Differentials of a Lagrangian in Fluid Mechanics

Question

My question relates to the paper by Foures et al. The paper presents a Lagrangian approach to finding an adjoint Poiseuille flow equation. It also defines the notion of an inner product (equation (3.10) in the paper) and proceeds to outline the method in which differentials are taken with respect to the componentwise velocity, $u_i$.

In the second term on the right-hand side of equation (A.5) below, how was the differential constructed? In particular, where do terms such as $\partial_t \delta u_i$ and $\delta u_j\partial_j U_i$ come from?

I'm struggling to see how these terms were derived, so if anyone could offer an explanation (as to how the differential was computed), I'd be very grateful.

EDIT: Using the Gateaux derivative on the terms $U_j\partial_j u_i$ and $u_j\partial_j\bar{u}_i = u_j\partial_j(U_i - u_i)$ gives $U_j\partial_j\delta u_i + \delta u_j \partial_j(U_i-u_i-\epsilon \delta u_i) + (u_j + \epsilon \delta u_j)\partial_j(-\delta u_i)\vert_{\epsilon=0}$. Clearly something has gone wrong here, but I'm not sure what...

Answer 1

That is simply the first variation of the Lagrangian functional with respect to $u_i$, or simply, the directional (Gateaux) derivative; see wikipedia for the definition. Note:

1. It took me a while to figure out because I didn't know what $\bar u$ is and that $U$ depends on $u$ as well.
2. You might also want to look at functional derivative if you are still confused.

I think some of the terms in (A5) are incorrect, either I don't understand their symbols (indices) or I am missing something else. Regardless, I managed to arrive at (A6) with all the correct indices. The functional $\mathscr{L}$ is actually written using Einstein summation convention and some terms have double sums which means we just need to be a little careful when taking the differentials. In what follows, I will show how to take differentials of $\langle u_i^\dagger, \partial_tu_i + U_j\partial_ju_i + u_j\partial_j\bar u_i\rangle$ with respect to $u_1$ but the idea is the same for taking differentials of other terms in $\mathscr{L}$ with respect to all the other direct variables; $\delta u_1$ is an arbitrary function for the $u_1$-variable.

Firstly, the term $\langle u_i^\dagger, \partial_tu_i\rangle$ is actually $$ G(u_1,u_2) = \langle u_1^\dagger, \partial_tu_1\rangle + \langle u_2^\dagger,\partial_tu_2\rangle. $$ The differential of $G$ with respect to $u_1$ is by definition \begin{align*} \left\langle \frac{\delta G(u_1,u_2)}{\delta u_1}, \delta u_1\right\rangle & \colon = \lim_{\varepsilon\to 0} \frac{G(u_1 + \varepsilon\delta u_1,u_2) - G(u_1,u_2)}{\varepsilon} \\ & = \lim_{\varepsilon\to 0} \frac{\langle u_1^\dagger, \partial_t(u_1 + \varepsilon\delta u_1) - \partial_tu_1\rangle}{\varepsilon} \\ & = \lim_{\varepsilon\to 0} \langle u_1^\dagger, \partial_t\delta u_1\rangle \\ & = \langle u_1^\dagger, \partial_t\delta u_1\rangle \\ & = \langle u_i^\dagger, \partial_t\delta u_i\rangle \bigg|_{i=1}. \end{align*}

The term $\langle u_i^\dagger, U_j\partial_ju_i\rangle$ means $$ H(u_1,u_2) = \underbrace{\langle u_1^\dagger, U_j\partial_ju_1\rangle}_{hh(u_1,u_2)} + \underbrace{\langle u_2^\dagger, U_j\partial_ju_2\rangle}_{HH(u_1,u_2)}. $$ Note crucially that both terms depends on $u_1$ as opposed to the functional $G$ where only the first term depends on $u_1$. Now, the differential of $H$ with respect to $u_1$ is \begin{align*} \left\langle \frac{\delta H(u_1,u_2)}{\delta u_1}, \delta u_1\right\rangle & \colon = \lim_{\varepsilon\to 0} \frac{H(u_1 + \varepsilon\delta u_1,u_2) - H(u_1,u_2)}{\varepsilon}. \end{align*} Two facts are used below:

1. $U_j = u_j + \bar u_j$ and varying $u_j$ doesn't change $\bar u_j$, i.e. "$\delta U_j = \delta u_j$
2. This also means that terms with $\bar u_j$ should be treated as "constants".

Expanding $H$ by summing over $j$ yields \begin{align*} H(u_1,u_2) & = \langle u_1^\dagger, U_j\partial_ju_1\rangle + \langle u_2^\dagger, U_j\partial_ju_2\rangle \\ & = \langle u_1^\dagger, U_1\partial_1u_1 + U_2\partial_2u_1\rangle + \langle u_2^\dagger, U_1\partial_1u_2 + U_2\partial_2u_2\rangle \\ & = \langle u_1^\dagger, (u_1 + \bar u_1)\partial_1u_1 + U_2\partial_2u_1\rangle + \langle u_2^\dagger, (u_1 + \bar u_1)\partial_1u_2 + U_2\partial_2u_2\rangle. \end{align*} and \begin{align*} & H(u_1 + \varepsilon\delta u_1,u_2) - H(u_1,u_2) \\ & = \langle u_1^\dagger, (u_1 + \varepsilon\delta u_1 + \bar u_1)\partial_1(u_1 + \varepsilon\delta u_1) + U_2\partial_2(u_1 + \varepsilon\delta u_1)\rangle - \Big[\langle u_1^\dagger, (u_1 + \bar u_1)\partial_1u_1 + U_2\partial_2u_1\rangle\Big] \\ & \qquad+ \langle u_2^\dagger, (u_1 + \varepsilon\delta u_1 + \bar u_1)\partial_1u_2 + U_2\partial_2u_2\rangle - \Big[\langle u_2^\dagger, (u_1 + \bar u_1)\partial_1u_2 + U_2\partial_2u_2\rangle\Big] \\ & = \varepsilon\Big[\langle u_1^\dagger, \delta u_1\partial_1u_1 + (u_1 + \bar u_1)\partial_1\delta u_1 + U_2\partial_2\delta u_1\rangle + \langle u_2^\dagger, \delta u_1\partial_1u_2\rangle\Big] + \mathcal{O}(\varepsilon^2) \\ & = \varepsilon\Big[\langle u_1^\dagger, \delta u_1\partial_1u_1 + U_1\partial_1\delta u_1 + U_2\partial_2\delta u_1\rangle + \langle u_2^\dagger, \delta u_1\partial_1u_2\rangle\Big] + \mathcal{O}(\varepsilon^2) \\ & = \varepsilon\Big[\langle u_1^\dagger, U_j\partial_j\delta u_1\rangle + \langle u_1^\dagger, \delta u_1\partial_1u_1\rangle + \langle u_2^\dagger, \delta u_1\partial_1u_2\rangle\Big] + \mathcal{O}(\varepsilon^2) \\ & = \varepsilon\Big[\langle u_1^\dagger, U_j\partial_j\delta u_1\rangle + \langle u_j^\dagger, \delta u_1\partial_1u_j\rangle\Big] + \mathcal{O}(\varepsilon^2). \end{align*} Thus it follows that $$ \lim_{\varepsilon\to 0} \frac{H(u_1 + \varepsilon\delta u_1,u_2) - H(u_1,u_2)}{\varepsilon} = \langle u_1^\dagger, U_j\partial_j\delta u_1\rangle + \langle u_j^\dagger, \delta u_1\partial_1u_j\rangle. $$

Similarly, the term $\langle u_i^\dagger, u_j\partial_j\bar u_i\rangle$ is \begin{align*} I(u_1,u_2) & = \langle u_1^\dagger, u_j\partial_j\bar u_1\rangle + \langle u_2^\dagger, u_j\partial_j\bar u_2\rangle \\ & = \langle u_1^\dagger, u_1\partial_1\bar u_1 + u_2\partial_2\bar u_1\rangle + \langle u_2^\dagger, u_1\partial_1\bar u_2 + u_2\partial_2\bar u_2\rangle. \end{align*} You should convinced yourself that \begin{align*} \left\langle \frac{\delta I}{\delta u_1}, \delta u_1\right\rangle & = \langle u_1^\dagger, \delta u_1\partial_1\bar u_1\rangle + \langle u_2^\dagger, \delta u_1\partial_1\bar u_2\rangle \\ & = \langle u_j^\dagger, \delta u_1\partial_1\bar u_j\rangle. \end{align*}

Finally, the differential of $G + H + I = \langle u_i^\dagger, \partial_tu_i + U_j\partial_ju_i + u_j\partial_j\bar u_i\rangle$ with respect to $u_1$ is \begin{align*} \left\langle \frac{\delta(G + H + I)}{\delta u_1}, \delta u_1\right\rangle & = \langle u_1^\dagger, U_j\partial_j\delta u_1\rangle + \langle u_j^\dagger, \delta u_1\partial_1u_j\rangle + \langle u_j^\dagger, \delta u_1\partial_1\bar u_j\rangle \\ & = \langle u_1^\dagger, U_j\partial_j\delta u_1\rangle + \langle u_j^\dagger, \delta u_1\partial_1U_j\rangle \\ & = \langle u_i^\dagger, U_j\partial_j\delta u_i\rangle + \langle u_j^\dagger, \delta u_i\partial_iU_j\rangle \bigg|_{i=1} \\ & = -\langle \partial_ju_i^\dagger, U_j\delta u_i\rangle + \langle u_j^\dagger, \delta u_i\partial_iU_j\rangle \bigg|_{i=1} \\ & = \langle u_j^\dagger\partial_iU_j - U_j\partial_ju_i^\dagger, \delta u_i\rangle\bigg|_{i=1}. \end{align*}