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I am following a similar derivation as found in the beginning of this paper "Quantitative aspects of the growth of (charged) silica spheres" by A.P. Philipse.

This paper calculates the growth of a charged silica sphere due to silica monomers in the solution overcoming the interparticle potential and diffusing to the surface of the sphere where it contributes to the growth. They do this by solving for the monomer particle distribution around the sphere:

$$J=4\pi r^2 D_m \left[ \frac{dn(r)}{dr} + \frac{n(r)}{kT} \frac{dV_{12}(r)}{dr} \right]$$

Here $D_m$ is the diffusivity of the monomer in the solution, and $n(r)$ is the monomer concentration surrounding the sphere. The derivative $dn(r)/dr$ is from Fick's law, $V_{12}(r)$ is the potential field (i.e. electrostatic + van der Waals as per DLVO theory) between the sphere and a monomer particle, and so $-dV_{12}/dr$ is the force exerted by the sphere on the monomer. Here the "flux" $J$ is not actually a flux, but is in units of $t^{-1}$, or number of monomers that merge with the sphere per unit of time, instead of a true flux which would be units of $t^{-1} L^{-2}$ or monomers that merge with the sphere per area per time. Of course the true flux can be determined by dividing by the $4\pi r^2$ term in the equation, but by keeping the equation like this the $J$ is independent of radius and so can be treated as a constant while solving for $n(r)$.

This equation can be solved with the boundary conditions of $\lim_{r\to\infty} n(r)=n_\infty$ and $\lim_{r\to\infty} V_{12}(r)=0$. Using the standard integrating factor method for solving 1st order ODEs but directly solving with a definite integral, the integrating factor $\mu$ is

$$\mu = \exp \left( \int \frac{1}{kT} \frac{dV_{12}(r)}{dr} dr \right) = \exp \left( \frac{V_{12}(r)}{kT} \right)$$

And the solution to the ODE is

$$n(r) = n_\infty \exp{\left(-\frac{V_{12}(r)}{kT}\right)}\left[1+\frac{J}{n_\infty D_m} \int_{\infty}^{r} \frac{\exp{(V_{12}(\xi)/kT)}}{4\pi \xi^2}d\xi \right]$$


Anyway, what I'm trying to do is a very similar derivation, but instead solve for the distribution $n(z)$ of monomer particles near a planar wall instead of a sphere. The equation should be very similar:

$$J=D_m \left[ \frac{dn(z)}{dz} + \frac{n(z)}{kT} \frac{dV(z)}{dz} \right]$$

In this case the flux $J$ is a true flux with units of $t^{-1} L^{-2}$, or monomers that merge with the planar surface per area per time. Because there is no generation or consumption of the monomer spheres within the doman $[0,\infty)$ the flux $J$ must also be constant and so not a function of $z$. Solving this equation for the monomer concentration $n(z)$ the result is:

$$n(z) = n_\infty \exp{\left(- \frac{V(z)}{kT} \right)} \left[ 1+ \frac{J}{n_\infty D_2} \int_{\infty}^{z} \exp{\left(\frac{V(\xi)}{kT}\right)}d\xi \right]$$

So the answer is identical to the original problem in spherical coordinates, the only difference being that there is no $4 \pi r^2$ in the denominator of the integrand. That does pose a problem though: does this equation converge for finite values of $z$? It seems to me that the integrand will blow up so long as z is finite.


The actual potential field $V(z)$ is a DLVO potential field, the sum of van der Waals potential and electrostatic potential from solving the Poisson-Boltzmann equation, which gives the electrical field in a screening electrolytic medium. The vad der Waals potential for a spherical particle and a planar surface is normally given as

$$V_W = -\frac{AR}{6z}$$

Where $A$ is the Hamaker constant, $R$ is the radius of the spherical particle, and $z$ is the distance between the sphere and the wall surface. However this is only accurate in the case that $z \ll R$, which is not true in my case. Instead I am using a more general formulation accurate for all values of $R$ and $z$:

$$V_w = -\frac{A}{12}\left( \frac{2R}{z} + \frac{2R}{z+2R} + 2\ln{\left( \frac{z}{z+2R} \right)} \right)$$

For the electrostatic potential I am using the equation given in the Handbook of Surface and Colloid Chemistry, 3rd ed. 2009 Taylor & Francis, Eq. 7.206 (pp. 254), which gives the electrostatic potential between two spheres as

$$V_{el}=\frac{\pi \varepsilon \varepsilon_0 R_1 R_2 }{R_1 + R_2}\left( \left( \psi_1 + \psi_2 \right)^2 \ln{\left( 1+e^{-\kappa z} \right)} + \left( \psi_1 - \psi_2 \right)^2 \ln{\left( 1-e^{-\kappa z} \right)}\right)$$

Where $\psi_i$ is the surface voltage potential of the sphere surface, $R_i$ is the sphere radius, $\kappa^{-1}$ is the Debye length, and $\varepsilon$ the relative permittivity and $\varepsilon_0$ the vacuum permittivity. In my case I am dealing with a sphere and a plane and both the plane and sphere have the same surface potential, so this simplifies to:

$$V_{el}=4 \pi \varepsilon \varepsilon_0 R \psi^2 \ln{\left( 1+e^{-\kappa z} \right)}$$

So the overall potential is $V(z) = V_w(z) + V_{el}(z)$.

A plot of the potential field for typical parameters I am dealing with is here:

DLVO potential example

So the potential field $V(z)$ certainly does die out for large $z$, as is obvious from the equations.

So since the potential field $V(z)$ is obviously going to $0$ as $z\rightarrow \infty$, does it physically make sense for

$$\int_{z}^{\infty}\exp{\left( \frac{V(\xi)}{kT} \right)}d\xi$$

to diverge for a planar/Cartesian coordinate space and finite values of $z$? Is this implying that it's impossible to have a finite concentration of particles in this field or something?

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    $\begingroup$ 1. "How exactly does the antiderivative become a standard integral here?" The Fundamental Theorem of Calculus. 2. "how I can resolve this problem and solve for the particle distribution?" Use the definite integral on the rhs, $\int_\infty^z{\exp\left(\frac{V(\zeta)}{kT}\right)d\zeta}$, it goes to 0 as $z\rightarrow \infty$. $\endgroup$ – udrv Apr 27 '16 at 4:02
  • $\begingroup$ @urdv, thanks for taking a look at this. For (1.) I read up some more on the fundamental theorem of calculus (specifically the second one) which states that $$\frac{d}{dx} \int_{a}^{x} f(t)dt = f(x)$$ Which means for a general IVP $y'=f(x)$, $y(x_0)=y_0$, then the solution is $y(x) = y_0 + \int_{x_0}^{x}f(t)dt$. Thanks for pointing me to that. For (2.), that enabled me to work the problem better, I'm adding an edit to the main post, but I still have some questions there. $\endgroup$ – Derek Apr 27 '16 at 16:31
  • $\begingroup$ Welcome. As for convergence problems, that may still be an issue. Do you have some general expression for $V(z)$? Or at least general features, asymptotic behavior, etc? $\endgroup$ – udrv Apr 27 '16 at 18:06
  • $\begingroup$ @urdv, yes, the $V(z)$ is a DLVO potential, which is the sum of van der Waals and electrostatic potentials. The VDW potential has an asymptote going to $-\infty$ at $z=0$, but since molecular contact occurs at $z \approx 3 Å$ that isn't an issue. Both the VDW and ES potentials go to $0$ as $z\rightarrow \infty$ as you would expect. I can post the full form of the $V(z)$ potential field if you like, but I'm guessing that knowing the asymptotic behavior as it goes to infinity is sufficient. $\endgroup$ – Derek Apr 27 '16 at 20:49
  • $\begingroup$ I wouldn't mind the full form, but I definitely have to know more about its behavior at large $z$ before I can draw any conclusions: is it negative, does it vary like $z^{-a}$ for some $a>0$, etc? Does it have any local extrema? $\endgroup$ – udrv Apr 28 '16 at 0:36

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