Vertical diffusion in EPIC GCM

[rc_nmt]

I am running a general circulation model (EPIC, which uses isentropic vertical coordinate system, potential temperature) and need to add vertical diffusion to the zonal velocity U. The vertical coordinate is theta (Θ, potential temperature) and there is no vertical height coordinate in z but I need to find ∂²U / ∂z².

I solve for ∂U / ∂z by using the relation ∂Θ / ∂z = N*Θ / g. So then I can use the chain rule and use finite difference approximations to solve for ∂U / ∂z = ∂U / ∂Θ * ∂Θ / ∂z which is also equal to ∂U / ∂z = ∂U / ∂Θ * N*Θ / g. This equation works correctly for the first derivative of U with respect to z.

The problem I am facing now is another derivative of this expression which gives me ∂ / ∂z [
∂U / ∂Θ * N*Θ / g ] . So what I would like to know is, is there some symmetry property of second derivatives I can use here? How can I solve this without creating a vertical coordinate in z and just use Θ as I did for the first derivative in ∂U / ∂z? Is it possible? How if I apply the product rule does the first term work out to have no z dependence and be valid?

Thank you to who ever can assist me.

Cheers

Rick

 

[Mark44]

I am running a general circulation model (EPIC, which uses isentropic vertical coordinate system, potential temperature) and need to add vertical diffusion to the zonal velocity U. The vertical coordinate is theta (Θ, potential temperature) and there is no vertical height coordinate in z but I need to find ∂²U / ∂z².

I solve for ∂U / ∂z by using the relation ∂Θ / ∂z = N*Θ / g. So then I can use the chain rule and use finite difference approximations to solve for ∂U / ∂z = ∂U / ∂Θ * ∂Θ / ∂z which is also equal to ∂U / ∂z = ∂U / ∂Θ * N*Θ / g. This equation works correctly for the first derivative of U with respect to z.

I'm confused. To calculate $\frac{\partial U}{\partial z}$ (you're not really solving for it), you need an equation with U as a function of z, and whatever other variables are in the equation. I don't see anywhere in this post an equation that gives U in terms of the other variables, so I don't see how you're going to find the partial you want.

The problem I am facing now is another derivative of this expression which gives me ∂ / ∂z [
∂U / ∂Θ * N*Θ / g ] . So what I would like to know is, is there some symmetry property of second derivatives I can use here? How can I solve this without creating a vertical coordinate in z and just use Θ as I did for the first derivative in ∂U / ∂z? Is it possible? How if I apply the product rule does the first term work out to have no z dependence and be valid?

Thank you to who ever can assist me.

Cheers

Rick

Also, please don't post the same message twice... I have deleted the older copy.

 

[rc_nmt]

I'm confused. To calculate $\frac{\partial U}{\partial z}$ (you're not really solving for it), you need an equation with U as a function of z, and whatever other variables are in the equation. I don't see anywhere in this post an equation that gives U in terms of the other variables, so I don't see how you're going to find the partial you want.

Also, please don't post the same message twice... I have deleted the older copy.

Again apologies for the duplicate post. I was trying to edit my original and thought I removed it but I guess not. Thanks for your help and reply.

Ok let me clarify. I need a first order derivative of U wrt to z and performed this transformation which I believe is correct given the relation in isentropic coordinates of $\frac{\partial \theta}{\partial z} = \frac{N \theta}{g}$

$\frac{\partial U}{\partial z} = \frac{\partial U}{\partial \theta} \frac{\partial \theta}{\partial z} = \frac{\partial U}{\partial \theta} \frac{N \theta}{g}$

So now I am looking for a way to find $\frac{\partial ^2 U}{\partial z^2}$.

The problem I am having is solving for this.

$\frac{\partial}{\partial z}[\frac{\partial U}{\partial \theta} \frac{N \theta}{g}]$

Any thoughts or ideas?

 

[Mark44]

Again apologies for the duplicate post. I was trying to edit my original and thought I removed it but I guess not. Thanks for your help and reply.

Ok let me clarify. I need a first order derivative of U wrt to z and performed this transformation which I believe is correct given the relation in isentropic coordinates of $\frac{\partial \theta}{\partial z} = \frac{N \theta}{g}$

$\frac{\partial U}{\partial z} = \frac{\partial U}{\partial \theta} \frac{\partial \theta}{\partial z} = \frac{\partial U}{\partial \theta} \frac{N \theta}{g}$

So now I am looking for a way to find $\frac{\partial ^2 U}{\partial z^2}$.

The problem I am having is solving for this.

$\frac{\partial}{\partial z}[\frac{\partial U}{\partial \theta} \frac{N \theta}{g}]$

Any thoughts or ideas?

You didn't answer the question I asked before: How are U and z related?

 

[rc_nmt]

U and z are not directly related in this model. First there is no z coordinate, $\theta$ is the vertical coordinate that is used. U is the dynamical zonal wind variable. The way to relate U to z is to use the relation above that relates $\theta$ and z

 

[Mark44]

Ok let me clarify. I need a first order derivative of U wrt to z and performed this transformation which I believe is correct given the relation in isentropic coordinates of $\frac{\partial \theta}{\partial z} = \frac{N \theta}{g}$

Is $\theta$ a function only of z or are there other independent variables? If $\theta$ is a function of z alone, then the equation above can be written as $\frac{d \theta}{dz} = \frac N g \theta$. I'm assuming that N and g are constants.

This differential equation is easy to solve.

If $\theta$ is a function of variables other than z, then the chain rule is more complicated than what you show below.

$\frac{\partial U}{\partial z} = \frac{\partial U}{\partial \theta} \frac{\partial \theta}{\partial z} = \frac{\partial U}{\partial \theta} \frac{N \theta}{g}$

So now I am looking for a way to find $\frac{\partial ^2 U}{\partial z^2}$.

 

[rc_nmt]

I am meeting to discuss with my advisor at length this afternoon. N is the Brunt Vaiasala frequency which for the part of the atmosphere I am simulating I think I could say it is a constant or at least approximate it as one. g is acceleration of gravity, so yes the differential equation is nearly trivial to solve for $\theta$ or z. The matter of $\frac{d^2U}{dz^2}$ is still the issue.

I have been trying to expand it out in numerous ways and use the symmetry of a mixed or double derivative which I am not sure if that is allowed in this case or how it would work. So this is what I am trying to expand with the product rule:

$\frac{d^2U}{dz^2}$ = $\frac{d}{dz}[\frac{dU}{dz}]$ = $\frac{d}{dz}[\frac{dU}{d\theta}\frac{d\theta}{dz}]$ = $\frac{d}{dz}[\frac{dU}{d\theta}\frac{N \theta}{g}]$ = $[\frac{d}{dz}\frac{dU}{d\theta}]\frac{N \theta}{g} + \frac{dU}{d\theta}[\frac{d}{dz}\frac{N \theta}{g}]$

does this last statement then equate to this?

$[\frac{d}{d\theta}\frac{dU}{dz}]\frac{N \theta}{g} + \frac{dU}{d\theta}[\frac{d}{dz}\frac{N \theta}{g}]$

if the last is true, I can approach the second term as you mentioned with solving for $\theta$ or approximating it as constant and throw it away completely and then this simplifies to something I have tested which seems to vertically diffusing

$[\frac{d}{d\theta}\frac{dU}{d\theta}\frac{N \theta}{g}]\frac{N \theta}{g} + 0$ = $\frac{d^2U}{d\theta^2}[\frac{N \theta}{g}]^2$

 

[Mark44]

I am meeting to discuss with my advisor at length this afternoon. N is the Brunt Vaiasala frequency which for the part of the atmosphere I am simulating I think I could say it is a constant or at least approximate it as one. g is acceleration of gravity, so yes the differential equation is nearly trivial to solve for $\theta$ or z. The matter of $\frac{d^2U}{dz^2}$ is still the issue.

I have been trying to expand it out in numerous ways and use the symmetry of a mixed or double derivative which I am not sure if that is allowed in this case or how it would work. So this is what I am trying to expand with the product rule:

$\frac{d^2U}{dz^2}$ = $\frac{d}{dz}[\frac{dU}{dz}]$ = $\frac{d}{dz}[\frac{dU}{d\theta}\frac{d\theta}{dz}]$ = $\frac{d}{dz}[\frac{dU}{d\theta}\frac{N \theta}{g}]$ = $[\frac{d}{dz}\frac{dU}{d\theta}]\frac{N \theta}{g} + \frac{dU}{d\theta}[\frac{d}{dz}\frac{N \theta}{g}]$

does this last statement then equate to this?

Yes, that looks OK to me.
I noticed that you switched from partial derivatives to ordinary derivatives. Does this mean that all of the functions are functions of a single variable?

$[\frac{d}{d\theta}\frac{dU}{dz}]\frac{N \theta}{g} + \frac{dU}{d\theta}[\frac{d}{dz}\frac{N \theta}{g}]$

if the last is true, I can approach the second term as you mentioned with solving for $\theta$ or approximating it as constant and throw it away completely and then this simplifies to something I have tested which seems to vertically diffusing

$[\frac{d}{d\theta}\frac{dU}{d\theta}\frac{N \theta}{g}]\frac{N \theta}{g} + 0$ = $\frac{d^2U}{d\theta^2}[\frac{N \theta}{g}]^2$

What bothers me is that you're trying to take the derivative of U with respect to z or with respect to $\theta$, but I don't see any relationship between U and either of these variables. Without that relationship I don't see the point of writing $\frac{dU}{dz}$ or $\frac{dU}{d \theta}$ or continuing on to the second derivative.

 

[rc_nmt]

Yes I did switch derivatives only because I am not sure of the role of each variable now.

N is definitely a function of both z and $\theta$ but like I said might be able to be consider constant over the domain of the model I am running. The same is true with g, it is also a function of z and $\theta$ but is nearly constant, just as 9.81m/s^2 is pretty good for Earth up to 8km or so.

U is the wind however. There is no relationship that ties U to these other quantities except the primitive hydrodynamic equations. In most analytical senses the horizontal winds and rarely coupled to the vertical because the equations and not unique and computational methods are needed, which is what the GCM is very good and I might add it is also a finite element GCM, not spectral. So a velocity like U is connected to the other variables like how velocity is related to F = ma, but otherwise it would be relations from mass continuity, momentum and energy equations in full 3D glory. Not sure if that is where we want to go with that...

I need the change in U with respect to some vertical coordinate to paramterize the effects of atmospheric waves.

 

[Mark44]

U is the wind however. There is no relationship that ties U to these other quantities except the primitive hydrodynamic equations. In most analytical senses the horizontal winds and rarely coupled

Should that be "are rarely coupled..."?

to the vertical because the equations and not unique

And "are not unique..."?

and computational methods are needed, which is what the GCM is very good and I might add it is also a finite element GCM, not spectral.

So on the one hand, if the horizontal component of wind speed has no relationship to z and $\theta$, there's no point in calculating $\frac{dU}{dz}$ or $\frac{dU}{d\theta}$ (or the partial derivative equivalents).

On the other hand, if U is related to z $\theta$ by the primitive hydrodynamic equations, then it seems to me that those are the equations you need to work with to calculate the derivatives you're looking for.

So a velocity like U is connected to the other variables like how velocity is related to F = ma

Velocity has no direct connection in this equation. If the velocity is constant, a = 0, so the force F is 0.

, but otherwise it would be relations from mass continuity, momentum and energy equations in full 3D glory. Not sure if that is where we want to go with that...

I need the change in U with respect to some vertical coordinate to paramterize the effects of atmospheric waves.

What I'm saying is that before you can get this change, you need to establish some relationship between U and either z or $\theta$.

 

[rc_nmt]

Here is the paper about the model. Note that the vector v is the wind velocity vector with components u,v,w and you can see the relationships it has to other variables in equations 3, 4b, and 10. Maybe take a look at figure 2. It might better explain what I mean in the vertical setup of the model. This is a very specific question and maybe should be directed to someone with experience with GCM's like this. But thank you for your replies and I spent some time yesterday with my advisor going through some of these ideas. It is not very straight forward. Apologies

http://epicwiki.atmos.louisville.edu/images/Dowling98.pdf