(an easy problem which will be graded). later in the quarter we will spend some time solving the diffusion equation op(r, t) = dv4p(r, t), at where d is called the diffusion constant. for now, let us consider a solution which represents a steady state solution (i. e apat = 0) and which has spherical symmetry. in that case the above reduces to 1d (20p) = 0 p2 dr dr / a) obtain the solution of this second-order differential equation. it will have two arbitrary constants. b) application: diffusion from a large distance to a spherical absorber. consider p(r) to represent the density of some particles which are diffusing through some media. think of them as food for a cell. their density at large distance (infinity) is po. there is a spherical absorber of radius r centered at the origin, which means p(r) = 0. (this absorber represents the cell.) show that the particular solution that satisfies these two condition is p(r) = po(1 – r/r). an interesting result: the current density (the number of particles passing through unit area per unit time) is given by j(r) = -ddp/dr = -dpor/r2. (the minus sign means that the flow is inward.) the magnitude of the current at a distance r is i = 47r2|j(r)= 47 dpor. note that this result says that the rate at which particles are absorbed is proportional to the radius of the sphere, not to the area of the sphere as would be the case if we were dealing with light being absorbed. of course light does not travel by diffusion!