User:Julius

Current focus

 * Build the bulk knowledge on calculus of variations based on Gelfand's Calculus of Variations, then recheck with a couple other books and slowly improve proofs.

Theorem
Hilbert cube is compact in $\ell^2$

Construction of subsequence
Let $C$ be the Hilbert cube.

Let $\sequence {\mathbf x_m}_{m \mathop \in \N}$ be a sequence in $C$.

Let $\mathbf x_m = \tuple {x_m^{\paren n}}_{n \mathop \in \N}$.

By definition:


 * $\ds \forall n,m \in \N : 0 \le x_m^{\paren n} \le \frac 1 n$

Consider the case $n = 1$.

Then:


 * $\ds \forall m \in \N : 0 \le x_m^{\paren 1} \le 1$

By Closed Real Interval is Compact, $\closedint 0 1$ is compact.

Hence, $\sequence {x^{\paren 1}_m}_{m \mathop \in \N}$ has a subsequence $\sequence {x^{\paren 1}_{\map {m_1} j}}_{j \mathop \in \N}$ that is convergent with limit, say, $x^{\paren 1} \in \closedint 0 1$.

Consider the case $n = 2$.

By analogous arguments, $\sequence {x^{\paren 2}_{\map {m_1} j}}_{j \mathop \ge 2}$ has a subsequence $\sequence {x^{\paren 2}_{\map {m_2} j}}_{j \mathop \ge 2}$ that is convergent with limit, say, $x^{\paren 2} \in \closedint 0 {\frac 1 2}$.

Proceeding in this manner, for all $k \in \N$ a sequence $\sequence {x^{\paren k}_{\map {m_{k - 1}} j}}_{j \mathop \ge k}$ has a convergent subsequence $\sequence {x^{\paren k}_{\map {m_k} j}}_{j \mathop \ge k}$ with a limit $x^{\paren k} \in \closedint 1 {\frac 1 k}$.

From all these sequences we construct a subsequence $\sequence {\mathbf x_{\map {m_j} j}}_{j \mathop \in \N}$.

Let $\mathbf x := \tuple {x^{\paren n}}_{n \mathop \in \N}$.

$\mathbf x \in C$ because $\ds \forall n \in \N : 0 \le x^{\paren n} \le \frac 1 n$.

Convergence of the subsequence
The goal is to prove that for all sufficiently large $j$'s $\ds \norm {\mathbf x_{\map {m_j} j} -\mathbf x}_2$ is small.

Split $\ds \norm {\mathbf x_{\map {m_j} j} -\mathbf x}_2^2$ as follows:


 * $\ds \norm {\mathbf x_{\map {m_j} j} -\mathbf x}_2^2 = \sum_{n \mathop = 1}^N \paren {x^{\paren n}_{\map {m_j} j} - x^{\paren n}}^2 + \sum_{n \mathop = N + 1}^\infty \paren {x^{\paren n}_{\map {m_j} j} - x^{\paren n}}^2$

Let $\epsilon \in \R_{> 0}$.

Let $N \in \N$ be such that:


 * $\ds \sum_{n \mathop = N + 1}^\infty \frac 1 {n^2} < \epsilon$

Then:

$\sequence {\map {m_N} j}_{j \mathop \ge n}$ is a subsequence of $\sequence {\map {m_n} j}_{j \mathop \ge n}$.

Likewise, $\sequence {x^{\paren n}_{\map {m_N} j}}_{j \mathop \ge n}$ is a subsequence of $\sequence {x^{\paren n}_{\map {m_n} j}}_{j \mathop \ge n}$.

By Limit of Subsequence equals Limit of Sequence, $\sequence {x^{\paren n}_{\map {m_N} j}}_{j \mathop \ge n}$ converges to the same limit as $\sequence {x^{\paren n}_{\map {m_n} j}}_{j \mathop \ge n}$:


 * $\ds \forall n \le N : \lim_{j \to \infty} x^{\paren n}_{\map {m_N} j} = x^{\paren n}$

We have that $\sequence {\map {m_j} j}_{j \mathop \ge N}$ is a subsequence of $\sequence {\map {m_N} j}_{j \mathop \ge N}$.

However, $N \ge n$.

Hence, $\sequence {\map {m_j} j}_{j \mathop \ge N}$ is also a subsequence of $\sequence {\map {m_j} j}_{j \mathop \ge n}$ for each $n \le N$.

Hence:


 * $\ds \forall n \le N : \lim_{j \to \infty} x^{\paren n}_{\map {m_j} j} = x^{\paren n}$


 * $\ds \exists J \in \N : \forall j > J : \sum_{n = 1}^N \paren {x^{\paren n}_{\map {m_j} j} - x^{\paren n} }^2 < \epsilon$

Lemmas and theorems for Bernstein's Theorem on Unique Extrema (1978)
Raw material

Example 1
Suppose that:


 * $J \sqbrk y = \int_1^2 \frac {\sqrt {1+y'^2} } {x} \rd x$

with the following boundary conditions:


 * $\map y 1 = 0$


 * $\map y 2 = 1$

Then the smooth minimizer of $J$ is a circle of the following form:


 * $\paren {y - 2}^2 + x^2 = 5$

Proof
$J$ is of the form


 * $J \sqbrk y = \int_a^b \map F {x, y'} \rd x$

Then we can use the "no y theorem":


 * $F_y = C$

i.e.


 * $\frac {y'} {x \sqrt {1 + y'^2} } = C$

or


 * $y' = \frac {C x} {\sqrt {1 - C^2 x^2} }$

The integral is equal to


 * $y = \frac {\sqrt {1 - C^2 x^2} } C + C_1$

or


 * $\paren {y - C_1}^2 + x^2 = C^{-2}$

From the conditions $\map y 1 = 0$, $\map y 2 = 1$ we find that


 * $C = \frac 1 {\sqrt 5}$


 * $C_1 = 2$

Example 3

 * $J \sqbrk = \int_a^b \paren {x - y}^2$

is minimized by


 * $\map y x = x$

Proof
Euler' equation:


 * $F_y = 0$

i.e.


 * $2 \paren {x - y} = 0$.

Example p31
Suppose:


 * $J \sqbrk r = \int_{\phi_0}^{\phi_1} \sqrt{r^2 + r'^2} \rd \phi$

Euler's Equation:


 * $\displaystyle \frac r {\sqrt{r^2 + r'^2} } - \dfrac \d {\d \phi} \frac {r'} {\sqrt{r^2 + r'^2} }$

Apply change of variables:


 * $x = r \cos \phi, y = r \sin \phi$

The integral becomes:


 * $\displaystyle \int_{x_0}^{x_1} \sqrt{1 + y'^2} \rd x$

Euler's equation:


 * $y'' = 0$

Its solution:


 * $y = \alpha x + \beta$

or


 * $r \sin \phi = \alpha r \cos \phi + \beta$

Example

 * $J \sqbrk = \int_{x_0}^{x_1} \map f {x,y} \sqrt {1+y'^2}\rd x$


 * $F_{y'} = \map f {x,y} \frac {y'} {\sqrt{1 + y'^2} }=\frac {y' F} {1 + y'^2}$


 * $F + \paren {\phi' - y'}F_{y'} = \frac {\paren{1+y'\phi'}F} {1+y'^2} = 0$


 * $F + \paren {\psi' - y'}F_{y'} = \frac {\paren{1+y'\psi'}F} {1+y'^2} = 0$

i.e.


 * $y' = -\frac 1 {\phi'}$


 * $y' = - \frac 1 {\psi'}$

Transversality reduces to orthogonality

Example: points on surfaces

 * $J \sqbrk {y,z} = \int_{x_0}^{x_1} \map F {x,y,z,y',z'} \rd x$

Transversality conditions:


 * $\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$


 * $\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$


 * $\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$


 * $\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$

Example: Legendre transformation

 * $\map f \xi = \frac {\xi^a} a, a>1$


 * $\map {f'} \xi = p = \xi^{a-1}$

i.e.


 * $\xi = p^{\frac {1} {a-1} }$


 * $H = - \frac {\xi^a} {a} + p\xi = - \frac {p^{\frac {a} {a-1} } } a + p p^{\frac {a} {a-1} } = p^{\frac {a} {a-1} } \paren{1 - \frac 1 a}$

Hence:


 * $\map H p = \frac {p^b} b$

where:


 * $\frac 1 a + \frac 1 b = 1$

Example

 * $J \sqbrk y = \int_a^b \paren {Py'^2 + Q y^2} \rd x$


 * $p = 2 P y', H = P y'^2 - Q y^2$

Hence:


 * $H = \frac {p^2} {4 P} - Q y^2$

Canonical equations:


 * $\dfrac {\d p} {\d x} = 2 Q y$


 * $\dfrac {\d y} {\d x} = \frac p {2 P}$

Euler's Equation:


 * $2 y Q - \dfrac \d {\d x} \paren {2 P y'} = 0$

Example: Noether's theorem 1

 * $J \sqbrk y = \int_{x0}^{x1} y'^2 \rd x$

is invariant under the transformation:


 * $x^* = x + \epsilon, y^* = y$


 * $y^* = \map y {x^* - \epsilon} = \map {y^*} {x^*}$

Then:


 * $J \sqbrk {\gamma^*} = \int_{x0^*}^{x1^*} \sqbrk { \dfrac {\d \map {y^*} {x^*} } {\d x^*} } \rd x^* = \int_{x0+\epsilon}^{x_1 + \epsilon} \sqbrk { \dfrac {\d \map y {x^* - \epsilon} } {\d x^*} }^2 \rd x^* = \int_{x0}^{x1} \sqbrk { \dfrac {\d \map y x} {\d x} }^2 \rd x = J \sqbrk \gamma$

Example: Neother's theorem 2

 * $J \sqbrk y = \int_{x_0}^{x_1} x y'^2 \rd x$

Example: Noether's theorem 3

 * $J \sqbrk y = \int_{x_0}^{x_1} \map F {y, y'} \rd x$

Invariant under $x^* = x + \epsilon, y_i^* = y_i$

I.e. $\phi = 1, \psi_i = 0$

reduces to $H = \const$

Momentum of the system:

 * $P_x = \sum_{y = 1}^n p_{ix}, P_y = \sum_{y = 1}^n p_{iy}, P_z = \sum_{z = 1}^n p_{iz}$

(Examples: attraction to a fixed point, attraction to a homogenous distribution on an axis)

Geodetic distance:Examples
If $J$ is arclength, $S$ is distance.

If $J$ is a moment of time to pass a segment of optical medium, then $S$ is the time needed to pass the whole optical body.

If $J$ is action, then $S$ is the minimal action.

Examples of quadratic functionals
1) $B \sqbrk {x, y} = \int_{t_0}^{t_1} \map x t \map y t \rd t$

Corresponding quadratic functional

$A \sqbrk x = \int_{t_0}^{t_1} \map {x^2} t$

2) $B \sqbrk {x, y} = \int_{t_0}^{t_1} \map \alpha t \map x t \map y t \rd t$

Corresponding quadratic functional

$A \sqbrk x = \int_{t_0}^{t_1} \map \alpha t \map {x^2} t \rd t$

3)

$A \sqbrk x = \int_{t_0}^{t_1} \paren {\map \alpha t \map {x^2} t + \map \beta t \map x t \map {x'} t+ \map \gamma t \map {x'^2} t} \rd t$

4)

$B \sqbrk {x, y} = \int_a^b \int_a^b \map K {s, t} \map x s \map y t \rd s \rd t$