Notation for Multiple Scale Expansion

Question

I'm reading through a set of lecture notes on waves in random media and on pages 16-17, a multi-scale expansion is worked out. I'm having some trouble with the notation and verifying a calculation.

$\textbf{QUESTIONS}$

1. What is meant by (cf. page 17, equation 2.19)

$$\langle \frac{1}{\rho}(\nabla_{\textbf{y}}\pmb{\theta} + I_{3}) \rangle$$

? I know that this is a $3 \times 3$ matrix, but what are the entries?

2. Integrating equation 2.18, page 17 over $Y = (0, 1)^{3}$ in the variable $\textbf{y}$ yields

\begin{align} \langle \kappa \rangle \frac{\partial^{2}p_{0}}{\partial t^{2}} - \int_{Y}\nabla_{\textbf{x}}\cdot \frac{1}{\rho(\textbf{y})}\nabla_{\textbf{x}}p_{0}(t, \textbf{x})d\textbf{y} &= -\int_{Y}\nabla_{\textbf{x}}\cdot \frac{\textbf{F}(t, \textbf{x})}{\rho(\textbf{y})}d\textbf{y} \\ \implies \langle \kappa \rangle \frac{\partial^{2}p_{0}}{\partial t^{2}} - \int_{Y}\nabla_{\textbf{x}}\cdot \frac{1}{\rho(\textbf{y})}(\nabla_{\textbf{x}}p_{0}(t, \textbf{x}) - \textbf{F}(t, \textbf{x}))d\textbf{y} &= 0 \end{align}

as all the divergence terms with $\nabla_{\textbf{y}}$ vanish. This is supposed to be equation 2.19 on page 17 in the notes, i.e.

$$ \langle \kappa \rangle \frac{\partial^{2}p_{0}}{\partial t^{2}} - \nabla_{\textbf{x}} \cdot \langle \frac{1}{\rho}(\nabla_{\textbf{y}}\pmb{\theta} + I_{3}) \rangle (\nabla_{\textbf{x}}p_{0} - \textbf{F}(t, \textbf{x})) = 0$$

How do we show

$$ \int_{Y}\nabla_{\textbf{x}}\cdot \frac{1}{\rho(\textbf{y})}(\nabla_{\textbf{x}}p_{0}(t, \textbf{x}) - \textbf{F}(t, \textbf{x}))d\textbf{y}= \nabla_{\textbf{x}} \cdot \langle \frac{1}{\rho}(\nabla_{\textbf{y}}\pmb{\theta} + I_{3}) \rangle (\nabla_{\textbf{x}}p_{0} - \textbf{F}(t, \textbf{x}))$$

?

Answer 1

To answer each question individually

1. The notation $\langle \cdot \rangle$ denotes an average over the spatial volume. In one dimension

$$\langle f(s) \rangle = \frac{1}{b - a} \int_{a}^{b} f(s) ds$$

In your case, it is a triple integral over $y_{i} \in [0,1]$ and the fraction out the front is $1/\lvert \text{vol } Y \rvert$. Next, $\nabla_{y} = (\partial_{y_{1}}, \partial_{y_{2}}, \partial_{y_{3}})$ is a gradient vector, $\boldsymbol{\theta} = (\theta_{1}, \theta_{2}, \theta_{3})$ is also a vector, so $\nabla_{y} \boldsymbol{\theta}$ is a rank 2 tensor (see page 7, part c of this for the exact form).

2. Taking the average of the equation, $p_{0}$ is independent of $t$ but $\kappa$ isn't, so the first term is $\langle \kappa \rangle \partial_{t}^{2} p_{0}$. By the divergence theorem, the terms

$$\nabla_{y} \cdot \frac{1}{\rho} \nabla_{y} p_{2}, \quad \nabla_{y} \cdot \frac{1}{\rho} \nabla_{x} p_{1}$$

vanish identically i.e the average of each is zero. Substituing the solution $p_{1} = \boldsymbol{\theta} (y)·(\nabla_{x} p_{0} − \boldsymbol{F})$ into the equation gives

\begin{align} &-\nabla_{x} \cdot \frac{1}{\rho} \nabla_{y} (\boldsymbol{\theta} (y) \cdot (\nabla_{x} p_{0} − \boldsymbol{F})) - \nabla_{x} \cdot \frac{1}{\rho} \nabla_{x} p_{0} + \nabla_{x} \cdot \frac{1}{\rho} \boldsymbol{F} \\ &= -\nabla_{x} \cdot \frac{1}{\rho} \nabla_{y} (\boldsymbol{\theta} (y) \cdot (\nabla_{x} p_{0} − \boldsymbol{F})) - \nabla_{x} \cdot \frac{1}{\rho} (\nabla_{x} p_{0} - \boldsymbol{F}) \\ &= -\nabla_{x} \cdot \frac{1}{\rho} (\nabla_{y} \boldsymbol{\theta} + I_{3})(\nabla_{x} p_{0} - \boldsymbol{F}) \end{align}

and then averaging, noting that $p_{0}$ and $\boldsymbol{F}$ are independent of $y$, yields the result.