My gradient function wont update in my for loop

Hello!

Im trying to use Newtons method for two gradient function dU/dx= f1(x,y) and dU/dy=f2(x,y) of the function U(x,y) and find where they cross eachother for when they are zero, f1(x,y)=f2(x,y)=0, i have that code and it works!

However, i would also like to change a variabel f for one of the gradient functions and run Newtons method for each new value of f:

f2=@(x,y) E * A *( 1/L - (1./Lm1(x,y) )).*(y-y1) + E * A *( 1/L - 1./Lm2(x,y)).*(y-y2) - E * A * f ;

So, i will use values between 0 to 1 for the variable f and run the Newtons method for each f som im bascially changing the gradient function f2(x,y) for each new value of f but im doing something wrong because its like its running with f=0 for the whole loop. I havent been using matlab for a while and cant wrap my head around what im doing wrong. Here is my code and thanks alot in advance:

%Test1001

clear all
clc
clf

f=0.0;

h=3;E=1;A=1;
L=sqrt(1+h^2) ;  x1=-1; y1=h; x2=1; y2=h; x3=0; y3=0;

Lm1=@(x,y)sqrt((x-x1).^2+(y-y1).^2);
Lm2=@(x,y)sqrt((x-x2).^2+(y-y2).^2);


U=@(x, y)  (E*A/2*L) * ( ( Lm1(x,y) - L) ).^2 + (E*A/2*L) * ( ( Lm2(x,y) - L) ).^2 - E*A*f*(y-y3); % Potential energy function

f1=@(x,y) E*A*( 1/L - (1./Lm1(x,y) )).*(x-x1) + E*A*( 1/L - 1./Lm2(x,y)).*(x-x2)  ; % Governing equations = 0
f2=@(x,y) E*A*( 1/L - (1./Lm1(x,y) )).*(y-y1) + E*A*( 1/L - 1./Lm2(x,y)).*(y-y2) - E*A*f  ;

A11=@(x)  1/L - (x(2)-y1).^2 ./ (Lm1(x(1),x(2)).^3) ;
A12=@(x) (x(1)-x1).*(x(2)-y1) ./ (Lm1(x(1),x(2)).^3) ;
A21=@(x) (x(1)-x1).*(x(2)-y1) ./ (Lm1(x(1),x(2)).^3) ;
A22=@(x) 1/L - (x(1)-x1).^2 ./ (Lm1(x(1),x(2)).^3) ;

B11=@(x)  1/L - (x(2)-y2).^2 ./ (Lm2(x(1), x(2)).^3) ;
B12=@(x) (x(1)-x2).*(x(2)-y2) ./ (Lm2(x(1), x(2)).^3) ;
B21=@(x) (x(1)-x2).*(x(2)-y2) ./ (Lm2(x(1), x(2)).^3) ;
B22=@(x) 1/L - (x(1)-x2).^2 ./ (Lm2(x(1), x(2)).^3) ;

A=@(x)[A11(x) A12(x);A21(x) A22(x)];
B=@(x)[B11(x) B12(x);B21(x) B22(x)];

Df=@(x) A(x) + B(x); % The jacobian matrix

% Find roots with Newtons method

fv=@(x)[f1(x(1),x(2))            % Function vector
    f2(x(1),x(2))];

for f=0:0.05:1
    
    x=[0.1;0.10];
    kmax=10; tol=0.5e-8;
    for k=1:kmax
        d=-Df(x)\fv(x);
        x=x+d;
        disp([x' norm(d)])
        if norm(d)<tol, break, end
    end
    
end

Answer 1

[Hassaan]

clear all
clc
clf
% Constants
h=3; E=1; Area=1;
L=sqrt(1+h^2); x1=-1; y1=h; x2=1; y2=h; x3=0; y3=0;

% Define the length functions
Lm1=@(x,y) sqrt((x-x1).^2 + (y-y1).^2);
Lm2=@(x,y) sqrt((x-x2).^2 + (y-y2).^2);

% Potential energy function
U=@(x, y) (E*Area/2*L) * ((Lm1(x,y) - L)).^2 + (E*Area/2*L) * ((Lm2(x,y) - L)).^2 - E*Area*f*(y-y3);
% Governing equations = 0 (f1 does not depend on f so we can define it outside the loop)
f1=@(x,y) E*Area*(1/L - (1./Lm1(x,y))).*(x-x1) + E*Area*(1/L - 1./Lm2(x,y)).*(x-x2);
% Define the Jacobian matrix and Function vector components outside the loop
A11=@(x) 1/L - (x(2)-y1).^2 ./ (Lm1(x(1),x(2)).^3);
A12=@(x) (x(1)-x1).*(x(2)-y1) ./ (Lm1(x(1),x(2)).^3);
A21=@(x) (x(1)-x1).*(x(2)-y1) ./ (Lm1(x(1),x(2)).^3);
A22=@(x) 1/L - (x(1)-x1).^2 ./ (Lm1(x(1),x(2)).^3);
B11=@(x) 1/L - (x(2)-y2).^2 ./ (Lm2(x(1), x(2)).^3);
B12=@(x) (x(1)-x2).*(x(2)-y2) ./ (Lm2(x(1), x(2)).^3);
B21=@(x) (x(1)-x2).*(x(2)-y2) ./ (Lm2(x(1), x(2)).^3);
B22=@(x) 1/L - (x(1)-x2).^2 ./ (Lm2(x(1), x(2)).^3);

% Assemble the Jacobian matrix and Function vector
A=@(x) [A11(x) A12(x); A21(x) A22(x)];
B=@(x) [B11(x) B12(x); B21(x) B22(x)];
Df=@(x) A(x) + B(x);

% Main loop over f values
for f_value=0:0.05:1
    % Redefine f2 with the current value of f
    f2=@(x,y) E*Area*(1/L - (1./Lm1(x,y))).*(y-y1) + E*Area*(1/L - 1./Lm2(x,y)).*(y-y2) - E*Area*f_value;
    % Initial guess
    x=[0.1;0.10];

    % Newton's method parameters
    kmax=10; tol=0.5e-8;
    for k=1:kmax
        % Function vector
        fv=@(x) [f1(x(1),x(2)); f2(x(1),x(2))];
        
        % Update step
        d=-Df(x)\fv(x);
        x=x+d;
        
        % Display current iteration results
        disp([x' norm(d)])
        if norm(d)<tol, break, end
    end
end

   -0.0274   -0.0025    0.1635

   -0.0001   -0.0001    0.0273

   1.0e-03 *

   -0.0000   -0.0000    0.1687

   1.0e-07 *

    0.0000   -0.0000    0.2646

   1.0e-15 *

    0.6924   -0.1745         0

   -0.0030    0.0869    0.1038

   -0.0000    0.0882    0.0032

   -0.0000    0.0882    0.0000

   -0.0000    0.0882    0.0000

    0.0214    0.1764    0.1096

    0.0001    0.1773    0.0213

    0.0000    0.1774    0.0001

    0.0000    0.1774    0.0000

   -0.0000    0.1774    0.0000

    0.0458    0.2658    0.1744

    0.0005    0.2672    0.0453

    0.0000    0.2675    0.0005

    0.0000    0.2675    0.0000

   -0.0000    0.2675    0.0000

    0.0701    0.3552    0.2569

    0.0030    0.3582    0.0672

    0.0000    0.3588    0.0030

    0.0000    0.3588    0.0000

   -0.0000    0.3588    0.0000

    0.0945    0.4446    0.3446

    0.2470    0.4568    0.1529

    0.0588    0.4470    0.1884

    0.8145    0.4644    0.7559

   89.8051   21.5534   91.4553

    3.0940    4.0488   88.4603

    7.7704   -7.2869   12.2625

    1.9337    0.9120   10.0642

    0.8409    0.2629    1.2710

    0.9068    0.5626    0.3068

    0.1189    0.5340    0.4344

   -0.0170    0.5432    0.1362

    0.0005    0.5458    0.0177

   -0.0000    0.5458    0.0005

    0.0000    0.5458    0.0000

    0.0000    0.5458    0.0000

    0.1433    0.6234    0.5252

   -0.0156    0.6384    0.1595

    0.0003    0.6421    0.0163

   -0.0000    0.6421    0.0003

    0.0000    0.6421    0.0000

    0.0000    0.6421    0.0000

    0.1676    0.7128    0.6165

   -0.0175    0.7355    0.1865

    0.0003    0.7407    0.0185

   -0.0000    0.7408    0.0003

    0.0000    0.7408    0.0000

    0.0000    0.7408    0.0000

    0.1920    0.8022    0.7082

   -0.0206    0.8351    0.2152

    0.0004    0.8423    0.0222

   -0.0000    0.8424    0.0004

    0.0000    0.8424    0.0000

   -0.0000    0.8424    0.0000

    0.2164    0.8916    0.8001

   -0.0246    0.9380    0.2454

    0.0005    0.9477    0.0269

   -0.0000    0.9478    0.0005

    0.0000    0.9478    0.0000

    0.0000    0.9478    0.0000

    0.2408    0.9810    0.8922

   -0.0291    1.0449    0.2773

    0.0006    1.0581    0.0325

   -0.0000    1.0583    0.0006

    0.0000    1.0583    0.0000

    0.0000    1.0583    0.0000

    0.2651    1.0704    0.9844

   -0.0339    1.1573    0.3114

    0.0007    1.1754    0.0390

   -0.0000    1.1757    0.0008

    0.0000    1.1757    0.0000

    0.0000    1.1757    0.0000

    0.2895    1.1598    1.0766

   -0.0383    1.2771    0.3482

    0.0009    1.3030    0.0469

   -0.0000    1.3035    0.0010

    0.0000    1.3035    0.0000

   -0.0000    1.3035    0.0000

    0.3139    1.2492    1.1690

   -0.0416    1.4070    0.3890

    0.0009    1.4477    0.0589

   -0.0000    1.4491    0.0016

    0.0000    1.4491    0.0000

   -0.0000    1.4491    0.0000

    0.3383    1.3386    1.2614

   -0.0422    1.5512    0.4358

    0.0003    1.6274    0.0873

    0.0000    1.6346    0.0072

    0.0000    1.6347    0.0001

   -0.0000    1.6347    0.0000

   -0.0000    1.6347         0

    0.3626    1.4281    1.3538

   -0.0373    1.7161    0.4928

   -0.0046    1.9185    0.2050

   -0.0276    4.2510    2.3326

    0.2360   15.0140   10.7662

    0.0516    7.4022    7.6141

    0.0011    7.3469    0.0748

    0.0000    7.3467    0.0011

    0.0000    7.3467    0.0000

   -0.0000    7.3467    0.0000

    0.3870    1.5175    1.4462

   -0.0217    1.9116    0.5678

   -0.2594    6.7462    4.8404

    0.1197    7.4493    0.7987

    0.0010    7.4295    0.1203

    0.0000    7.4286    0.0014

    0.0000    7.4286    0.0000

   -0.0000    7.4286    0.0000

    0.4114    1.6069    1.5387

    0.0138    2.1539    0.6762

   -0.9579    1.5676    1.1349

   -0.0167    1.7060    0.9513

   -0.0070    2.4814    0.7754

   -0.0108    2.0807    0.4007

    0.0919    1.3049    0.7826

   -0.0508    1.9748    0.6850

    0.5123   -0.6990    2.7324

    0.7919    1.5596    2.2759

    0.4358    1.6963    1.6312

    0.0865    2.4691    0.8481

    0.1627    2.0143    0.4612

   -2.9856   -2.7197    5.6853

   -1.8751    1.9840    4.8330

   -0.7489    0.7151    1.6966

    3.0101    1.0238    3.7717

    7.8054    9.5966    9.8228

    2.0385    6.9595    6.3412

   -0.0535    7.9443    2.3122

    0.4601    1.7857    1.7237

    0.2290    2.9002    1.1382

    0.0961    2.2975    0.6172

    0.5193    1.7289    0.7088

    0.1113    2.6525    1.0098

    0.1370    2.1290    0.5241

   -5.8026   -0.9538    6.6920

   -3.1425    3.5879    5.2634

   -7.6101  -14.6201   18.7481

   -1.3534    1.7319   17.5081

In this revised code, f2 is redefined for each new value of f_value in the loop, allowing the Newton's method to run for each value of f from 0 to 1 in increments of 0.05. This should fix the issue of it running as if f=0 for the entire loop.

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[Torsten]

Either rename the variable A

A=1;

or the function handle A

A=@(x) [A11(x) A12(x); A21(x) A22(x)];

There names are in conflict in the definition of f2.

[Hassaan]

@Torsten Thanks for pointing out. Resolved the confilit in the above code. Thank you.

[Daniel]

Thanks alot @Muhammad Hassaan Shah ! =)

Thanks for your input, @Torsten =)

Answer 2

[Steven Lord]

Changing the value of a variable included in an anonymous function when it was created does not change the value the anonymous function "remembers".

n = 2;
f = @(x) n*x;
f(1:5)

ans = 1×5

2 4 6 8 10

n = 3;
f(1:5)

ans = 1×5

2 4 6 8 10

You can see this "remembered" value using the functions function. To anticipate your next question no, you cannot modify the remembered value. See the "Variables in the Expression" section on this documentation page for a description of this "remembering" behavior.

info = functions(f);
info.workspace{1}

ans = struct with fields:
    n: 2

Either recreate the function handle in the loop so it "remembers" the new value of your variable or define the function handle so it accepts the value of that variable as an additional input argument and call it with the loop variable as that additional input argument.

g = @(x, n) x*n;
g(1:5, 2)

ans = 1×5

2 4 6 8 10

g(1:5, 3)

ans = 1×5

3 6 9 12 15

[Daniel]

Thanks for your pointers and help! =) @Steven Lord