Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/98

as in the ordinary calculus, but we must not apply these equations to cases in which the values of $\Phi$ are not homologous.

183. If, however, $\Gamma$ is any constant dyadic, the variations of $t\Gamma$ will necessarily be homologous with $t\Gamma$ and we may write without other limitation than that $\Gamma$ is constant,

${\frac {d\{e^{t\Gamma }\}}{dt}}=\Gamma .e^{t\Gamma }$

(1)

${\frac {d\sin\{t\Gamma \}}{dt}}=\Gamma .\cos\{t\Gamma \},$

(2)

${\frac {d\cos\{t\Gamma \}}{dt}}=-\Gamma .\sin\{t\Gamma \},$

(3)

${\frac {d\log\{t\Gamma \}}{dt}}={\frac {I}{t}}\cdot$

(4)

A second differentiation gives

${\frac {d^{2}\{e^{t\Gamma }\}}{dt^{2}}}=\Gamma ^{2}.e^{t\Gamma }$

(5)

${\frac {d^{2}\sin\{t\Gamma \}}{dt^{2}}}=-\Gamma ^{2}.\sin\{t\Gamma \},$

(6)

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

(7)

184. It follows that if we have a differential equation of the form

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

the integral equation will be of the form

$\rho =e^{t\Gamma }.\rho ',$

$\rho '$ representing the value of $\rho$ for $t=0.$ For this gives

${\frac {d\rho }{dt}}=\Gamma .e^{t\Gamma }.\rho '=\Gamma .\rho ,$

and the proper value of $\rho$ for $t=0.$

185. Def.—A flux which is a linear function of the position-vector is called a homogeneous-strain-flux from the nature of the strain which it produces. Such a flux may evidently be represented by a dyadic.

In the equations of the last paragraph, we may suppose $\rho$ to represent a position-vector, $t$ the time, and $\Gamma$ a homogeneous-strain-flux. Then $e^{t\Gamma }$ will represent the strain produced by the flux $\Gamma$ in the time $t.$

In like manner, if $\Lambda$ represents a homogeneous strain, $\{\log \Lambda \}/t$ will represent a homogeneous-strain-flux which would produce the strain $\Lambda$ in the time $t.$