MaplePrimes - answers and comments on Question, How do I solve a non-linear equation system

Use fsolve

alsholm — Thu, 11 Mar 2010 21:32:55 Z

If you know something about the ranges in which the answers should lie then you could use that information in fsolve.

(What are "the real" alfa values you mention?)

However, fsolve does come up with a solution without being given ranges or starting values:

EQ1:= 0.9868421053+74479.54250*a^4-1.*alfa-33391.41112*P*a^3-.5000000000*C2*a^2;

EQ2:= 2.9791817*10^5*a^3-0.4044127197e-1-1.001742334*10^5*P*a^2-1.*C2*a+0.4098048893e-1*A;

EQ3:= -2.003484667*10^5*P*a-0.3358800946e-2*A-1.*C2+8.937545100*10^5*a^2;

EQ4:= -2.003484667*10^5*P+0.1376453050e-3*A+0.1113310242e-4*a+0.1358341826e-3;

fsolve(eval({EQ1,EQ2,EQ3,EQ4},a=7),{C2,alfa,A,P});

Preben Alsholm

solving nonlinear systems

Doug Meade — Thu, 11 Mar 2010 21:35:04 Z

Your system of equations leaves a few questions. First, as you've written them, they depend on a, A, alfa, P, and C2. I assume that a and A are the same variable. (You also mention a2, but I don't see where that appears in your equations. Is this a? If so, then my specific solutions will be meaningless, but the general approach is still correct.) Second, you've given only expressions, not equations; I assume you make these into equations by equating each expression to zero.

If these assumptions are correct, then you have:

EQ1:=0.9868421053+74479.54250*a^4-1.*alfa-33391.41112*P*a^3-.5000000000*C2*a^2=0;
EQ2:=2.9791817*10^5*a^3-0.4044127197e-1-1.001742334*10^5*P*a^2-1.*C2*a+0.4098048893e-1*a=0;
EQ3:=-2.003484667*10^5*P*a-0.3358800946e-2*a-1.*C2+8.937545100*10^5*a^2=0;
EQ4:=-2.003484667*10^5*P+0.1376453050e-3*a+0.1113310242e-4*a+0.1358341826e-3=0;
                                     4                            3
         0.9868421053 + 74479.54250 a  - 1. alfa - 33391.41112 P a 

                               2    
            - 0.5000000000 C2 a  = 0
                   5  3                                 5    2          
     2.979181700 10  a  - 0.04044127197 - 1.001742334 10  P a  - 1. C2 a

        + 0.04098048893 a = 0
                 5                                                5  2    
  -2.003484667 10  P a - 0.003358800946 a - 1. C2 + 8.937545100 10  a  = 0
                       5                                            
        -2.003484667 10  P + 0.0001487784074 a + 0.0001358341826 = 0

From here, Maple does not have to think too hard to find solutions:

solve( {EQ1,EQ2,EQ3,EQ4}, {a,alfa,P,C2} );
    /                                   
   { C2 = -7.394704137 + 12.87893418 I, 
    \                                   

                       -10                 -12    
     P = 6.795062889 10    + 2.619700571 10    I, 

     a = 0.002042369542 + 0.003527749771 I, 

                                             \    /                  
     alfa = 0.9868939398 + 0.00008879760094 I }, { C2 = 14.91234974, 
                                             /    \                  

                       -10                                          \    /
     P = 6.749563134 10   , a = -0.004084733333, alfa = 0.9867384331 }, { 
                                                                    /    \

     C2 = -7.394704137 - 12.87893418 I, 

                       -10                 -12    
     P = 6.795062889 10    - 2.619700571 10    I, 

     a = 0.002042369542 - 0.003527749771 I, 

                                             \ 
     alfa = 0.9868939398 - 0.00008879760094 I }
                                             /

There are three solutions. The first and last are complex conjugates. The second has only P with a (very small) imaginary part.

Normally, I would say that the imaginary parts of order 10^(-12) are neglible and could be discarded. But, the real parts are only two orders of magnitude larger, so some more work will be necessary before I assume the imaginary parts can be ignored in this problem.

If you use fsolve instead of solve, then you find:

fsolve( {EQ1,EQ2,EQ3,EQ4}, {a,alfa,P,C2} );
       /                                    -10                       
      { C2 = 14.91234974, P = 6.749563134 10   , a = -0.004084733333, 
       \                                                              

                           \ 
        alfa = 0.9867384331 }
                           /

This gives us more confidence that the imaginary parts returned by solve can, in fact, be ignored.

You'll have to decide if all three solutions are appropriate for your problem, or whether you want only the real-valued solution.

I hope this is helpful,

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu

a2?

Robert Israel — Thu, 11 Mar 2010 21:42:23 Z

The equations (or rather expressions: I assume the equations are EQ1 = 0, ..., EQ4 = 0) you gave us don't have any a2 in them: the variables are a, P, A, C2 and alfa. Moreover, they are linear in P, A, C2 and alfa.

> solve({EQ1,EQ2,EQ3,EQ4},{P,A,C2,alfa});

{A = .2000000000e-1*(.5968744856e29*a^3+.4051173415e22-.6803542549e19*a^2)/(.1378851290e18*a^2+.8210378122e20+.6729306195e19*a), C2 = -.4000000000e-13*(-.5011956467e38*a^3+.6803542550e31-.1834515619e40*a^2+.5576250004e30*a+.1026962134e37*a^4)/(.1378851290e18*a^2+.8210378122e20+.6729306195e19*a), P = .4000000000e-9*(.2050351869e28*a^3+.2783275608e21+.9348468735e18*a^2+.2281199904e20*a)/(.1378851290e18*a^2+.8210378122e20+.6729306195e19*a), alfa = .4000000000e-14*(.6803542549e32*a^2+.2025586708e35+.1660190673e34*a+.8558017830e36*a^6-.7643815078e40*a^4-.1252989117e39*a^5+.1858750001e31*a^3)/(.1378851290e18*a^2+.8210378122e20+.6729306195e19*a)}

I am quite confident that this is correct (up to roundoff error). Yes, there will be some large values for the variables, e.g. for a = 40 you get

{A = .1335914871e12, C2 = 245770363.6, P = 91.78129128, alfa = -.2020898993e12}

But if these results are very different from the "real ones", it can only be because the equations are wrong.

Hello!Here are

hirnyk — Thu, 11 Mar 2010 22:15:17 Z

Hello!

Here are the commands of Maple 13 which solve your system:

> assume(a>0);additionally(a<80);

> solve({0.9868421053+74479.54250*a^4-1.*alfa-33391.41112*P*a^3-.5000000000*C2*a^2=0,2.9791817*10^5*a^3-0.4044127197e-1-1.001742334*10^5*P*a^2-1.*C2*a+0.4098048893e-1*A=0,-2.003484667*10^5*P*a-0.3358800946e-2*A-1.*C2+8.937545100*10^5*a^2=0,-2.003484667*10^5*P+0.1376453050e-3*A+0.1113310242e-4*a+0.1358341826e-3=0},{A,P,C2,alfa});

The outtput is somewhat large to be cited here, but you can reproduce it. All the values A, P, C2, alfa are real numbers. In fact, the coefficients are very big.

again how to solve non linear systems

JPA — Thu, 11 Mar 2010 22:56:47 Z

In fact the unknown "A" and "a" are different things. Since I was using a simplified case, probably I was not clear describing it. Sorry for making you loose time thinking on it....!

So, is better if I put here the complete non linear system I am interested to solve(already with a2 replaced on it).

Please consider only these case and ignore the previous one: (12 EQs vs 12 unknowns{A1,B1,A2,B2,C2,D2,A3,B3,D3,P,alfa,a3}

EQ1:=-16.75+(37.45*10^(-3))*A2+(961.07*10^(-3))*B2+(40.48*10^(-3))*C2+1.04*D2-1.00*B1=0;

EQ2:=(7.19*10^(-3))*A2-(7.78*10^(-3))*B2+(8.41*10^(-3))*C2+(7.78*10^(-3))*D2-(8.12*10^(-3))*A1+(8.12*10^(-3))*B1=0;

EQ3:=(2.27*10^(-6))*A2+(5.83*10^(-5))*B2-(2.46*10^(-6))*D2+(6.30*10^(-5))*C2-(6.59*10^(-5))*B1=0;

EQ4:=(4.36*10^(-7))*A2-(4.72*10^(-7))*B2+(4.72*10^(-7))*C2-(5.10*10^(-7))*D2-(5.35*10^(-7))*A1+(5.35*10^(-7))*B1=0;

EQ5:=B1-1.53*10^0=0;

EQ6:=-(1.55*10^(-8))*a3^(4.00*10^0)+(3.33*10^(-1))*A3*a3^(3.00*10^0)+(5.00*10^(-1))*B3*a3^(2.00*10^0)+D3+1.67*10-(1.00*10^0)*B2-(1.00*10^0)*D2=0;

EQ7:=-(6.21*10^(-8))*a3^(3.00*10^0)+A3*a3^(2.00*10^0)+B3*a3-(7.79*10^(-3))*A2+(7.79*10^(-3))*B2-(7.79*10^(-3))*C2-(7.79*10^(-3))*D2=0;

EQ8:=-(1.86*10^(-7))*a3^(2.00*10^0)+(2.00*10^0)*A3*a3+B3-(1.21*10^(-4))*B2-(1.21*10^(-4))*C2=0;

EQ9:=-(3.73*10^(-7))*a3+(2.00*10^0)*A3-(9.45*10^(-7))*A2+(9.45*10^(-7))*B2-(9.45*10^(-7))*C2+(9.45*10^(-7))*D2=0;

EQ10:=-2.52*10+B2+D2=0;

EQ11:=D3-1.*alfa=0;

EQ12:=A3-(6.24*10^(-7))*P=0;

thus, solving the system:

> solve(sys, {A1, A2, A3, B1, B2, B3, C2, D2, D3, P, a3, alfa});

I got multiple solutions: {A1 = -164.3765226, A2 = -166.1535085, A3 = 0.0001929630612, B1 = 1.533000000, B2 = 17.74633654, B3 = -0.1507247679, C2 = -8.532505619, D2 = 7.499026172, D3 = 31202.14335, P = 309.2788275, a3 = 1541.895107, alfa = 31202.14335}, { A1 = -164.3765226, A2 = -166.1535085, A3 = -0.00009648153046 + 0.00002277073987 I, B1 = 1.533000000, B2 = 17.74633654, B3 = 0.001854207963 + 0.02356930294 I, C2 = -8.532505619, D2 = 7.499026172, D3 = -3.444079163 + 117.2116020 I, P = -154.6394136 + 36.49666255 I, a3 = -10.70944769 + 122.1441183 I, alfa = -3.444079163 + 117.2116020 I}, {A1 = -164.3765226, A2 = -166.1535085, A3 = -0.00009648153046 - 0.00002277073987 I, B1 = 1.533000000, B2 = 17.74633654, B3 = 0.001854207963 - 0.02356930294 I, C2 = -8.532505619, D2 = 7.499026172, D3 = -3.444079163 - 117.2116020 I, P = -154.6394136 - 36.49666255 I, a3 = -10.70944769 - 122.1441183 I, alfa = -3.444079163 - 117.2116020 I}

using the command "fsolve" I have no solution.

My doubt is if I was doing the right approach using "solve" for non-linear equations and if yes, which solution por P and alfa is reasonable to choose? For instance, physically I can not expect an alfa=31202. ..

Thanks again.

complete system

hirnyk — Fri, 12 Mar 2010 00:45:31 Z

I have obtained the somewhat different solutions.

> solve({-16.75+(37.45*10^(-3))*A2+(961.07*10^(-3))*B2+(40.48*10^(-3))*C2+1.04*D2-1.00*B1=0,
(2.27*10^(-6))*A2+(5.83*10^(-5))*B2-(2.46*10^(-6))*D2+(6.30*10^(-5))*C2-(6.59*10^(-5))*B1=0,
(4.36*10^(-7))*A2-(4.72*10^(-7))*B2+(4.72*10^(-7))*C2-(5.10*10^(-7))*D2-(5.35*10^(-7))*A1+
(5.35*10^(-7))*B1=0,B1-1.53*10^0=0,-(1.55*10^(-8))*a3^(4.00*10^0)+
(3.33*10^(-1))*A3*a3^(3.00*10^0)+(5.00*10^(-1))*B3*a3^(2.00*10^0)+D3+1.67*10-
(1.00*10^0)*B2-(1.00*10^0)*D2=0,-(6.21*10^(-8))*a3^(3.00*10^0)+A3*a3^(2.00*10^0)+B3*a3-
(7.79*10^(-3))*A2+(7.79*10^(-3))*B2-(7.79*10^(-3))*C2-(7.79*10^(-3))*D2=0,-
(1.86*10^(-7))*a3^(2.00*10^0)+(2.00*10^0)*A3*a3+B3-(1.21*10^(-4))*B2-
(1.21*10^(-4))*C2=0,-(3.73*10^(-7))*a3+(2.00*10^0)*A3-(9.45*10^(-7))*A2+
(9.45*10^(-7))*B2-(9.45*10^(-7))*C2+(9.45*10^(-7))*D2=0,-2.52*10+B2+D2=0,D3-1.
*alfa=0,A3-(6.24*10^(-7))*P=0,(7.19*10^(-3))*A2-(7.78*10^(-3))*B2+(8.41*10^(-3))*C2+
(7.78*10^(-3))*D2-(8.12*10^(-3))*A1+(8.12*10^(-3))*B1=0},
{A1, A2, A3, B1, B2, B3, C2, D2, D3, P, a3, alfa});
{A1 = -163.3338772, A2 = -165.1041526, A3 = 0.0001899817798, B1 = 1.530000000, B2 = 17.71865478, B3 = -0.1462399898, C2 = -8.555238703, D2 = 7.481345216, D3 = 29516.12212, P = 304.4579805, a3 = 1522.481728, alfa = 29516.12212}, { A1 = -163.3338772, A2 = -165.1041526, A3 = -0.00009596652396 + 0.00002277688809 I, B1 = 1.530000000, B2 = 17.71865478, B3 = 0.001855552024 + 0.02344172909 I, C2 = -8.555238703, D2 = 7.481345216, D3 = -3.424534645 + 116.6085038 I, P = -153.7925063 + 36.50142322 I, a3 = -10.75314510 + 122.1280863 I, alfa = -3.424534645 + 116.6085038 I}, {A1 = -163.3338772, A2 = -165.1041526, A3 = -0.00009596652396 - 0.00002277688809 I, B1 = 1.530000000, B2 = 17.71865478, B3 = 0.001855552024 - 0.02344172909 I C2 = -8.555238703, D2 = 7.481345216, D3 = -3.424534645 - 116.6085038 I, P = -153.7925063 - 36.50142322 I, a3 = -10.75314510 - 122.1280863 I, alfa = -3.424534645 - 116.6085038 I}.

These solutions should be verifyed. Try to solve your complete system by Mathematica and to compare the answers. In my opinion, the usage of "solve" is correct. I have no idea concerning" the right solution".