Eliminate doesn't work in Mathematica

Question

Hello Mathematica users,

I have problems trying to eliminate variable $Q(x,y)$ froma PDE, but Mathematica output is just True. The problem is that I already found how to eliminate it. We can apply $d/dx$ on the second equation and then from the first equation we can determine and substitute $dQ[x,y]/dx$ in the second.

Why can't Mathematica do that? My output is in that case a PDE of fourth order, but I don't have Q[x,y], which is the aim, and I solved it by hand.

 Eliminate[{A0*D[w[x,y],{y,2}]-B0*D[w[x,y],{x,2}]-
   C0*D[Q[x,y],{x,1}]+C0*D[w[x,y],{x,2}]==0,
   E0*D[Q[x,y],{y,2}]-FF0*D[Q[x,y],{x,2}]-CC0*D[w[x,y],{x,1}]-Q[x,y]==0},
    {(Q^(0,2))[x,y],(Q^(1,0))[x,y],Q[x,y]}]

Answer 1

Eliminate, and its cousin GroebnerBasis, work with algebraic equations. If you require differential elimination you will need to take derivatives (prolongations, that is). Here is a blind approach: just take some derivatives, sort the variables into two sets, and eliminate all the Q stuff.

dpolys = {A0*D[w[x, y], {y, 2}] - B0*D[w[x, y], {x, 2}] - 
    C0*D[Q[x, y], {x, 1}] + C0*D[w[x, y], {x, 2}], 
   E0*D[Q[x, y], {y, 2}] - FF0*D[Q[x, y], {x, 2}] - 
    CC0*D[w[x, y], {x, 1}] - Q[x, y]};
derivs = {D[dpolys[[1]], x], D[dpolys[[1]], y]};
allpolys = Join[dpolys, derivs];
bigger = Join[allpolys, D[allpolys, x], D[allpolys, y]];

In[111]:= params = {A0, B0, C0, CC0, E0, FF0};
vars = Complement[Variables[bigger], params];
qvars = Select[Variables[bigger], ! FreeQ[#, Q] &];
wvars = Complement[vars, qvars];

In[110]:= GroebnerBasis[bigger, wvars, qvars, 
 MonomialOrder -> EliminationOrder]


{(-A0)*Derivative[0, 2][w][x, y] + A0*E0*Derivative[0, 4][w][x, y] + 
     B0*Derivative[2, 0][w][x, y] - C0*Derivative[2, 0][w][x, y] - 
     C0*CC0*Derivative[2, 0][w][x, y] - 
  B0*E0*Derivative[2, 2][w][x, y] + 
     C0*E0*Derivative[2, 2][w][x, y] - 
  A0*FF0*Derivative[2, 2][w][x, y] + 
     B0*FF0*Derivative[4, 0][w][x, y] - 
  C0*FF0*Derivative[4, 0][w][x, y]}

Answer 2

Generally if you have n variables you need to have at least n+1 equations to get meaningful elimination and end up with at least one equation. With m equations and k variables you get m-k resulting equations after elimination (in a well defined system). Your case is two variables and two equations which results of course in this:

 Eliminate[{x == 2 + y, y == z - 5 x}, {x, y}]

True

While meaningful request for Eliminate would be for example

 Eliminate[{x == 2 + y, y == z - 5 x}, x]

-10 + z == 6 y

My suspicion is you just would like to express your variables via the rest of stuff, which can be done with Solve:

eq={A0*Derivative[0, 2][w][x, t] + C0*Derivative[2, 0][w][x, y] == 
  C0*Derivative[1, 0][Q][x, t] + B0*Derivative[2, 0][w][x, t], 
 Q[x, y] + CC0*Derivative[1, 0][w][x, y] + FF0*Derivative[2, 0][Q][x, y] == 
  E0*Derivative[0, 2][Q][x, t]};

eq // Column // TraditionalForm

Solve[eq, {Derivative[0, 2][Q][x, t], Derivative[1, 0][Q][x, t]}] // 
   First // Column // TraditionalForm