Uniqueness in Bernstein's theorem of calculus of variations

Question

I'm working through Gelfand and Fomin's book on calculus of variations. One of the book's exercises is to prove the uniqueness portion of a result called "Bernstein's theorem" on solutions to equations of the form $y'' = F(x, y, y')$. The book states the theorem thus:

If the functions $F$, $F_y$, and $F_{y'}$ are continuous at every finite point $(x, y)$ for every finite $y$, and if a constant $k > 0$ and functions $$\alpha = \alpha(x, y) \geq 0, \qquad \beta = \beta(x, y) \geq 0$$ (which are bounded in every finite region of the plane) can be found such that $$F_y(x, y, y') > k, \quad |F(x, y, y')| < \alpha y'^2 + \beta,$$ then one and only one integral curve satisfying $y'' = F(x, y, y')$ passes through any two points $(a, A)$ and $(b, B)$ with different abscissas ($a \neq b$).

(Subscripts on $F$ mean partial derivatives.) The hint for the exercise is:

Let $\Delta(x) = \varphi_2(x) - \varphi_1(x)$, where $\varphi_1(x)$ and $\varphi_2(x)$ are two solutions of $y'' = F(x, y, y')$, write an expression for $\Delta''$ and use the condition $F_y(x, y, y') > k$.

Following the hint, I got the expression $$\Delta''(x) = F(x, \varphi_2(x), \varphi'_2(x)) - F(x, \varphi_1(x), \varphi_1'(x)).$$

I thought that I could use the condition on $F_y$ to get some sort of lower bound on the magnitude of the RHS of this equation, and then try to turn that into some sort of proof that $\Delta(a)$ and $\Delta(b)$ cannot both be zero. But because $\varphi_1'(x) \neq \varphi_2'(x)$, I don't know what I can conclude about $ F(x, \varphi_2(x), \varphi'_2(x)) - F(x, \varphi_1(x), \varphi_1'(x))$ unless I also know something about $F_{y'}$ as well as $F_y$, and the theorem imposes only a very weak hypothesis, continuity, on $F_{y'}$.

Answer 1

Fix $x$ and use the mean value theorem applied to the function $$g(t):=F(x, t\varphi_2(x)+(1-t)\varphi_1(x), t\varphi'_2(x)+(1-t)\varphi_1'(x))$$ to find \begin{align}\Delta''(x)&=g(1)-g(0)=g'(c)1=F_y(x,f_c(x),f'_c(x))(\varphi_2(x)-\varphi_1(x))\\ &\quad+F_{y'}(x,f_c(x),f'_c(x))(\varphi_2'(x)-\varphi_1'(x)) \\:&=-G(x)\Delta(x)-H(x)\Delta'(x),\end{align} where $f_c(x):=c\varphi_2(x)+(1-c)\varphi_1(x)$, $G(x):=-F_y(x,f_c(x),f'_c(x))$, and $H(x):=-F_{y'}(x,f_c(x),f'_c(x))$. So now you have the linear equation $$\Delta''(x)+H(x)\Delta'(x)+G(x)\Delta(x)=0,$$ where you know that $\Delta(a)=0$, $\Delta(b)=0$ and $G(x)\le -k<0$. Now you have to apply the maximum principle, which says that if $H$ and $G$ are bounded, $H\le0$, and $\Delta$ achieves a nonnegative maximum value $M$ at an interior point $d$ then $\Delta(x)\equiv M$. Assume by contradiction that $\Delta>0$ somewhere in $(a,b)$, then by continuity it must have a maximum value $M>0$ at some $d\in (a,b)$ and so $\Delta(x)\equiv M$ by the maximum principle, which contradicts the fact that $\Delta(b)=0$. This shows that $\Delta\le 0$. By interchanging $\varphi_1$ with $\varphi_2$ you get that $\Delta=0$.

Are you familiar with the maximum principle? You can find it in the book of Protter and Weinberger. Theorem 3. The trick is to take the function $$z(x):=\Delta (x)+\varepsilon (e^{\alpha (x-d)}-1),$$ where $\varepsilon$ is small and $\alpha$ very large. Let me know if you want more details.

Answer 2

First of all, we may assume without loss of generality that $\Delta$ does not change sign on the interval $(a,b)$. For if it does, we instead prove the theorem on a smaller interval $(a,b_*)$ where $b_*$ is the first time after $a$ that $\Delta$ vanishes. We conclude from the proof in the special case that $\Delta$ is identically $0$ on that shorter interval, and by the uniqueness theorem for the initial value problem---at $b_*$---both solutions are identical on the rest of the interval, too.

So let us assume that $\Delta>0$ on that interval---otherwise we swap the two solutions. You prove$$\Delta''>k\Delta + h(x)\Delta'$$with some function $h(x)$. We get $h(x)$ from the mean value theorem applied to $F$ as a function of $y'$. Can you justify this step?

Let us skip a technicality and assume without proof that $h$ is at least good enough to have a definite integral$$H(x):= \int_a^x h(t)\,dt.$$Of course $h$ may not be uniquely defined, so we mean to assume that some choice of $h$ is good enough to have a definite integral.

Then multiply the inequality$$\Delta'' - h(x)\Delta' >k\Delta$$with $\exp[-H(x)]$ and obtain$$[ \exp(-H(x)) \Delta'(x) ]' > k \exp(-H(x))\Delta(x) >0.$$Can you finish up from here?

Now to fix the gap with that technical point about the construction of $H(x)$, we may say that we can write $h(x)$ as an integral expression from which it is clear that $h$ is continuous:$$F(\varphi_2')-F(\varphi_1') = \int_0^1 \frac{d}{dt} F((1-t)\varphi_1'+t\varphi_2') \, dt$$from the Fundamental Theorem of Calculus---and we omitted the other variables $x$, $\varphi_1(x)$ here.

Are you familiar with the maximum principle? You can find it in the book of Protter and Weinberger. Theorem 3. The trick is to take the function $$z(x):=\Delta (x)+\varepsilon (e^{\alpha (x-d)}-1),$$ where $\varepsilon$ is small and $\alpha$ very large.

The argument I intended is tantamount to the $1$-dimensional maximum principle.

The trick with the integrating factor $\exp[-H(x)]$ is a standard trick in ordinary differential equations---normally used with $1$st order, but here towards the first-order term in $\Delta'$.

The reasoning I had in mind when arguing from$$[\exp(-H(x))\Delta'(x)]' > k\exp(-H(x))\Delta(x) > 0$$is that the expression $\exp(-H(x)) \Delta'(x)$ is strictly increasing over the interval $[a, b]$ because it has positive derivative. On the other hand, at $x = a$, this expression is nonnegative because $\Delta > 0$ for $x > a$ and $\Delta(a) = 0$, whereas at $x = b$, it is nonpositive for an analogous reason. This is a contradiction.

Of course this very argument can be used to prove the maximum principle in $1$ dimension.