### Derivation of the PDE

Thomson starts by defining the variables that he uses throughout the work. I shall define the same variables using more modern lettering conventions. Let $C$ be the capacitance of the wire, $R$ the resistance of the wire, $V$ the potential at a point $P$ on the wire at a time $t$, and $I$ the current at the same point in the same instant.

The charge $Q$ at $P$, called "quantity of electricity" by Thomson, is given by

\[Q = VC = It\]

Thus, in an infinitesimal length $dx$ of wire at $P$, we have a charge of $VCdx$ and, in an infinitesimal period of time, the charge that leaves $P$ is equal to

\[dtdI = dt\frac{dI}{dx}dx\]

Using two equations for the electromotive force, Thomson showed that $RI = -\frac{\partial V}{\partial x}$. Combining these formulae gives the all-important simple form of the telegraphist's equation:

\begin{equation}

CR\frac{\partial V}{\partial t} = \frac{\partial^2V}{\partial x^2}

\end{equation}

Thomson notes instantly that this is the heat equation.

In a real wire, there would be some electrical leakage into the surrounding water. By introducing some coefficient $h$ to measure this loss, Thomson recovers the equation:

\[CR\frac{\partial V}{\partial t} = \frac{\partial^2V}{\partial x^2} -hV\]

but with the change of variable:

\[V = e^{-\frac{h}{RC}t}\phi\]

he obtains a PDE in the original form, i.e:

\[CR\frac{\partial \phi}{\partial t} = \frac{\partial^2\phi}{\partial x^2}\]

### Solutions

where the wire is understood to be infinitely-long, $z:=x\sqrt{RC}, \tilde{V}$ is the voltage to which the end $O$ is instantaneously raised and $T$ is the time for which this potential is maintained. By cleverly noting that:

\begin{align}

\int_{t-T}^t \cos(2n\theta - zn^\frac{1}{2})d\theta &= \left[\frac{1}{2n}\sin(2n\theta-zn^\frac{1}{2})\right]_{t-T}^t \\

&= \frac{1}{n}\left[\sin(2nt-zn^\frac{1}{2})-\sin(2n(t-T)-zn^\frac{1}{2})\right]

\end{align}

and taking $T$ to be infinitesimal, so that $\sin(T) \approx T$ with $t > 0$, Thomson could conclude that:

\begin{align}

V &= T\frac{2\tilde{V}}{\pi}\int_0^\infty e^{-zn^\frac{1}{2}}\cos(2nt-zn^\frac{1}{2})\\

&= T\frac{\tilde{V}z}{2\pi^\frac{1}{2}t^\frac{3}{2}}e^{-\frac{z^2}{4t}}

\end{align}

Thomson now moves to finding the solution to the problem when a charge $Q$ is communicated instantaneously to the wire at $O$. He states the strength of the current $I$ at any point in the wire to be equal to $-\frac{1}{k}\frac{dV}{dx}$ and deduces that the "maximum electrodynamic effect" of an impulse into the wire will therefore be found by finding the value of t making $\frac{\partial V}{\partial z}$ (this being proportional to $\frac{\partial V}{\partial x}$) maximal: with $V$ given by an equation similar to the above:

\[V = \frac{e^{-\frac{z^2}{4t}}}{\sqrt{t}}\frac{Q}{\sqrt{\pi}}\sqrt{\frac{R}{C}}\]

Upon finding $\frac{\partial^2 V}{\partial z^2}$, we find that $\frac{\partial V}{\partial z}$ is maximal when:

\[\frac{1}{2t^\frac{3}{2}} = \frac{z^2}{4t^\frac{5}{2}}\]

giving the value:

\[t = \frac{z^2}{2} = \frac{RCx^2}{2}\]

Note that this is different to the value of t which Thomson gives, being $\frac{z^2}{6}$. This is however entirely irrelevant to the remainder of the exposition; the point of the calculation was to show that the time that one should leave between message pulses due to the "retardations of signals" was proportional to the square of the length of the wire, now taking x to be the length of the cable. This is a deduction which Thomson’s erroneous calculation still makes.

Thomson then goes on to discuss the "velocity of transmission" of the signal. He states that, if the potential at $O$ is varied in a sinusoidal way (namely with $\sin(2nt)$), then this velocity is equal to $2\sqrt{\frac{n}{RC}}$ . It is interesting that the velocity of transmission and the retardations of the signals bear no mathematical relation to one another: showing again that this problem is, perhaps counter-intuitively, one regarding the length of the wire

### Application to the cable

###
Conclusion

*sure*beforehand that the American telegraph will succeed"