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.

Theorem

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

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

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

Proof

If $Y$ is Banach Space then $\map {CL} {X, Y}$ is Banach Space

Let $Y$ be a Banach space.

Let $\sequence {T_n}_{n \mathop \in \N} \in \map {CL} {X, Y}$ be a Cauchy sequence.

Let $x \in X$.

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

$\sequence {T_n x}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {Y, \norm {\, \cdot \,}_Y}$

We have that:

\(\ds \forall x \in X : \forall n, m \in \N: \, \)

\(\ds \norm {T_n x - T_m x}_Y\)

\(=\)

\(\ds \norm {\paren {T_n - T_m} x}_Y\)

Linear Transformations form Vector Space

\(\ds \)

\(\le\)

\(\ds \norm {T_n - T_m} \norm x_X\)

Supremum Operator Norm as Universal Upper Bound

By definition of Cauchy sequence:

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

Suppose $m, n \ge N$.

Then:

$\forall \epsilon \in \R_{>0} : \forall x \in X : \norm {T_n x - T_m x}_Y < \epsilon \norm x_X$

Let $\epsilon' = \epsilon \norm x_X$

$\epsilon \in \R_{>0}$ and $x \in X$ were arbitrary.

Hence $\epsilon' \in \R_{> 0}$ is also arbitrary.

Therefore:

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

By definition, $\sequence {T_n x}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {Y, \norm {\, \cdot \,}_Y}$.

$\Box$

$\sequence {T_n x}_{n \mathop \in \N}$ converges in $\struct {\map {CL} {X, Y}, \norm {\, \cdot \,}}$

$Y$ is Banach.

$\sequence {T_n x}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {Y, \norm {\, \cdot \,}_Y}$.

Hence, $\sequence {T_n x}_{n \mathop \in \N}$ converges in $Y$ with limit, say, $Tx \in Y$.

$\Box$

$T$ is a linear transformation

Let $x_1, x_2 \in X$.

Then:

$\ds \lim_{n \mathop \to \infty} \paren{T_n x_1} = T x_1$

$\ds \lim_{n \mathop \to \infty} \paren{T_n x_2} = T x_2$

$Y$ is a vector space.

Thus, $x_1 + x_2 \in Y$.

Hence:

$\ds \lim_{n \mathop \to \infty} \paren {T_n \paren {x_1 + x_2}} = T \paren {x_1 + x_2}$

By combination of limits:

$\ds \lim_{n \mathop \to \infty} \paren{T_n x_1 + T_n x_2} = T x_1 + T x_2$

By linearity of $T_n$:

$\sequence {T_n x_1 + T_n x_2}_{n \mathop \in \N} = \sequence {T\paren {x_1 + x_2}}_{n \mathop \in \N}$

By uniqueness of limits:

$T \paren {x_1 + x_2} = T x_1 + T x_2$.

Let $\alpha \in \set {\R, \C}$.

Let $x \in X$.

Then:

$\ds \lim_{n \mathop \to \infty} \paren {T_n x} = T x$

By Multiple Rule for Sequences:

$\ds \lim_{n \mathop \to \infty} \paren {\alpha \cdot T_n x} = \alpha \cdot T x$

By linearity of $T_n$:

$\sequence{\alpha \cdot \paren{T_n x}}_{n \mathop \in \N} = \sequence{T_n\paren{\alpha \cdot x}}_{n \mathop \in \N}$

Since $X$ is a vector space:

$\alpha \cdot x \in X$

Then:

$\ds \lim_{n \mathop \to \infty} \paren {T_n\paren{\alpha \cdot x}}_{n \mathop \in \N} = T \paren {\alpha \cdot x}$.

Altogether:

$\alpha \cdot \map T x = \map T {\alpha \cdot x}$.

Altogether, by definition of linear transformation:

$T \in \map \LL {X, Y}$

$\Box$

$T$ is a continuous transformation

Let $\sequence {T_n}_{n \mathop \in \N} \in \map {CL} {X, Y}$ be a Cauchy sequence.

Then:

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

Hence:

$\exists N \in \N : \forall n > N : \norm {T_n - T_{N + 1} } < \epsilon$

Therefore:

\(\ds \forall n > N : \forall x \in X: \, \)

\(\ds \norm {T_n x - T_{N + 1} x}_Y\)

\(\le\)

\(\ds \norm {T_n - T_{N + 1} } \norm x_X\)

Supremum Operator Norm as Universal Upper Bound

\(\ds \)

\(<\)

\(\ds \epsilon \cdot \norm x_X\)

Take the limit $n \to \infty$.

Then:

$\forall x \in X : \norm {T x - T_{N + 1} x} < \epsilon \norm x_X$

Thus:

\(\ds \forall x \in X: \, \)

\(\ds \norm {Tx}_Y\)

\(=\)

\(\ds \norm {Tx - T_{N+1} x + T_{N+1} x }_Y\)

\(\ds \)

\(\le\)

\(\ds \norm {Tx - T_{N + 1} x}_Y + \norm {T_{N + 1} x}_Y\)

Definition of Norm on Vector Space

\(\ds \)

\(<\)

\(\ds \epsilon \norm x_X + \norm {T_{N + 1} } \norm x_X\)

Supremum Operator Norm as Universal Upper Bound

\(\ds \)

\(=\)

\(\ds \paren {\epsilon + \norm {T_{N + 1} } } \norm x_X\)

\(\ds \)

\(=\)

\(\ds M \norm x_X\)

$\epsilon + \norm {T_{N + 1} } = M \in \R$

By continuity of linear transformations:

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

$\Box$

$T_n$ converges to $T$ in $\struct {\map {CL} {X, Y}, \norm {\, \cdot \,}}$

By definition of Cauchy sequence.

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

Hence:

\(\ds \forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n, m > N : \forall x \in X: \, \)

\(\ds \norm {T_n x - T_m x}_Y\)

\(\le\)

\(\ds \norm {T_n - T_m} \norm x_X\)

Supremum Operator Norm as Universal Upper Bound

\(\ds \)

\(<\)

\(\ds \epsilon \norm x_X\)

We have that Norm on Vector Space is Continuous Function.

Take the limit $m \to \infty$.

By Limit of Composite Function:

\(\ds \forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n > N : \forall x \in X: \, \)

\(\ds \norm {T_n x - T x}_Y\)

\(\le\)

\(\ds \norm {T_n - T} \norm x_X\)

\(\ds \)

\(<\)

\(\ds \epsilon \norm x_X\)

Hence:

$\forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n > N : \norm {T_n - T} < \epsilon$

By definition, $T$ is continuous.

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

$J \sqbrk y = \int_{x_0}^{x_1} x y'^2 \rd x$

\(\ds J \sqbrk {y^*}\)

\(=\)

\(\ds \int_{x_0^*}^{x_1^*} x^* \sqbrk {\dfrac {\d \map {y^*} {x^*} } {\d x^*} }^2 \rd x^*\)

\(\ds \)

\(=\)

\(\ds \int_{x_0 + \epsilon}^{x_1 + \epsilon} x^* \sqbrk {\dfrac {\d \map y {x^* - \epsilon} } {\d x^*} }^2 \rd x^*\)

\(\ds \)

\(=\)

\(\ds \int_{x_0}^{x_1} \paren {x + \epsilon} \sqbrk {\dfrac {\d \map y x} {\d x} }^2 \rd x\)

\(\ds \)

\(=\)

\(\ds J \sqbrk \gamma + \epsilon \int_{x_0}^{x_1} \sqbrk {\dfrac {\d \map y x} {\d x} }^2 \rd x\)

\(\ds \)

\(\ne\)

\(\ds J \sqbrk \gamma\)

$\blacksquare$

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)

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