Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, Florida, USA. Division of Mathematics, National University of Ireland, Galway, ...
Differential Equations, Vol. 41, No. 5, 2005, pp. 739–743. Translated from Differentsial’nye Uravneniya, Vol. 41, No. 5, 2005, pp. 702–705. c 2005 by Agarwal, Perera, O’Regan. Original Russian Text Copyright
SHORT COMMUNICATIONS
Positive Solutions of Higher-Order Singular Problems R. P. Agarwal, K. Perera, and D. O’Regan Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, Florida, USA Division of Mathematics, National University of Ireland, Galway, Ireland Received November 8, 2004
1. INTRODUCTION Consider the boundary value problem (−1)m y (2m) = f (t, y),
0 < t < 1; i = 0, . . . , m − 1,
y (i) (0) = y (i) (1) = 0,
(1.1)
where m ≥ 1 and f : [0, 1] × (0, +∞) → (0, +∞) is a continuous function satisfying the inequality (2m)!ε ≤ f (t, y) ≤ Cy −γ ,
(t, y) ∈ (0, 1) × (0, ε),
(1.2)
for some constants ε > 0, C > 0, and γ ∈ (0, 1/m). Condition (1.2) is satisfied if, say, f (t, y) = y −γ + g(t, y), where g : [0, 1] × [0, +∞) → [0, +∞) is a continuous function. Let y + = max{0, y} and + (1.3) fε (t, y) = f t, (y − ϕε (t)) + ϕε (t) , where ϕε (t) = εtm (1 − t)m is the solution of the problem (−1)m y (2m) = (2m)!ε,
0 < t < 1; i = 0, . . . , m − 1.
y (i) (0) = y (i) (1) = 0, Along with (1.1), we consider the auxiliary problem (−1)m y (2m) = fε (t, y), (i)
0 < t < 1; i = 0, . . . , m − 1,
(i)
y (0) = y (1) = 0,
(1.4)
where, as follows from (1.2) and (1.3), the function fε satisfies the condition (2m)!ε ≤ fε (t, y) ≤ C (ϕε (t))
−γ
(t, y) ∈ (0, 1) × (−∞, ε).
,
(1.5)
Let us show that if y is a solution of problem (1.4), then y(t) ≥ ϕε (t),
0 < t < 1,
(1.6)
and hence y is a solution of problem (1.1). Suppose the contrary: y (t0 ) < ϕε (t0 )
(1.7)
for some t0 ∈ (0, 1). By the Agarwal–O’Regan Theorem 1.2 in [1], we have y(t) ≥ tm (1 − t)m |y|0 ,
t ∈ [0, 1],
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where |y|0 = max{|y(t)| : 0 ≤ t ≤ 1}. Therefore, it follows from (1.7) that |y|0 < ε. Then, with regard to (1.5), we obtain (−1)m y (2m) (t) ≥ (2m)!ε = (−1)m ϕ(2m) (t), ε
0 < t < 1,
and consequently, y (t0 ) ≥ ϕε (t0 ), which contradicts (1.7). This completes the proof of the estimate (1.6). In a similar way, we can show that each positive solution of problem (1.1) satisfies inequality (1.6) and hence is a solution of problem (1.4). ∈ L(0, 1), it follows from (1.5) that the solutions of problem (1.4) are critical points Since ϕ−γ ε of the functional 1 1 (m) 2 y (t) − Fε (t, y(t)) dt, y ∈ H = H0m (0, 1), (1.8) Φ(y) = 2 where Fε (t, y) =
y ε
0
fε (t, x)dx and H0m (0, 1) is the Sobolev space with norm 1/2 1 (m) 2 y = y (t) dt . 0
The aim of the present paper is to prove the existence of a positive solution of problem (1.1) with the use of this variational approach. We shall show that, under additional conditions on the behavior of f at infinity, the functional Φ satisfies the Palais–Smale compactness condition: an arbitrary sequence {yj } of elements of H such that lim Φ (yj ) = 0 (1.9) sup {|Φ (yj )| : j = 1, 2, . . . , n} < +∞, j→+∞
contains a convergent subsequence. Detailed information can be found in [2] (singular problems) and [3] (variational methods). 2. EXISTENCE PRINCIPLE We say that condition (f ) is satisfied if there exists a positive constant M such that, for an arbitrary λ ∈ (0, 1], each positive solution y of the boundary value problem (−1)m y (2m) = λf (t, y),
0 < t < 1;
y (i) (0) = y (i) (1) = 0,
i = 0, . . . , m − 1,
(2.1)
satisfies the inequality y = M . Proposition 2.1. If conditions (1.2) and (f) are satisfied, then problem (1.1) has at least one positive solution. Proof. Let ε > 0 be so small that
Cε1−γ < M 2 .
(2.2)
˚ = {y ∈ H : y < M } and B = {y ∈ H : y ≤ M }. Then inf Φ(B) > −∞ by (1.8). Let B Since problems (1.1) and (1.4) are equivalent, to prove the proposition, it suffices to show that ˚ inf Φ(B) is attained at some point of the open ball B. Let {yj } be a minimizing sequence. Without loss of generality, we can assume that {yj } converges to some y0 ∈ B weakly in H and strongly in C(0, 1). Then 1 2 Φ (y0 ) = y0 − 2
1
1 2 Fε (t, y0 (t)) dt ≤ lim inf yj − lim j→∞ 2 j→∞
0
1 Fε (t, yj (t)) dt 0
= lim Φ (yj ) = inf Φ(B), j→∞
and hence Φ (y0 ) = inf Φ(B). DIFFERENTIAL EQUATIONS
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˚ Suppose the contrary: y0 ∈ ∂B. Then y0 is also a minimizer Our aim is to show that y0 ∈ B. of Φ|∂B , and hence the gradient of Φ at y0 is directed along the interior normal to ∂B. Therefore, there exists a ν ≥ 0 such that 1 (2m) = (2.3) fε (t, y0 ) . (−1)m y0 1+ν If y0 (t) ≥ ϕε (t) for 0 < t < 1, then y0 is also a solution of problem (2.1) with λ = (1 + ν)−1 ∈ (0, 1]. By condition (f), y0 = M ; i.e., y0 ∈ ∂B. It remains to consider the case in which y0 (t0 ) < ϕε (t0 ) for some t0 ∈ (0, 1). In this case, y0 (t) < ε for 0 ≤ t ≤ 1 by the above-mentioned Agarwal–O’Regan theorem. Now if we multiply (2.3) by y0 (t) and integrate the resulting relation from 0 to 1, then, with regard to (1.5) and (2.2), we obtain 1 M = 1+ν
1
1 y0 (t)fε (t, y0 (t)) dt ≤ Cε
2
0
dt 1−γ < M 2. γ < Cε (ϕε (t))
0
This contradiction completes the proof of Proposition 2.1. By way of example, consider the boundary value problem (−1)m y (2m) = µf (t, y), (i)
0 < t < 1;
(i)
y (0) = y (1) = 0,
i = 0, . . . , m − 1,
(2.4)
with parameter µ > 0, where f : [0, 1] × (0, ∞) → (0, ∞) is a continuous function satisfying condition (1.2). If y is a positive solution of the problem (−1)m y (2m) = λµf (t, y),
0 < t < 1;
i = 0, . . . , m − 1, y (0) = y (1) = 0, 1 and y = M , then M 2 = λµ 0 y(t)f (t, y(t))dt ≤ µf (M ), where (i)
(i)
f (M ) = sup{yf (t, y) : 0 ≤ t ≤ 1, 0 ≤ y ≤ M }. This, together with Proposition 2.1, implies that if µ < sup {M 2 /f (M ) : M > 0}, then problem (2.4) has at least one positive solution. Likewise, if γ ∈ (0, 1/m) and g : [0, 1] × [0, +∞) → [0, +∞) is a continuous function, then a sufficient condition for the existence of a positive solution of the problem (−1)m y (2m) = y −γ + µg(t, y), (i)
(i)
y (0) = y (1) = 0,
0 < t < 1; i = 0, . . . , m − 1,
is that the parameter µ > 0 satisfies the inequality µ < sup {M 1−γ (M 1+γ − 1)/g (M ) : M > 0}. 3. THE SUBLINEAR CASE Consider the case in which the conditions f (t, y) ≤ ay + C, 0 ≤ a < λ1 ,
(t, y) ∈ (0, 1) × [ε, ∞),
(3.1) (3.2)
are satisfied along with (1.2), where 1 1 y (m) (t)2 dt |y(t)|2 dt : y ∈ H\{0} > 0 λ1 = inf 0
0
is the first eigenvalue of the problem (−1)m y (2m) = λy, DIFFERENTIAL EQUATIONS
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Theorem 3.1. If conditions (1.2), (3.1), and (3.2) are satisfied, then problem (1.1) has at least one positive solution. Proof. By (1.5) and (3.1), Fε (t, y) ≤
0 (a/2)y 2 + Cy
for y < ε for y ≥ ε.
This inequality, together with condition (3.2) and the representation (1.8), implies that the functional Φ is coercive and bounded below. Consequently, Φ satisfies the Palais–Smale condition and admits a global minimizer. The proof of the theorem is complete. 4. THE SUPERLINEAR CASE Theorem 4.1. If, in addition to (1.2) and condition (f), the condition 0 < θFε (t, y) ≤ yf (t, y),
(t, y) ∈ (0, 1) × [A, ∞),
(4.1)
is satisfied, where θ > 2 and A > ε, then problem (1.1) has two positive solutions. Proof. First, we show that Φ satisfies the Palais–Smale condition. To this end, it suffices to show that an arbitrary sequence {yj } of elements of H satisfying condition (1.9) is bounded. Indeed, by (1.5) and (4.1),
1 θ 2 − 1 yj = (θFε (t, yj (t)) − yj (t)fε (t, yj (t))) dt + θΦ (yj ) − Φ (yj ) , yj 2 0
≤ C0 (yj + 1) ,
j = 1, 2, . . . ,
where C0 is a sufficiently large positive number; consequently, yj ≤ C0 + 1, j = 1, 2, . . . As was mentioned in the proof of Proposition 2.1, problem (1.1) has a solution y0 such that ˚ and inf Φ(∂B) ≥ Φ (y0 ). It remains to prove the existence of a positive solution of proby0 ∈ B lem (1.1) other than y0 . First, we note that the function Fε satisfies the inequality −γ −C (ϕε (t)) if 0 ≤ y < ε (4.2) Fε (t, y) ≥ θ Fε (t, A)(y/A) if y ≥ A, which follows from conditions (1.5) and (4.1). Let y1 ∈ H be an arbitrary function satisfying the conditions y1 (t) > 0, 0 < t < 1, and y1 = 1. Then 1 1 2 (m) 2 R y1 (t) − Fε (t, Ry1 (t)) dt → −∞ as R → ∞ Φ (Ry1 ) = 2 0
by (4.2). Consequently, there exists an R1 > M such that Φ (R1 y1 ) < Φ (y0 ) ≤ inf Φ(∂B).
(4.3)
By Z we denote the set of continuous mappings z : [0, 1] → H such that z(0) = y0 and z(1) = R1 y1 . Let c = inf max Φ(z(s)) : z ∈ Z . 0≤s≤1
By the Palais–Smale condition and inequality (4.3), there exists a y ∈ H such that y = y0 , Φ(y) = c, and y is a positive solution of problem (1.1). The proof of the theorem is complete. DIFFERENTIAL EQUATIONS
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By way of example, consider the boundary value problems y (i) (0) = y (i) (1) = 0, (−1)m y (2m) = µ y −γ + y β , 0 < t < 1; m (2m)
(−1) y
=y
−γ
y
+ µe ,
0 < t < 1;
(i)
(i)
y (0) = y (1) = 0,
where γ ∈ (0, 1/m), µ > 0, and β > 1. It follows from Theorem 4.1 that if 1/(γ+β) (γ + 1)γ+1 (β − 1)β−1 µ< γ+β
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i = 0, . . . , m − 1,
(4.4)
i = 0, . . . , m − 1,
(4.5)
M 1+γ − 1 µ < sup M γ eM M>0
,
then problem (4.4) [respectively, problem (4.5)] has two positive solutions. REFERENCES 1. Agarwal, R.P. and O’Regan, D., J. Diff. Equations, 2001, vol. 170, no. 1, pp. 142–156. 2. Agarwal, R.P. and O’Regan, D., Singular Differential and Integral Equations with Applications, Dordrecht, 2003. 3. Rabinowitz, P.H., Minimax Methods in Critical Point Theory with Applications to Differential Equations, CBMS Regional Conference Series in Mathematics, Washington, 1986, vol. 65.
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