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.

Normal Bundle Theorem

Let $\tilde M$ be an $m$-dimensional Riemannian manifold.

Let $M \subseteq \tilde M$ be an immersed or embedded $n$-dimensional submanifold with or without boundary.

Let $\valueat {T \tilde M} M$ be the ambient tangent bundle.

Then $NM$ is a of rank-$\paren {m - n}$ smooth vector subbundle of $\valueat {T \tilde M} M$.

Theorem

Let $T \in \map {\DD'} \R$ be a distribution.

Then there exists a distribution $S \in \map {\DD'} \R$ such that in the distributional sense:

$S' = T$

Furthermore, $S$ is unique up to an arbitrary constant.

Proof

Let $\mathbf 0 : \R \to 0$ be the zero mapping.

Let $\phi_0 \in \map \DD \R \setminus \set {\mathbf 0}$ be a test function.

Let $\psi \in \map \DD \R$ be a test function.

Let $\phi : \R \to \R$ be a real function such that:

$\ds \phi := \psi - \frac {\int_{-\infty}^\infty \map \psi x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$

Existence of Unique Primitive of $\phi$

Since $\phi_0, \psi \in \map \DD \R$, by the closure of the test function vector space, $\phi \in \map \DD \R$ too.

Moreover:

\(\ds \int_{-\infty}^\infty \map \phi x \rd x\)

\(=\)

\(\ds \int_{-\infty}^\infty \map \psi x \rd x - \frac {\int_{-\infty}^\infty \map \psi x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \int_{-\infty}^\infty \map {\phi_0} x \rd x\)

\(\ds \)

\(=\)

\(\ds 0\)

Thus:

$\ds \phi \in Y := \set {\phi \in \map \DD \R : \int_{-\infty}^\infty \map \phi x \rd x = 0}$

By Characterization of Derivative of Test Function, there exists a unique $\Phi \in \map \DD \R$ such that $\Phi' = \phi$.

$\Box$

Let $T \in \map {\DD'} \R$ be a distribution.

Let $S: \map \DD \R \to \C$ be a mapping such that $\map S \psi = - \map T \Phi$.

Linearity of $S$

Let $\psi_1, \psi_2 \in \map \DD \R$.

Let $\Phi_1, \Phi_2 \in \map \DD \R$ be such that:

$\ds \Phi_1' = \phi_1 := \psi_1 - \frac {\int_{-\infty}^\infty \map {\psi_1} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$

$\ds \Phi_2' = \phi_2 := \psi_2 - \frac {\int_{-\infty}^\infty \map {\psi_2} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$

Then:

$\ds \paren {\Phi_1 + \Phi_2}' = \psi_1 + \psi_2 - \frac {\int_{-\infty}^\infty \paren {\map {\psi_1} x + \map {\psi_2} x} \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$

Thus:

\(\ds \map S {\psi_1 + \psi_2}\)

\(=\)

\(\ds - \map T {\Phi_1 + \Phi_2}\)

\(\ds \)

\(=\)

\(\ds - \map T {\Phi_1} - \map T {\Phi_2}\)

Definition of Distribution

\(\ds \)

\(=\)

\(\ds \map S {\psi_1} + \map S {\psi_2}\)

Similarly:

\(\ds \forall \alpha \in \C : \forall \psi \in \map \DD \R: \, \)

\(\ds \map S {\alpha \psi}\)

\(=\)

\(\ds - \map T {\alpha \Phi}\)

\(\ds \)

\(=\)

\(\ds - \alpha \map T \Phi\)

Definition of Distribution

\(\ds \)

\(=\)

\(\ds \alpha \map S \psi\)

$\Box$

Continuity

Let $\sequence {\psi_n}_{n \mathop \in \N}$ be a sequence in $\map \DD \R$ such that $\psi_n \stackrel {\DD} {\longrightarrow} \mathbf 0$.

By definition of the convergent sequence in the test function space, there exists $a \in \R_{> 0}$ such that:

$\forall n \in \N : \forall x \notin \closedint {-a} a : \map {\psi_n} x = 0$

and $\sequence {\psi_n}_{n \mathop \in \N}$ converges uniformly to $\mathbf 0$.

Hence:

\(\ds \size {\int_{-\infty}^\infty \map {\psi_n} x \rd x}\)

\(\le\)

\(\ds \size {\int_{-a}^a \map {\psi_n} x \rd x}\)

\(\ds \)

\(=\)

\(\ds 2 a \norm {\psi_n}_\infty\)

\(\ds \)

\(\stackrel {n \mathop \to \infty} {\longrightarrow}\)

\(\ds 0\)

Let:

$\ds \phi_n := \psi_n - \frac {\int_{-\infty}^\infty \map {\psi_n} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0$

$\phi_n$ is composed of $\phi_n$, which is compactly supported by its convergence conditions, and $\phi_0$ which has its own compact support.

Then:

$\exists b \in \R_{> 0} : \forall x \notin \closedint {-b} b : \map {\phi_n} x = 0$

Furthermore, for each $k \ge 0$ we have that $\sequence {\phi_n^{\paren k}}_{n \mathop \in \N}$ uniformly converges to $\mathbf 0$.

Thus, $\phi_n$ converges to $\mathbf 0$ in the test function space:

$\phi_n \stackrel {\DD} {\longrightarrow} \mathbf 0$

For every $n \in \N$ let $\Phi_n$ be the unique element in $\map \DD \R$ such that $\Phi'_n = \phi_n$.

From Conditions for Preservation of Covergence in Test Function Space under Differentiation we conclude that:

$\Phi_n \stackrel {\DD} {\longrightarrow} \mathbf 0$

Consequently, $\map S {\psi_n} = - \map S {\Phi_n} \to 0$ as $n \to \infty$.

Thus, $S \in \map {\DD'} \R$.

$S' = T$

Let $\Phi \in \map \DD \R$.

Then:

\(\ds \phi\)

\(:=\)

\(\ds \Phi' - \frac {\int_{-\infty}^\infty \map {\Phi'} x \rd x} {\int_{-\infty}^\infty \map {\phi_0} x \rd x} \phi_0\)

\(\ds \)

\(=\)

\(\ds \Phi' - 0\)

\(\ds \)

\(=\)

\(\ds \Phi'\)

Thus:

\(\ds \map {S'} \Psi\)

\(=\)

\(\ds - \map S {\Psi'}\)

\(\ds \)

\(=\)

\(\ds - \paren {- \map T \Psi}\)

\(\ds \)

\(=\)

\(\ds \map T \Psi\)

Hence, $S' = T$.

$\blacksquare$

Theorem(Derivative Operator on Continuously Differentiable Function Space with Supremum Norm is not Continuous)

Let $I = \closedint 0 1$ be a closed real interval.

Let $\map \CC I$ be the real-valued, continuous on $I$ function space.

Let $\map {\CC^1} I$ be the continuously differentiable function space.

Let $x \in \map {\CC^1} I$ be a continuoulsly differentiable real-valued function.

Let $D : \map {\CC^1} I \to \map \CC I$ be the derivative operator such that:

$\forall t \in \closedint 0 1 : \map {Dx} t := \map {x'} t$

Suppose $\map \CC I$ and $\map {\CC^1} I$ are equipped with the supremum norm such that

Then $D$ is not continuous.

Proof

Aiming for a contradiction, suppose $D$ is continuous.

By definition:

$\exists M \in \R_{> 0} : \forall x \in \map {\CC^1} I : \norm {\map D x}_\infty \le M \norm x_\infty$

Suppose $x = t^n$ with $n \in \N$.

Then:

$\norm {x}_\infty = \norm {t^n}_\infty = 1$

$\norm {x'}_\infty = \norm {n t^{n-1}}_\infty = n$

Hence:

\(\ds \norm{Dx}_\infty\)

\(=\)

\(\ds \norm{x'}_\infty\)

\(\ds \)

\(=\)

\(\ds n\)

\(\ds \)

\(\le\)

\(\ds M \norm {x}_\infty\)

\(\ds \)

\(=\)

\(\ds M \cdot 1\)

In other words:

$\forall n \in \N : n \le M$

But $M$ is finite.

This is a contradiction.

Hence, $D$ is not continuous.

$\blacksquare$

Theorem(Continuity of derivative operator)

$x \in \CC^1 \sqbrk {0, 1}$

$D : \CC^1 \sqbrk {0, 1} \to \CC \sqbrk {0, 1}$

$\map {Dx} t := \map {x'} t, t \in \sqbrk {0, 1}$

$D$ not continuous if equipped with $\norm {\, \cdot \,}_\infty$

Let $x = t^n, n \in \N$

$\norm {x}_\infty = \norm {t^n}_\infty = 1$

$\norm {x'}_\infty = \norm {n t^{n-1}}_\infty = n$

$\norm{Dx}_\infty = \norm{x'}_\infty = n \le M \norm {x}_\infty = M \cdot 1$

Hence, $D$ is not continuous.

$\norm {Dx}_\infty = \norm {x'}_\infty \le \norm {x}_\infty + \norm {x'}_\infty = \norm {x}_{1, \infty}$

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)

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