Question

A heart pacemaker, shown in Figure 3.1.15, consists of a switch, a battery, a capacitor, and the heart as a resistor. When the switch Sis at P, the capacitor charges; when Sis atQ, the capacitor discharges, sending an electrical stimulus to the heart. In Problem 58 in Exercises 2.3 we saw that during this time the electrical stimulus is being applied to the heart, the voltageE across the heart satisfies the linearDE.

$\frac{\mathbf{d}\mathbf{E}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{R}\mathbf{C}}\mathit{E}$

(a) Let us assume that over the time interval of length the switch S is at position P shown in Figure 3.1.15 and the capacitor is being charged. When the switch is moved to position Q at time ${\mathit{t}}_{\mathbf{1}}$ the capacitor dis-charges, sending an impulse to the heart over the time interval of length ${\mathit{t}}_{\mathbf{2}}\mathbf{,}{\mathit{t}}_{\mathbf{1}}\mathbf{⩽}\mathit{t}\mathbf{<}{\mathit{t}}_{\mathbf{1}}\mathbf{+}{\mathit{t}}_{\mathbf{2}}$. Thus over the initial charging/discharging interval $\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{<}{\mathit{t}}_{\mathbf{1}}\mathbf{+}{\mathit{t}}_{\mathbf{2}}$the voltage to the heart is actually modelled by the piecewise-linear differential equation

$\frac{\mathbf{d}\mathbf{E}}{\mathbf{d}\mathbf{t}}\mathbf{=}\{\begin{array}{l}0,0⩽t<t_1\\ -\frac{1}{\mathrm{RC}}E,t_1⩽t<t_1+t_2\end{array}$

By moving S between P and Q, the charging and discharging over time intervals of lengths ${\mathit{t}}_{\mathbf{1}}$ and ${\mathit{t}}_{\mathbf{2}}$is

Repeated indenitely. Suppose ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathit{s}\mathbf{,}{\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{s}\mathbf{,}{\mathit{E}}_{\mathbf{0}}\mathbf{=}\mathbf{12}\mathit{V}$, and $\mathit{E}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{E}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{=}\mathbf{12}\mathbf{,}\mathit{E}\mathbf{\left(}\mathbf{6}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{E}\mathbf{\left(}\mathbf{10}\mathbf{\right)}\mathbf{=}\mathbf{12}\mathbf{,}\mathit{E}\mathbf{\left(}\mathbf{12}\mathbf{\right)}\mathbf{=}\mathbf{0}$and so on. Solve for $\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}$for $\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{⩽}\mathbf{24}$.

(b) Suppose for the sake of illustration that $\mathit{R}\mathbf{=}\mathit{C}\mathbf{=}\mathbf{1}$. Use a graphing utility to graph the solution for the IVP in part (a) for$\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{⩽}\mathbf{24}\mathbf{.}$

Short answer

$$\begin{gathered} E=\left\{12e^{\frac{4-t}{RC}},t\in [4,6) \\ 12e^{\frac{10-t}{RC}},t\in [10,12) \\ 12e^{\frac{16-t}{RC}},t\in [16,18) \\ 12e^{\frac{22-t}{RC}},t\in [22,24) \\ 0,otherwise\right. \end{gathered}$$

Step-by-step solution

Find the E(t)

(a) For$t_1=4$ and $t_1=2$, the differential equation will be zero on $[0,4]$and then nonzero on $[4,6]$, then again zero on $[0,6]$, then nonzero on $[10,12]$and so on. Let’s first write our differential equation on $[0,24]$.

$$\begin{gathered} \begin{array}{rcl}\frac{dE}{dt}& =& \left\{-\frac{1}{RC}E,t\in [4,6)\cup [10,12)\cup [16,18)\cup [22,24) \\ 0,t\in [0,4)\cup [6,10)\cup [12,16)\cup [18,22)\right.\\ \frac{dE}{dt}& =& 0\end{array} \end{gathered}$$

Has the solution

$E=c$

For some constant c and the initial conditions

$E\left(0\right)=E\left(6\right)=E\left(12\right)=E\left(18\right)=0$

Imply that

$E=0$

On $[0,4)\cup [6,10)\cup [12,16)\cup [18,22)$.

Solve for E(t),0⩽t⩽24, 

On $[0,4)\cup [6,10)\cup [12,16)\cup [18,22)$we have

$\begin{array}{rcl}\frac{dE}{dt}& =& -\frac{1}{RC}E\\ \frac{1}{E}dE& =& -\frac{1}{RC}dt\\ \int \frac{1}{E}dE& =& \int -\frac{1}{RC}dt\\ InE& =& -\frac{1}{RC}t+c\\ E& =& d{e}^{-\frac{1}{RC}dt}\end{array}$

On $[4,6]$, $E\left(4\right)=12\Rightarrow 12=d{e}^{-\frac{4}{RC}}\Rightarrow d=12e^{\frac{4}{RC}}$

On $[10,12]$, $E\left(10\right)=12\Rightarrow 12=d{e}^{-\frac{10}{RC}}\Rightarrow d=12e^{\frac{10}{RC}}$

On $[16,18]$, $E\left(16\right)=12\Rightarrow 12=d{e}^{-\frac{16}{RC}}\Rightarrow d=12e^{\frac{16}{RC}}$

On $[22,24]$, $E\left(22\right)=12\Rightarrow 12=d{e}^{-\frac{22}{RC}}\Rightarrow d=12e^{\frac{22}{RC}}$

Now we can fully write our equation on $[0,24]$.

$$\begin{gathered} E=\left\{12e^{\frac{4-t}{RC}},t\in [4,6) \\ 12e^{\frac{10-t}{RC}},t\in [10,12) \\ 12e^{\frac{16-t}{RC}},t\in [16,18) \\ 12e^{\frac{22-t}{RC}},t\in [22,24) \\ 0,otherwise\right. \end{gathered}$$

To graph the solution for the IVP

Let’s graph the equation for $R=C=1$on $[0,24]$.