Math 19. Lecture 25Summary of Advection/DiffusionT. JudsonFall 20051 Advection and DiffusionThe advection equation is given by∂u∂t= −c∂u∂x+ f(u). (1)The diffusion equation is given by∂u∂t= µ∂2u∂x2+ f(u). (2)• c and µ are determined exp erimentally.• The advection equation is used when the particle motion is due to themotion of the ambient fluid.• The diffusion model is used when the particles move randomly.• Sometimes advection and diffusion are both at work:∂u∂t= µ∂2u∂x2− c∂u∂x+ f(u).• For either the advection or diffusion equation, there may be many solu-tions. The solution for a particular problem depends heavily on initialand boundary conditions.12 Equilibrium SolutionsAn equilibrium solution to (1) or (2), is a solution u = u(x, t) that is inde-pendent of time. Thus, u = ue(x) must either obey−cduedx+ f(ue) = 0orµd2uedx2+ f(ue) = 0,plus the relevant boundary conditions.3 StabilityLet uebe an equilibrium solution to∂u∂t= µ∂2u∂x2+ f(u)∂∂xu(t, 0) =∂∂xu(t, L) = 0.The solution ue(x) is a linearly stable solution toµd2uedx2+ f(ue) = 0ddxue(0) =ddxue(L) = 0if and only if there is no pair (g, λ) , where g(x) is some function that isnot identically zero for 0 ≤ x ≤ L, where λ ∈ R, and where the followingconstraints are satisfied.• λ ≥ 0• λg = µd2dx2+ z(x)g•dgdxx=0=dgdxx=L= 0A solution is unstable if there is even one such pair (g, λ) that obeys theabove conditions.24 An Example for Adve ctionThe equation∂u∂t= −∂u∂x−u(1 − u)100has a time independent solutionue(x) =1ex/100+ 1that models bacteria concentrations in a river that is downstream from asewage treatment plant, where x is the distance downstream from the plant.5 An Example for DiffusionThe equation∂u∂t=∂2u∂x2+ u − u2might model the fish concentration in a square lake that is stocked at bothends. Consider a rectangular lake of constant width and depth. Suppose thatfish are dumped into the lake at either end such that the concentration ofthe fish at the ends of the lake are three fish per 100 m3. If −π/2 ≤ x ≤ π/2andu(t, −π/2) = u(t, π/2) = 3, (3)then we have an equilibrium solutionue(x) =3cos x + 1.In order to verify stability, we must check solutions to the equationλg =d2gdx2+cos x − 51 + cos xg, (4)withg( −π/2) = g( π/2) = 0.If g satisfies these conditions, thenu(t, x) = eλtg( x) + ue(x)will satisfy (3). Given the complexity of ue, solving ( 4) may be difficult oreven impossible.36 The Maximum PrincipleWe can use the Maximum Princ iple to analyze (4). Here is the idea. Assumethat you have fo und a pair (λ, g) satisfying (4) such that g is not identicallyzero. We will argue that λ < 0. If this is the case, then we have a stablesolution.• First, g must have a maximum and a minimum on [−π/2, π/2]. Fur-thermore, the maximum and the minimum cannot be the same. If theywere, then g would have to be a constant f unction. Since we know gat the endpoints, g must b e identically zero. In this case, which weassumed could not be the case.• Suppose that g > 0 at its maximum.1Then g must b e concave downat this point and d2g/dx2≤ 0 here. Since g > 0 and −1 ≤ cos x ≤ 1,we know thatcos x − 51 + cos xg < 0.Thus, λg < 0, which tells us that λ < 0.• On the other hand, suppo se that g ≤ 0 and has no positive maxi-mum. Then g must have a negative minimum, and d2g/dx2≥ 0 at theminimum. Since g < 0 andcos x − 51 + cos x< 0,we know that t he right side of (4) is positive. Therefore, λ must be lessthan zero in order for the left side to be positive.Our argumen t depends hea v i l y on the boundary conditions and f(u).Readings and References• C. Taubes. Modeling Differen tial Eq uations in Biology. Prentice Hall,Upper Saddle River, NJ, 2001. Chapter 20.1We divided this into cases: g > 0 and g ≤
View Full Document
Unlocking...