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.

Theorem

Let $I := \closedint a b$ be a closed real interval.

Let $\map \CC I$ be a space of continuous on closed interval real-valued functions.

Let $\map {\CC^k} I$ be a space of continuous functions of differentiability class k on closed interval $I$.

Let $\struct {\R, +_\R, \times_\R}$ be the field of real numbers.

Let $\paren +$ be the pointwise addition of real-valued functions.

Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of real-valued functions.

Then $\struct {\map \CC I, +, \, \cdot \,}_\R$ is a vector space.

Proof

Let $f, g, h \in \map \CC I$ such that:

$f, g, h : I \to \R$

Let $\lambda, \mu \in \R$.

Let $\map 0 x$ be a real-valued function such that:

$\map 0 x : I \to 0$.

Let us use real number addition and multiplication.

$\forall x \in I$ define pointwise addition as:

$\map {\paren {f + g}} x := \map f x +_\R \map g x$.

Define pointwise scalar multiplication as:

$\map {\paren {\lambda \cdot f}} x := \lambda \times_\R \map f x$

Let $\map {\paren {- f}} x := - \map f x$.

Closure

By Sum Rule for Continuous Functions, $f + g \in \map \CC I$

Commutativity

By Pointwise Addition on Real-Valued Functions is Commutative, $f + g = g + f$

Associativity

By Pointwise Addition is Associative, $\paren {f + g} + h = f + \paren {g + h}$.

Identity

\(\displaystyle \map {\paren {0 + f} } x\)

\(=\)

\(\displaystyle \map 0 x +_\R \map f x\)

definition of pointwise addition of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle 0 +_\R \map f x\)

definition of $\map 0 x$

\(\displaystyle \)

\(=\)

\(\displaystyle \map f x\)

Inverse

\(\displaystyle \map {\paren {f + \paren {-f} } } x\)

\(=\)

\(\displaystyle \map f x +_\R \map {\paren {-f} } x\)

definition of pointwise addition of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \map f x +_\R \paren {-1} \times_\R \map f x\)

definition of $\map {\paren {-f} } x$

\(\displaystyle \)

\(=\)

\(\displaystyle 0\)

Scalar distributivity

\(\displaystyle \map {\paren { \paren {\lambda +_\R \mu} f} } x\)

\(=\)

\(\displaystyle \paren {\lambda +_\R \mu} \times_\R \map f x\)

definition of pointwise scalar multiplication of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \lambda \times_\R \map f x +_\R \mu \times_\R \map f x\)

Real Multiplication Distributes over Addition

\(\displaystyle \)

\(=\)

\(\displaystyle \map {\paren {\lambda \cdot f} } x +_\R \map {\paren {\mu\cdot f} } x\)

definition of pointwise scalar multiplication of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \map {\paren {\lambda \cdot f + \mu \cdot f} } x\)

definition of pointwise addition of real-valued functions

Vector distributivity

\(\displaystyle \lambda \times_\R \map {\paren {f + g} } x\)

\(=\)

\(\displaystyle \lambda \times_\R \paren {\map f x +_\R \map g x}\)

definition of pointwise addition of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \lambda \times_R \map f x +_\R \lambda \times_\R \map g x\)

Real Multiplication Distributes over Addition

\(\displaystyle \)

\(=\)

\(\displaystyle \map {\paren{\lambda \cdot f} } x +_\R \map {\paren{\lambda \cdot g} } x\)

definition of pointwise scalar multiplication of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \map {\paren {\lambda \cdot f + \mu \cdot f} } x\)

definition of pointwise addition of real-valued functions

Scalar associativity

\(\displaystyle \map {\paren {\paren {\lambda \times_\R \mu} \cdot f} } x\)

\(=\)

\(\displaystyle \paren {\lambda \times_\R \mu} \times_\R \map f x\)

definition of pointwise scalar multiplication of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \lambda \times_\R \paren {\mu \times_\R \map f x}\)

Real Multiplication is Associative

\(\displaystyle \)

\(=\)

\(\displaystyle \lambda \times_\R \map {\paren {\mu \cdot f} } x\)

definition of pointwise scalar multiplication of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \map {\paren {\lambda \cdot \paren {\mu \cdot f} } } x\)

definition of pointwise scalar multiplication of real-valued functions

Scalar identity

\(\displaystyle \map {\paren {1 \cdot f} } x\)

\(=\)

\(\displaystyle 1 \times_\R \map f x\)

definition of pointwise scalar multiplication of real-valued functions

\(\displaystyle \)

\(=\)

\(\displaystyle \map f x\)

Let $\map {\paren {- f}} x := - \map f x$

$\map f x$

$\map {f + g} x$

$\map {\paren {f + g} } x$

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:

\(\displaystyle \forall n > N\)

\(:\)

\(\displaystyle \norm {x_n - x}\)

\(\displaystyle \)

\(=\)

\(\displaystyle \norm {x_n - x_{n_K} + x_{n_K} - x}\)

\(\displaystyle \)

\(\le\)

\(\displaystyle \norm {x_n - x_{n_K} } + \norm {x_{n_K} - x}\)

\(\displaystyle \)

\(<\)

\(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\)

\(\displaystyle \)

\(=\)

\(\displaystyle \epsilon\)

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 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.

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)

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