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{SECT 0 {PARA 260 "" 0 "" {TEXT -1 24 "Passage Across Membranes" }}
{PARA 256 "" 0 "" {TEXT 261 9 "Jim Herod" }}{PARA 257 "" 0 "" {TEXT
262 21 "School of Mathematics" }}{PARA 258 "" 0 "" {TEXT 263 12 "Georg
ia Tech" }}{PARA 259 "" 0 "" {TEXT 264 21 "herod@math.gatech.edu" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 172 " In \+
many biological processes, the passage of substances across thin membr
anes is critical for sustaining life. For example, there is the passag
e of oxygen through the " }{TEXT 256 7 "alveoli" }{TEXT -1 85 " of the
lungs into the circulatory system, and the passage of nourishment wit
hin the " }{TEXT 257 5 "villi" }{TEXT -1 382 " of the intestines. Inde
ed, the day-to-day processes that a cell must perform require that nut
rients and oxygen move into the cell and that waste and carbon dioxide
move out. As another example, recall that a bi-product of the metabol
ism of protein is ammonia. In terrestrial animals, ammonia is converte
d to urea and removed as waste by passage across membranes within the \+
kidneys." }}{PARA 0 "" 0 "" {TEXT -1 424 " Transport of substances
across a membrane may be entirely passive and dependent upon purely p
hysical factors, or it may be an active process which is dependent upo
n energy transforming reactions. Consider the passage of water over a \+
camping filter, with retention on the filter of small (and large) soli
d material, or there is the passage of oxygen across the placenta from
the mother's hemoglobin to that of the fetus." }}{PARA 0 "" 0 ""
{TEXT -1 297 " In this work sheet, we assume that we have a thin w
alled membrane which allows the passage of a solute and that the fluid
transporting the solute does not pass across the membrane. We suppose
that the membrane forms the wall of a vessel in the shape of a unifor
m cylindrical vessal of length " }{TEXT 266 1 "L" }{TEXT -1 12 " and r
adius " }{TEXT 265 1 "a" }{TEXT -1 30 ". These assumptions lead to a \+
" }{TEXT 268 11 "toy problem" }{TEXT -1 211 ". We call this a toy prob
lem for it is unlikely that the assumptions are realistic in nature. R
ather, toy problems serve to give understanding and suggest appropriat
e questions to be asked in the real situation." }}{PARA 0 "" 0 ""
{TEXT -1 119 " The reader may provide a picture of how the situati
on could be visualized with the following syntax using Maple 6." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 192 "J:=tubeplot([t,t+1,0],t=0..1,radiu
s=1,orientation=[75,80],\n color=BLUE):\nK:=tubeplot([t,t+1,0]
,t=0..1,radius=0.5,orientation=[75,80],\n tubepoints=20,color=
RED):\ndisplay(\{J,K\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 " \+
Before describing how a model for this situation may be constructed,
we discuss units. Suppose the cylinder has length L with units " }
{XPPEDIT 18 0 "meters" "6#%'metersG" }{TEXT -1 80 ". Let C(x) represen
t the concentration of the solute in the cylinder at distant " }{TEXT
267 1 "x" }{TEXT -1 62 " along the tube from one end. Units for concen
tration will be " }{XPPEDIT 18 0 "grams" "6#%&gramsG" }{TEXT -1 1 "/"
}{XPPEDIT 18 0 "meter^3" "6#*$%&meterG\"\"$" }{TEXT -1 171 ". The con
centration of the solute as it enters the cylinder is C(0), and is C(L
) as it exits. Suppose that the concentration of the solute in the sur
rounding medium is S " }{XPPEDIT 18 0 "grams" "6#%&gramsG" }{TEXT -1
1 "/" }{XPPEDIT 18 0 "meter^3" "6#*$%&meterG\"\"$" }{TEXT -1 219 " and
that S is greater than C(x). For this model, suppose the concentratio
n in the surrounding medium is not affected by losses across the membr
ane. We suppose the solution passes along the vessel at a steady rate \+
of V " }{XPPEDIT 18 0 "meters^3" "6#*$%'metersG\"\"$" }{TEXT -1 1 "/"
}{XPPEDIT 18 0 "sec" "6#%$secG" }{TEXT -1 192 " and suppose that the \+
fluid in the cylinder will gain solute from the surrounding medium by \+
way of osmotic forces. The derivation of the equation is accomplished
by describing the amount as " }{XPPEDIT 18 0 "grams" "6#%&gramsG" }
{TEXT -1 1 "/" }{XPPEDIT 18 0 "sec" "6#%$secG" }{TEXT -1 40 " of the s
olute moving into a segment x (" }{XPPEDIT 18 0 "meters" "6#%'metersG
" }{TEXT -1 18 ") of the cylinder." }}{PARA 0 "" 0 "" {TEXT -1 5 " \+
" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 26 "Derivation of the equation"
}}{PARA 0 "" 0 "" {TEXT -1 71 " The rate of diffusion across the w
alls of the cylinder depends on " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 15 "" 0 "" {TEXT -1 79 "the area of the wall of the vessel acros
s which the diffusion takes place, and " }}{PARA 15 "" 0 "" {TEXT -1
118 "the difference between the concentration of the solute inside the
cylinder and the concentration outside the cylinder." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 " To begin, conside
r a small portion of the cylinder of length " }{XPPEDIT 18 0 "Delta" "
6#%&DeltaG" }{TEXT -1 83 "x. We approximate the amount A of diffusion \+
into this portion. Units for A will be " }{XPPEDIT 18 0 "grams" "6#%&g
ramsG" }{TEXT -1 1 "/" }{XPPEDIT 18 0 "sec" "6#%$secG" }{TEXT -1 1 ":
" }}{PARA 0 "" 0 "" {TEXT -1 22 " A(x + " }{XPPEDIT 18
0 "Delta" "6#%&DeltaG" }{TEXT -1 21 "x) - A(x) = k 2 " }{TEXT
259 1 "a" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT
-1 15 "x ( S - C(x) )," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 9 "where 2 " }{TEXT 260 1 "a" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 465 "x is the surface are
a of the small portion of the cylinder under consideration, and k is a
constant characteristic of the properties of the wall of the vessel a
nd of the particular diffusing substance that is being considered. The
difference S - C(x) is the difference in the concentration of the so
lute across the membrane. At the same time a volume V of the solution \+
passes through the small section and as it does, the concentration wil
l increase by the amount " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 21 " C(x + " }{XPPEDIT 18 0 "Delta" "6#%
&DeltaG" }{TEXT -1 14 "x) - C(x) = " }{XPPEDIT 18 0 "(A(x+Delta*x)-A
(x))/V" "6#*&,&-%\"AG6#,&%\"xG\"\"\"*&%&DeltaGF*F)F*F*F*-F&6#F)!\"\"F*
%\"VGF/" }{TEXT -1 12 " = k 2 " }{TEXT 258 1 "a" }{TEXT -1 1 " "
}{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 17 "x ( S - C(x) )/V." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "In the l
imit, we may write" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 27 " " }{XPPEDIT 18 0 "diff(C(x),x)
= (2*Pi*a*k/V*(S-C(x))" "6#/-%%diffG6$-%\"CG6#%\"xGF**.\"\"#\"\"\"%#P
iGF-%\"aGF-%\"kGF-%\"VG!\"\",&%\"SGF--F(6#F*F2F-" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "This is t
he differential equation we are to solve." }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Solution for the equati
on" }}{PARA 0 "" 0 "" {TEXT -1 27 " We solve the equation " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+
" }{XPPEDIT 18 0 "diff(C(x),x) = (2*Pi*a*k/V*(S-C(
x))" "6#/-%%diffG6$-%\"CG6#%\"xGF**.\"\"#\"\"\"%#PiGF-%\"aGF-%\"kGF-%
\"VG!\"\",&%\"SGF--F(6#F*F2F-" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "This equation was derived
in the previous section." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56
"dsolve(\{diff(C(x),x)=2*Pi*a*k*(S-C(x))/V,C(0)=Co\},C(x));" }}}{PARA
0 "" 0 "" {TEXT -1 58 "To illustrate, we take S=5, a=1/2, Co=1, V=3, a
nd k = 1/5." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "C:=unapply(rh
s(%),x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 ""
{TEXT -1 48 "We expect the graph of the solution to begin at " }
{XPPEDIT 18 0 "C[0]" "6#&%\"CG6#\"\"!" }{TEXT -1 75 " and rise to the \+
level S. A graph could be drawn with the following syntax." }{TEXT -1
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "plot(subs(\{S=5,a=1/2
,Co=1,V=3,k=1/5\},C(x)),x=0..20);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 74 "The level of the con
centration approaches that of the surrounding medium. " }}}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Exercise for
the reader " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 73 "1. What is the total area of the cylinder, with a = 1/2 and wi
th L = 20?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 124 "2. To what extent must V be increased in order that the level \+
of the concentration at x = 20 is half what it is with V = 3? " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 173 "3. To wh
at extent must the radius be increased so that when V is at the level \+
you computed, then the level of the concentration is the same as it wa
s with a = 1/2 and V = 3?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 108 "4. Draw a graph of the solution for the different
ial equation with V as was computed above and with a = 1/2." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 22 "Absorption and release" }}
{PARA 0 "" 0 "" {TEXT -1 742 "In this next section, we mimic the gathe
ring of solute over a long stretch of the vessel and the removal of th
e solute over a shorter section. One might view the purpose of this se
ction to be the acquisition of an elementary understanding of the acti
on of the kidneys in removing urea from a terrestrial animal. (The blo
od absorbs this ammonia by-product as waste while circulating through \+
the body, and the urea is removed in the kidneys.) We diagram this sit
uation by having a cylinder alternately having a concentrations 5 and \+
0 for the medium outside the cylinder. In our model, we suppose ammoni
a is absorbed in a segment of length 20 and a removal of ammonia in a \+
segment of length 2. This absorption/removal repeats over a long perio
d." }}{PARA 0 "" 0 "" {TEXT -1 443 " We will see that there will b
e an oscillation of the concentration inside the vessel. The oscillati
on of the concentration will approach a periodic solution, increasing \+
when it lies in the section with a high concentration, and decreasing \+
when it lies in the section with a low concentration. For this toy pro
blem, we initially take a zero concentration within the cylinder and a
llow the concentration to move to the periodic equilibrium." }}{PARA
0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 273 " We illus
trate how the solution evolves by drawing a graph of the solution for \+
the differential equation with the following initial values. First we \+
restart to re-initialize all variables. Here, the concentration outsid
e the the long portion of the cylinder is 5 units." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 35 "S:=5: a:=1/2: Co:=0: V:=3: k:=1/5: " }}}{PARA 0 "" 0
"" {TEXT -1 63 " Here is the solution over the first interval from
0 to 20." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "dsolve(\{diff(C
(x),x)=2*Pi*a*k*(S-C(x))/V,C(0)=Co\},C(x));\nC1:=unapply(rhs(%),x);" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 241 "
Over the next interval, we take the initial value Co to be the va
lue of the previous solution at the end -- at x = 20 -- and take the s
urrounding medium to have concentration 0 for the solute. We assume th
is length of the cylinder is 2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 24 "Co:=evalf(C1(20));\nS:=0;" }}}{PARA 0 "" 0 "" {TEXT -1 78 " \+
Here is the solution for the differential equation on this next inte
rval." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "dsolve(\{diff(C(x),
x)=2*Pi*a*k*(S-C(x))/V,C(20)=Co\},C(x));\nC2:=unapply(rhs(%),x);" }}}
{PARA 0 "" 0 "" {TEXT -1 102 " In essentially the same process, we
cross five more intervals of input and removal of the solute." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Co:=evalf(C2(22));\nS:=5;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "dsolve(\{diff(C(x),x)=2*Pi
*a*k*(S-C(x))/V,C(22)=Co\},C(x));\nC3:=unapply(rhs(%),x);" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
25 "Co:=evalf(C3(42));\nS:=0; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 80 "dsolve(\{diff(C(x),x)=2*Pi*a*k*(S-C(x))/V,C(42)=Co\},C(x));\nC
4:=unapply(rhs(%),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Co
:=evalf(C4(44));\nS:=5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "
dsolve(\{diff(C(x),x)=2*Pi*a*k*(S-C(x))/V,C(44)=Co\},C(x));\nC5:=unapp
ly(rhs(%),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Co:=evalf(
C5(64));\nS:=0; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "dsolve(
\{diff(C(x),x)=2*Pi*a*k*(S-C(x))/V,C(64)=Co\},C(x));\nC6:=unapply(rhs(
%),x);" }}}{PARA 0 "" 0 "" {TEXT -1 25 " We now draw a graph." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 162 "plot(\{[x,C1(x),x=0..20],[x
,C2(x),x=20..22],[x,C3(x),x=22..42],\n [x,C4(x),x=42..44],[x,C5(x
),x=44..64],[x,C6(x),x=64..66]\},\n x=0..66,y=0..5,color=BLACK);
" }}}{PARA 0 "" 0 "" {TEXT -1 76 " Notice that we quickly moved to
a solution that appears to be periodic." }}}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Exercise for the reader."
}}{PARA 0 "" 0 "" {TEXT -1 71 " Observe what was the value of the \+
solution at the final end point." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 14 "evalf(C6(66));" }}}{PARA 0 "" 0 "" {TEXT -1 269 " Repeat t
he above calculations but begin at x = 0 with a positive number less t
han 5. Show that C6(66) is about the same value as with the above calc
ulation. This may convince you that we quickly come to a periodic solu
tion independently of where we start at x = 0." }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 30 "An analytic, peri
odic solution" }}{PARA 0 "" 0 "" {TEXT -1 99 " We seek a periodic \+
solution for the equation. Take the same values for the physical situa
tion." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "a:=1/2: V:=3: k:=1/5:" }}}{PARA 0 "
" 0 "" {TEXT -1 121 " Make a piecewise function that has the value
5 on the interval [0, 20] and has the value 0 on the interval [20, 22
]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "outsd:=x->piecewise(0 \+
< x and x < 20,5,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plo
t(outsd(x),x=0..22);" }}}{PARA 0 "" 0 "" {TEXT -1 235 " We do not \+
know what the initial value should be so that the value of the solutio
n at x = 22 is same as the value at x = 0, That is, we want C(22) = C(
0). To find how to choose C(0) so that this happens, we let C(0) be th
e generic " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 24 " and sol
ve the equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "dsolve(
\{diff(C(x),x)=2*Pi*a*k*(outsd(x)-C(x))/V,C(0)=alpha\},C(x));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Cper:=unapply(rhs(%),x);" }}
}{PARA 0 "" 0 "" {TEXT -1 90 " Now, we require that the solution C
has the property that C(22) = C(0) and solve for " }{XPPEDIT 18 0 "al
pha" "6#%&alphaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 28 "solve(Cper(22)=alpha,alpha);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 6 "Co:=%;" }}}{PARA 0 "" 0 "" {TEXT -1 109 " Using th
is initial value, we solve the differential equation confident that we
have a periodic solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
63 "dsolve(\{diff(C(x),x)=2*Pi*a*k*(outsd(x)-C(x))/V,C(0)=Co\},C(x));
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Cper:=unapply(rhs(%),x)
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "plot(Cper(x),x=0..22,v
iew=[0..22,0..5]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 23 "E
xercise for the reader" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 128 "Plot the graph of the solution for the equation beg
inning at C(0) = 0 as above and the periodic solution on the same grap
h for " }{XPPEDIT 18 0 "0 <= x;" "6#1\"\"!%\"xG" }{XPPEDIT 18 0 "` ` \+
<= 66;" "6#1%\"~G\"#m" }{TEXT -1 0 "" }{TEXT -1 1 "." }}}{PARA 0 "" 0
"" {TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }