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Matrix Analysis of the Branched Dendrite Nerve Fiber

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

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Summary: In this module we will undertake a real-world example of the electrical modeling scheme we've built up in previous modules (see introductory paragraph). Modeling an actual 3-branched dendrite fiber with the Strang Quartet, we will solve the resulting system with the Backward-Euler method.

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Introduction

In the prior modules on static and dynamic electrical systems, we analyzed basic, hypothetical one-branch nerve fibers using a modeling methodology we dubbed the Strang Quartet. You may be asking yourself whether this method is stout enough to handle the real fiber of our minds. Indeed, can we use our tools in a real-world setting?

Presentation

Figure 1: A pyramidal neuron from the CA3 region of a rat's hippocampus, scanned at (FIX ME) X magnification.
An Actual Nerve Fiber
An Actual Nerve Fiber (neuron.png)

To answer your question, the above is a rendering of a neuron from a rat's hippocampus. The tools we have refined will enable us to model the electrical properties of a dendrite leaving the neuron's cell body. A three-branch model of such a dendrite, traced out with painstaking accuracy, appears in the diagram below.

Figure 2: Multi-compartment electrical model of a rendered dendrite fiber.
3-branch Dendrite Model
3-branch Dendrite Model (dendrite_model.png)

Our multi-compartment model reveals a 3 branch, 10 node, 27 edge structure to the fiber. Note that we have included the Nernst potentials, the nervous impulse as a current source, and the additional leftmost edges depicting stimulus current shunted by the cell body.

We will continue using our previous notation, namely: R i R i and R m R m denoting cell body and membrane resistances, respectively; xx representing the vector of potentials x 1 … x 10 x 1 … x 10 , and yy denoting the vector of currents y 1 … y 27 y 1 … y 27 . Using the typical value for a cell's membrane capacitance, c= 1 ( μ F / cm 2 ) , c 1 ( μ F / cm 2 ) , we derive (see variable conventions):

Definition 1: Capacitance of a Single Compartment
C m =2πalNc C m 2 a l N c
This capacitance is modeled in parallel with the cell's membrane resistance. Additionally, letting A cb A cb denote the cell body's surface area, we recall that its capacitance and resistance are
Definition 2: Capacitance of cell body
C cb = A cb c C cb A cb c
Definition 3: Resistance of cell body
R cb = A cb ρ m R cb A cb ρ m .

Applying the Strang Quartet

Step (S1')--Voltage Drops

Let's begin filling out the Strang Quartet. For Step (S1'), we first observe the voltage drops in the figure. Since there are a whopping 27 of them, we include only the first six, which are slightly more than we need to cover all variations in the set: e 1 = x 1 e 1 x 1 e 2 = x 1 − E m e 2 x 1 E m e 3 = x 1 − x 2 e 3 x 1 x 2 e 4 = x 2 e 4 x 2 e 5 = x 2 − E m e 5 x 2 E m e 6 = x 2 − x 3 … e 6 x 2 x 3 … e 27 = x 10 − E m e 27 x 10 E m

In matrix for, letting bb denote the vector of batteries, e=b−Ax       where       b=- E m 010010010001001001001001001 e b A x       where       b E m 01001 00100 01001 00100 10010 01 and A=-1000000000-1000000000-11000000000-1000000000-1000000000-11000000000-1000000000-1000000000-110000000001-1000000000-1000000000-1000000000-11000000000-1000000000-1000000000-11000000000-1000000000-1000000-10001000000000-1000000000-1000000000-11000000000-1000000000-1000000000-11000000000-1000000000-1 A -10000 00000 -10000 00000 -11000 00000 0-1000 00000 0-1000 00000 0-1100 00000 00-100 00000 00-100 00000 00-110 00000 0001-1 00000 0000-1 00000 0000-1 00000 0000-1 10000 00000 -10000 00000 -10000 00000 -11000 00000 0-1000 00000 0-1000 000-10 00100 00000 00-100 00000 00-100 00000 00-110 00000 000-10 00000 000-10 00000 000-11 00000 0000-1 00000 0000-1 Although our adjacency matrix AA is appreciably larger than our previous examples, we have captured the same phenomena as before.

Applying (S2): Ohm's Law Augmented with Voltage-Current Law for Capacitors

Now, recalling Ohm's Law and remembering that the current through a capacitor varies proportionately with the time rate of change of the potential across it, we assemble our vector of currents. As before, we list only enough of the 27 currents to fully characterize the set: y 1 = C cb ddt e 1 y 1 C cb t e 1 y 2 = e 2 R cb y 2 e 2 R cb y 3 = e 3 R i y 3 e 3 R i y 4 = C m ddt e 4 y 4 C m t e 4 y 5 = e 5 R m y 5 e 5 R m y 27 = e 27 R m y 27 e 27 R m In matrix terms, this compiles to y=Ge+Cddte , y G e C t e , where

Conductance matrix

G=00000000000000000000000000001 R cb 0000000000000000000000000001 R i 00000000000000000000000000000000000000000000000000000001 R m 0000000000000000000000000001 R i 00000000000000000000000000000000000000000000000000000001 R m 0000000000000000000000000001 R i 0000000000000000000000000001 R i 00000000000000000000000000000000000000000000000000000001 R m 0000000000000000000000000001 R i 00000000000000000000000000000000000000000000000000000001 R m 0000000000000000000000000001 R i 00000000000000000000000000000000000000000000000000000001 R m 0000000000000000000000000001 R i 000000000000000000000000000000000000000000000000000000001 R m 0000000000000000000000000001 R i 00000000000000000000000000000000000000000000000000000001 R m 0000000000000000000000000001 R i 00000000000000000000000000000000000000000000000000000001 R m G 00000 00000 00000 00000 00000 00 0 1 R cb 000 00000 00000 00000 00000 00 00 1 R i 00 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 0000 1 R m 00000 00000 00000 00000 00 00000 1 R i 0000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00 1 R m 00 00000 00000 00000 00 00000 000 1 R i 0 00000 00000 00000 00 00000 0000 1 R i 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 0 1 R m 000 00000 00000 00 00000 00000 00 1 R i 00 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 0000 1 R m 00000 00000 00 00000 00000 00000 1 R i 0000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00 1 R m 00 00000 00 00000 00000 00000 000 1 R i 00 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 1 R m 0000 00 00000 00000 00000 00000 0 1 R i 000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 000 1 R m 0 00 00000 00000 00000 00000 0000 1 R i 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 0 1 R m (1)
and

Capacitance matrix

C= C cb 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 C m 0000000000000000000000000000 . C C cb 0000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 000 C m 0 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 0 C m 000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 C m 0000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 000 C m 0 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 0 C m 000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 0000 C m 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00 C m 00 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 00 00000 00000 00000 00000 00000 C m 0 00000 00000 00000 00000 00000 00 . (2)

Step (S3): Applying Kirchoff's Law

Our next step is to write out the equations for Kirchoff's Current Law. We see: i 0 − y 1 − y 2 − y 3 =0 i 0 y 1 y 2 y 3 0 y 3 − y 4 − y 5 − y 6 =0 y 3 y 4 y 5 y 6 0 y 6 − y 7 − y 8 − y 9 =0 y 6 y 7 y 8 y 9 0 y 9 − y 10 − y 19 =0 y 9 y 10 y 19 0 y 10 − y 11 − y 12 − y 13 =0 y 10 y 11 y 12 y 13 0 y 10 − y 11 − y 12 − y 13 =0 y 10 y 11 y 12 y 13 0 y 13 − y 14 − y 15 − y 16 =0 y 13 y 14 y 15 y 16 0 y 16 − y 17 − y 18 =0 y 16 y 17 y 18 0 y 19 − y 20 − y 21 − y 22 =0 y 19 y 20 y 21 y 22 0 y 22 − y 23 − y 24 − y 25 =0 y 22 y 23 y 24 y 25 0 y 25 − y 26 − y 27 =0 y 25 y 26 y 27 0 Since the BB coefficient matrix we'd form here is equal to AT A , we can say in matrix terms: ATy=-f A y f where the vector ff is composed of f 1 = i 0 f 1 i 0 and f 2 ... 27 =0 f 2 ... 27 0 .

Step (S4): Stirring the Ingredients Together

Step (S4) directs us to assemble our previous toils together into a final equation, which we will then endeavor to solve. Using the process derived in the dynamic Strang module, we arrive at the equation

ATCAddtx+ATGAx=ATGb+f+ATCddtb , A C A t x A G A x A G b f A C t b , (3)
which is the general form for RC circuit potential equations. As we have mentioned, this equation presumes knowledge of the initial value of each of the potentials, x0=X x 0 X .

Observing our circuit, and letting 1 R foo = G foo 1 R foo G foo , we calculate the necessary quantities to fill out Equation 3's pieces (for these calculations, see dendrite.m): ATCA= C cb 0000000000 C m 0000000000 C m 000000000000000000000 C m 0000000000 C m 0000000000 C m 0000000000 C m 0000000000 C m 0000000000 C m A C A C cb 0000 00000 0 C m 000 00000 00 C m 00 00000 00000 00000 0000 C m 00000 00000 C m 0000 00000 0 C m 000 00000 00 C m 00 00000 000 C m 0 00000 0000 C m ATGA= G i + G cb - G i 00000000- G i 2 G i + G m - G i 00000000- G i 2 G i + G m - G i 00000000- G i 3 G i - G i 00- G i 00000- G i 2 G i + G m - G i 00000000- G i 2 G i + G m - G i 00000000- G i G i + G m 000000- G i 0002 G i + G m - G i 00000000- G i 2 G i + G m - G i 00000000- G i G i + G m A G A G i G cb G i 00000 000 G i 2 G i G m G i 00000 00 0 G i 2 G i G m G i 00000 0 00 G i 3 G i G i 00 G i 00 000 G i 2 G i G m G i 0000 0000 G i 2 G i G m G i 000 00000 G i G i G m 000 000 G i 000 2 G i G m G i 0 00000 00 G i 2 G i G m G i 00000 000 G i G i G m ATGb= E m G cb G m G m 0 G m G m G m G m G m G m A G b E m G cb G m G m 0 G m G m G m G m G m G m ATCddtb=0 , A C t b 0 , and an initial (rest) potential of x0= E m 1111111111 . x 0 E m 11111 11111 .

Applying the Backward-Euler Method

Since our system is so large, the Backward-Euler method is the best path to a solution. Looking at the matrix ATCA A C A , we observe that it is singular and therefore non-invertible. This singularity arises from the node connecting the three branches of the fiber and prevents us from using the simple equation x ′=Bx+g , x B x g , we used in earlier Backward-Euler-ings. However, we will see that a modest generalization to our previous form yields Equation 4:

D x ′=Ex+g D x E x g (4)
capturing the form of our system and allowing us to solve for xt x t . We manipulate Equation 4 as follows: D x ′=Ex+g D x E x g D x ˜ t− x ˜ t−dtdt=E x ˜ t+g D x ˜ t x ˜ t dt dt E x ˜ t g D−Edt x ˜ t=D x ˜ t−dt+gdt D E dt x ˜ t D x ˜ t dt g dt x ˜ t=D−Edt-1D x ˜ t−dt+gdt , x ˜ t D E dt D x ˜ t dt g dt , where in our case D=ATCA , D A C A , E=-ATGA , and E A G A , and g=ATGb+f . g A G b f . This method is implemented in dendrite.m with typical cell dimensions and resistivity properties, yielding the following graph of potentials.

Figure 3
Graph of Dendrite Potentials
(a) Large view of potentials.
Figure 3(a) (potential_graph1.png)
(b) Zoomed view of potentials showing maxima.
Figure 3(b) (potential_graph2.png)

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