I am trying to derive the Green's function of 2D brinkman flow while I encounter a general questions as following:
Given a tensor function $$\mathbf F(\rho)=g(\rho)\delta_{ij}+h(\rho)x_{i}x_{j},$$ where $(\rho,\theta)$ is polar coordinates. My question is how to derive $\nabla^{2}\mathbf F$ ?
It should be simple but I got stuck at some points. Thank you!
I am so happy that you didn't ask for the Laplacian in ND generalized spherical polar coordinates . I don't think I could handle the typing.
There are many equivalent ways of doing this. I will show one that I like below.
I am assuming that by $x_1$,$x_2$ you mean the Cartesian coordinates $x$,$y$ with respect to the basis vectors ${\bf i}$, ${\bf j}$
The general position vector describing points is ${\bf r}=\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\underbrace{\left[\begin{array}{c}x\\y\end{array}\right]}_{Cartesian}=\underbrace{\left[\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\end{array}\right]}_{Polar}$ where $r=\sqrt{x^2+y^2}$ and $\tan(\theta)=\frac{y}{x}$
When writing the Tensor down, make sure to use the correct notation. I think that you are mixing subscript notation with vector notation. It's not a huge deal, but it makes a difference when trying to parse what is written down, especially for more complicated Tensorial expressions.
You should either write it purely in subscript notation, i.e, $$F_{ij}=g(r)\delta_{ij}+h(r)x_ix_j,$$ where $i=1,2$ and $j=1,2$, and $\delta_{ij}$ is the Kronecker Delta Function, $\delta_{ij}=\left\{\begin{array}{cc}1,&i=j\\0,&i\neq j\end{array}\right.,$ which gives the 4 components of the tensor in component form
Or you could write it purely in Matrix form $$\begin{align*} \underline{\underline{F}}=\left[\begin{array}{cc}F_{11}&F_{12}\\ F_{21}&F_{22}\end{array}\right]&=g(r)\mathbb{I}+h(r){\bf r}\otimes {\bf r}\\ &=g(r)\left[\begin{array}{cc}1&0\\0&1\end{array}\right]+h(r)\left[\begin{array}{cc}x^2&xy\\yx&y^2\end{array}\right]\\ &=\underbrace{\left[\begin{array}{cc}g(r)+h(r)x^2&h(r)xy\\h(r)yx&g(r)+h(r)y^2\end{array}\right]}_{Cartesian}\\ &=\underbrace{\left[\begin{array}{cc}g(r)+h(r)r^2\cos^2(\theta)&h(r)r^2\cos(\theta)\sin(\theta)\\h(r)r^2\cos(\theta)\sin(\theta)&g(r)+h(r)r^2\sin^2(\theta)\end{array}\right]}_{Polar}\\ \end{align*}$$ where ${\bf r}\otimes {\bf r}$ is the dyadic tensor product (Dyadic Product) ${\bf r}{\bf r}^T=\left[\begin{array}{c}x\\y\end{array}\right]\left[\begin{array}{cC}x&y\end{array}\right]=\left[\begin{array}{cc}x\cdot x&x\cdot y\\y\cdot x&y\cdot y\end{array}\right]$
I like the full dyadic expansion notation best. It gives a very nice intuition for what is going on with any operators (i.e., the Laplacian) that are acting on the Tensor field. $$\begin{align*} \underline{\underline{F}}&=F_{11}{\bf i}{\bf i}^T+F_{12}{\bf i}{\bf j}^T+F_{21}{\bf j}{\bf i}^T+F_{22}{\bf j}{\bf j}^T \end{align*}$$
The main thing to notice is that the basis vectors here are Cartesian ${\bf i}$, ${\bf j}$ and thus the derivative operator in the Laplacian only act on the functions $F_{ij}$, this means that the Laplacian only hits the functions themselves
\begin{align*} \Delta\underline{\underline{F}}=\nabla^2\underline{\underline{F}}=\nabla^2F_{11}{\bf i}{\bf i}^T+\nabla^2F_{12}{\bf i}{\bf j}^T+\nabla^2F_{21}{\bf j}{\bf i}^T+\nabla^2F_{22}{\bf j}{\bf j}^T \end{align*}
Lastly, we need to get the formula for the Laplacian of a function in polar coordinates. This is a good derivation to carry out and it relys on the multivariable chain rule from multivariable calculus. I can show you how to get it if you want to see just lmk. For a general multivariable function $f$ it is $$\Delta f=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial f}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2f}{\partial \theta^2}$$
All that is left is to calculate this for each of the components of the Tensor. Luckily, it is symmetric $F_{12}=F_{21}$ so we only have to do 3 calculations.
$$\begin{align*} \Delta F_{11}&=\left(g''(r)+\frac{g'(r)}{r}+2h(r)\right) + \left(h''(r)+\frac{5h'(r)}{r}\right)r^2\cos^2(\theta) \end{align*} $$
$$\begin{align*} \Delta F_{12}=\left(h''(r)+\frac{5h'(r)}{r}\right)r^2\cos(\theta)\sin(\theta) \end{align*} $$
$$\Delta F_{22}=\left(g''(r)+\frac{g'(r)}{r}+2h(r)\right) + \left(h''(r)+\frac{5h'(r)}{r}\right)r^2\sin^2(\theta)$$
This means that the full Tensor takes the following form (don't worry, it looks more monstrous than it is $$ \begin{align*} \underline{\underline{T}}&=\left[\begin{array}{cc}F_{11}&F_{12}\\ F_{21}&F_{22}\end{array}\right]\\ &=\left[\begin{array}{cc}\left(g''(r)+\frac{g'(r)}{r}+2h(r)\right)+\left(h''(r)+\frac{5h'(r)}{r}\right)x^2&\left(h''(r)+\frac{5h'(r)}{r}\right)xy\\ \left(h''(r)+\frac{5h'(r)}{r}\right)xy&\left(g''(r)+\frac{g'(r)}{r}+2h(r)\right) + \left(h''(r)+\frac{5h'(r)}{r}\right)y^2\end{array}\right]\\ &=(g''(r)+\frac{g'(r)}{r}+2h(r))\left[\begin{array}{cc} 1&0\\ 0&1\end{array}\right]+\left(h''(r)+\frac{5h'(r)}{r}\right)\left[\begin{array}{cc} x^2&xy\\ xy&y^2\end{array}\right]\\ &=\boxed{\left(g''(r)+\frac{g'(r)}{r}+2h(r)\right)\mathbb{I}+\left(h''(r)+\frac{5h'(r)}{r}\right){\bf r}\otimes{\bf r}} \end{align*}$$
In index notation this is $$\boxed{F_{ij}=\left(g''(r)+\frac{g'(r)}{r}+2h(r)\right)\delta_{ij}+\left(h''(r)+\frac{5h'(r)}{r}\right)x_ix_j}$$ Which is actually pretty nice all things considered.
I didn't do the Laplacians partial derivative calculations out because I got tired of typing but I think you can definitely get the rest. Let me know if you need clarification on anything or if you need help with the other calculations and I can type the steps in more detail.
Another approach might be to start the Calculation using Cartesian Coordinates and then convert to Polar afterwords, this will give you the same result but it might be more messy.
I hope this helps!!!
