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APh 162 – Biological Physics Laboratory Diffusion of Solid Particles Confined in a Viscous Fluid1 I realize that some of you have little or no experience with partial differential equations (PDEs). However we think it is important to understand the nature of the diffusion process, especially as it relates to biology, to this end I would like to go through the theory behind the experiment you areabout to do. This reading is certainly of the crash-course variety, so feel free to ask Rob, Hernan, or me any questions. Solution of the 2D Diffusion Equation: The 2D diffusion equation allows us to talk about the statistical movements of randomly moving particles in two dimensions. By random, we mean that we cannot correlate the movement at one moment to movement at the next, or in other wordsthere is no deterministic/predictive power over the exact motion of the particle. The movement of each individual particle moving in a Brownian (diffuse) way does not follow the diffusion equation. However, many identical particles each obeying the same boundary and initial conditions share some statistical properties dealing with their spatial and temporal evolution. It is those statisticalproperties that the diffusion equation captures. The function P(x,y,t) gives the probability of finding a perfectly average particle in the small vicinity of the point (x,y) at time t (we call this a probability distribution). Such properties are for instance the average position of a statistical particle as a function of time. The evolution of some systems does follow the diffusion equation outright:when you put a drop of dye in a beaker, there are millions of dye molecules each exhibiting Brownian motion, but as a group they exhibit the smooth, well-behaved statistical features of the diffusion equation. Consider the two-dimensional diffusion equation in Cartesian coordinates:
∇2P − 1 ∂P ∂ 2 P ∂ 2 P 1 ∂P =0 → + − =0 D ∂t ∂x 2 ∂y 2 D ∂t

The diffusion equation can be derived from theprobabilistic nature of Brownian motion described as random walks (speak with me if you really want to see the derivation). The constant D is the diffusion coefficient whose nature we will explore in a moment, but for now we are solving a math problem. Notice we have no powers or complicated functions of any derivative of the function P. Since the derivatives of P only appear to first power, we callthis a linear PDE. Also notice that the equation does not mix x’s, y’s, or time within a single term, that is there are no multiplicative, divisional, or functional terms in the equation between the independent variables, we call this feature separability of the PDE. These two properties, linearity and separability, make the problem vastly more tractable than many other PDE’s encountered in physics.I am going to attempt to walk you through a challenging but very powerful method for solving PDE’s using the Fourier Transform (FT). Many of you may know that the FT is used in signal analysis and manipulation, but it was first used by Fourier to solve this problem. Consider the ˆ following integral relations that define the 2D FT in Cartesian coordinates. We will call the function P the FT of ouroriginal function P:
∞ ∞

ˆ P(k x , k y , t ) = P ( x, y , t ) =

−∞ −∞ ∞ ∞ i 2π ( k x ⋅ x + k y ⋅ y ) −∞ −∞

∫ ∫e

− i 2π ( k x ⋅ x + k y ⋅ y )

P( x, y, t )dxdy

∫ ∫e

ˆ P (k x , k y , t )dk x dk y

Notice the symmetry in going forward and backward in the transform. This is because switching between the normal form of the problem and what we call Fourier Space, where theproblem exists after the FT, are physically identical. We will require one more fact familiar from calculus, namely integration by parts:

∫ udv = uv − ∫ vdu
b a a a



Using these facts, let’s examine the spatial derivatives of the diffusion equation, where we consider the second derivative to be the function of interest. We can integrate these second derivatives by parts, identifying u...
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