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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:

  

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.

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

with(Physics); with(Tetrads); Setup(auto = true, tetradmetric = null, signature = `+---`)

`Setting lowercaselatin letters to represent tetrad indices `

 

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

 

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

 

`* Partial match of  'auto' against keyword 'automaticsimplification'`

 

[automaticsimplification = true, signature = `+ - - -`, tetradmetric = {(1, 2) = 1, (3, 4) = -1}]

(1)

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

gt, mt, tt, rt, thetat, phit := `𝔤`, `𝔪`, `𝔱`, `𝔯`, `ϑ`, `ϕ`

`𝔤`, `𝔪`, `𝔱`, `𝔯`, vartheta, varphi

(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)

Coordinates(Y = [tt, rt, thetat, phit])

`Default differentiation variables for d_, D_ and dAlembertian are: `*{Y = (`𝔱`, `𝔯`, `ϑ`, `ϕ`)}

 

`Systems of spacetime Coordinates are: `*{Y = (`𝔱`, `𝔯`, `ϑ`, `ϕ`)}

 

{Y}

(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*`𝔯`+`𝔪`)^2*Physics:-d_(`𝔱`)^2/(2*`𝔯`+`𝔪`)^2-(1/16)*(2*`𝔯`+`𝔪`)^4*(Physics:-d_(`𝔯`)^2+`𝔯`^2*Physics:-d_(vartheta)^2+`𝔯`^2*sin(vartheta)^2*Physics:-d_(varphi)^2)/`𝔯`^4

(4)

Set this to be the metric

Setup(metric = `#msup(mi("ds"),mn("2"))`)

Check it out

g_[]

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.

PetrovType()

"D"

(6)

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = -64*`𝔯`^3*`𝔪`/(2*`𝔯`+`𝔪`)^6, psi__3 = 0, psi__4 = 0

(7)

We see that the Weyl scalars are already in canonical form (see post in Mapleprimes about canonical forms): only `&Psi;__2` <> 0 and the important thing: it depends on only one coordinate, `&rfr;` .

 

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

g_[sc]

`Systems of spacetime Coordinates are: `*{X = (t, r, theta, phi), Y = (`&tfr;`, `&rfr;`, `&vartheta;`, `&varphi;`)}

 

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

 

`The Schwarzschild metric in coordinates `[t, r, theta, phi]

 

`Parameters: `[m]

 

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

(8)

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

TR := {phi = Phi(Y), r = R(Y), t = Tau(Y), theta = Theta(Y)}

{phi = Phi(Y), r = R(Y), t = Tau(Y), theta = Theta(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 `&mfr;`(m), Phi(Y), R(Y), Theta(Y) and Tau(Y).

 

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 

PetrovType()

"D"

(10)

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = -m/r^3, psi__3 = 0, psi__4 = 0

(11)

The fact that the Weyl scalars in both cases ((7) and (11)) are in canonical form (only `&Psi;__2` <> 0 ) 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 r = R(`&rfr;`) 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 Y = (`&tfr;`, `&rfr;`, `&vartheta;`, `&varphi;`)

tr := r = R(Y)

r = R(Y)

(12)

PDEtools:-declare(r = R(Y))

R(`&tfr;`, `&rfr;`, vartheta, varphi)*`will now be displayed as`*R

(13)

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

TransformCoordinates(tr, g_[mu, nu])

Matrix(%id = 18446744078873927542)

(14)

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

"convert(rhs(?)=? ,setofequations)"

{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])

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

(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 `&mfr;` <> m. For example:

"subs(mt = 1/(mt^(2)),?)"

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

(17)

Tackle now the same problem

"convert(rhs(?)=? ,setofequations)"

{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 theta (because there is cos(theta) around and that would take us to solve differential equations for Theta(theta), that involve things like cos(Theta(theta))), and also to speed up matters let's remove the change in phi, consider an arbitrary change in r and t

TR := select(has, {phi = Phi(Y), r = R(Y), t = Tau(Y), theta = Theta(Y)}, {r, t})

{r = R(Y), t = Tau(Y)}

(20)

PDEtools:-declare({r = R(Y), t = Tau(Y)})

R(`&tfr;`, `&rfr;`, vartheta, varphi)*`will now be displayed as`*R

 

Tau(`&tfr;`, `&rfr;`, vartheta, varphi)*`will now be displayed as`*Tau

(21)

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

TransformCoordinates(TR, g_[mu, nu])

Matrix(%id = 18446744078309268046)

(22)

Construct the set of Partial Differential Equations to be tackled

"convert(rhs(?)=?,setofequations)"

{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])

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

(24)

Consider for instance the first one

sol[1]

`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])

(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 Tau(Y) = -`&tfr;`+_C1 involving an arbitrary constant is expected: the time does not enter the metric, it only enters through derivatives of Tau(Y) 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

Hello,

In order to improve the readability of a worksheet, I would like to insert in text lines the equations that I calculate after with a input maple.

Question:

Is it possible to add in a text line equations and symbols as we can make with Latex or MathType ? For example, i would be interested to write vectors in a text line.

Thank you for your help.


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

• 

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

• 

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

• 

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

• 

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

• 

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

More mathematics

 

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

Examples

   

More powerful symbolic differentiation (nth order derivative)

 

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

Examples

   

Mathematical handling of symbolic sequences

 

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

Examples

   

Visualization of mathematical functions

 

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

Examples

   

Section and subsections displaying properties of mathematical functions

 

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

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

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

Examples

   

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


Download MathematicalFunctionsAndFunctionAdvisor.mw

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

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

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

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

plot(f, 0..0.9999);

 

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

        0.5714285714

identify(%);

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

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

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

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

We are sugested to consider

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

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

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

 

Hi!

In Mathematica 10.0 were introduced regions with functions like TransformedRegion, ReginIntersection, etc. Moreover, it is easy to check if a point is inside a region, etc.

I would like to ask if in Maple I could use some API with similar functionality?. For instance, I would like to get integer points which lay inside an intersection of two cubes. How I could do this in Maple?

 

 


The PDE & BC project , a very nice and challenging one, also one where Maple is pioneer in all computer algebra systems, has restarted, including now also the collaboration of Katherina von Bülow.

Recapping, the PDE & BC project started 5 years ago implementing some of the basic methods found in textbooks to match arbitrary functions and constants to given PDE boundary conditions of different kinds. At this point we aim to fill gaps, and the first one we tackled is the case of 1st order PDE that can be solved without boundary conditions in terms of an arbitrary function, and where a single boundary condition (BC) is given for the PDE unknown function, and this BC does not depend on the independent variables of the problem. It looks simple ... It can be rather tricky though. The method we implemented is a simple however ingenious use of differential invariants  to match the boundary condition.


The resulting new code, the portion already tested, is available for download in the Maplesoft R&D webpage for Differential Equations and Mathematical Functions (the development itself is bundled within the library that contains the new developments for the Physics package, in turn within the zip linked in the webpage).


The examples that can now be handled, although restricted in generality to "only one 1st order linear or nonlinear PDE and only one boundary condition for the unknown function itself", illustrate well how powerful it can be to use more advanced methods to tackle these tricky situations where we need to match an arbitrary function to a boundary condition.


To illustrate the idea, consider first a linear example, among the simplest one could imagine:

PDEtools:-declare(f(x, y, z))

f(x, y, z)*`will now be displayed as`*f

(1)

pde := diff(f(x, y, z), x)+diff(f(x, y, z), y)+diff(f(x, y, z), z) = f(x, y, z)

diff(f(x, y, z), x)+diff(f(x, y, z), y)+diff(f(x, y, z), z) = f(x, y, z)

(2)

Input now a boundary condition (bc) for the unknownf(x, y, z) such that this bc does not depend on the independent variables {x, y, z}; this bc can however depend on arbitrary symbolic parameters, for instance

bc := f(alpha+beta, alpha-beta, 1) = alpha*beta

f(alpha+beta, alpha-beta, 1) = alpha*beta

(3)

With the recent development, this kind of problem can now be solved in one go:

sol := pdsolve([pde, bc])

f(x, y, z) = (1/4)*(x-2*z+2+y)*(x-y)*exp(z-1)

(4)

Nice! And how do you verify this result for correctness? With pdetest , which actually also tests the solution against the boundary conditions:

pdetest(sol, [pde, bc])

[0, 0]

(5)

And what has been done to obtain the solution (4)? First the PDE was solved regardless of the boundary condition, so in general, obtaining:

pdsolve(pde)

f(x, y, z) = _F1(-x+y, -x+z)*exp(x)

(6)

In a second step, the arbitrary function _F1(-x+y, -x+z) got determined such that the boundary condition f(alpha+beta, alpha-beta, 1) = alpha*beta is matched. Concretely, the mapping _F1 is what got determined. You can see this mapping reversing the solving process in two steps. Start taking the difference between the general solution (6) and the solution (4) that matches the boundary condition

(f(x, y, z) = _F1(-x+y, -x+z)*exp(x))-(f(x, y, z) = (1/4)*(x-2*z+2+y)*(x-y)*exp(z-1))

0 = _F1(-x+y, -x+z)*exp(x)-(1/4)*(x-2*z+2+y)*(x-y)*exp(z-1)

(7)

and isolate here _F1(-x+y, -x+z)

PDEtools:-Solve(0 = _F1(-x+y, -x+z)*exp(x)-(1/4)*(x-2*z+2+y)*(x-y)*exp(z-1), _F1(-x+y, -x+z))

_F1(-x+y, -x+z) = (1/4)*exp(-x+z-1)*(x^2-2*x*z-y^2+2*y*z+2*x-2*y)

(8)

So this is the value _F1(-x+y, -x+z) that got determined. To see now the actual solving mapping _F1, that takes for arguments -x+y and -x+z and returns the right-hand side of (8), one can perform a change of variables introducing the two parameters `&tau;__1` and `&tau;__2` of the _F1 mapping:

{tau__1 = -x+y, tau__2 = -x+z, tau__3 = z}

{tau__1 = -x+y, tau__2 = -x+z, tau__3 = z}

(9)

solve({tau__1 = -x+y, tau__2 = -x+z, tau__3 = z}, {x, y, z})

{x = -tau__2+tau__3, y = -tau__2+tau__1+tau__3, z = tau__3}

(10)

PDEtools:-dchange({x = -tau__2+tau__3, y = -tau__2+tau__1+tau__3, z = tau__3}, _F1(-x+y, -x+z) = (1/4)*exp(-x+z-1)*(x^2-2*x*z-y^2+2*y*z+2*x-2*y), proc (u) options operator, arrow; simplify(u, size) end proc)

_F1(tau__1, tau__2) = -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2)

(11)

So the solving mapping _F1 is

_F1 = unapply(rhs(_F1(tau__1, tau__2) = -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2)), tau__1, tau__2)

_F1 = (proc (tau__1, tau__2) options operator, arrow; -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2) end proc)

(12)

Wow! Although this pde & bc problem really look very simple, this solution (12) is highly non-obvious, as is the way to get it just from the boundary condition f(alpha+beta, alpha-beta, 1) = alpha*beta and the solution (6) too. Let's first verify that this mapping is correct (even when we know, by construction, that it is correct). For that, apply (12) to the arguments of the arbitrary function and we should obtain (8)

(_F1 = (proc (tau__1, tau__2) options operator, arrow; -(1/4)*exp(tau__2-1)*tau__1*(tau__1-2*tau__2+2) end proc))(-x+y, -x+z)

_F1(-x+y, -x+z) = -(1/4)*exp(-x+z-1)*(-x+y)*(x-2*z+2+y)

(13)

Indeed this is equal to (8)

normal((_F1(-x+y, -x+z) = -(1/4)*exp(-x+z-1)*(-x+y)*(x-2*z+2+y))-(_F1(-x+y, -x+z) = (1/4)*exp(-x+z-1)*(x^2-2*x*z-y^2+2*y*z+2*x-2*y)))

0 = 0

(14)

Skipping the technical details, the key observation to compute a solving mapping is that, given a 1st order PDE where the unknown depends on k independent variables, if the boundary condition depends on k-1 arbitrary symbolic parameters alpha, beta, one can always seek a "relationship between these k-1parameters and the k-1differential invariants that enter as arguments in the arbitrary function _F1 of the solution", and get the form of the mapping _F1 from this relationship and the bc. The method works in general. Change for instance the bc (3) making its right-hand side be a sum instead of a product

bc := f(alpha+beta, alpha-beta, 1) = alpha+beta

f(alpha+beta, alpha-beta, 1) = alpha+beta

(15)

sol := pdsolve([pde, bc])

f(x, y, z) = (x-z+1)*exp(z-1)

(16)

pdetest(sol, [pde, bc])

[0, 0]

(17)

An interesting case happens when the boundary condition depends on less than k-1 parameters, for instance:

bc__1 := subs(beta = alpha, bc)

f(2*alpha, 0, 1) = 2*alpha

(18)

sol__1 := pdsolve([pde, bc__1])

f(x, y, z) = ((x-z+1)*_C1+x-y)*exp(((z-1)*_C1+y)/(1+_C1))/(1+_C1)

(19)

As we see in this result, the additional difficulty represented by having few parameters got tackled by introducing an arbitrary constant _C1 (this is likely to evolve into something more general...)

pdetest(sol__1, [pde, bc__1])

[0, 0]

(20)

Finally, consider a nonlinear example

PDEtools:-declare(u(x, y))

u(x, y)*`will now be displayed as`*u

(21)

pde := 3*(u(x, y)-y)^2*(diff(u(x, y), x))-(diff(u(x, y), y)) = 0

3*(u(x, y)-y)^2*(diff(u(x, y), x))-(diff(u(x, y), y)) = 0

(22)

Here we have 2 independent variables, so for illustration purposes use a boundary condition that depends on only one arbitrary parameter

bc := u(0, alpha) = alpha

u(0, alpha) = alpha

(23)

All looks OK, but we still have another problem: check the arbitrary function _F1 entering the general solution of pde when tackled without any boundary condition:

pdsolve(pde)

u(x, y) = RootOf(-y^3+3*y^2*_Z-3*y*_Z^2+_Z^3-_F1(_Z)-x)

(24)

Remove this RootOf to see the underlying algebraic expression

DEtools[remove_RootOf](u(x, y) = RootOf(-y^3+3*y^2*_Z-3*y*_Z^2+_Z^3-_F1(_Z)-x))

-y^3+3*y^2*u(x, y)-3*y*u(x, y)^2+u(x, y)^3-_F1(u(x, y))-x = 0

(25)

So this is a pde where the general solution is implicit, actually depending on an arbitrary function of the unknown u(x, y) The code handles this problem in the same way, just that in cases like this there may be more than one solution. For this very particular bc (23) there are actually three solutions:

pdsolve([pde, bc])

u(x, y) = x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)-((1/2)*I)*3^(1/2)*x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)+((1/2)*I)*3^(1/2)*x^(1/3)+y

(26)

Verify these three solutions against the pde and the boundary condition

map(pdetest, [u(x, y) = x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)-((1/2)*I)*3^(1/2)*x^(1/3)+y, u(x, y) = -(1/2)*x^(1/3)+((1/2)*I)*3^(1/2)*x^(1/3)+y], [pde, bc])

[[0, 0], [0, 0], [0, 0]]

(27)

:)


Download PDEs_and_Boundary_Conditions.mw

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


In connection with recent developments for symbolic sequences, a number of improvements were implemented regarding symbolic differentiation, that is the computation of n^th order derivatives were n is a symbol, the simplest example being the n^th derivative of the exponential, which of course is the exponential itself. This post is about these developments, done in collaboration with Katherina von Bülow, and available for download as usual from the Maplesoft R&D web page for Differential Equations and Mathematical functions (the update itself is bundled with the official updates of the Maple Physics package).

 

It is important to note that Maple is pioneer in having an actual implementation of symbolic differentiation, something that works for real, since several releases.  The development, however, was somewhat stuck because we were unable to compute the symbolic n^th derivative of a composite function f(g(z)). A formula for this problem is actually known, it is the Faà di Bruno formula, but, in order to implement it, first we were missing the incomplete Bell functions , that got implemented in Maple 15, nice, but then we were still missing differentiating symbolic sequences, and functions whose arguments are symbolic sequences (i.e. the number of arguments of the function is n, a symbol, of unknown value at the time of differentiating). All this got implemented now within the new MathematicalFunctions:-Sequence package, opening the door widely to these improvements in n^th differentiation.

 

The symbolic differentiation code works as mostly all other computer algebra code, by mapping complicated problems into a composition of simpler problems all of which are tractable; what follows is then an illustration of these basic cases.

 

Among the simplest new case that can now be handled there is that of a power where the exponent is linear in the differentiation variable. This is actually an easy problem

(%diff = diff)(f^(alpha*z+beta), `$`(z, n))

%diff(f^(alpha*z+beta), `$`(z, n)) = alpha^n*f^(alpha*z+beta)*ln(f)^n

(1)

More complicated, consider the k^th power of a generic function; the corresponding symbolic derivative can be mapped into a sum of symbolic derivatives of powers of g(z) with lower degree

(%diff = diff)(g(z)^k, `$`(z, n))

%diff(g(z)^k, `$`(z, n)) = k*binomial(n-k, n)*(Sum((-1)^_k1*binomial(n, _k1)*g(z)^(k-_k1)*(Diff(g(z)^_k1, [`$`(z, n)]))/(k-_k1), _k1 = 0 .. n))

(2)

In some cases where g(z) is a known function, the computation can be carried on furthermore. For example, for g = ln the result can be expressed using Stirling numbers of the first kind

(%diff = diff)(ln(alpha*z+beta)^k, `$`(z, n))

%diff(ln(alpha*z+beta)^k, `$`(z, n)) = alpha^n*(Sum(pochhammer(k-_k1+1, _k1)*Stirling1(n, _k1)*ln(alpha*z+beta)^(k-_k1), _k1 = 0 .. n))/(alpha*z+beta)^n

(3)

The case of sin and cos are relatively simpler, but then assumptions on the exponent are required in order to proceed further ahead from (2), for example

`assuming`([(%diff = diff)(sin(alpha*z+beta)^k, `$`(z, n))], [k::posint])

%diff(sin(alpha*z+beta)^k, `$`(z, n)) = (-1)^k*piecewise(n = 0, (-sin(alpha*z+beta))^k, alpha^n*I^n*(Sum(binomial(k, _k1)*(2*_k1-k)^n*exp(I*(2*_k1-k)*(alpha*z+beta+(1/2)*Pi)), _k1 = 0 .. k))/2^k)

(4)

The case of functions of arbitrary number of variables (typical situation where symbolic sequences are required) is now handled properly. This is the pFq hypergeometric function of symbolic order p and q 

(%diff = diff)(hypergeom([`$`(a[i], i = 1 .. p)], [`$`(b[j], j = 1 .. q)], z), `$`(z, n))

%diff(hypergeom([`$`(a[i], i = 1 .. p)], [`$`(b[j], j = 1 .. q)], z), `$`(z, n)) = (product(pochhammer(a[i], n), i = 1 .. p))*hypergeom([`$`(a[i]+n, i = 1 .. p)], [`$`(b[j]+n, j = 1 .. q)], z)/(product(pochhammer(b[j], n), j = 1 .. q))

(5)

The case of the MeijerG function is more complicated, but in practice, for the computer, once it knows how to handle symbolic sequences, the more involved problem becomes computable

(%diff = diff)(MeijerG([[`$`(a[i], i = 1 .. n)], [`$`(b[i], i = n+1 .. p)]], [[`$`(b[i], i = 1 .. m)], [`$`(b[i], i = m+1 .. q)]], z), `$`(z, k))

%diff(MeijerG([[`$`(a[i], i = 1 .. n)], [`$`(b[i], i = n+1 .. p)]], [[`$`(b[i], i = 1 .. m)], [`$`(b[i], i = m+1 .. q)]], z), `$`(z, k)) = MeijerG([[-k, `$`(a[i]-k, i = 1 .. n)], [`$`(b[i]-k, i = n+1 .. p)]], [[`$`(b[i]-k, i = 1 .. m)], [0, `$`(b[i]-k, i = m+1 .. q)]], z)

(6)

Not only the mathematics of this result is correct: the object returned is actually computable to the end (if you provide the values of n, p, m and q), and the typesetting is actually fully readable, as in textbooks, including copy and paste working properly; all this is new.

The n^th derivative of a number of mathematical functions that were not implemented before, are now also implemented, covering the gaps, for example:

(%diff = diff)(BellB(a, z), `$`(z, n))

%diff(BellB(a, z), `$`(z, n)) = Sum(Stirling2(a, _k1)*pochhammer(_k1-n+1, n)*z^(_k1-n), _k1 = 0 .. a)

(7)

(%diff = diff)(bernoulli(z), `$`(z, n))

%diff(bernoulli(z), `$`(z, n)) = pochhammer(nu-n+1, n)*bernoulli(nu-n, z)

(8)

(%diff = diff)(binomial(z, m), `$`(z, n))

%diff(binomial(z, m), `$`(z, n)) = (Sum((-1)^(_k1+m)*Stirling1(m, _k1)*pochhammer(_k1-n+1, n)*(z-m+1)^(_k1-n), _k1 = 1 .. m))/factorial(m)

(9)

(%diff = diff)(euler(a, z), `$`(z, n))

%diff(euler(a, z), `$`(z, n)) = pochhammer(a-n+1, n)*euler(a-n, z)

(10)

In the same way the fundamental formulas for the n^th derivative of all the 12 elliptic Jacobi functions  as well as the four elliptic JacobiTheta functions,  the LambertW , LegendreP  and some others are now all implemented.

Finally there is the "holy grail" of this problem: the n^th derivative of a composite function f(g(z)) - this always-unreachable implementation of Faa di Bruno formula. We now have it :)

(%diff = diff)(f(g(z)), `$`(z, n))

%diff(f(g(z)), `$`(z, n)) = Sum(((D@@k)(f))(g(z))*IncompleteBellB(n, k, `$`(diff(g(z), [`$`(z, j)]), j = 1 .. n-k+1)), k = 0 .. n)

(11)

Note the symbolic sequence of symbolic order derivatives of lower degree, both of of f and g, also within the arguments of the IncompleteBellB function. This is a very abstract formula ... And does this really work? Of course it does :). Consider, for instance, a case where the n^th derivatives of f(z) and g(z) can both be computed by the system:

sin(cos(alpha*z+beta))

sin(cos(alpha*z+beta))

(12)

This is the n^th derivative expressed using Faa di Bruno's formula, in turn expressed using symbolic sequences within the IncompleteBellB  function

(%diff = diff)(sin(cos(alpha*z+beta)), `$`(z, n))

%diff(sin(cos(alpha*z+beta)), `$`(z, n)) = Sum(sin(cos(alpha*z+beta)+(1/2)*k*Pi)*IncompleteBellB(n, k, `$`(cos(alpha*z+beta+(1/2)*j*Pi)*alpha^j, j = 1 .. n-k+1)), k = 0 .. n)

(13)

These results can all be verified. Take for instance n = 3

eval(%diff(sin(cos(alpha*z+beta)), `$`(z, n)) = Sum(sin(cos(alpha*z+beta)+(1/2)*k*Pi)*IncompleteBellB(n, k, `$`(cos(alpha*z+beta+(1/2)*j*Pi)*alpha^j, j = 1 .. n-k+1)), k = 0 .. n), n = 3)

%diff(sin(cos(alpha*z+beta)), z, z, z) = Sum(sin(cos(alpha*z+beta)+(1/2)*k*Pi)*IncompleteBellB(3, k, `$`(cos(alpha*z+beta+(1/2)*j*Pi)*alpha^j, j = 1 .. 4-k)), k = 0 .. 3)

(14)

Compute now the inert functions: on the left-hand side this is just the (now explicit) 3rd order derivative, while on the right-hand side we have a sum of IncompleteBellB  functions, where the number of arguments, expressed in (13) using symbolic sequences that depend on the summation index k and the differentiation order n, now in (14) depend only on k, and get transformed into explicit sequences of arguments when the summation is performed and k assumes integer values

value(%diff(sin(cos(alpha*z+beta)), z, z, z) = Sum(sin(cos(alpha*z+beta)+(1/2)*k*Pi)*IncompleteBellB(3, k, `$`(cos(alpha*z+beta+(1/2)*j*Pi)*alpha^j, j = 1 .. 4-k)), k = 0 .. 3))

alpha^3*sin(alpha*z+beta)*cos(cos(alpha*z+beta))-3*alpha^3*cos(alpha*z+beta)*sin(alpha*z+beta)*sin(cos(alpha*z+beta))+alpha^3*sin(alpha*z+beta)^3*cos(cos(alpha*z+beta)) = alpha^3*sin(alpha*z+beta)*cos(cos(alpha*z+beta))-3*alpha^3*cos(alpha*z+beta)*sin(alpha*z+beta)*sin(cos(alpha*z+beta))+alpha^3*sin(alpha*z+beta)^3*cos(cos(alpha*z+beta))

(15)

Take left-hand side minus right-hand side

simplify((lhs-rhs)(alpha^3*sin(alpha*z+beta)*cos(cos(alpha*z+beta))-3*alpha^3*cos(alpha*z+beta)*sin(alpha*z+beta)*sin(cos(alpha*z+beta))+alpha^3*sin(alpha*z+beta)^3*cos(cos(alpha*z+beta)) = alpha^3*sin(alpha*z+beta)*cos(cos(alpha*z+beta))-3*alpha^3*cos(alpha*z+beta)*sin(alpha*z+beta)*sin(cos(alpha*z+beta))+alpha^3*sin(alpha*z+beta)^3*cos(cos(alpha*z+beta))))

0

(16)

NULL

:)


Download SymbolicOrderDifferentiation.mw


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

Hi,

I am trying to use implicit plot. The plot is OK but I want to put labels such as what Latex produces :

$\frac{\Omega}{\omega_n}$ for x axis and $a_0 \mathrm{(m)}$ for y axis

How can I apply this in my maple code as below:

plot1:=implicitplot(a3, Omega_r=1.5..2.5, a=0.00000001..0.1, labeldirections=[horizontal, horizontal], axes=boxed, labels=["W/w_n",typeset("a_0 (m)")], labelfont=[SYMBOL]):

What I have put as bold does not work for me, it is making everything in Greek :) . I want combination of Greek and math.

 

Thasnks,

Bahareh

 

 

 

 

 

Symbolic sequences enter in various formulations in mathematics. This post is about a related new subpackage, Sequences, within the MathematicalFunctions package, available for download in Maplesoft's R&D page for Mathematical Functions and Differential Equations (currently bundled with updates to the Physics package).

 

Perhaps the most typical cases of symbolic sequences are:

 

1) A sequence of numbers - say from n to m - frequently displayed as

n, `...`, m

 

2) A sequence of one object, say a, repeated say p times, frequently displayed as

 "((a,`...`,a))"

3) A more general sequence, as in 1), but of different objects and not necessarily numbers, frequently displayed as

a[n], `...`, a[m]

or likewise a sequence of functions

f(n), `...`, f(m)

In all these cases, of course, none of n, m, or p are known: they are just symbols, or algebraic expressions, representing integer values.

 

These most typical cases of symbolic sequences have been implemented in Maple since day 1 using the `$` operator. Cases 1), 2) and 3) above are respectively entered as `$`(n .. m), `$`(a, p), and `$`(a[i], i = n .. m) or "`$`(f(i), i = n .. m)." To have computer algebra representations for all these symbolic sequences is something wonderful, I would say unique in Maple.

Until recently, however, the typesetting of these symbolic sequences was frankly poor, input like `$`(a[i], i = n .. m) or ``$\``(a, p) just being echoed in the display. More relevant: too little could be done with these objects; the rest of Maple didn't know how to add, multiply, differentiate or map an operation over the elements of the sequence, nor for instance count the sequence's number of elements.

 

All this has now been implemented.  What follows is a brief illustration.

restart

First of all, now these three types of sequences have textbook-like typesetting:

`$`(n .. m)

`$`(n .. m)

(1)

`$`(a, p)

`$`(a, p)

(2)

For the above, a$p works the same way

`$`(a[i], i = n .. m)

`$`(a[i], i = n .. m)

(3)

Moreover, this now permits textbook display of mathematical functions that depend on sequences of paramateters, for example:

hypergeom([`$`(a[i], i = 1 .. p)], [`$`(b[i], i = 1 .. q)], z)

hypergeom([`$`(a[i], i = 1 .. p)], [`$`(b[i], i = 1 .. q)], z)

(4)

IncompleteBellB(n, k, `$`(factorial(j), j = 1 .. n-k+1))

IncompleteBellB(n, k, `$`(factorial(j), j = 1 .. n-k+1))

(5)

More interestingly, these new developments now permit differentiating these functions even when their arguments are symbolic sequences, and displaying the result as in textbooks, with copy and paste working properly, for instance

(%diff = diff)(hypergeom([`$`(a[i], i = 1 .. p)], [`$`(b[i], i = 1 .. q)], z), z)

%diff(hypergeom([`$`(a[i], i = 1 .. p)], [`$`(b[i], i = 1 .. q)], z), z) = (product(a[i], i = 1 .. p))*hypergeom([`$`(a[i]+1, i = 1 .. p)], [`$`(b[i]+1, i = 1 .. q)], z)/(product(b[i], i = 1 .. q))

(6)

It is very interesting how much this enhances the representation capabilities; to mention but one, this makes 100% possible the implementation of the Faa-di-Bruno  formula for the nth symbolic derivative of composite functions (more on this in a post to follow this one).

But the bread-and-butter first: the new package for handling sequences is

with(MathematicalFunctions:-Sequences)

[Add, Differentiate, Map, Multiply, Nops]

(7)

The five commands that got loaded do what their name tells. Consider for instance the first kind of sequences mentione above, i.e

`$`(n .. m)

`$`(n .. m)

(8)

Check what is behind this nice typesetting

lprint(`$`(n .. m))

`$`(n .. m)

 

All OK. How many operands (an abstract version of Maple's nops  command):

Nops(`$`(n .. m))

m-n+1

(9)

That was easy, ok. Add the sequence

Add(`$`(n .. m))

(1/2)*(m-n+1)*(n+m)

(10)

Multiply the sequence

Multiply(`$`(n .. m))

factorial(m)/factorial(n-1)

(11)

Map an operation over the elements of the sequence

Map(f, `$`(n .. m))

`$`(f(j), j = n .. m)

(12)

lprint(`$`(f(j), j = n .. m))

`$`(f(j), j = n .. m)

 

Map works as map, i.e. you can map extra arguments as well

MathematicalFunctions:-Sequences:-Map(Int, `$`(n .. m), x)

`$`(Int(j, x), j = n .. m)

(13)

All this works the same way with symbolic sequences of forms "((a,`...`,a))" , and a[n], `...`, a[m]. For example:

`$`(a, p)

`$`(a, p)

(14)

lprint(`$`(a, p))

`$`(a, p)

 

MathematicalFunctions:-Sequences:-Nops(`$`(a, p))

p

(15)

Add(`$`(a, p))

a*p

(16)

Multiply(`$`(a, p))

a^p

(17)

Differentation also works

Differentiate(`$`(a, p), a)

`$`(1, p)

(18)

MathematicalFunctions:-Sequences:-Map(f, `$`(a, p))

`$`(f(a), p)

(19)

MathematicalFunctions:-Sequences:-Differentiate(`$`(f(a), p), a)

`$`(diff(f(a), a), p)

(20)

For a symbolic sequence of type 3)

`$`(a[i], i = n .. m)

`$`(a[i], i = n .. m)

(21)

MathematicalFunctions:-Sequences:-Nops(`$`(a[i], i = n .. m))

m-n+1

(22)

Add(`$`(a[i], i = n .. m))

sum(a[i], i = n .. m)

(23)

Multiply(`$`(a[i], i = n .. m))

product(a[i], i = n .. m)

(24)

The following is nontrivial: differentiating the sequence a[n], `...`, a[m], with respect to a[k] should return 1 when n = k (i.e the running index has the value k), and 0 otherwise, and the same regarding m and k. That is how it works now:

Differentiate(`$`(a[i], i = n .. m), a[k])

`$`(piecewise(k = i, 1, 0), i = n .. m)

(25)

lprint(`$`(piecewise(k = i, 1, 0), i = n .. m))

`$`(piecewise(k = i, 1, 0), i = n .. m)

 

MathematicalFunctions:-Sequences:-Map(f, `$`(a[i], i = n .. m))

`$`(f(a[i]), i = n .. m)

(26)

Differentiate(`$`(f(a[i]), i = n .. m), a[k])

`$`((diff(f(a[i]), a[i]))*piecewise(k = i, 1, 0), i = n .. m)

(27)

lprint(`$`((diff(f(a[i]), a[i]))*piecewise(k = i, 1, 0), i = n .. m))

`$`((diff(f(a[i]), a[i]))*piecewise(k = i, 1, 0), i = n .. m)

 

 

And that is it. Summarizing: in addition to the former implementation of symbolic sequences, we now have textbook-like typesetting for them, and more important: Add, Multiply, Differentiate, Map and Nops. :)

 

The first large application we have been working on taking advantage of this is symbolic differentiation, with very nice results; I will see to summarize them in a post to follow in a couple of days.

 

Download MathematicalFunctionsSequences.mw

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

Hi is there a way to identify curves in your plot? Especially when you dont know it yourself?

 

For example, I have 10 polynomials all order 5 and up, whilst I can trial and error identify which one is which it is very inefficient.

I used to use Mathematica, and I could manually label the curves with color which allows to see which curve is which. I assume there is something of the sort for Maple?

I am writing here because of a problem with writing mathematical expressions in Maple T.A.. I have been using it since two years, and I have a lot if questions created in version 9.0 and 9.5. A few months ago my university installed T.A. 10. At the beginning there were problems with the connections between T.A. and the Maple server. After the administrators got over these, there were another problems.

There are a lot of questions where I use greek letters, for example Sigma. Earlier it was easy, I wrote

in the 'Algorithm' section, and I could write

in the Text of the question. The letter had a perfect italic and bold style, like as it would be created with Equation Editor.

Now I have to write

sig2=maple("MathML:-ExportPresentation($sig1)");

I have the following mathematical expression

where a=0.2, b=0.09, c=0.57, p=0.3 and q=1-p=0.7, Now i want to find the value of n for which the value of the expression will be 1, i.e., for what value of n , A_n will be 1?

I am attempting to maximize Planck's intesnity function for any given temperature T with respect to lambda. The function is as follows:

Iy = (2*h*c^2)/(lambda^5*(e^(hc/(lambda*k*T))-1))

 

My initial attempt was to simply take the derivative of this function with respect to labda, set it equal to zero, and then solve for lamda. Theoretically this should give me the value of lambda that maximizes the intesity with respect to the constants / values h, c, k, and T (where T is temperature).

 

However doing this in maple just leads to the result "warning, solutions may have been lost".

 

Does anyone know how I can go about maximizing this?

 

Thanks!

we always have subscript variable in the math book, but how could this be natral done in maple

I want to get a seq aaa3

seq(a[i],i=1..3)

but how could I get a  aij

seq(a[i_j],i=1..3);

and

seq(a[ij],i=1..3);  both was not right

 

Is there a way with Maple 18 to place a bracket as shown below to contain two DE's? I am not trying to solve them. Only for note-taking purposes. If there is can you please share how to do it?

 

 

DE

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