Examples: Unconstrainted Optimization on Calulus of Variations
Example 1
Question
Consider \(Q+scc^T\) where \(Q\) is positive definite symmetric, prove that the largest eigenvalue \(\mu_n(s)\) of \(Q+scc^T\) is bounded by \(a_2 + \beta s \leq \mu_n(s) \leq a_1 + \beta s\).
proof.
\[\begin{align*} \mu_n(s) &= \max_{\|x\|=1}\big\{x^T(Q+scc^T)x\big\}\\ &= \max_{\|x\|=1} \big\{x^TQx + x^Tscc^Tx\big\} \end{align*}\]
Because \(Q\) is positive definite, \(x^TQx \geq 0\),
\[\begin{align*} \mu_n(s) &\geq \max_{\|x\|=1} \big\{x^Tscc^Tx\big\}\\ &= \max_{\|x\|=1} \big\{x^Tcc^Tx\big\}s &s\text{ is a scalar} \end{align*}\]
Also, we have
\[\begin{align*} \mu_n(s) &\leq \max_{\|x\|=1}\big\{x^TQx\big\} + \max_{\|x\|=1} \big\{x^Tscc^Tx\big\}\\ &= \max_{\|x\|=1}\big\{x^TQx\big\} + \max_{\|x\|=1} \big\{x^Tcc^Tx\big\}s \end{align*}\]
Let \(\alpha_2 = 0, \alpha_1 = \max_{\|x\|=1}\big\{x^TQx\big\}, \beta = \max_{\|x\|=1} \big\{x^Tcc^Tx\big\}\), since \(c\neq 0, \beta > 0\). Therefore,
\[a_2 + \beta s \leq \mu_n(s) \leq a_1 + \beta s\]
Example 2
Question
Derive the formula \(p(t):\mathbb R\rightarrow \mathbb R^2 := (x(t), y(t))\) where \(p(t)\) is the position of point on the rim of a circle of radius \(R\), the circle is rolling along the \(x\)-axis at a constant speed \(a\). Assume \(c(t) = (at, R)\) is the position of the center of the circle at \(t\) and \(p(0) = (0, 0)\).
Let \(f(\theta):\mathbb R\rightarrow \mathbb R^2 := (R\cos\theta, R\sin\theta)\) be the parametric equation of the circle.
Consider the point \(q(t) = (at, 0)\) on the rim of the circle at time \(t\), let
\[q_c(t) = q(t) - c(t) = (at, 0) - (at, R) = (0, -R)\]
be its position relative to the center of the circle, note that
\[q_c(t) = f(-\pi/2) = (\cos(-\frac{\pi}{2}), \sin(-\frac{\pi}{2}))\]
Then, let \(L\) be the arc of the circle between \(q_c\) and \(p_c\) and \(\theta(t)\) be the angle subtended by arc \(L\) from the center of the circle. Since the arc length is \(at\), we have
\[\begin{align*} \theta(0) - \theta(t) &= \frac{at}{R}\\ \theta(t) &= \theta(0) - \frac{at}{R}\\ \theta(t) &= -\frac{\pi}{2} - \frac{at}{R} \end{align*}\]
Therefore, let \(q_c(t)\) be \(p\)'s position relative to the center of the circle,
\[q_c(t) = f(\theta(t)) = \big(R\cos(-\frac{\pi}{2} - \frac{at}{R}), R\sin(-\frac{\pi}{2} - \frac{at}{R})\big)\]
so that
\[p(t) = p_c(t) + c(t) = \big(at + R\cos(-\frac{\pi}{2} - \frac{at}{R}), R\sin(-\frac{\pi}{2} - \frac{at}{R})\big)\]
Example 3
Question
\[F[\mu[\cdot] = \int_a^b L(x, \mu(x), \mu'(x))dx, x(a)=A, x(b)=B\]
\(L, \mu \in C^2\), partition \([a,b]\) into \(x_0, ..., x_{n+1}\) where \(x_i = a + \frac{b-a}{n+1}i\) so that we can approximate the functional by
\[F:\mathbb R^n\rightarrow \mathbb R, F(\mu_1, ..., \mu_n) :=\sum_{i=1}^{n+1}L(x_i, \mu_i, \frac{\mu_i - \mu_{i-1}}{h})h\]
Part (a)
Calculate \(\frac{dF}{d\mu_i}(\mu)\).
\[\begin{align*} \frac{dF}{d\mu_i}(\mu) &= \sum_{j=1}^{n+1}\frac{d}{d\mu_i}L(x_j, \mu_j, \frac{\mu_j - \mu_{j-1}}{h})h\\ &= \frac{d}{d\mu_i}L(x_{i}, \mu_{i}, \frac{\mu_i - \mu_{i-1}}{h})h + \frac{d}{d\mu_i}L(x_{i+1}, \mu_{i+1}, \frac{\mu_{i+1} - \mu_{i}}{h})h\\ &= L_z(x_{i}, \mu_{i}, \frac{\mu_i - \mu_{i-1}}{h})h\frac{d}{d\mu_i}\mu_i + L_p(x_{i}, \mu_{i}, \frac{\mu_i - \mu_{i-1}}{h})h\frac{d}{d\mu_i}\frac{\mu_i - \mu_{i-1}}{h}\\ &\quad+L_p(x_{i+1}, \mu_{i+1}, \frac{\mu_{i+1} - \mu_{i}}{h})h\frac{d}{d\mu_i}\frac{\mu_{i+1} - \mu_{i}}{h}\\ &= hL_z(x_{i}, \mu_{i}, \frac{\mu_i - \mu_{i-1}}{h}) - (L_p(x_{i+1}, \mu_{i+1}, \frac{\mu_{i+1} - \mu_{i}}{h})- L_p(x_{i}, \mu_{i}, \frac{\mu_i - \mu_{i-1}}{h})) \end{align*}\]
Part (b)
Take \(h\rightarrow 0\), show that \(\frac{dF}{d\mu_i}(\mu) = 0\Rightarrow L_z(x, \mu(x), \mu'(x) - \frac{d}{dx}L_p(x, \mu(x), \mu'(x))) = 0\).
proof. Since \(\frac{dF}{d\mu_i}(\mu) = 0\)
\[\begin{align*} \lim_{h\rightarrow 0} \frac{1}{h}\frac{dF}{d\mu_i}(\mu) &= \lim_{h\rightarrow 0}L_z(x_i, \mu_i, \frac{\mu_i-\mu_{i-1}}h) \\ &\quad- \lim_{h\rightarrow 0}\frac{L_p(x_{i+1}, \mu_{i+1}, \frac{\mu_{i+1} - \mu_{i}}{h})- L_p(x_{i}, \mu_{i}, \frac{\mu_i - \mu_{i-1}}{h})}{h}\\ &= L_z(x_i, \mu(x_i), \mu(x_i)')\\ &\quad - \lim_{h\rightarrow 0}\frac{L_p(x_i+h, \mu(x_i+h), \mu'(x_i+h))- L_p(x_{i}, \mu(x_i), \mu'(x_i))}{h}\\ &\text{by definition of derivative}\\ &= L_z(x_i, \mu(x_i), \mu(x_i)') - \frac{d}{dx}L_p(x_i, \mu(x_i), \mu(x_i)') \end{align*}\]
Therefore, for some \(x\in[a,b]\), since \(h\rightarrow 0\), we can partition \([a,b]\) s.t. \(x\in \text{partition}(a, b)\), so that we have
\[L_z(x_i, \mu(x_i), \mu(x_i)') - \frac{d}{dx}L_p(x_i, \mu(x_i), \mu(x_i)') = \lim_{h\rightarrow 0} \frac{1}{h}\frac{dF}{d\mu_i}(\mu) = 0\]
Example 4
Question
Find FONC of
\[\begin{align*} \text{minimize}\quad &F[\mu(\cdot)] := \int_a^b L(x, u(x), u'(x))dx\\ \text{subject to}\quad &\mu\in \mathcal A:= \{\mu:[a,b]\rightarrow\mathbb R\mid \mu\in C^2, \mu(a) = A\} \end{align*}\]
Let \(\mu_*\) be the minimizer of the problem.
Let \(\mathcal T:= \{v:[a, b]\rightarrow\mathbb R\mid v\in C^2, v(a) = 0\}\) be the set of test functions so that \(\forall v\in \mathcal T. \forall s\in\mathbb R. u_*+sv \in \mathcal A\).
Then as what have been done in the class till the integrations by parts
\[\begin{align*} \frac{d}{ds}F[u_*+sv] &=\int_a^b \big[L_z(...) - \frac{d}{dx}L_p(...)\big]v(x)dx + \big[L_p(...)\mid_a^b\big] \end{align*}\]
Since \(v(a) = 0\), note that
\[\begin{align*} \big[L_p(x, u_*(x), u_*'(x))v(x)\mid_a^b\big] &= L_p(b, u_*(b), u_*'(b))v(b) - L_p(a, u_*(a), u_*'(a))v(a)\\ &= L_p(b, u_*(b), u_*'(b))v(b) \end{align*}\]
Consider \(v_1\in\mathcal T\) s.t. \(v(b) = 0\), hence
\[\big[L_p(x, u_*(x), u_*'(x))v(x)\mid_a^b\big] = 0\]
Given \(u_*\) is the minimizer,
\[0 = \frac{d}{ds}F[u_*+sv] =\int_a^b \big[L_z(...) - \frac{d}{dx}L_p(...)\big]v(x)dx\]
follows the original Euler-Lagrange equation, i.e.
\[L_z(x, u_*(x), u_*'(x)) - \frac{d}{dx}L_p(x, u_*(x), u_*'(x)) = 0\]
Consider \(v_2\in\mathcal T, v_2(b)\neq 0\), since we have to fulfill the necessary conditions for cases like \(v_1\), we have
\[L_z(x, u(x), u'(x)) - \frac{d}{dx}L_p(x, u(x), u'(x)) = 0\]
so that
\[0 = \frac{d}{ds}F[u_*+sv] = 0 + L_p(b, u(b), u'(b))v(b)\]
Therefore, there are two FONC as
\[L_z(x, u_*(x), u_*'(x)) - \frac{d}{dx}L_p(x, u_*(x), u_*'(x)) = 0\]
\[L_p(b, u(b), u'(b))v(b) = 0\]