Chapter 1.11, Problem 19E

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 19E

The inequality $c\le 0.6667$

is not satisfied and it is not the case of the wenchback behaviour of 2:1 AV block nor the Wenckebach phenomenon bears a behaviour that is random in nature and means the heart beat not follow a sequence but it beats the random way that is in accordance to the signal by the AV node from the SA mode.

Explanation of Solution

Given:

The parameters are,

  $\begin{array}{l}{V}_c=20.0\text{mV}\\ u=\text{10}\text{.0}\text{mV}\\ c=\text{0}\text{.4,}0.9,1\\ V_t=30.0\text{mV}\end{array}$

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,

  $\begin{array}{c}{\widehat{V}}_t=0.4\left(30.0\text{mV}\right)\\ =12.0\text{mV}\end{array}$

For ${\widehat{V}}_t<V_c$ now setting $V^{*}$ to the equilibrium point we have,

  $\begin{array}{c}{V}^{*}=\frac{u}{1-c}\\ =\frac{10}{1-0.4}\\ =16.66\text{mV}\end{array}$

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

  $\begin{array}{c}cV*=c\frac{u}{1-c}\le V_c\\ =0.4\left(16.66\right)\\ =6.664\text{mV}\end{array}$

Thus, the equilibrium exists only if $cV*=6.664\text{mV}$ is less than $V_c=20.0$ as it is not the heart that will not beat every time with the voltage decaying from 20 to 6.664 beats and increase back to 6.664 on the beat.

For $c=0.09$ ,

  $\begin{array}{c}{\widehat{V}}_t=c{V}_t\\ =0.9\left(30.0\text{mV}\right)\\ =27\text{mV}\end{array}$

This gives $V_t^{*}>V_c$ .

Now set $V*$ to the equilibrium that is,

  $\begin{array}{c}V*=\frac{u}{1-c}\\ =\frac{10.0}{1-0.9}\\ =100\text{mV}\end{array}$

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

  $cV*=c\frac{u}{1-c}\le V_c$

The above equation is,

  $\begin{array}{c}cV*=c\frac{u}{1-c}\le V_c\\ =0.9\left(100\right)\\ =90\text{mV}\end{array}$

Thus, the equilibrium exists only if $cV*=90\text{mV}$ is less than $V_c=20.0$ as it shows that heart cannot beat every time. Therefore, the heart fails to beat with every signal.

The updated potential after two cycles matches the original potential if $V_t$ is equal to some value $\overline{V}$ that satisfies the equation is,

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

Then, the solution has,

  $\begin{array}{c}\overline{V}=\frac{u}{1-c^2}\\ =\frac{10.0}{1-{0.9}^2}\\ =52.63\text{mV}\end{array}$

The dynamic through the complete cycle finding is,

  $c^2\overline{V}+u=48.36$

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

The equation for the equilibrium.

  $V^{+}=\frac{u}{1-c}$

The requirement that $c{V}^{*}\le V_c$ , the equilibrium exists only when,

  $c\frac{u}{1-c}\le V_c$

From the values of $u$ and $V_c$ is,

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

The inequality is not satisfied as $c\le 0.6667$ so there is not a case of Wenckebach phenomenon.

Consider the value of $c=1$ then,

  $\begin{array}{c}\widehat{V}=c{V}_t\\ =1\left(30.0\text{mV}\right)\\ =30\text{mV}\end{array}$

Set the value of $V^{}*$ for the equilibrium point as,

  $\begin{array}{c}{V}^{*}=\frac{u}{1-c}\\ =\frac{10}{1-1}\\ =0\text{mV}\end{array}$

The equilibrium exists only when the heart is ready to beat at the arrival time of the new signal arrives or if,

  $\begin{array}{c}c{V}^{*}=c\frac{u}{1-c}\le V_c\\ =1\left(0\right)\\ =0\text{mV}\end{array}$

The equilibrium is there only if $c{V}^{*}=0$ is less than $V_c=20.0$ . It is not because the heart cannot beat every time but because the heart fails to beat with every signal.

The value of the potential after these two cycles come back exactly to where it started, thus the heart beat with the potential with every signal producing 2:1 AV block.

The updated potential after two cycles matches the original potential of $V_t$ is equal to some value of $\overline{V}$ that satisfies the following equation, this gives,

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

Thus, the solution is,

  $\begin{array}{c}\overline{V}=\frac{u}{1-c^2}\\ =\frac{10}{1-1}\\ =0\text{mV}\end{array}$

The equation for the equilibrium is,

  $\begin{array}{l}{V}^{*}=\frac{u}{1-c}\le V_c\\ 10\frac{c}{1-c}\le 2\\ c\le \frac{2}{3}\\ c\le 0.6667\end{array}$

Thus, the inequality $c\le 0.6667$

is not satisfied and it is not the case of the wenchback behaviour of 2:1 AV block nor the Wenckebach phenomenon bears a behaviour that is random in nature and means the heart beat not follow a sequence but it beats the random way that is in accordance to the signal by the AV node from the SA mode.