I have two implicit functions as f1(x,y1)=0 and f2(x,y2)=0. Is there any possible way to plot y1/y2 vs x?

Suppose you have two any form of implicit functions f1(x,y1)=0 and f2(x,y2)=0. You can't write y1, y2 in terms of x explicitely. Then is it possible to plot any function of y1, y2 versus x, in particular y1/y2 vs x?

Answer 1

[Torsten]

f1 = @(x,y1) ...;    % insert f1
f2 = @(x,y2) ...;    % insert f2
f = @(y) [f1(x,y(1)),f2(x,y(2))]
X = 0:0.1:1;         % you might change this x-interval
y0 = [1,1];          % you might change this initial guess for y1 and y2
for i = 1:numel(X)
  x = X(i);
  y = fsolve(f,y0);
  z(i) = y(1)/y(2);
  y0 = y;
end
plot(x,z)

[Dip]

It's showing some error.

To check the code, I took very easy function y1=x^3 and y2=x^2.

Expected to plot z=x. But some error is coming.

clc;
f1 = @(x,y1) y1-x^3;    % insert f1
f2 = @(x,y2) y2-x^2;    % insert f2
f = @(y) [f1(x,y(1)),f2(x,y(2))];
X = 0:0.1:1;         % you might change this x-interval
y0 = [1,1];          % you might change this initial guess for y1 and y2
for i = 1:numel(X)
    x = X(i);
    y = fsolve(f,y0);
    z(i) = y(1)/y(2);
    y0 = y;
end

Unrecognized function or variable 'x'.

Error in solution (line 4)
f = @(y) [f1(x,y(1)),f2(x,y(2))];

Error in fsolve (line 264)
fuser = feval(funfcn{3},x,varargin{:});

Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue.

plot(x,z);

Can you please validate this code by taking any easy example?

[Torsten]

f1 = @(x,y1) y1-x^3;    % insert f1
f2 = @(x,y2) y2-x^2;    % insert f2
f = @(x,y) [f1(x,y(1)),f2(x,y(2))];
X = 0:0.1:1;         % you might change this x-interval
y0 = [1,1];          % you might change this initial guess for y1 and y2
for i = 1:numel(X)
    x = X(i);
    y = fsolve(@(y)f(x,y),y0);
    z(i) = y(1)/y(2);
    y0 = y;
end

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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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.

plot(X,z)