Recall we wish to simulate the crystallization of droplets with the presence of flux. Let us simplify a bit, and suppose the droplet is hemispherical of radius . Let’s describe the concentration of in the droplet by a function , where we have used spherical coordinates to describe the location inside the hemisphere (so that our domain is ).

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

subject to boundary conditions

- , i.e. constant flux on the surface of the droplet;
- , i.e. no concentration on the bottom of the droplet.

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

Applying the Laplacian we get

.

Separating variables, we get:

for some constant . This gives us our ODEs:

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

If , we have

and given the boundary conditions , we see that for and the corresponding eigenvector is of the form:

.

Now if , we have a general solution of the form

However, only the trivial solution satisfies the boundary conditions.

Lastly if , we have , but once again only the trivial solution satisfies boundary conditions.

Therefore the only eigenvalues of the above operator are of the form for with corresponding eigenvectors

Then we have corresponding solutions to the first ODE: , where satisfies

which is what is obtained if we plug in this form of into the above ODE. That is, and . We require is bounded near , and so we ignore solutions .

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

We now apply the final boundary condition :

By Fourier Inversion, we can solve

and so

Plugging this in to the expression for we get:

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

That is:

and when so , i.e. when , we get:

I don’t know of a closed form for this, but we may plot truncated sums (for e.g. ) and examine the flux vs . We may also vary the flux . Here are the plots of for , and varying from to .

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

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[…] we do not get the distribution of crystal inside the droplet that we expected in this post. That is, we had previously calculated that there should be more atoms diffusing to the edges of […]

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