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I am looking for a proof of the following statement which is known to be true as far as I heard.

Let $g\colon [a,b]\to \mathbb{R}$ be a smooth function. Assume that $$b-a< \pi.$$ Assume also $$g(a)\geq 0,g(b)\geq 0,$$ $$g''+g\leq 0 \mbox{ on } [a,b].$$ Then $g\geq 0$ on $[a,b]$.

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    $\begingroup$ False. Take $g(x)=\sin x$, $a=\pi$, $b=2\pi$. $\endgroup$ – Mizar Jul 13 at 14:41
  • $\begingroup$ @Mizar: Thanks, corrected. The inequality $b-a<\pi$ has to be strict. $\endgroup$ – makt Jul 13 at 15:22
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Write $g=g^+-g^-$ in $[0,\ell]$, multiply $g''+g \le 0$ by $g^-$ (which vanishes at the the endpoints) and integrate. Then we get with $v=g^-$ $$ \int_0^l v'^2- \int_0^l v^2 \le 0.$$ Since the first eigenvalue of the Dirichlet laplacian in $[0,\ell]$ is $\pi^2/\ell^2 $ we have also $$ \int_0^l v^2 \le \frac{\ell^2}{\pi^2} \int_0^l v'^2 $$ and then $v=g^-=0$, since $\ell <\pi$.

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Here's another argument using Sturm-Liouville theory.

Given $b - \pi < a \le x \le b$, let $$ f(x) = \sin (x - b + \pi). $$ Observe that $f > 0$ and $f'' + f = 0$ on $[a,b)$, $f(a) > 0$, $f(b) = 0$, and $f'(b) = -1$.

Since \begin{align*} (fg'-f'g)' &= (g'' + g)f - (f''+f)g \le 0\\ (fg'-f'g)(b) &= f(b)g'(b) - f'(b)g(b) = g(b) \ge 0, \end{align*} it follows that, on $[a,b]$, $$ fg' - f'g \ge 0. $$ and therefore, on $[a,b)$, \begin{align*} \left(\frac{g}{f}\right)' &= \frac{fg' - gf'}{f^2} \ge 0. \end{align*} Since $$ \frac{g(a)}{f(a)} \ge 0, $$ it follows that, on $[a,b)$, $$ \frac{g}{f} \ge 0. $$

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    $\begingroup$ What is "Sturm comparison theorem"? $\endgroup$ – Fedor Petrov Jul 13 at 17:46
  • $\begingroup$ @FedorPetrov, it compares solutions to self-adjoint 2nd order ODEs. I never actually remember the precise statement of the theorem, so I adapt the relatively elementary argument to each particular situation. $\endgroup$ – Deane Yang Jul 13 at 17:51
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    $\begingroup$ And where is it used? The argument looks self-contained. $\endgroup$ – Fedor Petrov Jul 13 at 17:52
  • $\begingroup$ @FedorPetrov, variants of this argument are commonly used to compare the behavior of geodesics on a Riemannian manifold to those on a manifold with constant curvature. But the argument itself was originally developed in what's known as Sturm-Liouville theory. It appears in many ODE textbooks. $\endgroup$ – Deane Yang Jul 13 at 17:56
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    $\begingroup$ my question was: where is it used in your answer? Probably the answer is that you include the proof but not the formulation. $\endgroup$ – Fedor Petrov Jul 13 at 19:31
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Suppose the contrary, so that $g(s)<0$ for some $s\in(a,b)$. Replacing now $a$ and $b$ by $\max\{t\in[a,s)\colon g(t)\ge0\}$ and $\min\{t\in(s,b]\colon g(t)\ge0\}$, respectively, we see that without loss of generality (wlog) \begin{equation} g(a)=0=g(b). \end{equation} Also, by a horizontal shift, wlog \begin{equation} a=0,\quad b\in(0,\pi). \end{equation}

Let \begin{equation} h:=-(g''+g)\ge0. \end{equation} Then \begin{equation} G(t):=G_b(t):=g(t)\sin b=\sin t\,\int_0^b du\,h(u)\sin(b-u) -\sin b\,\int_0^t du\,h(u)\sin(t-u). \end{equation} We have to show that $G\ge0$ on $[0,b]$. Because the nonnegative function $h$ can be however closely approximated in $L^1$ by conical combinations of the indicators of intervals, wlog $h=1_{[c,d]}$ for some $c,d$ such that $0<c<d<b$, in which case \begin{equation} G(t)=\sin t\, (\cos (b-d)-\cos (b-c))-\sin b\, (\cos (t-\max (c,\min (d,t)))-\cos (c-t)). \end{equation} So, if $0\le t\le c$, then $G(t)=\sin t\, (\cos (b-d)-\cos (b-c))\ge0$. The case $d\le t\le b$ is similar, by the left-to-right symmetry.

It remains to consider the case when $c\le t\le d$. Then \begin{equation} G''(t)=\sin t\, (\cos b\, (\cos c-\cos d)-\sin b\, \sin d)-\sin b \cos c\,\cos t =A\sin(t+C)=:f(t), \end{equation} where $A,C$ depend only on $b,c,d$. The function $f$ will be $\le0$ on the interval $[c,d]$ of length $<\pi$ iff $f(c)\le0$ and $f(d)\le0$. In our case, we have \begin{equation} 2f(c)=\sin (b-c-d)-\sin (b+c-d)-\sin (b-2 c)-\sin (b) \end{equation} and hence $(f(c))'_d=-\sin c\, \sin (b-d)\le0$, so that $f(c)$ decreases in $d$. So, wlog $d=c$, in which case $f(c)=-\sin b\le0$. Thus, $f(c)\le0$ always, that is, for all $d\in[c,b]$. Similarly, by the left-to-right symmetry, $f(d)\le0$, which completes the proof.

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This is sometimes called a maximum principle "on thin domains". My answer below is technically equivalent to @Deane Yang's answer in this specific context, but the scope is more general so I thought I'd still give it a shot (in particular the argument applies to higher dimensions as well, see the comments below).

Note first that if the eliptic operator under consideration had nonnegative zeroth-order coefficient we could immediately conclude from the standard maximum principle that $-g''+g\geq 0$ implies $g\geq 0$, given that $g\geq 0$ on the boundaries. The problem here is of course that your operator $$ \mathcal L[g]=-g''-g $$ has negative zeorth-order coefficient, $c(x)\equiv -1$.


Here comes the trick now: assume that you can find some particular function $f(x)\geq C>0$ (up to the boundary) such that $\mathcal L[f]\geq 0$. (The fact that you can find such a function relies on the thinness of the domain, I will come back to this later on. In your specific example you can take $f(x)=\sin(x-b+\pi)$ as in @Deane Yang's answer.) Think then of $g$ as $uf$ for some $u$. (Here the strict positivity of $f$ is important so that $u$ is smooth enough, no funny business can arise from this change of variables) Define then $$ \tilde{\mathcal{L}}[u]:=\mathcal{L}[uf]. $$ Here you can compute expliticly $\mathcal L$, but the key is that in whole generality this new elliptic operator $$ \tilde{\mathcal{L}}[u] =\Big(\mbox{1st & 2nd order}\Big) + \Big(\mathcal L[f]\Big) u $$ always has a zeroth-order coefficient with the right sign, i-e $\tilde c(x)=\mathcal L[f](x)\geq 0$ . Now if $g$ is a supersolution of $\mathcal L[g]\geq 0$ you have by definition that $u:=\frac{g}{f}$ is a supersolution of $\tilde{\mathcal L}[u]\geq 0$. Applying the usual maximum principle (with $\tilde c(x) \geq 0$) you conclude that $u\geq 0$, hence $g\geq 0$.


Comment 1: here you see that the key point is the existence of a well-chosen $f(x)$, which is not to be taken for granted. The reason why you can actually find such a function is that your domain is small enough in terms of the lowest eigenvalue of the Dirichlet problem for the homogeneous problem, without zeroth-order terms: notice obviously that this mysterious function $f(x)=\sin(x-b+\pi)$ is indeed the principal eigenvalue of $-\frac{d^2}{dx^2}$ on the domain $(b-\pi,b)$ with zero boundary conditions. The trick is here that your domain $(a,b)\subset (b-\pi,b)$ is $b-a>0$ is small enough. (This reminds me of a classical exercise in elliptic PDEs where one is asked to apply the Lax-Milgram theorem for operator whose zeroth-order term has the wrong sign, but not too large in modulus compared to the first eigenvalue, see also @Giorgio Metafune's answer)

Comment 2: The trick works exactly the same in higher dimensions. For example if you were working in two dimensions on an infinite but thin enough strip, then plugging in a well-chosen $\sin$ function depending only on the thin coordinate automatically gives a suitable $f$. I can't remember where I learnt this trick, and also I'm pretty sure that I also read somewhere a completely general statement that, regardless of the initial coefficients, sufficiently narrowing down the domain always gives a suitable $f$ (this is natural if one thinks of resonant frequencies: in the absence of zeroth-order terms, thinness in any direction of a given object/domain gives a high-pitch natural frequency, hence a large principal eigenvalue that can effectively dominate any zeroth-order coefficient that you might want to add to the free resonance). The trick also works sometimes for nonlinear operators, but the "change of variables" to go from $g$ to $u$ can be very delicate to find, it's not just products and quotients anymore.

Comment 3: this is called "thin domains" because the argument is usually applied as above (in a systematic way, since thinness is a classical sufficient condition), but in fact everything relies on the existence of a "nice" $f$. So even in the broader context of not necessarily thin domains (balls, or whatever) one might get lucky by guessing the right $f$ (if any, of course!)

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    $\begingroup$ I like your discussion. I think basic elliptic PDE theory be presented first in the 1-dimensional case. $\endgroup$ – Deane Yang Jul 14 at 21:02
  • $\begingroup$ What do you mean by "think of $g$ as $uf$"? $\endgroup$ – UserA Jul 14 at 21:58
  • $\begingroup$ I mean "change variables" and write $g=uf$, so the new primary variable is $u$ $\endgroup$ – leo monsaingeon Jul 15 at 5:06
  • $\begingroup$ @Deane Yang: Thanks, me too, and this is how I always try to introduce and motivate the theory when I teach elliptic equations. But I guess this is really a matter of taste. $\endgroup$ – leo monsaingeon Jul 15 at 5:09
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Another approach in the spirit of viscosity solutions/maximum principle:

Assume wlog $a>0$ and $b<\pi$. Let $I:=[a,b]$ and define $$f(x):=\sin x,\quad g_\lambda:=g+\lambda f.$$ If $g=g_0<0$ somewhere on $I$, then we can then find $\lambda>0$ such that $\min_I g_\lambda=0$ (observe that $f\ge c$ on $I$ for some $c>0$ and look at $\lambda\mapsto\min_I g_\lambda$).

But then $g_\lambda''\le g_\lambda''+g_\lambda\le 0$ on $I$, so $g_\lambda$ is concave. Since $g_\lambda>g$ is positive at the boundary of $I$, it must be positive on all of $I$, contradiction.

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