Substituting solution into the equation

Question

I need to solve 6*6 equations over and over and put the coefficient into the equations again. But I don not know how? Can anyone help me?

u[x_, y_] := a1 + a2*x + a3*y + a4*x^2 + a5*x*y + a6*y^2

Solve[{u1 == u[0, 0], u2 == u[20, 0], u3 == u[10, 5], 
  u4 == u[10, 0], u5 == u[15, 2.5], u6 == u[5, 2.5]},
  {a1, a2, a3, a4, a5, a6}]

I need to backsubstitue a1,a2,... into the original equation and rewrite it for further use.

Possible duplicate - see this answer: http://mathematica.stackexchange.com/questions/18393/what-are-the-most-common-pitfalls-awaiting-new-users/18706#18706 — Michael E2, May 31 '15 at 01:28

score 3 · Answer 1 · answered May 31 '15 at 01:08

u[x_, y_] := a1 + a2*x + a3*y + a4*x^2 + a5*x*y + a6*y^2

solution = 
 Solve[{u1 == u[0, 0], u2 == u[20, 0], u3 == u[10, 5], u4 == u[10, 0],
     u5 == u[15, 2.5], u6 == u[5, 2.5]}, {a1, a2, a3, a4, a5, a6}] // Last;

uNew[x_,y_] := u[x, y] /. solution

uNew[x,y]//Rationalize

u1 + 1/20 (-3 u1 - u2 + 4 u4) x + 1/200 (u1 + u2 - 2 u4) x^2 + 
 1/10 (-3 u1 + u2 - 2 u3 - 4 u4 + 8 u6) y + 
 1/50 (u1 - u2 + 2 u5 - 2 u6) x y + 
 1/50 (u1 + u2 + 4 u3 + 2 u4 - 4 u5 - 4 u6) y^2

score 2 · Answer 2 · answered May 31 '15 at 01:12

A very slight variation on @Ivan solution using rationals

u[x_, y_] = a1 + a2*x + a3*y + a4*x^2 + a5*x*y + a6*y^2;

soln1 = Solve[{u1 == u[0, 0], u2 == u[20, 0], u3 == u[10, 5], u4 == u[10, 0], 
       u5 == u[15, 2.5], u6 == u[5, 2.5]}, {a1, a2, a3, a4, a5, a6}][[1]] // 
    Rationalize[#, 10^-9] & // Simplify;

u[x_, y_] = u[x, y] /. soln1 // Simplify

(1/200)*(200*u1 - 10*(3*u1 + u2 - 4*u4)*x + (u1 + u2 - 2*u4)x^2 + 20(-3*u1 + u2 - 2*u3 - 4*u4 + 8*u6)y + 4(u1 - u2 + 2*u5 - 2*u6)xy + 4*(u1 + u2 + 4*u3 + 2*u4 - 4*u5 - 4*u6)*y^2)

{u1 == u[0, 0], u2 == u[20, 0], u3 == u[10, 5], u4 == u[10, 0], 
  u5 == u[15, 5/2], u6 == u[5, 5/2]} // Simplify

{True, True, True, True, True, True}

Substituting solution into the equation

2 Answers2