i have been trying to solve 50 nonlinear equations simultaneously

this is my function file

function F=my_fin(X)
%parameters
L=1; %length of fin
%W=0.0598; %width og fin
%T=50;
%T_b=100;   %base temperature of fin
%T_a=20;   %ambient temperature of fin
% k_a=30;   %ka is the thermal conductivity at the ambient fluid temperature of the fin
%alpha=0.01; %? is the parameter describing the variation of thermal conductivity 0r temperature coefficient of thermal conductivity
% h=11;  %heat transfer coefficient 
%t=0.001; %thickness of fin
%P=2*(W+t); %fin perimeter...m(unit)
% A_c=(W*t); %cross-sectional area of the fin
%M=sqrt((h*P*L^2)/(A_c*k_a)); %thermo-geometric fin parameter
M=0.5;
n=50; % no of nodes
dx=(L/(n-1)); %space between nodes
%B=alpha*(T_b-T_a); % B dimensionless parameter describing variation of the thermal conductivity 
B=0.2;
X(1)=0;
X(51)=1;
F=zeros(1,50);
R=1;
%X(N)=(T-T_b)/(T_a-T_b);
F(1)=(1+B*(X(R+1)))*(X(R+2)-2*X(R+1)+X(R))+((B/4)*(X(R+2))^2)-((B/2)*(X(R+2)*X(R)))+((B/4)*(X(R))^2)-((M)*(X(R+1))*(dx)^2); %let (X(0))=X(1))=1 which is base temp 

for j=1:n-3
    %X(j)=(T-T_b)/(T_a-T_b);
    F(j+1)=(1+B*(X(j+2)))*(X(j+3)-2*X(j+2)+X(j+1))+((B/4)*(X(j+3))^2)-((B/2)*(X(j+3)*X(j+1)))+((B/4)*X(j+1)^2)-((M)*(X(j+2))*(dx)^2); 
end
k=1;
% X(k)=(T-T_b)/(T_a-T_b);
F(50)=(1+B*X(k+50))*(X(k+48)*X(k+50))+(B/2)*(X(k+50)-X(k+49))^2-(M*(2*(dx)^2)*X(k+50)); %X(20)=X(21)=0 insulated boundary 
end

this is my fsolve command file

clc
clear all
mata=zeros(1,50);
N=100;
OPTs = optimoptions('fsolve', 'MaxIter', N, 'Display','iter');
fun=@my_fin;
x=fsolve(fun,mata);
[Solution, Fval] = fsolve(fun, mata , OPTs)

plot(x)

and output is

Solver stopped prematurely.

fsolve stopped because it exceeded the iteration limit,

options.MaxIterations = 4.000000e+02.

Norm of First-order Trust-region

Iteration Func-count f(x) step optimality radius

0 51 0.00991687 0.12 1

1 102 0.00781698 0.0172541 0.0707 1

2 153 0.0070298 0.0103857 0.0477 1

3 204 0.00649362 0.0084817 0.0461 1

4 255 0.00606564 0.00754073 0.0329 1

5 306 0.00570042 0.00690803 0.0347 1

6 357 0.00537765 0.00649566 0.0281 1

7 408 0.00508647 0.00612239 0.028 1

8 459 0.00482033 0.00586997 0.0244 1

9 510 0.00457487 0.00559464 0.0234 1

10 561 0.00434705 0.00541385 0.0213 1

11 612 0.00413458 0.00518789 0.0202 1

12 663 0.00393573 0.00504675 0.0189 1

13 714 0.0037491 0.00485059 0.019 1

14 765 0.00357353 0.00473455 0.0168 1

15 816 0.00340806 0.00455874 0.0182 1

16 867 0.00325188 0.00446011 0.016 1

17 918 0.00310427 0.00429948 0.0174 1

18 969 0.0029646 0.00421372 0.0154 1

19 1020 0.00283233 0.00406518 0.0166 1

20 1071 0.00270694 0.00398941 0.0149 1

21 1122 0.00258801 0.00385093 0.0159 1

22 1173 0.00247512 0.0037832 0.0143 1

23 1224 0.00236789 0.00365338 0.0152 1

24 1275 0.002266 0.00359227 0.0138 1

25 1326 0.00216913 0.00347009 0.0145 1

26 1377 0.00207699 0.00341458 0.0132 1

27 1428 0.00198932 0.00329925 0.0138 1

28 1479 0.00190587 0.00324854 0.0127 1

29 1530 0.00182639 0.00313944 0.0132 1

30 1581 0.0017507 0.0030929 0.0122 1

31 1632 0.00167857 0.00298952 0.0126 1

32 1683 0.00160983 0.00294665 0.0116 1

33 1734 0.00154429 0.00284855 0.012 1

34 1785 0.00148179 0.00280892 0.0112 1

35 1836 0.00142217 0.00271574 0.0115 1

36 1887 0.0013653 0.00267902 0.0107 1

37 1938 0.00131103 0.00259042 0.011 1

38 1989 0.00125922 0.0025563 0.0102 1

39 2040 0.00120977 0.00247201 0.0105 1

40 2091 0.00116255 0.00244025 0.0098 1

41 2142 0.00111745 0.00235999 0.01 1

42 2193 0.00107437 0.00233038 0.00938 1

43 2244 0.00103322 0.00225392 0.00958 1

44 2295 0.000993894 0.00222626 0.00898 1

45 2346 0.000956312 0.00215339 0.00916 1

46 2397 0.000920391 0.00212751 0.00859 1

47 2448 0.00088605 0.00205804 0.00875 1

48 2499 0.000853216 0.0020338 0.00822 1

49 2550 0.000821818 0.00196753 0.00836 1

50 2601 0.000791789 0.0019448 0.00787 1

51 2652 0.000763065 0.00188157 0.00799 1

52 2703 0.000735586 0.00186024 0.00753 1

53 2754 0.000709294 0.00179989 0.00764 1

54 2805 0.000684135 0.00177985 0.0072 1

55 2856 0.000660056 0.00172222 0.0073 1

56 2907 0.000637008 0.00170338 0.00689 1

57 2958 0.000614945 0.00164835 0.00697 1

58 3009 0.000593822 0.00163062 0.00659 1

59 3060 0.000573595 0.00157806 0.00667 1

60 3111 0.000554226 0.00156136 0.00631 1

61 3162 0.000535674 0.00151114 0.00637 1

62 3213 0.000517905 0.00149541 0.00603 1

63 3264 0.000500881 0.00144742 0.00609 1

64 3315 0.000484571 0.0014326 0.00577 1

65 3366 0.000468943 0.00138672 0.00582 1

66 3417 0.000453966 0.00137274 0.00552 1

67 3468 0.000439611 0.00132889 0.00556 1

68 3519 0.000425852 0.0013157 0.00528 1

69 3570 0.000412662 0.00127377 0.00532 1

70 3621 0.000400015 0.00126132 0.00505 1

71 3672 0.000387889 0.00122122 0.00508 1

72 3723 0.00037626 0.00120947 0.00483 1

73 3774 0.000365107 0.00117112 0.00486 1

74 3825 0.000354408 0.00116003 0.00462 1

75 3876 0.000344145 0.00112333 0.00464 1

76 3927 0.000334299 0.00111286 0.00442 1

77 3978 0.000324851 0.00107776 0.00444 1

78 4029 0.000315784 0.00106787 0.00423 1

79 4080 0.000307082 0.00103428 0.00424 1

80 4131 0.000298729 0.00102494 0.00404 1

81 4182 0.00029071 0.000992789 0.00406 1

82 4233 0.000283011 0.00098397 0.00386 1

83 4284 0.000275619 0.0009532 0.00388 1

84 4335 0.000268519 0.000944873 0.00369 1

85 4386 0.000261701 0.000915419 0.0037 1

86 4437 0.000255151 0.000907557 0.00353 1

87 4488 0.000248858 0.000879359 0.00354 1

88 4539 0.000242812 0.000871937 0.00338 1

89 4590 0.000237002 0.000844939 0.00338 1

90 4641 0.000231418 0.000837935 0.00323 1

91 4692 0.000226051 0.000812082 0.00323 1

92 4743 0.000220892 0.000805475 0.00309 1

93 4794 0.000215931 0.000780716 0.00309 1

94 4845 0.000211161 0.000774486 0.00295 1

95 4896 0.000206573 0.000750773 0.00295 1

96 4947 0.00020216 0.0007449 0.00282 1

97 4998 0.000197915 0.000722186 0.00282 1

98 5049 0.00019383 0.000716654 0.00269 1

Solver stopped prematurely.

fsolve stopped because it exceeded the function evaluation limit,

options.MaxFunctionEvaluations = 5.000000e+03.

Solution =

Columns 1 through 10

0 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000

Columns 11 through 20

0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000

Columns 21 through 30

0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0001 0.0000

Columns 31 through 40

0.0001 -0.0001 -0.0001 -0.0004 -0.0004 -0.0006 -0.0001 0.0006 0.0022 0.0038

Columns 41 through 50

0.0054 0.0055 0.0032 -0.0030 -0.0139 -0.0299 -0.0503 -0.0733 -0.0958 -0.1136

Fval =

Columns 1 through 10

0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000

Columns 11 through 20

0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0001 0.0001 -0.0001 0.0001 -0.0001

Columns 21 through 30

0.0001 -0.0001 0.0001 -0.0001 0.0002 -0.0002 0.0002 -0.0002 0.0002 -0.0003

Columns 31 through 40

0.0002 -0.0003 0.0003 -0.0001 0.0006 0.0003 0.0008 0.0001 -0.0001 -0.0015

Columns 41 through 50

-0.0023 -0.0040 -0.0046 -0.0051 -0.0042 -0.0025 0.0006 0.0047 0 0.0086

now why my output show me it is not fully solved prematurely stop how i can increase iteration

[Torsten]

Please show us the problem you are trying to solve in a mathematical notation (the differential equation or whatever is behind your system of 50 algebraic equations).

[Akasha]

its a straight convective fin with thermal coductivity depend on temperature

[Akasha]

i have applied fdm on it

Answer 1

[Torsten]

Does the approximate solution below look correct ?

Note that you don't set F(49) in your code.

clc
clear all
mata=zeros(1,50);
OPTs = optimoptions('fsolve', 'MaxIterations', 100000, 'MaxFunctionEvaluations',100000);
fun=@my_fin;
[x, Fval,errorflag] = fsolve(fun, mata , OPTs);

Solver stopped prematurely.

fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 1.000000e+05.

errorflag

errorflag = 0

plot(linspace(0,1,50),x)

function F=my_fin(X)
%parameters
L=1; %length of fin
%W=0.0598; %width og fin
%T=50;
%T_b=100;   %base temperature of fin
%T_a=20;   %ambient temperature of fin
% k_a=30;   %ka is the thermal conductivity at the ambient fluid temperature of the fin
%alpha=0.01; %? is the parameter describing the variation of thermal conductivity 0r temperature coefficient of thermal conductivity
% h=11;  %heat transfer coefficient 
%t=0.001; %thickness of fin
%P=2*(W+t); %fin perimeter...m(unit)
% A_c=(W*t); %cross-sectional area of the fin
%M=sqrt((h*P*L^2)/(A_c*k_a)); %thermo-geometric fin parameter
M=0.5;
n=50; % no of nodes
dx=(L/(n-1)); %space between nodes
%B=alpha*(T_b-T_a); % B dimensionless parameter describing variation of the thermal conductivity 
B=0.2;
X(1)=0;
X(51)=1;
F=zeros(1,50);
R=1;
%X(N)=(T-T_b)/(T_a-T_b);
F(1)=(1+B*(X(R+1)))*(X(R+2)-2*X(R+1)+X(R))+((B/4)*(X(R+2))^2)-((B/2)*(X(R+2)*X(R)))+((B/4)*(X(R))^2)-((M)*(X(R+1))*(dx)^2); %let (X(0))=X(1))=1 which is base temp 

for j=1:n-3
    %X(j)=(T-T_b)/(T_a-T_b);
    F(j+1)=(1+B*(X(j+2)))*(X(j+3)-2*X(j+2)+X(j+1))+((B/4)*(X(j+3))^2)-((B/2)*(X(j+3)*X(j+1)))+((B/4)*X(j+1)^2)-((M)*(X(j+2))*(dx)^2); 
end
k=1;
% X(k)=(T-T_b)/(T_a-T_b);
F(50)=(1+B*X(k+50))*(X(k+48)*X(k+50))+(B/2)*(X(k+50)-X(k+49))^2-(M*(2*(dx)^2)*X(k+50)); %X(20)=X(21)=0 insulated boundary 
end

[Akasha]

why is it stop prematurely not fully solve

[Walter Roberson]

clc
clear all
mata=zeros(1,50);
OPTs = optimoptions('fsolve', 'MaxIterations', 100000000, 'MaxFunctionEvaluations', 100000000);
fun=@my_fin;
[x, Fval,errorflag] = fsolve(fun, mata , OPTs);

Equation solved, inaccuracy possible.

The vector of function values is near zero, as measured by the value
of the function tolerance. However, the last step was ineffective.

errorflag

errorflag = 3

plot(linspace(0,1,50),x)

function F=my_fin(X)
%parameters
L=1; %length of fin
%W=0.0598; %width og fin
%T=50;
%T_b=100;   %base temperature of fin
%T_a=20;   %ambient temperature of fin
% k_a=30;   %ka is the thermal conductivity at the ambient fluid temperature of the fin
%alpha=0.01; %? is the parameter describing the variation of thermal conductivity 0r temperature coefficient of thermal conductivity
% h=11;  %heat transfer coefficient 
%t=0.001; %thickness of fin
%P=2*(W+t); %fin perimeter...m(unit)
% A_c=(W*t); %cross-sectional area of the fin
%M=sqrt((h*P*L^2)/(A_c*k_a)); %thermo-geometric fin parameter
M=0.5;
n=50; % no of nodes
dx=(L/(n-1)); %space between nodes
%B=alpha*(T_b-T_a); % B dimensionless parameter describing variation of the thermal conductivity 
B=0.2;
X(1)=0;
X(51)=1;
F=zeros(1,50);
R=1;
%X(N)=(T-T_b)/(T_a-T_b);
F(1)=(1+B*(X(R+1)))*(X(R+2)-2*X(R+1)+X(R))+((B/4)*(X(R+2))^2)-((B/2)*(X(R+2)*X(R)))+((B/4)*(X(R))^2)-((M)*(X(R+1))*(dx)^2); %let (X(0))=X(1))=1 which is base temp 

for j=1:n-3
    %X(j)=(T-T_b)/(T_a-T_b);
    F(j+1)=(1+B*(X(j+2)))*(X(j+3)-2*X(j+2)+X(j+1))+((B/4)*(X(j+3))^2)-((B/2)*(X(j+3)*X(j+1)))+((B/4)*X(j+1)^2)-((M)*(X(j+2))*(dx)^2); 
end
k=1;
% X(k)=(T-T_b)/(T_a-T_b);
F(50)=(1+B*X(k+50))*(X(k+48)*X(k+50))+(B/2)*(X(k+50)-X(k+49))^2-(M*(2*(dx)^2)*X(k+50)); %X(20)=X(21)=0 insulated boundary 
end

Answer 2

[Torsten]

epsilon = 0.2;
N = 0.5;
L = 1;
bvpfcn = @(x,y)[y(2);(-epsilon*y(2)^2+N^2*y(1))/(1+epsilon*y(1))];
bcfcn = @(ya,yb)[ya(2);yb(1)-1];
guess = @(x)[0;1];
xmesh = linspace(0,L,100);
solinit = bvpinit(xmesh, guess);
sol = bvp4c(bvpfcn, bcfcn, solinit);
figure(1)
plot(sol.x, sol.y(1,:))

Npts = 50;
x = linspace(0,L,Npts);
dx = x(2)-x(1);
T0 = ones(Npts,1);
T = fsolve(@(x)fun(x,Npts,x,dx,epsilon,N),T0);

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

figure(2)
plot(x,T)

function res = fun(T,Npts,x,dx,epsilon,N)
  res = zeros(size(T));
  res(1) = (1+epsilon*T(1))*2*(T(2)-T(1))/dx^2 + epsilon*((T(2)-T(1))/dx)^2 - N^2*T(1);
  res(2:Npts-1) = (1+epsilon*T(2:Npts-1)).*(T(3:Npts)-2*T(2:Npts-1)+T(1:Npts-2))/dx^2 + epsilon*((T(3:Npts)-T(1:Npts-2))/(2*dx)).^2 - N^2*T(2:Npts-1);
  res(Npts) = T(Npts)-1;
end