Question 1

In order to refresh your matrix-vector multiply skills please calculate, by hand, the product ATGA A G A in the 3 compartment case and write out the 4 equations in the vector equation we arrived at in step (S4): ATGAx=f A G A x f .

Question 2

We began our discussion with the 'hope' that a multicompartment model could indeed adequately capture the fiber's true potential and current profiles. In order to check this one should run fib1.m with increasing values of NN until one can no longer detect changes in the computed potentials.

Let us now interpret this convergence. The main observation is that the difference equation, Equation 1, approaches a differential equation. We can see this by noting that ⅆz≡lN ⅆ z l N acts as a spatial 'step' size and that x k x k , the potential at k−1ⅆz k 1 ⅆ z , is approximately the value of the true potential at k−1ⅆz k 1 ⅆ z . In a slight abuse of notation, we denote the latter xk−1ⅆz x k 1 ⅆ z Applying these conventions to Equation 1 and recalling the definitions of R i R i and R m R m we see Equation 1 become πa2 ρ i -x0+2xⅆz−x2ⅆzⅆz+2πaⅆz ρ m xⅆz=0 a 2 ρ i x 0 2 x ⅆ z x 2 ⅆ z ⅆ z 2 a ⅆ z ρ m x ⅆ z 0 or, after multiplying through by ρ m πaⅆz ρ m a ⅆ z , a ρ m ρ i -x0+2xⅆz−x2ⅆzⅆz2+2xⅆz=0 a ρ m ρ i x 0 2 x ⅆ z x 2 ⅆ z ⅆ z 2 2 x ⅆ z 0 . We note that a similar equation holds at each node (save the ends) and that as N→∞ N and therefore ⅆz→0 ⅆ z 0 we arrive at

We shall determine αα and ββ by paying attention to the ends of the fiber. At the near end we find πa2 ρ i x0−xⅆzⅆz= i 0 a 2 ρ i x 0 x ⅆ z ⅆ z i 0 which, as ⅆz→0 ⅆ z 0 becomes