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 (Akhiezer)
Proof consists of three parts: existence of only one curve, rectangular bounds and the theorem itself.

A. Suppose there are two integral curves: $y = \phi_1(x)$ and $y = \phi_2(x)$.

Define mapping $\delta : \delta(x) = \phi_2(x) - \phi_1(x)$

Then

$\delta'' = F(x, \phi_2, \phi_2') - F(x, \phi_1, \phi_1') = \delta F_y^* + \delta' F_{y'}^*$

where $F_y^* = F_y(x, \phi_1 + \theta \delta, \phi_1' + \theta \delta')$ and $F_{y'}^* = F_{y'}(x, \phi_1 +\theta \delta, \phi_1'+\theta \delta')$

and $ 0 < \theta < 1$.

If $ \phi_2(x) \ne \phi_1(x) $ then there are two possibilities for $\delta$ vanishing at the ends of $[a,b]$: either $\delta(x)$ is a positive maximum or a negative minimum within $(a,b)$. Denote this point by $\xi$

In the first case at the point $\xi$ we have $\delta''\le 0$, $\delta >0$, $\delta'=0$.

It is not allowed by the condition $F_y^* > k > 0$.

In the second case at the point $\xi$ we have $\delta''\ge 0$, $\delta < 0$, $\delta'=0$, which is impossible for the same reason.

Thus the uniqueness has been proven.

B. Boundedness

If $y''(x) = F(x,y,y')$ over $[a,c]$, and $y(a)=a_1$, $y(c)=c_1$, then

$\left\vert y(x) - \frac{a_1(c-x)+c_1(x-a)}{c-a} \right\vert \le max_{a \le x \le c} \left \vert F(x, \frac{a_1(c-x)+c_1(x-a)}{c-a}, \frac{c_1-a_1}{c-a}) \right \vert$ and $\left \vert y'(x) - \frac{c_1 - a_1}{c - a} \right \vert \le M $

The constant $M$ depends on the rectangle with the base $a \le x \le c$ and the upper bound of the functions $\alpha(x,y)$, $\beta(x,y)$.

Introduce the following equation which is a consequence of $y'' = F$:

$y''(x) = F(x, \frac{a_1(c-x)+c_1(x-a)}{c-a},y'(x))+[y(x)-\frac{a_1(c-x)+c_1(x-a)}{c-a}]F_y(x,\psi,y'(x))$

where $\psi = \frac{a_1(c-x)+c_1(x-a)}{c-a}+\theta[y(x) - \frac{a_1(c-x)+c_1(x-a)}{c-a}]$ with $0 < \theta < 1$

Function $y(x) - \frac{a_1(c-a)+c_1(x-a)}{c-a}$ vanishes at $x=a$ and $x=c$.

Its maximum happens in the interior of $(a,b)$, unless it is identically zero.

At $\xi$, where the maximum is obtained, the function will have either a positive maximum or a negative minimum.

In the first case $y''(\xi) \le 0$, $y'(xi) = \frac{c_1 - a}{c - a}$

which implies $F(\xi, \frac{a_1(c - \xi) + c_1(\xi - a)}{c - a}, \frac{c_1 - a_1}{c - a}) + k[ y(\xi) - \frac{a_1(c-\xi)+c_1(\xi - a)}{c - a}]\le 0$

Hence, $y(\xi) - \frac{a_1(c-\xi)+c_1(\xi - a)}{c - a} \le -\frac{1}{k} F(\xi, \frac{a_1(c-\xi)+c_1(\xi-a)}{c-a},\frac{c_1-a_1}{c-a})$

In the second case it is shown that everywhere in $[a,c]$ it holds that $\vert y(x) - \frac{a_1(c-x)+c_1(x-a)}{c-a} \vert \le \frac{1}{k} max_{a\le x\le c}\vert F(x, \frac{a_1(c-x)+c_1(x-a)}{c-a},\frac{c_1-a_1}{c-a}) \vert$

Bound M:

Denote by $\mathfrak{A}$ and $\mathfrak{B}$ the least upper bounds of $\alpha(x,y)$ and $\beta(x,y)$ in the rectangle $a\le x \le c$, $\vert y \vert \le m + max \{\vert a_1 \vert, \vert c_1 \vert \}$

Assume, that $\mathfrak{A} \ge 1$ and define functions $u$, $v$ such that $y(x)-\frac{a_1(c-x)+c_1(x-a)}{c-a} +m = \frac{ln u}{2 \mathfrak{A}}$, $-y(x)+ \frac{a_1(c-x)+c_1(x-a)}{c-a} + m =\frac{ln v}{\mathfrak{2A}}$

Since the left sides are not negative in $[a,c]$, it follows that $u \le 1$, $v \le 1$.

Differentiate these equations: $y'(x) - \frac{c_1-a_1}{c-a}=\frac{u'}{2 \mathfrak{A}u}$, $-y'(x) + \frac{c_1 - a_1}{c - a} = \frac{v'}{2 \mathfrak{A}v}$

Differentiate again: $y(x)= \frac{u}{2 \mathfrak{A}u}-\frac{u'^2}{2 \mathfrak{A}u^2}$, $-y(x)=\frac{v}{2\mathfrak{A}v}-\frac{v'^2}{2\mathfrak{A}v^2}$

From inequality in the theorem statement: $\vert y''(x) \vert \le 2 \mathfrak{A} \{ y'(x) - \frac{c_1 - a_1}{c - a} \}^2 + \mathfrak{B}_1$, where $\mathfrak{B}_1 = \mathfrak{B} + 2 \mathfrak{A}(\frac{c_1 - a_1}{c - a})^2 $

Furthermore, $y(x) \le -2 \mathfrak{A} \{ y'(x) - \frac{c_1 - a_1}{c - a} \}^2 - \mathfrak{B}_1$ as well as $y(x)\le 2 \mathfrak{A} \{ y'(x) - \frac{c_1 - a_1}{c - a} \}^2 + \mathfrak{B}_1$

From second derivatives it follows that $\frac{u}{2 \mathfrak{A}u} - \frac{u'^2}{2 \mathfrak{A}u^2} \ge - 2 \mathfrak{A}\frac{u'^2}{4 \mathfrak{A}^2 u^2} - \mathfrak{B}_1$ and since $u \ge 1$ we have $ u \ge -2 \mathfrak{A} \mathfrak{B}_1 u$

Similarly, $v'' \ge -2\mathfrak{A}\mathfrak{B}_1 v$

Focus on the points, where $y'(x) - \frac{c_1-a_1}{c-a}$ vanishes. These points exist, because $u(a)=u(c)$ and $v(a)=v(c)$.

These points divide $[a,c]$ into subintervals.

$u'(x)$, $v'(x)$ do not change sign in the subintervals and vanish at one or both endpoints of each subinterval.

Let $I$ be on of the subintervals, and let functions $u'(x)$, $v'(x)$ be zero at $\xi$, the right end point. The quantity has to be either positive or negative.

Suppose it is positive in $I$. $u'$ is not negative so if we multiply both sides by $u'$, we obtain $u'' u' \ge -2 \mathfrak{A}\mathfrak{B}_1 u u'$.

Integrating this from $x \in I$ to $\xi$, and recalling that $u'(\xi) = 0$, we have $-u'^2(x) \ge -2 \mathfrak{A}\mathfrak{B}_1 \{ u^2(\xi) - u^2(x) \}$

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