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(Fundamental Solution of 2d Laplacian)
Let $\delta_{\tuple {0, 0}} \in \map {\DD'} {\R^2}$ be the Dirac delta distribution.

Let $\Delta = \dfrac {\partial^2} {\partial x^2} + \dfrac {\partial^2} {\partial y^2}$

Then in the distributional sense:


 * $\ds \map \Delta {\frac {\ln r} {2 \pi}} = \delta_{\tuple {0, 0}}$

where $r = \sqrt {x^2 + y^2}$.

Local integrability
Let $u : \R^2 \to \R$ be a radial function, say $\map u {\mathbf x} = \map f r$, where $r = \norm {\mathbf x}_2$.

Then:


 * $\ds \Delta f = \map {f''} r + \frac {\map {f'} r} r$

Thus:


 * $\ds \Delta \map \ln r = - \frac {1} {r^2} + \frac 1 r \cdot \frac 1 r = 0$

Furthermore, for all $R > 0$ we have:


 * $\ds \int_0^{2 \pi} \int_0^R r \ln r \d r \d \theta = 2 \pi \int_0^R r \ln r \d r < \infty$

Hence:


 * $\mathbf x \mapsto \ln r \in \map {L^1_{loc}} {\R^2}$

Fundamental Solution
Let $\phi \in \map \DD {\R^2}$ be a test function with support on open ball $\map {B_R} {\mathbf 0}$.

Then:

Green's formula on anulus $\Omega := \set {\mathbf x \in \R^2 : \epsilon < \norm {\mathbf x}_2 < R}$ with boundary $\partial \Omega = \map S {\epsilon} \cup \map S {R}$ where


 * $\map S \epsilon = \set {\mathbf x : \norm {\mathbf x}_2 = \epsilon}$


 * $\map S {R} = \set {\mathbf x : \norm {\mathbf x}_2 = R}$

We have that:


 * $\dfrac {\partial \phi} {\partial n} = \nabla \phi \cdot \map {\mathbf n} {\mathbf x}$


 * $\norm {\map {\mathbf n} {\mathbf x}}_2 = 1$

By Cauchy-Schwarz inequality:


 * $\size {\dfrac {\partial \phi}{\partial n}} = \norm {\nabla \phi \cdot \map {\mathbf n}{\mathbf x}} \le \norm {\dfrac {\partial \phi} {\partial n}} \le M$

where


 * $M := \sup_{\norm {\mathbf x}_2 \le R} \norm {\nabla \phi}_2 < \infty$

Finally:


 * $\int_{\norm {\mathbf x}_2 = \epsilon} \rd \map s {\mathbf x} = 2 \pi \epsilon$

Also:


 * $\ds \lim_{\epsilon \mathop \to 0} 2 \pi \epsilon \ln \epsilon = 0$

As for the second integral:


 * $\ds \dfrac {\partial \ln r} {\partial n} = - \dfrac {\partial \ln r} {\partial r} = - \frac 1 r = - \frac 1 \epsilon$


 * $\ds -\int_{\norm{\mathbf x}_2 = \epsilon} \paren{\dfrac {\partial \ln r}{\partial n}} \phi \rd s = \frac 1 \epsilon \int_{\norm{\mathbf x}_2 = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} = 2 \pi \paren {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x}}$

Let $\eta > 0$.


 * $\ds \size {\frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 = \epsilon} \map \phi {\mathbf x} \rd \map s {\mathbf x} - \map \phi {\mathbf 0}} = \frac 1 {2 \pi \epsilon} \int_{\norm {\mathbf x}_2 = \epsilon} \size {\map \phi {\mathbf x} - \map \phi {\mathbf 0}} \rd \map s {\mathbf x} \le \frac 1 {2 \pi \epsilon} \cdot \eta 2 \pi \epsilon = \eta$

where first $\epsilon_0 > 0$ is chosen small enough so that $\size {\map \phi {\mathbf x} - \map \phi {\mathbf 0}} \le \eta$ if $\norm {\mathbf x}_2 \le \epsilon_0$ and $\epsilon$ satisfies $0 < \epsilon \le \epsilon_0$.

Thus:


 * $\ds \lim_{\epsilon \to 0} \int_{\epsilon < \norm{\mathbf x}_2 < R} \ln r \Delta \phi \rd \mathbf x = 2 \pi \map \phi {\mathbf 0} = 2 \pi \map {\delta_{\tuple {0, 0}}} \phi$

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.

Then $D$ is not continuous.

Proof
$D$ 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:

In other words:


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

But $M$ is finite.

This is a contradiction.

Hence, $D$ is not continuous.

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


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


 * $C_1 = 2$

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)

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