Gradient vector field not perpendicular to contour map
9 次查看(过去 30 天)
显示 更早的评论
I have the following code:
function [ ] = kermack( I0,r,p,c,N,numSteps )
I(1)=I0;
S(1)=N-I0;
for t = 1:numSteps % construct I and S vectors
I(t+1) = I(t) + (p*c*I(t)*S(t) - r*I(t));
S(t+1) = S(t) - p*c*I(t)*S(t);
end
% create contour
x = min(I):0.2:max(I);
y = min(S):0.2:max(S);
[xx,yy] = meshgrid(x,y);
zz = xx + yy - (r/(p*c))*log(yy); % f(I,S)
[DX,DY] = gradient(zz);
figure % new window
contour(xx,yy,zz)
hold on % superimpose graphs
scale=10; % to make the arrows visible
quiver(xx,yy,DX,DY,scale) % vector field of f(I,S)
xlim([0 60]); % set bounds on the window
ylim([0,80]);
xlabel('I(t) (infected/infectious population)');
ylabel('S(t) (susceptible population)');
hold off
Now, as made clear by the title of my question, the generated gradient vector field doesn't seem to be perpendicular with the superimposed contour mapping. How can this be? What am I doing wrong?
5 个评论
采纳的回答
David Goodmanson
2017-10-16
Hi Cole,
I believe this is an artifact of the zoom process. If you zoom in by drawing a rectangle, the axes get scaled differently and right angles no longer appear to be right angles. If you add
axis equal
at the end of the code and then zoom in simply by repeated clicking on a point on a contour line (which does 2x on each axis), the results appear to be good.
更多回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Read, Write, and Modify Image 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!