Existence of positive radial solutions for superlinear, semipositone problems on the exterior of a ball

Accepted Manuscript Existence of positive radial solutions for superlinear, semipositone problems on the exterior of a ball

R. Dhanya, Q. Morris, R. Shivaji

PII: DOI: Reference:

S0022-247X(15)00651-4 http://dx.doi.org/10.1016/j.jmaa.2015.07.016 YJMAA 19641

To appear in:

Journal of Mathematical Analysis and Applications

Received date:

10 February 2015

Please cite this article in press as: R. Dhanya et al., Existence of positive radial solutions for superlinear, semipositone problems on the exterior of a ball, J. Math. Anal. Appl. (2015), http://dx.doi.org/10.1016/j.jmaa.2015.07.016

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Existence of positive radial solutions for superlinear, semipositone problems on the exterior of a ball R. Dhanyaa , Q. Morrisb , R. Shivaji

b,1,∗

a Universidad

de Concepci´ on, Facultad de Ciencias F´ısicas y Matem´ aticas, Avda. Esteban Iturra s/n - Barrio Universitario,Concepci´ on - Chile b The University of North Carolina at Greensboro, Department of Mathematics & Statistics, 116 Petty Building, PO Box 26170, Greensboro, NC 27402-6170

Abstract We study positive radial solutions to −Δu = λK(|x|)f (u);  x ∈ Ωe where λ > 0 is a parameter, Ωe = x ∈ RN | |x| > r0 , r0 > 0, N > 2 , Δ is the Laplacian operator, K ∈ C ([r0 , ∞), (0, ∞)) satisfies K(r) ≤ rN1+μ ; μ > 0 for r >> 1, and f ∈ C 1 ([0, ∞), R) is a class of non-decreasing functions satisfying lims→∞ f (s) s = ∞ (superlinear) and f (0) < 0 (semipositone). We consider solutions, u, such that u → 0 as |x| → ∞, and which also satisfy the nonlinear boundary conditon ∂u ∂ ˜(u)u = 0 when |x| = r0 , where ∂η is the outward normal derivative, ∂η + c and c˜ ∈ C ([0, ∞), (0, ∞)). We will establish the existence of a positive radial solution for small values of the parameter λ. We also establish a similar result for the case when u satisfies the Dirichlet boundary conditon (u = 0) for |x| = r0 . We establish our results via variational methods, namely using the Mountain Pass Lemma. Keywords: Existence, superlinear semipositone problems, nonlinear and Dirichlet boundary conditons, exterior domain

1. Introduction We study positive, radial solutions to steady state reaction diffusion equations of the forms ⎧ ⎨ −Δu = λK(|x|)f (u); u = 0; ⎩ u → 0;

x ∈ Ωe |x| = r0 |x| → ∞

(1)

∗ Corresponding

author Email addresses: [email protected] (R. Dhanya), [email protected] (Q. Morris), [email protected] (R. Shivaji ) 1 This work was partially supported by a grant from the Simon’s Foundation (# 317872) to Ratnasingham Shivaji.

Preprint submitted to Elsevier

July 23, 2015

and ⎧ ⎨ −Δu = λK(|x|)f (u); ∂u ˜(u)u = 0; ∂η + c ⎩ u → 0;

x ∈ Ωe |x| = r0 |x| → ∞,

(2)

  where λ > 0 is a parameter, Ωe = x ∈ RN | |x| > r0 , r0 > 0, N > 2 , Δ is the Laplacian operator, K ∈ C ([r0 , ∞), (0, ∞)) satisfies K(r) ≤ rN1+μ ; μ > 0 for ∂ is the outward normal derivative, and c˜ ∈ C ([0, ∞), (0, ∞)). Here, r >> 1, ∂η the reaction term f : [0, ∞) → R is a nondecreasing, C 1 function such that (F1) lims→∞

f (s) s

= ∞ (superlinear at ∞), and

(F2) f (0) < 0 (semipositone). The case when f (0) < 0 is referred to in the literature as a semipositone problem, and it has been well documented (see [22], [6]) that they pose considerably more challenges in the study of positive solutions than the case where f (0) > 0 (positone problems). For a rich history of superlinear, semipositone problems on bounded domains with Dirichlet boundary conditions, see [2], [3], [4], [5], [7], [8], [10], [11], [12], [13], [14], [15], [16], [20], [21], and [26]. The main focus of this paper is to extend an important existence result for λ ≈ 0 obtained in the case of bounded domains with Dirichlet boundary conditions to a domain exterior to a ball, and also to problems involving classes of nonlinear boundary conditions on the boundary of the ball. Recently, the authors in [1] studied (1) via degree theory arguments. In this paper we establish existence of positive radial solutions for (1) as well as (2) via variational methods. The nonlinear boundary conditions discussed above occur very naturally in applications. See [19], [23], and [25] for a detailed description. Here, the authors study a model arising in combustion theory involving a sublinear reaction term. See also [9] for an extension of this work to an exterior domain.  2−N transforms (1), Note that the change of variables r = |x| and S = rr0 (2) respectively into the following boundary value problems (see Appendix 8.1 in [9] for details): −u = λh(x)f (u); x ∈ (0, 1) (3) u(0) = 0 = u(1) and

⎧ ⎨ −u = λh(x)f (u); x ∈ (0, 1) u(0) = 0 ⎩ u (1) + c(u(1)) · u(1) = 0 

where h(t) = assume K(r)

−2(N −1) r02 N −2 K (2−N )2 t 1 ≤ rN +μ for r >>



1

r0 t 2−N

and c(s) =

(4)

r0 ˜(s). N −2 c

We will only

1 and for some μ ∈ (0, N − 2). Then h ∈

2

C ((0, 1], (0, ∞)) could be singular at t = 0. (If μ ≥ N − 2, h will be nonsingular at t = 0 and will therefore be an easier case to study.) Note that h is still an L1 (0, 1) function with h(t) > 0 ∀t ∈ (0, 1]. We will study positive solutions of (3) and (4) in the space C 2 (0, 1)∩C 1 [0, 1] when f satisfies the additional conditions (F3) there exists A, B ∈ (0, ∞) and q ∈ (1, ∞) such that for s > 0 sufficiently large, Asq ≤ f (s) ≤ Bsq , and (F4) there exists θ > 2 such that for s sufficiently large, sf (s) > θF (s), where

s F (s) = f (t) dt. 0

Remark 1. We will often make use of (F3) and (F4) in slightly different forms. ˜ B ˜ > 0 such that Asq − A˜ ≤ If f satisfies (F3), then there exist constants A, q ˜ ∀s ≥ 0. Similarly, if f satisfies (F4), then there exists a constant f (s) ≤ Bs + B θ˜ > 0 such that sf (s) > θF (s) − θ˜ ∀s ≥ 0. In the Dirichlet case, we will establish the following result: Theorem 1. Assume (F1)-(F4). Then (3) has a positive solution for λ ≈ 0. Remark 2. The example f (s) = sq − 1, for some q > 1, clearly satisfies (F1) and (F2). We further note that by choosing A = 12 and B = 1, (F3) is satisfied. Finally, choosing θ = q+3 2 satisfies (F4) for s sufficiently large. In the case of the nonlinear boundary condition, we will need an additional condition on the function c(s), namely that (C1) For θ satisfying (F4), c(s)s2 < θ



s

c(t)t dt 0

for s sufficiently large. Then, in that case, we establish the following result: Theorem 2. Assume (F1)-(F4) and (C1). Then (4) has a positive solution for λ ≈ 0. Remark 3. Again f (s) = sq − 1, q > 1, with c(s) = sp + 1 for 0 < p < satisfies (F1)-(F4) and (C1).

q−1 2

We will prove Theorem 1 in Section 2 and Theorem 2 in Section 3. Extending the ideas in [10] and [26] , we employ variational methods, namely the Mountain Pass Lemma, to prove our results. Since f (0) < 0, it is a very challenging problem to establish that the mountain pass solution is in fact positive. Crucial estimates of the solutions when λ ≈ 0 are established to overcome this difficulty. In the autonomous case, h(t) ≡ 1, one could study (3) and (4) completely via time map analysis. For such examples, see [17] and [18]. 3

2. Proof of Theorem 1 We will work primarily with three spaces, H = W01,2 (0, 1), C[0, 1], and Lp (0, 1) for p = 1, 2 with the standard norms on each space, denoted  · H , 1  · ∞ , and  · p respectively, where u2H = 0 |u |2 dx. Let J : H → R be defined by

1 1 1  2 J(u) = (u ) dx − λ hF (u) dx. (5) 2 0 0 The second term in the definition of J is well-defined, since H → C[0, 1] and 1 ≤ λh1 λ hF (u) dx max |F (s)| where M1 = u∞ . −M1 ≤s≤M1

0

1

Since f is a C map, we find that the map J is continuous, differentiable and

1

1 J  (u)(v) = u v  − λ hf (u)v Next we will show that J is a C u1 − u2 H < , then

0

0

1

map. Let Lu (v) :=



|Lu1 (v) − Lu2 (v)| = |

0 1

≤

1

0

1 0

h(x)f (u)v dx. If

h(x)(f (u1 ) − f (u2 ))v dx|

|h(x)||f  (η)||u1 − u2 ||v| dx,

where η(x) is such that min{u1 (x), u2 (x)} < η(x) < max{u1 (x), u2 (x)} for any fixed x. Since u1 , u2 ∈ H and f  is continuous we have |Lu1 (v) − Lu2 (v)| ≤ Ch1 vH , for some C > 0 and hence J is C 1 . Now it is well known that the critical points of the functional are weak solutions of (3). We will first establish the existence of a solution for (3) using the Mountain Pass Lemma and then prove that the solution thus obtained is positive. 2.1. Existence of a Mountain Pass Solution We wish to apply the standard Mountain Pass Lemma, as in [24], which is stated below. Lemma 3 (see [24]). Let H be a Hilbert space, and let J˜ ∈ C 1 (H; R) satisfy: ˜ n ) is bounded and J˜ (un ) → 0 as (PS) any sequence {un } ⊂ H such that J(u n → ∞ possesses a convergent subsequence, ˜ = 0, (MP1) J(0) ˜ (MP2) there exist α, r > 0 such that J(u) ≥ α ∀u = r, and ˜ < 0. (MP3) there exists v ∈ H such that v > r and J(v) If we define, Γ := {γ ∈ C[(0, 1); H] : γ(0) = 0, γ(1) = v} and ˜ ˜ c := inf max J(γ(t)) then c is a critical value of the functional J. γ∈Γ t∈[0,1]

4

2.1.1. J satisfies (PS) Lemma 4. The map J satisfies the Palais-Smale condition. Proof. First, we wish to show that any sequence, {un } ⊂ H, satisfying the hypotheses of (PS) must be bounded. Assume to the contrary that {un } is such that J  (un ) → 0, there exists some M > 0 such that |J(un )| < M θJ(un )− J  (un ),un  ∀n ≥ 1, and un H → ∞. Then consider the quantity , un H where θ > 2 is chosen as in (F3). Taking a limit as n → ∞, we see that θJ(un )− J  (un ),un  limn→∞ = 0, since J(un ) is bounded and J  (un ) → 0. Also un H we can write  1

θ  −1 θJ(un ) − J (un ), un  = (un )2 dx 2 0

1 h(x) (θF (un ) − f (un )un ) dx. −λ 0

Note that when un ≥ 0, θF (un ) − f (un )un ≤ θ˜ and when un < 0, θF (un ) − f (un )un = (θ − 1)f (0)un . Hence  1

θ  ˜ 1 −1 θJ(un ) − J (un ), un  ≥ (un )2 dx − λθh 2 0

≥

− λ(θ − 1)|f (0)|un ∞ h1  θ ˜ 1 − 1 un 2H − λθh 2 − λk(θ − 1)|f (0)|un H h1 ,

where k > 0 satisfies z∞ ≤ kzH ∀z ∈ H.. But by dividing both sides through by un H and taking a limit as n → ∞, we get a contradiction. Hence, {un } is bounded in H. Since it is bounded in H, there exists a subsequence, call it again {un }, which converges weakly in H and strongly in C[0, 1]. Now, since J  (un ) → 0, it is easy to show that {un } is Cauchy in H and therefore, converges strongly in H. This proves the Palais-Smale Compactness Condition. 2.1.2. Geometry of J First, note that J(0) = 0. Now for any v ∈ H such that vH = 1, v(x) > 0 ∀x ∈ (0, 1) and any parameter s > 0, we have

λ 1 s2 − J(sv) = h(x)F (sv) dx 2 2 0  





λA 1 s2 λA˜ 1 +s h(x)v dx − sq+1 h(x)v q+1 dx , ≤ 2 2 0 2 0 ˜s ∀˜ since F (˜ s) ≥ A(˜ s)q + 1 − A˜ s > 0. 5

Now, letting s → ∞, we note that lims→∞ J(sv) = −∞ since q > 1. Choose s∗ >> 1 such that J(s∗ v) < 0. Now, in order to apply the Mountain Pass Lemma we need a lemma which will show that there exists an r > 0 and an α > 0 such that ∀u = r, J(u) > α. Later, however, we will also need information on how J grows when r → 0+ in order to show that the mountain pass solution is positive. We prove: Lemma 5. There exists λ > 0 such that if λ ∈ (0, λ), then for any u ∈ H such −1 that uH = λ q−1 , J(u) ≥

2 1 − q−1 λ . 4

−1

Proof. Let uH = r, where r = λ q−1 . Now, rewriting J(u) as



1 J(u) = r2 − λ hF (u) dx − λ hF (u) dx − λ hF (u) dx 2 Ω1 Ω2 Ω3

(6)

˜ and where Ω1 := {x ∈ (0, 1) : u(x) < 0}, Ω2 := {x ∈ (0, 1) : 0 ≤ u(x) ≤ β}, ˜ ˜ ˜ = 0, Ω3 := {x ∈ (0, 1) : β < u(x)}, where β > 0 is the unique value where F (β) we obtain,



1 2 hf (0)u dx − λ hF (u) dx. (7) J(u) ≥ r − λ 2 Ω1 Ω3 Hence, applying (F3) to (7), we obtain, λB 1 2 ˜ r + λf (0)h1 kuH − h1 k q+1 uq+1 H − λBh1 kuH 2 q+1

 q 2q+2 −2 1 ˜ q−1 − B h1 k q+1 λ q−1 . − h1 k(|f (0)| + B)λ = λ q−1 2 q+1

J(u) ≥

−2

Hence, for λ sufficiently small, J(u) ≥ 14 λ q−1 . Therefore, by Lemma 1, we have the existence of at least one weak solution uλ of (3). 2.2. Positivity of Solution uλ for λ ≈ 0. We first establish an upper bound on uλ ∞ . ˆ ∈ (0, λ) and c4 > 0, independent of λ, such that for Lemma 6. There exists λ 1 − q−1 ˆ . λ ∈ (0, λ), uλ ∞ ≤ c4 λ Proof. Let v1 denote the eigenfunction corresponding to the principal eigenvalue, λ1 , of −u with Dirichlet boundary conditions with v1 >  0 and v1 H = 1. 1 q ˆ Now since f (s) ≥ As for s >> 1, it follows that A = maxs≥0 q+1 Asq+1 − F (s) <∞ 6

Futher, for s ≥ 0, J(sv1 ) =

1 2 s −λ 2



1

hF (sv1 ) dx  1 A(sv1 )q+1 1 2 ˆ − A dx h ≤ s −λ 2 q+1 0

1 1 λAsq+1 1 q+1 = s2 − hv1 dx + λAˆ h dx 2 q+1 0 0 1 λAc1 sq+1 ˆ ≤ s2 − + λAh 1 2 q+1 = p(s) (say),

where c1 =

1 0

0



hv1q+1 dx.

−1

Now p(s) is maximized when s = (λAc1 ) q−1 , and hence for λ ≈ 0, 

−2 −2 −2 1 1 ˆ q−1 , − (Ac1 ) q−1 λ q−1 + λAh J(sv1 ) ≤ 1 ≤ c2 λ 2 q+1 −2

for some c2 > 0 independent of λ. Hence, J(uλ ) ≤ c2 λ q−1 , for λ ≈ 0. Now, recalling that θF (s) − f (s)s ≤ θ˜ for s ≥ 0,



hF (uλ ) dx + 2λ hF (uλ ) dx uλ 2H = 2J(uλ ) + 2λ Ωc1

Ω1

≤ 2c2 λ

−2 q−1

−2

= 2c2 λ q−1

−2

≤ 3c2 λ q−1

 uλ f (uλ ) θ˜ + dx + 2λ huλ f (0) dx + 2λ h θ θ Ω1 0  

uλ f (0) θ˜ + dx h − 2λ θ θ Ω1 



1 θ˜ + 2λ 1 − huλ f (0) dx − 2λ h dx θ Ω1 Ω1 θ

1 2λ θ˜ huλ f (uλ ) dx + 2λ h1 + θ 0 θ 2 + 2λk|f (0)|h1 uλ H + uλ 2H , θ



1



for λ > 0 small. −2 Now, this implies that auλ 2H + bλuλ H − 3c2 λ q−1 < 0, for a = 1 − θ2 > 0 and b = −2k|f (0)|h1 < 0. So, uλ H must be less than thelargest root of the −2

quadratic as2 +bλs−3c2 λ q−1 . In other words, uλ H ≤ −1 q−1

−bλ+

c3 λ for some constant c3 , for λ > 0 small. Hence, uλ ∞ ≤ c4 λ c4 = kc3 .

7

−2

b2 λ2 +12ac2 λ q−1 2a −1 q−1

≤

where

2.2.1. Proof of Main Result First we note that,

1 hf (uλ )uλ dx = 2J(uλ ) + 2λ λ 0

1

hF (u) dx 0

2 1 − q−1 λ + 2λF (β) 2 2 1 ≥ λ− q−1 4

≥



1

h dx

(8)

0

1 q+1 ˆ for λ > 0 sufficiently small. Now, choose γ > 0 such that Bh = 16 , 1γ 1 − q−1 ˆ ˜ }. Then for λ where B = max{B, B}, and define Ωλ := {x|uλ (x) ≥ γλ sufficiently small, uλ (x) will be sufficiently large on Ωλ and hence f (uλ (x)) < Buλ (x)q on Ωλ . Then





1

λ 0

hf (uλ )uλ dx = λ

≤λ

Ωλ

Ωλ

hf (uλ )uλ dx + λ hBuq+1 dx + λ λ



Ωcλ

Ωcλ

hf (uλ )uλ dx

  q ˜ |uλ | dx, h B |uλ | + B

(9)

1 ˜ ∀s. Now, recalling that on Ωc , uλ (x) ≤ γλ− q−1 and, by since f (s) ≤ B|s|q + B λ 1 Lemma 6, on Ωλ , uλ (x) ≤ c4 λ− q−1 , from (8) and (9) for λ ≈ 0 we have, 2 2 2 1 − q−1 λ ≤ B|Ωλ |h1 cq+1 λ− q−1 + B(1 − |Ωλ |)h1 γ q+1 λ− q−1 4 4 −1 ˜ − |Ωλ |)h1 γλ q−1 + B(1   2 1 − q−1 ˆ |Ωλ |cq+1 ≤ Bh + γ q+1 + γλ q−1 1λ 4   2 − q−1 q+1 ˆ ≤ Bh + 2γ |Ωλ |cq+1 1λ 4

Hence, by the definition of γ, we may conclude that |Ωλ | ≥ 1 2 ),

1 q+1 ˆ 8Bh 1 c4

= K,

(say). Let N := [0, ) ∪ (1 − , 1] for  ∈ (0, where  is chosen sufficiently small such that |N | ≤ K . Letting K := Ω − N , we also have that |Kλ | ≥ K λ λ  2 2 . Recall that the Green’s function for the second derivative operator with Dirichlet boundary conditions is given by (1 − x)ξ; 0 ≤ ξ ≤ x ≤ 1 G(x, ξ) = . (1 − ξ)x; 0 ≤ x ≤ ξ ≤ 1 ˆ = inf t∈(0,1] h(t). Then for x ∈ Kλ and Define d(ξ) = min{ξ, 1 − ξ} and h ξ ∈ N , we have that G(x, ξ) ≥ d(ξ). So, for any ξ such that d(ξ) < , for

8

λ ≈ 0, we have

uλ (ξ) = λ

1 0

G(x, ξ)hf (uλ ) dx



q

≥λ



1

G(x, ξ)hA(uλ ) dx + λf (0) G(x, ξ)h dx Kλ 0

1 ˆ d(ξ)γ q dx + λf (0)h1 ≥ Aλ− q−1 h Kλ

qK ˆ + λf (0)h1 hd(ξ)γ 2 1 ≥ c5 d(ξ)λ− q−1 ,

≥ Aλ

1 − q−1

(10)

for some c5 > 0. We define wλ and zλ such that −wλ = λhf + (uλ ); x ∈ (0, 1) wλ (0) = 0 = wλ (1) and



−zλ = λhf − (uλ ); x ∈ (0, 1) zλ (0) = 0 = zλ (1)

where f + (s) = max{f (s), 0} and f − (s) = min{f (s), 0}. 1 Clearly, uλ = wλ + zλ , and also zλ (ξ) = λ 0 G(x, ξ)h(x)f − (uλ (x)) dx ≤ 0 since f − (uλ (x)) ≤ 0 and G(x, ξ), h(x) ≥ 0. Furthermore, since f − (uλ (x)) ≥ 1 f (0), we see that zλ (ξ) = λ 0 hG(x, ξ)f − (uλ (x)) dx ≥ λf (0)h1 . So λf (0)h1 ≤ zλ (ξ) ≤ 0. Also, for ξ such that d(ξ) = , we have wλ (ξ) = uλ (ξ) − 1 zλ (ξ) ≥ uλ (ξ) ≥ c5 λ− q−1 . Hence, by the maximum principle, we have wλ (ξ) ≥ 1 1 − q−1 ∀ξ ∈ Ω−N . Therefore, uλ (ξ) = wλ (ξ)+zλ (ξ) ≥ c5 λ− q−1 +λf (0)h1 c5 λ and hence, for λ > 0 small enough, uλ > 0 on Ω − N . This, combined with 1 the earlier proof that uλ (ξ) ≥ c5 d(ξ)λ− q−1 ∀ξ ∈ N , completes the proof of the theorem. 3. Proof of Theorem 1.2 Here we establish the existence result for λ ≈ 0 for the boundary value problem (4) which involves a nonlinear boundary condition at x = 1. 3.1. Variational Formulation ˜ is a Hilbert space with the ˜ := {u ∈ H 1 (0, 1) : u(0) = 0}. Clearly H Let H ˜ by standard H 1 norm. Let E be defined on H E(u) = J(u) + g(u(1)).

9

(11)

t where g(t) = 0 c(s)s ds and J(u) is defined as before. We note that if we extend the function c by letting c(s) = c(−s) for s < 0, then g(t) ≥ 0 ∀t ∈ R. We further note that (C1) implies that c(s)s2 < θg(s) for s >> 1 sufficiently large, where θ is as given in (F4). Remark 4. We will use this, usually in the form that θ˜1 < θg(s)−c(s)s2 ∀s ≥ 0. Since g is an even function, this implies that θ˜1 < θg(s) − c(s)s2 ∀s ∈ (−∞, ∞). Remark 5. It is easy to establish that  · H as previously defined is equivalent ˜ to the standard H 1 norm on H. Now, we wish to establish a regularity result to show that critical points of E are classical solutions to (4). Definition 1. We say u is a critical point of E if

1

1 u ϕ  − λ h(x)f (u)ϕ + g  (u(1))ϕ(1) = 0 0

0

˜ ∀ϕ ∈ H.

Lemma 7. If u is a critical point of E, then u satisfies (4) almost everywhere in (0, 1) and the boundary conditions in the classical sense. Additionally, if we know that h(x) is locally Holder continuous in (0, 1), then the solution u ∈ C 2 (0, 1) ∩ C[0, 1] and the equation is satisfied in the classical sense. ˜ If ϕ ∈ Cc∞ (0, 1), then Proof. Clearly u(0) = 0 since the critical point u ∈ H. we have

1

1   uϕ −λ h(x)f (u)ϕ = 0. (12) 0

0

In other words, u is a weak soluton of −u = λh(x)f (u). But from the as˜ ⊂ C[0, 1] we have λh(x)f (u(x)) ∈ L∞ ((0, 1]). sumptions on h and since u ∈ H loc 2.2 (0, 1), and from the definition of weak By standard elliptic regularity, u ∈ Wloc 1 1 second derivative, we have − 0 u ϕ dx = 0 u ϕ dx, ∀ϕ ∈ Cc∞ (0, 1). Now, from (12), we have

−

1 0





u ϕ dx − λ

1

h(x)f (u)ϕ dx = 0 0

∀ϕ ∈ Cc∞ (0, 1).

Since we now know that u + λh(x)f (u) ∈ L1loc (0, 1) from the previous expression, then we have that, −u = λh(x)f (u) a.e in (0, 1).

(13)

˜ we have u and u both in L1 (0, 1). Now from the previous repreSince u ∈ H, sentation, u = −λh(x)f (u) ∈ L1 (0, 1). Thus we have improved regularity, with 2,2 (0, 1). u ∈ W 2,1 (0, 1) ∩ Wloc

10

Now we know that u ∈ W 2,1 (0, 1). So u ∈ W 1,1 (0, 1) and hence u is an absolutely continuous function in [0, 1]. Therefore, it now makes sense to talk about the pointwise value u (1). Finally, we will show that u (1) + c(u(1))u(1) = 0. Let C = {ϕ ∈ C ∞ (0, 1) ∩ C 1 [0, 1] : support(ϕ) ⊂⊂ (0, 1], ϕ(1) = 0}. ˜ From Definition 1 applied for a ϕ ∈ C we have, Clearly C ⊂ H.

1

1   u ϕ dx − λ h(x)f (u)ϕ dx + g  (u(1))ϕ(1) = 0. 0

0

Using the integration by parts formula on W 2,1 (0, 1), we have

1 0

u ϕ dx = −



1 0

u ϕ dx + ϕu ]10 .

That is,

1 (u ϕ + λh(x)f (u)) ϕ dx + ϕ(1)u (1) + g  (u(1))ϕ(1) = 0 − 0

∀ϕ ∈ C.

Using (12) we have u (1) + g  (u(1)) = 0, i.e u satisfies the boundary condition at x = 1. 3.2. Existence of a Mountain Pass Solution 3.2.1. E is C 1 Recall that E(u) = J(u) + g(u(1)). Since J has already been shown to be a C 1 functional, we need only show that g(u(1)) is C 1 to conclude that E is C 1 . ˜ and consider the functional H(u) := g(u(1)). For any v ∈ H, ˜ Fix u ∈ H

H  (u), v = g  (u(1))v(1). It is clear that the fuction g(s), as previously defined, is differentiable. Further, since pointwise evaluation is a continuous operation, we may conclude that the derivative is also continuous. Hence, E(u) is a C 1 functional. 3.2.2. E satisfies (PS) Again, we first wish to show that any sequence, {un } satisfying the hypotheses of (PS) must be bounded. Assume to the contrary that {un } is such that E  (un ) → 0, there exists some M > 0 such that |E(un )| < M ∀n ≥ 1, and θE(un )− E  (un ),un  where θ > 2 is un H → ∞. Then consider the quantity un H chosen as in (F3). θE(un )− E  (un ),un  Taking a limit as n → ∞, we see that limn→∞ = 0 since un H

11

J(un ) is bounded and J  (un ) → 0. However, θE(un ) − E  (un ), un  = (θJ(un ) − J  (un ), un )   2 + θg(un (1)) − c(un (1)) (un (1)) ˜ 1 ≥ cun 2H − λθh   2 + θg(un (1)) − c(un (1)) (un (1)) ˜ 1 ≥ cun 2H − λθh − λk(θ − 1)|f (0)|un H h1 + θ˜1 . for some c > 0. But by dividing both sides through by un H and taking a ˜ Since it limit as n → ∞, we get a contradiction. Hence, {un } is bounded in H. ˜ there exists a subsequence, call it again {un }, which converges is bounded in H, ˜ and strongly in C[0, 1]. weakly in H Since {un } converges strongly in C[0, 1], for any  > 0, there exists an N1 > 0 such that ∀n, m > N1 , un − um ∞ < . Further, since E  (un ) → 0, for any  > 0, there exists an N2 > 0 such that ∀n, m > N2 , E  (un ) − E  (um )∗ < . ˜ > 0 so that Furthermore, since un converges in C[0, 1], there exists an M ˜ . Hence, we may choose N = max{N1 , N2 }, and, ∀n, m > |f (un ) − f (um )| ≤ M N, un − um 2H = E  (un ) − E  (um ), un − um 

1 h(x)(f (un ) − f (um ))(un − um ) dx +λ 0

2

− c (un (1) − um (1)) · (un (1) − um (1))   ≤ J (un ) − J (um )∗ un − um L2

+ λh1 f (un ) − f (um )∞ un − um ∞ 2

≤  + λh1 f (un ) − f (um )∞ . ˜ and therefore {un } converges strongly Hence {un } is a Cauchy sequence in H, ˜ This proves the Palais-Smale Compactness Condition. in H. 3.2.3. Geometry of E Again, we wish to show that the function E satisfies the appropriate geometric conditions of Lemma 3. It is again clear that E(0) = 0. We again take v1 to be the principle eigenfunction of the operator −u with Dirichlet boundary conditions such that v1 H = 1 and v(x) > 0 ∀x ∈ (0, 1), and note that since ˜ then v1 ∈ H. ˜ Then we note that E(sv1 ) = J(sv1 ) + g(v1 (1)) = H01 (0, 1) ⊂ H, J(sv1 ) and hence, as before, E(sv1 ) → −∞ as s → ∞. Hence, we may choose s∗ >> 1 sufficiently large so that E(s∗ v1 ) < 0. Finally, we establish a lemma similar to Lemma 5. ˜ Lemma 8. There exists λ > 0 such that if λ ∈ (0, λ), then for any uλ ∈ H −1 2 1 − q−1 q−1 , E(uλ ) ≥ 4 λ . such that uλ H = λ 12

Proof. Recall again that E(uλ ) = J(uλ ) + g (uλ (1)). But recall that as in 2 Lemma 5, J(uλ ) ≥ 14 λ− q−1 . But since g(s) ≥ 0 by definition, then E(uλ ) ≥ 2 J(uλ ) ≥ 14 λ− q−1 . Hence, E satisfies the hypotheses of the Mountain Pass Lemma, and therefore, there exists a solution uλ to (4). 3.3. Positivity of Solution uλ for λ ≈ 0. We again need a lemma similar to Lemma 6 in order to establish the result. ˆ ∈ (0, λ) and c6 > 0, independent of λ, such that for Lemma 9. There exists λ 1 ˆ uλ ∞ ≤ c6 λ− q−1 . λ ∈ (0, λ), Proof. Let v1 be as before (see Section 3.2.3). We note that for any s > 0, −2 since v1 (1) = 0, for λ ≈ 0, E(sv1 ) = J(sv1 ) + g(sv1 (1)) = J(sv1 ) ≤ c2 λ q−1 as −2 before. Hence, E(uλ ) ≤ c2 λ q−1 for λ ≈ 0. Now,



hF (uλ ) dx + 2λ hF (uλ ) dx − 2g(uλ (1)) uλ 2H = 2E(uλ ) + 2λ Ωc1

Ω1

≤ 2c2 λ

−2 q−1

−2

= 2c2 λ q−1

−2

≤ 2c2 λ q−1

−2

= 2c2 λ q−1

−2

≤ 2c2 λ q−1 −2

≤ 3c2 λ q−1



 uλ f (uλ ) θ˜ + + 2λ huλ f (0) dx + 2λ h dx θ θ Ω1 0  

uλ f (0) θ˜ + dx − 2g(uλ (1)) h − 2λ θ θ Ω1 



1 θ˜ + 2λ 1 − huλ f (0) dx − 2λ h dx θ Ω1 Ω1 θ

1 2λ θ˜ huλ f (uλ ) dx + 2λ h1 − 2g(uλ (1)) + θ 0 θ 2 2 + 2λ|f (0)|h1 uλ ∞ + uλ 2H + c(uλ (1))(uλ (1))2 θ θ ˜ θ + 2λ h1 − 2g(uλ (1)) θ 2 θ˜ + 2λ|f (0)|h1 uλ ∞ + uλ 2H + 2λ h1 θ θ  2 c(uλ (1))(uλ (1))2 − θg(uλ (1)) + θ 2 θ˜ θ˜1 + 2λ|f (0)|h1 uλ ∞ + uλ 2H + 2λ h1 − 2 θ θ θ 2 2 + 2λ|f (0)|h1 uλ ∞ + uλ H , θ



for λ > 0 small. 13

1

Now, similar to the Dirichlet case, this implies that auλ 2H + bλuλ H − −2 3c2 λ q−1 < 0 for a = 1 − θ2 > 0 and b = −2k|f (0)|h1 < 0. So, uλ H must be −2

less than the largest root of the quadratic as2 + bλs − 3c2 λ q−1 . In other words,  uλ H ≤

−bλ+

−2

b2 λ2 +12ac2 λ q−1 2a −1

−1

≤ c5 λ q−1 , for some constant c5 > 0, for λ > 0

small. Hence uλ ∞ ≤ c6 λ q−1 where c6 = kc5 . 3.3.1. Proof of Main Result First, we note that

1 2 hf (uλ )uλ dx = uλ 2H + c(uλ (1)) (uλ (1)) λ 0

= 2J(uλ ) + 2λ

1 0

2

hF (uλ ) dx + c(uλ (1)) (uλ (1))

2 c21 − q−1 λ + 2λF (β) 2 2 c2 ≥ 1 λ− q−1 4

≥



1

h dx

(14)

0

2

c q+1 ˆ for λ > 0 sufficiently small. Now, choose γ > 0 such that Bh = 161 , 1γ 1 ˆ = max{B, B}, ˜ and define Ωλ := {x|uλ (x) ≥ γλ− q−1 }. Then for λ where B sufficiently small, uλ (x) will be sufficiently large on Ωλ and hence f (uλ (x)) < Buλ (x)q on Ωλ . Hence,



1 hf (uλ )uλ dx = λ hf (uλ )uλ dx + λ hf (uλ )uλ dx λ 0



Ωλ

=λ Ωλ

hBuq+1 dx + λ λ



Ωcλ

Ωcλ

  q ˜ |uλ | dx. (15) h B |uλ | + B

1

Now, recalling that on Ωcλ , uλ (x) ≤ γλ− q−1 and, by Lemma 9, on Ωλ , 1 uλ (x) ≤ c4 λ− q−1 , from (14) and (15), for λ ≈ 0 we have, 2 2 2 c21 − q−1 λ ≤ B|Ωλ |h1 cq+1 λ− q−1 + B(1 − |Ωλ |)h1 γ q+1 λ− q−1 4 4 −1 ˜ − |Ωλ |)h1 γλ q−1 + B(1   2 1 − q−1 ˆ |Ωλ |cq+1 ≤ Bh + γ q+1 + γλ q−1 1λ 4   2 − q−1 ˆ |Ωλ |cq+1 ≤ Bh + 2γ q+1 1λ 4

Hence, by the definition of γ, we may conclude that |Ωλ | ≥ Defining N and Kλ as before, we again see that |Kλ | ≥ ˜ ) to be new Green’s function, G(x, ξ; 0 ≤ ξ ≤ x ≤ 1 ˜ G(x, ξ) = . x; 0 ≤ x ≤ ξ ≤ 1 14

K 2 .

1 q+1 ˆ 8Bh 1 c4

≡ K.

Now, we define the

so that

uλ (ξ) = λ

1 0

˜ ξ)h(x)f (uλ (x)) dx − c(uλ (1))uλ (1)ξ. G(x,

(16)

Using the boundary condition that uλ (1) + c(uλ (1))uλ (1) = 0, we may rewrite (16) as (see appendix),

uλ (ξ) = λ

1

˜ ξ)h(x)f (uλ (x)) dx + uλ (1)ξ. G(x,

0

(17)

1 1 Further, since uλ (1) = uλ (1)− 0 xu (x) dx = uλ (1)−λ 0 xh(x)f (uλ (x)) dx, by substituting we obtain,

1 G(x, ξ)h(x)f (uλ (x)) dx + uλ (1)ξ, (18) uλ (ξ) = λ 0

where G is as before. Now, proceeding as before, we recall that by (10), −1

uλ () ≥ c5 d(ξ)λ q−1 + uλ (1)ξ, for ξ ∈ N . Hence, if uλ (1) were nonnegative for λ > 0 sufficiently small, then −1 we could conclude that uλ (ξ) ≥ c5 d(ξ)λ q−1 for ∀ξ ∈ N . Assume, to the contrary, that uλ (1) < 0. Then by (16) and the fact that c(s) > 0 for s < 0,

uλ (1) = λ

≥λ ≥λ

1 0

0

xh(x)f (uλ (x)) dx − c(uλ (1))uλ (1)

1

xh(x)f (uλ (x)) dx

Kλ

xh(x)f (uλ (x)) dx + λf (0)h1

−1 ˆ q λ q−1 ≥ Ahγ + λf (0)h1 >0

for λ > 0 sufficiently small. Hence, we have a contradiction, and uλ (1) ≥ 0. −1 Therefore, uλ (ξ) ≥ c5 d(ξ)λ q−1 ∀ξ ∈ N . Now, let wλ and zλ be defined as ⎧ ⎨ −wλ = λh(x)f + (uλ ); x ∈ (0, 1) wλ (0) = 0 ⎩ wλ (1) + c(uλ (1)) · wλ (1) = 0

15

(19)

and

⎧ ⎨ −zλ = λh(x)f − (uλ ); x ∈ (0, 1) zλ (0) = 0 . ⎩ zλ (1) + c(uλ (1)) · zλ (1) = 0

(20)

Then clearly uλ = wλ + zλ . Further, note that since zλ (x) ≥ 0 and zλ (1) = −c(uλ (1))zλ (1), then zλ (x) < 0 ∀x ∈ (0, 1]. 1 ˜ ξ)h(x)f − (uλ (x)) dx − c(uλ (1))zλ (1)ξ ≥ λf (0)h1 . Also, zλ (ξ) = λ 0 G(x, So λf (0)h1 ≤ zλ (ξ) ≤ 0. Further, for ξ such that d(ξ) = , we have wλ (ξ) = −1 uλ (ξ) − zλ (ξ) ≥ uλ (ξ) ≥ c5 λ q−1 . Hence, by the maximum principle, we have −1 that wλ (ξ) ≥ c5 λ q−1 ∀ξ ∈ (0, 1) − N . Therefore,uλ (ξ) = wλ (ξ) + zλ (ξ) ≥ −1 c5 λ q−1 + λf (0)h1 , and hence for λ sufficiently small, we have that uλ (ξ) > 0 on (0, 1) − N . This, combined with the estimate of uλ (ξ) on N completes the proof. References [1] A. Abebe, M. Chhetri, L. Sankar, and R. Shivaji. Positive solutions for a class of superlinear semipositone systems on exterior domains. Bound. Value Probl. Accepted for Publication. [2] I.l Ali, A. Castro, and R. Shivaji. Uniqueness and stability of nonnegative solutions for semipositone problems in a ball. Proc. Amer. Math. Soc., 117(3):775–782, 1993. [3] W. Allegretto, P. Nistri, and P. Zecca. Positive solutions of elliptic nonpositone problems. Differential Integral Equations, 5(1):95–101, 1992. [4] A. Ambrosetti, D. Arcoya, and B. Buffoni. Positive solutions for some semipositone problems via bifurcation theory. Differential Integral Equations, 7(3-4):655–663, 1994. [5] V. Anuradha, D. Hai, and R. Shivaji. Existence results for superlinear semipositone bvp’s. Proc. Amer. Math. Soc., 124(3):757–763, 1996. [6] H. Berestycki, L. A. Caffarelli, and L. Nirenberg. Inequalities for secondorder elliptic equations with applications to unbounded domains. I. Duke Math. J., 81(2):467–494, 1996. A celebration of John F. Nash, Jr. [7] K. J. Brown, A. Castro, and R. Shivaji. Non-existence of radially symmetric nonnegative solutions for a class of semi-positone problems. Differential Integral Equations, 2(4):541–545, 1989. [8] K.J. Brown and R. Shivaji. Instability of nonnegative solutions for a class of semipositone problems. Proc. Amer. Math. Soc., 112(1):121–124, 1991.

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[9] D. Butler, E. Ko, E. Lee, and R. Shivaji. Positive radial solutions for elliptic equations on exterior domains with nonlinear boundary conditions. Commun. Pure Appl. Anal., 13(6):2713—-2731, 2014. [10] S. Caldwell, A. Castro, R. Shivaji, and S. Unsurangsie. Positive solutions for classes of multiparameter elliptic semipositone problems. Electron. J. Differential Equations, pages No. 96, 10 pp. (electronic), 2007. [11] A. Castro, S. Gadam, and R. Shivaji. Branches of radial solutions for semipositone problems. J. Differential Equations, 120(1):30–45, 1995. [12] A. Castro and R. Shivaji. Non-negative solutions for a class of radially symmetric non-positone problems. Proc. Amer. Math. Soc., 106(3):735– 740, 1989. [13] A. Castro and R. Shivaji. Nonnegative solutions to a semilinear dirichlet problem in a ball are positive and radially symmetric. Comm. Partial Differential Equations, 14(8-9):1091–1100, 1989. [14] M. Chhetri and P. Girg. Existence and nonexistence of positive solutions for a class of superlinear semipositone systems. Nonlinear Anal., 71(10):4984– 4996, 2009. [15] M. Chhetri and P. Girg. Existence of positive solutions for a class of superlinear semipositone systems. J. Math. Anal. Appl., 408(2):781–788, 2013. [16] H. Dang, S. Oruganti, and R. Shivaji. Nonexistence of positive solutions for a class of semilinear elliptic systems. Rocky Mountain J. Math., 36(6):1845– 1855, 2006. [17] A. K. Drame and D. G. Costa. On positive solutions of one-dimensional semipositone equations with nonlinear boundary conditions. Appl. Math. Lett., 25(12):2411–2416, 2012. [18] J. Goddard, E. K. Lee, and R Shivaji. A double s-shaped bifurcation curve for a reaction-diffusion model with nonlinear boundary conditions. Bound. Value Probl., 2010(357542), 2010. [19] P. V. Gordon, E. Ko, and R. Shivaji. Multiplicity and uniqueness of positive solutions for elliptic equations with nonlinear boundary conditions arising in a theory of thermal explosion. Nonlinear Anal. Real World Appl., 15:51– 57, 2014. [20] D. D. Hai, R. Shivaji, and M. Chhetri. An existence result for a class of superlinear p-Laplacian semipositone systems. Differential Integral Equations, 14(2):231–240, 2001. [21] J. Jacobsen and K. Schmitt. Radial solutions of quasilinear elliptic differential equations. In Handbook of differential equations, pages 359–435. Elsevier/North-Holland, Amsterdam, 2004. 17

[22] P.-L. Lions. On the existence of positive solutions of semilinear elliptic equations. SIAM Rev., 24(4):441–467, 1982. [23] A. Miyake, M. Wakakura, T. Uchida, and A. Ushikubo. Investigation of accidental explosion of raw garbage composting system. J. Therm. Anal. Calorim., 85(3):643–649, 2006. [24] P. H. Rabinowitz. Minimax methods in critical point theory with applications to differential equations, volume 65 of CBMS Regional Conference Series in Mathematics. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 1986. [25] A. S. Shteinberg. Fast reactions in energetic materials. Springer, 2008. [26] S. Unsurangsie. Existence of a solution for a wave equation and an elliptic Dirichlet problem. PhD thesis, University of North Texas, 1988.

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