find a zero of a two-variable function

Question

Douw Gerbrand 2024-4-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2109311-find-a-zero-of-a-two-variable-function

移动：Bruno Luong 2024-4-22

my main.m is:

clear
clc
close all
%
X0 = [9.609 , 32.288]; %initial value close to zero of function
%
f = Trajectory(X0);

The script Trajectory.m is a messy think that returns a value of f. I want to find values of tht two input variables (X0) is the initial vbalues that give f=0

8 个评论
显示 6更早的评论隐藏 6更早的评论

Sam Chak 2024-4-21

在 MATLAB Online 中打开

Is 'Trajectory' a script or a function? If it is a function, what is the output of the function?

function output = Trajectory()
    output  = ...;
end

Douw Gerbrand 2024-4-22

Hi Sam:

Oh wow! I have not heard the term Brachistochrone since my second year university course on applied mathematics. Yes, in some senses it is that, but with additional complications. My forces include gravity, but also two aerodunamic forces, parallel to motion and perpendicular to motion. This m,akes the script "Trajecrtory" quite complicated, and non-linear. Anyway, Bruno spotted my error and I now have a working code.

T^hanks to all!

Douw

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Bruno Luong 2024-4-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2109311-find-a-zero-of-a-two-variable-function#answer_1444866

编辑：Bruno Luong 2024-4-20

在 MATLAB Online 中打开

If you are not unlucky

Trajectory = @(x) sum(x.^2,2)-34^2-1; % Test function, use use rather own

X0 = [9.609 , 32.288]; %initial value close to zero of function

f1 = @(dx1) Trajectory(X0+[dx1 0]);

dx1 = fzero(f1,0);

X = X0+[dx1 0];

disp(X)

10.6998 32.2880

Trajectory(X)

ans = 0

fimplicit(@(x,y) Trajectory([x(:) y(:)]).')

hold on

plot(X(1),X(2),'or')

2 个评论
显示无隐藏无

Douw Gerbrand 2024-4-20

Hi Bruno:

My "Trajectory" is a fairly complicated thing. It involves numerical solution of a pair of non-linear ODE's. I believe it would therefore not be possible to specify it using an "@" as you suggest. Is there a way to use your approach but not using the "@"?

Bruno Luong 2024-4-20

编辑：Bruno Luong 2024-4-20

在 MATLAB Online 中打开

As I said I gave a simple Trajectory as example, you don't have to use it but plug in your complicated code.on this

X0 = [9.609 , 32.288]; %initial value close to zero of function
f1 = @(dx1) Trajectory(X0+[dx1 0]);
dx1 = fzero(f1,0);
X = X0+[dx1 0];
Trajectory(X)
disp(X)

请先登录，再进行评论。

Answer 2

John D'Errico 2024-4-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2109311-find-a-zero-of-a-two-variable-function#answer_1444896

在 MATLAB Online 中打开

A two variable function is a surface. (@Sam Chak said this, but did not take the idea to the point of completion.)

But if a surface has any solution, where it becomes zero, then almost always it will have infinitely many solutions. In fact, the tool which draws the zeros of such a function is usually called contour. That is, you want to effectively find a contour plot.

Yes, a contour plot usually is drawn with many contours at different levels, but a contour plot is what you want, all the same. You want to see only ONE contour drawn, at z == 0. Again, a contour plot can be thought of as the set of values (x,y), suze that z(x,y) is a level surface, so z(x,y) is equal to some given constant, here zero.

For an example, I'll use a symbolic function, as there are nice tools we can use. But a function handle will also work, or just an m-file. ANY function of two independent variables will work. Of course, you want it to be a continuous function, as if not, things will get very messy.

syms x y

z = x^3 - y^3 + x^2*y^2 - 3

z =

again, we cannot simply solve for a zero of that function, as there will be infintiely many such zeros, if any exist. We might do this:

H = fcontour(z);

H.LevelList = 0;

grid on

The curve drawn in cyan is the locus of points that solve the problem z(x,y)==0, so the level set of z, at z==0. As I said, not one solution. The problem is, a tool like fsolve will produce only one solution, it will not understand there are infinitely many such solutions, just finding only one point on one of those curves, and the solution it does find will be a function of the starting value you give it.

And of course, finding the "equation" of those curves here will be a very complicated looking thing, and that is for a rather simple problem to write. We can actually find at least a set of points that will approximate those curves, using the contourc function.

So what is it you want to see? Just a plot of the solutions? Then use tools like fcontour, or contour. I could also have used fimplicit. If your goal is a set of solutions, then we could use contourc. For example...

Xv = linspace(-5,5);
Yv = linspace(-5,5);
[X,Y] = meshgrid(Xv,Yv);
% Note my careful use of the dotted operators here to vectorize the code
% and make it properly evaluated for arrays X and Y.
zfun = @(x,y) x.^3 - y.^3 + x.^2.*y.^2 - 3;
Z = zfun(X,Y);
xy = contourc(Xv,Yv,Z,[0 0])
xy = 2x212
         0   -2.3787   -2.3737   -2.3603   -2.3420   -2.3238   -2.3058   -2.2881   -2.2727   -2.2706   -2.2528   -2.2354   -2.2183   -2.2017   -2.1856   -2.1717   -2.1700   -2.1547   -2.1402   -2.1267   -2.1143   -2.1032   -2.0939   -2.0865   -2.0817   -2.0800   -2.0825   -2.0906   -2.1062   -2.1328
   64.0000    5.0000    4.9712    4.8990    4.7980    4.6970    4.5960    4.4949    4.4054    4.3939    4.2929    4.1919    4.0909    3.9899    3.8889    3.7978    3.7879    3.6869    3.5859    3.4848    3.3838    3.2828    3.1818    3.0808    2.9798    2.8788    2.7778    2.6768    2.5758    2.4747
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

These are the solutions, split into two curves. Unfortunately contourc puts them into one array. Looking at the first column, we see [0;64]. That tells us the first contour had z==0, and there were 64 points found on that curve. The second branch of solutions had 146 points on it that were found.

So, if you want a set of solutions, then the simple approach is to use contourc. If you want only a plot, then just use one of the contour plotting tools.

6 个评论
显示 4更早的评论隐藏 4更早的评论

Douw Gerbrand 2024-4-21

Hi John: I tried this:

clear

clc

close all

%

X0 = [9.609 , 32.288]; %initial value close to zero of function

%

X = fminsearch(Trajectory,X

%

In the command window I see:

Undefined function or variable 'X0'.

Error in Trajectory (line 11)

velocity_init = X0(1);

Error in main (line 7)

X = fminsearch(Trajectory,X0)

what did do wrong?

Douw Gerbrand 2024-4-21

the line was:

X = fminsearch(Trajectory,X0)

请先登录，再进行评论。

Answer 3

Bruno Luong 2024-4-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2109311-find-a-zero-of-a-two-variable-function#answer_1445566

移动：Bruno Luong 2024-4-22

@Douw Gerbrand If your objective function f is pd quadratic form of 2 variable X, then I believe you should minimize f rather than solve for f = 0.

Also if you use MATLAB ode solver, be careful that it can be an issue of optimizing parameters, see an example here : https://www.mathworks.com/matlabcentral/answers/2105041-using-fmincon-to-solve-objective-function-in-integral-form?s_tid=srchtitle

If you have only 2 parameters to estimate you could use fminsearch.

2 个评论
显示无隐藏无

Bruno Luong 2024-4-21

移动：Bruno Luong 2024-4-22

在 MATLAB Online 中打开

Try

X0 = [9.609 , 32.288]; %initial value close to zero of function
X = fminsearch(@(X)Trajectory(X),X0);
% or alternatively a simpler
X = fminsearch(@Trajectory,X0);

Or if it throw an error, please explain us at least the declaration signature of Trajectory and what does it suppose to do precisely with the input/output arguments.

Douw Gerbrand 2024-4-22

移动：Bruno Luong 2024-4-22

Hi Bruno

You have spotted my error! I was not paying attention to how the variable vector X was being handled in my script "Trajectrory" I fixed that, andd not my code works.

Many thanks.

请先登录，再进行评论。