Level curves to two functions for solving nonlinear system of equations.

1 回表示 (過去 30 日間)
Snirisa  Gödel
Snirisa Gödel 2014 年 2 月 26 日
編集済み: John D'Errico 2014 年 2 月 26 日
I want to solve a system of nonlinear equations:
x^2 + y*sin(y) - 3 = 0 x*y + exp(x) + y -5 = 0
So I introduced the two functions f1 = x^2 + y*sin(y) - 3 and f2 = x*y + exp(x) + y -5. Now I am trying to plot the level curves(at 0) so I can see how many solutions exists. But I am unable to plot a meaningful result(i.e finding the intersections).
if true
% %%
x=linspace(-15,15,400);
y=linspace(-15,15,400);
f=@(x,y)(x.^2+y.*sin(y)-3);
[X,Y]=meshgrid(x,y);
Z=f(X,Y);
contour(X,Y,Z,10)
hold on
f=@(x,y)(x*y+exp(x)+y-5);
[X,Y]=meshgrid(x,y);
Z=f(X,Y);
contour(X,Y,Z,100)
end

回答 (1 件)

John D'Errico
John D'Errico 2014 年 2 月 26 日
編集済み: John D'Errico 2014 年 2 月 26 日
It looks like you are almost there.
Use contourc to get the contours as a set of polygons. Then use a tool from the file exchange to generate the intersections between the polygons. Doug's curve intersections tool is the one I'd choose.
In the event that there are several broken polygons produced by contourc, you will need to be careful to ensure you find all the intersections.

カテゴリ

Help Center および File ExchangeSurface and Mesh Plots についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by