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.

Theorem
Let $ F, F_y, F_{ y' } $ be functions.

Let $ F, F_y, F_{ y' } $ be continuous at every point $ \left ( { x, y } \right ) $ for all finite $ y' $.

Suppose a constant $ k > 0 $ and functions $ \alpha = \alpha \left ( { x, y } \right ) \ge 0 $, $ \beta = \beta \left ( { x, y } \right ) \ge 0 $ bounded in every finite region of the plane exist such that


 * $ F_y \left ( { x, y, y' } \right ) > k $


 * $ \left \vert { F \left ( { x, y, y' } \right ) } \right \vert \le \alpha y'^2 + \beta $

Then one and only one integral curve of equation $ y'' = F \left ( { x, y, y' } \right ) $ passes through any two points $ \left ( { a, A } \right ) $ and $ \left ( { b, B } \right) $ such that $ a \ne b $.

Proof
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 $-u'^2(x) \ge -2 \mathfrak{A}\mathfrak{B}_1 \{ u^2(\xi) - u^2(x) \}$

From which it follows that $u'^2(x)\le 2 \mathfrak{A}\mathfrak{B}_1 u^2(\xi) \le 2 \mathfrak{A}\mathfrak{B}_1 e^{8 \mathfrak{A}m}$.

Then $ \frac{u'^2(x)}{u^2(x)} \le 2 \mathfrak{A} \mathfrak{B}_1 e^{8 \mathfrak{A}m} $

Then $ 4 \mathfrak{A}^2 \{ y'(x) - \frac{c_1 - a_1}{c - a} \}^2 \le 2 \mathfrak{A} \mathfrak{B}_1 e^{8 \mathfrak{A}m}$

or in the interval $ I $ we have $ \vert y'(x) - \frac{c_1 - a_1}{c - a} \vert \le \sqrt{ \frac{\mathfrak{B}_1 }{2 \mathfrak{A} } } e^{4 \mathfrak{A}m} $

Similarl arguments for aforementioned function being negative.

C. The theorem itself (geometrical approach)

Draw an element of the integral curve for which $y'(a) = 0$ through the point $A(a,a_1)$.

On this curve choose a point $D(d,d_1)$ and join it with a point $B(b,b_1)$ with a broken line consisting of two segments parallel to the coordinate axes.

Choose a point anywhere on the broken line $DQB$ and denote it by $P(\xi,\xi_1)$.

A family of integral curves $y=\phi(x,a)$ can be constructed through the point $A$ where the parameter $\alpha$ represents $y'(a)$.

For $\alpha = 0$ the integral curve corresponds with $AD$.

If the point is sufficiently close to $D$, we can find a curve of the given family, which passes through $P$, i.e. if $\xi$ is sufficiently close to $d$, then the equation $d_1 = \phi(\xi,\alpha)$ can be solved for $\alpha$.

By uniqueness, there can be only one solution.

Consequently, equation $d$ implies $\xi$ as a monotonic function of $\alpha$.

This implies that $\alpha$ is a monotonic function of $\xi$.

Now we will show that an arbitrary point of $DQ$ can be reached by an integral curve in this way.

Suppose that there exists another point $R$ which can also be reached.

When $\xi$ increases to the value of the abscissa of $R$, the angle $\alpha$ varies monotonically, and hence approaches a limit.

If this limit is different from $\pm \frac{1}{2}\pi$ the point $R$ is attained, hence $\alpha \righarrow \pm \frac{1}{2}\pi$.

That is, as the point $P$ approaches the point $R$ the derivative at the point $a$ of an integral curve $y = y(x)$ joining $A$ to $P$ will not remain bounded.

This contradicts the evaluation for the derivative of the solution which provides that $\vert y'(a) \vert$ cannot be as large as we want, because the endpoints $A$ and $P$ of an integral curve are in a bounded region, and the difference of their abscissas does not approach zero, so the point can be reached.

Now we let the point $P$ move from $Q$ to $B$.

Here a point $P$ placed close to $Q$ can certainly be reached by an integral curve, i.e. the equation $\xi_1 = \phi(b,a)$ defines $\alpha$ as a single-valued function of $\xi_1$. Again we have the monotonic property. To finish the proof, repeat the argument that had just been made.

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