CoV24

Determine the function $\hat{y}\isin C^2[0,1]\,$ that minimizes the functional $J(y)=\int_0^1\left[y'(x)\right]^2dx+[y(1)]^2, y(0)=1, h(0)=0\,$ .

First, compute the first variation so that it can be set to zero:

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

To get this only in terms of $h\,$ , integrate by parts. Let

$u = 2y^\prime(x)$		$du = 2y^{\prime\prime}(x) \ dx$
$dv = h^\prime(x)\ dx$		$v = h(x)\,$

Thus, setting the first variation to zero

$2y^\prime(x)h(x)\left.\right\|^1_0 - \int_0^1 2y^{\prime\prime}(x)h(x)\ dx + 2y(1)h(1)$	$= 0\,$
$2y^\prime(1)h(1) - \int_0^1 2y^{\prime\prime}(x)h(x)\ dx + 2y(1)h(1)$	$= 0\,$

Since $h(1)\,$ is unspecified, suppose $h(1) = 0\,$ . Then

$-2\int_0^1 y^{\prime\prime}(x)h(x)\ dx$

$= 0\,$

By the fundamental lemma, for all $h(x)\,$

From the initial condition, $y(0) = 1 \Longrightarrow c_2 = 1$ . Now, suppose $h(1) = 1\,$ . Then, the first variation gives

$2y^\prime(1)\cdot 1 - (0) + 2y(1)\cdot 1$	$= 0\,$
$2y^\prime(1) + 2y(1)$	$= 0\,$

Substituting in gives

Therefore, the function that minimizes the functional is $\hat{y} = -\frac{x}{2} + 1$ .

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