# Kris's Research Notes

## October 12, 2010

### Diffusion through a Droplet

Filed under: GaAs Simulations — Kris Reyes @ 6:14 pm

Recall we wish to simulate the crystallization of $Ga$ droplets with the presence of $As$ flux. Let us simplify a bit, and suppose the $Ga$ droplet is hemispherical of radius $R$. Let’s describe the concentration of  $As$ in the droplet by a function $u(r, \theta)$, where we have used spherical coordinates to describe the location inside the hemisphere (so that our domain is $r \in [0, R], \theta \in [0, \pi]$).

We wish to solve Laplace’s equation (WHY??):

$\displaystyle \nabla^2 u =\frac{1}{r}\frac{\partial}{\partial r}\left( r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta ^2} = 0$

subject to boundary conditions

• $\displaystyle \frac{\partial u}{\partial r}\big|_{r = R} = c$, i.e. constant flux on the surface of the droplet;
• $u(r, 0) = u(r,\pi)= 0$, i.e. no concentration on the bottom of the droplet.

To solve this, we assume a separable solution of the form:

$u(r, \theta) = F(r)G(\theta).$

Applying the Laplacian we get

$\displaystyle \nabla^2\left[F(r)G(\theta)\right] = \frac{1}{r} \frac{dF}{dr} G + \frac{d^2F}{dr^2}G + \frac{1}{r^2}F\frac{d^2G}{d\theta^2} = 0$.

Separating variables, we get:

$\displaystyle \frac{r^2}{F}\left(\frac{1}{r}\frac{dF}{dR} + \frac{d^2F}{dr^2}\right) = \lambda = -\frac{1}{G} \frac{d^2G}{d\theta^2},$

for some constant $\lambda$. This gives us our ODEs:

$\displaystyle \lambda F - r\frac{dF}{dr} - r^2\frac{d^2F}{dr^2} = 0.$

$\displaystyle \lambda G + \frac{d^2G}{d\theta^2} = 0.$

We consider the second equation first, which is an eigenvalue problem.

If $\lambda > 0$, we have

$G(\theta) = c_1 \cos(\sqrt{\lambda} \theta) + c_2 \sin(\sqrt{\lambda}\theta)$

and given the boundary conditions $G(0) = G(\pi) = 0$, we see that $\lambda = n^2$ for $n = \mathbb Z^+$ and the corresponding eigenvector is of the form:

$G_n(\theta) = c_n \sin(n \theta)$.

Now if $\lambda < 0$, we have a general solution of the form

$G(\theta) = b_1\cosh(\sqrt{|\lambda|}\theta) + b_2\sinh(\sqrt{|\lambda|}\theta)$

However, only the trivial solution satisfies the boundary conditions.

Lastly if $\lambda = 0$, we have $G(\theta) = a_1 + a_2\theta$, but once again only the trivial solution satisfies boundary conditions.

Therefore the only eigenvalues of the above operator are of the form $\lambda = n^2$ for $n \in \mathbb Z^+$ with corresponding eigenvectors $G_n(\theta) = c_n \sin(n\theta).$

Then we have corresponding solutions to the first ODE: $F_n(r) = a_n r^p_n$, where $p$ satisfies

$n^2 r^{p_n} - p_ nr^{p_n} - p_n(p_n-1)r^{p_n} = 0,$

which is what is obtained if we plug in this form of $F$ into the above ODE. That is, $p_n = \pm n$ and $F(r) = a_nr^{\pm n}$. We require $F(r)$ is bounded near $r = 0$, and so we ignore solutions $F(r) = a_nr^{-n}$.

Therefore we have a general solution of the form (incorporating all undetermined coefficients)

$\displaystyle u(r,\theta) = \sum_{n=1}^\infty A_n r^{n}\sin(n\theta).$

We now apply the final boundary condition $\frac{\partial u}{\partial r}\big|_{r = R} = c$:

$\displaystyle \frac{\partial u}{\partial r}\big|_{r = R} = \sum_{n=1}^\infty nA_n R^{n-1} \sin(n\theta) = c.$

By Fourier Inversion, we can solve

$\displaystyle n A_n R^{n-1} = \frac{c}{\pi} \int_0^\pi \sin(n\theta) d\theta = \frac{c}{n\pi} \left(1-\cos(n\pi)\right)$

and so

$\displaystyle A_n = \frac{c}{\pi}\frac{(1-\cos(n\pi))}{n^2R^{n-1}}.$

Plugging this in to the expression for $u(r, \theta)$ we get:

$\displaystyle u(r, \theta) = \frac{c}{\pi} \sum_{n=1}^\infty \frac{1-\cos(n\pi)}{n^2R^{n-1}} r^{n}\sin(n\theta).$

We may compute the flux through the bottom of the crystal by calculating:

$\displaystyle \frac{\partial u}{\partial y} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial y}.$

That is:

$\displaystyle \frac{\partial u}{\partial y} = \frac{c}{\pi} \sum_{n=1}^\infty \frac{1-\cos(n\pi)}{n R^{n-1}} \left[\sin\theta \sin(n\theta) r^{n-1} + \frac{\cos(n\theta)}{\cos\theta} r^{n-1} - \frac{\cos(n\theta)\sin^2\theta}{\cos\theta}r^{n-1}\right]$

and when so $\theta = 0$, i.e. when $y=0, x \geq 0$, we get:

$\displaystyle \frac{\partial u}{\partial y}\bigg|_{\theta = 0} = \frac{c}{\pi} \sum_{n=1}^\infty \frac{1-\cos(n\pi)}{n R^{n-1}} r^{n-1} = \frac{2c}{\pi}\sum_{k=1}^\infty \frac{1}{2k+1}\left(\frac{r}{R}\right)^{2k}.$

I don’t know of a closed form for this, but we may plot truncated sums (for e.g. $n \leq 10000$) and examine the flux vs $r$. We may also vary the flux $c$. Here are the plots of $\frac{\partial u}{\partial y}\big|_{\theta = 0}$ for $R = 1$, $0 \leq r \leq R$ and $c$ varying from $0.1$ to $1.0$.

Note this is what we want: the amount of material being deposited is greater near the edge of the droplet than the middle.