The ``kinematic equation'' of fluid flows

[hunt_mat]

TL;DR Summary

I saw this in a fluid book and I want to get a clean derivation

I saw this in a textbook and I thought it is a corollary of Reynold's transport theorem. Let $\mathbf{F}$ be a smooth vector field Consider the surface integral:
$$\int_{S}\mathbf{F}\cdot d\mathbf{S}$$ and now take the derivative of it, then the expression can be written as:
$$\frac{d}{dt}\int_{S}\mathbf{F}\cdot d\mathbf{S}=\int_{S}\frac{\partial\mathbf{F}}{\partial t}\cdot d\mathbf{S}-\int_{\partial S}(\mathbf{u}\times\mathbf{F})\cdot d\mathbf{r}$$.

Is anyone familiar with this equation? Does anyone know of a nice clear vector analysis way of deriving it? I've looked in Batchelor but couldn't find a derivation.

 

[ergospherical]

This equation will hold so long as $\mathbf{F}(t,\mathbf{r})$ is divergence-free, $\nabla \cdot \mathbf{F} = 0$. Begin with the definition of the derivative, \begin{align*}
\dfrac{d}{dt} \int_{S(t)} \mathbf{F}(t) \cdot d\mathbf{S} = \lim_{h \rightarrow 0} \dfrac{1}{h} \left[ \int_{S(t+h)} \mathbf{F}(t+h) \cdot d\mathbf{S} - \int_{S(t)} \mathbf{F}(t) \cdot d\mathbf{S} \right]
\end{align*}Expand $\mathbf{F}(t+h) = \mathbf{F}(t) + h \dfrac{\partial \mathbf{F}}{\partial t} + O(h^2)$, then\begin{align*}
\dfrac{d}{dt} \int_{S(t)} \mathbf{F}(t) \cdot d\mathbf{S} &= \int_{S(t)} \dfrac{\partial \mathbf{F}}{\partial t} \cdot d\mathbf{S} + \lim_{h \rightarrow 0} \dfrac{1}{h} \left[ \int_{S(t+h)} \mathbf{F}(t) \cdot d\mathbf{S} - \int_{S(t)} \mathbf{F}(t) \cdot d\mathbf{S} \right] \\ \\
&\equiv \int_{S(t)} \dfrac{\partial \mathbf{F}}{\partial t} \cdot d\mathbf{S} + \lim_{h \rightarrow 0} \dfrac{1}{h} \left( \int_{S(t+h)} - \int_{S(t)} \right) \mathbf{F}(t) \cdot d\mathbf{S}
\end{align*}Between times $t$ and $t+h$, the boundary $\partial S$ of $S$ sweeps out a surface $B(t)$ such that $S(t+h)$, $B(t)$ and $S(t)$ together make a closed surface. Given that $\nabla \cdot \mathbf{F} = 0$, then keeping in mind the orientations of $S(t)$ and $S(t+h)$:\begin{align*}
\left(\int_{S(t+h)} + \int_{B(t)} -\int_{S(t)} \right) \mathbf{F}(t) \cdot d\mathbf{S} &= 0 \\ \\
\left(\int_{S(t+h)} -\int_{S(t)} \right) \mathbf{F}(t) \cdot d\mathbf{S} &= -\int_{B(t)} \mathbf{F}(t) \cdot d\mathbf{S}
\end{align*}If the boundary $\partial S$ of $S$ moves with a velocity $\mathbf{u}(t)$, then the surface element of $B(t)$ is nothing but the vector area $d\mathbf{r} \times h\mathbf{u}(t)$ where $d\mathbf{r}$ is the line element along $\partial S$. Therefore\begin{align*}
\left(\int_{S(t+h)} -\int_{S(t)} \right) \mathbf{F}(t) \cdot d\mathbf{S} &= -\int_{B(t)} \mathbf{F}(t) \cdot d\mathbf{r} \times h\mathbf{u}(t) \\ \\
&= -h\int_{B(t)} \mathbf{u}(t) \times \mathbf{F}(t) \cdot d\mathbf{r}
\end{align*}which finishes the derivation.

 

[hunt_mat]

Thanks for this but it wasn't quite the "clean" method I was asking. If I represent $d\mathbf{S}=\hat{\mathbf{n}}(t)dS$, then can't I use a change of variables to transform the surface into a "fixed" surface $S_{0}$ then we can take the derivative directly into the integral via the dominated convergence theorem as a partial derivative.

 

[pasmith]

Thanks for this but it wasn't quite the "clean" method I was asking. If I represent $d\mathbf{S}=\hat{\mathbf{n}}(t)dS$, then can't I use a change of variables to transform the surface into a "fixed" surface $S_{0}$ then we can take the derivative directly into the integral via the dominated convergence theorem as a partial derivative.

Not quite. You can use the Lagrangian view of the fluid, in which $\mathbf{X}$ represents the fluid parcel which started at $\mathbf{X}$ at $t = 0$. A surface which moves with the fluid is then fixed in this view, and if $\mathbf{x}$ is position in the Eulerian depiction (in which $\mathbf{x}$ is fixed and the fluid moves past it) then $$\left( \frac{\partial \mathbf{F}}{\partial t}\right)_{\mathbf{X}} =\left(\frac{\partial \mathbf{F}}{\partial t}\right)_{\mathbf{x}} + (\mathbf{u} \cdot \mathbf{\nabla})\mathbf{F}$$ and $$(\mathbf{x},t) = \left(\mathbf{X} + \int_0^t \mathbf{u}(\mathbf{X},t')\,dt',t\right).$$

 

[hunt_mat]

I've started the calculation but I've gotten stuck:
$$\begin{gather*} \frac{d}{dt}\int_{S_{t}}\mathbf{f}\cdot\hat{\mathbf{n}} = \frac{d}{dt}\int_{S_{0}}J\mathbf{F}\cdot\hat{\mathbf{N}}dS_{0} \\ = \int_{S_{0}}\frac{\partial J}{\partial t}\mathbf{F}\cdot\hat{\mathbf{N}}+J\frac{\partial\mathbf{F}}{\partial t}\cdot\hat{\mathbf{N}}+J\mathbf{F}\cdot\frac{\partial\hat{\mathbf{N}}}{\partial t}dS_{0} \\ = \int_{S_{t}}\mathbf{f}\cdot\hat{\mathbf{n}}\nabla\cdot\mathbf{u}+\frac{D\mathbf{f}}{Dt}\cdot\hat{\mathbf{n}}dS+\int_{S_{0}}J\mathbf{F}\cdot\frac{\partial\hat{\mathbf{N}}}{\partial t}dS_{0} \end{gather*}$$

I think the first two terms combine to yield:
$$\frac{\partial\mathbf{f}}{\partial t}+\nabla\cdot((\mathbf{f}\cdot\hat{\mathbf{n}})\mathbf{u})$$
I don't know what to do with the final term.

 

[wrobel]

The general fact is as follows
$$\frac{d}{dt}\Big|_{t=0}\int_{g_v^t(S)}\omega =\int_SL_v\omega,$$
where $g^t_v$ is a flow of the vector field $v(x)$ and $L_v$ is the Lie derivative.
One can also employ the formula $L_v\omega=i_vd\omega+di_v\omega$ and then use the Stokes theorem

 

[pasmith]

Parametrize $S(t)$ as $\mathbf{x}(s_1,s_2,t)$ for $(s_1,s_2) \in \Omega \subset \mathbb{R}^2$. Then $$\frac{d}{dt} \int_{S(t)} \mathbf{F} \cdot d\mathbf{S} = \frac{d}{dt} \int_{\Omega} \mathbf{F} \cdot \mathbf{n}\,ds_1\,ds_2.$$ Now $\Omega$ is independent of $t$ so we can pull the derivative inside the integral, but it becomes a partial derivative at fixed $s = (s_1,s_2)$. We then have $$\left(\frac{\partial\mathbf{F}}{\partial t}\right)_{s} = \left(\frac{\partial\mathbf{F}}{\partial t}\right)_{\mathbf{x}} + (\mathbf{u} \cdot \nabla) \mathbf{F}.$$ We also have $$\begin{split} \left(\frac{\partial n_i}{\partial t}\right)_s &= \epsilon_{ijk} \frac{\partial}{\partial t} \left( \frac{\partial x_j}{\partial s_1}\frac{\partial x_k}{\partial s_2}\right) \\ &= \epsilon_{ijk} \left(\frac{\partial u_j}{\partial s_1} \frac{\partial x_k}{\partial s_2} + \frac{\partial u_k}{\partial s_2} \frac{\partial x_j}{\partial s_1}\right) \\ &= \epsilon_{ijk} \left(\frac{\partial u_j}{\partial s_1} \frac{\partial x_k}{\partial s_2} - \frac{\partial u_j}{\partial s_2} \frac{\partial x_k}{\partial s_1}\right) \\ &= \epsilon_{ijk} \frac{\partial u_j}{\partial x_l}\left(\frac{\partial x_l}{\partial s_1} \frac{\partial x_k}{\partial s_2} - \frac{\partial x_l}{\partial s_2} \frac{\partial x_k}{\partial s_1}\right) \\ &= \epsilon_{ijk} \frac{\partial u_j}{\partial x_l} \left(\delta_{lp}\delta_{kq} - \delta_{lq}\delta_{kp}\right)\frac{\partial x_p}{\partial s_1} \frac{\partial x_q}{\partial x_2} \\ &= \epsilon_{ijk} \frac{\partial u_j}{\partial x_l} \epsilon_{lkm}\epsilon_{mpq} \frac{\partial x_p}{\partial s_1} \frac{\partial x_q}{\partial x_2} \\ &= -\epsilon_{ijk} \epsilon_{klm} \frac{\partial u_j}{\partial x_l} n_m \\ &= \left(\delta_{im}\delta_{jl} - \delta_{il}\delta_{jm}\right) \frac{\partial u_j}{\partial x_l} n_m\\ &= n_i\frac{\partial u_j}{\partial x_j} - n_j \frac{\partial u_j}{\partial x_i}. \end{split}$$ Hence $$\left(\frac{\partial}{\partial t} (\mathbf{F} \cdot \mathbf{n}) \right)_s = \left(\frac{\partial \mathbf{F}}{\partial t}\right)_{\mathbf{x}} \cdot \mathbf{n} + \mathbf{n} \cdot \left(((\mathbf{u} \cdot \nabla)\mathbf{F}) + \mathbf{F} (\nabla \cdot \mathbf{u}) - (\mathbf{F} \cdot \nabla)\mathbf{u}\right).$$ I think the last three terms can be rearranged into $-\mathbf{n} \cdot (\nabla \times(\mathbf{u} \times \mathbf{F}))$ on the assumption that $\nabla \cdot \mathbf{F} = 0$.

 

[hunt_mat]

The general fact is as follows
$$\frac{d}{dt}\Big|_{t=0}\int_{g_v^t(S)}\omega =\int_SL_v\omega,$$
where $g^t_v$ is a flow of the vector field $v(x)$ and $L_v$ is the Lie derivative.
One can also employ the formula $L_v\omega=i_vd\omega+di_v\omega$ and then use the Stokes theorem

This looks like a change of variables formula, am I right?

 

[wrobel]

sure

 

[hunt_mat]

sure

It's been a while since I've done integration on manifolds (~20 years), so you'll need to walk through this, I'm guessing that I would use the reverse flow of the vector field as the change of variable and that would mean I would have to pull back and taking the derivative inside the integral means that I get the Lie derivative?

 

[wrobel]

yes by definition it will be the Lie derivative and all the formulas above are the special cases of this construction

 

[hunt_mat]

yes by definition it will be the Lie derivative and all the formulas above are the special cases of this construction

I'm currently looking at the maths, I will want to write down a proof for this statement.

 

[hunt_mat]

Parametrize $S(t)$ as $\mathbf{x}(s_1,s_2,t)$ for $(s_1,s_2) \in \Omega \subset \mathbb{R}^2$. Then $$\frac{d}{dt} \int_{S(t)} \mathbf{F} \cdot d\mathbf{S} = \frac{d}{dt} \int_{\Omega} \mathbf{F} \cdot \mathbf{n}\,ds_1\,ds_2.$$ Now $\Omega$ is independent of $t$ so we can pull the derivative inside the integral, but it becomes a partial derivative at fixed $s = (s_1,s_2)$. We then have $$\left(\frac{\partial\mathbf{F}}{\partial t}\right)_{s} = \left(\frac{\partial\mathbf{F}}{\partial t}\right)_{\mathbf{x}} + (\mathbf{u} \cdot \nabla) \mathbf{F}.$$ We also have $$\begin{split} \left(\frac{\partial n_i}{\partial t}\right)_s &= \epsilon_{ijk} \frac{\partial}{\partial t} \left( \frac{\partial x_j}{\partial s_1}\frac{\partial x_k}{\partial s_2}\right) \\ &= \epsilon_{ijk} \left(\frac{\partial u_j}{\partial s_1} \frac{\partial x_k}{\partial s_2} + \frac{\partial u_k}{\partial s_2} \frac{\partial x_j}{\partial s_1}\right) \\ &= \epsilon_{ijk} \left(\frac{\partial u_j}{\partial s_1} \frac{\partial x_k}{\partial s_2} - \frac{\partial u_j}{\partial s_2} \frac{\partial x_k}{\partial s_1}\right) \\ &= \epsilon_{ijk} \frac{\partial u_j}{\partial x_l}\left(\frac{\partial x_l}{\partial s_1} \frac{\partial x_k}{\partial s_2} - \frac{\partial x_l}{\partial s_2} \frac{\partial x_k}{\partial s_1}\right) \\ &= \epsilon_{ijk} \frac{\partial u_j}{\partial x_l} \left(\delta_{lp}\delta_{kq} - \delta_{lq}\delta_{kp}\right)\frac{\partial x_p}{\partial s_1} \frac{\partial x_q}{\partial x_2} \\ &= \epsilon_{ijk} \frac{\partial u_j}{\partial x_l} \epsilon_{lkm}\epsilon_{mpq} \frac{\partial x_p}{\partial s_1} \frac{\partial x_q}{\partial x_2} \\ &= -\epsilon_{ijk} \epsilon_{klm} \frac{\partial u_j}{\partial x_l} n_m \\ &= \left(\delta_{im}\delta_{jl} - \delta_{il}\delta_{jm}\right) \frac{\partial u_j}{\partial x_l} n_m\\ &= n_i\frac{\partial u_j}{\partial x_j} - n_j \frac{\partial u_j}{\partial x_i}. \end{split}$$ Hence $$\left(\frac{\partial}{\partial t} (\mathbf{F} \cdot \mathbf{n}) \right)_s = \left(\frac{\partial \mathbf{F}}{\partial t}\right)_{\mathbf{x}} \cdot \mathbf{n} + \mathbf{n} \cdot \left(((\mathbf{u} \cdot \nabla)\mathbf{F}) + \mathbf{F} (\nabla \cdot \mathbf{u}) - (\mathbf{F} \cdot \nabla)\mathbf{u}\right).$$ I think the last three terms can be rearranged into $-\mathbf{n} \cdot (\nabla \times(\mathbf{u} \times \mathbf{F}))$ on the assumption that $\nabla \cdot \mathbf{F} = 0$.

Hi, this is exactly what I wanted, thank you.

 

[hunt_mat]

yes by definition it will be the Lie derivative and all the formulas above are the special cases of this construction

So we need a change of variables. Will this be the streamlines? This will lead to the pullback. The vector which is involved in the Lie derivative will be the velocity vector(tangent to the streamlines) at t=0. This is the basic calculation right?