DerivationSphericalCell

• Here is the differential equation shown in Equation 17, rewritten slightly differently: $$\tau \frac{dV}{dt} = -V$$

• If we want to find the solution as the system charges up to a final longterm value of $V_{\infty}$, which may be some value other than zero, we can rewrite this equation as $$\frac{dV}{dt} = \frac{-(V-V_{\infty})}{\tau}. \quad\text{(Equation 17.1)}$$ Adding a constant to the equation in this way does not affect its solution. Note how similar this is to the equation we developed for the gating variables of the Hodgkin/Huxley model (Equation 11 in unit Electrophysiology V).

• To integrate the equation, collect similar variables on either side of the equation: $$-\frac{dV}{V - V_{\infty}} = \frac{dt}{\tau}. \quad\text{(Equation 17.2)}$$

• Now integrate both sides, setting the limits of integration on the lefthand side from $V_{0}$ to $V_{m}(t)$, and the limits of integration on the righthand side from time 0 to time $t$. For simplicity, the constant $1/\tau$ is factored out of the righthand side, and the negative multiplier on the lefthand side of Equation 17.2 is moved over to the righthand side: $$\int\limits_{V_0}^{V_{m}(t)}\frac{1}{V - V_{\infty}} \, dV = -\frac{1}{\tau} \int\limits_{0}^{t} \, dt. \quad\text{(Equation 17.3)}$$

• Both integrations are very straightforward. The lefthand side evaluates to $\ln (V - V_{\infty})$, and the righthand side to $t$. When these functions are evaluated at their limits, the result is $$\ln (V_{m}(t) - V_{\infty}) - \ln (V_{0} - V_{\infty}) = -\frac{t}{\tau} + B, \quad\text{(Equation 17.4)}$$ where $B$ is a constant of integration.

• This can be simplified to $$\ln \left(\frac{V_{m}(t) - V_{\infty}}{V_{0} - V_{\infty}}\right) = -\frac{t}{\tau} + B, \quad\text{(Equation 17.5)}$$

• We can easily determine the constant of integration by realizing that at time $t = 0$, $$B = \ln \left(\frac{V_{0} - V_{\infty}}{V_{0} - V_{\infty}}\right) = \ln 1 = 0. \quad\text{(Equation 17.6)}$$

• Raising both sides of Equation 17.5 to the base of the natural logarithm $e$ leads to $$\frac{V_{m}(t) - V_{\infty}}{V_{0} - V_{\infty}} = e^{-t/\tau}, \quad\text{(Equation 17.7)}$$

• Multiplying both sides by the denominator of the lefthand side, we obtain $$V_{m}(t) - V_{\infty} = (V_{0} - V_{\infty}) e^{-t/\tau}, \quad\text{(Equation 17.8)}$$

• Solving for $V_{m}(t)$ yields $$V_{m}(t) = V_{\infty} - (V_{\infty} - V_{0}) e^{-t/\tau}, \quad\text{(Equation 17.9)}$$ which if differentiated and substituted back into Equation 17 satisfies that equation.

• A simpler way of writing this equation is obtained if $V_{0}$ is added to and subtracted from the righthand side, and terms are collected: $$V_{m}(t) = V_{\infty} + V_{0} - V_{0} - (V_{\infty} - V_{0}) e^{-t/\tau}, \quad\text{(Equation 17.10)}$$ which equals $$V_{m}(t) = V_{0} + (V_{\infty} - V_{0}) - (V_{\infty} - V_{0}) e^{-t/\tau}. \quad\text{(Equation 17.11)}$$

• Factoring out the term $(V_{\infty} -V_{0})$, one obtains $$V_{m}(t) = V_{0} + (V_{\infty} - V_{0}) (1 - e^{-t/\tau}). \quad\text{(Equation 18)}$$