Solving of vectorial equations

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Mepe
Mepe 2020 年 2 月 13 日
回答済み: John D'Errico 2020 年 2 月 17 日
Hi there,
I have the following problem:
Given are two equations:
y = 2*x.^2+2*x+5
x = 3*y.^3-C
C is a vector constant.
Now the solutions for x should be searched component by component.
I would like to avoid merging the equations beforehand.
How then can I best compute the solutions vectorially?
Thanks a lot
  2 件のコメント
Walter Roberson
Walter Roberson 2020 年 2 月 13 日
"Given are two equations: "
"C is a vector constant."
If you have a vector constant, you have one equation for each value in the vector. However, you only have two free variables, and that implies that there cannot be any solution unless C happens to be scalar instead of a vector.
Mepe
Mepe 2020 年 2 月 17 日
I see that ;-). I think I have not described my problem precisely enough.
I'll try again. In principle, the equations can be combined into one with an unknown one. Now I'm looking for solutions to x for the different values in c.

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回答 (1 件)

John D'Errico
John D'Errico 2020 年 2 月 17 日
You cannot do it, at least if these are completely general nonlinear equations, since there may be no unique solution for nonlinear equations, and even infinitely many such solutions. In your example, of course, it is simple enough, since you can visualize the problem by substituting y into the second equation. Thus we would implicitly have
x = 3*(2*x.^2+2*x+5) - C
so a 6'th degree polynomial in x, parameterized by C. For any given value of C, there will be 6 solutions, probably not all of which will be real. Of course, for any given C, there will be a differing number of real solutions, either 0, 2, 4, or 6 of them in this case.
Of course, if the relationships between x, y, and C are completely arbitrrily nonlinear (I assume this is a trivial example you gave), then there may indeed be infinitely many solutions, or at least some arbitrary number of them.
Of course, the solution is simple, at least in theory. For any given C, just solve for the set of ALL solutions of a pair of nonlinear equations. But I can easily provide a problem that has no easy solution for that, and no simple way to do it. So if you will insist on a magical answer to your problem, it ain't gonna happen.

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