My question is about math, no worries, I will just give you a bit of context. I'm actually reading a book for my work, and I need to understand how I can predict time for Ostwald ripening for emulsions.
Basically I want to know, if there is a drop of oil, with a radius $a$, in how much time it will completely diffuse through water.
Let's take an oil drop of radius $a$, $C(r,t)$ will represent the concentration of the oil in water at a radius $r$ from the centre of the particle at a given time. The law of mass conservation gives,
$$\frac{\partial C}{\partial t} + div\left(\vec{j}\right) =0$$
where $j$ is the density flux.
By Fick's law of diffusion one may find,
$$\frac{\partial C}{\partial t} + \nabla^2 C =0$$
Now here is my problem, in the book I'm reading, they give four references that I can't open (DOI...) while saying, when in steady state, the solution of the above equation is,
$$C(r)=\frac{a\left[C(a)-C(0)\right]}{r} + C(0)$$
The laplacian is supposed to be given in function of the radius $r$.
Have you any idea from where this solution (which makes sense physically meaning)$^{[1]}$ comes from, because for me an equation like the Laplace equation has a solution such as, $$C(r) = Ar +B$$ with $A$ and $B$ two constants.
$[1]$ : The concentration is maximum in the drop of oil (there is just oil) and around the drop the more you're far the less they will have oil.
The given references in the book for the solution in stationnary state are as follow,
Kabalnov A. S., Pertzov A. V., Shchukin E. D. (1987) Ostwald ripening in emulsions. 1. Direct observations of Ostwald ripening in emulsions J. Colloid Interface Sci. 118: 590
Kabalnov A. S., Makarov K. N., Pertzov A. V., Shchukin E. D. (1990) Ostwald ripening in emulsions. 2. Ostwald ripening in hydrocarbon emulsions : Experimental verification of equation for absolute rates. J. Colloid Interface Sci. 138 : 98
Kabalnov A. S., Shchukin E. D. (1992) Ostwald ripening theory : applications to fluorocarbon emulsion stability. Adv. Colloid Interface Sci. 38 : 69
Taylor P. (1998). Ostwald ripening in emulsions. Adv. Colloid Interface Sci. 75 : 107
Since it is steady-state, we have: $\dfrac{\partial C}{\partial t}=0$
So the following equation will lead to $ \nabla^2 C =0$.
For Laplace equation in spherical coordinates when changes are only happening by $r$, we have:
$\nabla^2 C = \dfrac{1}{r^2} \dfrac{\partial}{\partial r}(r^2 \dfrac{\partial C}{\partial r})=0 \to \dfrac{\partial}{\partial r}(r^2 \dfrac{\partial C}{\partial r})=0 \to r^2 \dfrac{\partial C}{\partial r}=c_1 \to \dfrac{\partial C}{\partial r}=\dfrac{c_1}{r^2}$
$ \to C(r)= \dfrac{c_1}{-r}+c_2 $
Or as $c_1$ is undetermined for now, simply:
$ \to C(r)= \dfrac{c_1}{r}+c_2 $
Substitute the boundary conditions values to find $c_1$ and $c_2$ and you will finally reach to: $C(r)=\frac{a\left[C(a)-C(0)\right]}{r} + C(0)$