2
$\begingroup$

I have solved below equation in Mathematica, but answers are very complicated and odd. For example for 'f2x' there are lots of terms like:

(-(((-4 g h (2 g^2 + 3 g h + h^2) + 
      3 a (2 g^3 + 3 g^2 h + g h^2 + h^3)) Cos[t]^2)/(
   24 a eta h (g + h) \[Pi])) - ((-4 g h (g^2 + 3 g h + 2 h^2) + 
     3 a (g^3 + g^2 h + 3 g h^2 + 2 h^3)) Cos[t]^2)/(
  24 a eta g (g + h) \[Pi])) 

or terms like:

 (-(((3 a - 2 H) Cos[tp]^2)/(6 a eta \[Pi])) - ((3 a - 2 H) Sin[tp]^2)/(6 a eta \[Pi]))

Which Mathematica can simply simplify it but it returns this complicated answer. Also, when I tried to simplify the answer Mathematica can not simplify it even after two hours! Could anyone help me? How can I avoid this form of complicated answer? Any answer i highy appreciated.

a11=((-3 a + 2 g) Cos[t])/(6 a eta g \[Pi]);
    a13=((-3 a + 2 g) Sin[t])/(6 a eta g \[Pi]);
    a14 =-(((-2 g h (g + h) + 3 a (g^2 + g h + h^2)) Cos[t])/(12 a eta g h (g + h) \[Pi]));
    a16 =-(((-2 g h (g + h) + 3 a (g^2 + g h + h^2)) Sin[t])/(12 a eta g h (g + h) \[Pi]));
    a17 =(g G (15 Sin[2 t - 4 the - tp] + 9 Sin[2 t - 2 the - tp] - 2 Sin[tp] - 3 Sin[2 t + tp] + 3 Sin[2 t - 2 the + tp] - 6 Sin[2 the + tp]))/(64 eta \[Pi] r12^3);
    a19 =(g G (15 Cos[2 t - 4 the - tp] - 9 Cos[2 t - 2 the - tp] - 2 Cos[tp] + 3 Cos[2 t + tp] + 3 Cos[2 t - 2 the + tp] + 6 Cos[2 the + tp]))/(64 eta \[Pi] r12^3);
    a110 =(g H (-15 Sin[2 t - 4 the - tp] - 9 Sin[2 t - 2 the - tp] + 2 Sin[tp] + 3 Sin[2 t + tp] - 3 Sin[2 t - 2 the + tp] + 6 Sin[2 the + tp]))/(64 eta \[Pi] r12^3);
    a112 =-((g H (15 Cos[2 t - 4 the - tp] - 9 Cos[2 t - 2 the - tp] - 2 Cos[tp] + 3 Cos[2 t + tp] + 3 Cos[2 t - 2 the + tp] + 6 Cos[2 the + tp]))/(64 eta \[Pi] r12^3));

    a21=((-2 g h (g + h) + 3 a (g^2 + g h + h^2)) Cos[t])/(12 a eta g h (g + h) \[Pi]);
    a23 =((-2 g h (g + h) + 3 a (g^2 + g h + h^2)) Sin[t])/(12 a eta g h (g + h) \[Pi]);
    a24 =((3 a - 2 h) Cos[t])/(6 a eta h \[Pi]);
    a26=((3 a - 2 h) Sin[t])/(6 a eta h \[Pi]);
    a27 =(G h (15 Sin[2 t - 4 the - tp] + 9 Sin[2 t - 2 the - tp] - 2 Sin[tp] - 3 Sin[2 t + tp] + 3 Sin[2 t - 2 the + tp] - 6 Sin[2 the + tp]))/(64 eta \[Pi] r12^3);
    a29 =(G h (15 Cos[2 t - 4 the - tp] - 9 Cos[2 t - 2 the - tp] - 2 Cos[tp] + 3 Cos[2 t + tp] + 3 Cos[2 t - 2 the + tp] + 6 Cos[2 the + tp]))/(64 eta \[Pi] r12^3);
    a210 =(h H (-15 Sin[2 t - 4 the - tp] - 9 Sin[2 t - 2 the - tp] + 2 Sin[tp] + 3 Sin[2 t + tp] - 3 Sin[2 t - 2 the + tp] + 6 Sin[2 the + tp]))/(64 eta \[Pi] r12^3);
    a212 =-((h H (15 Cos[2 t - 4 the - tp] - 9 Cos[2 t - 2 the - tp] - 2 Cos[tp] + 3 Cos[2 t + tp] + 3 Cos[2 t - 2 the + tp] + 6 Cos[2 the + tp]))/(64 eta \[Pi] r12^3));
    a31 =(g G (-2 Cos[t] + 3 (-2 Cos[t - 2 the] + 5 Cos[t - 4 the - 2 tp] + Cos[t + 2 tp] + 3 Cos[t - 2 (the + tp)] - Cos[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3);
     a33=-((g G (2 Sin[t] + 3 (-2 Sin[t - 2 the] + 5 Sin[t - 4 the - 2 tp] + Sin[t + 2 tp] - 3 Sin[t - 2 (the + tp)] + Sin[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3));
    a34=(G h (2 Cos[t] + 3 (2 Cos[t - 2 the] - 5 Cos[t - 4 the - 2 tp] - Cos[t + 2 tp] - 3 Cos[t - 2 (the + tp)] + Cos[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3);
    a36 =(G h (2 Sin[t] + 3 (-2 Sin[t - 2 the] + 5 Sin[t - 4 the - 2 tp] + Sin[t + 2 tp] - 3 Sin[t - 2 (the + tp)] + Sin[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3);
    a37=((-3 a + 2 G) Sin[tp])/(6 a eta G \[Pi]);
    a39 =((-3 a + 2 G) Cos[tp])/(6 a eta G \[Pi]);
    a310=-(((-2 G H (G + H) + 3 a (G^2 + G H + H^2)) Sin[tp])/(12 a eta G H (G + H) \[Pi]));
    a312=-(((-2 G H (G + H) + 3 a (G^2 + G H + H^2)) Cos[tp])/(12 a eta G H (G + H) \[Pi]));  

     a41=(g H (-2 Cos[t] + 3 (-2 Cos[t - 2 the] + 5 Cos[t - 4 the - 2 tp] + Cos[t + 2 tp] + 3 Cos[t - 2 (the + tp)] - Cos[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3);
         a43 =-((g H (2 Sin[t] + 3 (-2 Sin[t - 2 the] + 5 Sin[t - 4 the - 2 tp] + Sin[t + 2 tp] - 3 Sin[t - 2 (the + tp)] + Sin[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3));
        a44 =(h H (2 Cos[t] + 3 (2 Cos[t - 2 the] - 5 Cos[t - 4 the - 2 tp] - Cos[t + 2 tp] - 3 Cos[t - 2 (the + tp)] + Cos[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3);
        a46 =(h H (2 Sin[t] + 3 (-2 Sin[t - 2 the] + 5 Sin[t - 4 the - 2 tp] + Sin[t + 2 tp] - 3 Sin[t - 2 (the + tp)] + Sin[t + 2 (the + tp)])))/(64 eta \[Pi] r12^3);
         a47 =((-2 G H (G + H) + 3 a (G^2 + G H + H^2)) Sin[tp])/(12 a eta G H (G + H) \[Pi]);
        a49 =((-2 G H (G + H) + 3 a (G^2 + G H + H^2)) Cos[tp])/(12 a eta G H (G + H) \[Pi]);
        a410 =((3 a - 2 H) Sin[tp])/(6 a eta H \[Pi]);
        a412 =((3 a - 2 H) Cos[tp])/(6 a eta H \[Pi]);
        a61=-g Sin[t];
        a63=g Cos[t];
        a64 =h Sin[t];
        a66 =-h Cos[t];
        a77 =-G Cos[tp];
        a79=G Sin[tp];
        a710 =H Cos[tp];
        a712 =-H Sin[tp];
        ap11 =-(((-4 g h (g^2 + 3 g h + 2 h^2) + 3 a (g^3 + g^2 h + 3 g h^2 + 2 h^3)) Sin[t])/(24 a eta g h (g + h) \[Pi]));
         ap13 =((-4 g h (g^2 + 3 g h + 2 h^2) + 3 a (g^3 + g^2 h + 3 g h^2 + 2 h^3)) Cos[t])/(24 a eta g h (g + h) \[Pi]);
        ap14 =-(((-4 g h (2 g^2 + 3 g h + h^2) + 3 a (2 g^3 + 3 g^2 h + g h^2 + h^3)) Sin[t])/(24 a eta g h (g + h) \[Pi]));
        ap16=((-4 g h (2 g^2 + 3 g h + h^2) + 3 a (2 g^3 + 3 g^2 h + g h^2 + h^3)) Cos[t])/(24 a eta g h (g + h) \[Pi]);
        ap17=0;
        ap19 =0;
        ap110 =0;
        ap112=0;
        ap21 =0;
         ap23=0;
        ap24 =0;
        ap26 =0;
         ap27 =((-4 G H (G^2 + 3 G H + 2 H^2) + 3 a (G^3 + G^2 H + 3 G H^2 + 2 H^3)) Cos[tp])/(24 a eta G H (G + H) \[Pi]);
        ap29 =-(((-4 G H (G^2 + 3 G H + 2 H^2) + 3 a (G^3 + G^2 H + 3 G H^2 + 2 H^3)) Sin[tp])/(24 a eta G H (G + H) \[Pi]));
        ap210 =((-4 G H (2 G^2 + 3 G H + H^2) + 3 a (2 G^3 + 3 G^2 H + G H^2 + H^3)) Cos[tp])/(24 a eta G H (G + H) \[Pi]);
        ap212 =-(((-4 G H (2 G^2 + 3 G H + H^2) + 3 a (2 G^3 + 3 G^2 H + G H^2 + H^3)) Sin[tp])/(24 a eta G H (G + H) \[Pi]));

 sol = Solve[
   a11 f2x + a13 f2z + a14 f3x + a16 f3z + a17 f5x + a19 f5z + 
      a110 f6x + a112 f6z == gd && 
    a21 f2x + a23 f2z + a24 f3x  + a26 f3z + a27 f5x + a29 f5z + 
      a210 f6x + a212 f6z == hd && 
    a31 f2x + a33 f2z + a34 f3x  + a36 f3z + a37 f5x + a39 f5z + 
      a310 f6x + a312 f6z == Gd && 
    a41 f2x + a43 f2z + a44 f3x  + a46 f3z + a47 f5x + a49 f5z + 
      a410 f6x + a412 f6z == Hd &&

    a61 f2x + a63 f2z + a64 f3x  + a66 f3z == 0 &&

    a77 f5x + a79 f5z + a710 f6x + a712 f6z == 0 &&

    ap11 f2x + ap13 f2z + ap14 f3x + ap16 f3z + ap17 f5x + ap19 f5z + 
      ap110 f6x + ap112 f6z == 0 &&

    ap21 f2x + ap23 f2z + ap24 f3x + ap26 f3z + ap27 f5x + ap29 f5z + 
      ap210 f6x + ap212 f6z == 0 , {f2x, f2z, f3x, f3z, f5x, f5z, f6x,
     f6z}];
f2x = f2x /. sol[[1, 1]];
f2z = f2z /. sol[[1, 2]];
f3x = f3x /. sol[[1, 3]]; 
f3z = f3z /. sol[[1, 4]];
f5x = f5x /. sol[[1, 5]];
f5z = f5z /. sol[[1, 6]]; 
f6x = f6x /. sol[[1, 7]];
f6z = f6z /. sol[[1, 8]];

with:

$\endgroup$

closed as off-topic by Daniel Lichtblau, José Antonio Díaz Navas, MarcoB, Sektor, rhermans Jun 22 '18 at 19:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, José Antonio Díaz Navas, MarcoB, Sektor, rhermans
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Be patient and wait for the answer until MMA spits out the answer,or use computerworld.com/article/3086178/high-performance-computing/…. If you want a quick answer try NSolve. $\endgroup$ – Mariusz Iwaniuk Jun 9 '18 at 12:30
  • $\begingroup$ Mathematica solves it and gives answer, but the answer is very complicated. @MariuszIwaniuk $\endgroup$ – Holger Mate Jun 9 '18 at 12:31
  • $\begingroup$ Try FullSimplify and be patient and wait for the answer until MMA spits out the answer. $\endgroup$ – Mariusz Iwaniuk Jun 9 '18 at 12:33
  • $\begingroup$ I have not values of coefficient to use NSolve @MariuszIwaniuk $\endgroup$ – Holger Mate Jun 9 '18 at 12:33
  • 2
    $\begingroup$ What leads you to the (probably) assumption that these complicated equations can be solved symbolically? $\endgroup$ – Henrik Schumacher Jun 10 '18 at 6:34