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CAAM 335 Chapter 1 Exercises

Module by: Doug Daniels, Steven Cox

Summary: Excercises for Chapter 1.

Problem 1:

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 .

Problem 2:

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.

  • (a) Please run fib1.m with N = 8, 16, 32, and 64. Plot all of the potentials on the same (use hold) graph, using different line types for each. (You may wish to alter fib1.m so that it accepts NN as an argument).

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

d2dz2xz-2 ρ i a ρ m xz=0 z2 x z 2 ρ i a ρ m x z 0 (2)

  • (b) With μ≡2 ρ i a ρ m μ 2 ρ i a ρ m show that
    xz=αsinhμz+βcoshμz x z α 2 μ z β 2 μ z (3)
    satisfies Equation 2 regardless of αα and ββ.

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

ddzx0=- ρ i i 0 πa2 z x 0 ρ i i 0 a 2 (4)
At the far end, we interpret the condition that no axial current may leave the last node to mean
ddzxl=0 z x l 0 (5)

  • (c) Substitute Equation 3 into Equation 4 and Equation 5 and solve for αα and ββ and write out the final xz x z .
  • (d) Substitute into xx the ll, aa, ρ i ρ i , and ρ m ρ m values used in fib1.m, plot the resulting function (using, e.g., ezplot) and compare this to the plot achieved in part (a).

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