Why do I get no solution found in Fsolve despite the results match my nonlinear equations ?

Hi All

I am solving a set of Non Linear equations. when I put the solution results in each of my equations, the result is zero or very close to zero ( 1e-15 as residual)

the options I use :

options = optimoptions(@fsolve,'Display','iter','MaxFunEvals',1e6,'MaxIter',1e6,'TolFun',1e-1,'DerivativeCheck','on','Diagnostics','on');

but on the message board I get the no solution found, that does not change with any tolerance value :

Here I put a copy of the last iterations and the message :

    23         72     1.44467e+07    0.000238419            1.4        0.000238
    24         73     1.44467e+07    0.000238419            1.4        0.000238
    25         77     1.44467e+07    5.96046e-05          0.325        5.96e-05
    26         78     1.44467e+07    5.96046e-05          0.325        5.96e-05
    27         82     1.44467e+07    1.49012e-05          0.147        1.49e-05
    28         86     1.44467e+07    1.49012e-05         0.0306        1.49e-05

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
selected value of the function tolerance.


fsolve stopped because the sum of squared function values, r, has gradient with
relative norm 3.057666e-02; this is less than options.OptimalityTolerance = 1.000000e-01.
However, r = 1.444675e+07, exceeds sqrt(options.FunctionTolerance) = 3.162278e-01.

Optimization Metric                                          Options
norm(grad r) =   3.06e-02                        OptimalityTolerance =   1e-01 (selected)
r =   1.44e+07                               sqrt(FunctionTolerance) =  3.2e-01 (selected)

[Walter Roberson]

Can you provide your code for testing?

[farzad]

That's for sure not possible, very sorry.

I wish this problem is common and there is already a solution or procedure to verify the problem

I think the higher number of Nonlinear equations, the more divergent become the MATLAB output messages comparing with actual situation.

at the moment Matlab is giving an unreal output message

[Walter Roberson]

That's for sure not possible, very sorry.

Then we cannot be of assistance to you. You will need to hire someone that you can put under non-disclosure agreement.

I wish this problem is common and there is already a solution or procedure to verify the problem

The problem is common, and the solution is to repair your equations.

[farzad]

The equations are the exact analytical definition of the problem

what do you intend by repairing ?

[Walter Roberson]

If fsolve is telling you that the residual is about 1E15 and your code says that the function value is about 1e-15 then your testing code is wrong, or else your equations are numerically unstable, or you have a serious loss of precision.

[Bruno Luong]

What is the first guess x0 you provide to FSOLVE?

Yes if the problem is very-non linear or non-differentiable or ill-condition FSOLVE might converge toward a local minimum or not converge at all.

But as you do not show any code it's just anyone guess that you'll get as diagnostic.

[farzad]

Bruno my first guess is taken from some experimental data that I know they fit in the equations. So yes the whole problem is being ill conditioned and hard to differentiate. WhatsApp do you do generally to improve this case?

[Bruno Luong]

I asked for the first-guess of fit parameter x0, not the data.

If FSOLVE tells function value is as large as 1.44467e+07 then I doubt the data fits well the model as you claim (and this has nothing to do with non-linearity, since the function value reflects the maching - residual as MATT told you). Or your implementation is incorrect.

[farzad]

To test the results, I simply run each of the equations. Their value come very small

[Bruno Luong]

Sorry I certainly don't believe you more than the correctness of FSOLVE output of 1.44467e+07.

Your implementation or/and test is wrong.

[farzad]

Ok, apart from the method I mentioned, what's is your proposed method of implementation?

[Bruno Luong]

We can't advise the method if you refrain to provide any detail/desciption of the problem to be solved.

[farzad]

What is the difference in evaluation if the equations are a set of 3rd orders or a set of 2nd order equations?

[farzad]

@Bruno :

if you have 3 or 5 or 6 second order equations ,once you have the x matrix as the results of the solution , and if Matlab gives that large r = 1.444675e+07 in command prompt , what would you expect to get by running ( F9) each of your equations ?

[farzad]

@Bruno

is it an impossible question

[farzad]

please check my comment in Matt J answer

Answer 1

[Matt J]

You're asking for guesses? Because you are not plugging the solution into the same equations that were given to fsolve. fsolve thinks you have a residual norm of sqrt(1.44e+07).

[farzad]

What do you mean by plugging the solution into the same equations ?

what does this residual norm mean ?

[farzad]

if by plugging you mean tagging fun = @rood2d where your equations are and use it in Fsolve command , I have done it

[Matt J]

plugging in means substituting your results into your equations. You said you did that and it gave you residuals of 1e-15.

[farzad]

ok , So I don't understand why you should have written the first answer !!??

how is this possilbe then ?

I see my results are fine when even substituting manually in the equations

but MATLAB gives me this error !!

how come ?

[farzad]

I added the options I used and as you see the Tolerance is 1e-1

Is there anything else required to be modified ? should I change the method ?

[farzad]

Is it reasonable to use fzero when fsolve does not give good results ?

my only problem is how to make matlab read the command window message ,since there is incompatibility between the function residuals when trying the results manually and those messages of the command window

[farzad]

I feel part of the problem that I don't get good results is at zero point :

here

but what about the rest ?

[Matt J]

how is this possilbe then ? I see my results are fine when even substituting manually in the equations

Because you have not substituted manually into the same equations as fsolve has been given. fsolve is solving a different set of equations than what you truly intend. Why this occurs is not possible for us to know since we are blind to the code you are running.

[farzad]

Thats simply not true They are the SAME equations!!!!!

I wouldn't commit such an obvious mistake!!!!!!!!!!!!!!!!!!!!

[Torsten]

Honestly: How do you think anybody can help you from the information you've given us so far ?

[Matt J]

@farzad: Why don't you show us just the top-level code (not the objective function) that runs fsolve and the test of its output that you mention. If we are completely blind to what you are running, the only thing we can suspect are obvious mistakes.

[farzad]

the top level code is :

[x,fval,exitflag,output] = fsolve(fun,x0init,options);

F is a matrix containing all the equations. after the code ends, I manually rewrite the equations in the command prompt obviously with all of the x components calcualted ,so once typing F after all of these eqations , this is given out :

F =

   1.0e-08 *

    0.2954    0.2954    0.4458    0.2325    0.2492    0.2562    0.0531    0.0342    0.5142

Also exitflag= -2

[Torsten]

If you insert

fun(x)

after the line

[x,fval,exitflag,output] = fsolve(fun,x0init,options);

what do you get ?

[farzad]

 236.3596 -165.1109    5.0631  294.1734  286.7896   33.6382  179.1413 -109.6156   76.8316

this is what I get

[Torsten]

These are the errors in the equations for the x-vector fsolve returns as solution.

[farzad]

You are so correct, so may I ask how do I correct the errors ?

[farzad]

Specially if I am sure that my equations are analytically correct. could such an error be corrected by improving the solution by Jacobian matrix ,etc ?

[Torsten]

Since I don't know the equations you are trying to solve, the only thing that comes to mind is choosing different initial values (vector x0init) or the Multistart option.

[farzad]

do you mean this ? what should I write in the paranthesis after run(--) instead of the lines ?

[x,fval,exitflag,output,solutions] = run(___)

I mean how to plug this to @root2D where my equations are ?

[Matt J]

I wouldn't commit such an obvious mistake!!!!!!!!!!!!!!!!!!!!

Hah! This is me gloating!!!!

[farzad]

positive voted

[farzad]

so about my last question in multistart ?

[farzad]

I was wrong about the last time !!! fun(x) gives low value :

____________________________________________________________
   Diagnostic Information

Number of variables: 3

Functions 
Objective:                            rootcenter
Gradient:                             finite-differencing
 

Algorithm selected
   trust-region-dogleg


____________________________________________________________
   End diagnostic information

                                         Norm of      First-order   Trust-region
 Iteration  Func-count     f(x)          step         optimality    radius
     0          4     1.69272e+07                      2.24e+05               1
     1          8     1.64774e+07              1       2.02e+05               1
     2         12     1.55628e+07            2.5       1.48e+05             2.5
     3         16     1.44888e+07           6.25       2.25e+04            6.25
     4         17     1.44888e+07           6.25       2.25e+04            6.25
     5         21     1.44506e+07         1.5625       9.73e+03            1.56
     6         22     1.44506e+07         1.5625       9.73e+03            1.56
     7         26     1.44471e+07       0.390625       1.97e+03           0.391
     8         27     1.44471e+07       0.390625       1.97e+03           0.391
     9         31     1.44468e+07      0.0976562            876          0.0977
    10         35     1.44468e+07      0.0976562            820          0.0977
    11         36     1.44468e+07      0.0976562            820          0.0977
    12         40     1.44468e+07      0.0244141            175          0.0244
    13         41     1.44468e+07      0.0244141            175          0.0244
    14         45     1.44467e+07     0.00610352            107          0.0061
    15         49     1.44467e+07     0.00610352           29.7          0.0061
    16         50     1.44467e+07     0.00610352           29.7          0.0061
    17         54     1.44467e+07     0.00152588           10.2         0.00153
    18         55     1.44467e+07     0.00152588           10.2         0.00153
    19         59     1.44467e+07     0.00038147           5.32        0.000381
    20         63     1.44467e+07     0.00038147          0.435        0.000381
    21         64     1.44467e+07     0.00038147          0.435        0.000381
    22         65     1.44467e+07    9.53674e-05          0.435        9.54e-05
    23         69     1.44467e+07    2.38419e-05          0.214        2.38e-05
    24         73     1.44467e+07    2.38419e-05         0.0943        2.38e-05
    25         74     1.44467e+07    2.38419e-05         0.0943        2.38e-05
    26         78     1.44467e+07    5.96046e-06         0.0268        5.96e-06
    27         79     1.44467e+07    5.96046e-06         0.0268        5.96e-06
    28         83     1.44467e+07    1.49012e-06         0.0414        1.49e-06
    29         84     1.44467e+07    1.49012e-06         0.0414        1.49e-06
    30         85     1.44467e+07    3.72529e-07         0.0414        3.73e-07
    31         89     1.44467e+07    9.31323e-08         0.0107        9.31e-08
    32         90     1.44467e+07    9.31323e-08         0.0107        9.31e-08
    33         91     1.44467e+07    2.32831e-08         0.0107        2.33e-08
    34         92     1.44467e+07    5.82077e-09         0.0107        5.82e-09
    35         93     1.44467e+07    1.45519e-09         0.0107        1.46e-09
    36         94     1.44467e+07    3.63798e-10         0.0107        3.64e-10
    37         95     1.44467e+07    9.09495e-11         0.0107        9.09e-11
    38         96     1.44467e+07    2.27374e-11         0.0107        2.27e-11

No solution found.

fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the selected value of the function tolerance.

[farzad]

I don't know why the other time it gave me big numbers or I did not notice the 1e-8 value

[farzad]

This is the contradiction I am facing !!

by the way

the

funx = fun(x)

after the x is calculated

[Torsten]

funx can't be the output of fun(x) after the call to fsolve. Otherwise, it would have been printed after fsolve's iteration monitor and error message.

[farzad]

I don't understand you. so when should I call fun(x) ?

[Matt J]

The purpose of calling fun(x) is to test how well the result x returned by fsolve satisfies the equations. Therefore, you should call it immediately after fsolve has finished working and you must make sure that the x here is the x returned by fsolve.

[farzad]

so I am back to my question :

should I trust the fun(x) , or MATLAB output Error ???

[Matt J]

The output of fun(x) and the messages returned by fsolve should be telling you the same thing. They were telling you the same thing several Comments ago.

[Bruno Luong]

I answered you 6h ago:

"Sorry I certainly don't believe you more than the correctness of FSOLVE output of 1.44467e+07."

[farzad]

Bruno, shall we avoid this way of conversation ?

you asked me to test fun(x) and tell you the output and I did

and reported you.

[farzad]

@Matt J

why do you avoid checking my last report where I have a small fun(x)

and a big error

they are not telling the same thing

[farzad]

Sorry

I was doing something wrong

I had two fsolve loops

and I was reporting the error of the second one for the first one

sorry !

my mistake