Matlab Finite Boundary Conditons

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Julio Beatriz
Julio Beatriz 2018 年 3 月 8 日
回答済み: Julio Beatriz 2018 年 3 月 8 日
Hello, I wanted to ask if it was possible to have a finite boundary condition in matlab, for example u(0)=finite. If so, how would I go about writing it? Thank you
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John D'Errico
John D'Errico 2018 年 3 月 8 日
編集済み: John D'Errico 2018 年 3 月 8 日
As an ODE, a finite end condition is not a sufficient condition to allow a unique solution. There would be infinitely many solutions then, based on any number of infinitely many boundary values.
However, your use of the word boundary condition is ambiguous. In the context of u(0), that implies a one-dimensional problem, perhaps a boundary value problem, yet still an ODE.
If you have a PDE, "finite" still lacks definition, is still indecisive.
If you have some other form of problem that is not a PDE or an ODE, then you need to explain the mathematical context.
Finally, "finite" when used in a computational context is itself ambiguous and problematic, since is the number 1e18 merely large, but still finite? 1e100? 1e300? 1e1000? 1e100000000000? All of those numbers are mathematically finite, yet some of them will result in an overflow when double precision arithmetic is involved. Others are sufficiently large that even if no inf results, they may still result in mathematical garbage.
So I would suggest you need to describe the problem you want to solve, since "finite" is insufficient to achieve a solution. All it tells about your problem is what you don't want to happen.
Torsten
Torsten 2018 年 3 月 8 日
Consider for example the Laplace equation in cylindrical coordinates:
1/r d/dr (r*du/dr) = 0
The general solution is
u(r) = c1+c2*log(r)
To make the log-term cancel out, a condition that u(0) is finite is sufficient to get u(r)=constant.
Best wishes
Torsten.

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Julio Beatriz
Julio Beatriz 2018 年 3 月 8 日
Thank you very much for your help everyone, I really appreciate it. This has proven very helpful.

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