Page:Scientific Papers of Josiah Willard Gibbs.djvu/468

or, writing $T$ for the constant ${\frac {aAt}{v}},$

${\begin{aligned}\phi _{0}&=-Tk_{0}{\frac {d\gamma _{0}}{dx}}\mp k_{0}\gamma _{0}{\frac {dV}{dx}},\\{\frac {c_{0}\phi _{0}}{k_{0}}}&=-Tc_{0}{\frac {d\gamma _{0}}{dx}}\mp c_{0}\gamma _{0}{\frac {dV}{dx}},\\\textstyle \sum _{0}\displaystyle {\frac {c_{0}\phi _{0}}{k_{0}}}&=-T\textstyle \sum _{0}\displaystyle {\frac {c_{0}d\gamma _{0}}{dx}}\end{aligned}}$

[The terms $\mp {\frac {dV}{dx}}k_{0}\gamma _{0}$ disappear in the algebraic sum since $\textstyle \sum \displaystyle c_{1}\gamma _{1}=\textstyle \sum \displaystyle c_{2}\gamma _{2}$. For a similar reason]

$\textstyle \sum _{0}\displaystyle \mp {\frac {c_{0}\phi _{0}}{k_{0}}}=-{\frac {dV}{dx}}\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}.$

The first equation makes ${\frac {\textstyle \sum _{0}\displaystyle c_{0}\gamma _{0}}{dx}}$ constant throughout the tube, and since $\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}''=\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}',$ $\textstyle \sum _{0}\displaystyle c_{0}k_{0}\gamma _{0}$ must be constant throughout the tube. The second equation then makes ${\frac {dV}{dx}}$ constant throughout the tube. Let $X=-{\frac {dV}{dx}}\cdot$

Our original equation is

$\phi _{0}=-Tk_{0}{\frac {d\gamma _{0}}{dx}}\pm Xk_{0}\gamma _{0}.$

Now with $X$ constant this is easily integrated.

${\begin{aligned}{\frac {\phi _{0}}{Xk}}&=-{\frac {T}{X}}{\frac {d\gamma _{0}}{dx}}\pm \gamma _{0},\\\pm {\frac {T}{X}}{\frac {d\gamma _{0}}{dx}}&=\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}},\\{\frac {d\gamma _{0}}{\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}}}&=\pm {\frac {X}{T}}dx,\\\log \left(\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}\right)&=\pm {\frac {X}{T}}x+\log H_{0},\\\gamma _{0}\mp {\frac {\phi _{0}}{Xk_{0}}}&=H_{0}e^{\pm {\frac {X}{T}}x}.\end{aligned}}$

To determine $H_{0}$ we have

$\gamma _{0}''-\gamma _{0}'=H_{0}\left(e^{\mp {\frac {X}{T}}x''}-e^{\pm {\frac {X}{T}}x'}\right).$

If we put the origin of coordinates in the middle of the tube we have

$x'=-x''.$