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.

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 $Y$
We have that:

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:


 * $\norm {T_n x - T_m x}_Y < \epsilon \norm x_X$

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:


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

$Y$ is Banach.

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

Furthermore, $\sequence {T_n x_1 + T_n x_2}_{n \mathop \in \N} = \sequence {T_n \paren {x_1 + x_2}}_{n \mathop \in \N}$ converges to $T x_1 + T x_2$.

However, $\sequence {T_n \paren {x_1 + x_2}}_{n \mathop \in \N}$ converges to $T \paren {x_1 + x_2}$.

By uniqueness of limits, $T \paren {x_1 + x_2} = T x_1 + T x_2$.

$T$ is a linear transformation
Let $x_1, x_2 \in X$.

Then $\sequence {T_n x_1}_{n \mathop \in \N}$ converges to $T x_1$ and $\sequence {T_n x_1}_{n \mathop \in \N}$ converges to $T x_1$.

Hence, $\sequence {T_n x_1 + T_n x_2}_{n \mathop \in \N}$ converges to $T x_1 + T x_2$.

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

We have that $\sequence {T\paren {x_1 + x_2}}_{n \mathop \in \N}$ converges to $T \paren {x_1 + x_2}$.

By uniqueness of limits, $T \paren {x_1 + x_2} = T x_1 + T x_2$.

Let $\alpha \in \K$.

Let $x \in X$.

Then $\sequence {T_n x}_{n \mathop \in \N}$ converges to $T x \in Y$.

Furthermore, $\sequence{\alpha \cdot \paren{T_n x}}_{n \mathop \in \N}$ converges to $\alpha \cdot Tx \in Y$.

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

Then $\sequence{T_n\paren{\alpha \cdot x}}_{n \mathop \in \N}$ converges to $T \paren {\alpha \cdot x}$.

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

$T$ is a continuous transformation
Let $\epsilon = 1$.

Then:


 * $\exists N \in \N : \forall m, n \in \N : m, n > N : \norm {T_n - T_m} \le \epsilon = 1$

Hence:


 * $\forall n > N : \norm {T_n - T_{N + 1} } \le 1$

Therefore:


 * $\forall n > N : \forall x \in X : \norm {T_n x - T_{N + 1} x} \le \norm {T_n - T_{N + 1} } \norm {x} \le 1 \cdot \norm {x}$

Passin the limit $n \to \infty$ we obtain that $\forall x \in X : \norm {T x - T_{N + 1} x} \le x$

Thus:


 * $\forall x \in X : x \in X : \norm {Tx} \le \norm {Tx - T_{N + 1} x} + \norm {T_{N + 1} x} \le \paren {1 + \norm {T_{N + 1} } } \norm x$

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

$T_n$ converges to $T$
Let $\epsilon > 0$.

By definition of Cauchy sequence:


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

Hence:


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

Passing the limit $m \to \infty$ yields


 * $\forall n > N : \forall x \in X : \norm {T_n x - T x}_Y \le \epsilon \norm x$

Hence:


 * $\forall n > N : \norm {T_n - T} \le \epsilon$

Hence, $T$ 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


 * $C = \frac 1 {\sqrt 5}$


 * $C_1 = 2$

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)

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