User:Julius
Current focus
- Build the bulk knowledge on calculus of variations based on Gelfand's Calculus of Variations, then recheck with a couple of other books and slowly improve proofs.
Functional Analysis
Theorem (Open Real Interval is not Compact)
Let $I := \openint a b$ be an opean real interval.
$I$ is not compact
Proof
We have that $I$ is bounded by $a$ and $b$.
Consider the sequence $\ds \sequence {x_n}_{n \mathop \in \N}$ with $x_n = a + \frac {b - a} {2n}$.
It converges to $a$.
However, $a \notint I$.
Then every subsequence of $\sequence {x_n}_{n \mathop \in \N}$ converges to $a$ as well.
By definition $I$ is not compact.
Digestion of the following topics is in progress
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
- $J \sqbrk y = \int_{x_0}^{x_1} x y'^2 \rd x$
\(\ds J \sqbrk {y^*}\) | \(=\) | \(\ds \int_{x_0^*}^{x_1^*} x^* \sqbrk {\dfrac {\d \map {y^*} {x^*} } {\d x^*} }^2 \rd x^*\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{x_0 + \epsilon}^{x_1 + \epsilon} x^* \sqbrk {\dfrac {\d \map y {x^* - \epsilon} } {\d x^*} }^2 \rd x^*\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{x_0}^{x_1} \paren {x + \epsilon} \sqbrk {\dfrac {\d \map y x} {\d x} }^2 \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds J \sqbrk \gamma + \epsilon \int_{x_0}^{x_1} \sqbrk {\dfrac {\d \map y x} {\d x} }^2 \rd x\) | ||||||||||||
\(\ds \) | \(\ne\) | \(\ds J \sqbrk \gamma\) |
$\blacksquare$
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.
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)
$B \sqbrk {x, y} = \int_a^b \int_a^b \map K {s, t} \map x s \map y t \rd s \rd t$