Question

4. Advection-diffusion on a bounded domain: Suppose that a large population of ants is following a pheromone trail to a new nesting site, located at z = L. Ants originate frorn their nest at z-0, with flux q(0,t)-α-constant. Once an ant reaches the new nesting site, it stays there, and is removed from the domain. In the domain x E [0, L], the density of ants ρ(x, t) obeys the advection-diffusion equation (a) Explain why the boundary conditions are&L--:-Has-a (b) Find the equilibrium solution poo(x) (the solution as t-o). Use the boundary condi tions to solve for the arbitrary constants Ci and C2 (c) For α = 1, u = 0.01, D 0.005, and L 1, plot the equilibrium solution on a computer. (d) For the same set of parameter values, find the total number of ants in the domain at equilibrium

Answer 1

$\partial$p/$\partial$t = D$\partial$²p/$\partial$x² - u$\partial$p/$\partial$x

a). where D is a constant.  We can write down a general solution describing how it will disperse , but we need to specialise to our specific case to get a physical applicable answer.

Remember the tubes were sealed at both ends ; this means the flow substance at either end of the tube has to be zero. It turns out this is proportional to the first derivative, $\partial$_xc of the concentration. So we would like this to be zero at the both ends.

$\partial$c/$\partial$x$\mid$_x=0x = 0,  $\partial$c/$\partial$x$\mid$_x=L = 0.

In this case it is of the form , c(x,0) ~ $\delta$(x) since we injectb all point. A condition which a quantity that varies throughout a given space or enclosure must fulfill at every point on the boundary of that space especially when the velocity of a fluid at any point on the wall a rigid conduit is necessary parallel to the wall. For example, if the independent variable is time over the domain [0,1], a boundary value problem would specify values fpr y(t) at both t = 0 and t=1 , whereas an initial value problem would specify a vlaue of y(t) and y'(t) at time t = 0.

b). In other words it makes some sense that we should expect that as t$\rightarrow$$\infty$ our tempratue distribution, u (x,t) should behaves as,

u (x,t) = u_E (x)

liim

where, u_E (x) is called the equilibium temperature. Note as well that is should still satisfy the heat equation and boundary conditons. It would not satisfy the initial condition however because it is the temperature distribution as t $\rightarrow\infty$ whereas the initial condition is at t = 0. So, the eqilibirium temperature distribution should satisfy,

0 = d²uE/dx² uE (0) = T₁ uE (L) = T₂

This is a really ordinary differential equation. If we integrate twice we get

uE = (x) = c₁x + c₂

and applying Boundary condition

uE(x) = T₁ + (T₂ - T₁)x/L.