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.


 * 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.

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

Lagrangian mechanics
Lagrangian mechanics is a reformulation of mechanics, where the main object of the study is the Lagrangian function.

According to this principle, equations of motion are found by finding an extremum of the action of the given physical system.

Note, that not always a physical problem admits lagrangian formulation. Examples of such cases are inequality constraints, nonintegrable constraints and nonconservative forces.

Lagrangian
Let $P$ be a physical system composed of $n$ discrete particles.

Let $t$ be a real variable.

Let $\set {\map {\mathbf x_i} t, \map {\mathbf x_i'} t, \ldots}$ be real functions $\forall i: 1 \le i \le n$.

Let $L = \map L {t, \map {\mathbf x_i} t, \map {\mathbf x_i'} t, \ldots}$ be a real function.

Let the action of $P$ be a functional of the following form:


 * $S = \int_{t_1}^{t_2} \map L {t, \map {\mathbf x} t, \map {\mathbf x'} t, \ldots} \rd t$

Then $L$ is the Lagrangian of $P$.

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 $H = \const$

Kinetic energy

 * $T = \frac 1 2 \sum_{i = 1}^n m_i \paren {\dot {x_i}^2 + \dot {y_i}^2 + \dot {z_i}^2}$

Potential energy

 * $U = \map U {t, x_1, y_1, \ldots x_n, y_n, z_n}$

Force:


 * $X-i = - \dfrac {\partial U} {\partial x_i}$


 * $Y_i = - \dfrac {\partial U} {\partial y_i}$


 * $Z-i = - \dfrac {\partial U} {\partial z_i}$

Lagrangian Function of the system of particles

 * $L = T - U$

Principle of least action
The motion of a system of $n$ particles during the time interval $\sqbrk {t_0, t_1}$ is described by those functions $\map {x_i} t$, $\map {y_i} t$, $\map {z_i} t$, $1 \le i \le n$ for which the integral


 * $\int_{t_0}^{t_1} L \rd t$

called the action, is a minimum.

Proof
Euler's equations


 * $\dfrac L {x_i} - \dfrac \d {\d t} \dfrac {\partial L} {\partial \dot{x_i}}$


 * $\dfrac L {y_i} - \dfrac \d {\d t} \dfrac {\partial L} {\partial \dot{y_i}}$


 * $\dfrac L {z_i} - \dfrac \d {\d t} \dfrac {\partial L} {\partial \dot{z_i}}$

These can be rewritten as:


 * $- \dfrac {\partial U} {\partial x_i} - \dfrac \d {\d t} m_i \dot {x_i} = 0$


 * $- \dfrac {\partial U} {y_i} - \dfrac \d {\d t} m_i \dot {y_i} = 0$


 * $- \dfrac {\partial U} {z_i} - \dfrac \d {\d t} m_i \dot {z_i} = 0$

Since the derivatives are components of the force acting on the $i$th particle, the system reduces to


 * $m_i \ddot {x_i} = X_i$


 * $m_i \ddot {y_i} = Y_i$


 * $m_i \ddot {z_i} = Z_i$

Hamiltonian

 * $S = \int_{t_0}^{t_1} L \rd t = \int_{t_0}^{t_1} \paren {T - U} \rd t$


 * $p_{ix} = \dfrac L {\dot {x_i} } = m_i \dot {x_i}$


 * $p_{iy} = \dfrac L {\dot {y_i} } = m_i \dot {y_i}$


 * $p_{iz} = \dfrac L {\dot {z_i} } = m_i \dot {z_i}$


 * $H = \sum_{i = 1}^n \paren {\dot {x_i} p_{ix} + \dot {y_i} p_{iy} + \dot {z_i} p_{iz} } - L = 2 T - \paren {T - U} = T + U$

Conservation of momentum

 * $x^* = \map \Phi {x, y, y'; \epsilon} = x$


 * $y_i^* = \map {\Psi_i} {x, y, y'; \epsilon}$

implies the first integral


 * $\sum_{i = 1}^n$ F_{y_i} \psi_i = \const

where


 * $\map {\psi_i} {x, y, y'} = \dfrac {\partial \map {\Psi_i} {x, y, y'; \epsilon} } {\partial \epsilon} \vert_{\epsilon = 0}$

in this case:


 * $\map \phi {x, y, y'} = \dfrac {\partial \Phi {x, y, y'; \epsilon} } {\partial \epsilon} \vert_{\epsilon = 0} = 0$

The invariance of the functional under


 * $x_i^* = x_i + \epsilon, y_i^* = y_i, z_i^* = z_i$

implies that


 * $\sum_{i = 1}^n \dfrac {\partial L} {\partial \dot {x_i} } = \const$

or


 * $\sum_{i = 1}^n p_{i x} = \const$


 * $\sum_{i = 1}^n p_{i y} = \const$


 * $\sum_{i = 1}^n p_{i z} = \const$

Momentum of the system:


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

Conservation of angular momentum

 * $x_i^* = x_i \cos \epsilon + y_i \sin \epsilon$


 * $y_i^* = -x_i \sin \epsilon + y_i \cos \epsilon$


 * $z_i^* = z_i$

In this case:


 * $\psi_{ix} = \dfrac {\partial {x_i^*} } {\partial \epsilon} \vert_{\epsilon = 0} = y_i$


 * $\psi_{iy} = \dfrac {\partial {y_i^*} } {\partial \epsilon} \vert_{\epsilon = 0} = -x_i$


 * $\psi_{iz} = \dfrac {\partial {z_i^*} } {\partial \epsilon} \vert_{\epsilon = 0} = 0$

Noether's theorem implies


 * $\sum_{i = 1}^n \paren {\dfrac {\partial L} {\partial \dot {x_i} }y_i - \dfrac {\partial L} {\partial \dot {y_i} }x_i} = \const$

i.e.


 * $\sum_{i = 1}^n \paren {p_{ix}y_i - p_{iy}x_i} = \const$

(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)

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