Physics 301 12-Dec-2003 33-1Time Independent Solutions of the Diffusion EquationIn some cases, we’ll be interested in the time independent solution of the d iffusionequation. Why would be interested in this? Answer: when a s ystem reaches a steadystate, the temper ature distribution must satisfy∇2τ = 0 ,as well as whatever boundary con ditions exist. Suppos e for example that a high (τ1) andlow (τ2) tempera ture reservoir are connected by a bar with a non-uniform cr oss section.The transport equation applies, so Ju= −K∇τ . If we are in a steady state, then anyenergy entering a thi n slab of the bar must leave the other side of the s lab at the samerate. In other words, the energy flux (not flux density) must be constant along the lengthof the b ar. SoJ(x)A(x) ≈ constantwhere x measures pos ition along the bar and A(x) is the cross sectional area at positionx. This says that, approximately,∇τ ∝1A(x).As a specific example, suppose that the bar is in the form of a truncated con e, withA(x) = A1(x2/x21) where A1is the cross section at x1, and x1is the value of the xcoo rdinate at the τ1end of the bar and x2> x1is the value of the x coordina te at the τ2end of the bar. We also suppose that√A1 x1. Basically we have a bar that i s a wedgeof a spherical shell of inner and outer rad ii x1and x2. Th e temperature distribution mustsatisfy ∇2τ = 0 subject to the boundary conditions which a re: the temperature at x1isτ1, the temperature at x2is τ2and there in no energ y flux through the sides of the bar,only the ends . A little thought shows that if we actually had a complete spherical shell ofinner and outer r adii x1and x2at temperatures τ1and τ2, then the flux density would beradi al and the solution to this problem would a lso solve the present bar problem. So wewant the solutions for ∇2τ which are sphericall y symmetric. These are the same as thesolutions for a spherically symmetric electric potential in a charge free region. We knowwhat these are. There can be a po int charge at the center of the sphere (Φ = 1/r) and aconstant. So the temperature as fun ction of x must beτ (x) =C1x+ C2,A little algebra finds the constants,τ (x) =x1x2(τ1− τ2)x(x2− x1)+x2τ2− x1τ1x2− x1.Note that ∇τ ∝ 1/x2as we wanted .Copy rightc 2003, Princeton University Physics Depa rtment, Edward J. GrothPhysics 301 12-Dec-2003 33-2Continuity Equation for MassConsider a medium in which the mass density is ρ(r, t) and the medium moves witha bulk velocity v(r, t). Then the mass flux density a t r at time t (we’ll drop the explicitnotation) isJρ= ρv .Jus t like other quantities, this sa tisfies a continuity equation, or in this case a mass con-servation equation, which says that mass flux leaving a small volume is jus t the rate ofchange of mass in the volume∇ · Jρ+∂ρ∂t= ∇ · (ρv) +∂ρ∂t= 0 .Soun d Waves in a GasIn this section we’ll work ou t the wave equation for sound waves in an ideal gas. Tostart with, let’s consider a plane wave in the pressure. The change in pressure from itsequilibrium value isδp = δp0ei(k · r −ωt),where δp0is the amplitude of the wave. The wave is a ssumed to be small enoug h thatonly first o rder quantities n eed to be considered . Of, course, this means that there isa sinusoid ally varying pressure gradient which accelera tes the gas. Suppose there is apressure gradient in the x direction and consid er a small volume of gas in a box with crosssectional area A and length dx. T hen the net force on the gas in the box is −(∇p)xA dx.The momentum of the gas in the box at time t and position r is ρ(t, r)v(t, r)A dx. Therate of change of thi s momentum must be the force. (Note that we are ignorin g viscousforces and assuming smooth flow!) When calculating the rate of change, we must notethat the box is moving with velocity v, so at time dt, r → r + v dt. We cancel out theA dx and take the time derivative and find∂ρvx∂t+ v · ∇(ρvx) = −(∇p)x,or generalizing to a press ure gradient in an arbitrary direction,∂ρv∂t+ v · ∇(ρv) = −∇p .Now, in equili brium, ρ i s a cons tant, p i s a constant, a nd v = 0 . When a wave is present,all of these have sma ll oscillatory components. On the left hand side, we can ignore thev ·∇(ρv) term altogether since it has two powers of velocity and i s second order small. InCopy rightc 2003, Princet on University Physics Departme nt, Edward J. GrothPhysics 301 12-Dec-2003 33-3the other term, we can igno re the oscillatory wave in ρ since it i s multiplied by the firstorder small velocity. To ma ke all this explicit, let’s writeρ = ρ0+ δρ , p = p0+ δp , v = 0 + δv , τ = τ0+ δτ ,and so on. The quantities prefi xed by δ are the small o scillatory quantities and the othersare the equilibrium quantities and we wi ll drop products of oscillatory quantities in ourexpressions. In particular the force equation above becomesρ0∂(δv)∂t= −∇(δp) .If we have an ideal gas,p = nτ =ρτm,soρ0∂(δv)∂t= −τ0m∇(δρ ) −ρ0m∇(δτ ) .We also haveδU + p0δV = τ0δσ ,which can be rewritten in terms of unit volumeδUV0+p0V0δV = τ0δσV0= τ0δˆσ ,where ˆσ is the entropy per unit volume or the entropy density. Now divide by dt andremember that because we have an id eal gas, the energy density per unit volume isˆCVτ ,ˆCV∂(δτ )∂t+p0V0∂(δV )∂t= τ0∂(δˆσ)∂t.Note also tha t ρV = constant, so δρ/ρ0= −δV /V0andˆCV∂(δτ )∂t−p0ρ0∂(δρ)∂t= τ0∂(δˆσ)∂t.Also τ0∂(δˆσ)/∂t is the heat added per unit volume per unit time. This must equal theheat flow which is K∇2(δτ ). Note that earlier, in derivin g the thermal diffusion equation,we ignored the p dV term. This was OK, becau se p dV work is usually quite small for asolid.At this point, it may be useful to summarize the equations we’ve obtained so far. Wehave the continu ity equation for mass, wh ich written in terms of our small quantities, isρ0∇ · (δv) = −∂(δρ)∂t,Copy rightc 2003, Princet on University Physics Departme nt, Edward J. GrothPhysics 301 12-Dec-2003 33-4and the force equationρ0∂(δv)∂t= −τ0m∇(δρ ) −ρ0m∇(δτ ) ,and the thermal diffusion equation∂(δτ )∂t−p0ˆCVρ0∂(δρ)∂t= D∇2(δτ ) .Note that D = K/ˆCVand the thermal diffusion equation is the same as the one wediscussed ear lier without the density term. What we’re goin g to do is take the timederivative of the