# Way to solve these non-linear equations since fsolve isn't working fine.

조회 수: 2 (최근 30일)
Agrim Singh 2021년 6월 7일
댓글: Agrim Singh 2021년 6월 12일
function F = Glc_Gal_Lac(y,u,gal,lac)
Yx_Gal = 138000000;
KGal = 18.23;
Yx_Glc = 1010000000;
mGlc = 0.0000000000343;
fGal = 0.35;
KcGal = 5.27;
Yx_Lac = 54000000;
YLac_div_Glc = 1.56;
Lac_max1 = 21.20;
Lac_max2 = 16;
mLac = 0.000000000187;
F(1) = y(1) - (((-1*(u/Yx_Glc)) - mGlc)*((KcGal/(KcGal + gal)).^(1 - ((fGal*y(1))/y(2))))); % qGlc
F(2) = y(2) - (-1*(u/Yx_Gal)*(gal/(gal + KGal))); % qGal
F(3) = y(3) - (((u/Yx_Lac) - (YLac_div_Glc*y(2)))*((Lac_max1 - lac)/Lac_max1)) + ((mLac*(Lac_max2 - lac))/Lac_max2); % qLac
end
I believe the above function is understandable, now upon solving it what I get is
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
Error using fsolve (line 300)
Objective function is returning undefined values at initial point. FSOLVE cannot continue.
AND
Equation solved. The final point is the initial point.
The sum of squared function values, r = 6.140730e-22, is less than sqrt(options.FunctionTolerance) = 1.000000e-03.
The relative norm of the gradient of r, 4.417532e-11, is less than options.OptimalityTolerance = 1.000000e-06.
Now how to make this work, I have tried changing initial points. Plotting the function is difficult since other variables inputted are linked here and there. So, any alternate way you can suggest or any editing in code will be a great help for me to make it run successfully.
##### 댓글 수: 4이전 댓글 2개 표시이전 댓글 2개 숨기기
J. Alex Lee 2021년 6월 7일
your anonymous function looks messed up:
@(y) Glc_Gal_Lac(x,u,gal,lac)
should probably be
@(y) Glc_Gal_Lac(y,u,gal,lac)
Anyway you have tiny numbers and enormous numbers, which could be problematic and leading to numerically too-small residuals
Walter Roberson 2021년 6월 9일
When the input is ignored, the output would be constant which is curvature 0. J. Alex Lee's observation is probably the solution to the problem.

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### 답변 (1개)

Chidvi Modala 2021년 6월 9일
The objective function supplied to FSOLVE must return a vector without any Inf or NaN entries. In your case, it is possible that objfun(xo) returns a NaN or errors out (where "objfun" is objective function and "xo" is the initial point). To avoid this error, you need to choose a better initial point.
You may refer to documentation for FSOLVE: http://www.mathworks.com/help/optim/ug/fsolve.html
This applies to other functions as well, such as LSQNONLIN.
##### 댓글 수: 1이전 댓글 -1개 표시이전 댓글 -1개 숨기기
Agrim Singh 2021년 6월 12일
I used a better initial point that needed to be used but now I am struck here getting this
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
Equation solved. The final point is the initial point.
The sum of squared function values, r = 1.067653e-22, is less than sqrt(options.FunctionTolerance) = 1.000000e-03.
The relative norm of the gradient of r, 1.038534e-11, is less than options.OptimalityTolerance = 1.000000e-06.
So, I was wondering way to change the tolerance limits displayed in the above two line. So, how to go about it?

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