# User:Julius

## Theorem

$\norm {\, \cdot \,}_{L^p}$ on $\CC$ is a norm.

## Proof

If $\mathbf x \in \CC \closedint a b$ then $\size {\map {\mathbf x} t} \ge 0 \forall t \in \closedint 0 1$, hence $\int_a^b \size {\map {\mathbf x} t} \rd t \ge 0$.

Let $\mathbf x \in \CC \closedint a b : \norm {\mathbf x}_1 = 0$

If $\forall t \in \openint a b : \map {\mathbf x} t = 0$, then by continuity $\map {\mathbf x} t = 0 \forall t \in \closedint a b$.

Suppose that it is not the case, i.e. suppose $\exists t_0 \in \openint a b \map {\mathbf x} {t_0} \ne 0$.

Since $\mathbf x$ is continuous at $t_0$, $\exists \delta > 0 : a < t_0 - \delta, t_0 + \delta < b$ and $\forall t \in \closedint a b : t_0 - \delta < t < t_0 + \delta$, $\size {\map {\mathbf x} t - \map {\mathbf x} {t_0} } < \frac {\size {\map {\mathbf x} {t_0}}} 2$.

By reverse triangle inequality:

 $\displaystyle \size {\map {\mathbf x} t}$ $=$ $\displaystyle \size {\map {\mathbf x} t + \map {\mathbf x} {t_0} - \map {\mathbf x} {t_0} }$ $\displaystyle$ $\ge$ $\displaystyle \size {\map {\mathbf x} {t_0} } - \size {\map {\mathbf x} t - \map {\mathbf x} {t_0} }$ $\displaystyle$ $=$ $\displaystyle \size {\map {\mathbf x} {t_0} } - \frac {\size {\map {\mathbf x} {t_0} } } 2$ $\displaystyle$ $=$ $\displaystyle \frac {\size {\map {\mathbf x} {t_0} } } 2$ $\displaystyle$ $> 0$ $\displaystyle$

So:

 $\displaystyle 0$ $=$ $\displaystyle \norm {\mathbf x}_1$ $\displaystyle$ $=$ $\displaystyle \int_a^b \size {\map {\mathbf x} t} \rd t$ $\displaystyle$ $\ge$ $\displaystyle \int_{t_0 - \delta}^{t_0 + \delta} \size {\map {\mathbf x} t} \rd t$ $\displaystyle$ $\ge$ $\displaystyle 2 \delta \frac {\size {\map {\mathbf x} {t_0} } } 2$ $\displaystyle$ $=$ $\displaystyle \delta \size {\map {\mathbf x} {t_0} }$ $\displaystyle$ $>$ $\displaystyle 0$

Let $\mathbf x \in \CC \closedint a b$, $\alpha \in \R$.

Then:

 $\displaystyle \size {\alpha \mathbf x}_1$ $=$ $\displaystyle \int_a^b \size {\alpha \map {\mathbf x} t} \rd t$ $\displaystyle$ $=$ $\displaystyle \size \alpha \int_a^b \size {\map {\mathbf x} t} \rd t$ $\displaystyle$ $=$ $\displaystyle \size \alpha \norm {\mathbf x}_1$

Let $x, y \in \CC \closedint a b$

 $\displaystyle \norm {x + y}_1$ $=$ $\displaystyle \int_a^b \size {x + y} \rd t$ $\displaystyle$ $=$ $\displaystyle \int_a^b \size {x + y} \rd t$ $\displaystyle$ $\le$ $\displaystyle \int_a^b \paren {\size x + \size y} \rd t$ $\displaystyle$ $=$ $\displaystyle \int_a^b \size x \rd t + \int_a^b \size y \rd t$ $\displaystyle$ $=$ $\displaystyle \norm {x}_1 + \norm {y}_1$

### Convergent Subsequence of Cauchy Sequence

Let $\epsilon > 0$.

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

$\displaystyle \forall n, m > N : \norm {x_n - x_m} < \frac \epsilon 2$

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

$\displaystyle n_K > N \implies \norm {x_{n_K} - x} < \frac \epsilon 2$

Then:

 $\displaystyle \forall n > N$ $:$ $\displaystyle \norm {x_n - x}$ $\displaystyle$ $=$ $\displaystyle \norm {x_n - x_{n_K} + x_{n_K} - x}$ $\displaystyle$ $\le$ $\displaystyle \norm {x_n - x_{n_K} } + \norm {x_{n_K} - x}$ $\displaystyle$ $<$ $\displaystyle \frac \epsilon 2 + \frac \epsilon 2$ $\displaystyle$ $=$ $\displaystyle \epsilon$

## 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)

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