FUNCTIONAL
DIFFERENTIAL
EQUATIONS
VOLUME 23
2016, NO 1–2
PP. 3–8
ON ASYMPTOTIC BEHAVIOR OF SOLUTIONS TO THE
SECOND-ORDER EMDEN–FOWLER TYPE DIFFERENTIAL
EQUATIONS WITH UNBOUNDED NEGATIVE POTENTIAL
K. M. DULINA
∗
Abstract. In this paper second-order Emden–Fowler type differential equations with
regular nonlinearity are considered. Asymptotic behavior of non-extensible solutions to
equations near the domain boundaries is described.
Key Words. Ordinary second-order Emden–Fowler type differential equations, regular nonlinearity, asymptotic behavior.
AMS(MOS) subject classification. 34C11, 34E10
1. Introduction.
differential equation
(1)
Consider the second-order Emden–Fowler type
y 00 − p(x, y, y 0 ) |y|k sgn y = 0,
k > 1,
where the function p(x, u, v) defined on R × R2 is positive, continuous in x,
Lipschitz continuous in u, v.
Asymptotic behavior of all solutions to equation (1) in the case p = p(x)
was described by I.T. Kiguradze and T.A. Chanturia (see [1]). Results on
asymptotic classification of non-extensible solutions to third- and fourthorder similar differential equations for k > 0, k 6= 1 were given by I.V. Astashova (see [2, 3, 4]). Asymptotic classification of solutions to equation (1)
with regular (k > 1) and singular (0 < k < 1) nonlinearity for the bounded
function p(x, u, v) is contained in [5, 6].
In this paper asymptotic behavior of non-extensible solutions to equation
(1) for the unbounded function p(x, u, v) is considered. By using methods
∗
Lomonosov Moscow State University, Leninskiye gory, 1
3
4
K.M. DULINA
described in [4, 7, 8], conditions on the function p(x, u, v) and initial data
are obtained such that the related solutions have a vertical asymptote. Other
sufficient conditions on p(x, u, v) and the initial data provide
lim |y 0 (x)| = +∞,
x→a
lim |y(x)| < +∞,
x→a
where a < ∞ is a boundary point.
2. Preliminary results.
The following statement is used to prove
the main results of this paper.
Lemma 1. Suppose the function p(x, u, v) is continuous in x, Lipschitz continuous in u, v, and there exists a constant m > 0 such that p(x, u, v) ≥ m.
Let y(x) be a non-constant non-extensible solution to equation (1) satisfying
the condition y(x0 ) y 0 (x0 ) > 0 (respectively, y(x0 ) y 0 (x0 ) < 0) at some point
x0 . Then there exists x∗ ∈ (x0 , +∞) (respectively, x∗ ∈ (−∞, x0 )), such that
0
0
lim
|y (x)| = +∞
respectively, lim |y (x)| = +∞ .
∗
x→x −0
x→x∗ +0
If y(x0 )y 0 (x0 ) = 0, then there exist finite x∗ and x∗ , x0 ∈ (x∗ , x∗ ), such that
lim |y 0 (x)| = lim |y 0 (x)| = +∞.
x→x∗ −0
x→x∗ +0
Proof. Obviously follows from the results in [8, pp. 20–23].
3. Asymptotic behavior of solutions to Emden–Fowler type differential equations near the domain boundaries.
Using the substitutions x 7→ −x, y(x) 7→ −y(x) we obtain an equation of the same type
as (1). That is why we investigate asymptotic behavior of non-extensible
solutions to equation (1) near the right boundaries of their domains only.
Theorem 1. Suppose there exist constants u0 > 0, v0 > 0 such that for
u > u0 , v > v0 the function p = p(x, u, v) has the representation p = h(u)g(v),
where the functions h(u), g(v) are continuous and bounded below by a positive constant. Then for any non-extensible solution y(x) to equation (1) such
that y 0 (x) → +∞ as x → x∗ − 0, the line x = x∗ is a vertical asymptote if
and only if
Z+∞
(2)
v
dv = +∞.
g(v)
v0
Proof. Let y(x) be a non-extensible solution to equation (1) with the initial
data y(x0 ) = y0 ≥ u0 > 0, y 0 (x0 ) = y1 ≥ v0 > 0. Then from (1) we have
y 00 = h(y) g(y 0 ) y k .
5
ON ASYMPTOTIC BEHAVIOR
Separate the variables and integrate the above expression on the interval
(x0 , x), where x < x∗ :
y 0 y 00
= h(y) y k y 0 ,
g(y 0 )
Zx 0
Zx
y (x)y 00 (x)
dx = h(y) y k (x) y 0 (x) dx,
0
g(y )
x0
x0
y 0 (x)
Z
y0
dy 0 =
0
g(y )
y1
Zy(x)
h(y) y k dy.
y0
0
According to Lemma 1, y (x) → +∞ as x → x∗ − 0, hence
Z+∞
lim
x→x∗ −0
Z
v
dv =
g(v)
y1
y(x)
h(y) y k dy.
y0
First we prove sufficiency. If the integral from the left side of the above
equality diverges, then we have
lim
x→x∗ −0
y(x)
Z
h(y) y k dy = +∞.
y0
Suppose
lim y(x) < +∞. Since the function h(y) is continuous, the
x→x∗ −0
integrand is bounded above. Therefore the integral of the bounded function
on the finite interval converges and we obtain a contradiction. So, we have
y(x) → +∞ as x → x∗ − 0, i.e., x = x∗ is a vertical asymptote.
Further we prove necessity. If the integral from the left side of (2) converges, then we have
lim
x→x∗ −0
y(x)
Z
h(y) y k dy < +∞.
y0
Since the function h(y) is bounded below by a positive constant, there exists
a constant m > 0 such that h(y) ≥ m, whence
lim
x→x∗ −0
Z
y(x)
lim
x→x∗ −0
k
Z
h(y) y dy ≥ m
y0
y0
y(x)
m
y dy =
k+1
k
lim y
x→x∗ −0
k+1
(x) −
y0k+1
.
6
K.M. DULINA
So,
m
k+1
therefore,
lim y
k+1
x→x∗ −0
(x) −
y0k+1
< +∞,
lim y(x) < +∞.
x→x∗ −0
Theorem 2. Suppose there exist constants u0 > 0, v0 > 0 such that for
u > u0 , v > v0 the function p(x, u, v) satisfies the conditions of Lemma 1,
and the inequality p(x, u, v) ≤ f (x, u) g(v) is satisfied, where the function
f (x, u) is continuous, the function g(v) is continuous, bounded below by
+∞
R dv
diverges. Then for any nona positive constant, and the integral
g(v)
v0
extensible solution y(x) to equation (1) such that y 0 (x) → +∞ as x → x∗ −0,
the line x = x∗ is a vertical asymptote.
Proof. Let y(x) be a non-extensible solution to equation (1) with the initial
data y(x0 ) = y0 ≥ u0 > 0, y 0 (x0 ) = y1 ≥ v0 > 0. Then from (1) we have
y 00 ≤ f (x, y) g(y 0 ) y k .
Similarly to the proof of Theorem 1 we obtain
Z+∞
dv
≤
g(v)
Zx∗
f x, y(x) y k (x) dx.
x0
y1
The integral from the left side of the above equality diverges, whence we have
Zx∗
f x, y(x) y k (x) dx = +∞.
x0
Suppose
lim y(x) < +∞. Since the function f (x, u) is continuous,
x→x∗ −0
the integrand is bounded above. The integral of the bounded function on
the finite interval converges and we obtain a contradiction. So, we have
y(x) → +∞ as x → x∗ − 0, i.e., x = x∗ is a vertical asymptote.
Corollary 1. Let the function p = p(x, u) satisfy the conditions of Lemma 1.
Then for any non-extensible solution y(x) to equation (1) such that y 0 (x) →
+∞ as x → x∗ − 0, the line x = x∗ is a vertical asymptote.
Proof. Indeed, it is sufficient to consider f (x, u) = p(x, u), g(v) ≡ 1. Then
+∞
+∞
R dv
R
=
dv = +∞, the function p(x, u) satisfies the conditions of
g(v)
v0
v0
Theorem 2, and this concludes the proof of the corollary.
7
ON ASYMPTOTIC BEHAVIOR
Corollary 2. Suppose there exist constants u0 > 0, v0 > 0, C1 > 0 such
that for u > u0 , v > v0 the function p(x, u, v) satisfies the conditions of
Lemma 1, and the inequality p(x, u, v) ≤ C1 g(v) is satisfied, where the func+∞
R dv
diverges. Then for any nontion g(v) is continuous and the integral
g(v)
v0
extensible solution y(x) to equation (1) such that y 0 (x) → +∞ as x → x∗ −0,
the line x = x∗ is a vertical asymptote.
Proof. Similarly to the proof of Theorem 2.
Theorem 3. Suppose there exist constants u0 > 0, v0 > 0, C2 > 0 such that
for u > u0 , v > v0 the function p(x, u, v) satisfies the conditions of Lemma 1,
and the inequality p(x, u, v) ≥ C2 g(v) is satisfied, where the function g(v)
+∞
R dv
is continuous, bounded below by a positive constant and the integral
g(v)
v0
converges. Then any non-extensible solution y(x) to equation (1) with the
initial data y(x0 ) ≥ u0 , y 0 (x0 ) ≥ v0 , such that y 0 (x) → +∞ as x → x∗ − 0,
satisfies
0 < lim
y(x) < +∞
∗
x→x −0
and
1
x − x0 <
C2 y k (x0 )
Z+∞
∗
dv
.
g(v)
y 0 (x0 )
Proof. Let y(x) be a non-extensible solution to equation (1) with the initial
data y(x0 ) = y0 ≥ u0 > 0, y 0 (x0 ) = y1 ≥ v0 > 0. Then from (1) we have
y 00 ≥ C2 g(y 0 )y k .
Similarly to the proof of Theorem 1 we obtain
Z+∞
dv
≥ C2
g(v)
Zx∗
C2
y (x) dx =
k+1
k
lim y
x→x∗ −0
k+1
(x) −
y0k+1
.
x0
y1
The integral from the left side of the above equality converges, whence we
have
C2
k+1
k+1
< +∞
lim y (x) − y0
k + 1 x→x∗ −0
and lim
y(x) < +∞. Further,
∗
x→x −0
Z+∞
y1
dv
≥ C2
g(v)
Zx∗
x0
y k (x) dx ≥ C2 y0k (x∗ − x0 ).
8
K.M. DULINA
Since the integral from the left side of the above equality converges, we can
estimate the distance between x0 and x∗ :
1
x∗ − x0 <
C2 y0k
Z+∞
dv
,
g(v)
y1
where y0 = y(x0 ), y1 = y 0 (x0 ).
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