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Integral Curves of a Generalized-Homogeneous Differential Equation of the First Order
A. M. TEZIN
A qualitative study, in a neighborhood of a singular point, of the behavior of integral curves determined by a differential equation of the form
\[ y' = f(x,y), \]
whose right-hand side belongs to a very broad class of functions, is connected with great difficulties of both a theoretical and a practical character. Thus, for example, if one begins the study, by the widely known Frommer method [2], of the class of differential equations of the form
\[ y'=\frac{P_m(x,y)}{Q_n(x,y)}, \tag{1} \]
where \(P_m(x,y)\) and \(Q_n(x,y)\) are homogeneous polynomials of degrees \(m\) and \(n\), respectively, then one encounters an unnecessary complication of the investigation. Meanwhile, say, when \(m=n\), there exist quite rational ways of studying this equation. Differential equations whose right-hand side is the ratio of homogeneous polynomials of the same degree were studied by Poincaré [1] and Forster [3]. In recent years a number of works have appeared devoted to the qualitative study of the homogeneous equation of the form
\[ y'=F\left(\frac{y}{x}\right), \]
where \(F(k)\) is not necessarily a ratio of polynomials in \(k\). In the work of G. E. Shilov [4] a rather simple method was proposed for studying this equation when the right-hand side determines a differentiable field of directions. V. V. Potlov [6, 7], under broader assumptions concerning the right-hand side of the homogeneous equation, substantially supplemented G. E. Shilov’s method. For particular cases of homogeneous equations, detailed studies were carried out in the works of L. S. Lyagina [8] and V. F. Voronova [9].
In the present paper a method is proposed, close in idea to the method of G. E. Shilov, which makes it possible to study, in a neighborhood of the origin, the behavior of integral curves determined by an equation of the form
\[ y'=x^n \varphi(x)F\left(\frac{y}{x}\right). \tag{2} \]
where \(n\) is a nonnegative number such that the power \(x^n\) makes sense also for \(x<0\); the function \(F(k)\) has everywhere in its domain a continuous derivative; there may exist no more than a finite number of points \(k\) at which \(F(k)\) is not defined, but at these points there exist one-sided infinite limits \(F(k \pm 0)=\infty\); the function \(\varphi(x)\) has, in some neighborhood of zero, a continuous and nonvanishing derivative; \(0<\varphi(0)<\infty\); the equations \(\varphi(0)F(k)=k\), \(F(k)=0\), and \(F'(k)=0\) can have no more than a finite number of roots.
We shall not consider here the question of uniqueness and shall immediately assume that the right-hand side of equation (2) satisfies all the conditions of the uniqueness theorem in some neighborhood of the origin, with the possible exception of the origin itself.
Obviously, for \(m>n\) equation (1) can be written in the form
\[ y'=x^{m-n}\frac{P_m(1,k)}{Q_n(1,k)},\quad \text{where } k=\frac{y}{x}; \]
if \(m<n\), then one may consider the equation
\[ \frac{dx}{dy}=y^{\,n-m}\frac{Q_n(\bar k,1)}{P_m(\bar k,1)},\quad \text{where } \bar k=\frac{x}{y}. \]
All our subsequent arguments will be carried out with respect to the first coordinate quadrant. Everywhere below, “integral curves” and “direction field” are understood to be determined by the differential equation (2).
Consider a sector bounded by the rays \(y=k_1x\) and \(y=k_2x\) \((0<k_1<k_2)\) and by an arc of some circle with center at the origin. We shall call the indicated rays the lateral walls, and the arc of the circle the rear wall of this sector.
We shall say that a certain integral curve \(L\) enters the origin in the indicated sector if there exists an arc of this curve, lying in this sector, for which the origin is a limit point.
Let, in the sector under consideration, some integral curve \(L\) enter the origin. Since for equation (2) there can exist no more than a finite number of isoclines of infinity \(y=k_i x\), and the curve \(L\) cannot intersect any one of them at two points, there exists an arc of the integral curve \(L\) which can be regarded as a single-valued function \(y(x)\) on the interval \((0,x_0]\). This function can be extended to the origin by setting \(y(0)=0\), since \(y(x)\to 0\) as \(x\to 0\).
Definition. The ray \(k_0x\) which is tangent at the origin to some integral curve entering there shall be called a critical ray.
Theorem 1. If some integral curve \(y(x)\) enters the origin inside a sector whose lateral walls are the rays \(k_1x\) and \(k_2x\) \((0\le k_1<k_2)\), then there exists a critical ray \(k_0x\) \((k_1\le k_0\le k_2)\) such that \(y(x)\) enters the origin together with it.
Proof. Suppose first that \(y(x)=k_0x\) \((k_1\le k_0\le k_2)\); this is possible only in the case when \(\varphi(x)=c-\mathrm{const}\) and \(cF(k_0)=k_0\), \(n=0\). But then the invariant ray \(k_0x\), as a solution of the differential equation, will itself be a critical ray. Suppose that \(y(x)\ne k_0x\). Then it is not difficult to verify that the integral
the curve \(y(x)\) cannot intersect some ray \(k_0x\) in an infinite set of points for which the origin is a limit point. Indeed, suppose, on the contrary, that \(y(x)\) intersects the ray \(k_0x\) in an infinite set of points for which the origin is a limit point. Since between any two neighboring intersection points on the ray \(k_0x\) there exists a point of contact of some integral curve with this ray, it follows that on the ray \(k_0x\) there exists an infinite sequence of contact points converging to the origin. Let \(\{x_i\}\) be the corresponding sequence of abscissas of the contact points. At each point of contact the equality
\(x_i^n \varphi(x_i) F(k_0)=k_0\) holds. Passing to the limit as \(x_i\to 0\), we obtain a contradiction, since for \(n=0\), \(\varphi(x)\equiv c\), \(y(x)\) does not coincide with the ray \(k_0x\), and for \(\varphi(x)\ne\operatorname{const}\) equalities at the contact points are impossible. Therefore, as \(x\to 0\), the quantity \(y/x\) cannot equal the number \(k\) and differ from it infinitely many times. Divide the segment \([k_1,k_2]\) in half. Then, by what has just been said, all values of the continuous quantity \(y/x\) for all sufficiently small \(x\) remain in one of the halves of this segment. Divide this half in two and choose that fourth part of the segment in which all values of \(y/x\) lie for all sufficiently small \(x\). Continuing in this way indefinitely the process of bisecting the chosen segment, and each time choosing that one of the resulting halves in which all values of \(y/x\) lie for all sufficiently small \(x\), we obtain a contracting sequence of segments whose common point is the single point \(k_0\). By construction, the point \(k_0\) is characterized by the fact that, whatever the number \(\varepsilon>0\), always for sufficiently small \(x\) all values of the quantity \(y/x\) lie inside the \(\varepsilon\)-neighborhood of the point \(k_0\), i.e. \(|y/x-k_0|<\varepsilon\) for all sufficiently small \(x\). Hence it follows that
\[ \lim_{x\to 0}\frac{y(x)}{x}=k_0. \]
Consequently, the ray \(k_0x\) is critical for the integral curve \(y(x)\), since
\[ \lim_{x\to 0}\frac{y(x)}{x} = \lim_{x\to 0}\frac{y(0+x)-y(0)}{x} = y'(0). \]
The theorem is proved.
Theorem 2. If the ray \(k_0x\) is critical, then the number \(k\) is found either as a root of the equation
\[ \varphi(0)F(k)=k, \tag{3} \]
or as a point at which the function \(F(k)\) has a discontinuity with an infinite jump.
Proof. Let, together with the ray \(k_0x\), an integral curve enter the origin; this curve can be written in the form of a single-valued function \(y(x)\) on some segment \([0,x_0]\), with \(y(0)=0\). The isoclines of infinity, if such exist, are only certain rays \(kx\) for which \(F(k\pm0)=\infty\). The curve \(y(x)\) cannot intersect these rays more than at one point. Therefore, for all sufficiently small \(x\), the curve \(y(x)\) can lie only on one side of any isocline of infinity. Consequently, the functions \(k_0x\) and \(y(x)\), differentiable for all sufficiently small \(x\in(0,x_0)\), satisfy the hypotheses of Cauchy’s theorem on the segment \([0,x]\); therefore we have:
\[ \frac{y(x)}{k_0(x)} = \frac{y(x)-y(0)}{k_0x-0\cdot k_0} = \frac{y'(c)}{k_0}, \]
where \(0<c<x\). We obtain
\[ \frac{y(x)}{k_0x}=\frac{c^n\varphi(c)F\left[\dfrac{y(c)}{c}\right]}{k_0}. \tag{4} \]
The limit of the left-hand side of this equality as \(x\to 0\) is equal to one. Since \(y(c):c\to k_0\) as \(c\to 0\), in equality (4) one may pass to the limit as \(x\to 0\). In doing so, there exists a finite or infinite limit of the expression \(F\left[\dfrac{y(c)}{c}\right]\). Let us consider two cases. 1) \(n>0\). Since \(c^n\to 0\) as \(c\to 0\), the limit of the right-hand side of equality (4) can be equal to one only if \(F(k_0\pm0)=\infty\). Therefore the number \(k_0\) is a discontinuity point of the function \(F(k)\) and is found from the equation
\[ F^{-1}(k)=0. \tag{5} \]
2) \(n=0\). In this case, passing to the limit on both sides of equality (4) as \(x\to 0\), we obtain \(\varphi(0)F(k_0)=k_0\), i.e. the number \(k_0\) is a root of equation (3).
The theorem is proved.
§1. INVESTIGATION OF THE BEHAVIOR OF INTEGRAL CURVES IN THE CASE \(n>0\)
We shall consider the question of the behavior of integral curves in a sufficiently small neighborhood of the origin. For \(n>0\) the angular coefficients of possible critical rays are found from equation (5).
Let the ray \(k_0x\), \(k_0>0\), be critical. Then one of the following three cases may occur:
\[ \begin{aligned} &\text{a) } F(k_0-0)=+\infty,\quad F(k_0+0)=-\infty;\\ &\text{b) } F(k_0-0)=-\infty,\quad F(k_0+0)=+\infty;\\ &\text{c) } F(k_0-0)=F(k_0+0)=+\infty. \end{aligned} \]
Remark. A ray \(k_0x\) for which \(F(k_0-0)=F(k_0+0)=-\infty\) cannot be critical, since it lies in a sector inside which the right-hand side of the differential equation is not positive.
Let us consider the question of the behavior of integral curves in sectors containing critical rays.
- The behavior of integral curves in the case when \(F(k_0-0)=+\infty\), \(F(k_0+0)=-\infty\). Since the equation \(F(k)=0\) can have no more than a finite number of roots, it is obvious that there exists a left-hand neighborhood \((k_0-\varepsilon,k_0)\) of the point \(k_0\) in which \(F(k)>0\), and a right-hand neighborhood of this point \((k_0,k_0+\varepsilon)\) in which \(F(k)<0\). In the first neighborhood we take an arbitrary point \(k_1\), and in the second, \(k_2\), and consider the sector whose lateral sides are the rays \(k_1x\) and \(k_2x\), and whose rear side is an arc of a circle with center at the origin and sufficiently small radius. This sector is divided by the ray \(k_0x\) into two sectors. In the sector whose lower side is the ray \(k_0x\), for sufficiently small \(x\) the right-hand side of the differential equation is negative. Therefore in this sector no integral curve can enter the origin. Let us consider the question of the behavior of integral curves in the sector whose upper side is the ray \(k_0x\). For sufficiently small \(x\), on its lower side the inequality holds
\(0<x^n\varphi(x)F(k_1)<k_1\). This means that any integral curve \(L\) passing through an arbitrary point \(A\) of the lower wall of the sector under consideration, for small \(x\), enters this sector when continued in one of its two directions. Let us prove that the curve \(L\) enters the origin inside this sector.
Suppose the contrary: that it leaves the sector under consideration. It is clear that it cannot leave it by intersecting the ray \(k_1x\) a second time, since then on this ray, between the two points where it intersects \(L\), there would exist a point of contact of this ray with some integral curve. Draw through the point \(A\) a straight line parallel to the \(ox\)-axis, which intersects the ray \(k_0x\) at the point \(B\). It is clear that \(L\) cannot leave the sector through \(AB\), since then on \(AB\) there would exist a point at which the direction of the field would be parallel to the \(ox\)-axis. Finally, let us show that \(L\) cannot leave the sector \(AOB\) under consideration by intersecting its upper wall \(k_0x\) at some point \(C\) lying between \(O\) and \(B\). Suppose, on the contrary, that such a point \(C\) exists. Draw through the point \(C\) a straight line parallel to the \(ox\)-axis, which intersects the ray \(k_1x\) at the point \(D\). It is clear that the integral curve \(L\) passing through the point \(C\), when continued, enters the sector \(COD\). Consequently, it intersects the segment \(CD\) at some point \(E\) lying inside this segment. But this cannot be, since then between \(C\) and \(E\) there would exist a point of contact of some integral curve with the segment \(CD\), at which the right-hand side of the differential equation would be zero. Thus the integral curve \(L\) enters the origin inside the sector \(AOB\). Similarly one can show that the integral curve passing through the point \(B\) enters the origin inside the sector \(AOB\). Since inside the sector \(AOB\) there are no integral curves which leave the sector \(AOB\) with both their ends through the wall \(AB\), we have proved
Theorem 3. If \(F(k_0-0)=+\infty\), then there exists a sector whose upper wall is the ray \(k_0x\), inside which every integral curve enters the origin.
2. Behavior of integral curves in the case where \(F(k_0-0)=-\infty\), \(F(k_0+0)=+\infty\). There exists a neighborhood \((k_0-\varepsilon,\,k_0+\varepsilon)\), \(\varepsilon>0\), of the point \(k_0\), such that in \([k_0-\varepsilon,k_0)\) \(F(k)<0\), and in \((k_0,k_0+\varepsilon]\) \(F(k)>0\). It is clear that, for sufficiently small \(x\), in the sector whose side walls are the rays \((k_0-\varepsilon)x\) and \(k_0x\), the right-hand side of the differential equation is negative. Therefore integral curves can enter the origin only in a sector whose side walls are the rays \(k_0x\) and \(k_2x\), where \(k_0<k_2\le k_0+\varepsilon\).
Theorem 4. If \(F(k_0+0)=+\infty\), then there exists a sector whose lower wall is the ray \(k_0x\), such that any integral curve entering this sector through a side wall leaves it through the rear wall; however, on the rear wall there exists at least one point through which there passes an integral curve entering the origin inside this sector.
Proof. Let the side walls of the sector be the rays \(k_0x\) and \(k_2x\), where \(k_2\) is a point taken from the above-mentioned right-hand neighborhood of the point \(k_0\), and let the rear wall of the sector be an arc of a circle with center at the origin of sufficiently small radius, such that on the upper wall one has \(0<x^n\varphi(x)F(k_2)<k_2\), which is possible for sufficiently small \(x\). Take on the upper wall an arbitrary point \(C\) and draw through it a straight line \(CD\) parallel to the \(ox\)-axis. The written inequalities show that the integral curve \(L\) passing through the point \(C\), when continued, enters the sector under consideration, and inside this sector there are no points of the curve \(L\) lying below
the line \(CD\). This means that \(L\), when continued inside the sector, cannot enter the origin or leave this sector through the lower side wall. But \(L\) also cannot leave the sector through the upper side wall. Consequently, it leaves the sector under consideration through the rear wall. It is clear that in doing so \(L\) intersects the rear wall at only one point. It is proved similarly that an integral curve passing through an arbitrary point of the lower wall of the sector leaves the sector under consideration, crossing its rear wall once. We shall call the point of intersection of an integral curve leaving the sector with the rear wall the exit point. Obviously, under continuous motion of a point along the upper wall of the sector toward the origin, the corresponding exit point moves continuously along the rear wall in the clockwise direction. Conversely, under continuous motion of a point along the lower wall of the sector toward the origin, the corresponding exit point moves continuously along the rear wall in the counterclockwise direction. We have two classes of exit points, between which, obviously, there exists a point or an entire arc consisting of points through which pass integral curves entering the origin. The theorem is proved.
When the conditions of the theorem just proved are satisfied, the first problem of distinction arises; it is easily resolved by the following theorem.
Theorem 5. If on the interval \((k_0, k_2)\) \((0<k_0<k_2)\) the function \(F(k)\) does not increase, and if on this interval there are no points at which \(F(k)=\infty\), then in the sector whose side walls are the rays \(k_0x\) and \(k_2x\), no more than one integral curve can enter the origin.
Proof. Suppose that in the indicated sector the integral curves \(y_1(x)\) and \(y_2(x)\) enter the origin. It is not hard to see that for sufficiently small \(x\) we may regard these integral curves as single-valued functions of \(x\), since on the interval \((k_0,k_2)\) the function \(F(k)\) does not become infinite. Let \(y_1(x)>y_2(x)\). We have
\[ k'=\frac{y_1(x)}{x}>\frac{y_2(x)}{x}=k''. \]
Consider the difference \(u(x)=y_1(x)-y_2(x)\). Its derivative is
\[ \frac{du}{dx} = \frac{dy_1}{dx}-\frac{dy_2}{dx} = x^n\varphi(x)\left[ F\left(\frac{y_1}{x}\right)-F\left(\frac{y_2}{x}\right) \right] = \]
\[ = x^n\varphi(x)[F(k')-F(k'')]\le 0. \]
Consequently, for all sufficiently small \(x\le x_0\), \(u'(x)\le 0\). This means that for these \(x\) the function \(u(x)\) does not increase on \((0,x_0]\). Since \(u(0)=0\), it follows that \(u(x)\le 0\) on \([0,x_0]\). But the latter contradicts the fact that \(u(x)>0\) for all small \(x\). The theorem is proved.
Corollary. Under the conditions of Theorem 4, only one integral curve enters the origin in the sector indicated there.
Indeed, since \(F(k_0+0)=+\infty\) and the equation \(F'(k)=0\) can have no more than a finite number of roots, there exists an interval \((k_0,k_2)\) in which \(F(k)\) does not increase. Moreover, the point \(k_2\) is such that the conditions of Theorem 4 are not violated.
3. Behavior of integral curves in the case when \(F(k_0-0)=F(k_0+0)=+\infty\). There exists a neighborhood \((k_0-\varepsilon,\; k_0+\varepsilon)\) of the point \(k_0\), in wh-
for which, for \(k\ne k_0\), \(F(k)>0\). Let \(k_1\in(k_0-\varepsilon,k_0)\), and \(k_2\in(k_0,k_0+\varepsilon)\). Then, for sufficiently small \(x\), on the rays \(k_1x\) and \(k_2x\) the inequalities
\[ 0<x^n\varphi(x)F(k_i)<k_i,\qquad i=1,2. \]
will hold.
It is not difficult to verify that if one constructs a sector whose lateral sides are these rays, then the ray \(k_0x\) divides this sector into two sectors. Moreover, the sector whose upper side is the ray \(k_0x\) satisfies the conditions of Theorem 3, while the sector whose lower side is the ray \(k_0x\) satisfies the conditions of Theorem 4. These theorems, therefore, completely resolve the question of the behavior of integral curves inside the sector under consideration.
§ 2. METHOD OF INVESTIGATING THE BEHAVIOR OF INTEGRAL CURVES IN THE CASE \(n=0\)
If in the preceding case (§ 1) the critical rays were asymptotes at infinity, then in the present case the critical rays are, as follows from Theorem 2, those rays whose angular coefficients are found from the equation \(\varphi(0)F(k)=k\).
Graphically, in the Cartesian coordinate system \((k,0,p)\), the angular coefficients of the critical rays should be sought among the abscissae of the points of intersection or tangency of the curve \(p=\varphi(0)F(k)\) with the bisector \(p=k\).
Let us denote \(R(k)=\varphi(0)F(k)-k\), and let \(k_0\) be such a number that \(R(k_0)=0\). Since the equations \(R(k)=0\) and \(F'(k)=0\) have no more than a finite number of roots, we can always ensure that in some neighborhood of the point \(k_0\) there are no roots of the equations \(R(k)=0\), \(F(k)=0\), and \(F^{-1}(k)=0\). Then one of the following three cases may occur:
a) \(R(k)<0\) in \((k_0-\varepsilon,k_0)\), and in \((k_0,k_0+\varepsilon)\) \(R(k)>0\);
b) \(R(k)>0\) in \((k_0-\varepsilon,k_0)\), and in \((k_0,k_0+\varepsilon)\) \(R(k)<0\);
c) \(R(k)<0\) in \((k_0-\varepsilon,k_0)\) and in \((k_0,k_0+\varepsilon)\) \(R(k)<0\);
\[ R(k)>0\ \text{in }(k_0-\varepsilon,k_0)\ \text{and in }(k_0,k_0+\varepsilon)\ R(k)>0. \]
Corresponding to these three cases, we can classify into three types the sectors constructed about the rays \(k_0x\).
In each of the cases listed, take an arbitrary point \(k_1\) in \((k_0-\varepsilon,k_0)\), and a point \(k_2\) in \((k_0,k_0+\varepsilon)\), and call the sector whose lateral sides are the rays \(k_1x\) and \(k_2x\), respectively for cases a), b), c), a sector of the first, second, and third type.
Let us consider the question of the behavior of integral curves in a sector of each type.
1. Drawing integral curves in a sector of the first type
Theorem 6. Every integral curve passing through a point lying inside a sector of the first type or on its side enters the origin inside this sector.
Proof. Obviously, for all sufficiently small \(x\), on the lower side \(k_1x\) the inequality
\[ 0<\varphi(x)F(k_1)<k_1. \tag{6} \]
holds.
This means that the integral curve \(L\), passing through an arbitrary point \(A\) of the lower side of the sector, enters this sector when continued. Through the point \(A\), draw a straight line parallel to the \(oy\)-axis, which intersects the ray \(k_2x\) at the point \(B\). We shall show that \(L\) cannot leave the sec-
tor \(AOB\). Indeed, it cannot leave this sector either through the wall \(OA\) or through the rear wall \(AB\), since on the lower wall, by inequality (6), there are no points of contact of the ray \(k_1x\) with any integral curve, while on the rear wall \(AB\) there are no points of contact of the segment \(AB\) with any integral curves because, in the sector under consideration, the right-hand side of the differential equation is bounded. Next we shall show that \(L\) cannot leave the sector \(AOB\) through the upper wall \(OB\) either. Suppose, however, that it leaves this sector through a point \(C\) lying on the segment \(OB\). Through the point \(C\) draw a straight line parallel to the axis \(oy\), which intersects the ray \(k_1x\) at the point \(D\). Since at the point \(C\) the inequality \(\varphi(x)F(k_2)>k_2\) holds, the integral curve \(L\), passing through the point \(C\), enters the sector \(COD\). It is now clear that \(L\), passing from the point \(A\) into the interior of the sector \(COD\), must intersect the segment \(CD\) at some interior point of it, which is impossible, since then between this point and the point \(C\) there would exist a point of contact of the segment \(CD\) with some integral curve. Thus the integral curve \(L\) cannot leave the sector \(AOB\); consequently, it enters the origin inside this sector. It is proved analogously that an integral curve passing through an arbitrary point of the upper wall of the sector enters the origin. Since, obviously, there are no integral curves leaving the sector under consideration with both their ends through the rear wall, the theorem is proved.
- Behavior of integral curves in a sector of the second type.
Theorem 7. Every integral curve passing through an arbitrary point of the side wall of a sector of the second type enters this sector and leaves it through the rear wall; however, on the rear wall there exists at least one point through which passes an integral curve entering the origin inside this sector.
The proof of the theorem repeats the proof of Theorem 4 of the preceding section. As there, here too the first problem of discrimination arises. The following holds.
Theorem 8. If on the segment \([k_1,k_2]\) the function \(F(k)\) for any two values \(k'>k''\) satisfies the condition
\[ F(k')-F(k'')\leq N(k'-k''), \tag{a} \]
where
\[ N\leq \frac{1}{\varphi(0)},\qquad \varphi(x)\leq \varphi(0), \tag{b} \]
then into the origin, in a sector of the second type whose side walls are the rays \(k_1x\) and \(k_2x\), there enters no more than one integral curve.
Proof. Suppose that in a sector of the second type at least two integral curves \(y_1(x)\) and \(y_2(x)\) enter the origin with the critical ray \(k_0x\). Let \(y_1(x)>y_2(x)\). Since the ray \(k_0x\) is critical for these integral curves, whatever small number \(\varepsilon>0\) we choose, for all sufficiently small \(x\) the inequalities hold
\[ (k_0-\varepsilon)x<y_i(x)<(k_0+\varepsilon)x,\qquad i=1,2. \tag{c} \]
Consider the expression
\[ \frac{d(y_1-y_2)}{dx} = \varphi(x)\left[ F\left(\frac{y_1}{x}\right) - F\left(\frac{y_2}{x}\right) \right] = \varphi(x)[F(k')-F(k'')]\leq \]
\[ \leq \varphi(x)N(k' - k'') = N\frac{\varphi(x)}{x}(y_1 - y_2) \]
or
\[ \frac{d(y_1 - y_2)}{dx} \leq N\frac{\varphi(x)}{x}(y_1 - y_2), \]
where
\[ k' = \frac{y_1}{x}, \qquad k'' = \frac{y_2}{x}. \]
Integrating the last inequality over the interval from \(x\) to \(x_1\), where \(x_1\) is a sufficiently small number, we obtain
\[ \ln \frac{y_1(x_1)-y_2(x_1)}{y_1(x)-y_2(x)} \leq N\int_x^{x_1}\frac{\varphi(x)}{x}\,dx . \]
In view of the relations (b),
\[ N\int_x^{x_1}\frac{\varphi(x)}{x}\,dx \leq \frac{1}{\varphi(0)}\int_x^{x_1}\frac{\varphi(0)}{x}\,dx = \ln \frac{x_1}{x}. \]
Therefore we have
\[ \frac{y_1(x_1)-y_2(x_1)}{y_1(x)-y_2(x)} \leq \frac{x_1}{x} \]
or
\[ s\cdot x \leq y_1(x)-y_2(x), \qquad s>0. \tag{d} \]
With the aid of one of the inequalities (c) \((i=2)\), we strengthen inequality (d):
\(s\cdot x < y_1(x)-(k_0-\varepsilon)x\), whence
\(y_1(x)>(k_0+s-\varepsilon)x\). For \(\varepsilon<s\), from the inequality
\[ \frac{y_1(x)}{x} > k_0+s-\varepsilon \]
by passing to the limit as \(x\to 0\), we obtain a contradiction to the fact that the ray \(k_0x\) is critical for the integral curve \(y_1(x)\). The theorem is proved.
Corollary. If \(\varphi(x)\leq \varphi(0)\), then in a sector of the second type only one integral curve enters the origin.
Indeed, since on the half-segment \([k_1,k_0)\) \(\varphi(0)F(k)>k\), while on \((k_0,k_2]\) \(\varphi(0)F(k)<k\) and \(\varphi(0)F(k_0)=k_0\), it follows that \(\varphi(0)F'(k_0)\leq 1\). Since the equation \(F'(k)=1\) can have no more than a finite number of roots, in a sufficiently small neighborhood of the point \(k_0\)
\[ \varphi(0)F'(k)<1, \qquad F'(k)<\frac{1}{\varphi(0)}. \]
For any two points \(k'>k''\) from this neighborhood, by Lagrange’s theorem,
\[ F(k')-F(k'')=F'(\bar{k})(k'-k''), \]
where \(k''<\bar{k}<k'\) and \(F'(\bar{k})<\dfrac{1}{\varphi(0)}\).
Remark. If at the point \(k_0\) \(F'(k_0)<1\), then the theorem will remain true if the restriction \(\varphi(x)\leq\varphi(0)\) is removed.
3. Behavior of integral curves in a sector of the third type
Obviously, for sufficiently small \(x\), on the lateral sides of a sector of the third type the following relations hold:
\[ 0<\varphi(x)F(k_i)<k_i \quad \text{or} \quad \varphi(x)F(k_i)>k_i \quad (i=1,2). \]
Theorem 9. In a sector of the third type, either all integral curves passing through one of its lateral sides enter the origin, or not a single integral curve enters the origin inside this sector.
Proof of this theorem is based on the fact that inside the sector there are no points at which the direction of the field is parallel to the \(ox\)-axis.
In a sector of the third type there arises a second problem of distinction. The following theorem solves this problem under those assumptions concerning the right-hand side of the differential equation which were listed at the beginning of the article.
Theorem 10. In a sector of the third type, an infinite set of integral curves enters the origin.
Proof. The case when on the lateral sides the inequalities
\[ 0<\varphi(x)F(k_i)<k_i,\quad i=1,2 \]
hold. Consider the behavior of the integral curves in the sector whose lateral sides are the rays \(k_1x\) and \(k_0x\). In this sector, for all points \((x,y)\) the relations \(k_1\leq k\leq k_0\) hold, where
\[ k=\frac{y}{x}. \]
Subject the original differential equation to the transformation \(y=\bar{k}x\). We obtain
\[ x\frac{d\bar{k}}{dx}=\varphi(x)F(\bar{k})-\bar{k} \]
or
\[ \frac{dx}{d\bar{k}}=\frac{x/\varphi(x)}{F(\bar{k})-\bar{k}/\varphi(x)}. \]
Obviously, the axis \(x=0\) is a solution of this equation. Since for \(\bar{k}=k_0\) the denominator of the right-hand side of this equation vanishes together with \(x\), the point \((k_0,0)\) is a singular point for this equation. Make the shift \(\bar{k}=k+k_0\); the equation takes the form
\[ \frac{dx}{dk}= \frac{\dfrac{x}{\varphi(x)}}{F(k+k_0)-\dfrac{k+k_0}{\varphi(x)}}. \]
For this equation the singular point is the origin. Apply once more the Briot and Bouquet transformation \(x=kz\); the latter equation becomes
\[ \frac{dz}{dk}= \frac{\dfrac{kz}{\varphi(kz)}-z\left[F(k+k_0)-\dfrac{k+k_0}{\varphi(kz)}\right]} {k\left[F(k+k_0)-\dfrac{k+k_0}{\varphi(kz)}\right]}. \tag{7} \]
Obviously, the axis \(z=0\) is a solution of this equation, since for
\[ z=0 \qquad F(k+k_0)-\frac{k+k_0}{\varphi(0)}\neq 0. \]
We shall show that for sufficiently small
\(z\) and \(k \to 0\) the right-hand side of this equation tends to infinity. Using L’Hôpital’s theorem, we find
\[ \lim_{k\to 0} \frac{ F(k+k_0)-\dfrac{k+k_0}{\varphi(kz)} }{k} = \]
\[ = \lim_{k\to 0} \left\{ F'(k+k_0)-\frac{1}{\varphi(kz)} -(k+k_0)\frac{\partial}{\partial k} \left[\frac{1}{\varphi(kz)}\right] \right\} = \]
\[ = -k_0\frac{\partial}{\partial k} \left[\frac{1}{\varphi(kz)}\right]_{k=0} = k_0z\,\frac{\varphi'(0)}{[\varphi(0)]^2}. \]
Since \(z\) can be arbitrarily small, the numerator of the right-hand side of equation (7) does not vanish for small \(z\). Consequently, the axis \(k=0\) is a solution of equation (7). Under the transformations carried out, the sector considered at the beginning of the proof passed into a rectangle lying in the third quadrant, two sides of which coincide with the axes \(ok\) and \(oz\). In fact, first there was the domain \(k_1 \leq k \leq k_0,\ 0 \leq x \leq x_1\); then \(k=k-k_0 \leq 0,\ 0 \leq x \leq x_1\); finally, under the third transformation, one obtains the rectangle \(k' \leq k \leq 0,\ z' \leq z \leq 0\), since \(x=kz\) and \(x \geq 0\), while \(k \leq 0\).
Further, since in the case under consideration, for sufficiently small \(k\),
\[ \varphi(0)F(k+k_0)-(k+k_0)<0 \quad\text{and}\quad \frac{1}{\varphi(0)}>0, \]
then for sufficiently small \(k\) and \(z\)
\[ F(k+k_0)-(k+k_0)\frac{1}{\varphi(kz)}<0 \]
and the right-hand side of equation (7) becomes positive. This means that in the indicated rectangle there is a region adjacent to the axis \(ok\) in which the right-hand side of equation (7) is positive. Suppose that the integral curve \(L\), passing through some point of this region, does not intersect the axes \(ok\) and \(oz\), and suppose that this curve does not enter the origin. Then \(L\) must leave the region under consideration, crossing the zero isocline. But at the points of the zero isocline the right-hand side of equation (7) vanishes, i.e.
\[ \frac{1}{\varphi(kz)} - \frac{ F(k+k_0)-\dfrac{k+k_0}{\varphi(kz)} }{k} =0. \]
Since, in the left-hand side of this equality, the subtrahend for sufficiently small \(k\) and \(z\) can become arbitrarily small, as we established above, the whole expression on the left-hand side, for sufficiently small \(k\) and \(z\), cannot be equal to zero. Consequently, the integral curve \(L\) either enters the origin or intersects one of the coordinate axes (or merges with it). But it is not difficult to verify that the partial derivative with respect to \(z\) of the right-hand side of the equation is bounded at the points of the axis \(z=0\), and also that, at the points of the axis \(k=0\), the partial derivative with respect to \(k\) of the right-hand side of the equation
\[ \frac{dk}{dz}=\frac{1}{f(k,z)}. \]
where \(f(k,z)\) is the right-hand side of equation (7). Thus, on the coordinate axes the uniqueness conditions are fulfilled. Therefore the integral curve \(L\) enters the origin inside the indicated region, which corresponds to the fact that there exists an integral curve that enters the origin inside the sector considered at the beginning of the proof.
The case when on the lateral sides of the sector the inequality
\[ \varphi(x)F(k_i)>k_i,\quad i=1,2. \]
holds is proved analogously.
The theorem is proved.
§ 3. BEHAVIOR OF INTEGRAL CURVES IN SECTORS WHOSE LATERAL SIDE IS A SEMI-AXIS OF THE COORDINATE SYSTEM
1. The case when the lateral side of the sector is the positive semi-axis of abscissas.
Here a number of cases may arise, depending on the behavior of the function \(F(k)\) near the point \(k=0\) and on the number \(n\). Consideration of all possible cases is exhausted by the proofs of a series of the following theorems.
Theorem 11. If the differential equation satisfies the conditions 1) \(n>0\), 2) \(F(0)=0\), and 3) for small \(k>0\), \(F(k)>0\), then there exists a sector whose lower side is the semi-axis \(ox\), inside which no integral curve enters the origin.
Proof. Obviously, in this case the axis \(ox\) is a solution of the differential equation. Suppose that inside the sector mentioned some integral curve enters the origin. Since for small \(k\), \(F(k)>0\), there exists a segment \([0,k_2]\) on which the function \(F(k)\) will not have infinitely many discontinuities; if this were not so, the equation \(F'(k)=0\) would have infinitely many roots. Hence, the critical ray for \(L\) is the semi-axis \(ox\). This means that, whatever the ray \(kx\), there always exists such a neighborhood of the origin within which the curve \(L\) lies below this ray. Take on \(L\), sufficiently close to the origin, an arbitrary point \(A\), and draw through it the ray \(k_a x\). Let \(A\) move along \(L\) toward the origin. Obviously, in this process \(k_a\to 0\). The equality
\[ x^n\varphi(x)=\frac{k}{F(k)} \tag{8} \]
determines the abscissa of the point of contact of some integral curve with the ray whose angular coefficient is \(k\). Since \(F(0)=0\) and, for small \(k\), \(F(k)>0\), it follows that \(F'(k)>0\); if this were not so, the equation \(F'(k)=0\) would have infinitely many roots. By L’Hôpital’s theorem we have
\[ \lim_{k\to 0}\frac{k}{F(k)} = \lim_{k\to 0}\frac{1}{F'(k)} = \frac{1}{F'(0)}>0. \]
This means that there exists a neighborhood \(\delta(0,0)\) of the origin within which, for sufficiently small \(x\), equation (8) will have no solutions. Consequently, in this neighborhood there are no points of contact of the rays \(kx\) with integral curves. In its motion along the curve \(L\) toward the origin, the point \(A\) at some moment enters the neighborhood \(\delta(0,0)\). Let the point \(A\) be in this neighborhood. In view of what was said above, the arc \(OA\) of the integral curve \(L\), obviously, lies below the ray \(k_a x\). Take on the segment \(OA\) of the ray \(k_a x\) an arbitrary point \(M\). The integral curve
\(L_0\), passing through this point, upon being continued enters the region bounded by the segment \(OA\) and the arc \(OA\) of the curve \(L\). We note that \(L_0\) cannot enter the origin, but at the same time it cannot leave the indicated region by crossing the segment \(OA\) a second time, since then on the segment \(OA\) there would exist a point of contact of some integral curve with the ray \(k_{\alpha}x\). The contradiction obtained proves the theorem.
Theorem 12. If \(F(0)>0\), \(n>0\), then there exists a sector whose lower wall is the half-axis \(ox\), and into which only one integral curve enters the origin.
Proof. Since \(F(k)\) is continuous in the open domain of its definition, there exists a half-interval \((0,k_2]\) in which \(F(k)>0\). Consequently, on the upper wall of the sector, for all sufficiently small \(x\), we have
\[ 0<x^n\varphi(x)F(k_2)<k_2, \tag{9} \]
and on the lower wall (the \(ox\)-axis) \(x^n\varphi(x)F(0)>0\). It can be proved that any integral curve entering this sector through its lateral wall leaves it through the rear wall (see Theorem 7, § 2); however, on the rear wall there exists at least one point through which passes an integral curve entering the origin inside this sector. Consequently, the first separation problem applies. Suppose that inside the sector under consideration at least two integral curves \(y_1(x)\) and \(y_2(x)\) enter the origin, and let \(y_1(x)>y_2(x)\). The derivative is
\[ \frac{d(y_1-y_2)}{dx}=x^n\varphi(x)[F(k')-F(k'')], \]
where \(k'=\dfrac{y_1}{x}\), \(k''=\dfrac{y_2}{x}\) are points from \((0,k_2]\). Since on the segment \([k'',k']\) \(F(k)\) is differentiable, by Lagrange’s theorem we have
\[ \frac{d(y_1-y_2)}{dx} = x^n\varphi(x)F'(c)(k'-k'') = x^n\varphi(x)F'(c)\frac{y_1-y_2}{x}, \]
where \(k''<c<k'\). Assuming that \(F'(k)\) is continuous on \([0,k_2]\), and denoting \(\sup F'(k)=N\) on \([0,k_2]\), we have
\[ \frac{d(y_1-y_2)}{y_1-y_2}\leq N\cdot\varphi(x)x^{\,n-1}\,dx. \]
The theorem is proved by virtue of the convergence of the integral
\[ \int_x^{x_1}\varphi(x)x^{\,n-1}\,dx \quad \text{as } x\to 0. \]
Theorem 13. If \(n>0\), \(F(0+0)=+\infty\), then there exists a sector whose lower wall is the half-axis \(ox\), and into which only one integral curve enters the origin.
Proof. For sufficiently small \(k_2>0\), \(F(k)>0\) in \((0,k_2]\), and on the ray \(k_2x\), for all sufficiently small \(x\), inequality (9) holds (see Theorem 12). On the half-axis \(ox\), for small \(x\), the directions of the field will be parallel to the \(oy\)-axis. It can be proved that in the sector whose lateral walls are the ray \(k_2x\) and the half-axis \(ox\), at least one integral curve enters the origin. Since \(F(0+0)=+\infty\) and
the equation \(F'(k)=0\) can have no more than a finite number of roots, there exists an interval \((0,k_1)\) within which \(F(k)\) decreases monotonically with respect to \(k\). Consequently, by Theorem 5, § 1, in the sector under consideration only one integral curve enters the origin. The theorem is proved.
Theorem 14. If \(n=0\), \(F(0)=0\), and on some interval \((0,k_2)\) one has \(F(k)>k\), then in any sector whose lower wall is the half-axis \(ox\), an infinite set of integral curves enters the origin.
Proof. Since \(F(0)=0\), the axis \(y=0\) is a solution of the differential equation. Since \(F(k)>k\), for sufficiently small \(x\) on the ray \(k_2x\) one will have \(\varphi(x)F(k_2)>k_2\). This means that the integral curves passing through points of the upper wall of the sector enter this sector and cannot leave it either through the rear wall or through the side walls. Consequently, all such integral curves enter the origin inside the sector.
Theorem 15. If \(n=0\), \(F(0)=0\), and for all sufficiently small \(k\), \(0<F(k)<k\), \(0<k<k_2\), then inside the sector whose side walls are the half-axis \(ox\) and the ray \(k_2x\), no integral curve enters the origin.
Proof. Suppose, on the contrary, that in the indicated sector some integral curve \(L\) enters the origin. Then it is not hard to see that every integral curve passing through a point lying between \(ox\) and \(L\) enters the origin. On the other hand, since \(F(0)=0\) and in \((0,k_2)\) \(F(k)<k\), in a sufficiently small neighborhood of the point \(k=0\) one has \(F'(k)\leqslant 1\). Moreover, for small \(x\), \(\varphi(x)\leqslant \varphi(0)\). Thus the conditions of the corollary to Theorem 8, § 2, are satisfied. Hence, in the sector under consideration, no more than one integral curve enters the origin. The contradiction obtained proves the theorem.
Theorem 16. If \(n=0\), on the segment \([0,k_2]\) one has \(F(k)>k\), where \(k_2\) is a sufficiently small number, then in the sector whose side walls are the half-axis \(ox\) and the ray \(k_2x\), not a single integral curve enters the origin.
Proof. Suppose that in the indicated sector some integral curve \(L\) enters the origin. Since on any segment \([k'',k']\subset(0,k_2)\) one has \(F(k)>k\), by Theorem 2 the critical ray for \(L\) is the half-axis \(ox\). Take an arbitrary point \(A\) on \(L\) and draw through it the ray \(k_\alpha x\). Obviously, as \(A\) moves along the curve \(L\) toward the origin, \(k_\alpha\to0\). It is not hard to verify that on the arc \(OA\) of the curve \(L\) there exists a point at which the tangent to the curve has slope \(k_\alpha\). As \(A\) moves along \(L\) toward the origin, there exists a sequence of points on the curve \(L\), converging to the origin, at which the tangents to the curve have slopes \(k_\alpha\). Hence, at all points of this sequence,
\[ \varphi(x)F\left(\frac{y}{x}\right)=k_\alpha, \]
where \(k_\alpha\to0\) as the point \(A\) approaches the origin. But the last equality cannot hold because \(\varphi(0)>0\) and \(F(k)>0\) on \([0,k_2]\). The theorem is proved.
2. The case when the side wall of the sector is the half-axis of ordinates.
Theorem 17. If \(n>0\), \(x=0\) is a solution of the differential equation, on some half-interval \([k_1,\infty)\) one has \(F(k)>0\), and on the ray \(k_1x\), for all sufficiently small \(x\), \(0<x^n\varphi(x)F(k_1)<k_1\), then any inte-
an integral curve passing through the point of the ray \(k_1x\) at which the indicated inequality is satisfied enters the origin.
Proof. For sufficiently small \(x\) on the ray \(k_1x\), take a point \(A\) and draw through it a straight line parallel to the axis \(ox\), which intersects the axis \(oy\) at the point \(B\). The integral curve \(L\) passing through the point \(A\), when continued, enters the triangle \(AOB\) and can leave it only through the side \(AB\). But for this it is necessary that on \(AB\) there exist a point at which the right-hand side of the differential equation is equal to zero, which is impossible, since for small \(x\), \(\varphi(x)>0\), and on \([k_1,\infty)\), \(F(k)>0\). The theorem is proved.
Theorem 18. If on some half-interval \((0,y_1]\), which is part of the axis \(oy\), the right-hand side of the differential equation is bounded and, for sufficiently small \(x\), on every ray \(kx\) \((k_1\le k<\infty)\) the inequality \(x^n\varphi(x)F(k)<k\) holds, then there exists a sector, one of whose lateral sides is the semiaxis \(oy\), inside which there are no integral curves entering the origin.
Proof. Suppose, on the contrary, that in every sector whose lateral side is the semiaxis \(oy\), there is an integral curve entering the origin. Inside this sector, for small \(x\), the right-hand side of the differential equation is positive. In view of the accepted assumption, the semiaxis \(oy\) is tangent at the origin to some integral curve entering there. In this case it is easy to see that in any of the indicated sectors the right-hand side of the equation is unbounded. Let us prove that this cannot occur under the conditions given in the theorem. Draw through the point \(y_1\) a straight line parallel to the axis \(ox\), which intersects the ray \(k_1x\) at the point \(A\). We shall prove that in the triangle \(AOy_1\) the right-hand side of the differential equation is bounded. In the case when \(n=0\), this is obvious, since in this case one would have
\[ 0<\left|\lim_{k\to+\infty} F(k)\right|<\infty . \]
Let \(n>0\). It is clear that on the segment \(Ay_1\) the right-hand side of the equation is bounded. If this were not so, then in any arbitrarily small neighborhood of the point \(y_1\) on the segment \(Ay_1\) it would be bounded, which contradicts the continuity of the field of directions. Let \(M\) be an upper bound of the values of the function \(x^n\varphi(x)F\left(\dfrac{y}{x}\right)\) on the segment \(oy_1\), and \(C\) an upper bound of the function \(\varphi(x)\). Then on any ray \(kx\), \(k_1\le k<\infty\), inside the triangle \(AOy_1\), the right-hand side of the equation will be bounded by the number \(CM\). Suppose that within the segment \(oy_1\) the quantity under consideration is unbounded. Then there exists there a point \(S\) at which the right-hand side of the equation exceeds the number \(CM\) by some quantity \(N>0\). Since inside the triangle \(AOy_1\) the right-hand side of the equation is not greater than \(CM\), while at the point \(S\) the right-hand side of the equation exceeds the number \(CM+N\), we have a contradiction to the continuity of the field of directions. The theorem is proved.
Theorem 19. If \(n=0\), the axis \(x=0\) is a solution of the differential equation, on some half-interval \([k_1,\infty)\), \(F(k)>k\), and
\[ \lim_{k\to+\infty}\frac{k}{F(k)}\ne \varphi(0), \]
then in the sector whose lower side is the ray \(k_1x\), there are no integral curves entering the origin.
Proof. Suppose, on the contrary, that some integral curve \(L\) enters the origin, for which the semiaxis \(oy\) is tangent at the origin. Take on \(L\), sufficiently close to the origin, an arbitrary point \(A\) and draw through it the ray \(k_\alpha x\). Let the point \(A\) be the nearest of the points of intersection of \(L\) with the ray \(k_\alpha x\). It is clear that the arc \(OA\) of the curve \(L\) lies above the segment \(OA\) of the ray \(k_\alpha x\). For sufficiently small \(x\) on the segment \(OA\) the inequality \(k_\alpha < \varphi(x)F(k_\alpha)\) holds. This means that the integral curve \(\Gamma\), passing through one of such points of the segment \(OA\), upon being continued enters the region bounded by the arc \(OA\) of the curve \(L\) and the segment \(OA\) of the ray \(k_\alpha x\). At the same time \(\Gamma\) cannot enter the origin inside this region, since for this it would have to pass through a point of the region at which the right-hand side of the differential equation is equal to zero. Consequently, \(\Gamma\) leaves the region under consideration, intersecting the ray \(k_\alpha x\) a second time. But then on the ray \(k_\alpha x\) there exists a point of contact of some integral curve with the ray \(k_\alpha x\), at which
\[ \varphi(x)F(k_\alpha)=k_\alpha, \]
which is impossible for sufficiently small \(x\) and sufficiently large \(k_\alpha\). The theorem is proved.
In conclusion, we note that the method presented is applied analogously in the case when \(n<0\), and is transferred almost automatically to the case when the equation considered is
\[ y'=v(x)F\left[\frac{y}{u(x)}\right], \]
where \(u(x)\) is a function nonoscillating for small \(x\) and differentiable, \(u(0)=0\), while \(v(x)\) is a function nonoscillating for small \(x\) and differentiable everywhere except, possibly, at the origin.
References
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Received by the editors
November 9, 1964
Barnaul State
Pedagogical Institute