Edit detail for #170 Axiom fails to solve "separable" system of equations revision 3 of 3

Some examples:

(1) -> )set output algebra on
 

)set output tex off

L := [ A = 2*X+Y, B = 2*Y+X, C = 2*U+V, D = 2*V+U];

solve(L, [X,Y])

   (2)  []

However:

solve(L,[X,Y,U,V])

              - B + 2 A      2 B - A      - D + 2 C      2 D - C
   (3)  [[X = ---------, Y = -------, U = ---------, V = -------]]
                  3             3             3             3

Note that solution should satisfy all equations. Quantities which are not variables are treated as parameters and solution is valid for generic parameters. There are no values of X and V which solve the last two equations: generically C is different than 2*U+V

Simpler:

solve([a - b = 0, c - d = 0],[b])

   (4)  []

linSolve([a - b, c - d],[b])

   (5)  [particular = "failed", basis = []]

The operation solve calls linSolve, which sets up the corresponding matrix and vector and solves it using solve$LinearSystemMatrixPackage. This in turn returns "failed", since the last columns of the matrix contain zeros, the vector does not. In the example above, the matrix and vector are:

         +- 1+
   [mat= |   |,vec= [- a,d - c]]
         + 0 +

Note that

linSolve([a - b, 0],[b])

   (6)  [particular = [a], basis = []]

works. The same happens, if the equation is not linear:

L := [ A = 2*X^2+Y, B = 2*Y+X, C = 2*U+V, D = 2*V+U];

solve(L, [X, Y])

   (8)  []

)set output algebra on
 

)set output tex off

solve(L, [X, Y, U, V])

   (9)
   [
                        2                     2              - D + 2 C
     [X = - 2 Y + B, 8 Y  + (- 8 B + 1)Y + 2 B  - A = 0, U = ---------,
                                                                 3
          2 D - C
      V = -------]
             3
     ]

)set output algebra off
 

)set output tex on

So, very probably, a fix would need to do two things:

• seperate the equations into those that do and those that don't contain the given variables.
• check whether those that don't contain the variables are contradicting.
• solve the others.

The second point is necessary, since

L := [ A = P+Q, B = P-Q, C = 1, C = -1];

solve(L, [P,Q])

(1)

solve(L,[P,Q,C])

(2)

really has no solution. As far as I know, this would have to be done in the very last function defined in syssolp.spad, which is:

       -- general solver. Input in polynomial style  --
       solve(lr:L F,vl:L SE) ==
           empty? vl => empty()
           checkLinear(lr,vl) =>
                            -- linear system --
               soln := linSolve(lr, vl)
               soln case "failed" => []
               eqns: L EQ F := []
               for i in 1..#vl repeat
                   lhs := (vl.i::(P R))::F
                   rhs :=  rhs soln.i
                   eqns := append(eqns, [lhs = rhs])
               [eqns]

                         -- polynomial system --
           if R has GcdDomain then
             parRes:=triangularSystems(lr,vl)
             [[makeEq(map(makeR2F,f)$PP2,vl) for f in pr]
                                                        for pr in parRes]
           else [[]]

The letter F is a macro for FRAC POLY R here. To check whether an equation contains a variable we have to check numerator and denominator of both sides of the equation with variables$POLY R. I do not know however, how to find out whether the equations independent of vl are contradicting.