Page:Scientific Memoirs, Vol. 2 (1841).djvu/484

already arrived at its permanent state, and we have treated it at a length which it merits from the abundance and importance of the phænomena connected with it. However, not to let even here the other circuits pass entirely unnoticed, we will briefly indicate the method to be pursued for the most simple case, and thus point out the path to be followed, although only at a distance.

If it is intended to take into consideration the influence of

be the atmosphere on the galvanic circuit, the member ${\frac {bc}{\omega }}u$ must be added to the member $\chi {\frac {d^{2}u}{dx^{2}}}$ if the equation (a) in § 11, we then obtain for the circuit which has acquired a permanent state, for which ${\frac {du}{dt}}=0$, the equation

$0=\chi {\frac {d^{2}u}{dx^{2}}}-{\frac {bc}{\omega }}u;$

or if we put ${\frac {bc}{\chi \omega }}=\beta ^{2}$,

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

The integral of this equation is

$u=c\cdot e^{\beta x}+d\cdot e^{-\beta x},$

where $e$ represents the base of the natural logarithms, and $c$, $d$ any constants to be determined from the other circumstances of the problem.

If we now call $2l$ the length of the entire circuit, and fix the origin of the abscissæ in that place of the circuit which is equidistant from the point of excitation; if, further, we designate by $a$ the tension existing at the point of excitation, we obtain

$a=(c-d)\left(e^{\beta t}-e^{-\beta t}\right).$

If we write the previously found equation thus,

$u=(c-d)e^{\beta x}+d\left(e^{\beta x}+e^{-\beta x}\right),$

and substitute for $c-d$, the value just ascertained, we have

$u={\frac {a\cdot e^{\beta x}}{e^{\beta l}-e^{-\beta l}}}+d\left(e^{\beta x}+e^{-\beta x}\right).$

If we now suppose for the determination of the other constant, that the sum of the two electroscopic forces, situated at the