Solution of system of nonlinear equations

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I have following system of equations to find q1, q2, q3 and q4, which I am unable to solve using solve using matlab 'solve' or 'fsolve' commands. All the values should be positive and should not be exceeding 1.0. Are there any suggestions? Or should I implement newton method to solve it?
eqc1_pre = 91.907*q1 - 1.1978e-14*q3 + 3.534e-14*q1*q3 + 9.1309e+6*(0.007649*q1 + 0.03937*q3)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.819e-13*q3^2 == 0
eqc2_pre = 2914.4*q2 - 875.08*q3 + 3645.7*q2*q3 + 1.3054e+7*(0.007649*q2 + 0.03937*q3)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 18765.0*q3^2 == 0
eqc3_pre = 3.638e-13*q1*q3 - 875.08*q2 - 9.4963e+8*q3 - 18765.0*q4 - 1.1978e-14*q1 + 37530.0*q2*q3 + 9.1309e+6*(0.03937*q1 + 0.36373*q3 - 996.13)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.3054e+7*(0.03937*q2 + 0.36373*q3 - 996.23)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 1.767e-14*q1^2 + 1822.8*q2^2 + 260050.0*q3^2 + 7.7808e+12 == 0
eqc4_pre = - 1374.8*q1^2 - 14153.0*q1*q3 - 1965.6*q2^2 - 20235.0*q2*q3 - 158850.0*q3^2 + 8.701e+8*q3 + 34387.0*q4 - 1.4259e+13 == 0
  4 个评论
John D'Errico
John D'Errico 2024-10-16
Answers is not a job seeking forum. Please do not put requests for someone to hire you in your posts. I've removed them for you.
Aquatris
Aquatris 2024-10-16
编辑:Aquatris 2024-10-16
@Pavl M. Did you even solve the problem that was asked? The question askes for solutions within [0 1] space where as, if I understood your confusing comment, your solution does not put any constraints to the parameters.

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回答(3 个)

Alex Sha
Alex Sha 2024-10-17
@Swami Between the range [0,1] you specified, most likely, there should be no real numbers solution,an approximate solution is:
q1: -31830.1002324947
q2: -31835.6758682725
q3: 6184.74933332136
q4: 336421377.881206

John D'Errico
John D'Errico 2024-10-17
编辑:John D'Errico 2024-10-17
This will be literally impossible to do in double precision, given the huge dynamic range of those coefficients. But we can look to see if it is even likely a solution does exist.
syms q [4,1]
eqc_pre(1) = 91.907*q1 - 1.1978e-14*q3 + 3.534e-14*q1*q3 + 9.1309e+6*(0.007649*q1 + 0.03937*q3)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.819e-13*q3^2;
eqc_pre(2) = 2914.4*q2 - 875.08*q3 + 3645.7*q2*q3 + 1.3054e+7*(0.007649*q2 + 0.03937*q3)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 18765.0*q3^2;
eqc_pre(3) = 3.638e-13*q1*q3 - 875.08*q2 - 9.4963e+8*q3 - 18765.0*q4 - 1.1978e-14*q1 + 37530.0*q2*q3 + 9.1309e+6*(0.03937*q1 + 0.36373*q3 - 996.13)*(0.0038245*q1^2 + 0.03937*q1*q3 + 0.18186*q3^2 - 996.13*q3 - 0.03937*q4 + 1.6325e+7) + 1.3054e+7*(0.03937*q2 + 0.36373*q3 - 996.23)*(0.0038245*q2^2 + 0.03937*q2*q3 + 0.18186*q3^2 - 996.23*q3 - 0.03937*q4 + 1.6325e+7) + 1.767e-14*q1^2 + 1822.8*q2^2 + 260050.0*q3^2 + 7.7808e+12;
eqc_pre(4) = - 1374.8*q1^2 - 14153.0*q1*q3 - 1965.6*q2^2 - 20235.0*q2*q3 - 158850.0*q3^2 + 8.701e+8*q3 + 34387.0*q4 - 1.4259e+13;
vpa(eqc_pre(:),4)
UGH. But we can look at whether it is likely a solution even exists in the hyper-box [0,1]^4.
double(subs(eqc_pre,q,rand(4,1)))
ans = 1×4
1.0e+17 * 0.0001 0.0001 -3.6066 -0.0001
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double(subs(eqc_pre,q,rand(4,1)))
ans = 1×4
1.0e+17 * 0.0000 0.0000 -3.6070 -0.0001
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double(subs(eqc_pre,q,rand(4,1)))
ans = 1×4
1.0e+17 * 0.0000 0.0000 -3.6077 -0.0001
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However, I would point out that having substituted three sets of random numbers into those equations, almost always equation 3 tends to run about 3.6e17, and those computations were performed in high precision.
eq3 = matlabFunction(eqc_pre(3))
eq3 = function_handle with value:
@(q1,q2,q3,q4)q1.*(-1.1978e-14)-q2.*8.7508e+2-q3.*9.4963e+8-q4.*1.8765e+4+q1.*q3.*3.638e-13+q2.*q3.*3.753e+4+(q1.*3.59483533e+5+q3.*3.321182257e+6-9.095563417e+9).*(q3.*(-9.9613e+2)-q4.*3.937e-2+q1.*q3.*3.937e-2+q1.^2.*3.8245e-3+q3.^2.*1.8186e-1+1.6325e+7)+q1.^2.*1.767e-14+q2.^2.*1.8228e+3+q3.^2.*2.6005e+5+(q2.*5.1393598e+5+q3.*4.74813142e+6-1.300478642e+10).*(q3.*(-9.9623e+2)-q4.*3.937e-2+q2.*q3.*3.937e-2+q2.^2.*3.8245e-3+q3.^2.*1.8186e-1+1.6325e+7)+7.7808e+12
eq3val = eq3(rand(1000000,1),rand(1000000,1),rand(1000000,1),rand(1000000,1));
min(eq3val)
ans = -3.6078e+17
max(eq3val)
ans = -3.6061e+17
This is enough to convince me that no solution exists in that region, since any of a set of 1e6 random points in the box NEVER see any deviation from a very narrow range, that is very far from zero.

Pavl M.
Pavl M. 2024-10-17
编辑:Pavl M. 2024-10-17
Here, I've solved it in more than 4 methods, for specific equations' system approximated solutions found with fixed interval constraints [a,b] added as per requirements plan:
A.1: Runs:
I've corrected, revamped functionality, amended the software program running codes presented in my comment above.
Working codes in Matlab compatible T.C.E. (fles SystemOfNonlinEqSolver1.m, newton_method.m, fronext.m, ...+):
Pictures are worth than thouthands words.I have the corresponding, according to specifications software program code maintaining running on file disk storage controller electronic unit processed machine. I developed them. They are for sale, if you need the software program codes running, contact me more + for more.

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