User:Julius

Current focus

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


 * So I just noticed that vector notation is being used in Gelfand's for higher dimensional functionals. This implies rewriting all multivariable functionals. Implement this gradually.

Lemmas and theorems for Bernstein's Theorem on Unique Extrema (1978)
Raw material

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)

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

Functional Analysis
$\paren{C \closedint a b,\norm{\cdot}_\infty }$ is a Banach space.

Let $\sequence{x_n}_{n \in \N}$ be a Cauchy sequence.


 * $\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n, m > N : \norm{x_n − x_m}_\infty < \epsilon$

Suppose, all the elements of $\sequence{x_n}_{n \in \N}$ are additionally indexed with $t$:


 * $\sequence{x_n}_{n \in \N} = \sequence{\map {x_n} t }_{n \in \N}$

Let $t \in \closedint a b$.

But

$\displaystyle \forall n, m > N : \norm {\map {x_n} t - \map {x_m} t}_\infty < \max_{\tau \in \closedint a b}\norm {\map {x_n} \tau - \map {x_m} \tau}_\infty = \norm {x_n - x_m}_\infty < \epsilon$

Hence, $\sequence{\map {x_n} t}_{n \in \N}$ is a Cauchy sequence in $\R$.

$\R$ is complete.

Therefore, $\sequence{\map {x_n} t}_{n \in \N}$ is convergent with limit $L = \map L t$.

Choose $N$ such that $\forall n,m > N : \norm{x_n - x_m} \le \frac \epsilon 3$

Let $\tau \in \closedint a b$.

Then $\forall n > N : \norm {\map {x_n} \tau - \map {x_{N + 1}} \tau } \le \norm {x_n - x_{N + 1} }_\infty \le \frac \epsilon 3$

Take the limit $n \to \infty$:


 * $\lim_{n \to \infty} \norm {\map {x_n} \tau - \map {x_{N + 1}} \tau } = \norm {\map x \tau - \map {x_{N + 1}} \tau } \le \frac \epsilon 3$

which holds for all $\tau \in \closedint a b$.

Now $\map {x_{N+1} } \tau \in C \closedint a b$

$\exists \delta > 0: \norm {\tau - t} < \delta \implies \norm {\map {x_{N+1} } t - \map {x_{N+1} } \tau} \le \frac \epsilon 3$

Thus:


 * $\norm {\map x \tau - \map x t} = \norm {\map x \tau - \map {x_{N+1}} \tau + \map {x_{N+1}} \tau - \map {x_{N+1}} t + \map {x_{N+1}} t - \map x t} \le$


 * $\norm {\map x \tau - \map {x_{N+1}} \tau} + \norm {\map {x_{N+1}} \tau - \map {x_{N+1}} t} + \norm {\map {x_{N+1}} t - \map x t} \le \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3 = \epsilon$

Hence $x$ is continuous at $t$.

Since $t \in C \closedint a b$, $t$ is continuous in whole interval.

Finally, show that $\sequence {x_n}_{n \in \N}$ converges to $x$.

Let $\epsilon > 0$.

Choose $N$ such that $\forall n,m > N : \norm{x_n - x_m}_\infty < \epsilon$

Fix $n > N$.

Let $t \in \closedint a, b$.

Then $\forall m > N: \norm {\map {x_n} t - \map {x_m} t} \le \norm {x_n - x_m}_\infty < \epsilon$

Thus $\norm{\map {x_n} t - \map x t} = \lim_{n \to \infty} \norm {\map {x_n} t - \map {x_m} t} \le \epsilon$

Since $t$ was arbitrary: $\norm {x_n - x}_\infty = \max_{t \in \closedint a b } \norm{\map {x_n} t - \map x t} \le \epsilon$

This could also have been achieved by fixing $n > N$.

So, $\forall n > N \norm {x_n - x}_\infty \le \epsilon$.

Therefore $\lim_{x \to \infty} x_n = x$ in $C \closedint a b$

Theorem
Let $f: \R \times \R \rightarrow \R$ be a real function.

Suppose, there exists $r > 0$ and $L > 0$ such that:


 * $\forall t \ge 0, \forall x, y \in \R : \size {\map f {x, t} - \map f {y, t}} \le L \size {x - y}$

Then:


 * $\forall x_0 \in \R : \exists T > 0: \exists x \in C^1 \closedint 0 T$

such that for all $t \in \closedint 0 T$ the following is satisfied.


 * $\map {\dfrac {dx}{dt}} t = \map f {\map x t, t}$


 * $\map x 0 = x_0$

Furthermore, the solution is unique.

Uniqueness

Let $x_1, x_2$ be solutions to IVP for $t \in \closedint 0 T$ for some $T > 0$.

Let $t_* := \max \set{t \in \closedint 0 T : \map {x_1} {\tau} = \map {x_2} {\tau}, \forall \tau \le t }$

Then:


 * $ \map {x_1} t - \map {x_1} {t_*} = \int_{t_*}^t \map {x_1'} \tau \rd \tau = \int_{t_*}^t \map {f_1} {\map {x_1} \tau, \tau} \rd \tau$


 * $ \map {x_2} t - \map {x_2} {t_*} = \int_{t_*}^t \map {x_2'} \tau \rd \tau = \int_{t_*}^t \map {f_2} {\map {x_2} \tau, \tau} \rd \tau$

So:


 * $\map {x_1} t - \map {x_2} t = \int_{t_*}^t \paren{\map {f_1} {\map {x_1} \tau, \tau} - \map {f_2} {\map {x_2} \tau, \tau}} \rd \tau$

Let

$N > \max \set {1, \frac 1 L, \frac 1 {L \paren{t - t_*}}}$

and

$\displaystyle M = \max_{t \in \closedint {t_*} {t_* + \frac 1 {LN}}} \size {\map {x_2} t - \map {x_1} t}$

Note that: $t_* + \frac 1 {LN} < T$

Then $\forall t \in \closedint {t_*} {t_* + \frac 1 {LN}}$

Thus $\forall t \in \closedint {t_*} {t_* + \frac 1 {LN}} : \size {\map {x_1} \tau - \map {x_2} \tau} \le \frac M N$

so $M \le \frac M N$ or $N \le 1$. Contradiction.

Sequence of recursively defined functions being increasingly better approximations.

Note that:


 * $x_{n + 1} = \Sigma_{k = 0}^n \paren {x_{n + 1} - x_n}$

For $0 \le t \le \frac {1}{2L}$

Therefore:


 * $\norm {x_{n+1} - x_n} \le \frac 1 2 \norm {x_n - x_{n-1}}$

and


 * $\norm {x_{n+1} - x_n} \le \frac 1 {2^n} \norm {x_1 - x_0}$

Therefore:


 * $\sum_{n = 0}^\infty \norm {x_{n+1} - x_n} \le \norm {x_1 - x_0} \sum_{n = 0}^\infty \frac 1 {2^n} < \infty$

and $x_0 + \sum_{n = 0}^\infty \paren{x_{n+1} -x_n}$ converges in $\struct{C \paren{0, T}, \norm{\cdot}_{\infty} }$ to $x \in C \paren{0, T}$.

We know that :$x_{n + 1} = x_0 + \int_0^t \map f {\map x \tau, \tau} \rd \tau$ for $t \in \sqbrk{0, T}$.

Define $\map {g_n} t = \map f {\map {x_n} t, t}$

Then the sequence $g_0, g_1, g_2$ is a sequence of partial sums $g_0 + \sum_{k = 0}^n \paren{g_{k + 1} - g_k}$.

We have $\norm{\map {g_{k+1}} t - \map {g_k} t} = \norm {\map f {\map {x_{k+1}} t, t} - \map f {\map {x_k} t, t}} \le L \norm{\map {x_{k+1}} t - \map {x_k} t} \le L \norm{x_{k+1} - x_k}_\infty \le \frac 1 {2^k} \norm{x_1 - x_0}_\infty$

So the partial sum converges to some $g$ in $\struct{C\paren{0, T}, \norm{\cdot}_\infty}$

We have :$\map g t = \lim_{n \to \infty} \map {g_n} t = \lim_{n \to \infty} \map f {\map {x_n} t, t}$

Since $\sum_{k = 1}^n \map {f_k} t$ converges in $\struct {C \paren{a,b}, \norm{\cdot}_\infty}$ then


 * $\sum_{k=1}^\infty \int_a^b \map {f_k} t \rd t = \int_a^b \map f t \rd t$

Using this, :$\lim_{n \to \infty} \int_0^T \map f {\map {x_n} t, t} \rd t = \lim_{n \to \infty} \int_0^T \paren {\map {g_0} t + \sum_{k = 0}^{n - 1} \paren {\map {g_{k+1}} t - \map {g_k} t}} \rd t = \int_0^T \map f {\map x t, t} \rd t$

Thus $\map x 0 = x_0 + 0 = x_0$ and by fundamental theorem of calculus $\map {x'} t = 0 + \map f {\map x t, t}$ for all $t \in \sqbrk{0, T}$.

-

Existence
Define the following sequence of recursively defined functions:

Note that:


 * $\displaystyle y_{n + 1} = y_0 + \sum_{k = 0}^n \paren {y_{k + 1} - y_k}$

Then, for $a \le x \le a + \frac {1}{2A}$ we have

By taking supremum norm of both sides, we get:


 * $\norm {y_{n+1} - y_n}_{\infty} \le \frac 1 2 \norm {y_n - y_{n-1}}_{\infty}$

By induction, the inequality can be extended:


 * $\displaystyle \norm {y_{n+1} - y_n} \le \frac 1 {2^n} \norm {y_1 - y_0}$

Therefore:

Hence, $\displaystyle y_0 + \sum_{n = 0}^\infty \paren{y_{n+1} - y_n}$ converges in $\paren{C \paren{a, a + \frac 1 {2A}}, \norm{\cdot}_{\infty}}$ to $y \in C \paren{a, a + \frac 1 {2A} }$ absolutely.


 * $\displaystyle \map y x = \lim_{n \to \infty} \map {y_{n + 1}} x = x_0 + \lim_{n \to \infty} \int_a^x \map f {\map {y_n} x, x} \rd x$

We know that :$y_{n + 1} = y_0 + \int_a^x \map f {\map y t, t} \rd t$ for $x \in \sqbrk{a - h, a + h}$.

To find the limit, consider the following sequence:

Define $\map {g_n} x = \map f {\map {x_n} x, x}$

Then the sequence $g_0, g_1, g_2$ is a sequence of partial sums $\displaystyle g_0 + \sum_{k = 0}^n \paren{g_{k + 1} - g_k}$.

We have $\norm{\map {g_{k+1}} x - \map {g_k} x} = \norm {\map f {\map {y_{k+1}} x, x} - \map f {\map {y_k} x, x}} \le L \norm{\map {y_{k+1}} x - \map {y_k} x} \le L \norm{y_{k+1} - y_k}_\infty \le \frac 1 {2^k} \norm{y_1 - y_0}_\infty$

So the partial sum converges to some $g$ in $\struct{C\paren{a - h, a + h}, \norm{\cdot}_\infty}$

We have :$\map g x = \lim_{n \to \infty} \map {g_n} x = \lim_{n \to \infty} \map f {\map {y_n} x, x}$

Since $\sum_{k = 1}^n \map {f_k} x$ converges in $\struct {C \paren{a,b}, \norm{\cdot}_\infty}$ then


 * $\sum_{k=1}^\infty \int_a^b \map {f_k} t \rd t = \int_a^b \map f t \rd t$

Using this, :$\lim_{n \to \infty} \int_0^T \map f {\map {x_n} t, t} \rd t = \lim_{n \to \infty} \int_0^T \paren {\map {g_0} t + \sum_{k = 0}^{n - 1} \paren {\map {g_{k+1}} x - \map {g_k} x}} \rd t = \int_0^T \map f {\map x t, t} \rd t$

Thus $\map x 0 = x_0 + 0 = x_0$ and by fundamental theorem of calculus $\map {y'} x = 0 + \map f {\map y x, x}$ for all $x \in \sqbrk{a - h, a + h}$.