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I was wondering if there was an identity concerning the material derivative of a line integral, e.g., the material derivative of the zonal barotropic wind $u_{\tau}$ (which itself is the vertical-average of the total zonal wind $u$)

$$\frac{\text{D} u_{\tau}}{\text{D}t} = \frac{\text{D}}{\text{D}t} \left[ \frac{1}{H}\,\int_{0}^{H} u\left(t,\ x,\ y,\ z\right)\,dz \right],$$

where the material derivative is

$$\frac{\text{D}}{\text{D}t} = \partial_t + u\,\partial_x + v\,\partial_y + w\,\partial_z.$$ The source here says that it is trivial and simply moves the material derivative into the integral, which I don't quite trust. What would the material derviative of $dz$ be?


The fluid I am working with is incompressible, which guarantees the existence of a stream function. In fact, the stream function for the barotropic wind $\tau$ is the vertical-average of the stream function for the total wind $\chi$

$$\tau\left(t,\ x,\ y\right) = \frac{1}{H} \int_{0}^{H} \chi\left(t,\ x,\ y,\ z\right)\,dz$$

and so the barotropic winds are entirely horizontal, and are given by

$$\overrightarrow{u}_{\tau} = \left[ -\partial_y \tau,\ \partial_x \tau,\ 0 \right]^T.$$


EDIT: I see from the preliminary responses that I can indeed just bring the material derivative into the integral. I guess I was hoping for some sort of identity relating the material derivative of $u_{\tau}$ to some function of $u_\tau$ i.e., eliminate the dependence on $z$. When you bring the material derivative into the integral, you must be certain that you don't integrate $u$ and $v$ with respect to $z$, i.e.,

$$ \begin{align} \frac{\text{D} u_{\tau}}{\text{D}t} &= \frac{\text{D}}{\text{D}t} \left[ \frac{1}{H}\,\int_{0}^{H} u\left(t,\ x,\ y,\ z\right)\,dz \right] \nonumber \\ &= \frac{1}{H}\,\int_{0}^{H} \partial_t u\left(t,\ x,\ y,\ z\right) + u\left(t,\ x,\ y,\ z'\right)\,\partial_x u\left(t,\ x,\ y,\ z\right) + v\left(t,\ x,\ y,\ z'\right)\,\partial_y u\left(t,\ x,\ y,\ z\right) + w\left(t,\ x,\ y,\ z'\right)\,\partial_z u\left(t,\ x,\ y,\ z\right)\,dz \nonumber \\ &\neq \frac{1}{H}\,\int_{0}^{H} \partial_t u\left(t,\ x,\ y,\ z\right) + u\left(t,\ x,\ y,\ z\right)\,\partial_x u\left(t,\ x,\ y,\ z\right) + v\left(t,\ x,\ y,\ z\right)\,\partial_y u\left(t,\ x,\ y,\ z\right) + w\left(t,\ x,\ y,\ z\right)\,\partial_z u\left(t,\ x,\ y,\ z\right)\,dz \end{align} $$

What I was really hoping for was an evolution equation for the barotropic voticity $\zeta_{\tau} = \partial_x v_{\tau} - \partial_y u_{\tau}$ dependent only on the barotropic wind speeds for the model given in Ogrosky, Stechmann, Hottovy 2019,

OSH19 System

I've been told the evolution equation is

$$ \partial_t \zeta_{\tau} = -\frac{1}{\tau_u} - \beta\,v_{\tau} - u_\tau\,\partial_x \zeta_\tau - v_{\tau}\,\partial_y \zeta_\tau $$

which looks a lot like a sort of "barotropic" material derivative. If that is the correct evolution equation, how did the total wind speeds in the advection terms become barotropic wind speeds?

  • The material derivative is linear since it is a linear combination of linear operators, namely the partial time derivative and covariant derivative. So it follows the same rules. – K.defaoite Mar 18 '21 at 22:41
  • If nothing depends upon $t$ in $H^{-1}\int_0^H u(t,x,y,z) , dz$ other than the integrand -- that is $x$, $y$, and $H$ are fixed -- then this is no different than switching an ordinary derivative with an integral. There is "an identity concerning the material derivative of a line integral" but it would pertain to something like $\frac{\partial}{\partial t}\int_{C(t)} \mathbf{u} \cdot d \mathbf{x}$ where $C(t)$ is a material curve moving in time. This does not appear to be the case here. – RRL Mar 18 '21 at 22:54
  • Thank you @K.defaoite! I made an edit to the original question with some follow-up. – jltorchinsky Mar 18 '21 at 23:45
  • Thank you @RRL! I made an edit to the original question with some follow-up. – jltorchinsky Mar 18 '21 at 23:45

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