aplicaciones a la biologia.pdf

8/9/2019 APLICACIONES A LA BIOLOGIA.pdf

1/141

UNIVERSIDAD DE CHILEFACULTAD DE CIENCIAS FISICAS Y MATEMATICASDEPARTAMENTO DE INGENIERIA MATEMATICA

PROBLEMAS INVERSOS Y CALCULO DE VARIACIONES:

APLICACIONES A LA BIOLOGIA Y AL ANALISIS DE IMAGENES

TESIS PARA OPTAR AL TITULO DE GRADO DE DOCTOR EN CIENCIAS DE LAINGENIERIA, MENCION MODELACION MATEMATICA

RODRIGO ANTONIO LECAROS LIRA

PROFESORES GUIAS:CARLOS CONCA ROSENDEJAIME H. ORTEGA PALMA

PROFESOR CO-GUIA:LIONEL ROSIER

MIEMBROS DE LA COMISION:MATIAS COURDURIER BETTANCOURT

ZIOMARA GERDTZEN HAKIM

Este trabajo ha sido parcialmente financiado por Conicyt

SANTIAGO DE CHILEAGOSTO 2012


2/141ii


3/141

Resumen

El tema central de esta tesis, para optar al grado de Doctor en Ciencias de la Ingeniera conmencion en Modelamiento Matematico, es el estudio de los llamados problemas inversos, elmodelamiento matematico de fenomenos de interaccion fluido estructura y la implementacionnumerica de un algoritmo de segmentacion.

Este trabajo se divide en tres partes. En primer lugar (Captulo2), estudiamos un pro-blema inverso en biologia. Nos interesamos en la posibilidad de recuperar la densidad deun canal ionico (canal CNG) en el cilio olfatorio, solo midiendo la corriente producida alactivar el sistema mediante la difusion de un agente en el interior del cilio. Este problemaes modelado como un problema inverso en que interviene una ecuacion de Fredholm lineal,de primer tipo, con un nucleo difusivo. Consideramos una aproximacion para el nucleo deloperador, obteniendo los siguientes resultados:

a) Problema de identificabilidad: Bajo diferentes regularidades de la funcion de densidad,

logramos establecer la inyectividad del operador bajo estudio.

b) Problema de estabilidad: Tanto en el caso general, usando la transformada de Mellin,como en un caso particular de los parametros del problema, logramos establecer laestabilidad, bajo diferentes normas, es decir, la continuidad del inverso del operador.

d) Problema de reconstruccion: En el caso de una eleccion apropiada de los parametrosdel problema, logramos construir la funcion inversa.

e) Reconstruccion numerica: Utilizando la inversa, construida en la parte anterior, imple-mentamos un algoritmo numerico de reconstruccion de la densidad.

Por otra parte, en el Captulo3investigamos la controlabilidad de un submarino inmersoen un volumen infinito de un fluido potencial. Consideramos como control, el flujo del fluido atraves de una parte de la frontera del movil. Se llega a un sistema finito dimensional, similaral sistema de Kirchhoff, en el cual los controles aparecen a traves de un termino lineal (conderivada temporal) y uno bilineal. Aplicando el metodo del retorno, establecimos bajo ciertascondiciones geometricas un resultado de controlabilidad local para la posicion y velocidaddel vehculo solo con cuatro controles.

Finalmente, en el Captulo4implementamos un algoritmo numerico que permite resolverel problema de segmentacion de una imagen, desde un punto de vista relajado, pues permite

segmentar la imagen en un numero de la forma 2N colores. Implementamos una variantedel esquema del funcional de Ambrosio & Tortorelli, el cual aproxima en el sentido de la-convergencia al funcional propuesto por Mumford & Shah, en esencia nuestra variante esinspirada por los trabajos de T. Chan.

iii


4/141

Abstract

This PhD thesis, to obtain the degree of Doctor in Engineering, speciality: AppliedMathematics, is devoted to the study of an inverse problem, the mathematical analysis ofa fluid-structure interaction system and the numerical implementation of a segmentationalgorithm.

This work is divided in three parts. First we investigate (in Chapter2) an inverse problemcoming from Biology. We are interested in recovering the density of the ion channel (CNG-channel) in the olfactory cilium, from the measurement of the current flux produced by thediffusion of the agent in the cilium. This problem is described by a linear Fredholm equationof the first kind with diffusive kernel. Considering an approximation of the kernel we obtainthe following results:

a) Identifiability Problem: Under different assumptions about of the regularity of thedensity function we establish the injectivity of the operator under study.

b) Stability Problem: We prove the stability (i.e. the continuity of the inverse of theoperator) in different norms in a particular case, as well as the stability in the generalcase in weighted L2-norms using Mellins transform.

c) The Reconstruction Problem: For an appropriate choice of the parameters, we recon-struct the inverse operator.

d) Numerical Reconstruction: Using the inverse operator in the reconstruction problem,we implement a numerical algorithm to reconstruct the density.

Next, in Chapter 3, we investigate the controllability of an under-actuated underwatervehicle immersed in an infinite volume of ideal fluid with a potential flow. Taking as controlinput the flow of the fluid through a part of the boundary of the vehicle, we obtain a finite-dimensional system similar to Kirchhoff laws in which the control input appears throughboth linear terms (with time derivative) and bilinear terms. Applying the return method,we establish under certain geometric conditions a local controllability result for the positionand velocities of the vehicle with four controls.

Finally, in Chapter4, we develop a numerical algorithm allowing to solve the segmentationproblem for a picture in a soft sense, limiting ourselves to numbers of colors of the form 2N.We implement some variant of Ambrosio & Tortorelli scheme, which is an approximation

in the sense of the -convergence of Mumford & Shah functional. Our approach is in partinspired by T. Chans works.

iv


5/141

Agradecimientos

Deseo agradecer a todos los que hicieron posible la realizacion de esta tesis. Han transcurrido

varios anos de trabajo y esfuerzo, como siempre me ha dicho mi amigo, este trabajo es laculminacion y a la vez el comienzo de una etapa de mi vida. Como en todo tipo de proyectoshan ocurrido cosas buenas y malas, pero siendo honesto he disfrutado este periodo de mivida y agradezco el haber podido vivirlo.

Soledad, mi companera, amiga y orientadora, te agradezco por tu incondicional apoyo yconsejos en los momentos difciles, te agradezco que me permitas disfrutar de tu compana.Corina, que puedo decir de un angel, gracias por comprender las innumerables ocasiones enlas que tuviste que esperar por tu papa o simplemente comprender y entretenerte sola concocodrilo, admiro tu dedicacion y preocupacion por los detalles, tienes muchos talentos yespero poder guiarte y cuidarte mucho. Bebe, no has llegado aun, pero te esperamos con

alegra, no has sido parte de esta etapa, pero seras participe de lo que viene, lo que puedodecirte, es que aun cuando las cosas se ven difciles antes de realizarlas, lo mas difcil escomenzar, una vez iniciado, tarde o temprano terminara.

Quisiera dar las gracias a mis padres Marta y Gonzalo, nunca pense en llegar a estepunto en mi vida, creo que ustedes tampoco lo pensaron. Mama, gracias por cuidarme ypreocuparse por mi y mi familia, por llamar y mantener el contacto, por haberme inculcadola superacion. Papa, sin duda mi heroe mi modelo, solo espero algun da poder llegar a serla mitad de la persona que usted es. Nunca me ha molestado estar a su sombra, solo me hainspirado, siempre he pensado que nunca hubiese podido realizar las incontables proezas queusted ha realizado a lo largo de su vida, gracias.

Agradezco tambien a mis profesores por su constancia y dedicacion en mostrarme y en-senarme la forma de afrontar esta profesion.

A mis amigos, quisiera expresarles mi gratitud por los valores que han sembrado en mi,ademas de aprender que los suenos se pueden hacer realidad, ya que no hay cosas imposiblessolo seres incapaces, Juanito, gracias por mostrarme de lo que soy capaz, extrano nuestrasconversaciones, las cuales en cierto sentido me han motivado a seguir esta carrera. No tengosuficientes palabras para dar las gracias y enumerar las cosas que han hecho por mi, graciasPablo, Jaime, Lionel, Fernando, Takeshi, Alexis, Erica, Andres, Alejandra, Duvan, Miguel,Senores Darmois, Dugan, Ivonne, Rafa, Alejandro, Marcia, ..., y mil disculpas a quienes he

olvidado mencionar, pero a todos muchas gracias.

Por ultimo quisiera agradecer al Centro de Modelamiento Matematico CMM de la U. deChile.

v


6/141

Contents

1 Introduction 9

1.1 Inverse problem in biology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.1.1 Linear problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.2 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.1.3 Polynomial approximation. . . . . . . . . . . . . . . . . . . . . . . . 151.1.4 A non-linear inverse problem. . . . . . . . . . . . . . . . . . . . . . . 16

1.2 Control of underwater vehicles in potential fluids. . . . . . . . . . . . . . . . 17

1.3 Image problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.3.1 The Mumford and Shah functional . . . . . . . . . . . . . . . . . . . 21

1.3.2 Approximations of the Mumford and Shah functional . . . . . . . . . 23

2 Inverse problem in biology 24

2.1 Setting the problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.3 Introduction to inverse problem . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.3.1 Basic definitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4 Mathematical modeling of signal transduction in olfactory cilia . . . . . . . . 34

2.4.1 The experimental procedure . . . . . . . . . . . . . . . . . . . . . . . 35

2.4.2 Mathematical model . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.5 Useful tools: Mellins transform . . . . . . . . . . . . . . . . . . . . . . . . . 402.6 Results for the operator m . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.7 Proof of stability result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.8 Case j =0j . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.9 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.10 Polynomial approximation of the Hills function . . . . . . . . . . . . . . . . 61

2.10.1 Proof of the identifiability Theorem 2.10.1 . . . . . . . . . . . . . . . 62

vi


7/141


8/141

CONTENTS

4.3 Proposed algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

4.3.1 Framework for the 2 color case. . . . . . . . . . . . . . . . . . . . . . 126

4.3.2 Setup for the optimization scheme. . . . . . . . . . . . . . . . . . . . 127

4.3.3 The 4 color segmentation setup . . . . . . . . . . . . . . . . . . . . . 128

4.3.4 Further remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

4.4 Numerical experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

5 Conclusiones 131

5.1 Problema inverso en biologa . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

5.2 Control de un vehculos submarino en un fluido potencial . . . . . . . . . . . 133

5.3 Segmentacion de una imagen multicolor. . . . . . . . . . . . . . . . . . . . . 134

References 135

viii


9/141

Chapter 1

Introduction

This thesis is divided in three parts.

In Chapter2we study a biological inverse problem and we focus on a linearization of thisproblem, considering the following issues:

a) Identifiability problem.

b) Stability problem.

c) Reconstruction problem.

d) Numerical algorithm for reconstruction.

Next, in Chapter3, we study a control problem coming from fluid-structure interactionsystems. More precisely, we investigate the controllability of the motion of a submarine,which is surrounded by an inviscid incompressible fluid.

Finally, in Chapter4, we develop a numerical algorithm to obtain the segmentation of apicture in the soft sense, as we consider a finite number of colors.

Now we present a short description of the problems and the main results obtained in thisthesis.

1.1 Inverse problem in biology

Identification of detailed features of neuronal systems is an important challenge in the bio-sciences today. Olfactory cilia are thin hair-like filaments that extend from olfactory receptorneurons into the nasal mucus. Transduction of an odor into an electrical signal is accom-plished by a depolarizing influx of ions through cyclic-nucleotide-gated (CNG) channels inthe membrane that forms the lateral surface of the cilium and are activated by adenosine 3,5-cyclic monophosphate (cAMP). In an experimental procedure developed by S. Kleene, a

9


10/141

CHAPTER 1. INTRODUCTION

cilium is detached at its base and drawn into a recording pipette. The cilium base is thenimmersed in a bath of a channel activating agent (cAMP) which is allowed to diffuse intothe cilium interior, opening channels as it goes and initiating a transmembrane current. Thetotal current is recorded as a function of time.

In[37], French et. al. propose a mathematical model consisting of two nonlinear differ-

ential equations and a constrained Fredholm integral equation of the first kind is developedto model this experiment. The unknowns in the problem are the concentrationc of cAMP,the membrane potential Vand, the quantity of most interest in this work, the distributionof CNG channels along the length of a cilium. Then the basic idea is to recover the distri-bution of CNG channels along the length of a cilium only measuring the electrical activityproduced by the diffusion of cAMP into cilia, this constitute an interesting inverse problem.

French et. al. in [37]present a simple numerical method to obtain estimates of the spatialdistribution of CNG ion channels along the length of a cilium. Certain computations indicatethat this mathematical problem is ill-conditioned.

Later, French and Edwards in [36], studied this inverse problem using perturbation tech-niques. A simple perturbation approximation is derived and used to solve this inverse problemand thus obtain estimates of the spatial distribution of CNG ion channels along the lengthof a cilium. A one-dimensional computer minimization and a special delay iteration are usedwith the perturbation formulas to obtain approximate channel distributions in the cases ofsimulated and experimental data.

French and Groetsch in[38] present some simplifications and approximations in the prob-lem, obtaining an analytical solution for this inverse problem and a numerical procedure isproposed for a class of integral equations suggested by this simplified model and numericalresults using simulated and laboratory data are presented.

In this work we consider the linear problem proposed in [38], but with a better ap-proximation of the operator kernel and we study the identifiability, stability and numericalreconstruction for this inverse problem.

1.1.1 Linear problem

Now we introduce the lineal model presented in [38]. In [37] a nonlinear integral equationmodel is developed for determining the spatial distribution of ion channels along the lengthof frog olfactory cilia. The essential nonlinearity in the model arises from the binding of the

channel activating ligand to the cyclic-nucleotide-gated ion channels as the ligand diffusesalong the length of the cilium. We investigate a linear model of this process in which thebinding mechanism is neglected, leading to a particular type of linear Fredholm integralequation of the first kind with a diffusive kernel. The linear mathematical model consists offinding = (x)> 0, such that

I[](t) =J0

L0

(x)K(t, x)dx, (1.1.1)

10 R. Lecaros


11/141


where the kernel isK(t, x) =F(c(t, x)), (1.1.2)

and c is the concentration the of the problem

c

t D

2c

x2 = 0, t >0, x(0, L)

c(0, x) = 0, x(0, L)

c(t, 0) = c0, t >0

c

x(t, L) = 0, t >0,

(1.1.3)

with

F(x) = xn

xn +Kn1/2. (1.1.4)

F(x) is a Hill function with constant K1/2 and exponent n, where K1/2 is the half-bulk concentration and n is an experimentally determined parameter. The function is anunknown ion channel density function, c is the concentration of a channel activating ligandthat is diffusing from left-to-right in a thin cylinder (the interior of the cilium) of length Lwith diffusivity constant D, I[](t) is a given total transmembrane current, the constant J0has units of current/length, and c0 is the maintained concentration at the open end of thecylinder (while x = L is considered the closed end).

This problem arises in models of diffusion in tiny tubular biological structures. In physicalexperiments a ligand (cAMP), which is held at constant concentration c0 at the open end,diffuses into the cilia and binds to ion channels (CNG) opening the channels and thereby

allowing an ionic influx which initiates a local transmembrane current. The model (1.1.1)-(1.1.3) does not include terms modeling the binding or changes in membrane potential.However, if the quantity of ion channels is small these effects will be negligible. Frog olfactorycilia have diameters of about 0.28m and length around 40m and are connected to theknobs at the ends of the dendrites of the olfactory receptor neurons. The Hill functionin the experiments modeled here typically has exponent n = 1.7 and half-concentrationK1/2= 1.7m. Here J0= 0.232 pA/channel is a positive constant.

Now we derive an analytical solution of a simplified version of problem (1.1.1)-(1.1.3). Tothat end we assume:

i) The influence of the no-flux boundary condition for c at the closed end is negligible (orequivalently that the cilium is very long). Neglecting the boundary condition at theclosed end, the diffusion equation has the closed form solution

c(t, x) =c0erfc

x

2

Dt

, (1.1.5)

where erfc is the complementary error function defined by

erfc(z) = 1 2

z0

exp(2)d. (1.1.6)

11 R. Lecaros


12/141


Therefore, curves of the form tx2 are level curves for the concentration function c.ii) The Hill function exponent n is large.

This assumption leads to the following simplification;

F(c(t, x)) = c(t, x)n

c(t, x)n +Kn1/2(1.1.7)

H(c(t, x) K1/2) (1.1.8)= H(2t x2) (1.1.9)

whereHis the Heaviside unit step function

H(u) =

0 ifu < 0,

1/2 ifu= 0,

1 ifu > 0,

(1.1.10)

and Dis implicitly given in terms of the half bulk concentration by the equation

K1/2= c0erfc(/(2

Dt)). (1.1.11)

This type of assumption has also been made in the study of calcium sparks (see Section8.11 by [28]).

Using these assumptions we have

I[](t) =J0 L0

H(2t x2)(x)dx= J0 t

0(x)dx.

Differentiating yields

I(t) =1

2t1/2J0(

t).

Lettingy=

t, we obtain

(y) =2I(y2/2)y

J02 .

In this simplified model, the channel density is given analytically in terms of the deriva-tive of the total current and a parameter which is obtained by solving equation (1.1.11)numerically.

Our idea to consider a more general situation. We replace (1.1.8) by the following ap-proximation of the Hill function (1.1.4)

F(x)Fm(x) =F(c0)mj=1

ajH(x j), x[0, c0], (1.1.12)

12 R. Lecaros


13/141


whereaj, j are positive constants verifying:

mj=1

aj = 1, (1.1.13)

and

0< 1 < 2 0.

1.1.2 Main results

Now we present the main results in this work. We begin by setting the notations. For any >0, we consider the weight function(x) =|x|, and introduce the following norms

f0,,b = fL2(0,b) ,f1,,b = fH1(0,b) ,

f1,,b = fH1(0,b) ,

and we set the constants

Lk = L/k, fork = 1,...,m, and L0 = 0, (1.1.17)

where

j = erfc1

(j/c0)2D forj= 1,...,m. (1.1.18)First, we present a result about the continuity of the operator Im.

Theorem 1.1.1. Let: [0, L] R be a function inL2(0, L). Then for 34

, there existsa positive constantC1> 0, such that,

Im[]1,,L2m C1 L2(0,L) , (1.1.19)

whereC1, only depends onL, 1, m1, m, am and.

13 R. Lecaros


14/141


The above theorem sets the functional framework forIm. Now we present our main resultsof the stability. First we define the constant

0= max

0, ln(am)

ln

mm1

12

. (1.1.20)

Theorem 1.1.2.Let: [0, L] R be a function inL2(0, L). Then, for all >max{0, 3/4},there exists a positive constantC2 > 0, such that,

1,+1,LC2 Im[]1,214,L2m

, (1.1.21)


This theorem corresponds to the observability inequality for the functionalIm, in partic-ular, this result proves the identifiability for our inverse problem. For the proof see Section2.2.

In the proof of the above result, we can see the role played by the mesh {j}mj=1, forj = 1,...,m 1. We will focus our analysis in the special case where

j =c0 erfc

0

j

2

D

j = 1,...,m, (1.1.22)

With (0, 1) and 0 > 0 constants. This simplification is not very restrictive because{j}mj=1defined a mesh in the interval [0, c0], therefore (1.1.22) only fixed a non-regular meshin which we want to have the approximation (1.1.12), but this choice allows us to obtain areconstruction algorithm for the density .

Firstly, we study the following functional

m[](t) = 1

J0F(c0)Im[](t

2) t >0, (1.1.23)

where

(x) =

x0

(z)dz. (1.1.24)

It is worth noting that the identifiability for m is equivalent to the identifiability for Im,for that reason we will focus in the study of identifiability on Im.

In what follows Im and m the functionals Im, m respectively, the choice of j as in(1.1.22).

Thus, for C0([0, L]), we can construct a function : [0, L] R depending only on ,by the equation define x

0

(z)dz= (x) +Im[](L

2m)

J0F(c0) x[0, L], (1.1.25)

14 R. Lecaros


15/141


A: approximation of. B: currentI(t) as defined by (1.1.26)

Figure 1.1: Approximation of the function (x) and with current I(t) as defined by (1.1.26)

therefore, deriving (1.1.25) we can recover (see Theorem 2.2.9). Moreover the equation(1.1.25) define a numerical algorithm for the reconstruction problem (see Section2.9on page59). If we consider the same example considered in [38], thus we define

I(t) =

0, t(0, tDelay),

IMax

1 +

KI

t tDelay

nI1, t > tDelay,

(1.1.26)

with tDelay = 30ms, nI = 2.2, IMax = 150pA and KI = 100ms. The current (1.1.26) hasa short delay (Figure2.5B), which is similar to the profiles in several applications (see [18],[33] or[58]).

Thus we obtain the numerical solution as shown in Figure 1.1A. In this figure one seesthat the main features of the solution of this model are consistent with those obtained in[33,38].

With our approach is possible to consider a better and more realistic approximation ofthe mathematical model (1.1.1). Furthermore we have developed a reconstruction methodand not only a numerical approach of the model. On other hand, we have obtained a stability

result with different norms.

1.1.3 Polynomial approximation

We study a second approximation for the kernel (1.1.2), in this case, we consider a polynomialapproximation of the Hills function (1.1.4), aroundc0.

We consider Pm, the Taylor expansion of degree m. For Hills function defined by (1.1.4)

15 R. Lecaros


16/141


at point c0> 0. Thus we have

F(x) =Pm(x c0) +O(|x c0|m+1). (1.1.27)Therefore, we define the polynomial kernel approximation by

P Km(t.x) =Pm(w(t, x) c0), (1.1.28)

wherew(t, x) is the solution of the problem (1.1.3).

The total current with polynomial approximation is given by

P Im[](t) =

L0

(x)P Km(t, x)dx, t >0. (1.1.29)

Now we present our principal result related with the above for L2(0, L).Theorem 1.1.3. Letm8, be a constant. Then we have

KerP Im ={0},

where KerP Im =f L2(0, L) P Im[f](t) = 0,t >0 .

Theorem1.1.3corresponds to the identifiability of the operator P Im.The proof is givenin Section2.10.

1.1.4 A non-linear inverse problem

We consider a variation of the olfactory inverse problem introduced by French et al. in [37]

and we will show that, this inverse problem corresponds to determine a potential q in aninterval ofR, from the Neumann data measured on one side of the boundary for the equation + q. This is correlated with the problem of determining a complex-valued potentialq ina bounded two dimensional domain, from the Cauchy data measured on an arbitrary opensubset of the boundary for the associated Schrodinger equation + q, the identifiabilityfor this problem was shown in [48]. The motivation for this problem comes from a classicalinverse problem of electrical impedance tomography. In this case one attempts to determinethe electrical conductivity of a body, considering measurements of voltage and current on theboundary of the body. This problem was proposed by A. Calder on[14] and is well known asCalderons problemand had been study by several authors. We proof the non-identifiabilityfor our problem, showing the importance of result [48].

Let us define q(x) = PRaF(c0) (x). We consider the stationary case, i.e. when thefunctions do not depend on time in the model develop in [37]. Thus, we can reduce thisinverse problem to the following form: We want to determine the positive potentialq(x) suchthat:

v qv = 0, x(0, L)

v(0) = v0,

v(L) = 0,

(1.1.30)

16 R. Lecaros


17/141


18/141


They aimed to control the position and velocity of the rigid body by the control inputs.For this system, the state lives in R6 (but there is a PDE in the dynamics) and the controlinputs lives in R2 and there is no control objective for the fluid flow, as it is an exteriordomain. It provides an accurate model for the motion of a boat with two propellers, onedisplayed in a transversal bow thruster at the bow of the ship, the other one placed at thestern of the boat (see Figure 1.3). A rigorous analysis of the control properties of such a

system is performed in [44].

Bow thruster Longitudinal thruster

Figure 1.3: Example of the a bow thruster and longitudinal thruster

In this thesis we aim to extend this work to the dimension three. We investigate themotion of under-actuated underwater vehicles (for instance submarines).

The fluid, still inviscid and incompressible, will no longer be assumed to be irrotational(i.e., the vorticity may not vanish everywhere), and the only geometric assumption for theshape of the submarine will be the existence of two planar symmetries.

To be more precise, our fluid-structure interaction problem can be described as follow.The underwater vehicle, represented by a rigid body occupying a connected compact setS(t) R3, is surrounded by an homogeneous incompressible perfect fluid filling the open set(t) := R3 \ S(t) (as for e.g. a submarine immersed in an ocean). We assume that (t) isC smooth and connected. Let S=S(0) and

= (0) = R3 \ S(0)

denote the initial configuration (t= 0). Then, the dynamics of the fluid-structure system is

18 R. Lecaros


19/141


governed by the following system of PDEs

u

t + (u )u+ p= 0, t(0, T), x(t) (1.2.1)

divu = 0, t(0, T), x(t) (1.2.2)u

n= (h+

(x

h))

n+w(t, x), t

(0, T), x

(t) (1.2.3)

lim|x|+u(t, x) = 0, (1.2.4)

mh=

(t)

pn d, t(0, T) (1.2.5)

d

dt(QJ0Q

) =

(t)

(x h) pn d, t(0, T) (1.2.6)

Q = S()Q, t(0, T) (1.2.7)u(0, x) =u0(x), x (1.2.8)

(h(0), Q(0), h(0), (0)) = (h0, Q0, h1, 0)

R3

SO(3)

R3

R3. (1.2.9)

In the above equations, u (resp. p) is the velocity field (resp. the pressure) of the fluid, hdenotes the position of the center of mass of the solid, denotes the angular velocity and Qthe orthogonal matrix giving the orientation of the solid. The positive constant m and thematrixJ0, which denote respectively the mass and the inertia matrix of the rigid body, aredefined as

m=

S

(x)dx, J 0=

S

(x)(|x|2Id xx)dx,

where () represents the density of the rigid body. n is the outward unit vector to (t),x

y is the cross product between the vectors x andy , andS(y) is the skew-adjoint matrix

such that S(y)x= y x, i.e.

S(y) =

0 y3 y2y3 0 y1y2 y1 0

.Finally, the term w(t, x), which stands for the flow through the boundary of the rigid

body, is taken as control input. Its support will be strictly included in (t), and actuallyonly a finite dimensional control input will be considered here (see below (3.2.16) for theprecise form of the control term w(t, x)).

Here, we are interested in the control properties of (1.2.1-1.2.9). The controllability ofEuler equations has been established in 2D (resp. in 3D) in [23] (resp. in [42]). Note,however, that there is no hope here to control both the fluid and the rigid body motion.Indeed, (t) is an exterior domain, and the vorticity is transported by the flow with a finitespeed propagation, so that it is not affected (at any given time) far from the boat. Therefore,we will deal with the control of the motion of the rigid body only. As the state of therigid body is described by a point in R3 SO(3) R3 R3, it is natural to consider afinite-dimensional control input.

19 R. Lecaros


20/141


Consider the matrixJ R66 introduced in3.3.31,which represents a generalization ofthe mass and the inertia matrix for the fluid-struct system (1.2.1-1.2.9).

Let (l, r) R3 R3 denote the part of linear and angular velocities of the body andintroduce the corresponding pulses (P, ) R3 R3 defined by

J lr = P . (1.2.10)Then Proposition3.3.2provides the following equations for the dynamics of the rigid body

dP

dt +CMw = (P+CMw) r

1pn

wp

LMp l+RMp r+W

Mp w

,

d

dt +CJw = ( + CJw) r+P l

1pn

wp

LJp l+RJp r+W

Jpw

, (1.2.11)

where w(t) := (w1(t),...,wn(t))

R

n denotes the control input and the matrixes CM, CJ,

LMp , RMp , WMp , LJp , RJp y WJp only depend on the geometry of the body and the position ofthe controls on the body surface.

If we linearize the above system, we obtain l

r

=J1

CM

CJ

w, (1.2.12)

which is controllable if the rank of the matrix CM

CJ

,

is six. Therefore, we need to consider at least six controls input.

The work is organized as follows. In Section 3.3, we simplify system (1.2.1-1.2.9) byassuming that the fluid is potential. We obtain a finite dimensional system (namely (3.3.64))similar to Kirchhoff laws (1.2.11), in which the control input w appears through both linearterms (with time derivative) and bilinear terms. The investigation of the control propertiesof (3.3.64) is performed in Section3.4. After noticing that the controllability of the linearizedsystem at the origin requires six control inputs, we apply the return methoddue to Jean-Michel Coron to take advantage of the nonlinear terms in (3.3.64). (We refer the reader to [24]for an exposition of that method for finite-dimensional systems and for PDEs.) We considerthe linearization along a certain closed-loop trajectory and obtain a local controllability

result (Theorem 3.4.9) assuming that two rank conditions are fulfilled, by using a variantof Silverman-Meadows test for the controllability of a time-varying linear system. Someexamples using symmetry properties of the rigid body are given in Section 3.5.

1.3 Image problem

This section is concerned with image segmentation, which plays a very important role inmany applications. The aim is to find a partition of an image into its constituent parts. As

20 R. Lecaros


21/141


we will see, the main difficulty is that one needs to manipulate objects of different kinds:Functions, domains in R2 and curves. We could say that segmenting an image means dividingit into its constituent parts.

Our research aims at image segmentation using the variational framework of Mumfordand Shah, following an approximation proposed by Ambrosio and Tortorelli. This technique

circumvents the use of parametric contours and implicit level-set techniques, where its solu-tion may be regarded as a soft segmentation, with a number of levels or colors being 2N. Onthe other hand, the implementation was based on an finite difference discretization, wheretwo- and four- color cases are described with their corresponding numerical results. SeeChapter4.

1.3.1 The Mumford and Shah functional

Let us present the model introduced by Mumford and Shah in 1989 [ 70]. In this section is a

bounded open set ofRN

,N= 2, 3,and u0(x) is the initial image. Without loss of generalitywe can always assume that 0 u0(x) 1 a.e. x . We search for a pair (u, K), whereK is the set of discontinuities, minimizing

F(u, K) =

\K

(u u0)2dx+

\K|u|2dx+HN1(K),

where and are nonnegative constants andHN1(K) corresponds to N 1 dimensionalHausdorff measure ofK, which is the most natural way of extending the notion of length tonon-smooth sets and u belongs to the Sobolev space W1,2( \ K).

The difficulty in studyingFis that it involves two unknownsuandKof different natures:u is a function defined on anN-dimensional space, while K is an (N 1)-dimensional set.

In order to apply the direct method of the calculus of variations, it is necessary to finda topology that ensures at the same time lower semicontinuity ofFand compactness of theminimizing sequences. The difficulty comes fromHN1(K). Indeed, let Ebe a Borel set ofRN with topological boundaryE. It is easy to convince oneself that the mapEHN1(E).is not lower semicontinuous with respect to any compact topology.

This shows the necessity of finding another formulation ofF(u, K). The new formulationinvolves the space B V() of functions of bounded variation in . The idea is to identify theset of edges Kwith the jump set Su ofu, which allows us to eliminate the unknown K. So

the idea is to consider the functional

G(u) =

(u u0)2dx+

|u|2dx+HN1(Su).

Now, it is tempting to minimize G on the spaceBV(). Unfortunately, the space BV()may contain pathological non-constant functions that are continuous and have approximatedifferential equal to zero almost everywhere (a well-known example is the Cantor-Vitali func-tion). For such a function v we have

21 R. Lecaros


22/141


G(v) =

(v u0)2dx infuBV()

G(u),

and since these pathological functions are dense in L2(), we get

infuBV()

G(u) = 0,

which implies that the infimum ofG cannot be achieved in B V() in general.

To avoid this phenomenon we must eliminate these pathological functions, which havethe peculiarity that their distributional derivatives are measures concentrated on Cantor sets.Let us recall that the distributional derivative Du of aBV() function can be split into threemutually singular measures:

Du=udx+ (u+ u)nuHN1|Su +Cu,

where J(u) = (u+

u)nu

HN1|Su

is the jump part and Cu the Cantor part. Following Di

Giorgi [40, 39] we define SBV() as the space of special functions of bounded variation,which is the space ofBV() functions such that Cu = 0. The natural question is now toestablish the relation between the following two problems:

infu,K

F(u, K), uW1,2( \ K) L()

K, K closed, HN1


23/141


1.3.2 Approximations of the Mumford and Shah functional

The lack of differentiability of the functional for a suitable norm does not allow us to use,as is classical, Euler-Lagrange equations. Moreover, the discretization of the unknown dis-continuity set is a very complex task. A commonly used method is to approximate F(u, K)(orG(u)) by a sequence Fof regular functionals defined on Sobolev spaces, the convergenceofF to F as 0 being understood in the convergence framework. Of course, if wewant to get an efficient approximation, the set Kmust not appear inF. Approximations byelliptic functionals were introduced in[5]. In this approach the set Su (or K) is replaced byan auxiliary variable v (a function) that approximates the characteristic function (1 Su),i.e., v(x) 0 ifx Su and v(x) 1 otherwise. Ambrosio and Tortorelli [5] proposed thefollowing sequence of functionals:

F(u, v) =

|u u0|2dx+

v2|u|2dx+

|v|2dx+ 14

(v 1)2dx.

23 R. Lecaros


24/141

Chapter 2

Inverse problem in biology

The understanding of transduction in the olfactory system is an important challenge inneuroscience. It is worth noting that the olfactory receptor neuron is bipolar, the one endextending dendrites to the surface of the neuroepithelium and the other end sending axonsthat terminate in the olfactory bulb in the brain. The dendrites form knobs from whichspecialized cilia emanate, the transduction begins when an odorant molecule (odor) binds toreceptors on the cytoplasmic membrane of a cilium, of an olfactory neuron (for reviews ofolfactory transduction see [79,11,81]).

Each olfactory neuron has several of these thin cilia into the nasal mucus. When theodorant binds with the receptor, this causes and increased of the level of Cyclic adenosinemonophosphate (cAMP) inside the cilium. cAMP is one of the intracellular messengersinvolved in olfactory signal transduction. This cAMP is synthesized by an adenylate cyclase,and hydrolyzed by a phosphodiesterase. This is achieved through a GTP-binding protein,see [77,86,85,13,10]. The cAMP opens cyclic nucleotide-gated (CNG) channels, the CNGchannels allow a depolarizing the cell through a influx of Na+ and Ca2+ into the cilia[31,32].

Although the single-channel properties have been well studied and described, less is knownabout the distribution of these channels along the cilia. Actually it is in general unknown,and it is therefore assumed to by uniform at first approximation, even if its variations couldsignificantly have an influence on the sensitivity of the neuron to odor stimuli. It is importantto note that Ca2+ has a main role in olfactory transduction[81], opening Ca2+-activated Cl

channels (ClCa), that allows Clflux from the cilia. This flux is one of those most responsiblefor depolarizing the cell. This generates an excitatory response, characterized by an increase

in the frequency of action potential[53,59,64].Experiments can be performed (see [18]) on single cilia; recordings are made of current as

cAMP diffuses and activates the CNG channels. In this work, we intend to see whether CNGchannel distributions can be derived from the experimental current data and hence if we canrecover some properties of the cilia. Therefore, this problem can be represented by a math-ematical model, and CNG channel distributions can be computed, in an approximate way,from the current data, considering suitable approximations (see [37]). The model involvesa system of nonlinear equations that corresponds to an inverse problem in Partial Differen-tial Equations (PDE). The perturbation theory allows to find an approximate solution (see

24


25/141

CHAPTER 2. INVERSE PROBLEM IN BIOLOGY

[36]), and in [38], under certain approximations and simplifications, was derived an integralequation model.

2.1 Setting the problem

In this section we will set the mathematical model related to the inverse problem arisingin olfaction experimentation.

In order to simplify our problem, we will consider the domain R3 as a cylinder, closedat one end and open at the other end. Note that the boundary of the cylinder correspondsto the plasmatic membrane. Thanks to the symmetry of the domain, the dynamics aredescribed by a one dimensional model for the olfactory cilium, and we consider a diffusion-reaction model for the cAMP molecules, immersed in a solution. We denote by c(t, x) the(volume) concentration of cAMP molecules. Those molecules diffuse in thex-direction in aone dimensional channel of lengthL. We have the Neumann boundary condition cx (t, L) = 0for t > 0. Moreover, in our situation the concentration of cAMP is constant at the otherside, that is, c(t, 0) =c0 > 0 for t >0, and initially the channel does not have cAMP.

Therefore, the conservation of mass is given by

c

t D

2c

x2 = 0, t >0, x(0, L)

c(0, x) = 0, x

(0, L)

c(t, 0) = c0, t >0

c

x(t, L) = 0, t >0,

(2.1.1)

where the constant D >0 is the diffusion coefficient.

The binding of the cAMP on the cilium surface produces a change in the current flux(J(t, x)) through the channels. This fact is modeled by

J(t, x) = J0(x)F(c(t, x)), t >0, x

(0, L), (2.1.2)

where (x) is the density of channels along the surface of the cilium and F is the classicaldimensionless Hill function, given by

F(x) = |x|n|x|n +n . (2.1.3)

The constant >0 is termed the half-maximal, because F() = 12 . The experiment willbe considered thereafter with a parameter ntypically greater than 1 (for instance n = 1.7).

25 R. Lecaros


26/141


The total current I(t) corresponds to the integral ofJin the interval [0, L]:

I(t) =I[](t) =

L0

J(t, x)dx= J0

L0

(x)F(c(t, x))dx. (2.1.4)

Thus, the inverse problem consists in obtainingfrom the measurement ofI[](t), t0.We note that this is a Fredholm integral equation of the first kind, that is

I[](t) =

L0

K(t, x)(x)dx, (2.1.5)

where K(t, x) = J0F(c(t, x)) is called the kernel of the operator. This kind of problem islinear, but it is often ill-posed. For example, ifK is sufficiently smooth, the operator definedabove is compact on reasonable Banach spaces such as Lp spaces. Even if the mapping isinjective, its inverse will not be continuous. Indeed, ifI is compact and I1 is continuous,then it follows that the identity map in Lp is compact, which is clearly false.

We study a simplified version of the above problem under more general assumptions thatthose considered in[38].

We assume the following

i)

c(t, x) =c0erfc

x

2

Dt

, (2.1.6)

where erfc is the complementary error function:

erfc(z) = 1 2

z0

exp(2)d. (2.1.7)

ii) We consider the following approximation of the Hill function (2.1.3)

F(x)Fm(x) =F(c0)mj=1

ajH(x j), x[0, c0], (2.1.8)

whereHis the Heaviside unit step function i.e.

H(u) = 1 ifu0,

0 ifu


27/141


With the above assumptions we define the approximate total current

Im[](t) =J0

L0

(x)Km(t, x)dx, (2.1.12)

where

Km(t.x) =Fm(c(t, x)) = Fmc0erfc x

2Dt . (2.1.13)Therefore, our inverse problem is to recoverfrom the measurementIm[](t), for allt0.

2.2 Main results

In this section we present the main results in this paper. We began to set the notation. For

any >0, we consider the function (x) =|x|, and introduce the following normsf0,,b = fL2(0,b) ,

f1,,b = fH1(0,b) ,

f1,,b = fH1(0,b) ,and we set

Lk =L/k for k = 1,...,m, andL0 = 0, (2.2.1)

where

j = erfc1(j/c0)2D forj= 1,...,m. (2.2.2)First, we present a result about the continuity of the operator Im.

Theorem 2.2.1. Let: [0, L] R be a function inL2(0, L). Then for 34 there exists apositive constantC1 > 0 such that

Im[]1,,L2m C1 L2(0,L) , (2.2.3)whereC1, only depend ofL, 1, m1, m, am and.

Now we present our main results. First we define the constant

0= max

0, ln(am)ln mm11

2

. (2.2.4)Theorem 2.2.2.Let: [0, L] R be a function inL2(0, L). Then, for all >max{0, 3/4},there exists a positive constantC2 > 0 such that

1,+1,LC2 Im[]1,214,L2m

, (2.2.5)


27 R. Lecaros


28/141


29/141


Theorem 2.2.8. LetH1(0, L) be a function. Then for > 0, there exists a constantC2> 0 such that

0,,L C2 m[]H1(0,Lm) , (2.2.11)where the constantC2 only depends onL, 1, n1, n, an, and.

In the proof of Theorem 2.2.5we can see the role played by the mesh{j}mj=1, for j =1,...,m 1.Therefore we study the special case where we set

j =c0erfc

0

j

2

D

, j = 1,...,m. (2.2.12)

with (0, 1) and 0 > 0 constants. This simplification is not very restrictive because{j}mj=1 defines a mesh in the interval [0, c0], therefore (2.2.12) only prescribes a non-regularmesh in which we want to have the approximation (2.1.8). In what follows, Im and m aredenoted by Im and m, respectively, when j is given by (2.2.12).

On the other hand, for the reconstruction we first introduce the function

g(t) =Im[](t

2/20 ) Im[](L2m)J0F(c0)

, t[0, 0Lm) . (2.2.13)

We want to obtain a reconstruction algorithm and a numerical implementation to recoverthe function from the measurement ofIm[]. We begin by recovering : [0, L] R suchthat

m[](t/0) =g(t), t[0, 0Lm). (2.2.14)Next, we define g1, g2,...,gm in the following recurrent form

g1(x) =

1

amg x

m , ifx[L,L),

0, otherwise,

(2.2.15)

and

gk+1(x) =

1

am

g

x

m

kj=1

amk1+jgj

jx

k+1

, ifx[k+1L, kL),

0, otherwise,

(2.2.16)

for k = 1,..,m 1. Furthermore for km, we define

gk+1(x) =

1

am

g

x

m

m1j=1

ajgj+km+1

jx

m

, ifx[k+1L, kL),

0, otherwise.

(2.2.17)

With the above definitions we have the following reconstruction result.

29 R. Lecaros


30/141


Theorem 2.2.9. Let be a function inC0([0, L]),and letg be defined by(2.2.13)and{gj}j1be given by definitions (2.2.15)-(2.2.17). Then the function defined by

(x) =

+j=1

gj(x), x(0, L],

g(0), x= 0,

(2.2.18)

is well defined and satisfies the equation

m[](t/0) =g(t), t[0, 0Lm]. (2.2.19)Furthermore satisfies x

0

(z)dz= (x) +Im[](L

2m)

J0F(c0) , x[0, L]. (2.2.20)

Now we will present a stability result for m

for a more general norm.

Let us consider, for 0a < b 0, there exists a positive constant C() such that

g[a,b)C() f[a,b) , (2.2.22)

whereg(x) =f(x/), and C() is a increasing function.For example, we note that the Lp norms satisfies (2.2.21)-(2.2.22) for any 1p+,

with

C() =

1p p[1, +)

1 p=.But it is more interesting to consider the boundary variation norm defined by:

fBV =fL(a,b)+ supax1


31/141


2.3 Introduction to inverse problem

An inverse problem is a general framework that is used to convert observed measurementsinto information about a physical object or system that we are interested in. For example, ifwe have measurements of the Earths gravity field, then we might ask the question: given

the data that we have available, what can we say about the density distribution of the Earthin that area?. The solution to this problem (i.e. the density distribution that best matchesthe data) is useful because it generally tells us something about a physical parameter thatwe cannot directly observe. Thus, inverse problems are one of the most important, and well-studied mathematical problems in science and mathematics. Inverse problems arise in manybranches of science and mathematics, including computer vision, machine learning, statistics,statistical inference, geophysics, medical imaging (such as computed axial tomography andEEG/ERP), remote sensing, ocean acoustic tomography, nondestructive testing, astronomy,physics and many other fields.

The field of inverse problems was first discovered and introduced by Soviet-Armenian

physicist, Viktor Ambartsumian. While still a student, Ambartsumian thoroughly studiedthe theory of atomic structure, the formation of energy levels, and the Schrodinger equationand its properties, and when he mastered the theory of eigenvalues of differential equations,he pointed out the apparent analogy between discrete energy levels and the eigenvalues ofdifferential equations. He then asked: given a family of eigenvalues, is it possible to find theform of the equations whose eigenvalues they are? Essentially Ambartsumian was examiningthe inverse Sturm-Liouville problem, which dealt with determining the equations of a vibrat-ing string. This paper was published in 1929 in the German physics journal Zeitschrift furPhysik and remained in oblivion for a rather long time. Describing this situation after manydecades, Ambartsumian said, If an astronomer publishes an article with a mathematicalcontent in a physics journal, then the most likely thing that will happen to it is oblivion.

Inverse problems appear in several fields, including medical imaging, image processing,mathematical finance, astronomy, geophysics, nondestructive material testing and sub-surfaceprospecting. Typical inverse problems arise from asking simple questions backwards. Forinstance, the simple question might be If we know precisely the structure of the innerorgans of a patient, what kind of X-ray images would we get from her?. The same questionbackwards is Given a set of X-ray images of a patient, what is the three- dimensionalstructure of her inner organs?. This is the inverse problem of Computerized Tomography,or CT imaging.

Usually the inverse problem is more difficult than the simple question that it reverses.

For example, even though the Earths gravitational field is governed by Newtons law ofgravitation, the inverse problem of finding sub-surface structures from minor variations ofthe gravitational field on the surface is extremely hard. Successful solution of inverse problemsrequires specially designed algorithms that can tolerate errors in measured data.

Inverse problems research concentrates on the mathematical theory and practical inter-pretation of indirect measurements. The study of inverse problems is an active area of modernapplied mathematics and one of the most interdisciplinary field of science.

31 R. Lecaros


32/141


33/141


For sufficiently smoothk the operator defined above is compact on reasonable Banach spacessuch asLp spaces. Even if the mapping is injective its inverse will not be continuous. (How-ever, by the bounded inverse theorem, if the mapping is bijective, then the inverse will bebounded (i.e. continuous)). Thus small errors in the data d are greatly amplified in thesolution p. In this sense the inverse problem of inferring p from measured d is ill-posed.To obtain a numerical solution, the integral must be approximated using quadrature, and

the data sampled at discrete points. The resulting system of linear equations will be ill-conditioned.

Another example of inverse problem is the well-known problem, about electrical impedance.This problem called Calderons problem was introduced by A. P. Calderon in 1980 (see theseminal articleOn an inverse boundary value problem[14]). The Calderons problem consistsin recovering the electrical conductivity of the medium from its map over the boundary (formore reference see [92]).

Calderon asked the question of whether it is possible to determine the electrical con-ductivity of a body by making current and voltage measurements at the boundary. Put in

mathematical terms, given a bounded and non-zero function(x) (possibly complex-valued),which models the electrical conductivity of . Then a potential u H1() satisfies theDirichlet problem

div(u) = 0 in ,u = f on,

(2.3.1)

wheref H12 () is a given boundary voltage potential. The Dirichlet-to-Neumann (DN)map is defined by

[](f) =u

|, (2.3.2)

and if we assume that

[1](f) = [2](f), fH1

2 (),implies that1and 2are equal? The answers have been given in many cases (see for instance[57,90,71,6]).

Moreover, it is well-known that this problem can be reduced to studying the set of Cauchydata for the Schrodinger equation

u+qu = 0, (2.3.3)

with the potentialqgiven by:q=

. (2.3.4)

In the case when the dimension is greater than 3 and all conductivities are in C2(), wehave the following positive result.

Theorem 2.3.1. (Sylvester-Uhlmann 1987, see [90])

If1 = 2 , then1= 2 in.

33 R. Lecaros


34/141


35/141


36/141


A photograph of an olfactory receptor neuron

Figure 2.2: (C) Cilia, (D) Dendrite with a knob at the top, (S) Soma and (A) The axon.

current that is measured with electrodes placed inside and outside the pipette. The insideof the cilium is held at50 mV, causing Na+ ions to flow from the outside of the ciliumto the inside. (Ca2+ is eliminated from all of the solutions and so is unavailable to gate theCa2+-activated Cl channels which are also known to exist in these cilia. Thus there is noCl current.) See[18] for examples of experimental current versus time data.

The diffusion and binding of cAMP into the cilium is modeled by a nonlinear time-dependent partial differential equation which also depends on the channel distribution .

The membrane potential satisfies a second-order boundary value problem which dependson and on the concentration of cAMP. The final equation in our model is the Fredholmintegral equation which has a kernel that depends on the membrane potential and concen-tration of cAMP. It is constrained by the fact that the unknown (the channel distributionfunction ) must be nonnegative. Experimental work has shown that spatial variations inthe concentration ofN a+ are insignificant.

2.4.2 Mathematical model

The following mathematical model was introduced by D. French, R. Flannery, C. Groetsch,W. Krantz and S. J. Kleene in their seminal article [37].

Frog olfactory cilia are thin tubes (diameter 2.8105 cm [69] with lengths L in therange from 1.5 103 cm to as much as 20.0 103 cm) that are connected to the ends ofthe dendrites of the olfactory receptor neurons. We are mainly interested in the distributionof CNG channels = (x). The units of are number of CNG channels per unit length (incm) and x is the distance from the base of the cilium where it is attached to the rest of theolfactory receptor neuron.

36 R. Lecaros


37/141


There are two key assertions about the binding of cAMP to the CNG channels in thepresence of a local concentration of cAMP inside the cilium. These assumptions allow us toquantify the number of bound cAMP molecules and the current from the entrance ofNa+

ions. The radial dynamics are hypothesized to be insignificant; thus the problem is spatiallyone-dimensional in the length variable x, see Figure 2.3. The first assumption is

S(t, x) =Bs(x)F(c(t, x)), (2.4.1)

whereS(t, x) is the number of bound cAMP per unit length of the cilia on a thin ring at xand timet in the presence of cAMP concentration c(t, x) andF is the classical dimensionlessHills function, given by

F(x) = |x|n|x|n +n , (2.4.2)

where the constant >0 is the half-maximal, becauseF() = 12

. In generalt[0, T] whereTis a time in seconds when the experimental measurements are stopped. Typically the Hillsfunction parameters are the exponent n= 1.7 and half-maximal concentration = 1.7 M.The function c is the concentration of cAMP in the thin ring in the cilium at (t, x) and has

units ofM. The number Bs (its units are molecules/channel) measures the binding sites.Even though there are 4 cAMP binding sites on each CNG channel [68] the exponent n inthe Hills function suggests only 1 2 sites are needed; we have chosen to take Bs= n.

Figure 2.3: We approximated the cilium as a line segment.

The current flow assumption is

J(t, x) =J0(x)F(c(t, x))V(t, x), (2.4.3)

(see [18]) whereJ(t, x) is theN a+ current through the CNG channels at point x on timet, and has units of pA/cm. The functionV(t, x) is the membrane potential, really part ofthe driving force with equilibrium potential equal to zero and has units of mV. The constant

J0 = gCNGP,

37 R. Lecaros


38/141


where the conductance is gCNG = 8.31012 S/channel and the constant P = .70 is themaximum open probability for the channels.

Now we derive a key equation relating the total current at t, which we labelI(t), toc(x, t)and (x). We assume that currents due to capacitive effects are negligible. Using the factthat

I(t) = L0

J(t, x)dx= J0 L0

(x)V(t, x)F(c(t, x))dx. (2.4.4)

If is the only unknown, this is a Fredholm integral equation of the first kind. These equationsare well understood (see, e.g., [27,47]), but the more complicated nonlinear system that arisesin this way of modeling is non-standard. We must also impose the natural constraint that >0.Assuming that the rate of change of concentration of cAMP is balanced by Ficks lawof diffusion of the cAMP and the binding, which is directly proportional to S

t (see [25]), it

follows, after matching units, that

c

t D

2c

x2 =S

t,

where the constant = 1NAA 2.7 106 mol cm

liter molecule and D = 2.7 106cm2/s. Here, NA isAvogadros number andA is the cross-sectional area of the cilium. We have chosen D to besmaller than the diffusion constant for ATP, a similar molecule to cAMP, and bigger thanthe values predicted in [18]. Using F0 and (2.4.1), we obtain

c

t D

1 +BsF(c)2c

x2 = 0, t >0, x(0, L)

c(0, x) = 0, x(0, L)

c(t, 0) = c0, t >0

c

x(t, L) = 0, t >0,

(2.4.5)

The final equation, for the membrane potential, is derived using standard cable theory (see,for instance, [52]). Current between the two electrodes passes through two significant resis-tances in series: the resistance of the ciliary membrane, and a longitudinal resistance alongthe length of the solution filling the cilium. This second resistance varies with distance alongthe length of the cilium, as does the transmembrane voltage. We find that V = V(x, t)satisfies

1ra

2

Vx2

(t, x) = J(t, x), t >0, x(0, L)

V(t, 0) = V0, t >0

V

x(t, L) = 0, t >0,

(2.4.6)

where ra = 1.491011 (S cm)1 is the intracellular resistance to longitudinal current ofthe saline solution in the cilium (here, ra is formed by dividing the intracellular resistivity

38 R. Lecaros


39/141


Ri= 91.7 cm by the cross-sectional areaA). We have assumed the usual capacitance termis negligible as well as the background conductance.

We note that this problem can be transformed into a different form. By integrating thedifferential equation involving the membrane potential (2.4.6) in x from 0 to L we obtain,using the fact that the flux at the right boundary is zero and incorporating the integral

equation (2.4.4),1

ra

V

x(t, 0) =I(t). (2.4.7)

Thus, we can view our problem as having unknowns c, V and and consisting of thesystem (2.4.5,2.4.6) and the new condition above (2.4.7). In this reformulation we wouldview the new condition (2.4.7) as compensating for the extra unknown.

French et al. [37]develops a mathematical model for this inverse problem and a numericalapproximation of the solution. In [36], they found a solution in an approximate sense, thatis, they consider a perturbation of the problem, considering the parameter

=

c0

nsmall enough and the approximation

F(x)1 xn

, x(0, 1).

In [38], under some approximations, the authors propose a linear inverse problem in whichthey find an explicit solution and propose a method to find a numerical solution to the inverseproblem, the main approach corresponds to

c(t, x)w(t, x) =c0erfc x

2

Dt

, (2.4.8)

where erfc is the complementary error function, defined by

erfc(z) = 1 2

z0

exp(2)d. (2.4.9)

V and Sare constant, andF(w(t, x))

H(2t

x2),

where w(t, x) represents the concentration of cAMP, H is the Heaviside unit step functioncentered at the origin

H(s) =

0 , s 0

and = 2

D erfc1(/c0).

39 R. Lecaros


40/141


41/141


2.6 Results for the operator m

This section is devoted to the proof of Theorems 2.2.3,2.2.5,2.2.7and2.2.8. These are theprincipal results for the operator m. First we find a more easy expression for m, using(2.1.12) we have

Im[](t) = J0 L

0

(x)Km(t, x)dx (2.6.1)

= J0F(c0)mj=1

aj

L0

(x)H(c(x, t) j)dx (2.6.2)

= J0F(c0)mj=1

aj

Gj(t)(0,L)

(x)dx, (2.6.3)

withGj(t) ={x R : c(x, t)j}, since erfc function is decreasing, we have

Gj(t) = 0, jt, (2.6.4)where{j}mj=1 are given by (2.2.2) and 1> 2> > m. Thus we have

Im[](t) =J0F(c0)

mj=1

aj

hj(t)

0

(x)dx

, (2.6.5)wherehj(s) = min{L, js}.

Replacing (2.6.5) in (2.2.6) and using (2.2.7) we obtain

m[](t) = 1

J0F(c0)Im[](t

2)

=mj=1

aj (hj(t)) t0.(2.6.6)

In the rest of the paper we use (2.6.6) for the operator m.

Using the linearity of the operator m and since mj=1 aj = 1,in the equation (2.6.39) we

obtainm[f] =f, for any constant functionf. (2.6.7)

Proof of the Theorem2.2.3.

Let be an analytic function such that

m[](t) =n

j=1

aj(hj(t)) = 0, t(0, ),

41 R. Lecaros


42/141


43/141


Proof. We need to consider two cases: s= 0 and s >0.

Ifs >0, actuallyk0, such that k0 < s, and using (2.6.13) we obtain the result, becausek =s, k > k0.

Otherwise if k

s,

k, replacing in (2.6.13), we have

k+1= k,

then k = 0k10, but that is impossible because s >0.

Now ifs = 0,replacing in (2.6.13) we obtain

k =0k.

Then using (2.6.12) we havef() = 0,(0, L).

That completes the proof.

Lemma 2.6.3. Let f : [0, L] R be a function, let s, 0 [0, 1) , 1,...n (0, 1) andak >0, k= 0,..,n,such that1 > 2 > > n0, and

a0f(t) +n

j=1

ajf(jt) = 0, t[sL,L), (2.6.14)

andf()0, [0L, L). (2.6.15)

Thenf() = 0, [L,L), (2.6.16)

where= ns.

Proof. The proof is done by induction on n. In fact:

Case n= 1. In this case, from (2.6.14) we have the following equations

a0f(t) +a1f(1t) = 0t[sL,L), (2.6.17)

f() = 0, [0L, L), (2.6.18)and 01. Then if we apply Lemma2.6.1,with g = f, we obtain

f() = 0 [1L, 1L),where1 = 1max{0, s}, and we have

f() = 0 [1L, L),because 01.

43 R. Lecaros


44/141


45/141


wherej = j1

, j = 2,..,n+ 1, and we have 1 > 0/n+1, then using (2.6.22) we have

() = 0, [1 0n+1

L, L). (2.6.23)

On the other hand, from (2.6.21), we have

a0f() +(1) = 0 [sL,L),

then from (2.6.22) and Lemma2.6.1, with g = , we conclude that

() = 0, [1max{0, s}L, 1L).

Now we set s1= 1max{0, s}, and using (2.6.23) we have 0, on [s1L, 1L)[1 0n+1L, L),therefore

() =a1f() +n+1

i=2aif(i) = 0, [s1L, L). (2.6.24)

We can note that, 1 > 2 >3 > > n+1, and 0n+1 < n+11 =n+1. Then if we usethe induction hypothesis in (2.6.24) and (2.6.22), we obtain

f() = 0, [1L, L),

where 1 = s1n+1 = n+1max{s, 0} < n+1. Then we can repeat the last argumentreplacing0 by 1, and we have

f() = 0, [kL, L), k1,

where k =n+1max{s,k1}, k1, (2.6.25)with 0 = 0 given. If s = 0, we replace t = 0, in (2.6.21) we have f(0) = 0, and usingLemma2.6.2we obtain

f() = 0, [L,L),where = n+1s, which completes the proof.


Let : [0, L]

R be a function such that

m[](t) =mj=1

aj(hj(t)) = 0, t[0, Lm],

then ift = Lm, we obtainhj(Lm) =L, j= 1,...,m,

thus0 = m[](Lm) =(L). (2.6.26)

45 R. Lecaros


46/141


47/141


and 0= mmi

=m, then again Lemma2.6.3, we obtain

() = 0, [mLmi1, L].

Then() = 0, [mLk1, L],

for k = 1,..,m, that implies() = 0, [0, L].

This concludes the proof of Theorem2.2.5.

Now we will prove Theorems2.2.7.


We do some estimates,

hj2L2(0,Lm) = Lm

0

2(hj(t))dt

=

Lj0

2(jt)dt+2(L)L

1

m 1

j

1

j

L0

2(t)dt+2(L)L

m

1m

2L2(0,L)+2(L)L

1

m 1 + (Tr(L))2 L 2H1(0,L) , (2.6.32)where

Tr(u), u > 0, (2.6.33)is the norm of the trace operator in H1(0, u).

Now, if we set

c1= 1

m

1 + (Tr(L))2 L

12 ,

Then using (2.6.32) we obtain

m[]L2(0,Lm)mj=1

aj hjL2(0,Lm)c1 H1(0,L) . (2.6.34)

47 R. Lecaros


48/141


49/141


Lemma 2.6.5. Let: [0, L] R, andf : [0, Lm]R be functions such that

m[](t) =f(t), t[0, Lm]. (2.6.39)Then

M[f](s) = mj=1

ajsj M[](s) fors in the fundamental strip associated with,(2.6.40)

where

f(t) =

f(t) f(Lm) 0tLm,

0 t0,, and (t) =

(t) (L) 0tL,

0 tL.(2.6.41)

Proof. If we replace t = Lm in (2.6.39), we have

(L) =f(Lm).

Using (2.6.7), we obtainm[](t) = f(t), t0. (2.6.42)

On the other hand, using Lemma2.6.4,we have

m[](t) =mj=1

aj(jt), t0,

and, for any b >0, by Lemma2.5.1, we have

M[(bt)](s) =bsM[](s),thus

M [m[]] (s) = mj=1

ajsj

M[](s), s< , > .

Using this and applying Mellin transform in (2.6.39) we obtain (2.6.40).

Lemma 2.6.6. LetH1(0, L) be a function. Then for any > 0, there exists a positiveconstantC>0, such that

0,,LCm[]0,,Ln + |m[](Lm)| , (2.6.43)

whereConly depends onL, 1, m1, m, am and.

Proof. Let >0 be a fixed constant. Using Lemma 2.6.5and Lemma2.5.1we obtain mj=1

aj(+ 1

2is)

j

M[x(x)](s) = M [xm[](x)] (s), s R, (2.6.44)

49 R. Lecaros


50/141


51/141


Thus if we set

C= 1

cmax

1

n

+1/2+c

L+1/2

2+ 1, 1

,

we obtain (2.6.43). The proof of Lemma2.6.6is complete.


Using Lemma2.6.6, we have that for any > 0, there exists a constant C > 0, suchthat

0,,LCm[]0,,Lm + |m[](Lm)|

C((Lm) + Tr(Lm)) m[]H1(0,Lm) ,

where Tr(Lm) has been defined in (2.6.33) . Thus if we set

C2 = C

L

n

+ Tr(Lm)

,

we obtain (2.2.11). The proof of Theorem2.2.8is complete.

2.7 Proof of stability result

In this section we will proof Theorems2.2.1and 2.2.2.


Let us fix any >0 and : [0, L] R be a function in L2(0, L). From (2.2.6) we obtainthat

(xIm[](x)) = x1Im[](x) +x(Im[](x))

= x1Im[](x) +x

12 J0F(c0)

2 (m[])

(

x), (2.7.1)

where(x) =x

0

()d.

Since L2m0

x21

(m[])(

x)2

dx= 2

Lm0

41 ((m[])())2

d= 2 (m[])20,2 12,Lm

,

51 R. Lecaros


52/141


53/141


Derivating, it follows that

(x+1(x))= (+ 1)x(x) +g+1(x),

thus

< g+1, > = L0

g+1(x)(x)dx= L0(x+1(x)) (+ 1)x(x)(x)dx

= L

0

x+1(x)(x) + (+ 1)x(x)(x)

dx,

then we have

|< g+1, >| 0,+1,L+ (+ 1) 0,,L

H1(0,L)

(L ++ 1) 0,,L H1(0,L) .

Therefore

g+1H1(0,L)(L ++ 1) 0,,L . (2.7.4)

Thus, using Lemma2.6.6we have that for any > max{0, 34}, there exists a constantC>0, such that

1,+1,L =g+1H1(0,L)(L ++ 1) Cm[]0,,Lm + |m[](Lm)|

. (2.7.5)

Using (2.2.6), we have

m[](Lm) =

1

F(c0)J0 Im[](L2

m).

Replacing in (2.7.5) and using (2.7.3) change 2 3/2 by, we obtain

1,+1,L (L ++ 1)

|J0F(c0)| C

12

Im[]0,214,L2m

+ |Im[](L2m)|

= (L ++ 1)

2|J0F(c0)|C

Im[]0,

2 1

4,L2m

+

2

L1

2m

|(L2m)214 Im[](L

2m)|

(L ++ 1)

2|J0F(c0)|C1 +

2

Tr(L2m)

L

1

2m Im[]1,214 ,L2m ,where Tr has been defined in (2.6.33). Therefore if we set

C2 = (L++ 1)

2|J0F(c0)|C

1 +

2

Tr(L2m)

L1

2m

,

we obtain (2.2.5). The proof of Theorem2.2.2is complete.

53 R. Lecaros


54/141


2.8 Case j =0j

This section is devoted to the proof of Theorems 2.2.9and 2.2.10.


Let be a function inC0([0, L]), and we consider{gj}j1 have been defined in (2.2.15-2.2.17).

First we note thatSupp(gk+1)[k+1L, kL), k0. (2.8.1)

Then we can define the sequence{p}pN as

p(x) =

p

j=1gj(x), x0.

Using (2.8.1) we have thatx R \ (0, L),

p(x) = 0, p N,

thenlim

p+p(x) = 0, x R \ (0, L).

On the other hand, we consider the ceiling function

x= min{k Z

kx},

i.e., it is the smallest integer not less thanx.

Now we define

k(x) =

ln(x/L)

ln()

, x(0, L). (2.8.2)

We have the following relation

x[k(x)L, k(x)1L), x(0, L),

therefore we obtain for x(0, L)

p(x) =gk(x)(x), pk(x),thus

limp+

p(x) =gk(x)(x), x(0, L). (2.8.3)

This prove that defined in (2.2.18) is well defined.

On the other hand, replacing (2.2.12) in (2.2.2) we obtain

j =0j, j= 1,...,m.

54 R. Lecaros


55/141


Using (2.8.1) we havesupp()[0, L),

then with Lemma2.6.4we have

m[](t/0) =m

j=1aj(

jt), t0. (2.8.4)

Replacingt = 0 and t = 0Lm in (2.2.13), (2.8.4) and using (2.2.18), (2.2.15) we obtain

m[](0) =g(0), and m[](Lm) =g(0Lm) = 0.

Now, if we take t(0, 0Lm) = (0,L/m), we havejt(0, jmL), forj ={1, 2,...,m}.

We need consider two cases when t < LandtL:

Case t < L. We havejt(0, L), forj={1, 2,...,m},

andk(jt) =j +k(t),

thus replacing (2.8.3) in (2.8.4) we obtain

m[](t/0) =mj=1

ajgj+k(t)(jt),

therefore using (2.2.17) withk+ 1 =m+k(t) we obtain

m[](t/0) =m1j=1

ajgj+k(t)(jt) +amgm+k(t)(

mt)

=m1j=1

ajgj+k(t)(jt) +

g (t)

m1j=1

ajgj+k(t)

jt

= g(t).

Case tL. We setk(x) =

ln(x/L)ln(1/)

, xL,

wherex= max{k Z

kx},is the floor function, i.e., it is the largest integer not greater thanx.

Thus we havek(x)+1x < Lk(x)x, xL, (2.8.5)

55 R. Lecaros


56/141


57/141


58/141


We note thatC(u)1, u(0, 1), (2.8.12)

because C() is decreasing. Therefore

Ck+1

j 1, j {1,...,k},Thus we obtain

gk+1(0,L)C(m)

ak+1mg(k+10Lm,0Lm) .

That prove (2.8.10) for all k ={0,...,m 1.}

Case k+ 1 > m. We supposed (2.8.11) for all n k. Replacing (2.2.17) and using theproperty (2.2.21-2.2.22) and induction hypothesis, we obtain

gk+1[k+1L,kL)

1

am

C(m) g[k+10Lm,k0Lm)+

m1j=1

ajC

m

j

C(m)

aj+km+1mg[j+km+10Lm,0Lm)

=C(m)

ak+1m

akm g[k+10Lm,k0Lm)+

m1j=1

ajC

m

j

g[j+km+10Lm,0Lm)

C(m)

ak+1m akm+m1

j=1 ajCmj g[k+10Lm,0Lm) ,and using (2.8.12) we obtain C(m/j) 1, for all j {1,...,m 1}. This complete theproof of (2.8.10).

On the other hand, using (2.2.19) and (2.2.22), we have

g[k+10Lm,0Lm) = C(0)m[]

[k+1Lm,Lm),

replacing (2.8.8) in (2.8.10) we obtain

gk+1[k+1L,kL)C(m)

ak+1mC(0)

m[]() m[](Lm)[k+1Lm,Lm)

,

and replacing in (2.8.9) we have (2.2.24). This completes the proof of Theorem2.2.10.

58 R. Lecaros


59/141


2.9 Numerical results

Now we will discuss the numerical implementation of the scheme developed in the proof ofTheorem2.2.9.

First we choose{

}mj=1 as in the equation (2.2.12), and we define

Fj =

0 j = 0,

F

j+j+1

2

j= 1,...,m 1,

1 j = m,

then we fixaj as follows:aj =Fj Fj1, j= 1,...,m, (2.9.1)

using thatFis increasing and 0F(x)< 1,for allx0,we haveaj0,for allj = 1,...,m,and m

j=1

aj = 1.

We can see the approximation Hills function in Figure 2.4.

Figure 2.4: Hills function and its approximation

We define a non-regular mesh in the interval [0, L]. We begin by defining

Pq,1 =

x0< x1


60/141


ThusPq,j =

x j1x Pq,1 ,

and their corresponding representative vector

Pj =j1P1 = Pj1 R1+q.

Thus if we fixed p, we want to recover in the mesh

p,q =pj=1Pq,j ,and their corresponding representative vector is given by

P=

P1P2

Pp

R1+q+p.

Using (2.2.15-2.2.17) we define the vectors G1, G2,..., Gm Rq+1, in the following recur-rent form

(G1)s= g1((P1)s) , s= 1,...,q+ 1, (2.9.2)and

(Gk+1)s= 1

am

g

(Pk+1)s

m

kj=1

amk1+j(Gj)s

, s= 1,...,q+ 1, (2.9.3)

for k = 1,..,m 1. Furthermore we define the vectors Gk R1+q for k = m,..., p 1,as

(Gk+1)s= 1

am

g

(Pk+1)s

m

m1j=1

aj(Gj+km+1)s

. (2.9.4)

Thus we set the vector

G=

G1G2

Gp

R1+q+p,

this vector represent a discretization of the function given by Theorem2.2.9on the meshdefined by p,q, i.e. ((P)s, (G)s)

p+q+1s=1 is a discretization of the curve (x, (x)), x(0, L).

Therefore, using (2.2.20) and apply a forward difference schema we obtain an approximateof, i.e. we defined the vectors X, Y Rp+q

(X)s= (P)s, (Y)s= max

(G)s+1 (G)s(P)s+1 (P)s , 0

, s= 1,...,p+q,

60 R. Lecaros


61/141


the maximum appears because we are looking for a density, i.e. a positive function. Thus((X)s, (Y)s)

p+qs=1 is a approximation of the curve (x, (x)), x(0, L).

We consider the example considered in [38], thus we define

I(t) =

0, t

(0, tDelay),

IMax

1 +

KI

t tDelay

nI1, t > tDelay,

(2.9.5)

with tDelay = 30ms, nI 2.2, IMax = 150pA and KI 100ms. The current (2.9.5) is asigmoidal function with short delay (Figure 2.5B), which is similar to the profiles in someapplications (see[18], [33] or[58]).

A: approximation of. B: current I(t) as defined by (2.9.5)

Figure 2.5: Approximation of the function (x) and with current I(t) as defined by (2.9.5)

Thus we obtain the numerical solution as shown in Figure 2.5A. In this figure one seesthat the main features of the solution of this model are consistent with those obtained in[33].

2.10 Polynomial approximation of the Hills functionIn this section we study a different approximation of the kernel (2.1.13), we will consider apolynomial approximation of the Hills function (2.1.3), aroundc0.

We consider Pm the Taylors polynomial of degree m, from Hills function defined by(2.1.3), at point c0 > 0. Thus we have

F(x) =Pm(x c0) +O(|x c0|m+1). (2.10.1)

61 R. Lecaros


62/141


Therefore we define the polynomial kernel approximation by

P Km(t.x) =Pm(w(t, x) c0), (2.10.2)

wherew(t, x) is the solution of (2.1.1), given by

w(t, x) =c0 2c02L+k=0

e2kDt

kk(x) , (2.10.3)

with

k =2k+ 1

2L , (2.10.4)

and

k(x) =

2

L

1/2sin(kx). (2.10.5)

Thus we define the total current with polynomial approximation, given by

P Im[](t) =

L0

(x)P Km(t, x)dx, t >0. (2.10.6)

Now we present our principal result released with the operator (2.10.6).

Theorem 2.10.1. Letm8 be a constant. Then we have

KerP Im ={0},

where

KerP Im = f L2(0, L) P Im[f](t) = 0,t >0 .Remark 2.10.2. Theorem2.10.1 corresponds to the identifiability for the operatorP Im.

2.10.1 Proof of the identifiability Theorem2.10.1

First we note that{k}k0 is a orthonormal basis in L2(0, L). Thus for anyf L2(0, L),wehave

f(x) = k0 < f, k > k(x), x(0, L), (2.10.7)where

< f, k >=

L0

f(x)k(x)dx.

If we set

G(t, x) =+k=0

e2kDt

kk(x), (2.10.8)

62 R. Lecaros


63/141


we have

G(t, )2L2(0,L) =+k=0

e22kDt

2kk2L2(0,L)

=+

k=0e2

2kDt

2k

e220Dt

+k=0

1

2k

,

using+k=0

1

(2k+ 1)2 =

2

8,

we obtain

G(t,

)

2L2(0,L)

2

8

e220Dt,

t

0. (2.10.9)

On the other hand we have the following

Lemma 2.10.3. Let{ak}k0 be a bounded sequence, i.e. exists a positive constantC > 0,such that|ak| C, for allk0. Then we have that the series

k0ake

2kt

is absolutely convergent for allt >0.

Proof. We need prove that the series is bounded, in factk0

ake2kt Ck0

e

2t4L2

(2k+1)2

Ck0

e

2t4L2

k=

C

1 e 2t

4L2

.

This completes the proof.

Now, we define for all t >0,

k(t) =e

2kDt

k, k0,

and using Lemma2.10.3we havek0

1

ke

2kDt =

k0

k(t),

is absolutely convergent for all t >0.

63 R. Lecaros


64/141


On other hand, we define the set

m=

nj=1

2nj

nj0, nm

. (2.10.10)

Let us setPm(z) =0+1z+ +mzm,

where j R for all j = 0, 1,...,m. Let >0 fixed constant, and for any L2(0, L), weobtain, for all s0, that

P Im[](+s) = 0

L0

(x)dx+mj=1

j(2c0

2L)j L

0

k0

e2kD( +s)

kk(x)

j(x)dx

= 0

L0

(x)dx

+mj=1

j(2c0

2L)j

k1,...,kj0e

jp=1

2kpDs L0

jp=1kp()kp(x)(x)dx.

Therefore we define{a(, )}m , such that

P Im[](+s) = 0

L0

(x)dx+m

a(, )eDs, s0. (2.10.11)

Lemma 2.10.4. Let

L2(0, L), be a function, such that

P Im[](t) = 0, t >0. (2.10.12)

Then we have L0

(x)dx= 0, (2.10.13)

anda(, ) = 0, m. (2.10.14)

Proof. First we see that the series defined in (2.10.11) is absolute convergent for all s > 0,

in fact using Lemma2.10.3, we only need to prove that a(, ) are bounded for all m,therefore we have

|a(, )| mj=1

|j(2c0

2L)j

k1,...,kj0

L0

jp=1kp()kp(x)(x)dx

mj=1

|j(2c0

2L)j

k0

|k()|j

L2(0,L)< +, m,

(2.10.15)

64 R. Lecaros


65/141


66/141


replacing (2.10.21) and (2.10.13) in (2.10.12) we obtain

an+1(, )en+1D +en+1DsSn+2(s) = 0, s >0, (2.10.22)

but using (2.10.18) we have

|en+1DsSn+2(s)| Ctee(n+2n+1)Ds,

thus note that n+1< n+2 and take a limit when s go to 0, in (2.10.22) we obtainan+1(, ) = 0.

Then we have (2.10.14).

This complete the proof of Lemma2.10.4.

Lemma 2.10.5. If we consider{k}k0, defined in (2.10.4) and we assumed that2n1+ +2nk =2n. (2.10.23)

Then we havek= 1 mod8. (2.10.24)

Proof. We have

2n= 2

4L2(4(n) + 1),

where(n) =n2 +n,and replacing in (2.10.23) we obtain

k+ 4ki=1

(ni) = 1 + 4(n),

using that (p) is even for all p N,we obtain (2.10.24).


LetL2(0, L) be a function, from Lemma2.10.4we have (2.10.14) and ifm8, usingLemma2.10.5we have that for all k0, doesnt exist n1,...,nj, such that

2k =

j1

2nj ,

for anyjm, then for = 2km, we obtain

au2k(, ) =e2kD

F(c0)(2c02L) L0 k(x)(x)dx,but F(c0)= 0,because Fis increasing, then we have

< k, >= 0, k0,therefore from (2.10.7) we obtain

= 0.

This complete the proof of the Theorem2.10.1.

66 R. Lecaros


67/141


68/141


where the constant >0 is the half-maximal, because F() = 12 . The constant n >0, is aparameter, typically greater than 1.

The binding produces a change in the current flux J(t, x) through the channels, which isgoverned by the following system:

J(t, x) = P

gCNGBS S(t, x)v(t, x), t >0, x(0, L) (2.11.4)

1

Ra

2v

x2 = J(t, x), t >0, x(0, L) (2.11.5)

v(t, 0) = v0, t >0 (2.11.6)v

x(t, L) = 0, t >0, (2.11.7)

where the constant P > 0, represents the probability of to find open the CNG channel,and the constants Ra and gCNG are the resistance density of the longitudinal current in thecilium and the conductivity of CNG channel, respectively. We can see that the above systemis parametric in the variable t, and the constant v0 > 0 represents the voltage outside thecilium, furthermore the boundary condition at the end of the cilium is consistent with thefact that it is closed.

We can measure the quantityI(t), which corresponds to the total current, which is simplythe integral ofJ in the interval [0, L]:

I(t) =

L0

J(t, x)dx= 1Ra

v

x(0, t). (2.11.8)

Our inverse problem consists in given the positive constants L,D,w0, Bs,n,,P, gCNG, Ra, v0,which define constitutive properties of the medium, if we measured I(t) for all t > 0 (see2.11.8), we are interested in to recover , which is the density of channels along the surfaceof the cilium.

Let us define q(x) = PRaF(w0) (x), and replace it in (2.11.5), then we obtain

v = qv.

Thus, we can reduce this inverse problem to following form: we want to determine thepositive potentialq(x) such that:

v qv = 0, x(0, L)

v(0) = v0,

v(L) = 0,

(2.11.9)

and we measured v (0), where q(x) = PRaF(w0) (x), and with this form, we obtain .

68 R. Lecaros


69/141

aplicaciones a la biologia.pdf

Documents