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.

Definition: Differential $k$-Form
Let $M$ be a smooth manifold with or without boundary.

Let $T^* M$ be the cotangent bundle of $M$.

Let $T^k T^*M$ be the space of contravariant $k$-tensors on $T^* M$.

Let $\Lambda^k T^* M$ the subbundle of $T^k T^*M$ consisting of alternating tensors.

Then a section of $\Lambda^k T^* M$ is called a differential $k$-form, or just a $k$-form.

Theorem(Completion of Nondegenerate Bases)
Let $\struct {V, q}$ be an $n$-dimensional scalar product space.

Let $\tuple {v_1, \ldots, v_k}$ be a nondegenerate $k$-tuple in V with $0 \le k < n$.

Then there exist vectors $v_{k + 1}, \ldots v_n \in V$ such that $\tuple {v_1, \ldots, v_n}$ is a nondegenerate basis for $V$.

Theorem
Let $\norm {\, \cdot \,} : \map {CL} {X, Y} \to \R$ such that:


 * $\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm {Tx} : x \in X : \norm x \le 1}$

Supremum operator norm $\norm {\, \cdot \,}$ is norm.

$N1$
Let $T \in \map {CL} {X, Y}$.

Then $\ds \norm T = \sup_{x \in X, \norm x \le 1} \norm {T x} \ge 0$ since $\norm {T x} \ge 0$.

If $\norm T = 0$, then $\norm {T x} \le \norm T \norm x = 0 \norm x = 0$.

Hence, $\norm {T x} = 0$ and $\forall x \in X : Tx = \mathbf 0_Y$, so $T = \mathbf 0$.

$N2$
Let $\alpha \in \KK$.

Let $T \in \map {CL} {X, Y}$.

Then:

N3

Theorem
Let $T \in \map {CL} {X, Y}$.

Suppose:


 * $\exists M \in \R_{> 0} : \forall x \in X : \norm {T x} \le M \norm x$

Then:


 * $\norm {T} \le M$

Proof
Let $x \in X$ such that $\norm x \le 1$.

Then:

Hence, $M$ is an upper bound for $S = \set {\norm {T x} : x \in X : \norm x \le 1}$.

Thus $\sup S \le M$.

That is, $\norm T \le M$.

Theorem
Let $T \in \map {CL} {X, Y}$.

Then:


 * $\forall x \in X : \norm {Tx} \le \norm T \norm x$

Proof
Suppose $x = \mathbf 0$.

Then:

Suppose $x \ne \mathbf 0$.

Let $y \in X$ such that:


 * $y = \frac x {\norm x}$

Then:


 * $\norm y = \frac {\norm x}{\norm x} = 1$.

Thus, $\norm {T y} \in S$ and $\norm {T y} \le \sup S = \norm T$.

In other words:

Rearranging, we get:


 * $\norm {T x} \le \norm T \norm x$

Theorem
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.

Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:


 * $\norm T := \map \sup {\norm {Tx} : x \in X : \norm x \le 1}$

Then $T$ is a norm.

Proof
Let $S = \set {\norm {Tx} : x \in X : \norm x \le 1}$

By definition of the norm:


 * $S \subseteq \R$

$S$ is non-empty
Let $x = \mathbf 0_X \in X$.

Then:

and:

Thus:


 * $0 \in S$

and


 * $S \ne \empty$

$S$ is bounded above
Let $\map {CL} {X, Y}$.

Then:

Let $x \in X : \norm x \le 1$.

By continuity of linear transformations on normed vector spaces:

By the least upper bound property of $\R$, a supremum of $S$ exists.

Hence:


 * $\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm {Tx} : x \in X : \norm x \le 1} < \infty$

Lemma A
Let $T \in \map {CL} {X, Y}$.

Suppose:


 * $\exists M \in \R_{> 0} : \forall x \in X : \norm {Tx} \le M \norm x$

Then:


 * $\norm {T} \le M$

Proof
If $x \in X$ and $\norm x \le 1$, then $\norm {Tx} \le M \norm x \le M \cdot 1 = M$.

So $M$ is an upper bound of $S$.

Thus $\sup S \le M$, that is $\norm T \le M$.

Theorem
Space of continuous linear transformations is a subspace of the space of linear transformations.

Proof
Let $\struct {X, \norm \cdot }$ and $\struct {Y, \norm \cdot }$ be normed vector spaces.

Closure under vector addition
Let $S, T \in \map {CL} {X, Y}$.

By continuity of linear transformations in normed vector space:


 * $\exists M_S \in \R : M_S > 0 : \forall x \in X : \norm {Sx} \le M_S \norm x$


 * $\exists M_T \in \R : M_T > 0 : \forall x \in X : \norm {Tx} \le M_T \norm x$

Furthremore:

By continuity of linear transformations in normed vector space:


 * $S + T \in \map {CL} {X, Y}$

Closure under scalar multiplication
Let $\alpha \in \R$.

Let $T \in \map {CL} {X, Y}$.

By continuity of linear transformations in normed vector space:


 * $\exists M \in \R_{> 0} : \forall x \in X : \norm {Tx} \le M \norm {x}$

Hence:

By continuity of linear transformations in normed vector space:


 * $\alpha T \in \map {CL} {X, Y}$

Existence of identity element under vector addition
Let $\mathbf 0 : X \to Y$ be the zero mapping.

Then:

Hence:


 * $\mathbf 0 \in \map {CL} {X, Y}$

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 $H = \const$

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)

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