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 $F, F_y, F_{y'} : \R \rightarrow \R$ be functions defined on an interval $I = \sqbrk{a,b}$.

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

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


 * $F_y \paren {x,y,y'} > k$


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

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

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

Lemma 1
There exists only one integral curve.

Proof
This proof will use Proof by Contradiction.

Suppose there are two integral curves $y = \phi_1(x)$ and $y = \phi_2(x)$ such that


 * $y''(x)=F\paren{x,y,y'}$.

Define a mapping $\delta : \R \rightarrow \R$:


 * $\delta(x) = \phi_2(x) - \phi_1(x)$

From definition it follows that


 * $\delta\paren a=\delta\paren b=0$

Then the second derivative of $\delta$ yields

where


 * $F_y^* = F_y \paren{x,\phi_1 + \theta \delta,\phi_1' + \theta\delta'},$


 * $F_{y'}^* = F_{y'}\paren{x,\phi_1 + \theta\delta, \phi_1' + \theta\delta'}$

and


 * $\theta \in \R : 0< \theta <1$.

Suppose


 * $ \forall x\in I:\phi_2(x) \ne \phi_1(x)$.

Then there are two possibilities for $\delta$ vanishing at the ends of $\sqbrk{a,b}$:
 * $\delta(x)$ attains a positive maximum within $\paren{a,b}$;
 * $\delta(x)$ attains a negative minimum within $\paren{a,b}$.

Denote this point by $\xi$.

Then for maximum at $x=\xi$ we have


 * $\delta''(\xi)\le 0$
 * $\delta(\xi) > 0$
 * $\delta'(\xi)=0$.

From $\delta''$, this implies that $F_y^* \le 0$.

This is in contradiction with the assumption $F_y^* > k > 0$.

For the minimum inequalities are reversed, but the last equality is the same.

Therefore, it must be so that


 * $\phi_1(x)=\phi_2(x)$

Lemma 2
Suppose for $x\in \sqbrk{a,c}$


 * $y''(x) = F\paren{x,y,y'}$

where


 * $y(a)=a_1,$
 * $y(c)=c_1.$

Then

$\size{ y(x) - \dfrac{a_1 \paren{c-x}+c_1 \paren{x-a} }{c-a} } \le \max\limits_{a \le x \le b} \size{ F\paren{x, \dfrac{a_1 \paren{c-x}+c_1 \paren{x-a} }{c-a}, \dfrac{c_1-a_1}{c-a}} }$

Proof
Let $\psi : \R \rightarrow \R$ be a real mapping such that


 * $\psi = \dfrac{a_1 \paren{c-x}+c_1 \paren{x-a} }{c-a}+\theta\sqbrk{y(x) - \dfrac{a_1 \paren{c-x}+c_1 \paren{x-a} }{c-a} }$

where


 * $\theta \in \R : 0 < \theta < 1$

As a consequence of $y''=F\paren{x,y,y'}$ we have

$y''(x) = F \paren{x, \dfrac{a_1 \paren{c-x}+c_1 \paren{x-a} }{c-a},y'(x) }+\sqbrk{y(x)-\dfrac{a_1 \paren{c-x}+c_1 \paren{x-a} }{c-a}}F_y \paren{x,\psi,y'(x)}$

Note that the term $\chi$, defined as


 * $\chi=y(x) - \dfrac{a_1 \paren{c-a}+c_1 \paren{x-a} }{c-a}$

vanishes at $x=a$ and $x=c$.

Unless $\chi$ is identically zero, $\size{\chi}$ is maximised at a point $\xi\in \paren{a,b}$.

At this point $\chi$ will have either a positive maximum or a negative minimum.

In the first case


 * $y''\paren{\xi}\le 0$,


 * $y'\paren{\xi}=\dfrac{c_1-a}{c-a}$

which implies


 * $F \paren{\xi, \dfrac{a_1 \paren{c - \xi} + c_1 \paren{\xi - a} }{c - a}, \dfrac{c_1 - a_1}{c - a} } + k\sqbrk{ y(\xi) - \dfrac{a_1 \paren{c-\xi}+c_1 \paren{\xi - a} }{c - a}}\le 0$

Hence,


 * $y(\xi) - \dfrac{a_1(c-\xi)+c_1(\xi - a)}{c - a} \le -\dfrac{1}{k} F\paren{\xi, \dfrac{a_1(c-\xi)+c_1(\xi-a)}{c-a},\dfrac{c_1-a_1}{c-a}}$

In the second case it is shown that everywhere in $\sqbrk{a,c}$ it holds that


 * $\size{ y(x) - \dfrac{a_1(c-x)+c_1(x-a)}{c-a} } \le \dfrac{1}{k} \max\limits_{a \le x \le b} \size{F\paren{x, \dfrac{a_1(c-x)+c_1(x-a)}{c-a},\dfrac{c_1-a_1}{c-a}}}$

Lemma 3
Suppose for $x\in \sqbrk{a,c}$


 * $y''(x) = F\paren{x,y,y'}$

where


 * $y(a)=a_1,$
 * $y(c)=c_1.$

Then


 * $\size{ y'(x) - \dfrac{c_1-a_1}{c-a} }\le M $

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

Proof
Let $\mathfrak{A}$ and $\mathfrak{B}$ be the least upper bounds of $\alpha(x,y)$ and $\beta(x,y)$ respectively in the rectangle $a\le x \le c$, $\size{y} \le m + max \{\size{ a_1 }, \size{ c_1 } \}$

Suppose, that $\mathfrak{A} \ge 1$.

Let $u$, $v$ be real mappings such that


 * $y(x)-\dfrac{a_1(c-x)+c_1(x-a)}{c-a} +m = \dfrac{\Ln u}{2 \mathfrak{A}}$,


 * $-y(x)+ \dfrac{a_1(c-x)+c_1(x-a)}{c-a} + m =\dfrac{\Ln v}{\mathfrak{2A}}$

For $x\in I$ the lefthand sides are not negative.

Thus,


 * $\forall x\in I:u,v\le 1$.

Differentiate these equations wrt $x$:


 * $y'(x) - \dfrac{c_1-a_1}{c-a}=\dfrac{u'}{2 \mathfrak{A}u}$,


 * $-y'(x) + \dfrac{c_1 - a_1}{c - a} = \dfrac{v'}{2 \mathfrak{A}v}$

Differentiate again:


 * $y(x)=\dfrac{u}{2 \mathfrak{A}u}-\dfrac{u'^2}{2 \mathfrak{A}u^2}$,


 * $-y(x)=\dfrac{v}{2\mathfrak{A}v}-\dfrac{v'^2}{2\mathfrak{A}v^2}$

By assumption:

where


 * $\mathfrak{B}_1 = \mathfrak{B} + 2 \mathfrak{A}\paren{\dfrac{c_1 - a_1}{c - a}}^2 $

From the absolute value of $y''$ it follows, that

Multiply the inequality by $2 \mathfrak{A}u$:

Similarly:

Multiply the inequality by $-2\mathfrak{A}v$:

Since $u(a)=u(c)$ and $v(a)=v(c)$, it follows that


 * $\exists A \subset I : \forall x_0\in A : y'(x_0) - \dfrac{c_1-a_1}{c-a} = 0$

Points $x_0$ divide $I$ into subintervals.

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

Let $J$ be one of the subintervals.

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 $J$.

From first derivative of logarithmic equations, $u'$ is not negative.

Multiply both sides by $u'$:


 * $u'' u' \ge -2 \mathfrak{A}\mathfrak{B}_1 u u'$.

Integrating this from $x \in J$ to $\xi$, together with $u'(\xi) = 0$, yields


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

Then:

Hence,


 * $\forall x\in J:\size{ y'(x) - \dfrac{c_1 - a_1}{c - a} } \le \sqrt{ \dfrac{\mathfrak{B}_1 }{2 \mathfrak{A} } } e^{4 \mathfrak{A}m} $

Similar 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 $A\paren{a,a_1}$.

On this curve choose a point $D\paren{d,d_1}$.

Join it with a point $B\paren{b,b_1}$ with a broken line consisting of two segments parallel to the coordinate axes.

Choose a point anywhere on the broken line $DQB$.

Denote it by $P\paren{\xi,\xi_1}$.

A family of integral curves $y=\phi\paren{x,a}$ 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 $P$.

In other words, if $\xi$ is sufficiently close to $d$, then the equation


 * $d_1 = \phi \paren{\xi,\alpha}$

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 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.

Therefore, it approaches a limit.

If this limit is different from $\pm \dfrac{1}{2}\pi$ the point $R$ is attained.

Hence,


 * $\alpha \rightarrow \pm \dfrac{1}{2}\pi$.

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 $\xi_1 = \phi \paren{b,a}$ 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