Chapter 1.11, Problem 7E

To determine

To find: The long term dynamic in each of the cases and find one that will beat other every time and displays 2:1 AV block and which show some sort of Wenckebach phenomenon.

Answer to Problem 7E

The value of “c” when a little above the threshold value of 0.667 to maintain the equilibrium, the heart, beat 12 times and it builds up to a higher and higher potential. Eventually, the potential is so high that the AV node is not able to recover and the heart fails to beat. As soon as the beat fails the potential drops and the process starts again, this phenomenon is known as the Wenckebach phenomenon.

Explanation of Solution

Given:

The given parameter is,

  $\begin{array}{l}{V}_c=20.0\text{mV}\\ u=\text{10}\text{.0}\text{mV}\\ c=\text{0}\text{.7}\\ V_t=30.0\text{mV}\end{array}$

Calculation:

The potential of the AV node ${\widehat{V}}_t$ just before receiving the next signal from the SA node is given by,

  ${\widehat{V}}_t=c{V}_t$

Then,

  $\begin{array}{c}{\widehat{V}}_t=0.7\left(30.0\text{mV}\right)\\ =21.0\text{mV}\end{array}$

For ${\widehat{V}}_t>V_c$ the heart is not ready to beat and for $V_{t+1}=\widehat{V}$ if ${\widehat{V}}_t=V_c$ .

For ${\widehat{V}}_t\le V_c$ the AV node responds and tell the heart to beat and $V_{t+1}={\widehat{V}}_t+u$ if ${\widehat{V}}_t\le V_c$ .

To translate the description into discrete dynamical system we must write $V_{t+1}$ entirely in terms of $V_t$ eliminating the ${\widehat{V}}_t$ terms.

Thus, the two cases is summarized as,

  $V_{t+1}=\{\begin{array}{l}c{V}_t\text{if}c{V}_t>V_c\\ c{V}_t+u\text{if}c{V}_t<V_c\end{array}$

Since ${\widehat{V}}_t>V_c$ then,

  $\begin{array}{c}{V}_{t+1}=c{V}_t\\ =0.7V_t\end{array}$

Now setting $V^{*}$ to be equilibrium point so we have.

  $\begin{array}{c}{V}^{*}=\frac{u}{1-c}\\ =\frac{10}{1-0.7}\\ =33.33\text{mV}\end{array}$

The equilibrium exist only if the heart is indeed ready to beat when the next signal comes or if,

  $\begin{array}{c}c{V}^{*}=c\frac{u}{1-c}\\ =\frac{\left(0.7\right)10}{1-0.7}\\ =23.33\text{mV}\end{array}$

Thus, the equilibrium exists only if $c{V}^{*}=23.33\text{mV}$ is less than $V_c=20.0$ as it is not the heart will not beat every time.

Thus, there are two conditions, one is of 2.1 AV block where the heart beats only with every other stimulus and the other is the Wenckebach phenomenon where the heart beats normally for a while and then it skips the beat and then resumes normal beating and repeats cycle.

Consider the potential just after the beating is $V_t$ .

If the potential after the two full cycles comes back exactly to where it was started the heart beats with every signal that produces 2:1 AV block.

The expression for the updated potential after the two cycle’s matches the original potential if $V_t$ is equal to some value $\overline{V}$ that satisfies the following equation.

  $\begin{array}{l}{c}^2\overline{V}+u=\overline{V}\\ c\overline{V}>V_c\\ c^2\overline{V}a<V_c\end{array}$

Then, the solution is,

  $\overline{V}=\frac{u}{1-c^h}$

The case when the inequalities are satisfied.

  $\begin{array}{c}\overline{V}=\frac{10}{1-{\left(0.7\right)}^2}\\ =19.60784\end{array}$

As the inequality $c\overline{V}<V_c$ is not satisfied.

Thus, it is not a case of 2:1 AV block.

Compute the condition on c for the existence of the equilibrium.

The equilibrium is,

  $c\frac{u}{1-c}\le V_c$

Then,

  $\begin{array}{l}\frac{10c}{1-c}\le 20\\ c\le 2-2c\\ c\le \frac{2}{3}\\ c\le 0.6667\end{array}$

Thus, the value of “c” when a little above the threshold value of 0.667 to maintain the equilibrium, the heart, beat 12 times and it builds up to a higher and higher potential. Eventually, the potential is so high that the AV node is not able to recover and the heart fails to beat. As soon as the beat fails the potential drops and the process starts again, this phenomenon is known as the Wenckebach phenomenon.