Question

A Schematic Model of Red Blood Cell Production The following problem deals with the number of red blood cells (RBCs) circulating in the blood. Here we will present it as a discrete problem to be modeled by differ- ence equations, though a different approach is clearly possible. In the circulatory system, the red blood cells (RBCs) are constantly being de stroyed and replaced. Since these cells carry oxygen throughout the body, their num. ber must be maintained at some fixed level. Assume that the spleen filters out and destroys a certain fraction of the cells daily and that the bone marrow produces a number proportional to the number lost on the previous day. What would be the cell count on the nth day? To approach this problem, consider defining the following quantities: R. = number of RBCs in circulation on day n, M. = number of RBCs produced by marrow on day n, f = fraction RBCs removed by spleen, y = production constant (number produced per number lost). It follows that equations for R. and M, are R+1 = (1-P)R. + M.. M.-1 = yfR, (47) Red blood cell production (a) Explain the equations given in Section 1.9, and show that they can be re- duced to a single, higher-order equation. (b) Show that eigenvalues are given by $1,2 = (1 - ) = V(1 - )2 + 4yf (c) and determine their signs and magnitudes. For homeostasis in the red cell count, the total number of red blood cells, R. should remain roughly constant. Show that one way of achieving this is by letting 1 = 1. What does this imply about y? Using the results of part (c) show that the second eigenvalue is then given by Az = -f. What then is the behavior of the solution R = A + BA? (d)

Answer 1

Answer: Rati =(1-F) Rnt mn smoti - ny en Im=nf RAI .: Rn+1=(1-f]ent vf pri =) Let Ratl=x2 1 Rrah @ 12 (1-4)-NF-o so it is can be reduced in a single highen order equation © => 12-(1-6) X-rf=0 ax2+bx+c=0 > x=-bt bzuac za 1-C1-F) 4-vf=0 11.2 = (1-P) + VC-1) 2 Curt) 1112 - C-FDA CI-F)2+urF

= bis C1-F ).|(1-6)2 +urf 2 - x2 = (1-F)- 1-f24urf when li=1 1- C1-F ** TC1-F32401F 2 2-1-ftrCo-F)2+urF I+F = VC1-692 +urf squring on both sides => C1+4)2 = (1-P)24urf X + 2F+ f -/26+ ff turf UF=UTF rail » when 12=-f -f= C1-F)-1C1-)24urf - -2f=1-F- CI-F12 turf - VCI-F)24 UVF = 1+f Trail 1. Rn- AYÇı+){\C1-f92440f2"+Bq (1--|(1+)2}ufn"- 2