According to the entry "Differential inequality" of the Encyclopedia of Mathematics


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.


Chaplygin's theorem is true even for non-linear differential equations of the form $$y^{(n)}=F(x,y,y',\ldots,y^{(n-1)}),$$ under the only assumption that both $y$ and $z$ their derivatives $z$ exist on the considered interval, and partial derivatives of $f$ exist and are continuous in the required range. However, the number $b$ in general depends on $F$ and on the initial conditions in a complicated way.

To demonstrate the idea, consider the case $n=1$. Subtracting, we obtain $$y'-z'-p(y-z)>0,\quad (1)$$ where $$p(x)=\frac{F(x,y(x))-F(x,z(x))}{y(x)-z(x)},$$ which is well defined also at the points where $y$ equals $z$. Let $$s(x)=\exp\left(-\int_0^xp(t)dt\right).$$ Evidently $s>0$. Mulpiplying (1) on $s(x)dx$ and integrating from $0$ to $b$, we obtain $$(y-z)s\vert_{0}^b>0,$$ that is $y(b)>z(b)$.

In this special case, there is no further restriction on $b$, except the existence of solutions on $[0,b]$. For higher order cases (linear or non-linear), there are restrictions.

For the general case, the same idea is used with a little more writing. I was surprised that I could not find this simple argument in modern textbooks. Of course the case of linear equation of second order can be found everywhere as a corollary of Sturm's comparison theorem (with a different proof).

Edit. One can give a simpler proof of the general case when $F$ is analytic. In this case, $y$ and $z$ are also analytic, and because of the initial conditions the first terms of their Taylor series at $0$ coincide, up to the power $x^{n-1}$. So the lowest degree terms which are different are $$y^{(n)}(0)x^n/n!>z^{(n)}x^n/n!,$$ where the inequality holds by the assumption of the theorem. This implies that there exists $b>0$ such that $y(x)>z(x)$ for $x\in (0,b)$.

  • $\begingroup$ Thanks. In your proof to the analytic case it is not clear to me why y should be analytic. You also use the strict inequality, but from what I get from your first argument this does not seem really needed. $\endgroup$ Jan 16 '14 at 10:29
  • $\begingroup$ Every solution of Cauchy problem of an ODE with analytic RHS is an analytic function on some interval around the initial point. This is the fundamental existence theorem. Concerning strict and non-strict inequality, you are right. $\endgroup$ Jan 16 '14 at 15:11
  • $\begingroup$ So the existence theorem applies to z. The problem I see is that y is not a solution, for it satisfies $Ly>f$. $\endgroup$ Jan 16 '14 at 16:43
  • $\begingroup$ You are right. About $y$ we have to assume that it is analytic. In practice, we CHOOSE $y$, verify that it satisfies the inequality and use it to estimate $z$. $\endgroup$ Jan 16 '14 at 20:19

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 $[0,b]$. $\square$


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