CoV27

Compute the first variation $\delta J(y,h)\,$ for $y\isin C[0,1]\,$ : $J(y)=\int_0^1\int_0^1\sin(xt)y(x)y(t)dxdt\,$

$\delta J(y, h)\,$	$= \frac{d}{d\varepsilon} J(y + \varepsilon h)\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_0^1\int_0^1 \sin(xt)[(y + \varepsilon h)(x)][(y + \varepsilon h)(t)]\ dxdt\left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \frac{d}{d\varepsilon}\left[\sin(xt)(y(x) + \varepsilon h(x))(y(t) + \varepsilon h(t))\right] \ dxdt \left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \frac{d}{d\varepsilon}\left[\sin(xt)(y(x)y(t) + \varepsilon h(x)y(t) + \varepsilon h(t)y(x) + \varepsilon^2h(x)h(t))\right] \ dxdt \left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \sin(xt)[0 + h(x)y(t) + h(t)y(x) + 2\varepsilon h(x)h(t)] \ dxdt\left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \sin(xt)(h(x)y(t) + h(t)y(x))\ dxdt$

$\delta J(y, h)\,$	$= \frac{d}{d\varepsilon} J(y + \varepsilon h)\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_0^1\int_0^1 \sin(xt)[(y + \varepsilon h)(x)][(y + \varepsilon h)(t)]\ dxdt\left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \frac{d}{d\varepsilon}\left[\sin(xt)(y(x) + \varepsilon h(x))(y(t) + \varepsilon h(t))\right] \ dxdt \left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \frac{d}{d\varepsilon}\left[\sin(xt)(y(x)y(t) + \varepsilon h(x)y(t) + \varepsilon h(t)y(x) + \varepsilon^2h(x)h(t))\right] \ dxdt \left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \sin(xt)[0 + h(x)y(t) + h(t)y(x) + 2\varepsilon h(x)h(t)] \ dxdt\left.\right\|_{\varepsilon = 0}$
	$= \int_0^1\int_0^1 \sin(xt)(h(x)y(t) + h(t)y(x))\ dxdt$