CoV10

From Exampleproblems

Show that the first variation $\delta J(y_0,h)\,$ satisfies the homogeneity condition $\delta J(y_0, \alpha h) = \alpha \delta J(y_0, h), \alpha \isin \mathbb{R}\,$ .

Define the real-valued function $\phi(\varepsilon) := J(y_0 + \varepsilon h)$ . Then $\frac{d}{d\varepsilon} J(y_0 + \varepsilon h) = \phi^\prime(\varepsilon)$ . Consequently, $\phi^\prime(0) = \delta J(y_0, h)$ . So

$\delta J(y_0, \alpha h)\,$	$= \frac{d}{d\varepsilon} J(y_0 + \varepsilon\alpha h)$
	$= \frac{d}{d\varepsilon} \phi(\varepsilon\alpha)\left.\right\|_{\varepsilon = 0}$
	$= \phi^\prime(\varepsilon\alpha)\cdot\alpha\left.\right\|_{\varepsilon = 0}$
	$= \alpha\phi^\prime(0)$
	$= \alpha\delta J(y_0, h)\,$

A second method uses the limit definition. Here is an outline of that proof

$\delta J(y_0, \alpha h) = \lim_{\varepsilon \to 0} \frac{J(y_0 + \varepsilon \alpha h) - J(y_0)}{\varepsilon}$
Multiply by 1
$\alpha \cdot \frac{J(y_0 + \varepsilon \alpha h) - J(y_0)}{\alpha\varepsilon}$
With $t = \varepsilon\alpha$
$\delta J(y_0, \alpha h) = \alpha \cdot \lim_{t \to 0} \frac{J(y_0 + th) - J(y_0)}{t}$
$= \alpha \delta J(y_0, h)\,$