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Notes on symmetric and monotonic solutions of a nonlocal diffusive logistic equation
Applied Mathematics Letters 26 (2013) 804–806
Contents lists available at SciVerse ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Notes on symmetric and monotonic solutions of a nonlocal diffusive logistic equation✩
1. Introduction In [1], the authors studied the following problem:
u′′ + λu 1 −
1
u(y)f (x, y) dy
= 0,
x ∈ (−1, 1),
(1.1)
−1
u(−1) = u(1) = 0.
They established the existence, uniqueness and stability of the steady state solution of the nonlocal problem (1.1). In that paper, they also proposed a question. They conjecture that the monotonicity still holds without the following condition: (f2 ) f (x, y) = f1 (|x − y|), where f1 : [0, 2] → (0, +∞) is non-decreasing and piecewise continuous, and
2 0
f1 (y) dy > 0.
Using the well-known moving plane method, we shall show that the conjecture is right. In fact, the proof of Theorem 3.3 of [1] is wrong. In the proof of Theorem 3.3 of [1], the authors claim that from the condition (f2 ), one can show that
1
f1 (|x − y|) − f1 xθ − y
u(y) dy > 0,
−1
where xθ = 2θ − x with θ ∈ (0, 1) and some x ∈ (1 − θ , 1), and u is the positive solution of (1.1). We take θ = x=
9 . 10
1 5
and
Consider
9 − y 9 10 f1 − y = 7 10
9 y∈ ,1 \ 5 10 1 9 y ∈ −1, ∪ . 10 5 10 9 9 9 It is clear that the above function f1 10 − y satisfies (f2 ). However, f1 | 10 − y| < f1 | − 21 − y| for all 15 , 1 \ 10 and 9 1 9 f1 10 − y ≡ f1 − 2 − y for all −1, 51 ∪ 10 . Therefore, for this special function, we obtain 1 f1 (|x − y|) − f1 xθ − y u(y) dy ≤ 0. 1
−1
2. Main result and its proof In this note, we assume that a similar monotonicity condition in the classical paper [2] on x is required for the function f (x, y), i.e.,
(f1 ) f (x, y) = f (|x|, y) : [−1, 1] × [−1, 1] → (0, +∞) is a piecewise continuous function. Moreover, f (·, y) is nondecreasing for every y. Theorem 1. Suppose that f satisfies (f1 ). Then 1. any positive solution u(λ, x) of (1.1) is an even function for x ∈ (−1, 1), and u(λ, x) is strictly decreasing in x for x ∈ (0, 1]; ✩ Research supported by NNSF of China (No. 11261052, No. 11061030, No. 11101335, No. 11201378).
0893-9659/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2013.01.014
Correspondence / Applied Mathematics Letters 26 (2013) 804–806
2. for every λ >
π2 4
, (1.1) has a unique positive solution u(λ, x), and it satisfies
1
805
f (0, y)u(y) dy ≤ 1 − −1
π2 . 4λ
(2.1)
1
Proof. 1. For a positive solution u of (1.1), define k(x) = −1 f (|x|, y) dy. We use the well-known moving plane method. Step 1. Start moving the plane. For θ ∈ (0, 1), define xθ = 2θ − x and uθ (x) = u xθ . Let wθ = uθ (x) − u(x). Then it is easy to see that uθ (x) satisfies the same equation as u does. One can verify that wθ satisfies
θ
w +λ 1−k x ′′
w + λu
1
f (|x|, y) − f xθ , y
u(y) dy = 0.
−1
From the condition (f1 ), one can show that
1
f (|x|, y) − f xθ , y
u(y) dy ≥ 0.
−1
On the other hand,
wθ (θ ) = 0,
wθ (1) > 0.
Now we can apply the ‘‘Narrow Region principle’’ (see [3]) to conclude that, for θ close to 1, wθ (x) ≥ 0 for x ∈ [θ , 1] and
wθ (x) > 0 for x ∈ (θ , 1).
Step 2. Move the plane to its left limit. We now decrease the value of θ continuously. More precisely, let
θ = inf θ : wµ (x) > 0,
∀x ∈ (µ, 1), µ ≥ θ .
We claim that θ ≤ 0. Otherwise, we shall show that the plane can be further moved to the left by a small distance, and this would contradict the definition of θ . In fact, if θ > 0, then wθ (1) > 0, wθ (θ ) = 0. By the strong maximum principle (see [3]), we deduce that wθ (x) > 0 for x ∈ (θ , 1). Let d be the maximum width of narrow regions so that we can apply the ‘‘Narrow Region principle’’. Choose a small positive number δ , such that δ ≤ min θ , 2d . We consider the function wθ−δ (x) on the narrow region
Ωδ = θ − δ, θ +
d 2
.
It satisfies
1 θ −δ ′′ wθ−δ + λ 1 − k x f (|x|, y) − f xθ −δ , y u(y) dy = 0 in Ωδ , wθ−δ + λu −1 d wθ−δ θ − δ = 0, wθ−δ θ + ≥ 0. 2
The equation is obvious. To see the boundary condition, we first notice that wθ−δ θ − δ = 0 due to the definition of
wθ −δ . To see that wθ−δ θ + 2d ≥ 0, we use continuity argument. Note that wθ θ + 2d > 0. More precisely and more generally, there exists a constant c0 , such that wθ θ + 2d ≥ c0 . Since wθ is continuous in θ , for δ sufficiently small, we still have wθ−δ θ + 2d ≥ c0 . Now we can apply the ‘‘Narrow Region principle’’ to conclude that d d wθ−δ (x) ≥ 0 for x ∈ θ − δ, θ + and wθ−δ (x) > 0 for x ∈ θ − δ, θ + . 2
2
And therefore,
wθ−δ (x) > 0 for x ∈ θ − δ, 1 . This contradicts the definition of θ . Thus w0 (x) ≥ 0, i.e., u(−x) ≥ u(x). We then start from θ close to −1 and move the plane x = θ toward the right. Similarly, we obtain θ ≥ 0. Hence u(−x) ≤ u(x). Therefore u(−x) = u(x). x +x For any x1 , x2 ∈ (0, 1) with x1 < x2 , let θ = 1 2 2 . Then, θ ∈ (0, 1), x2 ∈ (θ , 1). Thus 0 < wθ (x2 ) = u (2θ − x2 ) − u (x2 ) = u (x1 ) − u (x2 ) ⇒ u (x1 ) > u (x2 ) . Therefore, u(x) is strictly decreasing on x for x ∈ (0, 1).
806
Correspondence / Applied Mathematics Letters 26 (2013) 804–806 f (1,ξ )
2. We claim that there exists a constant ξ ∈ [−1, 1] such that 0 < k(x) < f (0,ξ ) for x ∈ [−1, 1]. From (f1 ), it is clear that k(x) > 0. From the last paragraph, we know that u achieves its maximum at x = 0 and from the maximal principle, k(0) < 1. From (f1 ), we obtain that k(x) =
1
f (|x|, y)u(y) dy ≤
f (1, ξ ) f (0, ξ )
f (1, ξ )
k(0) <
f (0, ξ )
f (1, y)
−1 f (0, y)
−1
≤
1
f (0, y)u(y) dy
.
Multiplying (1.1) by u and integrating on [−1, 1], we obtain that
λ
1
u2 (x) dx =
1
2
u′ (x) dx + λ
1
k(x)u2 (x) dx.
(2.2)
−1
−1
−1
From the Poincaré inequality, Eq. (2.2) becomes
λ
1
u (x) dx ≥ 2
π2 4
−1
1
u (x) dx + λ 2
1
f (0, y)u(y) dy −1
−1
1
u2 (x) dx, −1
and this is equivalent to (2.1). From Theorem 2.3 andCorollary 3.2 of [1], we have known that there exists only one local positive solution curve S1+ bifurcating from π 2 /4, 0 . Now, we prove the existence of a positive solution u(λ, x) for all λ > π 2 /4 by showing that S1+ can be extended to λ = +∞. Suppose this is not true, then S1+ can be at most extended to some λ∗ > π 2 /4. We claim that if that is the case, then ∥u(λ, ·)∥L2 → +∞ as λ → λ+ ∗ . Indeed if ∥u(λ, ·)∥L2 is bounded, and we also know that k(x) is bounded, then ∥u(λ, ·)∥H 1 is bounded from (2.2), and a weak solution would exist at λ = λ∗ , so we can extend S1+ beyond λ∗ , which is a contradiction.
(
u λn ,x
)
We choose an increasing sequence λn → λ∗ so that ∥u (λn , ·)∥L2 → +∞, and we define vn (x) = ∥u(λn ,·)∥ . Also we L2
1
define kn (x) = −1 f (|x|, y)u (λn , y) dy. So vn satisfies
vn′′ + λn (1 − kn (x)) vn = 0,
x ∈ (−1, 1), vn (±1) = 0.
Since ∥vn ∥L2 = 1, {1 − kn (x)} is bounded, and λn → λ∗ , then {vn } is bounded in H01 (−1, 1). Hence {vn } has a subsequence converge strongly in L2 (−1, 1) and weakly in H01 (−1, 1) to a limit v ∈ H01 (−1, 1), and ∥v∥L2 = 1. Moreover, from the Sobolev embedding theorem, v ∈ C α [−1, 1] for some α ∈ (0, 1), v(x) is nonnegative, v(x) is even and v is decreasing for x ∈ (0, 1). Hence there exist ε and ρ > 0 such that v(x) > ε when |x| < ρ . This implies that for n large, kn (0) ≥
ρ
−ρ
f (|x|, y)u(y) dy ≥
∥u(λn ,·)∥L2 ρ 2
−ρ
f (0, y)vn (y) dy → +∞, which contradicts the boundedness of kn . Therefore, the
assumption that ∥u (λ , ·)∥L2 → +∞ as λ → λ∗ does not hold, and S1+ can be extended to λ = +∞. n
Remark 1. We note that, using a similar proof of Theorem 1, the properties of symmetric and monotonic solutions also hold on the unit ball for problem (1.1). References [1] L. Sun, J. Shi, Y. Wang, Existence and Uniqueness of Steady State Solutions of a Nonlocal Diffusive Logistic Equation, http://www.math.wm.edu/shij/ shi/paper.html. [2] B. Gidas, W.M. Ni, L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phys. 68 (1979) 209–243. [3] W. Chen, C. Li, Methods on nonlinear elliptic equations, AIMS Ser. Differ. Equ. Dyn. Syst. 4 (2010).
Guowei Dai ∗ Department of Mathematics, Northwest Normal University, Lanzhou, 730070, PR China E-mail address: [email protected]. 7 December 2012 ∗ Tel.: +86 931 7971124.
Contents lists available at SciVerse ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Notes on symmetric and monotonic solutions of a nonlocal diffusive logistic equation✩
1. Introduction In [1], the authors studied the following problem:
u′′ + λu 1 −
1
u(y)f (x, y) dy
= 0,
x ∈ (−1, 1),
(1.1)
−1
u(−1) = u(1) = 0.
They established the existence, uniqueness and stability of the steady state solution of the nonlocal problem (1.1). In that paper, they also proposed a question. They conjecture that the monotonicity still holds without the following condition: (f2 ) f (x, y) = f1 (|x − y|), where f1 : [0, 2] → (0, +∞) is non-decreasing and piecewise continuous, and
2 0
f1 (y) dy > 0.
Using the well-known moving plane method, we shall show that the conjecture is right. In fact, the proof of Theorem 3.3 of [1] is wrong. In the proof of Theorem 3.3 of [1], the authors claim that from the condition (f2 ), one can show that
1
f1 (|x − y|) − f1 xθ − y
u(y) dy > 0,
−1
where xθ = 2θ − x with θ ∈ (0, 1) and some x ∈ (1 − θ , 1), and u is the positive solution of (1.1). We take θ = x=
9 . 10
1 5
and
Consider
9 − y 9 10 f1 − y = 7 10
9 y∈ ,1 \ 5 10 1 9 y ∈ −1, ∪ . 10 5 10 9 9 9 It is clear that the above function f1 10 − y satisfies (f2 ). However, f1 | 10 − y| < f1 | − 21 − y| for all 15 , 1 \ 10 and 9 1 9 f1 10 − y ≡ f1 − 2 − y for all −1, 51 ∪ 10 . Therefore, for this special function, we obtain 1 f1 (|x − y|) − f1 xθ − y u(y) dy ≤ 0. 1
−1
2. Main result and its proof In this note, we assume that a similar monotonicity condition in the classical paper [2] on x is required for the function f (x, y), i.e.,
(f1 ) f (x, y) = f (|x|, y) : [−1, 1] × [−1, 1] → (0, +∞) is a piecewise continuous function. Moreover, f (·, y) is nondecreasing for every y. Theorem 1. Suppose that f satisfies (f1 ). Then 1. any positive solution u(λ, x) of (1.1) is an even function for x ∈ (−1, 1), and u(λ, x) is strictly decreasing in x for x ∈ (0, 1]; ✩ Research supported by NNSF of China (No. 11261052, No. 11061030, No. 11101335, No. 11201378).
0893-9659/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2013.01.014
Correspondence / Applied Mathematics Letters 26 (2013) 804–806
2. for every λ >
π2 4
, (1.1) has a unique positive solution u(λ, x), and it satisfies
1
805
f (0, y)u(y) dy ≤ 1 − −1
π2 . 4λ
(2.1)
1
Proof. 1. For a positive solution u of (1.1), define k(x) = −1 f (|x|, y) dy. We use the well-known moving plane method. Step 1. Start moving the plane. For θ ∈ (0, 1), define xθ = 2θ − x and uθ (x) = u xθ . Let wθ = uθ (x) − u(x). Then it is easy to see that uθ (x) satisfies the same equation as u does. One can verify that wθ satisfies
θ
w +λ 1−k x ′′
w + λu
1
f (|x|, y) − f xθ , y
u(y) dy = 0.
−1
From the condition (f1 ), one can show that
1
f (|x|, y) − f xθ , y
u(y) dy ≥ 0.
−1
On the other hand,
wθ (θ ) = 0,
wθ (1) > 0.
Now we can apply the ‘‘Narrow Region principle’’ (see [3]) to conclude that, for θ close to 1, wθ (x) ≥ 0 for x ∈ [θ , 1] and
wθ (x) > 0 for x ∈ (θ , 1).
Step 2. Move the plane to its left limit. We now decrease the value of θ continuously. More precisely, let
θ = inf θ : wµ (x) > 0,
∀x ∈ (µ, 1), µ ≥ θ .
We claim that θ ≤ 0. Otherwise, we shall show that the plane can be further moved to the left by a small distance, and this would contradict the definition of θ . In fact, if θ > 0, then wθ (1) > 0, wθ (θ ) = 0. By the strong maximum principle (see [3]), we deduce that wθ (x) > 0 for x ∈ (θ , 1). Let d be the maximum width of narrow regions so that we can apply the ‘‘Narrow Region principle’’. Choose a small positive number δ , such that δ ≤ min θ , 2d . We consider the function wθ−δ (x) on the narrow region
Ωδ = θ − δ, θ +
d 2
.
It satisfies
1 θ −δ ′′ wθ−δ + λ 1 − k x f (|x|, y) − f xθ −δ , y u(y) dy = 0 in Ωδ , wθ−δ + λu −1 d wθ−δ θ − δ = 0, wθ−δ θ + ≥ 0. 2
The equation is obvious. To see the boundary condition, we first notice that wθ−δ θ − δ = 0 due to the definition of
wθ −δ . To see that wθ−δ θ + 2d ≥ 0, we use continuity argument. Note that wθ θ + 2d > 0. More precisely and more generally, there exists a constant c0 , such that wθ θ + 2d ≥ c0 . Since wθ is continuous in θ , for δ sufficiently small, we still have wθ−δ θ + 2d ≥ c0 . Now we can apply the ‘‘Narrow Region principle’’ to conclude that d d wθ−δ (x) ≥ 0 for x ∈ θ − δ, θ + and wθ−δ (x) > 0 for x ∈ θ − δ, θ + . 2
2
And therefore,
wθ−δ (x) > 0 for x ∈ θ − δ, 1 . This contradicts the definition of θ . Thus w0 (x) ≥ 0, i.e., u(−x) ≥ u(x). We then start from θ close to −1 and move the plane x = θ toward the right. Similarly, we obtain θ ≥ 0. Hence u(−x) ≤ u(x). Therefore u(−x) = u(x). x +x For any x1 , x2 ∈ (0, 1) with x1 < x2 , let θ = 1 2 2 . Then, θ ∈ (0, 1), x2 ∈ (θ , 1). Thus 0 < wθ (x2 ) = u (2θ − x2 ) − u (x2 ) = u (x1 ) − u (x2 ) ⇒ u (x1 ) > u (x2 ) . Therefore, u(x) is strictly decreasing on x for x ∈ (0, 1).
806
Correspondence / Applied Mathematics Letters 26 (2013) 804–806 f (1,ξ )
2. We claim that there exists a constant ξ ∈ [−1, 1] such that 0 < k(x) < f (0,ξ ) for x ∈ [−1, 1]. From (f1 ), it is clear that k(x) > 0. From the last paragraph, we know that u achieves its maximum at x = 0 and from the maximal principle, k(0) < 1. From (f1 ), we obtain that k(x) =
1
f (|x|, y)u(y) dy ≤
f (1, ξ ) f (0, ξ )
f (1, ξ )
k(0) <
f (0, ξ )
f (1, y)
−1 f (0, y)
−1
≤
1
f (0, y)u(y) dy
.
Multiplying (1.1) by u and integrating on [−1, 1], we obtain that
λ
1
u2 (x) dx =
1
2
u′ (x) dx + λ
1
k(x)u2 (x) dx.
(2.2)
−1
−1
−1
From the Poincaré inequality, Eq. (2.2) becomes
λ
1
u (x) dx ≥ 2
π2 4
−1
1
u (x) dx + λ 2
1
f (0, y)u(y) dy −1
−1
1
u2 (x) dx, −1
and this is equivalent to (2.1). From Theorem 2.3 andCorollary 3.2 of [1], we have known that there exists only one local positive solution curve S1+ bifurcating from π 2 /4, 0 . Now, we prove the existence of a positive solution u(λ, x) for all λ > π 2 /4 by showing that S1+ can be extended to λ = +∞. Suppose this is not true, then S1+ can be at most extended to some λ∗ > π 2 /4. We claim that if that is the case, then ∥u(λ, ·)∥L2 → +∞ as λ → λ+ ∗ . Indeed if ∥u(λ, ·)∥L2 is bounded, and we also know that k(x) is bounded, then ∥u(λ, ·)∥H 1 is bounded from (2.2), and a weak solution would exist at λ = λ∗ , so we can extend S1+ beyond λ∗ , which is a contradiction.
(
u λn ,x
)
We choose an increasing sequence λn → λ∗ so that ∥u (λn , ·)∥L2 → +∞, and we define vn (x) = ∥u(λn ,·)∥ . Also we L2
1
define kn (x) = −1 f (|x|, y)u (λn , y) dy. So vn satisfies
vn′′ + λn (1 − kn (x)) vn = 0,
x ∈ (−1, 1), vn (±1) = 0.
Since ∥vn ∥L2 = 1, {1 − kn (x)} is bounded, and λn → λ∗ , then {vn } is bounded in H01 (−1, 1). Hence {vn } has a subsequence converge strongly in L2 (−1, 1) and weakly in H01 (−1, 1) to a limit v ∈ H01 (−1, 1), and ∥v∥L2 = 1. Moreover, from the Sobolev embedding theorem, v ∈ C α [−1, 1] for some α ∈ (0, 1), v(x) is nonnegative, v(x) is even and v is decreasing for x ∈ (0, 1). Hence there exist ε and ρ > 0 such that v(x) > ε when |x| < ρ . This implies that for n large, kn (0) ≥
ρ
−ρ
f (|x|, y)u(y) dy ≥
∥u(λn ,·)∥L2 ρ 2
−ρ
f (0, y)vn (y) dy → +∞, which contradicts the boundedness of kn . Therefore, the
assumption that ∥u (λ , ·)∥L2 → +∞ as λ → λ∗ does not hold, and S1+ can be extended to λ = +∞. n
Remark 1. We note that, using a similar proof of Theorem 1, the properties of symmetric and monotonic solutions also hold on the unit ball for problem (1.1). References [1] L. Sun, J. Shi, Y. Wang, Existence and Uniqueness of Steady State Solutions of a Nonlocal Diffusive Logistic Equation, http://www.math.wm.edu/shij/ shi/paper.html. [2] B. Gidas, W.M. Ni, L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phys. 68 (1979) 209–243. [3] W. Chen, C. Li, Methods on nonlinear elliptic equations, AIMS Ser. Differ. Equ. Dyn. Syst. 4 (2010).
Guowei Dai ∗ Department of Mathematics, Northwest Normal University, Lanzhou, 730070, PR China E-mail address: [email protected]. 7 December 2012 ∗ Tel.: +86 931 7971124.