User:Julius

Current focus

 * Build the bulk knowledge on calcul\us of variations based on Gelfand's Calculus of Variations, then recheck with a couple of other books and slowly improve proofs.

Continuous Linear Transformation from Normed Vector Space to Banach Space forms Banach Space
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.

Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a Banach space.

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

Let $\norm {\, \cdot \,}$ be the supremum operator norm.

Then $\map {CL} {X, Y}$ is a Banach space.

Proof
Let $\sequence {T_n}_{n \mathop \in \N}$ be a Cauchy sequence such that:


 * $\forall n \in N : T_n \in \map {CL} {X, Y}$

By definition of Cauchy sequence:


 * $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {T_n - T_m} < \epsilon$

Then:

Hence:


 * $\forall x \in X : \forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {T_n x - T_m x}_Y \le \epsilon \norm x_X$

Let $\map {\epsilon'} {\epsilon, x} \in \R$ be such that:


 * $\forall x \in X : \forall \epsilon \in \R_{> 0} : \epsilon \norm x_X < \map {\epsilon'} {\epsilon, x}$

We have that $\epsilon \in \R_{> 0}$ and $\forall x \in X : \norm x_X \in \R_{\ge 0}$.

Thus, $\map {\epsilon'} {\epsilon, x} \in \R_{> 0}$.

Therefore:


 * $\forall x \in X : \forall \epsilon' \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {T_n x - T_m x}_Y < \epsilon'$

By definition, the sequence $\sequence {T_n x}_{n \mathop \in \N}$ such that $\forall n \in \N : \forall x \in X : T_n x \in Y$ is a Cauchy sequence.

By assumption, $Y$ is a Banach space.

By definition, a Cauchy sequence in $Y$ converges in $Y$:


 * $\ds \exists y \in Y : \lim_{n \mathop \to \infty} T_n x = y$

Let $\ds \lim_{n \mathop \to \infty} T_n x = T x$ with a mapping $T : X \to Y : Tx = y$.

Then:

By definition, $T$ is a linear map.

Let $\epsilon' = 2 \epsilon$.

Then:


 * $\ds \forall \epsilon' > 0 : \exists N_{\epsilon'} \in \R_{> 0} : \forall n, m \in \N_{> 0} : n, m \ge N_{\epsilon'} \implies \norm {T_n - T_m} < \frac {\epsilon'} 2$

Suppose $n \ge N_{\epsilon'}$.

Then:


 * $\ds \forall x \in X : \exists \map m x \ge N_{\epsilon'} : \norm {T_{\map m x} x - T x} \le \frac {\epsilon'} 2$

If $\norm x = 1$ then:

Hence, $T$ is bounded.

Finally:


 * $\ds \lim_{n \mathop \to \infty} \norm {T_n - T} = 0$

Neumann Series Theorem
Let $X$ be a Banach space.

Let $\map {CL} X$ be the continous linear transformation space.

Let $\norm {\, \cdot \,}$ be the supremum operator norm.

Let $A \in \map {CL} X$ be such that $\norm A < 1$.

Let $\circ$ be the composition of mappings.

Then:


 * $I - A$ is invertible in $\map {CL} X$


 * $\ds \paren {I - A}^{-1} = \sum_{n \mathop = 0}^\infty A^n$


 * $\ds \norm {\paren{I - A}^{-1}} \le \frac 1 {1 - \norm A}$

Furthermore, $I - A : X \to X$ is bijective and:


 * $\ds \forall y \in X : \exists ! x \in X : x - A x = y$

as well as $\ds \norm x \le \frac 1 {1 - \norm A} \norm y$ so $x$ depends on $y$ continuously.

Proof
Let $\ds S_k := \sum_{n \mathop = 0}^k A^n$.

$\ds \sum_{n \mathop = 0}^\infty A^n$ converges absolutely
By Supremum Operator Norm on Continuous Linear Transformation Space is Submultiplicative:


 * $\forall n \in \N : \norm {A^n} \le \norm A^n$

By assumption, $\norm A < 1$.

By Sum of Infinite Geometric Sequence, $\ds \sum_{n \mathop = 0}^\infty \norm A^n$ converges.

By series comparison, $\ds \sum_{n \mathop = 0}^\infty \norm {A^n}$ converges too.

By definition, $\ds \sum_{n \mathop = 0}^\infty A^n$ is absolutely convergent.

$\ds \sum_{n \mathop = 0}^\infty A^n$ converges
By assumption, $X$ is Banach.

By Continuous Linear Map space from normed vector space to Banach Space is Banach, $\map {CL} X$ is Banach too.

Let $\ds S := \sum_{n \mathop = 0}^\infty A^n$.

We have that Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach.

Then $S$ converges in $\map {CL} X$.

In other words:


 * $\ds \lim_{k \mathop \to \infty} S_k = S$

Inverse of $\paren {I - A}$ is $S$
We have that:

Furthermore:

Hence:

That is:


 * $S A = A S = S - I$

It follows that:

Therefore, $I - A$ is invertible in $\map {CL} X$.

Furthermore:


 * $\ds \paren {I - A}^{-1} = S = \sum_{k \mathop = 0}^\infty A^k$

Moreover:

Theorem (Example of invertible linear system)
Let $\tuple {x_1, x_2} \in \R^2$.

Consider the system:


 * $x_1 = \frac 1 2 x_1 + \frac 1 3 x_2 + 1$


 * $x_2 = \frac 1 3 x_1 + \frac 1 4 x_2 + 2$

Let $I$ be the 2x2 identity matrix.

Let $\norm {\, \cdot \,}_2$ be the 2-norm on $\R^2$.

Let $\norm {\, \cdot \,}$ be the supremum operator norm.

Then:


 * $\norm K \le 1$


 * $\mathbf x_n = \paren {\sum_{k = 0}^n K^k} \mathbf y =$


 * $\frac {\norm{\mathbf x - \mathbf x_n}_2}{\norm{\mathbf x}_2} =$

P-Sequence Space is Subset of Bounded Sequence Space

 * $\ell^p \subseteq \ell^\infty$

$c$, $c_0$, $c_{00}$ are Subspaces of P-Sequence Space
$c_{00} \subseteq c_0 \subseteq c \subseteq \ell^p$

Definition(Algebra)
Let $\struct {K, +_K, \times_K}$ be a field.

Let $\struct {V, +_V, \circ}_K$ be a vector space over $K$.

An algebra is $V$ additionally equipped with a bilinear mapping $\otimes : V \times V \to V$ such that for all $x, y, z \in V$ and for all $\alpha \in K$ we have:


 * $\paren {x \otimes y} \otimes z = x \otimes \paren {y \otimes z}$


 * $\paren {x +_V y} \otimes z = x \otimes z +_V y \otimes z$


 * $z \otimes \paren {x +_V y} = z \otimes x +_V z \otimes y$


 * $\alpha \circ \paren {x \otimes y} = x \otimes \paren {\alpha \circ y} = \paren {\alpha \circ x} \otimes y$

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)

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