How to get numerical values of nonlinear implicit function?

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Romio
Romio 2023-3-19
编辑： John D'Errico 2023-3-19
I plotted the following implicit function
k = -1.5
F = @(t,y) log(abs(y)) - y^2/2 - t - k;
fimplicit(F,[0,2],'*')
How do I get the numerical values for every t between 0 and 1, for the function value greater than or equal to y = 1 (the upper part half of the hyperbole)?

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采纳的回答

Torsten
Torsten 2023-3-19
编辑：Torsten 2023-3-19
Ran in:
k = -1.5;
F = @(t,y) log(abs(y)) - y.^2/2 - t - k;
fimplicit(F,[0,4],'*')
t = 0:0.01:0.99;
y0 = 2.0;
y = zeros(size(t));
for i = 1:numel(t)
y(i) = fzero(@(y)F(t(i),y),y0);
y0 = y(i);
end
plot(t,y)

更多回答（1 个）

John D'Errico
John D'Errico 2023-3-19
编辑：John D'Errico 2023-3-19
Ran in:
This is far easier then you may think. Um, trivially so. Just solve for t, as a function of y. Pencil and paper suffice for that. But if you prefer, we can use MATLAB to do the complicated work.
syms t y
k = -1.5;
tsol = solve(log(abs(y)) - y^2/2 - t - k,t)
tsol = 
Yeah, I know, that was complicated. WHEW! You can see it is symmetric as a function of y. negative values of y will yield the same result due to the abs and y^2.
fplot(tsol,[0,5])
xlabel y
ylabel t
grid on
y = (0:0.25:5)';
tfun = matlabFunction(tsol)
tfun = function_handle with value:
@(y)log(abs(y))-y.^2./2.0+3.0./2.0
t_y = tfun(y);
table(y,t_y)
ans = 21×2 table
y t_y ____ ________ 0 -Inf 0.25 0.082456 0.5 0.68185 0.75 0.93107 1 1 1.25 0.94189 1.5 0.78047 1.75 0.52837 2 0.19315 2.25 -0.22032 2.5 -0.70871 2.75 -1.2696 3 -1.9014 3.25 -2.6026 3.5 -3.3722 3.75 -4.2095
Not unexpectedly, undefined at y==0, but very simply solved.
To go the other way, you need to recognize there are two solutions for every possible value of t. So that relationship is not single valued. If you allow negative solutions for y, then there are four solutions for any value of t.

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