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

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 minimzer 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


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

Propagation of light in inhomogeneous medium
Suppose 3d space is filled with an optically inhomogeneous medium such that at each point speed of light $v = \map v {x, y, z}$

If the curve joining two points $A$ and $B$ is specified by $y = \map y x$ and $z = \map z x$

then the time it takes to traverse the curve equals:


 * $\displaystyle \int_a^b \frac {\sqrt{1 + y'^2 + z'^2} } {\map v {x, y, z} }$

Euler's Equations:


 * $\displaystyle \dfrac {\partial v} {\partial y} \frac {\sqrt{1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \frac {y'} {v\sqrt {1 + y'^2 + z'^2} } = 0$


 * $\displaystyle \dfrac {\partial v} {\partial z} \frac {\sqrt{1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \frac {z'} {v \sqrt {1 + y'^2 + z'^2} } = 0$

Theorem
Let $y$ be a smooth curve, embedded in 2-dimensional Euclidean space.

Let $y$ have a total length of $l$.

Let it be contained in the upper halfplane with exception of endpoints, which are on the x-axis.

Suppose, $y$, together with a line segment connecting $y$'s endpoints, maximizes the enclosed area.

Then $y$ is a semicircle.

Proof
Without loss of generality, we choose our point of reference such that $y$ intersect x-axis at points $\tuple {-a, 0}$ and $\tuple {a, 0}$.

The area below the curve $y$ is a functional of the following form:


 * $A \sqbrk y = \int_{-a}^a y \rd x$

Furthermore, $y$ has to satisfy the following conditions:


 * $\map y {-a} = \map y a = 0$


 * $L \sqbrk y = \int_{-a}^a \sqrt{1 + y'^2} \rd x = l$

By Simplest Variational Problem with Subsidiary Conditions, there exists a constant $\lambda$ such that functional:


 * $A \sqbrk y + \lambda L \sqbrk y = \int_{-a}^a \paren {y + \lambda \sqrt {1 + y'^2} }\rd x$

as well as the area $A \sqbrk y$ are extremized by the function $y$.

Euler's Equation:


 * $1 + \lambda \dfrac \d {\d x} \frac {y'} {\sqrt{1 + y'^2} } = 0$

Implies:


 * $x + \lambda \frac {y'} {\sqrt{1 + y'^2} } = C_1$

Integration yields:


 * $\paren {x - C_1}^2 + \paren {y - C_2}^2 = \lambda^2$

To find $C_1, C_2, \lambda$, apply boundary conditions and the length constraint:


 * $ \paren {- a - C_1}^2 = \lambda^2$


 * $ \paren {a - C_1}^2 = \lambda^2$


 * $l = \pi \lambda$

Subtract the 2nd equation from the 1st one:

$\paren {- a - C_1}^2 - \paren {a - C_1}^2 = 4 a C_1 = 0 \rightarrow C_1 = 0$

Shortest path on a sphere
Sphere:


 * $x^2 + y^2 + z^2 = a^2$

Curve passes through $\paren {x_0, y_0, z_0}, \paren {x_1, y_1, z_1}$

Length of the curve:


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

Auxiliary functional:


 * $\int_{x_0}^{x_1} \sqbrk {\sqrt{1 + y'^2 + z'^2} + \map {\lambda} x \paren{x^2 + y^2 + z^2} } \rd x$

Euler's Equations


 * $2 y \map \lambda x - \dfrac \d {\d x} \frac {y'} {\sqrt{1 + y'^2 + z'^2} } = 0$


 * $2 z \map \lambda x - \dfrac \d {\d x} \frac {z'} {\sqrt{1 + y'^2 + z'^2} } = 0$

Minimize a functional when endpoints lie on curves
Suppose end points lie on curves $y = \map \phi x$, $y = \map \psi x$


 * $\displaystyle \delta J = F_{y'}|_{x=x_1}\delta y_1 + \paren {F-F_{y'}y'}|_{x=x_1}\delta x_1-F_{y'}|_{x=x_0}\delta y_0 - \paren {F - F_{y'}y'}|_{x=x0}\delta x_0$


 * $\displaystyle \delta J = \paren {F_{y'}\psi' + F - y' F_{y'} }|_{x=x_1} \delta x_1 - \paren {F_{y'}\phi' + F - y' F_{y'} }|_{x=x_0}\delta x_0 = 0$


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


 * $\sqbrk {F + \paren {\psi' - y'}F_{y'} }|_{x=x_1}=0$

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)

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