CoV13

Compute the first variation of $J(y)=\int_a^b yy' dx\,$

$\delta J(y, h)\,$	$= \frac{d}{d\varepsilon} J(y + \varepsilon h)\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b (y + \varepsilon h)(y^\prime + \varepsilon h^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b (yy^\prime + y\varepsilon h^\prime + y^\prime\varepsilon h + \varepsilon^2 hh^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b \frac{d}{d\varepsilon} (yy^\prime + y\varepsilon h^\prime + y^\prime\varepsilon h + \varepsilon^2 hh^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (yh^\prime + y^\prime h + 2\varepsilon hh^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (yh^\prime + y^\prime h) \ dx$

$\delta J(y, h)\,$	$= \frac{d}{d\varepsilon} J(y + \varepsilon h)\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b (y + \varepsilon h)(y^\prime + \varepsilon h^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b (yy^\prime + y\varepsilon h^\prime + y^\prime\varepsilon h + \varepsilon^2 hh^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b \frac{d}{d\varepsilon} (yy^\prime + y\varepsilon h^\prime + y^\prime\varepsilon h + \varepsilon^2 hh^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (yh^\prime + y^\prime h + 2\varepsilon hh^\prime) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (yh^\prime + y^\prime h) \ dx$