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Constraining equilibrium solver to positive values (fsolve, lsqnonlin)

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LS
LS 2011 年 5 月 16 日
From my understanding, it's impossible to constrain fsolve to only find positive zeros but that it is possible using lsqnonlin (see this thread: http://www.mathworks.com/matlabcentral/newsreader/view_thread/280353). I see how I can use bound constraints with lsqnonlin to only return positive values, but how do I "minimize the residuals to 0"? Thank you!

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Andrew Newell
Andrew Newell 2011 年 5 月 17 日
Suppose you started with a function myfun(x) that inputs a vector x and outputs a vector, for example:
myfun = @(x) x.^2-1;
x0 = [-1.5 1.5];
x1 = fsolve(myfun,x0)
x1 =
-1.0000 1.0000
To incorporate your constraints, you could do this:
lb = zeros(size(x0)); %lower bound of zero
ub = Inf*ones(size(x0));
x2 = lsqnonlin(myfun,x0,lb,ub)
x2 =
1.0000 1.0000
As John D'Errico implies, the minimization may give you an answer that is not a root, so you then need to check your answer:
myfun(x2)
ans =
1.0e-08 *
0.3056 0.0030
(Note: answer rewritten completely.)
  1 件のコメント
LS
LS 2011 年 5 月 19 日
Thanks very much for your help!!

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その他の回答 (1 件)

John D'Errico
John D'Errico 2011 年 5 月 17 日
There is NO assurance that you can minimize the residuals to zero. What is the root of y=x^2+1, with x restricted to real values?
Just wanting to solve a problem does not mean there will be a solution.

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