How to derive the 2D field and convergence based on xy gradients

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Jakob Sievers
Jakob Sievers 2015 年 4 月 1 日
回答済み: Jakob Sievers 2015 年 4 月 1 日
Hi all
This might be an easy question but I am a bit stuck on it. The function gradient yields xy gradients of a 2D field:
[px,py] = gradient(z);
But how do I move the other way? I.e.: z=f(px,py) where I have px and py and f is the unknown function/mechanism. My first thought revolves around using filter2 but I can't seem to figure it out.
Secondly: What would be an easy way to determine whether a region is subject to convergence/divergence?
Any thoughts?
Cheers
Jakob

採用された回答

Torsten
Torsten 2015 年 4 月 1 日
For your first question:
https://www.mathworks.com/matlabcentral/newsreader/view_thread/313599
For your second question:
I don't understand what you mean by: determine whether a region is subject to convergence/divergence ?
Best wishes
Torsten.

その他の回答 (2 件)

Jakob Sievers
Jakob Sievers 2015 年 4 月 1 日
編集済み: Jakob Sievers 2015 年 4 月 1 日
Hi Torsten
Thanks for the swift response!
Reg. the second question:
I would like to locate mathematically points of convergence and divergence in a vector field. For instance, given the below figure I would like to find determine automatically the region/point of divergence on the left and convergence on the right. I know it might be a silly question but it seems unclear to me, how to progress with this at the moment.

Jakob Sievers
Jakob Sievers 2015 年 4 月 1 日
I just realized that the function div=divergence(U,V) will do just that. Thanks for your response!

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