TITLE: Local positivity of solutions to linear differential inequalities (Chaplygin's theorem)
QUESTION [7 upvotes]: According to the entry "Differential inequality" of the Encyclopedia of Mathematics
http://www.encyclopediaofmath.org/index.php/Differential_inequality
the following result is due to Chaplygin (1919). Let $y: [0, a)\to \mathbb{R}$ satisfy the differential inequality
$$
y^{(m)}+\alpha_1(x) y^{(m-1)}+\cdots+ \alpha_{m}(x) y> f
$$
and let  $z: [0, a)\to \mathbb{R}$ be a solution of the same equation with the equality sign
$$
z^{(m)}+\alpha_1(x) z^{(m-1)}+\cdots+ \alpha_{m}(x) z= f
$$
both with the same initial condition at $x=0$.
Then there is some $b>0$, $b<a$, such that $y\ge z$ on $[0,b]$. Unfortunately, I do not have access to the sources listed in that article, and I do not read Russian. Can you give me some more modern references? If not, can you sketch the argument of proof? More specifically, is the strict  equality really needed? Can one conclude that also $y^{(i)}\ge z^{(i)}$, for some $1\le i<m$, on some subinterval? I stress that the signs of the $\alpha$s are not restricted. If they are negative the  interval of positivity exists  and is the whole $[0,a)$ accordingly to a theorem due to Wazewski.

REPLY [2 votes]: This is the answer I worked out in 
arXiv:1406.5909
Theorem
Let $y: [0,c] \to \mathbb{R}$, be any $C^n$ function which  satisfies the differential inequality
$$ 
y^{(n)}+a_1(x) y^{(n-1)}+\cdots+ a_{n}(x) y\ge 0, \qquad(1)
$$
and the initial conditions
$$
y^{(i)}=0, \textrm{ for } i=0,1,\cdots,n-1,
$$
where the coefficients $\{a_i(x)\}$ are Lipschitz.
Then there is some constant $b$,  $0<b\le c$, dependent on the coefficients $\{a_i(x)\}$ but independent of $y$, such that for every $y$  and every $i=0,\cdots, n-1$ we have  $y^{(i)}\ge 0$ on $[0,b]$. Moreover, the equality $y=0$ holds on the whole interval $[0,b]$ if and only if the equality in   (1) holds  on the whole interval $[0,b]$.
Proof
Let us denote with $L$ the linear differential operator on the left-hand side of (1). Let $K(x,\xi)$ be the Cauchy function, that is the function defined on the triangle $0\le \xi \le x < c/2$  which solves $L K=0$ for every $\xi \in[0,c/2)$ with initial condition
$$
K(\xi,\xi)=K^{(1)}(\xi,\xi)=\cdots=K^{(n-2)}(\xi,\xi)=0, \quad  K^{(n-1)}(\xi,\xi)=1,
$$
where these derivatives are with respect to $x$. This function is well defined because it is equivalently determined by the linear system of first-order differential equations in the dependent variables $(K^\xi_0, K^\xi_1, K^\xi_2,\cdots, K^\xi_{n-1}) : [0,c/2]\to \mathbb{R}$
\begin{align*}
\frac{d}{ds}{K^\xi}_0&=K^\xi_1, \\
\cdots &\\
\frac{d}{ds}{K^\xi}_{n-2}&= K^\xi_{n-1},\\
\frac{d}{ds} {K^\xi}_{n-1}&=- a_1(s+\xi) K^\xi_{n-1}-a_2(s+\xi) K^\xi_{n-2}\cdots-a_n(s+\xi) K^\xi_0 .
\end{align*}
where $s+\xi=x$, under the initial condition
$$
K^\xi_0(0)=K^\xi_1(0)=\cdots=K^\xi_{n-2}(0)=0, \quad  K^\xi_{n-1}(0)=1.
$$
Indeed, there is one and only one solution by the Picard-Lindel\"of theorem (Cor. 5.1 in Hartman 1964 Ordinary Differential Equations), and so we obtain our desired Cauchy function once we set $K(x,\xi)=K^\xi_0(x-\xi)$ (thus $K^{(i)}(x,\xi)=K^\xi_i(x-\xi)$). Moreover, the above system of first-order differential equations is Lipschitz also with respect to the external parameter $\xi$, thus its solution $(K^\xi_0, K^\xi_1, K^\xi_2,\cdots, K^\xi_{n-1}) $ has a Lipschitz dependence on $(s,\xi)$ (this is Peano's theorem; if the $a$s are $C^1$ then one can infer that the $K$s are $C^1$ too (Theor. 3.1 in Hartman (1964) Ordinary Differential Equations), for the Lipschitz case see (Lang 1995, Differential and Riemannian manifolds or Cartan 1971 Differential Calculus). The function $K^{(n-1)}(x,\xi)$ is continuous in both $(x,\xi)$ on the triangle $0\le \xi \le x < c/2$ and in particular at $(0,0)$. Since $K^{(n-1)}(0,0)=1>0$,  there is some neighborhood of $(0,0)$ (in the product topology) over which  $K^{(n-1)}$ is positive, and hence a triangle $0\le \xi \le x < b\le c/2$ over which  $K^{(n-1)}$ is positive. But on the diagonal $K^{(i)}$, $i=1,\cdots, n-2$, vanishes so upon integration on $x$ we obtain that $K^{(i)}$, is positive on $0\le \xi < x < b$ for every $i=1,\cdots, n-1$.
From the assumption we have that $Ly\ge 0$, where $Ly$ is continuous. The  uniqueness of the solution to the differential equation $L y= f$  implies the easily verifiable   formula
$$
y(x)=\int_0^x K(x,\xi) Ly(\xi) \, d \xi ,
$$
 which under differentiation gives  more generally
$$
y^{(i)}(x)=\int_0^x K^{(i)}(x,\xi) Ly(\xi)\, d \xi , \qquad i=0,1,\cdots, n-1,
$$
thus $y^{(i)}\ge 0$ on $[0,b]$, and the equality $y=0$ on $[0,b]$ is possible only if $Ly=0$ on $[0,b]$. $\square$