explict (pointwise) function from numerical solution to implicit equation

I can plot the numerical solutions to a complicated expression f(x,y)=0 over intervals for x and y using fimplicit(f,interval). However, how do I then define/specify an explicit relation y=g(x) pointwise using the numerical values computed from f(x,y)=0 for use in a separate expression? (There is a functional relation.)

 採用された回答

Ameer Hamza
Ameer Hamza 2020 年 6 月 6 日
編集済み: Ameer Hamza 2020 年 6 月 6 日
Try interp1 to create an explicit relation for 'y' in term of 'x' using the datapoints from fimplicit
f = @(x, y) y - x.^2;
fp = fimplicit(f, [-2 2 0 10]);
x = fp.XData;
y = fp.YData;
g = @(xq) interp1(x, y, xq);
Then run
>> g(1)
ans =
1.0002
>> g(0.5)
ans =
0.2501
Alternate solution:
Use fsolve() directly. The advantage is that you can control the spacing of x-points
f = @(x, y) y - x.^2;
x = -2:0.01:2;
y = fsolve(@(y) f(x, y), rand(size(x)));
g = @(xq) interp1(x, y, xq);

5 件のコメント

Laurence Kranich
Laurence Kranich 2020 年 6 月 9 日
The first method worked! (I had a problem applying the alternate solution in my case.) Thank you very much!
Ameer Hamza
Ameer Hamza 2020 年 6 月 11 日
Laurence's comment posted as answer moved here:
Ameer,
As I said, your first suggestion worked. (Thank you again!) However, I am now trying to generalize this method verbatim to the case of three variables and am having trouble. My attempt is included below. Even if the 3-D commands are not well-defined, it is clear what I am attempting to do. That is, I am trying to solve f(x,y,z)=0 pointwise for an explicit function z=g(x,y), for which I can then store the data and use in the expression for dtdp, below. Do you have any suggestions?
Laurence
f = @(x,y,z)(2.*sqrt(z).*(1+sqrt(z)).*((y+sqrt(z).^2).*((8.*(1+x))./((1-x).*(3+x).^2))+((8./9).*(1-y).^2)+(2.*(y+sqrt(z)).*(1-y).*(8./(3-x).^2))))-((1-z).*(1-y).*((y+sqrt(z)).*(8.*(1+x)./((1-x).*(3+x).^2))-((1-y).*(8./9))+((1-y).*(8./((3-x).^2)))-((y+sqrt(z)).*(8./((3-x).^2)))));
interval = [0.0001 .9999 0.0001 .9999 0.0001 .9999];
fp = fimplicit3(f, interval);
x = fp.XData ;
y = fp.YData ;
z = fp.ZData ;
g = @(xq, yq) interp2(x, y, z, Xq, Yq );
dtdp = (((1-z).*(1-y))-((1-z).*(y+sqrt(z)))-(4.*sqrt(z)).*(1+sqrt(z)).*(y+sqrt(z)).*(((8.*(1+x))./((1-x).*((3+x).^2)))-(8./((3-x).^2)))-(2.*(1-z).*(1-y))-((1-z)+(4.*sqrt(z)).*(1+sqrt(z)).*(1-y)).*((8./((3-x).^2))-(8./9)))./((((1+sqrt(z))./(sqrt(z))).*((y+sqrt(z)).^2).*((8.*(1+x))./((1-x).*((3+x).^2)))+((y+sqrt(z)).^2).*((8.*(1+x))./((1-x).*((3+x).^2)))+((2.*(1+sqrt(z)).*((y+sqrt(z))).*((8.*(1+x))./((1-x).*((3+x).^2)))+((1-y)).*(y+sqrt(z)).*((8.*(1+x))./((1-x).*((3+x).^2)))-(((1-z)./(2.*(sqrt(z))))).*(1-y).*((8.*(1+x))./((1-x).*((3+x).^2)))+(((1+sqrt(z))./(sqrt(z))).*(1-y).^2).*(8./9)+(((1+sqrt(z))./(sqrt(z)))).*2.*(y+sqrt(z)).*(1-y).*(8./((3-x).^2))+((y+sqrt(z)).*(1-y))*(8./((3-x).^2))+(2.*(1+sqrt(z)).*(1-y))*(8./((3-x).^2))+((y+sqrt(z)).*(1-y))*(8./((3-x).^2))+((1-y).^2).*(8./((3-x).^2))+((2.*(1+sqrt(z)).*((y+sqrt(z))).*(8./((3-x).^2))))))));
plot3 (x,y,dtdp);
xlabel('alpha')
ylabel('po')
zlabel('dtdp')
Ameer Hamza
Ameer Hamza 2020 年 6 月 11 日
You can do something like this in 3D case
x = fp.Triangulation.Points(:,1);
y = fp.Triangulation.Points(:,2);
z = fp.Triangulation.Points(:,3);
g = @(xq, yq) griddata(x, y, z, xq, yq );
There seems to be some issue in the equation of dtdp. See if it is written correctly.
Laurence Kranich
Laurence Kranich 2020 年 6 月 11 日
That worked. Thanks again!!
Ameer Hamza
Ameer Hamza 2020 年 6 月 11 日
I am glad to be of help!

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