Page:Scientific Memoirs, Vol. 2 (1841).djvu/486

modified by the cotemporaneous circumstances, must occur from that which is peculiar to the rapid current to that belonging to the state of perfect equilibrium. Here, then, is a wide field open for future researches.

23. In cases where the permanent state is not instantaneously assumed, as it usually is in dry piles, we should, in order to become acquainted with the changes of the circuit up to that period, proceed from the complete equation

$\gamma {\frac {du}{dt}}=\chi {\frac {d^{2}u}{dx^{2}}}-{\frac {bc}{\omega }}u$,

(✳)

because in this case we cannot consider ${\frac {du}{dt}}=0$, and the member ${\frac {bc}{\omega }}u$ must either remain in it, or be removed from it, according to whether it is considered worth while to take the influence of the atmosphere on the circuit into consideration or not. If we again place, as in the previous paragraph, $\beta ^{2}={\frac {bc}{\chi \omega }}$, and, also ${\frac {\chi }{\gamma }}=\chi '$, the preceding equation changes into the following:

${\frac {du}{dt}}=\chi '\left({\frac {d^{2}u}{dx^{2}}}-\beta ^{2}u\right)$,

and we immediately perceive, that on admitting $\beta =0$, the action of the atmosphere is left out of the question.

In the present case $u$ represents a function of $x$ and $t$, which, however, in proportion as the time $t$ increases, becomes gradually less dependent on $t$, and at last passes over into a mere function of $x$, which expresses the permanent state of the circuit, with the nature of which we have already become acquainted. If we designate this latter function by $u'$, and place $u=u'+v$, then $v$ is evidently a function of $x$ and $t$, which indicates every deviation of the circuit from its permanent state, and consequently after the lapse of a certain time entirely disappears. If we now substitute $u'+v$ for $u$ in the equation (✳), and bear in mind that $u'$ is independent of $t$, and of such nature that

$0={\frac {d^{2}u'}{dx^{2}}}-\beta ^{2}u'$,

the equation

${\frac {dv}{dt}}=\chi '\left({\frac {d^{2}v}{dx^{2}}}-\beta ^{2}v\right)$

(☽)