Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Assume that x[i] with 1≤i≤n are binary numbers. Let I_[k] be a subset with k elements of the set {1,2,⋯,n}.

Now Consider the following binary linear functions

It is clear that to obtain f[j]'s we need to compute ∑k[j] XOR bitwise operator. But it is possible to get f[j]'s with less than ∑k[j] XOR.

Example: Let n=8 and m=7 and suppose that

It follows from (1) that we need to do 20 XOR bitwise operator to get f[j]'s with 1≤j≤7. But set

which results in  f[j]'s are computed with just 9 XOR.

It is useful to mention that the relation (1) can be represented by a 8x8 binary matrix in the form of M*X=F. 

Question: Is it possible to implement a procedure in Maple such that by applying the procedure we get  f[j]'s with the minimum number of XOR bitwise operator.

One solution for this question is the "parr algorithm" that is given in the following paper

https://ieeexplore.ieee.org/document/613165

This edition is because of @Carl Love commnet.

The paar algorithm is provided in C++ in the following link

https://github.com/rub-hgi/shorter_linear_slps_for_mds_matrices/blob/master/paar.cpp

and is defined by 

My problem is that I couldn't implement the paar algorithm in Maple.

Thanks for any help

 

 

 

Hi,

How can I extract the coefficients of uij in Eq.(3) to form a system of PDEs in F, P and Q?

 

coeff_of_DE.mw

Hello,

I'm trying to verify a computation using maple. But the simplification of Physics doesn't return the right value.

The computation is

I wrote the follow code


 

J[x]^2*J[z]^2+(J[x]*J[z]*J[x])*J[z]+J[x]*J[z]^2*J[x]+J[x]^2*J[z]*J[z]+(J[x]*J[z]*J[z])*J[x]+J[x]^2*J[z]^2 = 3*J[x]^2*J[z]^2+3*J[x]^2*J[z]^2+2*J[x]^2-3*J[y]^2+2*J[z]^2

``

restart; with(Physics)*Setup(noncommutativepre = J)

CommuRules := seq(op([%Commutator(J[j[1]], J[j[2]]) = I*J[j[3]]]), j = [[x, y, z], [y, z, x], [z, x, y]])

NULL

NULL

Setup(CommuRules)

[algebrarules = {%Commutator(J[x], J[y]) = I*J[z], %Commutator(J[y], J[z]) = I*J[x], %Commutator(J[z], J[x]) = I*J[y]}]

(1)

NULL

with(combinat)

np := numbperm([J[x], J[x], J[z], J[z]], 4)

6

(2)

``

``

L := permute([J[x], J[x], J[z], J[z]], 4)

[[J[x], J[x], J[z], J[z]], [J[x], J[z], J[x], J[z]], [J[x], J[z], J[z], J[x]], [J[z], J[x], J[x], J[z]], [J[z], J[x], J[z], J[x]], [J[z], J[z], J[x], J[x]]]

(3)

``

NULL

L2 := seq(map(proc (x, y, z, t) options operator, arrow; x*y*z*t end proc, op(L[i])), i = 1 .. np)

Physics:-`*`(Physics:-`^`(J[x], 2), Physics:-`^`(J[z], 2)), Physics:-`*`(J[x], J[z], J[x], J[z]), Physics:-`*`(J[x], Physics:-`^`(J[z], 2), J[x]), Physics:-`*`(J[z], Physics:-`^`(J[x], 2), J[z]), Physics:-`*`(J[z], J[x], J[z], J[x]), Physics:-`*`(Physics:-`^`(J[z], 2), Physics:-`^`(J[x], 2))

(4)

S := add(j, j = L2)

Physics:-`*`(Physics:-`^`(J[x], 2), Physics:-`^`(J[z], 2))+Physics:-`*`(J[x], J[z], J[x], J[z])+Physics:-`*`(J[x], Physics:-`^`(J[z], 2), J[x])+Physics:-`*`(J[z], Physics:-`^`(J[x], 2), J[z])+Physics:-`*`(J[z], J[x], J[z], J[x])+Physics:-`*`(Physics:-`^`(J[z], 2), Physics:-`^`(J[x], 2))

(5)

NULL

NULL

S1 := Simplify(S, algebrarules)

-(3*I)*Physics:-`*`(J[x], J[y], J[z])-7*Physics:-`^`(J[z], 2)-(8*I)*Physics:-`*`(J[z], J[x], J[y])+6*Physics:-`*`(Physics:-`^`(J[z], 2), Physics:-`^`(J[x], 2))-I*Physics:-`*`(J[y], J[z], J[x])

(6)

NULL

NULL

``


 

Download MapleCode.mw

 


If someone knows, or have suggestions that what i can do to find the right result, plz let me know

I realize that it would be fairly straight forward for me to perform the necessary conversions on the raw data set elements to achieve the same result, but it would be helpful I think if I were able to load the transform into the plot command as I go, basically I want to implement the ability to make additions to the list that is on the coords help page, give the entry a name that plot()  will remember, because I will have commentary data on why I was looking at the plot, what ideas it was relevant to, references etc  and date of entry  etc etc that will also be associated with that name.

Essentially a hint as to where I should start is what I need 

When call_external is a part of a procedure, how does one retrive the code pertaining to the external function being called, and how does maple know what call_external refers to, I thought maybe the packages that have been loaded contain these instructions, how can I retrive the relevant code for a given procedure that contains call_external when 'showstat' is used to inspect a command's maple code?

 

The help page for call_external redirects me to define_external, which is really what I expected to need calling before call_external can take on it's necessary meaning.

I have a proc f(a,b)

 

For example

m := n->piecewise(type(n,even),0, 1):
f := proc(a,b)
    m(a*b);
end;

 

when I execute f(a,b) I get 1!!

 

for some reason the type is returning that it is not even, even though the expansion needs to be delayed.

I did this because mod was also doing this. I replaced it so I could undefine m and actually get a general expansion in terms of m.

 

Is there any way to get it to delay the expansion? I guess I need to check if the type is a "variable" or not an itneger and just return it.

 

I've created

 

m := n->piecewise(type(n,even),0, type(n, integer), 1, M(n)):

 

which seems to work but this is not ideal because sometimes M is returned even if an integer is used when used in other expressions. I suppose it is how ever maple figures out how evaluate things.

 

 

Hi,

Sorry to ask such a stupid question but I can't find out where my error is. Probably it's so huge it blinds me!

The double loop and the matrix product F^+ . F should give the same result, no? (it seems that F^+ . F has its rows reordered ?)


 

restart:

N   := 3:
P   := 2:
niv := [seq(Z[i], i=1..N)];
f   := Matrix(N^P, P, (i,j) -> `if`(j=P, niv[(i mod 3)+1], niv[iquo(i-1,3)+1]));

niv := [Z[1], Z[2], Z[3]]

 

f := Matrix(9, 2, {(1, 1) = Z[1], (1, 2) = Z[2], (2, 1) = Z[1], (2, 2) = Z[3], (3, 1) = Z[1], (3, 2) = Z[1], (4, 1) = Z[2], (4, 2) = Z[2], (5, 1) = Z[2], (5, 2) = Z[3], (6, 1) = Z[2], (6, 2) = Z[1], (7, 1) = Z[3], (7, 2) = Z[2], (8, 1) = Z[3], (8, 2) = Z[3], (9, 1) = Z[3], (9, 2) = Z[1]})

(1)

ds := subs(niv =~ [$0..N-1], f);

ds := Matrix(9, 2, {(1, 1) = 0, (1, 2) = 1, (2, 1) = 0, (2, 2) = 2, (3, 1) = 0, (3, 2) = 0, (4, 1) = 1, (4, 2) = 1, (5, 1) = 1, (5, 2) = 2, (6, 1) = 1, (6, 2) = 0, (7, 1) = 2, (7, 2) = 1, (8, 1) = 2, (8, 2) = 2, (9, 1) = 2, (9, 2) = 0})

(2)

vs := [ seq(V__||i, i=1..P)]:
es := unapply( sort( [ seq( mul(vs ^~ [entries(ds[i,..], nolist)]), i=1..N^P) ] ), vs);
 

proc (V__1, V__2) options operator, arrow; [1, V__1, V__2, V__1^2, V__2^2, V__1*V__2, V__1*V__2^2, V__1^2*V__2, V__1^2*V__2^2] end proc

(3)

ff := convert([ seq(es(entries(ffd[i,..], nolist)), i=1..N^P) ], Matrix);


UnityRoots := [solve(z^3=1, z)]:
F := simplify(subs(niv =~ UnityRoots, ff)) /~ sqrt(N^P):

ff := Matrix(9, 9, {(1, 1) = 1, (1, 2) = Z[1], (1, 3) = Z[2], (1, 4) = Z[1]^2, (1, 5) = Z[2]^2, (1, 6) = Z[1]*Z[2], (1, 7) = Z[1]*Z[2]^2, (1, 8) = Z[1]^2*Z[2], (1, 9) = Z[1]^2*Z[2]^2, (2, 1) = 1, (2, 2) = Z[1], (2, 3) = Z[3], (2, 4) = Z[1]^2, (2, 5) = Z[3]^2, (2, 6) = Z[1]*Z[3], (2, 7) = Z[1]*Z[3]^2, (2, 8) = Z[1]^2*Z[3], (2, 9) = Z[1]^2*Z[3]^2, (3, 1) = 1, (3, 2) = Z[1], (3, 3) = Z[1], (3, 4) = Z[1]^2, (3, 5) = Z[1]^2, (3, 6) = Z[1]^2, (3, 7) = Z[1]^3, (3, 8) = Z[1]^3, (3, 9) = Z[1]^4, (4, 1) = 1, (4, 2) = Z[2], (4, 3) = Z[2], (4, 4) = Z[2]^2, (4, 5) = Z[2]^2, (4, 6) = Z[2]^2, (4, 7) = Z[2]^3, (4, 8) = Z[2]^3, (4, 9) = Z[2]^4, (5, 1) = 1, (5, 2) = Z[2], (5, 3) = Z[3], (5, 4) = Z[2]^2, (5, 5) = Z[3]^2, (5, 6) = Z[2]*Z[3], (5, 7) = Z[2]*Z[3]^2, (5, 8) = Z[2]^2*Z[3], (5, 9) = Z[2]^2*Z[3]^2, (6, 1) = 1, (6, 2) = Z[2], (6, 3) = Z[1], (6, 4) = Z[2]^2, (6, 5) = Z[1]^2, (6, 6) = Z[1]*Z[2], (6, 7) = Z[1]^2*Z[2], (6, 8) = Z[1]*Z[2]^2, (6, 9) = Z[1]^2*Z[2]^2, (7, 1) = 1, (7, 2) = Z[3], (7, 3) = Z[2], (7, 4) = Z[3]^2, (7, 5) = Z[2]^2, (7, 6) = Z[2]*Z[3], (7, 7) = Z[2]^2*Z[3], (7, 8) = Z[2]*Z[3]^2, (7, 9) = Z[2]^2*Z[3]^2, (8, 1) = 1, (8, 2) = Z[3], (8, 3) = Z[3], (8, 4) = Z[3]^2, (8, 5) = Z[3]^2, (8, 6) = Z[3]^2, (8, 7) = Z[3]^3, (8, 8) = Z[3]^3, (8, 9) = Z[3]^4, (9, 1) = 1, (9, 2) = Z[3], (9, 3) = Z[1], (9, 4) = Z[3]^2, (9, 5) = Z[1]^2, (9, 6) = Z[1]*Z[3], (9, 7) = Z[1]^2*Z[3], (9, 8) = Z[1]*Z[3]^2, (9, 9) = Z[1]^2*Z[3]^2})

(4)

Scalar products of pairs of comumn vectors

F must be an orthogonal array

for i1 from 1 to N^P do
  for i2 from 1 to N^P do
    printf("%a ", simplify(add(F[..,i1] . F[.., i2])))
  end do:
  printf("\n"):
end do:
printf("\n");

1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1
 

 

or more simply:

simplify(F^+ . F)

Matrix([[1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 1, 0, 0, 0]])

(5)

 


 

Download Too_Blind_To_Find_My_Mistake.mw

I have several plots where I use display(a,b,c). Maple displays them with display(actual plot of a, actual plot of b, ...).
I've tried adding insequence but nothing and also tried passing them as sets or lists. In earlier versions I could use display(plots) and it would display all the plots on one plot together but now maple always separates them. I'd rather have the old behavior.

How do I get them to plot on the same plot and is there any way to set it as a global ooption?

what is the wrong in this operations for isentropic process ( isentropic-process) 

 

Restart: with(RealDomain) : interface(dispalyprecision=4) : ; Isentropical := proc(N,v) local K,PTR,KM1,KP1,GO, M,MS,C,PPT,TTT,RRT,AAS,eq,o ; K : = G() : KM1 : = k - 1 : KP1 : = K + 1 : GO := 1 : If N=1 then M := v Elif N=2 then PPT := v : M := sqrt ( 2 * (PPT^(-KM/K) -1 ) : Elif N=3 then MS := v : M := sqrt ( 1 / (KP1/ (2*MS^2) –KM1/2 ) : Elif N=4 then C := v : M := sqrt ( 2 / (KM1* (1 /(C^2) -1 )) ) : Elif N=5 then TTT := v : M := sqrt ( 2 * (1/TTT-1 ) / KM1 : Elif N=6 then RRT := v : M := sqrt ( 2 * (1/RRT^KM1 -1 ) / KM1 : Elif N=7 then AAS := v : eq :=AAS- (2*(1+KM1*M^2/2)/ KP1) ^ (KP1/ (2*KM1)) /M : M := fsolve(eq,M,0..1) : GO :=2 : Elif N=8 then AAS := v : eq :=AAS- (2*(1+KM1*M^2/2)/ KP1) ^ (KP1/ (2*KM1)) /M : M := fsolve(eq,M,1..infinity) : GO :=2 : Fi:fi GO<>2 then ASS := (2 *(1+ KM1*M^2/2/KP1) ^ (KP1/ (2*KM1)) /M fi: PTR := (1+ KM1*M^2/2/KP1) : PPT := PTR^(-K/KM1) : TTT:=PTR^ (-1) : RRT:+PTR^(-1/KM1) : If N <> 3 then MS := sqrt ( KP1 / ( 2 / ( M^2)+KM1) ) If N <> 3 then C := sqrt( 1 / (2 / (KM1*M^2) + 1 ) ) O := evalf([m,1./PPT,MS,C,1./TTT,1./RRT,AAS]) end : irp := proc (N, v) local f; f:=Isentropic1(N,v) : printf(cat(`% 9.4f `$7, ` \ n`) , f[ii$ii=1..7]) end : G := ( ) -> 1.4 : For mi form .1 to .5 do irp(1.mi) od : > > > Eq := table( [ (M2) = M2 , (MS2) = (K+1) /2 (2 / ( M2)+ (K-1) ) (C2) = 1 / (2 / (K-1)*M2+1 ) (PP0) = (1+(K-1)*M2/2)^(-K/(K-1) ) (RR0) = (1+(K-1)*M2/2)^(-1/(K-1) ) (TT0) = (1+(K-1)*M2/2)^(-1) (AAS) = (2 *(1 +(K-1)*M2/2 ) / (K+1) )^( (k-1) /(2*(k-1)))/ squt(M2) ]) : if vL=M then M2v := solve ( M2 -Eq[M2 ] , M2 ) : M2v := eval(M2v ,M2 = VR^2 ) : elif vL=MS then M2v := solve( M2 -Eq[MS2 ] , M2 ) : M2v := eval(M2v ,MS2 = VR^2 ) : elif vL=C then M2v := solve( M2 -Eq[C2 ] , M2 ) : M2v := eval(M2v ,C2 = VR^2 ) : elif vL=PP0 then M2v := solve( M2 -Eq[PP0] , M2 ) : M2v := eval(M2v ,PP0 = VR ) : elif vL=RR0 then M2v := solve( M2 -Eq[RR0] , M2 ) : M2v := eval(M2v ,RR0 = VR ) : elif vL=TT0 then M2v := solve( M2 -Eq[TT0] , M2 ) : M2v := eval(M2v ,TT0 = VR ) : elif vL=AASsub then M2v := fsolve( M2 -Eq[AAS] , M2 ) : M2 :=0..1 : elif vL=AASup then M2v := fsolve( M2 -Eq[AAS] , M2 ) : M2 :=1..infinity : fi: Eqsva1 := eva1(Eq, M2 = M2v) : Eqava1 [M] := sqrt(EqsVa1[M2] ) : Eqsva1[MS] := sqrt(EqsVa1[MS2] ) : Eqava1 [C] := sqrt(EqsVa1[C2] ) : 1var := [M, MS, C, PP0, RR0, TT0, AAS] : [seq(1var[i]=EqsVA1[1var[i] ] , I=1…nope(1var))] end: Egs :=[M=0.5, MS=0.53452, C=0.2182, P0P=0.84302, RR0=0.88517, TT0=0.95238, AASsub=1.33985, AASsup=0.95238] : For i from 1 to nops(Egs) do print(I,Isentropic2(Egs[i])) od : (process)

I have functions with sin(x) and  cos(x) terms with x values are degrees.

How to calculate sin(30 degree)?

How to insert degree symbol in maple?

Suppose evaluate  sin(Pi/2), my out put becomes sin(Pi/2)...How to solve this problem?

Hi,

Idea for trigonometric animation like this? Thanks

Please I need help with my proc.

 

> RK2skritt:=proc(FR::procedure,xo,yo,vxo,vyo,h) local x, y, vx, vy, r, kx1, kx2, kx3, kx4, ky1, ky2, ky3, ky4, l x1, l x2, l x3, l x4, ly1, ly2, ly3, ly4, tmp; r:=sqrt(xo*xo+yo*yo); tmp:=-(h*FR(r))/(r); lx1:=h*vxo; ly1:=h*vyo; kx1:=tmp*xo; ky1:=tmp*yo; lx2:=h*(vxo+0.5*kx1); ly2:=h*(vyo+0.5 ky1); r:=sqrt((xo+0.5*lx1)^(2)+(yo+0.5*ly1)^(2)); tmp:=-(h*FR(r))/(r); kx2:=tmp*(xo+0.5*lx1); ky2:=tmp*(yo+0.5*ly1); lx3:=h*(vxo+0.5*kx2); ly3:=h*(vyo+0.5*ky2); r:=sqrt((xo+0.5*lx2)^(2)+(yo+0.5*ly2)^(2)); tmp:=-(h*FR(r))/(r); kx3:=tmp*(xo+0.5*lx2); ky3:=tmp*(yo+0.5*ly2); lx4:=h*(vxo+kx3); ly4:=h*(vyo+ky3); r:=sqrt((xo+lx3)^(2)+(yo+ly3)^(2)); tmp:=-(h*FR(r))/(r); kx4:=tmp*(xo+lx3); ky4:=tmp(yo+ly3); x:=xo+(lx1+2*lx2+2*lx3+lx4)/(6); y:=yo+(ly1+2*ly2+2*ly3+ly4)/(6); vx:=vxo+(kx1+2*kx2+2*kx3+kx4)/(6); vy:=vyo+(ky1+2*ky2+2*ky3+ky4)/(6); [x,y,vx,vy]; end proc: Error, invalid local declaration Please someone help me!!!!!!!!

in maple after fitting an equation (A + B.x) in the experimental data, how to assign to parameters A and B the adjusted values will be assigned in the equation?

 

Thanks

Splitting PDE parameterized symmetries

and Parameter-continuous symmetry transformations

The determination of symmetries for partial differential equation systems (PDE) is relevant in several contexts, the most obvious of which is of course the determination of the PDE solutions. For instance, generally speaking, the knowledge of a N-dimensional Lie symmetry group can be used to reduce the number of independent variables of PDE by N. So if PDE depends only on N independent variables, that amounts to completely solving it. If only N-1 symmetries are known or can be successfully used then PDE becomes and ODE; etc., all advantageous situations. In Maple, a complete set of symmetry commands, to perform each step of the symmetry approach or several of them in one go, is part of the PDEtools  package.

 

Besides the dependent and independent variables, PDE frequently depends on some constant parameters, and besides the PDE symmetries for arbitrary values of those parameters, for some particular values of them, PDE transforms into a completely different problem, admitting different symmetries. The question then is: how can you determine those particular values of the parameters and the corresponding different symmetries? That was the underlying subject of a recent question in Mapleprimes. The answer to those questions is relatively simple and yet not entirely obvious for most of us, motivating this post, organized briefly around one example.

 

To reproduce the input/output below you need Maple 2019 and to have installed the Physics Updates v.449 or higher.

 

Consider the family of Korteweg-de Vries equation for u(x, t)involving three constant parameters a, b, q. For convenience (simpler input and more readable output) use the diff_table  and declare  commands

with(PDEtools)

U := diff_table(u(x, t))

pde := b*U[]*U[x]+a*U[x]+q*U[x, x, x]+U[t] = 0

b*u(x, t)*(diff(u(x, t), x))+a*(diff(u(x, t), x))+q*(diff(diff(diff(u(x, t), x), x), x))+diff(u(x, t), t) = 0

(1)

declare(U[])

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

(2)

This pde admits a 4-dimensional symmetry group, whose infinitesimals - for arbitrary values of the parameters a, b, q- are given by

I__1 := Infinitesimals(pde, [u], specialize_Cn = false)

[_xi[x](x, t, u) = (1/3)*_C1*x+_C3*t+_C4, _xi[t](x, t, u) = _C1*t+_C2, _eta[u](x, t, u) = (1/3)*((-2*b*u-2*a)*_C1+3*_C3)/b]

(3)

Looking at pde (1) as a nonlinear problem in u, a, b and q, it splits into four cases for some particular values of the parameter:

pde__cases := casesplit(b*u(x, t)*(diff(u(x, t), x))+a*(diff(u(x, t), x))+q*(diff(diff(diff(u(x, t), x), x), x))+diff(u(x, t), t) = 0, parameters = {a, b, q}, caseplot)

`========= Pivots Legend =========`

 

p1 = q

 

p2 = b*u(x, t)+a

 

p3 = b

 

 

`casesplit/ans`([diff(diff(diff(u(x, t), x), x), x) = -(b*u(x, t)*(diff(u(x, t), x))+a*(diff(u(x, t), x))+diff(u(x, t), t))/q], [q <> 0]), `casesplit/ans`([diff(u(x, t), x) = -(diff(u(x, t), t))/(b*u(x, t)+a), q = 0], [b*u(x, t)+a <> 0]), `casesplit/ans`([u(x, t) = -a/b, q = 0], [b <> 0]), `casesplit/ans`([diff(u(x, t), t) = 0, a = 0, b = 0, q = 0], [])

(4)

The legend above indicates the pivots and the tree of cases, depending on whether each pivot is equal or different from 0. At the end there is the algebraic sequence of cases. The first case is the general case, for which the symmetry infinitesimals were computed as I__1 above, but clearly the other three cases admit more general symmetries. Consider for instance the second case, pass the ignoreparameterizingequations to ignore the parameterizing equation q = 0, and you get

I__2 := Infinitesimals(pde__cases[2], ignore)

`* Partial match of  'ignore' against keyword 'ignoreparameterizingequations'`

 

[_xi[x](x, t, u) = _F3(x, t, u), _xi[t](x, t, u) = Intat(((b*u+a)*(D[1](_F3))(_a, ((b*u+a)*t-x+_a)/(b*u+a), u)-_F1(u, ((b*u+a)*t-x)/(b*u+a))*b+(D[2](_F3))(_a, ((b*u+a)*t-x+_a)/(b*u+a), u))/(b*u+a)^2, _a = x)+_F2(u, ((b*u+a)*t-x)/(b*u+a)), _eta[u](x, t, u) = _F1(u, ((b*u+a)*t-x)/(b*u+a))]

(5)

These infinitesimals are indeed much more general than I__1, in fact so general that (5) is almost unreadable ... Specialize the three arbitrary functions into something "easy" just to be able follow - e.g. take _F1 to be just the + operator, _F2 the * operator and _F3 = 1

eval(I__2, [_F1 = `+`, _F2 = `*`, _F3 = 1])

[_xi[x](x, t, u) = 1, _xi[t](x, t, u) = Intat(-(u+((b*u+a)*t-x)/(b*u+a))*b/(b*u+a)^2, _a = x)+u*((b*u+a)*t-x)/(b*u+a), _eta[u](x, t, u) = u+((b*u+a)*t-x)/(b*u+a)]

(6)

simplify(value([_xi[x](x, t, u) = 1, _xi[t](x, t, u) = Intat(-(u+((b*u+a)*t-x)/(b*u+a))*b/(b*u+a)^2, _a = x)+u*((b*u+a)*t-x)/(b*u+a), _eta[u](x, t, u) = u+((b*u+a)*t-x)/(b*u+a)]))

[_xi[x](x, t, u) = 1, _xi[t](x, t, u) = (b^3*t*u^4+((3*a*t-x)*u^3-u^2*x-t*x*u)*b^2+((3*a^2*t-2*a*x)*u^2-a*u*x-a*t*x+x^2)*b+a^2*u*(a*t-x))/(b*u+a)^3, _eta[u](x, t, u) = (b*u^2+(b*t+a)*u+a*t-x)/(b*u+a)]

(7)

This symmetry is of course completely different than [_xi[x](x, t, u) = (1/3)*_C1*x+_C3*t+_C4, _xi[t](x, t, u) = _C1*t+_C2, _eta[u](x, t, u) = ((-2*b*u-2*a)*_C1+3*_C3)/(3*b)]computed for the general case.

 

The symmetry (7) can be verified against pde__cases[2] or directly against pde after substituting q = 0.

[_xi[x](x, t, u) = (1/3)*_C1*x+_C3*t+_C4, _xi[t](x, t, u) = _C1*t+_C2, _eta[u](x, t, u) = (1/3)*((-2*b*u-2*a)*_C1+3*_C3)/b]

(8)

SymmetryTest([_xi[x](x, t, u) = 1, _xi[t](x, t, u) = (b^3*t*u^4+((3*a*t-x)*u^3-u^2*x-t*x*u)*b^2+((3*a^2*t-2*a*x)*u^2-a*u*x-a*t*x+x^2)*b+a^2*u*(a*t-x))/(b*u+a)^3, _eta[u](x, t, u) = (b*u^2+(b*t+a)*u+a*t-x)/(b*u+a)], pde__cases[2], ignore)

`* Partial match of  'ignore' against keyword 'ignoreparameterizingequations'`

 

{0}

(9)

SymmetryTest([_xi[x](x, t, u) = 1, _xi[t](x, t, u) = (b^3*t*u^4+((3*a*t-x)*u^3-u^2*x-t*x*u)*b^2+((3*a^2*t-2*a*x)*u^2-a*u*x-a*t*x+x^2)*b+a^2*u*(a*t-x))/(b*u+a)^3, _eta[u](x, t, u) = (b*u^2+(b*t+a)*u+a*t-x)/(b*u+a)], subs(q = 0, pde))

{0}

(10)

Summarizing: "to split PDE symmetries into cases according to the values of the PDE parameters, split the PDE into cases with respect to these parameters (command PDEtools:-casesplit ) then compute the symmetries for each case"

 

Parameter continuous symmetry transformations

 

A different, however closely related question, is whether pde admits "symmetries with respect to the parameters a, b and q", so whether exists continuous transformations of the parameters a, b and q that leave pde invariant in form.

 

Beforehand, note that since the parameters are constants with regards to the dependent and independent variables (here u(x, t)), such continuous symmetry transformations cannot be used directly to compute a solution for pde. They can, however, be used to reduce the number of parameters. And in some contexts, that is exactly what we need, for example to entirely remove the splitting into cases due to their presence, or to proceed applying a solving method that is valid only when there are no parameters (frequently the case when computing exact solutions to "PDE & Boundary Conditions").

 

To compute such "continuous symmetry transformations of the parameters" that leave pde invariant one can always think of these parameters as "additional independent variables of pde". In terms of formulation, that amounts to replacing the dependency in the dependent variable, i.e. replace u(x, t) by u(x, t, a, b, q)

 

pde__xtabq := subs((x, t) = (x, t, a, b, q), pde)

b*u(x, t, a, b, q)*(diff(u(x, t, a, b, q), x))+a*(diff(u(x, t, a, b, q), x))+q*(diff(diff(diff(u(x, t, a, b, q), x), x), x))+diff(u(x, t, a, b, q), t) = 0

(11)

Compute now the infinitesimals: note there are now three additional ones, related to continuous transformations of "a,b,"and q - for readability, avoid displaying the redundant functionality x, t, a, b, q, u on the left-hand sides of these infinitesimals

Infinitesimals(pde__xtabq, displayfunctionality = false)

[_xi[x] = (1/3)*(_F4(a, b, q)*q+_F3(a, b, q))*x/q+_F6(a, b, q)*t+_F7(a, b, q), _xi[t] = _F4(a, b, q)*t+_F5(a, b, q), _xi[a] = _F1(a, b, q), _xi[b] = _F2(a, b, q), _xi[q] = _F3(a, b, q), _eta[u] = (1/3)*((b*u+a)*_F3(a, b, q)-2*((b*u+a)*_F4(a, b, q)+(3/2)*u*_F2(a, b, q)+(3/2)*_F1(a, b, q)-(3/2)*_F6(a, b, q))*q)/(b*q)]

(12)

This result is more general than what is convenient for algebraic manipulations, so specialize the seven arbitrary functions of a, b, q and keep only the first symmetry that result from this specialization: that suffices to illustrate the removal of any of the three parameters a, b, or q

S := Library:-Specialize_Fn([_xi[x] = (1/3)*(_F4(a, b, q)*q+_F3(a, b, q))*x/q+_F6(a, b, q)*t+_F7(a, b, q), _xi[t] = _F4(a, b, q)*t+_F5(a, b, q), _xi[a] = _F1(a, b, q), _xi[b] = _F2(a, b, q), _xi[q] = _F3(a, b, q), _eta[u] = (1/3)*((b*u+a)*_F3(a, b, q)-2*((b*u+a)*_F4(a, b, q)+(3/2)*u*_F2(a, b, q)+(3/2)*_F1(a, b, q)-(3/2)*_F6(a, b, q))*q)/(b*q)])[1 .. 1]

[_xi[x] = 0, _xi[t] = 0, _xi[a] = 1, _xi[b] = 0, _xi[q] = 0, _eta[u] = -1/b]

(13)

To remove the parameters, as it is standard in the symmetry approach, compute a transformation to canonical coordinates, with respect to the parameter a. That means a transformation that changes the list of infinitesimals, or likewise its infinitesimal generator representation,

InfinitesimalGenerator(S, [u(x, t, a, b, q)])

proc (f) options operator, arrow; diff(f, a)-(diff(f, u))/b end proc

(14)

into [_xi[x] = 0, _xi[t] = 0, _xi[a] = 1, _xi[b] = 0, _xi[q] = 0, _eta[u] = 0] or its equivalent generator representation  proc (f) options operator, arrow; diff(f, a) end proc

That same transformation, when applied to pde__xtabq, entirely removes the parameter a.

The transformation is computed using CanonicalCoordinates and the last argument indicates the "independent variable" (in our case a parameter) that the transformation should remove. We choose to remove a

CanonicalCoordinates(S, [u(x, t, a, b, q)], [upsilon(xi, tau, alpha, beta, chi)], a)

{alpha = a, beta = b, chi = q, tau = t, xi = x, upsilon(xi, tau, alpha, beta, chi) = (b*u(x, t, a, b, q)+a)/b}

(15)

declare({alpha = a, beta = b, chi = q, tau = t, xi = x, upsilon(xi, tau, alpha, beta, chi) = (b*u(x, t, a, b, q)+a)/b})

` u`(x, t, a, b, q)*`will now be displayed as`*u

 

` upsilon`(xi, tau, alpha, beta, chi)*`will now be displayed as`*upsilon

(16)

Invert this transformation in order to apply it

solve({alpha = a, beta = b, chi = q, tau = t, xi = x, upsilon(xi, tau, alpha, beta, chi) = (b*u(x, t, a, b, q)+a)/b}, {a, b, q, t, x, u(x, t, a, b, q)})

{a = alpha, b = beta, q = chi, t = tau, x = xi, u(x, t, a, b, q) = (upsilon(xi, tau, alpha, beta, chi)*beta-alpha)/beta}

(17)

The next step is not necessary, but just to understand how all this works, verify its action over the infinitesimal generator proc (f) options operator, arrow; diff(f, a)-(diff(f, u))/b end proc

ChangeSymmetry({a = alpha, b = beta, q = chi, t = tau, x = xi, u(x, t, a, b, q) = (upsilon(xi, tau, alpha, beta, chi)*beta-alpha)/beta}, proc (f) options operator, arrow; diff(f, a)-(diff(f, u))/b end proc, [upsilon(xi, tau, alpha, beta, chi), xi, tau, alpha, beta, chi])

proc (f) options operator, arrow; diff(f, alpha) end proc

(18)

Now that we see the transformation (17) is the one we want, just use it to change variables in pde__xtabq

PDEtools:-dchange({a = alpha, b = beta, q = chi, t = tau, x = xi, u(x, t, a, b, q) = (upsilon(xi, tau, alpha, beta, chi)*beta-alpha)/beta}, pde__xtabq, [upsilon(xi, tau, alpha, beta, chi), xi, tau, alpha, beta, chi], simplify)

upsilon(xi, tau, alpha, beta, chi)*(diff(upsilon(xi, tau, alpha, beta, chi), xi))*beta+chi*(diff(diff(diff(upsilon(xi, tau, alpha, beta, chi), xi), xi), xi))+diff(upsilon(xi, tau, alpha, beta, chi), tau) = 0

(19)

As expected, this result depends only on two parameters, beta, and chi, and the one equivalent to a (that is alpha, see the transformation used (17)), is not present anymore.

To remove b or q we use the same steps, (15), (17) and (19), just changing the parameter to be removed, indicated as the last argument  in the call to CanonicalCoordinates . For example, to eliminate b (represented in the new variables by beta), input

CanonicalCoordinates(S, [u(x, t, a, b, q)], [upsilon(xi, tau, alpha, beta, chi)], b)

{alpha = b, beta = a, chi = q, tau = t, xi = x, upsilon(xi, tau, alpha, beta, chi) = (b*u(x, t, a, b, q)+a)/b}

(20)

solve({alpha = b, beta = a, chi = q, tau = t, xi = x, upsilon(xi, tau, alpha, beta, chi) = (b*u(x, t, a, b, q)+a)/b}, {a, b, q, t, x, u(x, t, a, b, q)})

{a = beta, b = alpha, q = chi, t = tau, x = xi, u(x, t, a, b, q) = (upsilon(xi, tau, alpha, beta, chi)*alpha-beta)/alpha}

(21)

PDEtools:-dchange({a = beta, b = alpha, q = chi, t = tau, x = xi, u(x, t, a, b, q) = (upsilon(xi, tau, alpha, beta, chi)*alpha-beta)/alpha}, pde__xtabq, [upsilon(xi, tau, alpha, beta, chi), xi, tau, alpha, beta, chi], simplify)

upsilon(xi, tau, alpha, beta, chi)*(diff(upsilon(xi, tau, alpha, beta, chi), xi))*alpha+chi*(diff(diff(diff(upsilon(xi, tau, alpha, beta, chi), xi), xi), xi))+diff(upsilon(xi, tau, alpha, beta, chi), tau) = 0

(22)

and as expected this result does not contain "beta. "To remove a second parameter, the whole cycle is repeated starting with computing infinitesimals, for instance for (22). Finally, the case of function parameters is treated analogously, by considering the function parameters as additional dependent variables instead of independent ones.

 


 

Download How_to_split_symmetries_into_cases_(II).mw

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

I have a surface described by the equation.

f := (x, y) -> 3*(1 - x)^2*exp(-x^2 - (y + 1)^2) - (2*x - 10*x^3 - 10*y^5)*exp(-x^2 - y^2) - 1/3*exp(-(x + 1)^2 - y^2)

with(plots):
plot3d(f(x,y),x=-3..3,y=-3..3,orientation=[-40,60,0])

If a drop of water were put at point f(0.2,1.4), how would we draw a line to follow it's path?  (is that the gradient curve?)

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