhede method, summarized in chapter 9.2 of the book Exact Solutions to Einstein Field Equations. whose solutions were all digitized within the Physics and DifferentialGeometry packages for Maple 2016. This method of Chapter 9.2 (see also Tetrads and Weyl scalars in canonical form, Mapleprimes post), however, is not the only approach to the problem, and sometimes simpler methods can handle the problem faster, or just in simpler forms.

The example worked out below is actually the example from Karlhede's paper just mentioned, on pages 704 - 706: "Show that the Schwarzschild metric and its form written in terms of isotropic spherical coordinates are equivalent, and derive the transformation that relates them". Because this problem happens to be simple for nowadays computer algebra, below I also tackle it modified, slightly more difficult variants of it. The approach shown works for more complicated cases as well.

Below we tackle Karlhede's paper-problem using: one PDEtools command, the Physics:-TransformCoordinates, the Physics:-Weyl command to compute the Weyl scalars and the Physics:-Tetrads:-PetrovType to see the Petrov type of the metrics involved. The transformation resolving the equivalence is explicitly derived.

Start loading the Physics and Tetrads package. To reproduce the computations below, as usual, update your Physics library with the one available for download at the Maplesoft R&D Physics webpage

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

(1)

To formulate the problem, set first some symbols to represent the changed metric, changed mass and changed coordinates - no mathematics at this point

(2)

Set now a new coordinates system, call it Y, involving the new coordinates (in the paper they are represented with a tilde on top of the letters)

$`Default differentiation variables for d_, D_ and dAlembertian are: `*{Y = (`&tfr;`, `&rfr;`, `ϑ`, `ϕ`)}$

(3)

According to eq.(7.6) of the paper, the line element of Schwarzschild solution in isotropic spherical coordinates is given by

`#msup(mi("ds"),mn("2"))` := ((1-mt/(2*rt))/(1+mt/(2*rt)))^2*d_(tt)^2-(1+mt/(2*rt))^4*(d_(rt)^2+rt^2*d_(thetat)^2+rt^2*sin(thetat)^2*d_(phit)^2)

(-2*`&rfr;`+`&mfr;`)^2*Physics:-d_(`&tfr;`)^2/(2*`&rfr;`+`&mfr;`)^2-(1/16)*(2*`&rfr;`+`&mfr;`)^4*(Physics:-d_(`&rfr;`)^2+`&rfr;`^2*Physics:-d_(vartheta)^2+`&rfr;`^2*sin(vartheta)^2*Physics:-d_(varphi)^2)/`&rfr;`^4

(4)

Set this to be the metric

Check it out

Physics:-g_[mu, nu] = Matrix(%id = 18446744078306516254)

(5)

In connection with the transformation used further below, compute now the Petrov type and the Weyl scalars for this metric, just to have an idea of what is behind this metric.

(6)

(7)

We see that the Weyl scalars are already in canonical form (see post in Mapleprimes about canonical forms): only and the important thing: it depends on only one coordinate, .

Now: we want to see if this metric (5) is equivalent to Schwarzschild metric in standard spherical coordinates

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

Physics:-g_[mu, nu] = Matrix(%id = 18446744078795590102)

(8)

The equivalence we want to resolve is regarding an arbitrary relationship between the masses used in (5) and (8) and a generic change of variables from X to Y

(9)

Using a differential equation mindset, the formulation of the equivalence between (8) and (5) under the transformation (9) is actually simple: change variables in (8), using (9) and the Physics:-TransformCoordinates command (this is the command that changes variables in tensorial expressions), then equate the result to (5), then try to solve the problem for the unknowns , and .

We note at this point, however, that the Weyl scalars for Schwarzschild metric in this standard form (8) are also in canonical form of Petrov type D and also depend on only one variable, r

(10)

(11)

The fact that the Weyl scalars in both cases ((7) and (11)) are in canonical form (only ) and in both cases this scalar depends on only one coordinate is already an indicator that the transformation involved changes only one variable in terms of the other one. So one could just search for a transformation of the form and resolve the problem instantly. Still, to make the problem slightly more general, consider instead a generic transformation for r in terms of all of

(12)

(13)

Transform the coordinates in the metric (because of having used PDEtools:-declare, derivatives of the unknowns R are displayed indexed, for compact notation)

Matrix(%id = 18446744078873927542)

(14)

Proceed equating (14) to (5) to obtain a set of equations that entirely formulates the problem

${0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+`&mfr;`)^2/(2*`&rfr;`+`&mfr;`)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+`&mfr;`)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}$

(15)

This problem, shown in Karlhede's paper as the example of the approach he summarized, is solvable using the differential equation commands of PDEtools (in this case casesplit) in one go and no time, obtaining the same solution shown in the paper with equation number (7.10), the problem actually admits two solutions

$PDEtools:-casesplit({0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+`&mfr;`)^2/(2*`&rfr;`+`&mfr;`)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+`&mfr;`)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}, [R, mt])$

(16)

By all means this does not mean this differential equation approach is better than the general approach mentioned in the paper (also in section 9.2 of the Exact Solutions book). This presentation above only makes the point of the paragraph mentioned at the beginning of this worksheet "... [in Maple 2016] you can also formulate and, depending on the the metrics also resolve, the equivalence problem; that is: to answer whether or not, given two metrics, they can be obtained from each other by a transformation of coordinates, as well as compute the transformation."

In any case this problem above is rather easy for the computer. Consider a slightly more difficult problem, where . For example:

Physics:-g_[mu, nu] = Matrix(%id = 18446744078854733566)

(17)

Tackle now the same problem

${0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}$

(18)

The solutions to the equivalence between (17) and (5) are then given by

$PDEtools:-casesplit({0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}, [R, mt])$

`casesplit/ans`([R(Y) = -(1/4)*(m-2*`&rfr;`)^2/`&rfr;`, `&mfr;`^2 = -1/m], [`&mfr;` <> 0]), `casesplit/ans`([R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, `&mfr;`^2 = 1/m], [`&mfr;` <> 0])

(19)

Moreover, despite that the Weyl scalars suggest that a transformation of only one variable is sufficient to solve the problem, one could also consider a more general transformation, of more variables. Provided we exclude (because there is around and that would take us to solve differential equations for , that involve things like ), and also to speed up matters let's remove the change in , consider an arbitrary change in r and t

(20)

(21)

So our transformation now involve two arbitrary variables, each one depending on all the four coordinates, and a more complicated function . Change variables (because of having used PDEtools:-declare, derivatives of the unknowns R and are displayed indexed, for compact notation)

Matrix(%id = 18446744078309268046)

(22)

Construct the set of Partial Differential Equations to be tackled

${0 = (-4*(diff(Tau(Y), `&rfr;`))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), vartheta))+(diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))^2+(diff(R(Y), `&tfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))^2+(diff(R(Y), `&rfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = (diff(Tau(Y), vartheta))^2*(R(Y)-2*m)/R(Y)-(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), varphi))^2+2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)^2)/(R(Y)*(-R(Y)+2*m))}$

(23)

Solve the problem running a differential elimination (actually without solving any differential equations): there are more than two solutions

$sol := PDEtools:-casesplit({0 = (-4*(diff(Tau(Y), `&rfr;`))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), vartheta))+(diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))^2+(diff(R(Y), `&tfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))^2+(diff(R(Y), `&rfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = (diff(Tau(Y), vartheta))^2*(R(Y)-2*m)/R(Y)-(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), varphi))^2+2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)^2)/(R(Y)*(-R(Y)+2*m))}, [R, mt])$

(24)

Consider for instance the first one

(25)

Compute the actual solution behind this case :

>	$pdsolve(`casesplit/ans`([R(Y) = (1/4)(2`&rfr;`+m)^2/`&rfr;`, `&mfr;`^2 = 1/m, diff(Tau(Y), `&tfr;`) = -1, diff(Tau(Y), `&rfr;`) = 0, diff(Tau(Y), vartheta) = 0, diff(Tau(Y), varphi) = 0], [`&mfr;` <> 0]), {R, Tau, mt})$

${`&mfr;` = -1/m^(1/2), R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, Tau(Y) = -`&tfr;`+_C1}, {`&mfr;` = 1/m^(1/2), R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, Tau(Y) = -`&tfr;`+_C1}$

(26)

The fact that the time t appears defined in terms of the transformed time involving an arbitrary constant is expected: the time does not enter the metric, it only enters through derivatives of entering the Jacobian of the transformation used to change variables in tensorial expressions (the metric) in (22).

Summary: the approach shown above, based on formulating the problem for the transformation functions of the equivalence and solving for them the differential equations using the commands in PDEtools, after restricting the generality of the transformation functions by looking at the form of the Weyl scalars, works well for other cases too, specially now that, in Maple 2016, the Weyl scalars can be expressed also in canonical form in one go (see previous Mapleprimes post on "Tetrads and Weyl scalars in canonical form"). Also important: in Maple 2016 it is present the functionality necessary to implement the approach of section 9.2 of the Exact solutions book as well.

Download Equivalence_-_Schwarzschild.mw

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

Residual invariance

Petrov type I

Petrov type II

Petrov type III

Petrov type D

remains invariant under rotations of Class III

Petrov type N

remains invariant under rotations of Class II

Posted: ecterrab 14540 Product: Maple

April 20 2016

4 0

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:

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.

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

(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:

(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)

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

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

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

(3)

The Weyl scalars

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).

(5)

The scalars are now simpler, although still not in "canonical form" because and .

(6)

The Petrov type

(7)

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

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 and .

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

(10)

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

Inded we now have and

[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)

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

"`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

(13)

The starting tetrad

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

(14)

results in Weyl scalars not in canonical form:

(15)

For Petrov type "II", the canonical form is as for type "I" but in addition Again let's assume positive, not necessary, but to get simpler formulas around

(16)

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

Matrix(%id = 18446744078835949430)

(17)

Set this tetrad and check the Weyl scalars again

(18)

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

An Example of Petrov type III

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

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)

(20)

The Petrov type and the original tetrad

(21)

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

(22)

This tetrad results in the following 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 .

Compute now a canonical form for the tetrad

Matrix(%id = 18446744078500057566)

(24)

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

(25)

Great!

An Example of Petrov type N

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

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

(26)

(27)

(28)

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

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

(29)

(30)

For Petrov type "N", the canonical form has and all the other .

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

Matrix(%id = 18446744078518486190)

(31)

(32)

All as expected.

An Example of Petrov type D

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

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

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The default tetrad and related Weyl scalars are not in canonical form, which for Petrov type "D" is with and all the other

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

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Transform the tetrad, set it and recompute the Weyl scalars

Matrix(%id = 18446744078814996830)

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Again the expected canonical form of the Weyl scalars, and 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"