invert_omega: Need to pass N2 and f0 as dataArrays about xinvert HOT 9 CLOSED

Hi, of course yes. f0 is specified on beta plane. So if the forcing is defined on a lat/lon grid, then you don't need to provide f0, which is overlooked in lat/lon case. N2(k) should be defined as the same as forcing function in the vertical.

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The true f will be calculated automatically according to your latitude definitions wrapped in the forcing function.

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"N2(k) should be defined as the same as forcing function in the vertical." - Sounds good.

Since the equation is solving for the following,

and not,

which is the desired omega equation, so I am planning to set $N^2$ as 1 and replace $f0$ with $f0/\sqrt(N^2)$ as a workaround even though it is not a perfect solution as $N^2$ gets differentiated in the vertical.

1. Is there a reason for this change in the equation solved by invert_omega?
2. Is it possible to solve the latter equation instead of the former?

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That depends on the case you choose. $f$ does not vary horizontally, so it can be inside or outside the vertical derivative. If you choose $N(z)$ to be a constant horizontally, the two forms are actually the same. Traditionally, there is no previous practice to allow $N$ to vary horizontally. Do you need to do this (which allows $N(x,y,z)$ ) right now?

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Yes, that makes sense. I had missed that you are using an array for the stability parameter in your example, so that solves my issue. Thank you.

Even though that wasn't my initial question, I would want to let $N^2$ vary with y.

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Yes, in your case of $N(y, z)$, one cannot derive the traditional form as your second equation above. The exact form when $N$ is a 3D variable is not as simple as one may expect. I guess the first one is not correct neither, except you can derive it from the very start (you may use $N(z)$ as a condition in the derivation). This is what I can recall right now.

But if you want a try with solving the first equation above, then $N$ can be varying both horizontally and vertically, which fits the solver of xinvert.

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Yes, I suspected that. Will see what resultant equation I get, deriving with $N^2(y,z)$.

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I have a notebook here and the derived Eq. (14) is:

So the $N^2$ (here it is $S$ ) is inside the Laplacian, which is slightly different from your first equation. To invert this, one may choose the general form rather than standard form. But I am also curious about how large the differences are when using your first equation above and the one shown here.

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Close this now and feel free to re-open it in any case.

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