A remark on incoherent feedforward circuits as change detectors and feedback controllers

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by  Eduardo D. Sontag

A remark on incoherent feedforward circuits
as change detectors and feedback controllers
EduardoD.Sontag
RutgersUninversity
Abstract
6 This note analyzes incoherent feedforward loops in signal processing and control. It studies the
1 responsepropertiesofIFFL’stoexponentiallygrowinginputs, bothforastandardversionoftheIFFL
0 and for a variation in which the output variable has a positive self-feedback term. It also considers a
2
negativefeedbackconfiguration,usingsuchadeviceasacontroller. Ituncoversasomewhatsurprising

n phenomenoninwhichstabilizationisonlypossibleindisconnectedregionsofparameterspace, asthe
a controlledsystem’sgrowthrateisvaried.
J

0
3
1 Introduction

]
Y
S This note derives several theoretical results regarding the use of incoherent feedforward loops (IFFL’s) in
. signalprocessingandcontrol. Wewillstudythesystem:
s
c
[
x˙ = −ax+bu (1a)

1 u
y˙ = c −δy (1b)
v x
2
u˙ = (λ−κy)u (1c)
6
1
0 aswellasamodifiedsysteminwhichthereisalsoanautocatalytictermin(1b):
0
. u Vyn
2 y˙ = c − δy + (1b’)
0 x Kn+yn
6
1 whichrepresentsapositivefeedbackofthey variableonitself. Theconstantsa,b,c,δ,κ,V,K arepositive
:
v (butλisallowedto benegative), dotindicatesd/dt, nistypicallyan integer> 1thatrepresentsmolecular
i cooperativity, and the scalar functions of time x = x(t), y = y(t), and u = u(t) take positive values.
X
(It is easy to verify that, for any positive initial conditions, solutions remain positive for all times.) Of
r
a course, setting V = 0 allows seeing (1b’) as a special case of (1b), but it is more interesting to treat the
non-autocatalyticcasebyitself.
Wewillseparatelystudythefirsttwoequations(1ab)(or(1ab’)whenthereisanautocatalyticterm),viewing
u = u(t)asanexternalinputtotheIFFLdescribedby(1ab)(or(1ab’)),andviewingy = y(t)asanoutput
or response of the system. Later, we “close the loop” by letting u be described by (1c), thinking of it as a
variablethatiscontrolledbyy throughanegativefeedbackwithgainκ,andwhich,conversely,feedsback
intotheIFFLthroughthexvariable. Inthatcontext,westudythefullsystem(1abc)(or(1ab’c)).
The motivation for this work is the potential role that these motifs might play in immunology [3]. In that
context,onemightviewthexvariableasrepresentingthelevelofactivityofaregulatoryinhibitorycompo-
nent(suchasapopulationofTregcellsataparticularinfectionsiteorinacertaintumormicroenvironment),
y as the level of activity of an immune response component (such as cytotoxic T cells), and u as a popula-
tion of pathogens or the volume of a tumor, which might grow exponentially (if λ > 0) in the absence of
1
immuneaction,butwhichiskilledatarateproportionaltotheimmuneresponse. Thefeedbackintoxand
yrepresentstheactivationofboththeresponseandoftheregulatorymechanisminresponsetotheinfection
or tumor. As remarked in [3], a very interesting feature of the IFFL controller is its capability of detecting
change as well as the fact that the level of activity is proportional to the rate of growth of the input, which
mayaccountfortoleranceofslow-growinginfectionsandcancersaswellasWeber-likelogarithmicsensing
and“foldchangedetection”ofinputs.
Inanimmunologicalcontext,autocatalyticfeedbackmightbeimplementedbyacytokine-mediatedrecruit-
ing of additional immune components, or by autocrine stimulation. This results in an excitable system,
which allows y to “lock” into a high state of activity given a sufficiently rapid rate of change in its input.
Changing the growth rate λ of the pathogen or tumor, while fixing all other parameters, results in elimina-
tionofuforsmallgrowthratesλ,andinproliferationasλincreases. Thisis,ofcourse,obvious. However,
and very surprisingly, it may happen in this model that further increase of the growth rate λ, that is, when
presented with a more aggressive pathogen or tumor, leads to the eventual elimination of the pathogen or
tumor. This might be intuitively interpreted as a higher growth rate triggering locking of the immune re-
sponseatahighervalue. Anevenlargerincreaseinλleadsagaintoproliferation. Inotherwords,thepattern
“elimination,proliferation,elimination,proliferation”canbeobtainedsimplybygraduallyincreasingλ.
Remark1.1 In the system (1abc), and in particular in the system (1ab), one may assume without loss of
generality that a = b = c = 1. This is because we may eliminate these parameters by rescaling variables.
Indeed,substituting
b c 1 δ λ cκ
x = x∗, y = y∗, t = t∗, δ∗ = , λ∗ = , κ∗ = , (2)
a b a a a ab
intosystem(1abc),oneobtains:
dx∗
= −x∗+u (3a)
dt∗
dy∗ u
= −δ∗y∗ (3b)
dt∗ x∗
du
= (λ−κy∗)u (3c)
dt∗
(cid:50)
2 IFFL’s responses to various classes of inputs
Let us consider the system (1ab), a differentiable function u = u(t) viewed as an external input or forcing
function, and any (positive) solution (x(t),y(t)) corresponding to this input. We are interested first in
understandinghowthegrowthrateoftheinputaffectstheasymptoticvaluesoftheoutputvariabley.
Wedenotethederivativeoflnu(t)withrespecttotasfollows:
u˙(t)
v(t) :=
u(t)
anditslimsupandliminfast → ∞
µ = liminfv(t), µ = limsupv(t).
t→∞ t→∞
We assume that v is bounded, and thus both of these numbers are finite. We also introduce the following
function:
u(t)
p(t) := .
x(t)
2
Since
p˙ = u˙/x−ux˙/x2 = (u/x)[u˙/u−x˙/x] = (u/x)[u˙/u−(−ax+bu)/x] = (u/x)[u˙/u+a−bu/x],
wehavethatpsatisfiesthefollowingODEwithinputv:
p˙ = p(a+v−bp). (4)
Lemma2.1 Letubeadifferentiableinputtosystem(1ab)witha = b = c = 1. Withtheabovenotations,
max{0,1+µ} ≤ liminfp(t) ≤ limsupp(t) ≤ max{0,1+µ} (5)
t→∞ t→∞
Proof.Sincea = b = c = 1,
p˙ = p(1+v−p).
To prove the upper bound, we consider two cases, 1 + µ < 0 and 1 + µ ≥ 0. In the first case, let
ε := −(1+µ) > 0; the definition of µ gives that, for some T ≥ 0, 1+v(t) < −ε/2 for all t ≥ T. It
follows that p˙ ≤ p(−ε/2 − p) for all t ≥ T. Thus, p˙ < 0 whenever p > 0, from which it follows that
limsup p(t) = lim p(t) = 0. Supposenowthat1+µ ≥ 0. Pickanyε > 0andaT = T(ε) ≥ 0
t→∞ t→∞
suchthatv(t) ≤ µ+εforallt ≥ T. Forsucht,p˙ = p(1+v−p) ≤ p(1+µ+ε−p). Thisimpliesthat
p˙ < 0 whenever p(t) > 1+µ+ε, which implies that limsup p(t) ≤ 1+µ+ε. Letting ε → 0, we
t→∞
concludethatlimsup p(t) ≤ 1+µ.
t→∞
We next prove the lower bound. Pick any ε > 0 and a T = T(ε) ≥ 0 such that v(t) ≥ µ − ε for all
t ≥ T. Thus p˙ = p(1 + v − p) ≥ p(1 + µ − ε − p) for all t ≥ T. This implies that p˙ > 0 whenever
p(t) < 1 + µ − ε (recall that p(t) > 0 for all t, since by assumption u(t) > 0 and x(t) > 0 for all t).
Therefore liminf p(t) ≥ 1 + µ − ε, and letting ε → 0 we have liminf p(t) ≥ 1 + µ. Since
t→∞ t→∞
p(t) ≥ 0forallt,wealsohaveliminf p(t) ≥ max{0,1+µ}. Thiscompletestheproof.
t→∞
Inparticular,ifv(t) → µast → ∞thenµ = µ = µ,sowehaveasfollows.
Corollary2.2 Ifv(t) → µast → ∞then lim p(t) = max{0,1+µ}. (cid:50)
t→∞
Fortheoriginalsystem(1ab),wehaveasfollows.
Proposition2.3 Considerasolutionof(1ab),withadifferentiableu(t) > 0asinputandx(t) > 0,y(t) > 0.
Assumingthatv = u˙/uisbounded,wehave:
c (cid:8) (cid:9) c
max 0,a+µ ≤ liminfy(t) ≤ limsupy(t) ≤ max{0,a+µ} (6)
bδ t→∞ t→∞ bδ
Proof. We first assume that a=b=c=1. Let p := liminf p(t) and p := limsup p(t). Equation
t→∞ t→∞
(1b) can be written as y˙ = p − δy. This is a linear system forced by the input p = p(t). Pick any
ε > 0. Then there is some T = T(ε) such that p − ε < p(t) < p + ε for all t ≥ T. For such t,
y˙(t) > 0 whenever y(t) < (1/δ)(p−ε) and y˙(t) < 0 whenever y(t) > (1/δ)(sup+ε). It follows that
(1/δ)(p−ε) ≤ y(t) ≤ (1/δ)(p+ε)forallt ≥ T. Lettingε → 0weconcludethat
p/δ ≤ liminfy(t) ≤ limsupy(t) ≤ p/δ (7)
t→∞ t→∞
and the inequalities (6) follow when a=b=c=1. To deal with general parameters, we recall that (2ab) are
obtainedwithx = bx∗,y = cy∗,t = 1t∗,andδ∗ = δ. Notethatt∗ → ∞ifandonlyift → ∞. Thus(7)
a b a a
holdsforp∗ = u/x∗ = (b/a)p,y∗,andδ(cid:63) inplaceofp,y,andδ. Similarly,(5)holdsforp∗ = u/x∗ and
µ∗ = liminfv∗(t∗), µ = limsupv∗(t∗),
t→∞ t→∞
3
wherev∗ = du/dt∗ = (1/a)v,soµ∗ = (1/a)µandµ∗ = (1/a)µ. Therefore,
u
liminfy(t) = liminf cy∗(t∗) ≥ cp∗ = c p∗ = ac p∗ = ac max(cid:8)0,1+µ∗(cid:9) = c max(cid:8)a+µ(cid:9) .
t→∞ t∗→∞ b bδ(cid:63) bδ/a bδ bδ bδ
Asimilarremarkappliestolimsup,andtheresultfollows.
c
Corollary2.4 Ifv(t) → µast → ∞then lim y(t) = max{0,a+µ}. (cid:50)
t→∞ bδ
Threeparticularcasesare:
ac
• Whenu(t)hassub-exponentialgrowth,meaningthatdlnu/dt ≤ 0,thenlimsupy(t) ≤ .
bδ
t→∞
ac
• Inparticular,ifu(t) = α+βtislinear,thenµ = 0andthus lim y(t) = .
t→∞ bδ
c
• Ifu(t) = βeµt isexponential,then lim y(t) = max{0,a+µ}.
t→∞ bδ
In conclusion, when u is constant, or even with linear growth, the value of the output y(t) converges to a
constant, which does not depend on the actual constant value, or even the growth rate, of the input. For
constantinputs,thisiscalledthe“perfectadaptation”property. If,instead,ugrowsexponentially,theny(t)
convergestoasteadystatevaluethatisalinearfunctionofthelogarithmicgrowthrate.
Remark2.5 ApossiblealternativeIFFLmodelisthatinwhichy followsthisequation:
y˙ = cu−δxy. (8)
insteadof(1b). Thismodelrepresentsadifferentwayofimplementingthenegativeeffectofxony,through
degradation instead of inhibition of production A reduction to a = b = c = 1 is again possible. Now the
substitutions
b c 1 bδ λ cκ
x = x∗, y = y∗, t = t∗, δ∗ = , λ∗ = , κ∗ = , (9)
a a a a2 a a2
into(1a-8-1c)transformthesysteminto:
dx∗
= −x∗+u (10a)
dt∗
dy∗
= u−δ∗x∗y∗ (10b)
dt∗
du
= (λ∗−κ∗y∗)u (10c)
dt∗
Consideramodelthatuses(8)insteadofequation(1b)andsupposethat,forsomeγ > 0,u(t) ≥ γ > 0for
all t ≥ 0 (for example, u(t) = βeµt or u(t) = α+βt). Then (6) again holds, as does Corollary 2.4. This
is because we one may rewrite y˙ = cu−cδy as y˙ = x(cu/x−δy), and, provided that, for some ξ > 0,
x(t) > ξ > 0forallt,solutionshavethesameasymptoticbehaviorasfor(1b). Ontheotherhand,fromthe
factthatp(t) = u(t)/x(t)isbounded, weknowthat, forsomeγ(cid:48) > 0, forallt,x(t) ≥ γ(cid:48)u(t) > γ(cid:48)γ > 0.
(cid:50)
4
3 IFFL’s as feedback controllers
c
Aswe remarked, inthecase ofexponentialinputsu(t) = βeµt, lim y(t) = y¯ = max{0,a+µ}. This
t→∞ bδ
holds both for (1ab) and for the combination (1a)-(8). Now suppose that, in turn, u(t) satisfies equation
(1c),whichmeansmeansthatv(t) = λ−κy(t),andthereforeµ = lim v(t) = λ−κy¯. Thisgivesan
t→∞
implicitequationfortherateµ:
cκ
µ = λ−κy¯= λ− max{0,a+µ}. (11)
bδ
Wenowsolvethisequation.
Supposefirstthata+µ ≥ 0. Inthatcase,asolutionhastosatisfyµ = λ− cκ(a+µ)andthereforethereis
bδ
auniqueµ ≥ −athatsolvestheequation,namely:
λbδ−cκa
µ = . (12)
bδ+cκ
Observethatµ ≥ −aimpliesthatλbδ−cκa ≥ (−a)(bδ+cκ) = −abδ−acκandthereforeλbδ ≥ −abδ,
or λ ≥ −a. (And conversely, λ ≥ −a implies λbδ ≥ −abδ and so λbδ −cκa ≥ −abδ −cκa and hence
µ ≥ −a.) So, if λ < −a, there is no such solution. Now we look for a solution with a+µ ≤ 0. Such an
µ must satisfy µ = λ−0 = λ. In summary, when λ < −a, the unique solution of (11) is (12), and when
λ ≥ −aitisµ = λ.
Notethatwhen
caκ > bδλ (13)
(which happens automatically when λ < 0) the formula (12) gives that m < 0, that is, u(t) → 0 as
t → +∞. Conversely, ifcaκ < bδλ, thenµ > 0andsou(t) → ∞ast → +∞. Qualitatively, thismakes
sense: alargefeedbackgainκ,orasmallgrowthrateλintheabsenceoffeedback,leadstotheasymptotic
vanishingoftheuvariable.
In addition, from the formula y¯ = c max{0,a+µ} we conclude the following piecewise linear formula
bδ
forthe dependenceof thelimitof theoutput ontheparameterλthatgives thegrowth rateofuwhenthere
isnofeedback:

0 if λ < −a

y¯ = c(a+λ) (14)
if λ ≥ −a.

bδ+cκ
Theseconsiderationsprovidehelpfulintuitionabouttheclosed-loopsystem,buttheydonotprovethat(13)
isnecessaryandsufficientforstability,nordotheyshowthevalidityof(14)fortheclosed-loopsystem. The
reason that the argument is incomplete is that there is no a priori reason for u(t) to have the exponential
formu(t) = βeµt. Wenextprovidearigorousargument.
3.1 Analysisoftheclosed-loopsystem
Theorem1 Supposethat(x(t),y(t),u(t)isa(positive)solutionof(1abc),anddefine
v(t) := u˙(t)/u(t) = λ−κy(t),
p(t) := u(t)/x(t),
y¯byformula(14),whichwerepeathere:

0 if a+λ < 0

y¯ = c(a+λ)
if a+λ ≥ 0

bδ+cκ
5
p¯:= (δ/c)y¯. and

λ if a+λ < 0

v¯ = c(a+λ)
λ−κ if a+λ ≥ 0.

bδ+cκ
Then:
lim y(t) = y¯
t→∞
lim p(t) = p¯
t→∞
lim v(t) = v¯.
t→∞
(cid:26)
0 if acκ > bδλ
lim u(t) =
t→∞ ∞ if acκ < bδλ.
Proof. Substituting v(t) = λ − κy(t) into (4), we have the surprising and very useful fact that there is a
closedsystemofjusttwodifferentialequationsforpandy:
p˙ = p(a+λ−κy−bp) (15a)
y˙ = cp−δy. (15b)
(Thissystemcouldbeviewedasanon-standardpredator-preyofsystem,whereybehavesasapredatorand
pasaprey.) Inalloftherealplane,therearetwoequilibriaofthissystem,oneatp = y = 0andtheother
δ(a+λ) c(a+λ)
at p = , y = . The second equilibrium point is in the interior of first quadrant if and only if
bδ+cκ bδ+cκ
a+λ > 0.
WestartbyevaluatingtheJacobianmatrixofthelinearizedsystem. Thisis:
(cid:18) (cid:19)
a+λ−κy−2bp −pκ
J =
c −δ
which,whenevaluatedatp = y = 0,hasdeterminant−δ(a+λ)andtracea+λ−δ,andwhenevaluated
at(p¯,y¯)hastrace
−bδ(a+λ)
−δ
cκ+bδ
and determinant δ(a+λ). Thus, when a+λ > 0, the trace is negative and the determinant is positive, so
theequilibrium(p¯,y¯)isstable,and(0,0)isasaddlebecausethedeterminantoftheJacobianisnegativeat
that point. When instead a+λ ≤ 0, the only equilibrium with non-negative coordinates is (0,0), and the
determinantoftheJacobianispositivethere,whilethetraceisnegative,sothisequilibriumisstable.
We note that, in general, if have shown that there is a limit v(t) → v¯as t → ∞ then u(t) → 0 as t → ∞
if v¯ < 0 and u(t) → ∞ as t → ∞ if v¯ > 0 Indeed, in the first case there is some T ≥ 0 so that for
t ≥ T, v = u˙/u < v¯/2, meaning that d(e−v¯t/2u(t))/dt ≤ 0, and hence e−v¯t/2u(t) ≤ e−v¯T/2u(T), so
u(t) ≤ ev¯(t−T)/2u(T) → 0(sincev¯< 0). Similarly,inthesecondcaseweusethatthereissomeT ≥ 0so
thatfort ≥ T,v = u˙/u > v¯/2,meaningthatd(e−v¯t/2u(t))/dt ≥ 0,andhencee−v¯t/2u(t) ≥ e−v¯T/2u(T),
sou(t) ≥ ev¯(t−T)/2u(T) → ∞(sincev¯> 0).
Consider first the case a + λ ≤ 0. Then p˙ = p(a + λ − κy − bp) ≤ p(−κy − bp) < 0 for all p > 0,
and therefore p(t) → p¯ = 0 as t → ∞. We may now view the linear system y˙ = cp − δy as a one-
dimensional system with input p(t) → 0, which implies that also y(t) → y¯ = 0. In turn, this implies that
v = λ−κy → v¯= λ < 0. Bythegeneralfactprovedearlieraboutlimitsforu(t),weknowthatu(t) → 0
ast → ∞. Thiscompletestheproofwhena+λ ≤ 0.
So we assume from now on that a+λ > 0. We will show that, in this case, all solutions with p(t) > 0
andy(t) > 0globallyconvergetotheuniqueequilibrium(p¯,y¯). Oncethatthisisproved,itwillfollowthat
6
v(t) → v¯= λ−κy¯. Now,thisvalueofv¯,fory¯pickedasin(14)(casea+λ ≥ 0),coincideswithµin(12),
λbδ−cκa. So v¯ < 0 if cκa > λbδ and v¯ > 0 if λbδ > cκa, and this provides the limit statement for u(t),
bδ+cκ
completingtheproof.
We next show global convergence. A sketch of nullclines (see Fig. 1 for a numerical example) makes
convergence clear, and helps guide the proof. Consider any P ≥ (a+λ)/b and any Y ≥ cP/δ and the
rectangle[0,P]×[0,Y](seeFig.1). Onthesidesofthisrectangle,thefollowingpropertieshold:
Figure 1: Phase plane for (15), with several representative trajectories plotted. Nullclines are the y axis,
correspondingtothestablemanifoldof(0,0),andthelinesgivenbyy = (a+λ−bp)/κ(dashedredline)
and y = cp/δ (dashed magenta line). In this plot, we picked a = b = c = λ = δ = 1 and κ = 2, but the
qualitative picture is similar for all valid parameter values. With these values, trajectories converge to the
equilibrium(p¯,y¯) = (2/3,2/3). Shownalsoisaninvariantregion[0,P]×[0,Y]withP = Y = 2.5(green
dash-dottedlinesandaxes).
1. Ontheset{0}×(0,Y),p˙ ≥ 0,becausep˙ = 0.
2. Ontheset{P}×(0,Y),p˙ ≤ 0,becausep˙ = p(a+λ−bP) ≤ 0,bythechoiceofP.
3. Ontheset(0,P)×{0},y˙ ≥ 0,becausey˙ = cp > 0.
4. Ontheset(0,P)×{Y},y˙ ≤ 0. becausey˙ = cp−δY ≤ cP −δY ≤ 0bythechoiceofY.
5. Atthecornerpoint(0,0),p˙ ≥ 0,y˙ ≥ 0,becausep˙ = y˙ = 0.
6. Atthecornerpoint(0,Y),p˙ ≥ 0,y˙ ≤ 0,becausep˙ = 0,y˙ = −δY < 0.
7. Atthecornerpoint(P,0),p˙ ≤ 0,y˙ ≥ 0,becausep˙ = p(a+λ−bP) ≤ 0,y˙ = cP > 0.
7
8. Atthecornerpoint(P,Y),p˙ ≤ 0,y˙ ≤ 0,becausep˙ = p(a+λ−bP −κY) < p(a+λ−bP) ≤ 0,
y˙ = cP −κy ≥ 0.
These properties imply that the vector field points inside the set at every boundary point and therefore it is
forward-invariant,meaningthateverytrajectorythatstartsinthissetremainsthereforallpositivetimes[1].
TherestoftheproofofstabilityusesthePoincare´-BendixsonTheoremtogetherwiththeDulac-Bendixson
criterion. Notethat, foranyinitialconditionξ = (p(0),y(0))onecanalwayspickalargeenoughvalueof
P andY sothat(p(0),y(0)) ∈ [0,P]×[0,Y]. Theinvariancepropertyguaranteesthattheomegalimitset
ω+(ξ) is a nonempty compact connected set, and the Poincare´-Bendixson Theorem insures that such a set
isoneofthefollowing: (a)theequilibrium(0,0),(b)aperiodicorbitintheinteriorofthesquare,or(c)the
equilibrium(p¯,y¯)[2]. Notethatahomoclinicorbitaround(0,0)cannotexist,becausetheunstablemanifold
of this equilibrium is the entire y axis. For the same reason, if ξ has positive coordinates, ω+(ξ) (cid:54)= (0,0).
Therefore, all that we need to do is rule out periodic orbits. Consider the function ϕ(p,y) = 1/p. The
divergenceofthevectorfield
(cid:32) (cid:33)
1(p(a+λ−κy−bp)) (cid:18) a+λ−κy−bp (cid:19)
p =
1(cp−δy) c−δy/p
p
is ∂a+λ−κy−bp + ∂c−δy = −b−δ/p,whichhasaconstantsign(negative). TheDulac-Bendixsoncriterion
∂p ∂y
[2]thenguaranteesthatnoperiodicorbitscanexist,andtheproofiscomplete.
4 Adding positive feedback
Wenowstudyamodelinwhichthereisanadditionalautocatalyticpositivefeedbackonyvariable. Wefirst
consider the open loop system (1ab’), and then discuss the full feedback system (1ab’c), which we repeat
hereforconvenience:
x˙ = −ax+bu (16a)
u Vyn
y˙ = c − δy + (16b)
x Kn+yn
u˙ = (λ−κy)u (16c)
4.1 Open-loopsystemwithautocatalysis
Wefirstconsideronlytheopen-loopsystem(16ab),inwhichu = u(t)isseenasaninputfunction(stimulus)
andy asanoutput(response).
Forappropriateparameters,andassumingthattheHillexponent(cooperativityindex)nisgreaterthanone,
thesystem
Vyn
y˙ = q − δy + (17)
Kn+yn
admitsmorethanonesteadystate. (Incontrast,ifthereisnoautocatalyticfeedback,V = 0,thenthereisa
uniquesteadystate,y¯= q/δ.) Letusfixallparametersexceptq,whichwetemporarilyviewasabifurcation
parameter. Adjusting the value of q, one may obtain a low steady state, multiple steady states, or a higher
steadystate. Asanillustration, picka = b = c = 1, δ = 3, n = 2, V = 10, andK = 2. Fig,2showsthe
right-handsideof(17)plottedforq = 0.8andq = 1.1. Forthelattervalueofq,thereislargersteadystate.
(Intermediatevaluestypicallygiveasystemwithtwostablestatesandoneunstablestate.)
u(t)
Let us now write q(t) = c in the system (16ab). Suppose that we consider an input u which has a
x(t)
step increase at time t = 0, from u(t) = u for t < 0 to u(t) = u for t ≥ 0. Suppose also that
− +
8
Figure 2: Plots of f(y) = q −δy + Vyn , with a = b = c = 1, δ = 3, n = 2, V = 10, and K = 2,
Kn+yn
comparingq = 0.8(brown)andq = 1.1(blue). Thesteadystatechangesfromalowtoahighvalue.
x(0) = x = (b/a)u , that is, that the system at time t = 0 has an internal steady state preadapted to u .
0 − −
Sincex(t)isacontinuousfunctionoftime,wehavethat,forsmalltimest > 0,x(t) ≈ x andu(t) = u ,
0 0
andthusq(t) ≈ αu /u ,whereα = ac/b. Thismeansthatthevalueofq(t)for0 ≤ t (cid:28) 1isproportional
+ −
to the “fold change” in the input. On the other hand, as t → ∞, x(t) → b/a, so q(t) → ac/b = α. In
thesystemwithnoautocatalyticeffect(V = 0),thedifferentialequationy˙ = q−δy hasauniqueglobally
asymptotically stable equilibrium, and therefore y(t) → q/δ = α/δ. That is to say, there is complete
adaptation: after a step increase in the input u, y responds in a way that transiently depends on the fold
change,butiteventuallyreturnstoitsadaptedvalue.
Ontheotherhand,ifthereisanautocatalyticfeedbackterm(V (cid:54)= 0),theinitialinputq(t)tothey-subsystem
may trigger an irreversible transition to a different state y than the adapted value. Since the initial value of
q(t)dependsonthefoldchangeoftheinput,thisimpliesthatfordifferentrangesoffold-changemagnitudes,
ymightswitchtodifferentstates,andremainthereevenaftertheexcitationgoesaway. Asanexample,using
the same parameters a = b = c = 1, δ = 3, n = 2, V = 10, and K = 2 as earlier, Fig, 3 shows how
a step change in the input can result in an irreversible locking to a higher activation state, for the system
withfeedback,comparedwiththesystemwithoutfeedback,whichdoesnotswitchbuthasonlyatransient
changeinactivity.
Figure 3: Response to an input stepping from u=1 to u=2 (fold change of input is 2). Comparing system
withnopositivefeedbacktosystemwithpositivefeedback. Statex(t)isthesameinbothsystems,soonly
one panel is shown. Parameters are a = b = c = 1, δ = 3, n = 2, V = 10, and K = 2 in system with
feedback,substitutingV = 0insystemwithoutfeedback.
4.2 Closed-loopsystemwithautocatalysis
We now turn to the full feedback system (16abc). Just as in the case in which there was no autocatalytic
terms,wemayagainreducetoatwo-dimensionalsystemwrittenintermsofp = u/xandy. Thesystemis
now:
p˙ = p(a+λ−κy−bp) (18a)
Vyn
y˙ = cp−δy+ . (18b)
Kn+yn
9
Forappropriateparameterregimes,thereisauniquepositivesteadystate(p¯,y¯). Specifically,forn > 1the
√
derivative of Vyn attains its maximum at y = (n−1)1/nK = K/ 3 when n = 2, and the derivative is
Kn+yn n+1
√
3V 3 there. Thus,thefunction
8K
(cid:18) Vyn (cid:19)
g(y) = a+λ−κy−(b/c) δy− ,
Kn+yn
√
whose roots determine the nonzero equilibrium values of y, has derivative ≤ −κ−bδ/c+ 3V 3. Thus,
8K
when √
3V 3
< κ+bδ/c
8K
the function g is strictly decreasing and therefore (in the nontrivial case a+λ > 0), since g(0) > 0 and
g(y) → −∞asy → ∞,thereisauniquezeroy¯. SeeforexamplethephaseplanedrawninFig.4.
Figure 4: Phase-plane for system (18), with a = 0.8, b = 1, c = 0.1, δ = 1, n = 2, V = 1.95, K = 1,
κ = 20 λ = 25. The y-nullcline is cp − δy + Vyn = 0 (dot-dashed orange). The p-nullcline has
Kn+yn
two components: p = 0 (the y-axis) and the line y = (a+λ−bp)/κ (dashed red). Three representative
trajectories are shown (solid blue). Notice the vertical-looking motion of one trajectory near the y-axis:
along such solutions, p(t) = u(t)/x(t) stays ≈ 0 for a time interval, after which this ratio converges to p¯.
Grayarrowsindicatedirectionsofmovementinphaseplane. Theequilibriumpoint(p¯,y¯)issuchthatp <
0.8andthus,sinceu˙ = (λ−κy)u,u(t)behaveslikeαeµtforlarget,whereµ = λ−κy¯= bp−a = p−0.8,
wehaveµ < 0(elimination).
Aremarkablefeatureemergesforthissystem. Whendoesu(t) → 0ast → ∞,correspondingtoelimination
of a pathogen or tumor, in the motivating context of immunology? When does u(t) → ∞ as t → ∞,
corresponding to proliferation? Note that, if (p(t),y(t)) → (p¯,y¯) as t → ∞, then, since u˙ = (λ−κy)u,
u(t)behaveslikeαeµt forlarget. Ontheotherhand,atsteadystatea+λ−κy¯−bp¯= 0,whichmeansthat
µ−λ−κy¯= bp¯−a. Therefore:
a
p¯ < ⇒ u(t) → 0 ast → ∞
b
a
p¯ > ⇒ u(t) → ∞ ast → ∞.
b
10