Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/99

186. If we have

${\frac {d^{2}\rho }{dt^{2}}}=\Gamma ^{2}.\rho ,$

where $\Gamma$ is complete, the integral equation will be of the form

$\rho =e^{t\Gamma }.\alpha +e^{-t\Gamma }.\beta .$

For this gives

${\frac {d\rho }{dt}}=\Gamma .e^{t\Gamma }.\alpha -\Gamma .e^{-t\Gamma }.\beta ,$

${\frac {d^{2}\rho }{dt^{2}}}=\Gamma ^{2}.e^{t\Gamma }.\alpha -\Gamma ^{2}.e^{-t\Gamma }.\beta =\Gamma ^{2}.\rho ,$

and $\alpha$ and $\beta$ may be determined so as to satisfy the equations

${\begin{aligned}\rho _{t=0}&=\alpha +\beta ,\\\left[{\frac {d\rho }{dt}}\right]_{t=0}=\Gamma .\{\alpha -\beta \}.\end{aligned}}$

187. The differential equation

${\frac {d^{2}\rho }{dt^{2}}}=-\Gamma ^{2}.\rho$

will be satisfied by

$\rho =\cos\{t\Gamma \}.\alpha +\sin\{t\Gamma \}.\beta ,$

whence

${\frac {d\rho }{dt}}=-\Gamma .\sin\{t\Gamma \}.\alpha +\Gamma .\cos\{t\Gamma \}.\beta ,$

${\frac {d^{2}\rho }{dt^{2}}}=-\Gamma ^{2}.\cos\{t\Gamma \}.\alpha -\Gamma ^{2}.\sin\{t\Gamma \}.\beta .=-\Gamma ^{2}.\rho .$

If $\Gamma$ is complete, the constants $\alpha$ and $\beta$ may be determined to satisfy the equations

${\begin{aligned}\rho _{t=0}&=\alpha ,\\\left[{\frac {d\rho }{dt}}\right]_{t=0}&=\Gamma .\beta .\end{aligned}}$

where $\Gamma ^{2}-\Lambda ^{2}$ is a complete dyadic, and

$\Gamma .\Lambda =\Lambda .\Gamma =0,$

we may set

$\rho =\{{\tfrac {1}{2}}e^{t\Gamma }+{\tfrac {1}{2}}e^{-t\Gamma }+\cos\{t\Lambda \}-{\text{I}}.\alpha +\{{\tfrac {1}{2}}e^{t\Gamma }-{\tfrac {1}{2}}e^{-t\Gamma }+\sin\{t\Lambda \}\}.\beta$

which gives

${\begin{aligned}{\frac {d\rho }{dt}}&=\{{\tfrac {1}{2}}\Gamma .e^{t\Gamma }-{\tfrac {1}{2}}\Gamma .e^{-t\Gamma }-\Lambda .\sin\{t\Lambda \}\}.\alpha +\{{\tfrac {1}{2}}\Gamma .e^{t\Gamma }+{\tfrac {1}{2}}\Gamma .e^{-t\Gamma }+\Lambda .\cos\{t\Lambda \}\}.\beta ,\\{\frac {d^{2}\rho }{dt^{2}}}&=\{{\tfrac {1}{2}}\Gamma ^{2}.e^{t\Gamma }+{\tfrac {1}{2}}\Gamma ^{2}.e^{-t\Gamma }-\Lambda ^{2}.\cos\{t\Lambda \}\}.\alpha +\{{\tfrac {1}{2}}\Gamma ^{2}.e^{t\Gamma }-{\tfrac {1}{2}}\Gamma ^{2}.e^{-t\Gamma }-\Lambda ^{2}.\sin\{t\Lambda \}\}.\beta \\&=\{\Gamma ^{2}-\Lambda ^{2}\}.\rho .\end{aligned}}$