# User:Julius

## Current focus

• So I just noticed that vector notation is being used in Gelfand's for higher dimensional functionals. This implies rewriting all multivariable functionals. Implement this gradually.

## 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

$C = \frac 1 {\sqrt 5}$
$C_1 = 2$

$\blacksquare$

## 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.

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

$\blacksquare$

## 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

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

$\blacksquare$

## 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

$\blacksquare$

## Example: points on surfaces

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

Transversality conditions:

$\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$
$\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$
$\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$
$\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$

$\blacksquare$

## 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:

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

$\blacksquare$

## 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:

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

$\blacksquare$

## 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

$J \sqbrk y = \int_{x_0}^{x_1} x y'^2 \rd x$
 $\displaystyle J \sqbrk {y^*}$ $=$ $\displaystyle \int_{x_0^*}^{x_1^*} x^* \sqbrk {\dfrac {\d \map {y^*} {x^*} } {\d x^*} }^2 \rd x^*$ $\displaystyle$ $=$ $\displaystyle \int_{x_0 + \epsilon}^{x_1 + \epsilon} x^* \sqbrk {\dfrac {\d \map y {x^* - \epsilon} } {\d x^*} }^2 \rd x^*$ $\displaystyle$ $=$ $\displaystyle \int_{x_0}^{x_1} \paren {x + \epsilon} \sqbrk {\dfrac {\d \map y x} {\d x} }^2 \rd x$ $\displaystyle$ $=$ $\displaystyle J \sqbrk \gamma + \epsilon \int_{x_0}^{x_1} \sqbrk {\dfrac {\d \map y x} {\d x} }^2 \rd x$ $\displaystyle$ $\ne$ $\displaystyle J \sqbrk \gamma$

$\blacksquare$

## 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 $H = \const$

$\blacksquare$

### 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.

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

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

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

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

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)

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

## Functional Analysis

$\paren{C \closedint a b,\norm{\cdot}_\infty }$ is a Banach space.

Let $\sequence{x_n}_{n \in \N}$ be a Cauchy sequence.

$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n, m > N : \norm{x_n − x_m}_\infty < \epsilon$

Suppose, all the elements of $\sequence{x_n}_{n \in \N}$ are additionally indexed with $t$:

$\sequence{x_n}_{n \in \N} = \sequence{\map {x_n} t }_{n \in \N}$

Let $t \in \closedint a b$.

But

$\displaystyle \forall n, m > N : \norm {\map {x_n} t - \map {x_m} t}_\infty < \max_{\tau \in \closedint a b}\norm {\map {x_n} \tau - \map {x_m} \tau}_\infty = \norm {x_n - x_m}_\infty < \epsilon$

Hence, $\sequence{\map {x_n} t}_{n \in \N}$ is a Cauchy sequence in $\R$.

$\R$ is complete.

Therefore, $\sequence{\map {x_n} t}_{n \in \N}$ is convergent with limit $L = \map L t$.

Choose $N$ such that $\forall n,m > N : \norm{x_n - x_m} \le \frac \epsilon 3$

Let $\tau \in \closedint a b$.

Then $\forall n > N : \norm {\map {x_n} \tau - \map {x_{N + 1}} \tau } \le \norm {x_n - x_{N + 1} }_\infty \le \frac \epsilon 3$

Take the limit $n \to \infty$:

$\lim_{n \to \infty} \norm {\map {x_n} \tau - \map {x_{N + 1}} \tau } = \norm {\map x \tau - \map {x_{N + 1}} \tau } \le \frac \epsilon 3$

which holds for all $\tau \in \closedint a b$.

Now $\map {x_{N+1} } \tau \in C \closedint a b$

$\exists \delta > 0: \norm {\tau - t} < \delta \implies \norm {\map {x_{N+1} } t - \map {x_{N+1} } \tau} \le \frac \epsilon 3$

Thus:

$\norm {\map x \tau - \map x t} = \norm {\map x \tau - \map {x_{N+1}} \tau + \map {x_{N+1}} \tau - \map {x_{N+1}} t + \map {x_{N+1}} t - \map x t} \le$
$\norm {\map x \tau - \map {x_{N+1}} \tau} + \norm {\map {x_{N+1}} \tau - \map {x_{N+1}} t} + \norm {\map {x_{N+1}} t - \map x t} \le \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3 = \epsilon$

Hence $x$ is continuous at $t$.

Since $t \in C \closedint a b$, $t$ is continuous in whole interval.

Finally, show that $\sequence {x_n}_{n \in \N}$ converges to $x$.

Let $\epsilon > 0$.

Choose $N$ such that $\forall n,m > N : \norm{x_n - x_m}_\infty < \epsilon$

Fix $n > N$.

Let $t \in \closedint a, b$.

Then $\forall m > N: \norm {\map {x_n} t - \map {x_m} t} \le \norm {x_n - x_m}_\infty < \epsilon$

Thus $\norm{\map {x_n} t - \map x t} = \lim_{n \to \infty} \norm {\map {x_n} t - \map {x_m} t} \le \epsilon$

Since $t$ was arbitrary: $\norm {x_n - x}_\infty = \max_{t \in \closedint a b } \norm{\map {x_n} t - \map x t} \le \epsilon$

This could also have been achieved by fixing $n > N$.

So, $\forall n > N \norm {x_n - x}_\infty \le \epsilon$.

Therefore $\lim_{x \to \infty} x_n = x$ in $C \closedint a b$