Some Explicit Solutions of the Cable Equation

Transcription

1 Some Explicit Solutions of the Cable Equation Marco Herrera-Valdéz and Sergei K. Suslov Mathematical, Computational and Modeling Sciences Center, Arizona State University, Tempe, AZ , U.S.A. W. Miller s S 4 Conference September 19, 2010

2 Introduction The Cable Equation Analytic, Asymptotic and/or Numerical Solutions for a Single Branch A Graphical Approach Transient Solutions Appendix: Modi ed Orthogonality Relation

3 1 Introduction Neurons are among the most important and interesting cells in the body. They are the fundamental building blocks of the central nervous system and hence responsible for motor control, cognition, perception, and memory, among other things. Although our understanding of how networks of neurons interact to form an intelligent system is extremely limited, one prerequisite for an understanding of the nervous system is an understanding of how individual nerve cells behave. A typical neuron consists of three principal parts: the dendrites; the cell body, or soma; and the axon. There is a great deal of experimental data indicating that parts of neurons (dendrites) conduct electricity in a passive manner. Thus, there has been developed a theory describing the ow of electricity in neurons using the theory of electrical ow in cables.

4 Graphical methods are very useful and popular in di erent branches of modern physics. It is worth noting, for example, Feynman diagrams in quantum mechanical or statistical eld theory, Vilenkin-Kuznetsov-Smorodinskii approach to solutions of n-dimensional Laplace equation, applications in solid-state theory, etc. Cable theory plays a central role in many areas of electrophysiology, but, to the best of our knowledge, graphical methods seem have not been widely applied yet in the mathematical modeling of neurons. A goal of this presentation is to make a modest step in this direction. We use explicit solutions from recent papers on variable quadratic Hamiltonians in nonrelativistic quantum mechanics.

5 2 The Cable Equation The cable equation was derived for the rst transatlantic telegraph cable, around 1855, by Professor William Thomson (Lord Lelvin). As a student in Paris, Thompson had mastered the mathematical methods pioneered by Fourier including PDEs and separation of variables. (In electrophysiology, the magnetic eld can be neglected.) For the axial symmetric pro le r = r (x) ; where x represents distance along the axis in the membrane variable cylinder, the cable theory implies the following set of equations: 2rI m ds ; ds dx = s 1 + dr 2 (1) dx I m = V R m + ; (2) I = r2 : (3)

6 Here, V represents the voltage di erence across the membrane (interior minus exterior) as a deviation from its resting value, I m is the membrane current density, I = I a is the total axial current, R m is the membrane resistance, R i is the intercellular resistivity and C m is membrane capacitance. Di erentiating equation (3) with respect to x and substituting the result into (1) with the help of (2) one gets V + C m R m 1 = R m r 2@V 2R dr 2 (4) which is the cable equation with tapering for a single branch of dendritic tree. ; dx Once again, V represents the voltage di erence across the membrane, R m is the membrane resistance, R i is the intercellular resistivity and C m is membrane capacitance.

7 We shall be particularly interested in solutions of the cable equation (4) corresponding to termination with a sealed end, namely, when at the end point x = x 1 the membrane cylinder is sealed with a disk composed of the same membrane. In this case, the corresponding boundary condition can be derived by setting I a = r 2 I m ; (5) at x = x 1 : Then, in view of (2) (3), one gets: V + C m + R m x=x1 = 0; t 0: In a similar fashion, at the somatic end one gets V + C s R R x=x0 = 0; t 0; (6) (7) where R s is the somatic resistance and C s is the somatic capacitance.

8 We shall mainly concentrate on steady-state solutions of the cable equation, when 0: Then V s 1 + (x 0 x x 1 ) and dr V j x=x0 = V 0 ; dx 2 = R m 2rR i d dx dv dx + B (x) V r 2dV dx (8) x=x 1 = 0; (9) B (x 1 ) = R i R m : This boundary value problem can be conveniently solved in terms of standard solutions of this second order ordinary di erential equation as follows V (x) = V (x 0 ) C (x) + B (x 1) S (x) C (x 0 ) + B (x 1 ) S (x 0 ) ; (10) where C (x) and S (x) are two linearly independent solutions of the stationary cable equation (8) that satisfy special boundary conditions C (x 1 ) = 1; C 0 (x 1 ) = 0 and S (x 1 ) = 0; S 0 (x 1 ) = 1:

9 Then dv dx + B (x) V = 0 (x 0 x x 1 ) (11) with the corresponding current density/voltage ratio function B (x) given by B (x) = V 0 V = C0 (x) + B (x 1 ) S 0 (x) C (x) + B (x 1 ) S (x) (12) in term of the standard solutions C (x) and S (x) : Throughout this talk, we shall refer to a case, when B (x) > 0 (x 0 < x < x 1 ) ; as the case of weak tapering. An opposite situation, when B () = 0 at a certain point x 0 < < x 1 and an inverse of the current may occur, shall be called a case of the strong tapering.

10 3 Tapering with Analytic, Asymptotic and/or Numerical Solutions We consider a general model of a dendrite as a (binary) directed tree (from the soma to its terminal ends) consisting of axially symmetric branches with the following types of tapering. 3.1 Cylinder (Standard Model) Here, r = r 0 =constant, 0 x L and the cable equation takes the simplest form 2d2 V dx 2 = V; with a familiar solution: 2 = r 0R m 2R i (13) V (x) = V 0 cosh ((L x) =) + B L sinh ((L x) =) cosh (L=) + B L sinh (L=) (14)

11 subject to boundary conditions Then V (0) = V 0 ; B L V (L) + dv dx (L) = 0: (15) B (x) = B L + tanh ((L x) =) + 2 B L tanh ((L x) =) : (16) 3.2 Frustum(Cone) Here, r = r (x) = r 0 + r 1 r 0 L x with 0 x L: The steady-state solution of the corresponding cable equation r d2 V dx 2 +2dV dx = 2 4 V; = s 8Ri R m 1 + r 1 r 0 L r 1 r 0 L subject to boundary conditions (15) are given by 2 1=4 (17) C (L x) + B V (x) = V L S (L x) 0 : (18) C (L) + B L S (L)

12 Here, the standard solutions (that satisfy S (0) = 0; S 0 (0) = 1 and C (0) = 1; C 0 (0) = 0) can be constructed as follows S (x) = and 2Lr 3=2 0 (r 1 r 0 ) p r h K 1 ( p p r 0 ) I 1 r (19) I 1 ( p r 0 ) K 1 p r i C (x) = [ p r 0 K 0 ( p r 0 ) + 2K 1 ( p r 0 )] (20) s r0 p I 1 r r 1 + [ p r 0 I 0 ( p r 0 ) 2I 1 ( p r 0 )] s r0 p K 1 r r 1 in terms of modi ed Bessel functions I (z) and K (z) of orders = 0; 1.

13 3.3 Hyperbola If r = a cosh x b a on an interval x0 x x 1 ; the cable equation (4) takes the x b V + C m R m cosh a = R " m x x 2 # 2 sinh 2R i + a cosh V 2 : This special case of tapering is integrable in terms of elementary functions (a similar problem occurs in a model of the dumped quantum oscillator). For the steady-state solutions one obtains the following equation with new parameters V tanh (x + ) V 0 = 2 0 V (22) = 1 a ; = b; 2 0 = 2R i ar m : (23)

14 The corresponding two linearly independent solutions, namely, V 1 (x) = V 2 (x) = ( = sinh (x + ) cosh (x + ) ; (24) cosh (x + ) cosh (x + ) q ) can be veri ed by a direct substitution for an arbitrary parameter : The required steady-state solution of the boundary value problem V (x 0 ) = V 0 ; is given by BV (x 1 ) + dv dx (x 1) = 0 (25) V (x) = V 0 C (x 1 x) + BS (x 1 x) C (x 1 x 0 ) + BS (x 1 x 0 ) ; (26)

15 where and C (x 1 x) (27) = cosh ( (b x 1 )) cosh ( (x 1 x)) cosh ( (x b)) + sinh ( (b x 1 )) sinh ( (x 1 x)) cosh ( (x b)) S (x 1 x) = cosh ( (b x 1 )) sinh ( (x 1 x)) cosh ( (x b)) : (28) 3.4 General Case of Axial Symmetry One can use numerical methods and/or WKB approximation in order to obtain standard solutions.

16 4 A Graphical Approach Graphical rules for voltages and currents in a model of dendritic tree with tapering are as follows. 4.1 Single Axially Symmetric Branch with Arbitrary Tapering For a single branch with tapering voltage and current density/voltage ratio are given by V (x) = C (x 1 x) + B (x 1 ) S (x 1 x) C (x 1 x 0 ) + B (x 1 ) S (x 1 x 0 ) V (x 0 ) (29) = (C (x 1 x) + B (x 1 ) S (x 1 x)) V (x 1 ) ; B (x) = C0 (x 1 x) + B (x 1 ) S 0 (x 1 x) C (x 1 x) + B (x 1 ) S (x 1 x) = V 0 (x) V (x) ; respectively (see Figure). (30)

17 4.2 Junction of Three Branches with Different Types of Tapering The internal potential and current are assumed to be continuous at all dendritic branch points and at the somadendritic junction. We consider a general case when each branch has its own tapering, say r = r (x) ; r 1 = r 1 (x) and r 2 = r 2 (x) (see Figure). Then V (x 12 ) = V 1 (x 12 ) = V 2 (x 12 ) ; (31) r 2 (x 12 ) B (x 12 ) (32) = r 2 1 (x 12) B 1 (x 12 ) + r 2 2 (x 12) B 2 (x 12 ) : The total ratio constant B (x 12 ) at the branching point x 12 is given by the following expression B (x 12 ) = r2 1 (x 12) r 2 (x 12 ) B 1 (x 12 ) + r2 2 (x 12) r 2 (x 12 ) B 2 (x 12 )(33) = B (B 1 (x 1 ) ; B 2 (x 2 )) = r2 1 (x 12) C1 0 (x 1 x 12 ) + B 1 (x 1 ) S1 0 (x 1 x 12 ) r 2 (x 12 ) C 1 (x 1 x 12 ) + B 1 (x 1 ) S 1 (x 1 x 12 ) + r2 2 (x 12) C2 0 (x 2 x 12 ) + B 2 (x 2 ) S2 0 (x 2 x 12 ) r 2 (x 12 ) C 2 (x 2 x 12 ) + B 2 (x 2 ) S 2 (x 2 x 12 ) :

18 Then the ratio constant B (x 0 ) is B (x 0 ) = C0 (x 12 x 0 ) + B (x 12 ) S 0 (x 12 x 0 ) C (x 12 x 0 ) + B (x 12 ) S (x 12 x 0 ) (34) with the coe cient B (x 12 ) found by the previous formula (33). 4.3 Junction of (n + 1)-Branches In a similar fashion, at the branching point x ; one gets V (x ) = V 1 (x ) = V 2 (x ) = ::: = V n (x ) ; (35) r 2 (x ) B (x ) (36) = r 2 1 (x ) B 1 (x ) + ::: + r 2 n (x ) B n (x ) = nx i=1 (See Figure.) r 2 i (x ) B i (x ) :

19 Combination of the above graphical rules results in a simple algorithm of evaluation of voltages and currents in the model of dendritic tree under consideration as follows: Evaluate constants B (x ) for all branching points of the tree: (a) rst apply formula (33) for all open notes; (b) remove the above nodes from the tree and keep repeating the previous step until you reach the root of tree (soma). In order to nd voltage at a point x of the dendritic tree, follow the path x 0! x and multiply the initial voltage V (x 0 ) by consecutive corresponding factors from formula (29) changing at each intersection of the tree. The ratio of voltages V (x ) and V x at two terminal points, can be determine in a graphic form by the previous rule applied to the shortest path x! x : (Examples are left to the reader.)

20 5 Transient Solutions Let us consider the cable equation (4) for a single branch with an arbitrary smooth tapering r = r (x) on the interval x 0 x x 1 : The separation of variables V (x; t) = e (1+2 )t= m u (x) ; m = C m R m (37) in results in 1 d r 2du +! 2 ds r dx dx dx u = 0; where is a separation constant.!2 = 2R i R m 2 ; (38) The boundary condition at the sealed end (6) takes the form du 1 dx 2!2 u = 0: (39) x=x 1 A general solution of this problem can be written (for each branch of the dendritic tree) as follows u (x) = u (x;!) = A C (x;!) + 1 2!2 S (x;!) ; (40)

21 where A is a constant and C (x;!) and S (x;!) are two linearly independent standard solutions of equation (38) that satisfy special boundary conditions C (x 1 ;!) = 1; C 0 (x 1 ;!) = 0 and S (x 1 ;!) = 0; S 0 (x 1 ;!) = 1: Then the boundary condition (7) at the somatic end x = x 0 ; namely, du Ri dx + s 1 + C s! u 2 = 0 R s m 2C m x=x0 (41) with s = C s R s results in a transcendental equation " s R!# m! 2 Ri (42) 2R i R s m = C0 (x 0 ;!) + 1 2!2 S 0 (x 0 ;!) C (x 0 ;!) + 1 2!2 S (x 0 ;!) for the eigenvalues!: (There are in nitely many eigenvalues.) The corresponding eigenfunctions u n = u (x;! n ) are orthogonal (u m ; u n ) = mn (u n ; u n ) (43)

22 with respect to an inner product that is given in terms of the Lebesgue Stieltjes integral (see Appendix) (u; v) := Z x1 x 0 u (x) v (x) r (x) ds (44) r2 (x 1 ) u (x 1 ) v (x 1 ) + C s r 2 (x 0 ) u (x 0 ) v (x 0 ) : 2C m A formal solution of the corresponding initial value problem takes the form "! t V (x; t) = X A n exp 1 + R m! 2 n (45) n 2R i m C (x;! n ) + 1 2!2 ns (x;! n ) + V (x; 1) ; where V (x; 1) is the steady-state solution and! =! n are roots of the transcendental equation (42). Coe - cients A n can be obtained with the help of the modi ed orthogonality relation (43) as follows # A n = (V (x; 0) V (x; 1) ; u n (x)) ; (46) (u n (x) ; u n (x)) u n (x) = C (x;! n ) + 1 2!2 ns (x;! n ) :

23 Substitution of (46) into (45) results in V (x; t) = V (x; 1) (47) + Z Supp G (x; y; t) (V (y; 0) V (y; 1)) d (y) ; where G (x; y; t) = X " exp 1 + R! # m! 2 t n (48) n 2R i m u n (x) u n (y) ku n k 2 is an analog of the heat kernel. The case of a piecewise tapering r = ( r0 (x) ; x 0 x x 01 r 1 (x) ; x 01 x x 1 (49) with r 0 x01 = r1 x + 01 is considered in a similar fashion.

24 6 Summary In this presentation, we have discussed a simple graphical approach to steady state solutions of the cable equation for a general model of dendritic tree with tapering. A simple case of transient solutions is also analyzed.

25 7 Appendix: Modi ed Orthogonality Relation We consider the Sturm Liouville type problem, Lu + u = 0; (50) for the second order di erential operator Lu = d k (x) du q (x) u; (51) dx dx where k; q and are continuous real-valued functions on an interval [x 0 ; x 1 ] ; k and are positive in [x 0 ; x 1 ] ; k 0 exists and is continuous in [x 0 ; x 1 ] ; subject to modi ed boundary conditions u 0 (x 0 ) + (a 0 + b 0 ) u (x 0 ) = 0; (52) u 0 (x 1 ) + (a 1 b 1 ) u (x 1 ) = 0; where a 0 ; b 0 0 and a 1 ; b 1 0 are constants.

26 With the help of the second Green s formula, Z x1 (vlu ulv) dx = k v du u dv x 0 dx dx x 1 x 0 ; (53) for two eigenfunctions u and v corresponding to di erent eigenvalues Lu + u = 0; Lv + v = 0; 6= (54) one gets the following orthogonality relation: Z x1 x 0 u (x) v (x) dx (55) +b 1 k (x 1 ) u (x 1 ) v (x 1 ) +b 0 k (x 0 ) u (x 0 ) v (x 0 ) = 0: Here, the modi ed inner product (u; v) := = Z x1 x 0 Z Supp u (x) v (x) dx uv d (56) +b 1 k (x 1 ) u (x 1 ) v (x 1 ) + b 0 k (x 0 ) u (x 0 ) v (x 0 ) is de ned in terms of the Lebesgue Stieltjes integral.