THE BILOCAL PROBLEM AND SA CIAPLÎGHIN'S THEOREM OF DIFFERENTIAL INEQUALITIES WITH CONFUSED NODES, FOR SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

by
OLEG ARAMĂ

Paper presented at the papers meeting of the Computing Institute of the RPR Academy,
Cluj Branch, on December 23, 1957

This paper answers a problem posed by Prof. T. Popoviciu, namely, to investigate the connection between the interpolation property of the family of integrals of a linear differential equation and the convexity property ¹ ) with respect to this family of functions that make the first member of the respective differential equation positive. This problem was posed in order to obtain necessary and sufficient conditions for the bilocal problem (with simple nodes) for the considered differential equation to admit a solution.

In the case where the convexity property mentioned above is designed for simple and distinct nodes, the problem was solved in a general case by E. Moldovan [1].

In this paper, we study the stated problem, in the case when the convexity property is defined with confused nodes, in the sense of that which occurs in SA Ciaplighin's "differential inequality theorem" for the linear differential equation of order 2, with variable coefficients. To facilitate the exposition, we introduce the following definitions and notations.

Definition 1. Let $F$ a family of functions $f(x)$ , of a real variable $x$ , defined in the interval ( $a,b$ ) and continue in this interval. We say that such a family $F$ , owns the property $I_{2}(a,b)$ , that is, it is an interpolator of order 2 in the open interval ( $a,b$ ), if any two distinct nodes $x_{1}$ and $x_{2}\operatorname{in}(a,b)$ , and whatever the real values are $y_{1}$ and $y_{2}$ , there is one and only one function $f(x)$ , belonging to the family $F$ , which satisfies the conditions $f\left(x_{1}\right)=y_{1}$ and $f\left(x_{2}\right)=y_{2}$ .

⁰⁰footnotetext:¹ ) See definitions 1 and 5 of this paper.

Definition 2. We say that a family $F$ of functions $f(x)$ defined and continuous in an interval $(a,b)$ own the property $N_{2}(a,b)$ (its elements are non-oscillatory of the second order among themselves in the interval ( $a,b$ )), if two arbitrary functions $f_{1}(x)$ and $f_{2}(x)$ distinct from $F$ cannot take equal values in $(a,b)$ than at most one point in that interval.

Definition 3. We consider a linear differential operator

L(y)=y^{\prime\prime}+p(x)y^{\prime}+q(x)y,

where $p(x)$ and $q(x)$ are continuous functions in $(a,b)$ , - defined, on the set $C_{2}(a,b)$ of continuous functions together with their 1st and 2nd order derivatives in ( $a,b$ ). Either $x_{0}$ a point in the interval ( $a,b$ ). We will say that the operator $L(y)$ own the property $T_{x_{0}}^{(2)}(a,b)$ , if whatever the function is $y(x)$ , belonging to the crowd $C_{2}(a,b)$ and satisfying within the range $(a,b)$ inequality $L(y)>0$ as well as the conditions $y\left(x_{0}\right)==y^{\prime}\left(x_{0}\right)=0$ , then that function checks throughout the interval $(a,b)$ inequality $y(x)\geq 0$ , the equal sign occurring only at the point $x_{0}$ .

We will say that the differential operator $L(y)$ own the property $T_{2}(a,b)$ , if he owns the property $T_{x_{0}}^{(2)}(a,b)$ , whatever it is $x_{0}\in(a,b)$ .

Note: Definitions 1, 2, 3 above can also be stated relative to a closed interval. $[a,b]$ In this case, we will write the respective properties as follows: $I_{2}[a,b],N_{2}[a,b],T_{x_{0}}^{(2)}[a,b]$ , where $x_{0}\in[a,b]$ and $T_{2}[a,b]$ In the following, when any of these properties are mentioned, it will always be understood that the functions $p(x)$ and $q(x)$ , which intervene in the operator expression $L(y)$ , are continuous in the closed interval $[a,b]$ .

In this paper, the following equivalence theorem is first established.
Theorem 1. Let be a second-order linear differential equation

L(y)=y+p(x)y+q(x)y=r(x)

(1)

having the coefficients $p(x),q(x),r(x)$ continuous in the open interval ( $a,b$ ). Either $Y$ family of integrals $y(x)$ of equation (1), in ( $a,b$ ). The necessary and sufficient condition that the family $Y$ to own the property $I_{2}(a,b)$ or $N_{2}(a,b)$ , is like the differential operator $L(y)$ to own the property $T_{2}(a,b)$ .

To prove this theorem, we will first state a few helpful properties.

We associate to the differential equation (1), the homogeneous differential equation

L(y)=y+p(x)y+q(x)y=0

(2)

to note with $Y_{0}$ the set of particular integrals of this equation. The following lemmas hold, the proof of which is immediate.

I love them. $1.{}^{1}$ ) The necessary and sufficient condition that the family $Y$ of the integrals of equation (1) to possess the property $I_{2}(a,b)$ respectively $I_{2}[a,b]$ , it's like family $Y_{0}$ of the homogeneous differential equation (2), to possess the property $I_{2}(a,b)$ , respectively $I_{2}[a,b]$ .

⁰⁰footnotetext:¹ ) In the following we will assume that the coefficients of equations (1) and (2) are continuous, either in the open interval (

a,b

) either in the closed interval [

a,b

], as the lemma refers to the open interval (

a,b

), or on the closed interval

[a,b]

Lemma 2. Necessary and sufficient condition that the family $Y_{0}$ to own the property $I_{2}(a,b)$ , respectively $I_{2}[a,b]$ , is for this family to own the property $N_{2}(a,b)$ , respectively $N_{2}[a,b]$ .

Lemma 3. Necessary and sufficient condition that the family $Y_{0}$ to have the property $I_{2}[a,b)$ , respectively $I_{2}[a,b]$ is that equation (2) admits at least one positive integral in ( $a,b$ ) respectively in $[a,b]^{1}$ ).

Proof. The condition is necessary.
We first establish the necessity of the condition for the case of a closed interval. $[a,b]$ We therefore assume that $Y_{0}$ has the property $I_{2}[a,b]$ . Either $y(x)$ integral of equation (2), satisfying the conditions $y(a)=y(b)=1$ This particular integral cannot vanish in the interval $[a,b]$ : Indeed, assuming by absurdity that it would cancel out at a point in the interval $[a,b]$ , no root can have a multiplicity order greater than 1 , because otherwise it would result that the considered integral is identically zero, which would contradict the boundary conditions it verifies. Then from the continuity of this integral, it would follow that the number of its roots (all are simple), from $[a,b]$ , is greater than or at least equal to 2 . From this it would follow that $y(x)$ does not have the property $I_{2}[a,b]$ , which would contradict the hypothesis.

Establishing the necessity of the condition expressed by the lemma, in the case of a semiclosed interval $[a,b)$ , is done as follows: Either $y_{a}(x)$ a particular integral that satisfies the conditions $y_{a}(a)=0$ and $y_{a}^{\prime}(a)>0$ This integral cannot be cancelled in $(a,b)$ Indeed, if $y_{a}(x)$ would cancel out at some point $\xi_{a}\in(a,b)$ , then for a positive number $\varepsilon$ sufficiently small, the particular integral $y_{a+\varepsilon}(x)$ , which satisfies the conditions $y_{a+\varepsilon}(a+\varepsilon)=0$ and $y_{a+\varepsilon}^{\prime}(a+\varepsilon)=y_{a}^{\prime}(a)$ , would still cancel out at one point $\xi_{a+\varepsilon}$ neighbor of the point $\xi_{a}$ , also located in the interval ( $a,b$ ) (when $\varepsilon$ is sufficiently small). Ultimately, it would follow that there exists a particular integral $y_{a+\varepsilon}(x)$ , which cancels out in at least two (distinct) points in ( $a,b$ ). It would follow from this that the family $Y_{0}$ does not have $[a,b)$ property $N_{2}[a,b)$ and according to Lemma 2, that it does not have the property either $I_{2}[a,b)$ , contrary to the hypothesis.

The condition is sufficient. Let $y_{1}(x)$ a particular integral of equation (2), such that $y_{1}(x)>0$ in the interval $[a,b]$ We will show that the family $Y_{0}$ has the property $N_{2}[a,b]$ , from which by virtue of Lemma 2 it will follow that $Y_{0}$ has the property $I_{2}[a,b]$ We assume by absurdity that there would be a particular integral $y_{2}(x)\neq 0$ which would cancel out at two distinct points $x_{1}$ and $x_{2}$ FROM $[a,b]$ Let us consider the integral $\lambda y_{2}(x)$ , where $\lambda$ is a constant. Let $\lambda_{0}$ constant value $\lambda$ , for which the representative curves of the integrals $y_{1}(x)$ and $\lambda_{0}y_{2}(x)$ are tangent at a point $\xi\in\left(x_{1},x_{2}\right)$ . Function $y(x)=y_{1}(x)-\lambda_{0}y_{2}(x)$ will be an integral of equation (2), which at the point $\xi$ satisfies the conditions $y(\xi)=y^{\prime}(\xi)=0$ It would follow from this that $y_{1}(x)-\lambda_{0}y_{2}(x)\equiv 0$ , that is $y_{1}(x)\equiv\lambda_{0}y_{2}(x)$ , from which it would follow that $y_{1}(x)$ is cancelled in $x_{1}$ and in $x_{2}$ , contrary to the hypothesis.

Proof of the sufficiency of the condition in the case of an open interval $(a,b)$ it is done the same way.

They are 4. If the family $Y$ of the integrals of equation (1) possesses the property $I_{2}(a,b)$ , respectively $I_{2}[a,b]$ , then the differential operator $L(y)$ appropriately own the property $T_{2}(a,b)$ respectively $T_{2}[a,b]$ .

We will give the proof of this lemma for the case of an open interval $(a,b)$ , it does the same in the case of a closed interval $[a,b]$ . We therefore assume that the family $Y$ of the integrals of the differential equation (1) possesses the property $I_{2}(a,b)$ According to Lemma 1

⁰⁰footnotetext:¹ ) The property expressed by this lemma also results from the work [2] of VA Kondratiev, as well as from the work [7] of G. Polya.

it turns out that the family $Y_{0}$ of the integrals of equation (2) possesses the property $I_{2}(a,b)$ Let us prove that under these conditions the differential operator $L(y)$ has the property $T_{2}(a,b)$ For this purpose, let us assume absurdly that there exists a function $\nu(x)\in C_{2}(a,b)$

which satisfies the conditions

\left.\begin{array}[]{l}L(v)>0,x\in(a,b),\\ v\left(x_{0}\right)=v^{\prime}\left(x_{0}\right)=0,\end{array}\right\}

where $x_{0}$ is a point in the interval ( $a,b$ ) and that this function $v(x)$ would have in the interval ( $a,b$ ) outside the root $x_{0}$ and another root $x_{1}$ different from $x_{0}$ We assume for the fixation of ideas that $x_{0}<x_{1}$ From (3) it follows that $v^{\prime\prime}\left(x_{0}\right)>0$ and whereas $v\left(x_{0}\right)=v^{\prime}\left(x_{0}\right)=0$ , it follows that in the vicinity of the point $x_{0}$ , the equation curve $z=v(x)$ is located above the axis $OX$ . Thus we can assume that the root $x_{1}$ is consecutive to the right of the root $x_{0}$ . Be it now $a_{1}$ and $b_{1}$ two real numbers satisfying the inequalities $a<a_{1}<x_{0}<x_{1}<b_{1}<b$ Since by hypothesis $Y_{0}$ has the property $I_{2}(a,b)$ it immediately follows that she will also have the property $I_{2}\left[a_{1},b_{1}\right]$ According to Lemma 3, the equation $L(y)=0$ will admit at least one integral $\eta(x)$ positive in $\left[a_{1},b_{1}\right]$ . Performing the function change

y(x)=\eta(x)z(x)

(4)

equation (2) is transformed into the differential equation

L(y)=\eta(x)\left[z^{\prime\prime}+\frac{2\eta^{\prime}+p\eta}{\eta(x)}z^{\prime}\right]=\eta(x)L^{*}(z)=0

(5)

defined in the range $\left[a_{1},b_{1}\right]$ , and the function $v(x)$ will correspond to it by transformation (4), the function $w(x)$ which will satisfy the conditions:

\left.\begin{array}[]{l}L^{*}(w)>0,x\in\left[a_{1},b_{1}\right]\\ w\left(x_{0}\right)=w^{\prime}\left(x_{0}\right)=0.\end{array}\right\}

From the fact that $x_{0}$ and $x_{1}$ are consecutive roots for the function $v(x)$ , it follows from (4) that $x_{0}$ and $x_{1}$ are consecutive roots for the function $w(x)$ . Then from (3') it follows that $w^{\prime\prime}\left(x_{0}\right)>0$ and whereas $w\left(x_{0}\right)=w^{\prime}\left(x_{0}\right)=0$ , it follows that in the vicinity of the point $x_{0}$ equation curve $z=w(x)$ is located above the axis $Ox$ (fig. 1). Let $\xi$ abscissa of a maximum point of the function $z=w(x)$ in the interval ( $x_{0},x_{1}$ ). From figure 1 it can be seen that at point $\xi$ relationships take place $w^{\prime}(\xi)=0,w^{\prime\prime}(\xi)\leqq 0$ , from which it follows that at the point $\xi$ inequality occurs $\left.L^{\star}(w)\right|_{x=\xi}\leqq 0$ . This inequality contradicts the corresponding inequality in (3'). This leads to a contradiction. It ultimately follows that the function $v(x)$ , which satisfies the conditions (3), cannot have in the interval ( $a,b$ ) another root besides $x_{0}$ . From here it follows, taking into account (3), that in the interval ( $a,b$ ) the inequality occurs $v(x)\geqq 0$ , the equal sign occurring only at the point $x_{0}$ , what

Lemma 5. If the differential operator $L(y)$ own the property $T_{2}(a,b)$ then this operator $L(y)$ own the property $T_{2}\left[a_{1},b_{1}\right]$ , whatever the closed subinterval is $\left[a_{1},b_{1}\right]$ contained in the interval ( $a,b$ ).

Demonstration. Let $\left[a_{1},b_{1}\right]$ some interval contained in $(a,b)$ and either $u(x)$ a function defined in the interval $\left[a_{1},b_{1}\right]$ , which satisfies the inequality $L(u)>0$ in $\left[a_{1},b_{1}\right]$ ,
as well as the conditions $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=0$ , where $x_{0}$ is a point in the interval $\left[a_{1},b_{1}\right]$ Let us show that under these assumptions the inequality holds $u(x)\geqq 0,x\in\left[a_{1},b_{1}\right]$ , the equal sign occurring only at the point $x_{0}$ .

Indeed it is easy to show that we can extend the function $u(x)$ throughout the interval ( $a,b$ ), so that the inequality is preserved $L(u)>0$ throughout this interval ( $a,b$ ) and also to preserve the continuity property of the 2nd order derivative of the function $u(x)$ in $(a,b)$ that is, the function obtained by extending the original function $u(x)$ , to belong to the class $C_{2}(a,b)$ .

For this, we consider the interval $\left[b_{1},b\right)$ and we consider with respect to this interval, the integral $\beta(x)$ of the differential equation
with boundary conditions

L(y)=\left.L(u)\right|_{x=b_{1}}

		$\displaystyle\beta\left(b_{1}\right)=u\left(b_{1}\right),$
		$\displaystyle\beta^{\prime}\left(b_{1}\right)=u^{\prime}\left(b_{1}\right).$

function $\beta(x)$ thus defined, it also has the property that $\beta^{\prime\prime}\left(b_{1}\right)=u^{\prime\prime}\left(b_{1}\right)$ , which is deduced from the differential equation that defines $\beta(x)$ This function $\beta(x)$ , by the way it was built, checks in the interval $\left[b_{1}b\right)$ differential inequality $L[\beta(x)]>0$ .

Analogously, we consider in the interval $\left(a,a_{1}\right]$ integral $\alpha(x)$ of the differential equation

L(y)=\left.L(u)\right|_{x=a_{1}}

with boundary conditions

		$\displaystyle\alpha\left(a_{1}\right)=u\left(a_{1}\right),$
		$\displaystyle\alpha^{\prime}\left(a_{1}\right)=u^{\prime}\left(a_{1}\right).$

This solution $\alpha(x)$ obviously it will still satisfy the condition $\alpha^{\prime\prime}\left(a_{1}\right)=u^{\prime\prime}\left(a_{1}\right)$ and $L[\alpha(x)]>0$ , when $x\in\left(a,a_{1}\right]$ .

Be it now $u^{*}(x)$ function defined as follows:

u^{\star}(x)=\left\{\begin{array}[]{l}\alpha(x)\text{ dacă }x\in\left(a,a_{1}\right),\\ u(x)\text{ dacă }x\in\left[a_{1},b_{1}\right],\\ \beta(x)\text{ dacă }x\in\left(b_{1},b\right).\end{array}\right.

This function admits derivatives up to and including the second order, continuous in the interval $(a,b)$ and satisfies in this interval the inequality $L\left(u^{*}\right)>0$ Since by hypothesis the function $u(x)$ satisfies the conditions $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=0$ , where $x_{0}\in\left[a_{1},b_{1}\right]$ , it follows that the function $u^{*}(x)$ will satisfy these conditions. According to the property $T_{2}(a,b)$ of the differential operator $L(y)$ , it follows that $u^{*}(x)\geqq 0$ in the interval ( $a,b$ ), the equal sign occurring only at the point $x_{0}$ . From here the stated lemma results.

In what follows we will establish another helpful property.
Lemma 6. If the differential operator $L(y)$ has the property $T_{2}(a,b)$ , then any function $u(x)$ , belonging to the class $C_{2}(a,b)$ and which satisfies in the interval $(a,b)$ differential inequality $L(u)>0$ , cannot be cancelled in ( $a,b$ ) in more than two distinct points.

Proof. We assume that the differential operator $L(y)$ enjoys the property $T_{2}(a,b)$ Let us show that under this assumption, any function $u(x)$ belonging

CLASSROOM $C_{2}(a,b)$ and satisfying in this interval the differential inequality $L(u)>0$ cannot have in ( $a,b$ ) more than two distinct roots. Suppose by the absurd converse, that there exists a function $u(x)$ belonging to the class $C_{2}(a,b)$ , which satisfies in the interval $(a,b)$ differential inequality $L(u)>0$ and which has at least 3 distinct roots $x_{1},x_{2},x_{3}$ (which we assume are consecutive and written in ascending order) ¹ ). Since the differential operator $L(y)$ possess by hypothesis the property $T_{2}(a,b)$ , it follows that none of the roots $x_{1},x_{2},x_{3}$ cannot be multiple and therefore the representative curve of the function $u(x)$ crosses the axis $Ox$ in each of the points $x_{1},x_{2},x_{3}$ (fig. 2). The following two cases can be presented, as $u^{\prime}\left(x_{1}\right)>0$ or $u^{\prime}\left(x_{1}\right)<0$ (case $u^{\prime}\left(x_{1}\right)=0$ cannot take place, as shown previously).

Case 1. $u^{\prime}\left(x_{1}\right)>0$ , (fig. 2). Let $\xi_{0}$ any point in ( $a,x_{1}$ ), so that there is no root of the function $u(x)$ in the interval $\left[\xi_{0},x_{1}\right)$ . Either $\eta(x)$ integral of the differential equation $L(y)=1$ , with the boundary conditions

\eta\left(\xi_{0}\right)=0,\quad\eta^{\prime}\left(\xi_{0}\right)=0

According to the property $T_{2}(a,b)$ of the differential operator $L(y)$ , it follows that $\eta(x)>0$ in the interval $\left(\xi_{0},b\right)$ Let's perform in the expression of the differential operator $L$ , change of function $y(x)=\eta(x)z(x)$ . We obtain

	$\displaystyle L(y)$	$\displaystyle\equiv\eta(x)\left[z^{\prime\prime}+\left(\frac{2\eta^{\prime}+p\eta}{\eta(x)}\right)z^{\prime}+\frac{L(\eta)}{\eta(x)}z\right]\equiv$
		$\displaystyle\equiv\eta(\mathrm{x})\left[z^{\prime\prime}+\left(\frac{2\eta^{\prime}+p\eta}{\eta(x)}\right)z^{\prime}+\frac{z}{\eta(x)}\right]\equiv\eta(x)L^{*}(z)$

function $u(x)$ will turn into a function $v(x)=\frac{u(x)}{\eta(x)}$ , which will have the same roots as $u(x)$ in $\left(\xi_{0},b\right)$ and in addition it will satisfy the condition $\lim_{x\rightarrow\xi_{0}\div 0}v(x)=-\infty$ . Also, the function $v(x)$ satisfy within the range $\left(\xi_{0},b\right)$ inequality $L^{*}(v)>0$ .

It is easy to see that the differential operator $L^{*}(z)$ has the property $T_{2}\left(\xi_{0},b\right)$ , since the operator $L(y)$ has this property as follows from Lemma 5, taking into account the assumptions made.

With these results obtained regarding the operator $L^{\star}(z)$ , let us show that the inequality $L^{\star}(v)>0$ in $\left(\xi_{0},b\right)$ is in contradiction with the shape of the representative curve of the function $v(x)$ (Fig. 3).

Indeed, either $x_{0}$ a point in the interval ( $x_{2},x_{3}$ ), in which the function $v(x)$ reaches its minimum value in that interval. Let $\lambda=-v\left(x_{0}\right)$ Obviously that $\lambda>0$ , since

⁰⁰footnotetext: ¹ ) It is found that the function

u(x)

cannot have an infinity of roots in the interval (

a,b

), having an accumulation point

\xi

in the interval (

a,b

), because in the affirmative case, from the continuity of the function

u(x)

as well as its derivative

u^{\prime}(x)

, it would result that

u(\xi)=u^{\prime}(\xi)=0

and property

T_{2}(a,b)

of the operator

L

would be contradicted.

in the interval ( $x_{2},x_{3}$ ) function $u(x)$ so and $v(x)$ takes negative values, which results from the assumptions made. Let us perform on the representative curve ( $v$ ) of the function $v(x)$ a translation in the positive direction of the axis $Oy$ , of parameter $\lambda$ . The curve ( $v$ ) will take the position ( $v^{*}$ ) indicated in figure 3 with a dotted line. This new curve will be tangent to the axis $Ox$ at the point $x_{0}$ and will cross the axis $Ox$ at a point located in the interval $\left(\xi_{0},x_{1}\right)$ The equation of this curve ( $\left.v^{\star}\right)$ will be $y=\nu^{*}(x)=\nu(x)+\lambda$ .
But as shown previously, it occurs in the interval ( $\xi_{0},b$ ) inequality

\begin{gathered}L^{\star}(v)=v^{\prime\prime}+\frac{\left(2\eta^{\prime}+p\eta\right)}{\eta(x)}v^{\prime}+\\ +\frac{v}{\eta(x)}>0,x\in\left(\xi_{0},b\right)\end{gathered}

It follows from this that the function $v^{*}(x)$ satisfies an inequality of the same kind in the interval $\left(\xi_{0},b\right)$ Indeed, taking into account that $\lambda$ is a positive constant, as well as from the previous inequality, we obtain

L^{\star}\left(v^{\star}\right)=L^{\star}(v)+\frac{\lambda}{\eta(x)}>0,x\in\left(\xi_{0},b\right).

function $v^{\star}(x)$ check the conditions again $v^{\star}\left(x_{0}\right)=0,\left(\frac{dv^{\star}}{dx}\right)_{x=x_{0}}=0$ Since the differential operator $L^{\star}(z)$ own the property $T_{2}\left(\xi_{0},b\right)$ , it follows that the function $v^{\star}(x)$ must satisfy the inequality $v^{\star}(x)\geqq 0$ in the interval $\left(\xi_{0},b\right)$ , which contradicts the fact that the curve ( $v^{\star}$ ) crosses the axis $Ox$ in the interval ( $\xi_{0},x_{1}$ ). The contradiction comes from the absurd assumption that the function $u(x)$ , satisfying in the interval $(a,b)$ inequality $L(u)>0$ , would have more than two distinct roots in this interval (in the specific hypothesis of case 1).

Case 2. $u^{\prime}\left(x_{1}\right)<0$ In this case, the point $\xi_{0}$ is chosen so that it is located in the interval ( $x_{3},b$ ) and not coincide with any other root of the function $u(x)$ . We proceed exactly as in case 1. We thus arrive at the following result:

If the differential operator $L(y)$ own the property $T_{2}(a,b)$ , then any function $u(x)$ belonging to the class $C_{2}(a,b)$ and satisfying within the range $(a,b)$ differential inequality $L(u)>0$ cannot have in the range ( $a,b$ ) more than two distinct roots, which means that it cannot take values equal to the values of any function in the family $Y_{0}$ in more than two distinct points in ( $a,b$ ).

Le ma 6'. If the coefficients of the differential equation (2) are continuous in the closed interval $[a,b]$ and if the differential operator $L(y)$ which occurs in the left-hand side of equation (2) has the property $T_{2}[a,b]$ , then any function $u(x)$ belonging to the class $C_{2}[a,b]$ and satisfying within the range $[a,b]$ differential inequality $L(u)>0$ , cannot have in the range $[a,b]$ more than two distinct roots.

The proof of this lemma is done in the same way as in the case of lemma 6, with the only difference that everywhere we must consider instead of the open interval ( $a,b$ ), the interval

closed $[a,b]$ and that apart from cases 1 and 2 where the roots are assumed to be $x_{1}$ , $x_{2},x_{3}$ of the function $u(x)$ , which satisfies in $[a,b]$ inequality $L(u)>0$ , are inside this interval, - we should also consider the cases when one or possibly two of the roots $x_{1},x_{2},x_{3}$ of this function $u(x)$ would coincide respectively with the ends of the interval $[a,b]$ For example, let's say that $x_{1}=a,x_{3}=b,a<x_{2}<b$ (fig. 4). We can reduce this case to one of cases 1 or 2 as follows: We extend the function $u(x)$ out of range $[a,b]$ with polynomials of degree 2, which coincide with $u(x)$ in the heads $a$ , respectively $b$ , the coincidence occurring up to the 2nd order derivatives. We thus obtain a function $\bar{u}(x)$ , defined on the entire axis and admitting derivatives of order 2 inclusive, continuous on the entire axis $Ox$ and in addition, $\bar{u}(x)$ coincides with $u(x)$ in the interval $[a,b]$ .

We perform on the equation curve $y=\bar{u}(x)$ - infinitesimal contraction towards the median of the segment $[a,b]$ (fig. 4), so as to maintain the strict inequality $L(\bar{u})>0$ in $[a,b]$ We can consider for example the transformation

\left\{\begin{array}[]{l}\xi=\frac{a+b}{2}+\lambda\left(x-\frac{a+b}{2}\right)\\ \eta=y\end{array}\right.

where $\lambda$ is a real number that satisfies the inequalities $0<\lambda<1$ and sufficiently close to unity. The equation of the transformed curve will be

y=\bar{u}^{*}(x)=\bar{u}\left[\frac{a+b}{2}+\frac{1}{\lambda}\left(x-\frac{a+b}{2}\right)\right]

This curve will have the shape indicated by the dotted line in figure 4 and will intersect the axis $Ox$ in three distinct points $x_{1},x_{2},x_{3}$ , all inside the interval $[a,b]$ If the parameter $\lambda$ is chosen close enough to unity (and smaller than 1), then for reasons of continuity of the coefficients of the differential equation, the inequality will hold $L\left(\bar{u}^{*}\right)>0$ in the interval $[a,b]$ Thus we have reduced this singular case to case 1 of Lemma 6.

For ease of expression in what follows, we will give the following:
Definition 4. Given a second-order linear differential operator

L(y)=y^{\prime\prime}+p(x)y^{\prime}+q(x)y

(6)

coefficients $p(x)$ and $q(x)$ being continuous functions on an interval $(a,b)$ We will say that the operator $L(y)$ own the property $P_{2}(a,b)$ if any real numbers $x_{0},y_{0}^{\prime},y_{0}^{\prime\prime}$ , so that

	$\displaystyle a<x_{0}<b,\quad y_{0}^{\prime}<0,$		(7)
	$\displaystyle y_{0}^{\prime\prime}+p\left(x_{0}\right)y_{0}^{\prime}>0$		(8)

and if however small the number $\varepsilon$ satisfying inequalities $0<\varepsilon\leqq b-x_{0}$ , then for the number system $x_{0},y_{0}^{\prime},y_{0}^{\prime\prime},\varepsilon$ , there is at least one number $\lambda\in\left(x_{0},x_{0}+\varepsilon\right)$
and at least one function $r_{\lambda}(x)$ , continues in $(a,b)$ , satisfying in the closed interval $\left[x_{0},\lambda\right]$ inequality

r_{\lambda}(x)>0,\quad x\in\left[x_{0},\lambda\right]

(9)

so that for this function $r_{\lambda}(x)$ , the differential equation

L(y)=y^{\prime\prime}+p(x)y^{\prime}+q(x)y=r_{\lambda}(x)

(10)

to admit within the range $(a,b)$ a particular integral $y_{\lambda}(x)$ which satisfies the conditions

	$\displaystyle y_{\lambda}\left(x_{0}\right)=y_{\lambda}(\lambda)=0$
	$\displaystyle y_{\lambda}^{\prime}\left(x_{0}\right)=y_{0}^{\prime}$		(11)
	$\displaystyle y_{\lambda}^{\prime\prime}\left(x_{0}\right)=y_{0}^{\prime\prime}$

where $x_{0},y_{0}^{\prime},y_{0}^{\prime\prime}$ are the numbers chosen according to conditions (7) and (8).
Le ma 7. If the coefficients $p(x)$ and $q(x)$ of the differential operator $L(y)$ from (6) are continuous in the interval ( $a,b$ ), then the operator $L(y)$ has the property $P_{2}(a,b)$ .

Proof. Before proceeding to the actual proof of this lemma, we will make some preliminary observations, which we will use in the proof.

Remark 1. To show that the operator $L(y)$ has the property $P_{2}(a,b)$ it is enough to show that that operator has the property $P_{2}\left(a_{1},b_{1}\right)$ whatever the subinterval ( $a_{1},b_{1}$ ) so that $a<a_{1}<b_{1}<b$ . Reducing the proof of Lemma 7 to the case of a subinterval $\left(a_{1},b_{1}\right)$ , we have the advantage of being able to have the continuity property of the coefficients $p(x)$ and $q(x)$ in the closed range $\left[a_{1},b_{1}\right]$ as well as all the properties that arise from it.

Observation 2. Let $\eta(x)$ some function belonging to the class $C_{2}(a,b)$ and satisfying the inequality $\eta(x)>0$ whatever $\mathrm{fi}x\in(a,b)$ . By performing the change of function in the operator expression (6)

y=\eta(x)z(x)

(12)

get

L(y)\equiv\eta(x)\left[z^{\prime\prime}+\frac{2\eta^{\prime}+p\eta}{\eta(x)}z^{\prime}+\frac{L(\eta)}{\eta(x)}z\right]\equiv\eta(x)\bar{L}(z)

(13)

because $\eta(x)\in C_{2}(a,b)$ , it follows that the differential operator $\bar{L}(z)$ has continuous coefficients in ( $a,b$ ). Let us denote by $\bar{p}(x)$ respectively $\bar{q}(x)$ operator coefficients $\bar{L}(z)$ , that is

\bar{p}(x)=\frac{2\eta^{\prime}+p\eta}{\eta(x)},\quad\bar{q}(x)=\frac{L(\eta)}{\eta(x)}

(14)

Taking into account the positivity of the function $\eta(x)$ in the interval $(a,b)$ it is verified that if the operator $L(y)$ has the property $P_{2}(a,b)$ , then the operator $\bar{L}(z)$ has the property $P_{2}(a,b)$ and vice versa. For illustration, let us prove for example the reciprocal of the above statement that if the operator $\bar{L}(z)$ has the property $P_{2}(a,b)$ , then the operator $L(y)$ has the property $P_{2}(a,b)$ Let for this purpose a number system $x_{0}$ ,
$y^{\prime}{}_{0},y^{\prime\prime}{}_{0}$ satisfying conditions (7) and (8) and either $x_{0},z^{\prime}{}_{0},z^{\prime\prime}{}_{0}$ the corresponding number system, obtained using relation (12), taking into account (11)

z_{0}^{\prime}=\frac{y_{0}^{\prime}}{\eta_{0}},z_{0}^{\prime\prime}=\frac{1}{\eta_{0}^{2}}\left(\eta_{0}y_{0}^{\prime\prime}-2\gamma_{10}^{\prime}y_{0}^{\prime}\right).

(15)

Here it was noted $\eta_{0}=\eta\left(x_{0}\right)$ . Taking into account relations (7), (8), (14), as well as the positivity of the function $\eta(x)$ , it is verified that the numbers $z_{0}^{\prime},z_{0}^{\prime\prime}$ from (15) satisfy the inequalities $z_{0}^{\prime}<0,z_{0}^{\prime\prime}+\ddot{p}\left(x_{0}\right)z_{0}^{\prime}>0$ , which shows us that the numbers $z_{0}^{\prime}$ and $z_{0}^{\prime\prime}$ satisfy conditions (7) and (8) relative to the operator $\bar{L}(z)$ Since by hypothesis the operator $\bar{L}(z)$ has the property $P_{2}(a,b)$ , it follows that whatever the real number $\varepsilon$ so that $0<\varepsilon\leqq\leqq b-x_{0}$ , there is at least one number $\lambda\in\left(x_{0},x_{0}+\varepsilon\right)$ and at least one function $\bar{r}_{\lambda}(x)$ continues in ( $a,b$ ), satisfying in the interval [ $x_{0},\lambda$ ] inequality

\bar{r}_{\lambda}(x)>0,\quad x\in\left[x_{0},\lambda\right]

(16)

so that for this function $\bar{r}_{\lambda}(x)$ , the differential equation in the unknown function $z(x)$

\bar{L}(z)=z^{\prime\prime}+\frac{2\eta^{\prime}+p\eta}{\eta(x)}z^{\prime}+\frac{L(\eta)}{\eta(x)}z=\bar{r}_{\lambda}(x)

(17)

to admit in the interval ( $a,b$ ) a particular integral $z_{\lambda}(x)$ satisfying the conditions

	$\displaystyle z_{\lambda}\left(x_{0}\right)=z_{\lambda}(\lambda)=0$
	$\displaystyle z_{\lambda}^{\prime}\left(x_{0}\right)=z_{0}^{\prime}$		(18)
	$\displaystyle z_{\lambda}^{\prime\prime}\left(x_{0}\right)=z_{0}^{\prime\prime}$

where $z_{0}^{\prime}$ and $z_{0}^{\prime\prime}$ are the numbers given by (15).
Referring now to the differential operator $L(y)$ , it is verified that the function

y_{\lambda}(x)=\eta(x)z_{\lambda}(x)

(19)

(where $z_{\lambda}(x)$ is the function highlighted previously), is an integral of the differential equation

L(y)=\eta(x)\vec{r}_{\lambda}(x),\text{ unde }\eta(x)\vec{r}_{\lambda}(x)>0\text{ în }\left[x_{0},\lambda\right]

(20)

and satisfies the boundary conditions

	$\displaystyle y_{\lambda}\left(x_{0}\right)=y_{\lambda}(\lambda)=0$
	$\displaystyle y_{\lambda}^{\prime}\left(x_{0}\right)=y_{0}^{\prime}$		(21)
	$\displaystyle y_{\lambda}^{\prime\prime}\left(x_{0}\right)=y_{0}^{\prime\prime}$

Verifying that the function $y_{\lambda}(x)$ given by (19) satisfies equation (20), it is immediate. The verification of conditions (21) is done taking into account relations (19), (18) and (15). Since the number system $x_{0},y_{0}^{\prime},y_{0}^{\prime\prime}$ was arbitrarily assumed to satisfy conditions (7) and (8) and since the number $\varepsilon$ was arbitrarily assumed to satisfy the inequalities $0<\varepsilon\leqq b-x_{0}$ , it follows that the operator $L(y)$ has the property $P_{2}(a,b)$ .

Observation 3. Let ( $a_{1},b_{1}$ ) an arbitrary subinterval of the interval ( $a,b$ ), so that $a<a_{1}<b_{1}<b$ For the interval ( $a_{1},b_{1}$ ) there is at least one function $\eta(x)\in C_{2}(-\infty,+\infty)$ , positive in the interval ( $-\infty,+\infty$ ) so that performing in the operator expression $L(y)$ change of function (12), the corresponding operator $\bar{L}(z)$ to have the coefficient $\bar{p}(x)$ negative in the closed range $\left[a_{1},b_{1}\right]$ .

Indeed, either $M=\max_{\left[a_{1},b_{1}\right]}p(x)$ and either $\eta(x)$ a positive integral of the equation

2\eta^{\prime}+(M+1)\eta=0

We can take $\eta(x)=e^{-\frac{M+1}{2}x}$ ; then according to formula (14), the expression of the coefficient $\bar{p}(x)$ will be
$\bar{p}(x)=\frac{2\eta^{\prime}+p\eta}{\eta}=p(x)-M-1\leqq-1$ , for $x\in\left[a_{1},b_{1}\right]$ .

Let us now move on to the actual proof of Lemma 7. Based on observations 1,2,3 made above, it follows that to prove Lemma 7 it is sufficient to show that any differential operator $L(y)$ of the form (6), having continuous coefficients in the closed interval $[a,b]$ and $p(x)<0$ in this interval, it has the property $P_{2}(a,b)$ We therefore assume that the coefficients of the operator $L(y)$ from (6) are continuous in $[a,b]$ and that $p(x)<0$ in $[a,b]$ . Either $x_{0},y_{0}^{\prime},y_{0}^{\prime\prime}$ a system of real numbers satisfying conditions (7) and (8). We can always consider $x_{0}=0$ , which can be achieved by performing the translation on the independent variable $X=x-x_{0}$ Such a variable change is allowed provided that instead of the property $P_{2}(a,b)$ to be considered property $P_{2}\left(a-x_{0},b-x_{0}\right)$ .

We will therefore assume in what follows that $x_{0}=0$ and that the interval ( $a,b$ ) contains the origin $x_{0}=0$ . In these hypotheses, let the differential equation be

L(y)=y^{\prime\prime}+p(x)y^{\prime}+q(x)y=f(x)

(22)

in which the free term $f(x)$ we regard it as an undetermined function in $[a,b]$ . Assuming that $f(x)$ is continuous in $[a,b]$ , we first propose to find the form of the integral of equation (22), which satisfies the conditions

y(0)=0,\quad y^{\prime}(0)=y_{0}^{\prime},\quad y^{\prime\prime}(0)=y_{0}^{\prime\prime}

(23)

For this purpose, let us consider the corresponding homogeneous equation

L(y)=y^{\prime\prime}+p(x)y^{\prime}+q(x)y=0

(24)

and either $y_{1}(x)$ and $y_{2}(x)$ two particular integrals of equation (24), forming a fundamental system in $[a,b]$ Then, as is known, the general integral of the homogeneous differential equation (24) can be written as

y(x)=C_{1}y_{1}(x)+C_{2}y_{2}(x)+\int_{0}^{x}G(x,s)f(s)ds

(25)

where was it noted

G(x,s)=-\frac{\left|\begin{array}[]{ll}y_{1}(x)&y_{2}(x)\\ y_{1}(s)&y_{2}(s)\end{array}\right|}{\left|\begin{array}[]{ll}y_{1}(s)&y_{2}(s)\\ y_{1}^{\prime}(s)&y_{2}^{\prime}(s)\end{array}\right|}.

Formula (25) can be easily obtained by applying the method of variation of constants to integrate the differential equation (22).

Let's take for $y_{1}(x)$ and $y_{2}(x)$ integrals that satisfy the conditions

\begin{array}[]{r}y_{1}(0)=0,y_{1}^{\prime}(0)=y_{0}^{\prime}<0\\ y_{2}(0)=1,\quad y_{2}^{\prime}(0)=0\end{array}

where $y_{0}^{\prime}$ is the number given previously. With these clarifications, taking into account (25), it is immediately obtained that the particular integral of equation (22), which satisfies the first two conditions in (23), is given by the formula

y(x)=y_{1}(x)+\int_{0}^{x}G(x,s)f(s)ds

(28)

From here we deduce

y^{\prime\prime}(x)=y_{1}^{\prime\prime}(x)+f(x)+\int_{0}^{x}\frac{\partial^{2}G(x,s)}{\partial x^{2}}f(s)ds

Replacing in this relationship $x=0$ and taking into account the fact that

y_{1}^{\prime\prime}(0)=-p(0)y_{0}^{\prime}

which is obtained from equation (24), it follows that

y^{\prime\prime}(0)=-p_{0}y_{0}^{\prime}+f(0),\quad p_{0}=p(0).

Writing that the third condition in (23) is verified, we obtain for the function $f(x)$ condition

f(0)=y_{0}^{\prime\prime}+p_{0}y_{0}^{\prime}

(29)

Be it now $\lambda$ some number in the range ( $x_{0}=0,b$ ). We impose on the integral $y(x)$ from (28) the condition $y(\lambda)=0$ , that is

y_{1}(\lambda)+\int_{0}^{\lambda}G(\lambda,s)f(s)ds=0

(30)

This equality is also a condition for $f(x)$ We try to satisfy conditions (29) and (30) by taking for $f(x)$ a first degree polynomial in the variable $x$ , that is

f(x)=\alpha x+\beta

(31)

In order for condition (29) to be fulfilled, we must take

\beta=y_{0}^{\prime\prime}+p_{0}y_{0}^{\prime}.

Replacing $f(x)$ from (31) to (30) and taking into account the value found for $\beta$ , the coefficient can be determined $\alpha$ . Substituting in (31) the coefficients $\alpha$ and $\beta$ thus determined, it is obtained for $f(x)$ expression

f_{\lambda}(x)=-\frac{\left(y_{0}^{\prime\prime}+p_{0}y_{0}^{\prime}\right)\int_{0}^{\lambda}G(\lambda,s)ds+y_{1}(\lambda)}{\int_{0}^{\lambda}G(\lambda,s)sds}x+y_{0}^{\prime\prime}+p_{0}y_{0}^{\prime}

(

\prime

)

Formula (31') makes sense since the denominator of the coefficient of $x$ is not canceled if $\lambda$ is sufficiently close to zero. This statement follows from an existence theorem relative to polylocal boundary value problems given by de la Vallée Poussin [8].

Let us now prove that if the positive number $\lambda$ is sufficiently close to zero, then the function $f_{\lambda}(x)\operatorname{din}\left(31^{\prime}\right)$ satisfies the inequality $f_{\lambda}(x)>0$ for $x\in[0,\lambda]$ To this end, taking into account that $f_{\lambda}(x)$ is a polynomial of degree 1, it suffices to verify that $f_{\lambda}(0)>0$ and $f_{\lambda}(\lambda)>0$ if $\lambda$ is close enough to zero.

Inequality $f_{\lambda}(0)>0$ immediately follows from the fact that the number system $x_{0}=0$ , $y_{0}^{\prime},y_{0}^{\prime\prime}$ previously chosen, satisfies condition (8).

Let us now prove that for positive values $\lambda$ , sufficiently close to zero the inequality occurs $f_{\lambda}(\lambda)>0$ To this end we will show that $\lim_{\lambda\rightarrow 0^{+}}f_{\lambda}(\lambda)>0$ The first term in (31') presents an indeterminacy of the form $\frac{0}{0}$ when $\lambda\rightarrow 0^{+}$ . Applying l'Hôspital's rule and noting for abbreviation $y_{0}^{\prime\prime}+p_{0}y_{0}^{\prime}=A^{2}$ , we obtain

\begin{gathered}J=\lim_{\lambda\rightarrow 0^{+}}\frac{-\lambda\left[A^{2}\int_{0}^{\lambda}G(\lambda,s)ds+y_{1}(\lambda)\right]}{\int_{0}^{\lambda}G(\lambda,s)sds}=\\ =\lim_{\lambda\rightarrow 0^{+}}\frac{-\left[A^{2}\int_{0}^{\lambda}G(\lambda,s)ds+y_{1}(\lambda)\right]-\lambda\left[\left.A^{2}\int_{0}^{\lambda}\frac{\partial G}{\partial x}\right|_{x=\lambda}ds+y_{1}^{\prime}(\lambda)\right]}{\left.\int_{0}^{\lambda}\frac{\partial G}{\partial x}\right|_{x=\lambda}sds}\end{gathered}

But the term on the right-hand side of the above equality also presents an indeterminacy of the form $\frac{0}{0}$ when $\lambda\rightarrow 0^{+}$ Applying l'Hôspital's rule once again, we obtain

J=\lim_{\lambda\rightarrow 0^{+}}\frac{-2\left[\left.A^{2}\int_{0}^{\lambda}\frac{\partial G}{\partial x}\right|_{x=\lambda}ds+y_{1}^{\prime}(\lambda)\right]-\lambda\left[A^{2}\left(1+\left.\int_{0}^{\lambda}\frac{\partial^{2}G}{\partial x^{2}}\right|_{x=\lambda}ds\right)+y_{1}^{\prime\prime}(\lambda)\right]}{\lambda+\left.\int_{0}^{\lambda}\frac{\partial^{2}G}{\partial x^{2}}\right|_{x=\lambda}sds}

(32)

Let us now show that there is a positive number $\delta$ (small enough) so that whatever $\lambda\in(0,\delta)$ , the denominator of the fraction in (32) must be positive. For this it is sufficient to show that the function $\left.\frac{\partial^{2}G}{\partial x^{2}}\right|_{x=\lambda}$ by the variables $\lambda$ and $s$ is positive in a triangular domain defined by the inequalities $0<s<\lambda,0<\lambda<\delta$ , or what is equivalent - that the function $\frac{\delta^{2}G}{\partial x^{2}}$ of variables $x$ and $s$ is positive in a field $0<s<\lambda$ and $0<\lambda<\delta$ ( $\delta$ being a sufficiently small positive number). Indeed, from (26), taking into account the conditions (27) we deduce

\left.\frac{\partial^{2}G}{\partial x^{2}}\right|_{\begin{subarray}{c}x=0\\ s=0\end{subarray}}=-\frac{\left|\begin{array}[]{cc}y_{1}^{\prime\prime}(0)&y_{2}^{\prime\prime}(0)\\ 0&1\end{array}\right|}{\left|\begin{array}[]{cc}0&1\\ y_{0}^{\prime}&0\end{array}\right|}=\frac{y_{1}^{\prime\prime}(0)}{y_{0}^{\prime}}.

(33)

From equation (24) we deduce that $y^{\prime\prime}{}_{1}(0)=-p_{0}y^{\prime}{}_{1}(0)-q_{0}y_{1}(0)$ and taking into account (27), we obtain $y^{\prime\prime}{}_{1}(0)=-p_{0}y^{\prime}{}_{0}$ . Substituting in (33) we obtain that $\left.\frac{\partial^{2}G}{\partial x^{2}}\right|_{\begin{subarray}{c}x=0\\ s=0\end{subarray}}=-p_{0}$ Because by hypothesis $p(x)<0$ , it results in particular $p_{0}<0$ and therefore that

\left.\frac{\partial^{2}G}{\partial x^{2}}\right|_{\begin{subarray}{c}x=0\\ s=0\end{subarray}}>0

(34)

Because the coefficients $p(x)$ and $q(x)$ are by hypothesis continuous functions in $[a,b]$ , it follows that and $y^{\prime\prime}{}_{1}(x)$ and $y^{\prime\prime}{}_{2}(x)$ are continuous in $[a,b]$ , as integrals of the differential equation (24) and therefore that the function $\frac{\partial^{2}G(x,s)}{\partial x}$ of variables $x$ and $s$ is continuous in the field $\{a\leqq x\leqq b,a\leqq s\leqq b\}$ , which contains by hypothesis the point ( $x=0,s=0$ ). It ultimately follows that the integral $\left.\int_{0}^{\lambda}\frac{\partial^{2}G}{\partial x^{2}}\right|_{x=\lambda}ds$ is positive when $\lambda$ is a positive number

sufficiently small, and also that the denominator of the fraction (32) tends to $0^{+}$ when $\lambda\rightarrow 0^{+}$ Thus we obtain for the limit in (32)

J=\frac{-2y_{1}^{\prime}(0)}{0^{+}}=\frac{-2y_{0}^{\prime}}{0^{+}}=+\infty

since by hypothesis $y_{0}^{\prime}<0$ It
follows that if the positive number $\lambda$ is small enough, then the function $f_{\lambda}(x)$ from (31') satisfies the inequality $f_{\lambda}(x)>0$ , when $x\in[0,\lambda]$ With this Lemma 7 is completely proven.

Property is established in exactly the same way. $P_{2}[a,b)$ in case the coefficients $p(x)$ and $q(x)$ of the differential operator $L(y)$ are continuous in the interval $[a,b)$ The previously established Lemma 7 can be stated in the following more suggestive form:

Lemma 7'. Let $L(y)$ a linear differential operator of the form (6), having the coefficients $p(x)$ and $q(x)$ continue in an interval ( $a,b$ ) and either $u(x)$ some function in the class $C_{2}(a,b)$ , satisfying at one point $x_{0}$ FROM $(a,b)$ conditions:

\begin{array}[]{ll}1^{0}&u\left(x_{0}\right)=0\\ 2^{0}&u^{\prime}\left(x_{0}\right)<0\\ 3^{0}&\left.L(u)\right|_{x=x_{0}}>0\end{array}

Then whatever positive number is $\varepsilon$ satisfying the inequality $\varepsilon<b-x_{0}$ , for this there is a number $\lambda\in\left(x_{0},x_{0}+\varepsilon\right)$ and a function $y_{\lambda}(x)\in C_{2}(a,b)$ satisfying the conditions

		$\displaystyle\text{ {$\alpha$}) }y_{\lambda}\left(x_{0}\right)=u\left(x_{0}\right)=0,\quad y_{\lambda}(\lambda)=0$
		$\displaystyle\text{ {$\beta$}) }y_{\lambda}^{\prime}\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)$
		$\displaystyle\text{ {$\gamma$}) }y_{\lambda}^{\prime\prime}(x)=u^{\prime\prime}\left(x_{0}\right)$
		$\displaystyle\text{ {$\delta$}) }L\left(y_{\lambda}\right)>0\text{ pentru orice }x\in\left[x_{0},\lambda\right]\text{. }$

Proof. Referring to the statement of Lemma 7, let us consider the number system $x_{0},y_{0}^{\prime}=u^{\prime}\left(x_{0}\right),y^{\prime\prime}{}_{0}=u^{\prime\prime}\left(x_{0}\right)$ According to the hypotheses $2^{\circ}$ and $3^{\circ}$ from the statement of Lemma 7 it follows that the numbers $x_{0},y_{0}^{\prime},y_{0}^{\prime\prime}$ thus chosen, satisfy conditions (7) and (8) relative to the operator $L(y)$ considered. Then according to Lemma 7, for any positive number $\varepsilon$ satisfying the inequality $\varepsilon<b-x_{0}$ there is at least one number $\lambda\in\left(x_{0},x_{0}+\varepsilon\right)$ and at least one function $r_{\lambda}(x)$ continues in $(a,b)$ , satisfying the inequality $r_{\lambda}(x)>0$ for $x\in\left[x_{0},\lambda\right]$ so that for this function $r_{\lambda}(x)$ , the differential equation

L(y)=y^{\prime\prime}+p(x)y^{\prime}+q(x)y=r_{\lambda}(x)

to admit in $(a,b)$ a particular integral $y_{\lambda}(x)$ , satisfying the conditions $\left.\left.\alpha\right),\beta\right),\gamma$ ). Obviously this integral belongs to the class $C_{2}(a,b)$ and also satisfies the condition $\delta$ ), since $r_{\lambda}(x)>0$ in the interval $\left[x_{0},\lambda\right]$ .

Definition 5. Let $F$ a family of functions $f(x)$ defined within a certain range $J$ , which can be open, closed or semi-closed. We say that a function $u(x)$ defined
in the range $J$ is convex in this interval with respect to the family $F$ , if it satisfies the following conditions:
$1^{\circ}$ It cannot take values equal to the values of any function $f(x)\in F$ at more than two distinct points in the interval $J$ .
$2^{\circ}$ If $u(x)$ takes values equal to the values of some function $f(x)\in F$ in two distinct points $x_{1}<x_{2}$ from the range $J$ , that is $u\left(x_{i}\right)=f\left(x_{i}\right),i=1,2$ , then the inequalities hold $u(x)<f(x)$ in the interval ( $x_{1},x_{2}$ ) and $u(x)>f(x)$ on the set of points in the interval $J$ which do not belong to the interval $\left[x_{1},x_{3}\right]$ .

A definition analogous to this, in a more general sense, was considered by E. Moldovan in the work « $O$ generalization of the notion of convexity", communicated at the 4th Congress of Romanian Mathematicians, held in Bucharest, May 27 - June 4, 1956.

With this definition , Lemma 8 holds . If the differential operator $L(y)$ from (6) has the property $T_{2}(a,b)$ then any function $u(x)\in C_{2}(a,b)$ , which satisfies in the interval ( $a,b$ ) differential inequality $L(u)>0$ , is in the interval ( $a,b$ ) convex with respect to the family of integrals of the homogeneous differential equation $L(y)=0$ .

Proof. To prove this lemma, we note that it is sufficient to show that under the assumption that $L(y)$ has the property $T_{2}(a,b)$ , then whatever the function is $u(x)$ belonging to the class $C_{2}(a,b)$ and satisfying in the interval ( $a,b$ ) differential inequality $L(u)>0$ , - function $u(x)$ cannot have in ( $a,b$ ) more than two distinct roots and in the case when it would cancel at two points $x_{1}<x_{2}$ from ( $a,b$ ), then the inequalities take place:

	$\displaystyle u(x)<0\text{ în }\left(x_{1},x_{2}\right)$		(35)
	$\displaystyle u(x)>0\text{ în }\left(a,x_{1}\right)\text{ și în }\left(x_{2},b\right).$

The fact that such a function $u(x)$ cannot be cancelled within $(a,b)$ in more than two points, follows from Lemma 6. Now suppose that a function $u(x)\in C_{2}(a,b)$ , satisfying in $(a,b)$ differential inequality $L(u)>0$ , it cancels out in the interval ( $a,b$ ) in two points $x_{1}<x_{2}$ . Let us show that under these conditions the inequalities (35) hold. For this purpose let us suppose by absurdity that the function $u(x)$ would not satisfy the inequalities (35). Then since the roots $x_{1}$ and $x_{2}$ are simple, according to the property $T_{2}(a,b)$ of the operator $L(y)$ , and since by hypothesis $x_{1}$ and $x_{2}$ are the only roots of the function $u(x)$ in ( $a,b$ ), it follows that $u(x)$ satisfies the following inequalities, opposite to inequalities (35)

	$\displaystyle u(x)>0\text{ în }\left(x_{1},x_{2}\right)$		(36)
	$\displaystyle u(x)<0\text{ în }\left(a,x_{1}\right)\text{ și în }\left(x_{2},b\right)$

and therefore has the form shown in figure 6. We observe that the function $u(x)$ satisfy at point $x_{2}$ the conditions $1^{\circ},2^{\circ},3^{\circ}$ from the statement of the lemma $7^{\prime}$ . Then according to the lemma $7^{\prime}$ , there is a number $\lambda\in\left(x_{2},b\right)$ and a function $y_{\lambda}(x)\in C_{2}(a,b)$ satisfying the conditions:

	$\displaystyle y_{\lambda}\left(x_{2}\right)=y_{\lambda}(\lambda)=0$
	$\displaystyle y_{\lambda}^{\prime}\left(x_{2}\right)=u^{\prime}\left(x_{2}\right)$		(37)
	$\displaystyle y_{\lambda}^{\prime\prime}\left(x_{2}\right)=u^{\prime\prime}\left(x_{2}\right)$
	$\displaystyle L\left(y_{\lambda}\right)>0\text{ pentru }x\in\left[x_{2},\lambda\right].$

We consider the function $U(x)$ defined in $(a,b)$ as follows:

U(x)=\left\{\begin{array}[]{l}u(x)\text{ dacă }x\in\left(a,x_{2}\right),\\ y_{\star}^{\star}(x)\text{ dacă }x\in\left[x_{2},b\right).\end{array}\right.

Taking into account (37), it follows that $U(x)\in C_{2}(a,b)$ and therefore that $L(U)$ is a continuous function in ( $a,b$ ) and in particular at the point $x_{2}\in(a,b)$ . Considering the definition of the function $U(x)$ as well as the last relation in (37), it follows that $L(U)>0$ in the half-closed interval $(a,\lambda]$ and how the function $L(U)$ of variable $x$ is continuous in $(a,b)$ , it follows that it continues to take positive values to the right of the point $x=\lambda$ in a neighborhood of the form $\left[\lambda,b_{1}\right)$ , where $b_{1}$ satisfies inequalities $\lambda<b_{1}<b$ It ultimately follows that the function $U(x)$ satisfies the inequality $L(U)>0$ throughout the interval ( $a,b_{1}$ ). Since by hypothesis the operator $L(y)$ has the property $T_{2}(a,b)$ , it follows that he will also have the property $T_{2}\left(a,b_{1}\right)$ , since the interval ( $a,b_{1}$ ) is a subinterval of the interval ( $a,b$ ). The proof of this statement is done exactly like the proof of Lemma 5. But the function $U(x)$ it cancels out in 3 distinct points in the interval $\left(a,b_{1}\right)$ , namely at the points $x_{1},x_{2},\lambda$ . This circumstance, however, contradicts the statement of lemma 6. It therefore follows that inequalities (36) must be excluded, while inequalities (35) remain valid (under the assumptions made), which proves the statement of lemma 8.

Remark. Lemma 8 also holds true in the case when instead of the open interval $(a,b)$ is considered a semi-closed interval $[a,b)$ .

Le ma 9. Given a linear and homogeneous differential equation

L(y)=y^{\prime\prime}+p(x)y^{\prime}+q(x)y=0

having the coefficients $p(x)$ and $q(x)$ continuous in an interval ( $a,b$ ), respectively $[a,b)$ The necessary and sufficient condition that the family $Y_{0}$ of the integrals of the differential equation considered to have the property $I_{2}(a,b)$ , respectively $I_{2}[a,b)$ it's like whatever the function is $u(x)$ belonging to the class $C_{2}(a,b)$ , respectively $C_{2}[a,b)$ and satisfying in $(a,b)$ respectively $[a,b)$ differential inequality $L(u)>0$ , to be convex (in the sense of Definition 5) with respect to the set $Y_{0}$ in $(a,b)$ respectively. $[a,b)^{1}$ ).

We will give the proof for the case of an open interval ( $a,b$ ), it remaining broadly the same for the case of a semi-closed interval $[a,b)$ .

The condition is necessary. This statement follows immediately from the successive application of Lemmas 4 and 8.

The condition is sufficient. We assume that the considered differential equation has the property: for any function $u(x)\in C_{2}(a,b)$ , the differential inequality $L(u)>0$ in ( $a,b$ ), entails convexity in ( $a,b$ ) of the function $u(x)$ towards the family $Y_{0}$ of the integrals of the considered differential equation. Let us show that in this hypothesis, $Y_{0}$ has the property $I_{2}(a,b)$ For this purpose let us suppose by absurdity that the equation

⁰⁰footnotetext: ¹ ) The idea of this property belongs to E1ena Mo1dovan and was communicated in [1] for the case of linear differential equations of order

n

, in a slightly different formulation than the one in the statement of Lemma 9. The author of this paper took the liberty of giving his own proof for the particular case

n=?

Differential equation $L(y)=0$ would admit a particular integral $\widetilde{y}(x)$ , non-identical to null, which would cancel out in the interval $(a,b)$ in two points $x_{1}<x_{2}$ .

Without restricting the generality of the reasoning, we can assume that the roots $x_{1}$ and $x_{2}$ of the particular integral $\widetilde{y}(x)$ are consecutive and that, in the interval $(x_{1},x_{2})$ , this integral satisfies the inequality

\widetilde{y}(x)>0,

because, if this condition is not met, it can be achieved by multiplying the particular integral considered by the constant $-1$ .

because $\widetilde{y}(x)$ is a non-zero integral of the equation $L(y)=0$ , it follows that the derivative $\widetilde{y}^{\prime}(x)$ cannot be canceled at points $x_{1}$ and $x_{2}$ ; therefore, the equation curve

y=\widetilde{y}(x)

crosses the axis $Ox$ in the points $x_{1}$ and $x_{2}$ .

It follows that there are three numbers $\xi_{1},\xi_{2},\xi_{3}$ which satisfy the inequalities

x_{1}<\xi_{1}<\xi_{2}<\xi_{3}<x_{2}.

a<\xi_{1}<\xi_{2}<\xi_{3}<b

and for which inequalities still occur

\tilde{y}\left(\xi_{1}\right)<0,\quad\tilde{y}\left(\xi_{2}\right)>0,\quad\tilde{y}\left(\xi_{3}\right)<0.

(38)

In other words, it is found that there are functions $\eta(x)\in C_{2}\left(\xi_{1},\xi_{3}\right)$ , satisfying the conditions

\eta(x)>0\text{ în }\left[\xi_{1},\xi_{3}\right]\text{ și }L(\eta)>0\text{ în }\left[\xi_{1},\xi_{3}\right]\text{. }

(39)

We can take for example $\eta(x)=e^{Cx}$ , where $C$ is a sufficiently large constant. Let's perform in the operator expression $L(y)$ change of function

y=\eta(x)z(x),

(40)

where $\eta(x)$ is the function considered previously. We obtain

L(y)\equiv\eta(x)\left[z^{\prime\prime}+\frac{2\eta^{\prime}+p\eta}{\eta}z+\frac{L(\eta)}{\eta}z\right].

(41)

Whether $\tilde{z}(x)=\frac{\tilde{y}(x)}{\eta(x)}$ . From (41) it follows in particular that

L(\widetilde{y})\equiv\eta(x)\left[\widetilde{z}^{\prime\prime}+\frac{2\eta^{\prime}+p\eta}{\eta}\widetilde{z}^{\prime}+\frac{L(\eta)}{\eta}\widetilde{z}\right]\equiv 0

(42)

Taking into account (38) and (39), it results

\widetilde{z}\left(\xi_{1}\right)<0,\quad\widetilde{z}\left(\xi_{2}\right)>0,\quad\widetilde{z}\left(\xi_{3}\right)<0.

(43)

Whether $\varepsilon$ a positive number small enough so that the function $\widetilde{z}(x)=\widetilde{z}(x)+\varepsilon$ to satisfy the following inequalities analogous to (43)

\widetilde{\widetilde{z}}\left(\xi_{1}\right)<0,\quad\widetilde{\widetilde{z}}\left(\xi_{2}\right)>0,\quad\widetilde{z}\left(\xi_{3}\right)<0.

(44)

Either then $\widetilde{y}=\eta(x)\widetilde{z}(x)$ . Substituting into (41) and taking into account the equality $\widetilde{z}(x)==\widetilde{z}(x)+\varepsilon$ as well as from relations (42) and (39), we obtain

	$\displaystyle L(\widetilde{\widetilde{y}})$	$\displaystyle=\eta(x)\left[\widetilde{\widetilde{z}}^{\prime\prime}+\frac{2\eta^{\prime}+p\eta}{\eta}\widetilde{\widetilde{z}}^{\prime}+\frac{L(\eta)}{\eta}\widetilde{z}\right]=\eta(x)\left[\widetilde{z}^{\prime\prime}+\frac{2\eta^{\prime}+p\eta}{\eta}\widetilde{z}^{\prime}+\right.$		(45)
		$\displaystyle\left.+\frac{L(\eta)}{\eta}\widetilde{z}\right]+\varepsilon L(\eta)=0+\varepsilon L(\eta)>0\quad\text{ pentru }x\in\left[\xi_{1},\xi_{3}\right]$

Then from (44), taking into account that $\eta(x)>0$ in the interval $\left[\xi_{1},\xi_{3}\right]$ , we deduce

\widetilde{\tilde{y}}\left(\xi_{1}\right)<0,\quad\widetilde{\tilde{y}}\left(\xi_{2}\right)>0,\quad\widetilde{\tilde{y}}\left(\xi_{s}\right)<0.

(46)

As previously shown in the proof of Lemma 5, the function $\widetilde{\mathcal{y}}(x)$ defined in the range $\left[\xi_{1},\xi_{2}\right]$ can be extended throughout the range $(a,b)$ , maintaining throughout this interval the continuity of the 1st and 2nd order derivatives as well as the differential inequality (45). In this $\bmod$ we get a function $\widetilde{y}(x)\in C_{2}(a,b)$ which satisfies the differential inequality $L(\widetilde{\widetilde{y}})>0$ in $(a,b)$ and which cancels out at two points in the interval ( $a,b$ ) satisfying inequalities (46). However, this contradicts the hypothesis made. It ultimately results that the differential equation $L(y)=0$ cannot have any integral $y(x)$ , which cancels out at two points in the interval ( $a,b$ ) and therefore that the family $Y_{0}$ of the integrals of the equation $L(y)=0$ has the property $N_{2}(a,b)$ . From here, according to Lemma 2, it follows that $Y_{0}$ has the property $I_{2}(a,b)$ , what

Proof of Theorem 1. The necessity of the condition expressed by this theorem is established in the statement of Lemma 4. The sufficiency of this condition results from the successive application of Lemmas 8 and 9.

Observation. Theorem 1 remains true even if in its statement we consider instead of the open interval ( $a,b$ ), one of the intervals [ $a,b$ ) or ( $a,b$ ]. This statement is established by extending Lemmas 4, 8, 9 to the case of intervals [ $a,b$ ) and ( $a,b$ ].

We mention that Theorem 1 might not hold if its statement refers to a closed interval. $[a,b]$ This can be seen from the example given by the differential equation $L(y)=y^{\prime\prime}+y=0$ The corresponding differential operator has the property $T_{2}[0,\pi]$ , and the family $Y_{0}$ of the integrals of the differential equation $L(y)=0$ does not have the property $I_{2}[0,\pi]$ The fact that the family $Y_{0}$ does not have the property $I_{2}[0,\pi]$ it is immediately observed, taking into account that the considered differential equation admits the particular integral $y=\sin x$ , which cancels out at the points $x=0$ and $x=\pi$ .

The fact that the differential operator $L(y)$ has the property $T_{2}[0,\pi]$ it is found by writing the function $u(x)$ which satisfies the differential inequality $L(u)>0$ as well as the boundary conditions $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=0$ (where $x_{0}$ is a number in the range $[0,\pi]$ ) in the form

u(x)=\int_{x_{0}}^{x}L[u(s)]\sin(x-s)ds=\int_{x_{n}}^{x}\left[u^{\prime\prime}(s)+u(s)\right]\sin(x-s)ds

Applications related to the theory of differential inequalities of SA Ciaplîghin

There are many criteria known regarding the coefficients of the differential equation (1), which ensure, in a given interval, the non-oscillation property or the interpolation property (in the sense of definitions 2 and 1 of the present paper) of the integrals of the differential equation (1). According to the previously established theorem 1, all these criteria become criteria that ensure the applicability of the differential inequality theorem of SA Ciaplîghin for equation (1) (in the sense of definition 3). In this context,
we allow ourselves to state the following criterion of applicability of the differential inequality theorem of SA Ciaplîghin, a criterion that immediately results from theorem 1 and lemma 3, established previously:

Theorem 2. Necessary and sufficient condition that the differential operator $L(y)$ from (1) having the continuous coefficients in the semi-closed interval $[a,b)$ to have the property $T_{2}[a,b)$ , is like the following Riccati equation

\sigma^{\prime}+\sigma^{2}+p(x)\sigma+q(x)=0

(47)

admit at least one particular integral $\sigma(x)$ continues in ( $a,b$ ).
Proof. We first establish the necessity of the condition. We assume that the differential operator $L(y)$ has the property $T_{2}[a,b)$ According to Theorem 1 it follows that the family $Y_{0}$ of the differential equation (2) possesses the property $I_{2}[a,b)$ Then according to Lemma 3, the differential equation (2) must admit an integral $\eta(x)$ , positive and continuous in ( $a,b$ ). Performing the change of function in equation (2)

y(x)=e^{\int_{x_{0}}^{x}\sigma(s)ds},\quad x_{0}\in(a,b)

(48)

equation (2) is transformed into equation (47). By transforming (48), the integral $\eta(x)>0$ in $(a,b)$ will correspond to the integral equation (47) $\sigma_{\eta}(x)=\frac{\eta^{\prime}(x)}{\eta(x)}$ which will obviously be continuous in ( $a,b$ ), since $\eta(x)$ does not cancel out within this interval. Thus, the necessity of the condition is established.

The sufficiency of the condition is immediately evident, taking into account that the integral $y_{\sigma}(x)$ of equation (2), which corresponds to an integral $\sigma(x)$ , assumed continuous in the interval ( $a,b$ ), of equation (47), is positive in ( $a,b$ ) as shown by the transformation formula (48). Then, the statements of Lemmas 3 and 4 are taken into account.

Observations. $1^{\circ}$ The differential equation (47) resembles the so-called "Riccati protective equation"

\sigma^{\prime}+\sigma^{2}-p(x)\sigma-p^{\prime}(x)+q(x)=0

(49)

highlighted by S. A. Tsiaplîghin on the occasion of developing his method of approximate integration of second-order linear differential equations (see for example [3]). The existence of a continuous solution in $[a,b)$ of equation (49) constitutes, as SA Ciaplîghin showed, a sufficient condition for the differential operator $L(y)$ to own the property $T_{a}^{(2)}[a,b)$ (at point $x=a$ ). This sufficient condition given by SA Ciaplîghin, assumes the derivability of the coefficient $p(x)$ in $[a,b)$ .
$2^{\circ}$ We mention that given the point $x_{0}$ , the problem of determining the largest interval of the form $\left[x_{0},x_{0}+h\right)$ , in which the family of integrals of a linear differential equation of order $n$ own the property $T_{x_{0}}^{(n)}\left[x_{0},x_{0}+h\right)$ (at point $x_{0}$ ), was recently solved by NA Kascheev in [4]. Analogous results for the case of particular linear and nonlinear second-order equations were obtained by VN Petrov in [5] and [6]. All these results refer to a property of the type $T_{a}^{(2)}[a,b)$ at a given point $a$ , without talking about properties of the type $T_{2}[a,b)$ (see definition 3).

Other characterization of the property $\Pi_{2}(a,b)$ of the family $\boldsymbol{Y}$ of the integrals of the differential equation (1)

In what follows we will try to give another characterization of the property $I_{2}(a,b)$ (and therefore also of the property $I_{2}[a,b)$ ) of the family $Y$ , than that given by Theorem 1. For this purpose we first give

Definition 6. We say that the differential operator $L(\mathrm{y})$ , which appears in the left-hand side of equation (1), defined on the set of functions in the class $C_{2}(a,b)$ , owns the property $\bar{T}_{x_{0}}^{(2)}(a,b),x_{0}$ being a point in the interval ( $a,b$ ), if any and the function $u(x)\in C_{2}(a,b)$ , satisfying the condition $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=0$ , from the inequality $L(u)\geqq 0$ valid throughout the entire range $(a,b)$ , the inequality results $u(x)\geqq 0$ in $(ab)$ We will say that the operator $L(y)$ own the property $\bar{T}_{2}(a,b)$ , if that operator possesses the property $\bar{T}_{x_{0}}^{(2)}(a,b$ , $)oricarearfipunctulx_{0}\in(a,b)$ .

Lemma 10. Necessary and sufficient condition that the differential operator $L(y)$ which appears on the left-hand side of equation (1) has the property $\bar{T}_{2}(a,b)$ respectively $\bar{T}_{2}[a,b]$ it's like that operator $L(y)$ to have the property $T_{2}(a,b)$ respectively $T_{2}[a,b]$ .

Proof. We assume that the operator $L(y)$ has the property $\bar{T}_{2}(a,b)$ Let us show that in this hypothesis, the operator $L(y)$ also has the property $T_{2}(a,b)$ . Indeed, either $u(x)$ a function belonging to the class $C_{2}(a,b)$ and satisfying the conditions $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=0$ , where $x_{0}$ is a point in the interval $(a,b)$ , as well as the differential inequality $L(u)>0$ in $(a,b)$ (these conditions intervene in the definition of property $T_{2}(a,b)$ ). By hypothesis, the property $T_{2}(a,b)$ , taking place, it results that $u(x)\geqq 0$ in $(a,b)$ We will demonstrate that in the assumptions made on the function $u(x)$ , the strict inequality takes place $u(x)>0$ in the intervals $\left(a,x_{0}\right)$ and $\left(x_{0},b\right)$ Indeed, let us suppose by absurdity that $u(x)$ is canceled outside the point $x_{0}$ , in one more point $x_{1}\in(a,b)$ , different from $x_{0}$ (fig. 8). Either $a_{1}$ and $b_{1}$ some numbers satisfying the inequalities $a<a_{1}<\min\left\{x_{0},x_{1}\right\}<\max\left\{x_{0},x_{1}\right\}<b_{1}<b$ . Then either $u_{1}(x)$ a function from the class $C_{2}\left(a_{1},b_{1}\right]$ , satisfying the condition $u_{1}\left(x_{0}\right)=u_{1}^{\prime}\left(x_{0}\right)=0$ and being positive in the intervals $\left[a_{1},x_{0}\right),\left(x_{0},b_{1}\right]$ .

We consider the linear combination

v(x)=u(x)+\lambda u_{1}(x)

where $\lambda$ is a constant. From the hypothesis $L(u)>0$ in $(a,b)$ and from the continuity property of functions $u(x),u_{1}(x)$ as well as the 1st and 2nd order derivatives of these functions in the closed interval $\left[a_{1},b_{1}\right]$ and also the coefficients $p(x)$ and $q(x)$ of the operator $L(y)$ , it follows that for negative and sufficiently small absolute values of the parameter $\lambda$ , occurs throughout the entire interval $\left[a_{1},b_{1}\right]$ strict inequality $L(v)>0$ . Either $\lambda_{0}\circ$ such value and $v_{0}(x)$ the corresponding function. $\lambda_{0}$ is a negative number and $u\left(x_{1}\right)=0$ , it follows that $v_{0}\left(x_{1}\right)<0$ . Then from the conditions that the functions satisfy $u(x)$ and $u_{1}(x)$ at the point $x_{0}$ , it follows
that $v_{0}\left(x_{0}\right)=v_{0}^{\prime}\left(x_{0}\right)=0$ . Ultimately for the function $v_{0}(x)$ the following conditions are met that affect the ownership $\bar{T}_{2}\left[a_{1},b_{1}\right]$ :

v_{0}\left(x_{0}\right)=v_{0}^{\prime}\left(x_{0}\right)=0\text{ și }L\left(v_{0}\right)>0\text{ în }\left[a_{1},b_{1}\right].

(50)

But since by hypothesis the operator $L(y)$ has the property $\bar{T}_{2}(a,b)$ , it follows that he will also have the property $T_{2}\left[a_{1},b_{1}\right]$ , whatever the subinterval $\left[a_{1},b_{1}\right]$ content in the range $(a,b)^{1}$ ). From here and from (50) it would follow that $v_{0}(x)\geqq 0$ in $\left[a_{1},b_{1}\right]$ what

contradicts inequality $v_{0}\left(x_{1}\right)<0$ previously established. It ultimately follows that the function $u(x)$ is nonnegative in the interval ( $a,b$ ) and is only canceled at the point $x_{0}$ , what

Proof of the lemma in the case when instead of the properties $T_{2}(a,b)$ and $T_{2}(a,b)$ the properties are considered respectively $T_{2}[a,b]$ and $\bar{T}_{2}[a,b]$ , is done as above, with the simplification that one can take $a_{1}=a$ and $b_{1}=b$ .

Let us now establish the sufficiency of the condition expressed by the lemma. For this purpose, we assume that the operator $L(y)$ has the property $T_{2}(a,b)$ and let us show that in this hypothesis he also has the property $T_{2}(a,b)$ . Indeed, either $u(x)$ a function satisfying the condition $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=0$ , where $x_{0}$ is a point in the interval $(a,b)$ , and also differential inequality $L(u)\geqq 0$ throughout the range $(a,b)$ , (these conditions intervene in the property $\bar{T}_{2}(a,b)$ ). We will demonstrate that under these assumptions the inequality holds $u(x)\geqq 0$ in the interval ( $a,b$ ). Indeed, let us suppose by absurdity that there would be a point $x_{1}$ from the interval ( $a,b$ ) so that the inequality holds

u\left(x_{1}\right)<0.

(51)

Obviously that $x_{1}\neq x_{0}$ (fig. 9). Either $u_{1}(x)$ integral in the interval ( $a,b$ ) of the differential equation $L(y)=1$ , satisfying the boundary conditions $u_{1}\left(x_{0}\right)=u_{1}^{\prime}\left(x_{0}\right)=0$ . Then either $\lambda$ a sufficiently small positive value so that the inequality is satisfied

\lambda\left|u_{1}\left(x_{1}\right)\right|<\left|u\left(x_{1}\right)\right|.

(52)

Then the function $v(x)=u(x)+\lambda u_{1}(x)$ has the properties

	$\displaystyle v\left(x_{0}\right)=v^{\prime}\left(x_{0}\right)=0$		(53)
	$\displaystyle L(v)=L(u)+\lambda L\left(u_{1}\right)=L(u)+\lambda\geqq\lambda>0\quad\text{ în }(a,b).$

But since by hypothesis the operator $L(y)$ has the property $T_{2}(a,b)$ , it follows from (53) that $v(x)>0$ in $(a,b)$ On the other hand, from (51) and (52) it follows that $v\left(x_{1}\right)<0$ . We thus obtained a contradiction. Ultimately it results that $u(x)\geqq 0$ in the interval $(a,b)$ , what

⁰⁰footnotetext: ¹ ) The proof of this statement is done exactly as in the case of Lemma 5.

Establishing the sufficiency of the condition expressed by the lemma, in the case of a closed interval $[a,b]$ it is done in an analogous way.

Definition 6'. Definition 6 given previously, is immediately extended to the case of linear and homogeneous differential operators of order $n$ ,

L_{n}(y)=y^{(n)}-\sum_{i=1}^{n}a_{i}(x)y^{(n-i)}

(54)

having the coefficients $a_{i}(x)$ continue within an interval $(a,b)$ We will say that the differential operator $L_{n}(y)$ own the property $\bar{T}_{n}(a,b)$ , if whatever the function is $u(x)$ belonging to the class $C_{n}(a,b)$ (i.e. having derivatives up to the order $n$ inclusive, continue in the interval $(a,b)$ ), satisfying the conditions

u(x)_{0}=u^{\prime}\left(x_{0}\right)=\ldots=u^{(n-1)}\left(x_{0}\right)=0

where $x_{0}$ is a point in the interval ( $a,b$ ), then from the inequality $L_{n}(u)\geqq 0$ , valid in the interval $(a,b)$ , the inequality results $u(x)\geqq 0$ in the same interval ( $a,b$ ).

Definition 7. Function $\varphi(x)$ is said to be the Cauchy function associated with the operator $L_{n}(y)$ and the node $\alpha\in(a,b)$ , if $\varphi(x)$ is an integral in the interval $(a,b)$ , of the differential equation $L_{n}(y)=0$ and if it still satisfies the boundary conditions

\varphi(\alpha)=\varphi^{\prime}(\alpha)=\ldots=\varphi^{(n-2)}(\alpha)=0,\quad\varphi^{(n-1)}(\alpha)=1.

We will denote such a function by $\varphi(x,\alpha)$ , highlighting the node where the boundary conditions written above occur.

With these two definitions, we establish the following property, analogous to that established by NA Kascheev in the paper [4]:

Le ma 11. The necessary and sufficient condition that the operator $L_{n}(y)$ to have the property $\bar{T}_{n}(a,b)$ is like the Cauchy function $\varphi(x,\alpha)$ associated with the operator $L_{n}(y)$ to be nonnegative in the triangle ( $D_{1}$ ), bounded by the lines $\alpha=x,\alpha=a,x=b$ , and at the same time be nonpositive in the triangle ( $D_{2}$ ), bounded by the lines $\alpha=x$ , $\alpha=b,x=a$ (fig. 10).

Proof. The condition is sufficient. Indeed, suppose that the function $\varphi(x,\alpha)$ satisfies the conditions in the lemma. Let $u(x)$ a function from the class $C_{n}(a,b)$ , satisfying at one point $x_{0}\operatorname{din}(a,b)$ conditions:

u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=\ldots=u^{(n-1)}\left(x_{0}\right)=0

(55)

In addition, we assume that the function $u(x)$ check the inequality

L(u)\geqq 0,\quad x\in(a,b)

(56)

Let us show that under these conditions the inequality holds $u(x)\geqq 0$ , whatever it is $x\in(a,b)$ Indeed, it is known that the integral of the equation $L_{n}(y)=f(x)$ , which satisfies at the point $x_{0}$ the conditions

y\left(x_{0}\right)=u\left(x_{0}\right),\quad y^{\prime}\left(x_{0}\right)=u^{\prime}\left(x_{0}\right),\ldots\quad y^{(n-1)}\left(x_{0}\right)=u^{(n-1)}\left(x_{0}\right)

has the expression

y(x)=u(x)-\int_{x_{0}}^{x}\left\{L_{n}[u(\alpha)]-f(\alpha)\right\}\varphi(x,\alpha)d\alpha

From this formula, taking $f(x)\equiv 0$ in $(a,b)$ and taking into account the fact that the function $u(x)$ verifying conditions (55), it follows that the integral $y(x)$ corresponding is identically null in ( $a,b$ ) and consequently the identity holds

u(x)=\int_{x_{0}}^{x}L_{n}[u(\alpha)]\varphi(x,\alpha)d\alpha

(57)

From this identity, taking into account inequality (56), as well as the assumptions made on the function $\varphi(x,\alpha)$ , it follows that $u(x)\geqq 0$ when $x\in(a,b)$ , what

The condition is necessary. Suppose that the operator $L_{n}(y)$ has the property $\bar{T}_{n}(a,b)$ Let's prove that the function $\varphi(x,\alpha)$ is nonnegative in the domain ( $D_{1}$ ) and non-positive in ( $D_{2}$ ). Indeed, let us suppose by absurdity that there would be a point $\left(x^{\star},\alpha^{\star}\right)\in\left(D_{1}\right)$ , so that the strict inequality holds $\varphi\left(x^{*},\alpha^{\star}\right)<0$ . Either $x_{0}$ any point, satisfying the inequalities $a<x_{0}<x^{\star}<b$ Then obviously one can choose a function $h(x)$ , positive in $(a,b)$ , so that the integral $\int_{x_{0}}^{x}h(\alpha)\varphi(x,\alpha)d\alpha$ , viewed as a function of the variable $x$ , to have a negative value for $x=x^{\star}$ . Thus choosing the function $h(x)$ , let us denote with $u(x)$ integral of the differential equation $L_{n}(y)=h(x)$ , satisfying the initial conditions

u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=\ldots=u^{(n-1)}\left(x_{0}\right)=0

Then according to formula (57), we have

u(x)=\int_{x_{0}}^{x}L_{n}[u(\alpha)]\varphi(x,\alpha)d\alpha=\int_{x_{0}}^{x}h(\alpha)\varphi(x,\alpha)d\alpha

and from here, according to how the function was chosen $h(x)$ , the inequality results $u\left(x^{*}\right)<0$ . On the other hand, $L_{n}(u)=h(x)>0$ when $x\in(a,b)$ and $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)==\ldots=u^{(n-1)}\left(x_{0}\right)=0$ . Then since it was assumed by hypothesis that $L_{n}(y)$ has the property $\bar{T}_{n}(a,b)$ , it follows that $u(x)\geqq 0$ when $x\in(a,b)$ This inequality, however, contradicts the inequality $u\left(x^{*}\right)<0$ , previously established. It follows that there cannot be any point ( $x^{\star},\alpha^{\star}$ ) from the field ( $D_{1}$ ), in which $\varphi(x,\alpha)$ to be negative.

Analogously it is shown that in the hypothesis that $L_{n}(y)$ has the property $\bar{T}_{n}(a,b)$ , function $\varphi(x,\alpha)$ is non-positive in the field $\left(D_{2}\right)$ .

Remark. Lemma 11 also extends to the case when instead of an open interval ( $a,b$ ), is considered a closed interval $[a,b]$ , or a semi-closed interval $[a,b)$ .

Taking into account Theorem 1 as well as Lemmas 10 and 11, we can state the following

Theorem 3. Let the differential equation (1), having the coefficients $p(x),q(x),r(x)$ continuous in the interval ( $a,b$ ). The necessary and sufficient condition that the family $Y$ of the integrals of this equation have the property $I_{2}(a,b)$ or $N_{2}(a,b)$ , is like the Cauchy function $\varphi(x,\alpha)$ , corresponding to the linear and homogeneous differential operator $L(y)$ associated with equation (1), to be nonnegative in the domain ( $D_{1}$ ) and non-positive in the domain ( $D_{2}$ ) (fig. 10).

Remark. It is known that the Cauchy function corresponding to the differential operator $L_{n}(y)$ considered in (54) admits the representation

\varphi(x,\alpha)=\frac{1}{(n-1)!}\int_{\alpha}^{x}\Gamma(t,\alpha)(x-t)^{n-1}dt+{\frac{(x-\alpha)^{n-1}}{(n-1)!}}^{n}

(58)

where $\Gamma(x,\alpha)$ is the resolving kernel of the kernel

K(x,\alpha)=a_{1}(x)+a_{2}(x)(x-\alpha)+\ldots+a_{n}(x)\frac{(x-\alpha)^{n-1}}{(n-1)!},

(59)

in the case of Volterra's equation:

\Gamma(x,\alpha)-\int_{\alpha}^{x}K(x,s)\Gamma(s,\alpha)ds=K(x,\alpha)

In the particular case of the differential operator $L_{2}(y)$ which occurs in (1), the corresponding Cauchy function admits the representation

\varphi(x,\alpha)=\int_{\alpha}^{x}\Gamma(t,\alpha)(x-t)dt+(x-\alpha)

(60)

where $\Gamma(x,\alpha)$ is the resolving kernel of the kernel

K(x,\alpha)=-p(x)-q(x)(x-\alpha)

(61)

in the case of Volterra's equation

\Gamma(x,\alpha)-\int_{\alpha}^{x}K(x,s)\Gamma(s,\alpha)ds=K(x,\alpha)

It is shown ¹ ) in the theory of integral equations that the solution of this integral equation admits the representation:

\Gamma(x,\alpha)=\sum_{i=1}^{\infty}K_{i}(x,\alpha)

(62)

⁰⁰footnotetext: ¹ ) See, for example, the textbook SG Mihlin, Integralnïe Uravnenia, Moscow-Leningrad, pp. 22-27.

where $K_{i}(s,\alpha)$ is defined from close to close, using the formula

K_{i}(x,\alpha)=\int_{\alpha}^{x}K(x,t)K_{i-1}(t,\alpha)dt;\quad K_{1}(x,\alpha)=K(x,\alpha)

Theorem 3, together with formulas (60) and (62), gives us a principal solution to the problem of determining exact limits of intervals $(a,b)$ of the axis $Ox$ , in which the family $Y$ of the integrals of equation (1) has the property $I_{2}(a,b)$ .

Taking into account formula (60), we observe that if $\Gamma(x,\alpha)$ is nonnegative in the domain ( $D_{1}$ ) and non-positive in the domain ( $D_{2}$ ) (fig. 10), then the function $\varphi(x,\alpha)$ will be nonnegative in ( $D_{1}$ ) and non-positive in ( $D_{2}$ ), and therefore the family $Y$ of the integrals of the differential equation (1) will have the property $I_{2}(a,b)$ It follows from this that the simultaneous fulfillment of the inequalities $\Gamma(x,\alpha)\geqq 0$ in $\left(D_{1}\right)$ and $\Gamma(x,\alpha)\leqq 0$ in $\left(D_{2}\right)$ constitutes a sufficient condition for the family $Y$ of the integrals of equation (1) to have the property $I_{2}(a,b)$ .

General observation. Theorems $1,2,3$ remain true even if instead of the finite interval ( $a,b$ ) respectively $[a,b)$ is considered an infinite interval ( $a,\infty$ ) respectively $[a,\infty)$ This statement results from the fact that the properties $I_{2}(a,\infty)$ , $N_{2}(a,\infty),T_{2}(a,\infty)$ attract the properties respectively $I_{2}(a,b),N_{2}(a,b)$ , $T_{2}(a,b)$ whatever the finite number $b>a$ and reciprocally.

Approximate methods for delimiting maximum intervals $[a,a+h)$ , for which the family $Y$ of the integrals of equation (1) has the property $\boldsymbol{I}_{2}\boldsymbol{(}\boldsymbol{a},\boldsymbol{a}+\boldsymbol{h}\boldsymbol{)}$

In what follows we will assume that the coefficients of equation (1) are continuous in a sufficiently large interval, containing the point $x=a$ From Lemma 3, the following immediately follows

Lemma $3^{\prime}$ Given the number $x=a$ , the maximum range of the form $[a,a+h)$ for which the family $Y$ of the integrals of equation (1) has the property $I_{2}[a,a+h)$ , is the semi-closed interval between two consecutive roots of any non-identically zero integral of the associated homogeneous equation (2), the integral being subject to the only condition of passing through the point $A(a,0)$ , - the first root of the pair is taken $x=a$ .

This lemma gives us the possibility to approximate with any precision the number $a+h$ , when the number is given $a$ Indeed, let us consider for example the particular integral $y(x)$ of the homogeneous equation (2), satisfying at the point $x=a$ the conditions $y(a)=0,y^{\prime}(a)=1$ . The first root of the equation $y(x)=0$ , located to the right of the point $x=a$ (if any), it will give us the number $a+h$ Let us suppose that through some approximation procedure it was possible to obtain an infinite series of functions $\left\{y_{n}(x)\right\}$ which converges uniformly to the function $y(x)$ , in an interval containing the interval $[a,a+h)$ inside it and also the derivatives of these functions converge to $y^{\prime}(x)$ (for example by the method of successive approximations of E. Picard). Then noting with $r_{n}$ the smallest root, located to the right of the point $x=a$ , of the equation $y_{n}(x)=0$ (if it exists), the string $\left\{r_{n}\right\}$ will converge to the number $a+h$ searched.

Remark. From Lemma 3', taking into account the fact that an integral curve of equation (2) that does not coincide with the axis $Ox$ , cannot have a contact with this axis greater than
or equal to 1 , and also from the fact that such a curve deforms uniformly continuously in a given finite interval, when the coefficients of the differential equation $p(x)$ , $q(x)$ undergoes continuous variations in that interval, but keeping the initial conditions, - it can be easily shown that by maintaining the number $x=a$ fixed and making the functions vary $p(x),q(x)$ in the space of continuous functions, then the number $h$ , which occurs in the statement of the lemma $3^{\prime}$ , is a continuous functional of the arguments $p(x)$ and $q(x)$ .

This property was established in a different way by C. Foiaş, G. Gussi and V. Poenaru in the paper [9].

In what follows we will deal with the bilateral approximation of the number $a+h$ , when the number is given $a$ We establish for this purpose the following

Lemma 12. Let $u(x)$ and $v(x)$ two functions belonging to the class $C_{2}[a,a+\lambda)$ where the interval $[a,a+\lambda)$ is large enough to contain the interval $\left.[a,a+h)^{1}\right)$ and satisfying the conditions:

u(a)=v(a)=0,\quad u^{\prime}(a)>0,\quad v^{\prime}(a)>0

as well as differential inequalities

L(u)\leqq 0\text{ şi }L(v)\geqq 0\text{ in intervalul }[a,a+\lambda)\text{. }

Then noting with $y_{u}(x)$ the particular integral of equation (2), which satisfies the conditions $y_{u}(a)=0,y_{u}^{\prime}(a)=u^{\prime}(a)>0$ , the inequality occurs

u(x)\leqq y_{u}(x),x\in[a,a+h)

Likewise noting with $y_{\nu}(x)$ the particular integral of equation (2), which satisfies the conditions $y_{\nu}(a)=0,y_{\nu}^{\prime}(a)=\nu^{\prime}(a)>0$ , the inequality occurs

y_{v}(x)\leqq v(x),x\in[a,a+h).

Noting with $r_{u}$ and $r_{\nu}$ the smallest roots located in the interval $(a,a+\lambda)$ of functions $u(x)$ respectively $v(x)$ , (if of course there are such roots), the delimitations take place:

r_{u}\leqq a+h\leqq r_{v}

The proof of this lemma is immediate if one takes into account Theorem 1 and Lemma $3^{\prime}$ .

Observation. Inequality $r_{u}\leqq a+h$ it also results from a result obtained by VA Kondratiev in the paper [2].

Computing Institute, RPR Academy-Cluj Branch

⁰⁰footnotetext: ¹ ) The interval

[a,a+h)

represents the maximum interval, having the left extremum

x=a

date, when the family

Y_{0}

and so does the family

Y

has the property

I_{2}

LE PROBLÈME BILOCAL ET LE THEORÈME DES INÉGALITÉS DIFFERENTIELLES À NOEUDS CONFONDUS DE SA TCHAPLYGUINE POUR DES ÉQUATIONS DIFFÉRENTIELLES LINÉAIRES DU $2^{\mathrm{e}}$ ORDER

SUMMARY

On considerar, dans ce travail, l'équation différentialelle linéaire du $2^{\mathrm{e}}$ ordre (1), aux coefficients continus dans un intervalle quelconque ( $a,b$ ). It is designated by $Y$ la famille des integrales de cette équation dans l'intervale ( $a,b$ ). We begin by giving the following definitions:

Definition 1. It is said that $Y$ enjoys ownership $I_{2}(a,b)$ and, quels que soient les nódes distincts $x_{1}$ and $x_{2}$ dance $(a,b)$ et quelles que soient les valeurs réals $y_{1}$ and $y_{2}$ , there exists a particular integral $y(x)\in Y$ , et une seule, qui puisse satisfy aux conditions $y\left(x_{1}\right)=y_{1},y\left(x_{2}\right)=y_{2}$ .

Definition 2. On dit que la famille $Y$ enjoys ownership $N_{2}(a,b)$ and, quelles que soient deux intégrales particuliers distinctes $y_{1}(x)$ and $y_{2}(x)$ dance $Y$ , celles-ci ne peuvent prendre de valeurs égales dans l’intervale ( $a,b$ ) que tout au plus en un point.

Définition 3. On dit que l'opérateur différentiel linear et homogène $L_{2}(y)$ defined on the class $C_{2}(a,b)$ des fonctions admettant des dérivées du second ordre, continue dans l'intervale ( $a,b$ ), enjoys ownership $T_{x_{0}}^{(2)}(a,b)$ and, quelle que soit la fonction $u(x)\in C_{2}(a,b)$ , in the conditions $L_{2}(u)>0$ dance $(a,b)$ and $u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=0$ , il en resulterera l'inégalité $u(x)\geqq 0$ dance $(a,b)$ , le signe égal ne pouvant se trouver qu'au point $x_{0}$ It is assumed here that $x_{0}$ is a point in the interval $(a,b)$ .

Definition 4. On dit que l'opérater différenceiel $L_{2}(y)$ enjoys ownership $T_{2}(a,b)$ , and cet opérateur a la propriété $T_{x_{0}}^{(2)}(a,b)$ whatever $x_{0}\in(a,b)$ .

These definitions permit to establish the following theories:
Theorem 1. The necessary and sufficient condition for the family $Y$ des intégrales de l'équation différentielle considéraire (1) jouisse de la propriété $I_{2}(a,b)$ or property $N_{2}(a,b)$ , est que l'operator differential, linear et homogène, $L_{2}(y)$ , respectively, ait la propriété $T_{2}(a,b)$ .

Theorem 2. The necessary and sufficient condition for the differential, linear and homogeneous operator, $L_{2}(y)$ , aux coefficients continuus dans l'intervale demifermé $[a,b)$ , enjoy the property $T_{2}[a,b)$ est que l'équation de Riccati (47) admits au moins une intégrale particulier $\sigma(x)$ continue in the meantime $(a,b)$ .

We then consider in (59) a differential operator, linear and homogeneous, $L_{n}(y)$ , by order $n$ , aux coefficients continus dans l'intervale ( $a,b$ ) and defined in the class $C_{n}(a,b)$ .

Definition 5. On dit que l'operatore $L_{n}(y)$ enjoys ownership $\bar{T}_{x_{0}}^{(n)}(a,b)$ and, quelle que soit la fonction $u(x)\in C_{n}(a,b)$ satisfactory to the conditions

u\left(x_{0}\right)=u^{\prime}\left(x_{0}\right)=\ldots=u^{(n-1)}\left(x_{0}\right)=0,L_{n}(u)\geqslant 0\text{ pour }x\in(a,b)

for such a function $u(x)$ , it results in inequality $u(x)\geqslant 0$ dance $(a,b)$ .
Definition 6. On dit que l'opérater différenceiel $L_{n}(y)$ enjoys ownership $\bar{T}_{n}(a,b)$ , and what operator owns the property $\bar{T}_{x_{0}}^{(n)}(a,b)$ , whatever $x_{0}\in(a,b)$ .

Definition 7. On dit que la fonction $\varphi(x)$ is the Cauchy function associated with the differential operator $L_{n}(y)$ and have knot $\alpha\in(a,b)$ , and $\varphi(x)$ is a particular integral dans l'intervale ( $a,b$ ) of the differential, linear and homogeneous equation, $L_{n}(y)=0$ , satisfying aux conditions à la limite

\varphi(\alpha)=\varphi^{\prime}(\alpha)=\ldots=\varphi^{(n-2)}(\alpha)=0;\varphi^{(n-1)}(\alpha)=1

Dans ce qui suit, on designera une telle fonction par $\varphi(x,\alpha)$ , en mettant ainsi en évidence le node $\alpha$ where are the conditions à la limite ci dessus.

These conditions permit to establish the following property, analogous to the one established by NA Kastchéév [4]:

Theorem 3. The necessary and sufficient condition for the differential operator $L_{n}(y)$ enjoyment of property $\bar{T}_{n}(a,b)$ is that the Cauchy function $\varphi(x,\alpha)$ , associated with this operator, soit non negative dans le domaine $D_{1}$ et non positive dans le domaine $D_{2}$ (the domains $D_{1}$ and $D_{2}$ are indicated on figure 10).

Dans ce travail, on montre aussi que, dans le cas de l'équation différenceielle (2), les properties $T_{2}(a,b)$ and $\bar{T}_{2}(a,b)$ they are equivalent. It results from this observation

Théorème 4. La condition necessaire et suffisante pour que la famille $Y$ des intégrales de l'équation différenceielle (1) jouisse de la propriété $I_{2}(a,b)$ egg $N_{2}(a,b)$ is that the Cauchy function $\varphi(x,\alpha)$ associated with the differential operator $L_{2}(y)$ respectively, soit non négative dans le domaine $D_{1}$ et non positive dans le domaine $D_{2}$ (fig. 10).

General observation. Les théorèmes 1,2,3,4 restent vrais dans des intervalsles demi-fermés $[a,b)$ et en outre s'étendent immediately à des intervalles infinis, from the forms $(a,\infty)$ , respectively $[a,\infty)$ .

Théorème 5. En supposant que l'opérateur différential $L_{2}(y)$ in (1) the coefficients $p(x)$ and $q(x)$ continuos dans l'intervale demi-fermé $[a,b)$ , are $u(x)$ and $v(x)$ two functions belonging to the class $C_{2}[a,b)$ et satisfactory aux conditions :

		$\displaystyle u(a)=0,u^{\prime}(a)>0,L_{2}(u)\leqslant 0\text{ pour }x\in(a,b)$
		$\displaystyle v(a)=0,v^{\prime}(a)>0,L_{2}(v)\geqslant 0\text{ pour }x\in(a,b).$

Soient $r_{u}$ and $r_{v}$ les plus petites racines situate dans l'intervale ( $a,b$ ) functions $u(x)$ and $v(x)$ (à conditions toutefois que de telles racines existent). Dans ce cas, en désignant par $[a,a+h)$ , l'intervale maximum contenu dans $[a,b)$ dans léquel l'intégrale générale de l'équation (1) est interpolatrice du deuxième ordre ( $c$ that is to say where $Y$ enjoys ownership $I_{2}$ ), we will have the inequalities

r_{u}\leqslant a+h\leqslant r_{v}.

C'e théorème permet l'approximation du nombre $h$ with the desired precision.

The bilocal problem and S.A. Ciaplȃghin’s theorem of differential inequalities with confused nodes for second-order linear

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THE BILOCAL PROBLEM AND SA CIAPLÎGHIN'S THEOREM OF DIFFERENTIAL INEQUALITIES WITH CONFUSED NODES, FOR SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

Applications related to the theory of differential inequalities of SA Ciaplîghin

Other characterization of the property $\Pi_{2}(a,b)$ of the family $\boldsymbol{Y}$ of the integrals of the differential equation (1)

Approximate methods for delimiting maximum intervals $[a,a+h)$ , for which the family $Y$ of the integrals of equation (1) has the property $\boldsymbol{I}_{2}\boldsymbol{(}\boldsymbol{a},\boldsymbol{a}+\boldsymbol{h}\boldsymbol{)}$

Computing Institute, RPR Academy-Cluj Branch

LE PROBLÈME BILOCAL ET LE THEORÈME DES INÉGALITÉS DIFFERENTIELLES À NOEUDS CONFONDUS DE SA TCHAPLYGUINE POUR DES ÉQUATIONS DIFFÉRENTIELLES LINÉAIRES DU $2^{\mathrm{e}}$ ORDER

SUMMARY

Related Posts

The bilocal problem and S.A. Ciaplȃghin’s theorem of differential inequalities with confused nodes for second-order linear

Abstract

Authors

Keywords

Paper coordinates

PDF

About this paper

Journal

Publisher Name

DOI

Print ISSN

Online ISSN

References

Paper (preprint) in HTML form

THE BILOCAL PROBLEM AND SA CIAPLÎGHIN'S THEOREM OF DIFFERENTIAL INEQUALITIES WITH CONFUSED NODES, FOR SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

Applications related to the theory of differential inequalities of SA Ciaplîghin

Other characterization of the propertyΠ2​(A,b)\Pi_{2}(a,b)of the family𝒀\boldsymbol{Y}of the integrals of the differential equation (1)

Approximate methods for delimiting maximum intervals[A,A+h)[a,a+h), for which the familyYYof the integrals of equation (1) has the propertyI2​(𝒂𝒂𝒏,𝒂𝒂𝒏+𝒉)\boldsymbol{I}_{2}\boldsymbol{(}\boldsymbol{a},\boldsymbol{a}+\boldsymbol{h}\boldsymbol{)}

Computing Institute, RPR Academy-Cluj Branch

LE PROBLÈME BILOCAL ET LE THEORÈME DES INÉGALITÉS DIFFERENTIELLES À NOEUDS CONFONDUS DE SA TCHAPLYGUINE POUR DES ÉQUATIONS DIFFÉRENTIELLES LINÉAIRES DU2it is2^{\mathrm{e}}ORDER

SUMMARY

Related Posts

Other characterization of the property $\Pi_{2}(a,b)$ of the family $\boldsymbol{Y}$ of the integrals of the differential equation (1)

Approximate methods for delimiting maximum intervals $[a,a+h)$ , for which the family $Y$ of the integrals of equation (1) has the property $\boldsymbol{I}_{2}\boldsymbol{(}\boldsymbol{a},\boldsymbol{a}+\boldsymbol{h}\boldsymbol{)}$

LE PROBLÈME BILOCAL ET LE THEORÈME DES INÉGALITÉS DIFFERENTIELLES À NOEUDS CONFONDUS DE SA TCHAPLYGUINE POUR DES ÉQUATIONS DIFFÉRENTIELLES LINÉAIRES DU $2^{\mathrm{e}}$ ORDER