Using barrier strip conditions, we study the existence of $C^2[0,1]$-solutions of the boundary value problem $(\phi_p(x^{\prime}))^{\prime}=f(t,x,x^{\prime}),$ $x(0)=A,\ x^{\prime}(1)=B,$ where $\phi_p(s)=s|s|^{p-2},\ p>2$. The  question of the existence of  positive monotone solutions is also affected

Bojerikov, Silvestar

Kelevedjiev, Petio

Electronic journal of qualitative theory of differential equations

Petio Kelevedjiev

Silvestar Bojerikov

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On the solvability of a boundary value problem for $p$-Laplacian differential equations

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Electronic Journal of Qualitative Theory of Differential Equations2017, No. 8, 1–9; doi: 10.14232/ejqtde.2017.1.8 http://www.math.u-szeged.hu/ejqtde/On the solvability of a boundary value problem forp-Laplacian differential equationsPetio KelevedjievB and Silvestar BojerikovTechnical University of Sofia, Branch Sliven59 Burgasko Shousse Blvd, Sliven, 8800, BulgariaReceived 13 September 2016, appeared 27 January 2017Communicated by Paul EloeAbstract. Using barrier strip conditions, we study the existence of C2[0, 1]-solutionsof the boundary value problem (φp(x′))′ = f (t, x, x′), x(0) = A, x′(1) = B, whereφp(s) = s|s|p−2, p > 2. The question of the existence of positive monotone solutions isalso affected.Keywords: boundary value problem, second order differential equation, p-Laplacian,existence, sign conditions.2010 Mathematics Subject Classification: 34B15, 34B18.1 IntroductionThis paper is devoted to the solvability of the boundary value problem (BVP)(φp(x′))′ = f (t, x, x′), t ∈ [0, 1], (1.1)x(0) = A, x′(1) = B. (1.2)Here φp(s) = s|s|p−2, p > 2, the scalar function f (t, x, y) is defined for (t, x, y) ∈ [0, 1]× Dx ×Dy, where the sets Dx, Dy ⊆ R may be bounded, and B ≥ 1. Besides, f is continuous on asuitable subset of its domain.The solvability of various singular and nonsingular BVPs with p-Laplacian has been stud-ied, for example, in [1–5, 7–12, 14]. Conditions used in these works or do not allow the mainnonlinearity to change sign, [2, 11], or impose a growth restriction on it, [3, 9, 11], or requirethe existence of upper and lower solutions, [1, 3, 5, 8, 9, 12]; other type conditions have beenused in [7], where the main nonlinearity may changes its sign. As a rule, the obtained resultsguarantee the existence of positive solutions.Another type of conditions have been used in [10] for studying the solvability of (1.1), (1.2)in the case p ∈ (1, 2). The existence of at least one positive and monotone C2[0, 1]-solution isestablished therein under the following barrier condition:BCorresponding author. Email: keleved@mailcity.com2 P. Kelevedjiev and S. BojerikovH. There are constants Li, Fi, i = 1, 2, and a sufficiently small σ > 0 such thatF1 ≥ F2 + σ, F1 − σ > 0, L2 − σ ≥ L1,[A− σ, L + σ] ⊆ Dx, [F2, L2] ⊆ Dy, where L = L1 + |A|,f (t, x, y) ≥ 0 for (t, x, y) ∈ [0, 1]× Dx × [L1, L2], (1.3)f (t, x, y) ≤ 0 for (t, x, y) ∈ [0, 1]× DA × [F2, F1], (1.4)where the constants m and M are, respectively, the minimum and the maximum off (t, x, p) on [0, 1]× [A− σ, L + σ]× [F1 − σ,L1 + σ] and DA = (−∞, L] ∩ Dx.Let us recall, the strips [0, 1]× [L1, L2] and [0, 1]× [F2, F1] are called “barrier” because they limitthe values of the first derivatives of all C2[0, 1]-solution of (1.1), (1.2) between themselves. Re-cently, it was shown in [13] that conditions of form (1.3) and (1.4) guarantee C1[0, 1]-solutionsto the φ-Laplacian equation(φ(x′))′ = f (t, x, x′), t ∈ (0, 1),with boundary conditions (1.2), where φ : R → R is an increasing homeomorphism andf : [0, 1]×R2 → R is continuous.It turned out that the cases 1 < p < 2 and p > 2 require different technical approaches forthe use of H for studying the solvability of (1.1), (1.2). So, in the present paper we show thatH with the additional requirementB−M ≥ F1 (1.5)guarantees the existence of at least one monotone, and positive in the case A ≥ 0, C2[0, 1]-solution to (1.1), (1.2) with p > 2. In fact, our main result is the following.Theorem 1.1. Let H and (1.5) hold, and f (t, x, y) be continuous on the set [0, 1]× [A− σ,L + σ]×[F1 − σ, L1 + σ]. Then BVP (1.1), (1.2) has at least one strictly increasing solution in C2[0, 1] foreach p ∈ (2,∞).The paper is organized as follows. In Section 2 we present preliminaries needed to for-mulate the Topological Transversality Theorem, which is our basic tool, and prove auxiliaryresults. In Section 3 we give the proof of Theorem 1.1, formulate a corollary and give anexample.2 Fixed point theorem, auxiliary resultsLet K be a convex subset of a Banach space E and U ⊂ K be open in K. Let L∂U(U, K) be theset of compact maps from U to K which are fixed point free on ∂U; here, as usual, U and ∂Uare the closure of U and boundary of U in K.A map F in L∂U(U, K) is essential if every map G in L∂U(U, K) such that G/∂U = F/∂Uhas a fixed point in U. It is clear, in particular, every essential map has a fixed point in U.The following fixed point theorem due to A. Granas et al. [6].On the solvability of a BVP for p-Laplacian differential equations 3Theorem 2.1 (Topological transversality theorem). Suppose:(i) F, G : U → K are compact maps;(ii) G ∈ L∂U(U, K) is essential;(iii) H(x,λ),λ ∈ [0, 1], is a compact homotopy joining G and F, i.e. H(x, 0) = G(x) and H(x, 1) =F(x);(iv) H(x,λ),λ ∈ [0, 1], is fixed point free on ∂U.Then H(x,λ),λ ∈ [0, 1], has at least one fixed point in U and in particular there is a x0 ∈ U such thatx0 = F(x0).The following results is important for our consideration. It can be found also in [6].Theorem 2.2. Let l ∈ U be fixed and F ∈ L∂U(U, K) be the constant map F(x) = l for x ∈ U. ThenF is essential.Further, we need the following fact.Proposition 2.3. Let the constants B and M be such that B ≥ 1 and B > M > 0. Then(B−M)r ≤ Br −M for r ∈ [1,∞).Proof. The inequality is evident for r = 1. For M ∈ (0, B) consider the function g(r) =(B−M)r − Br + M, r ∈ (1,∞). First, let B−M ∈ (0, 1). Then ln(B−M) < 0 and sog′(r) = (B−M)r ln(B−M)− Br ln B < 0 for r ∈ R.Next, assume B−M = 1. Now we getg′(r) = −(1+ M)r ln(1+ M) < 0 for r ∈ R.Finally, let B − M ∈ (1,∞). In this case from B > B − M > 0 we have Br ≥ (B − M)r forr ∈ [0,∞) and sog′(r) ≤ Br ln(B−M)− Br ln B = Br ln B−MB< 0 for r ∈ [0,∞).In summary, we have proved that g′(r) < 0 for each r ∈ [0,∞). Then, the result followsfrom the fact that g(1) = 0.Let us emphasize explicitly that we conduct the rest consideration of this section for anarbitrary fixed p > 2.For λ ∈ [0, 1] consider the family of BVPs{(φp(x′))′ = λ f (t, x, x′), t ∈ [0, 1],x(0) = A, x′(1) = B, B ≥ 1, (2.1)where f : [0, 1]× Dx × Dy → R, Dx, Dy ⊆ R. Sinceφp(s) = s|s|p−2 ={sp−1, s ≥ 0,−(−s)p−1, s < 0,4 P. Kelevedjiev and S. Bojerikovwe haveφ′p(s) ={(p− 1)sp−2, s ≥ 0(p− 1)(−s)p−2, s < 0 = (p− 1)|s|p−2and (φp(x′(t)))′ = (p− 1)|x′(t)|p−2x′′(t), if x′′(t) exists. So, we can write (2.1) as{(p− 1)|x′(t)|p−2x′′(t) = λ f (t, x, x′), t ∈ [0, 1],x(0) = A, x′(1) = B.(2.1′)For convenience setmp =m(p− 1)(F1 − σ)p−2 and Mp =M(p− 1)(F1 − σ)p−2 ,where F1, σ, m and M are as in H.The next result gives a priori bounds for the C2[0, 1]-solutions of family (2.1′) (as well as of(2.1)).Lemma 2.4. Let H hold and x ∈ C2[0, 1] be a solution to family (2.1′). ThenA ≤ x(t) ≤ L, F1 ≤ x′(t) ≤ L1 and mp ≤ x′′(t) ≤ Mp for t ∈ [0, 1].Proof. The proof of the bounds for x and x′ is the same as the corresponding part of the proofof [10, Lemma 3.1], but we will state it for completeness. So, assume on the contrary thatx′(t) ≤ L1 for t ∈ [0, 1] (2.2)is not true. Then, x′(1) = B ≤ L1 together with x′ ∈ C[0, 1] implies thatS+ = {t ∈ [0, 1] : L1 < x′(t) ≤ L2}is not empty. Moreover, there exists an interval [α, β] ⊂ S+ with the propertyx′(α) > x′(β). (2.3)Then, by the fundamental theorem of calculus applied to x′, (2.3) implies that there is a γ ∈(α, β) such thatx′′(γ) < 0.We have (γ, x(γ), x′(γ)) ∈ S+ × Dx × (L1, L2], which yieldsf (γ, x(γ), x′(γ)) ≥ 0,by (1.3). Then,0 > (p− 1)|x′(γ)|p−2x′′(γ) = λ f (γ, x(γ), x′(γ)) ≥ 0 for λ ∈ [0, 1],a contradiction. Thus, (2.2) is true.By the mean value theorem, for each t ∈ (0, 1] there exists ξ ∈ (0, t) such that x(t)− x(0) =x′(ξ)t, which yieldsx(t) ≤ L for t ∈ [0, 1].Arguing as above and using (1.4), we establish x′(t) ≥ F1 for all t ∈ [0, 1] and, as aconsequence, x(t) ≥ A on [0, 1].On the solvability of a BVP for p-Laplacian differential equations 5To reach the bounds for x′′(t) fromx′(t) > F1 − σ > 0, t ∈ [0, 1],we obtain firstly0 <1(p− 1)(x′(t))p−2 ≤1(p− 1)(F1 − σ)p−2 .Next, multiplying both sides of this inequality by λM ≥ 0 and λm ≤ 0, for t ∈ [0, 1] obtainrespectivelyλM(p− 1)(x′(t))p−2 ≤λM(p− 1)(F1 − σ)p−2 ≤M(p− 1)(F1 − σ)p−2 = Mp,andλm(p− 1)(x′(t))p−2 ≥λm(p− 1)(F1 − σ)p−2 ≥m(p− 1)(F1 − σ)p−2 = mp;from f (t, x, L1) ≥ 0 for (t, x) ∈ [0, 1] × [A − σ, L + σ] and f (t, x, F1) ≤ 0 for (t, x) ∈ [0, 1]×[A− σ, L + σ], it follows that M ≥ 0 and m ≤ 0.On the other hand,m ≤ f (t, x(t), x′(t)) ≤ M for t ∈ [0, 1],since (x(t), x′(t)) ∈ [A, L] × [F1, L1] for each t ∈ [0, 1]. Multiplying the last inequality byλ(p− 1)−1(x′(t))2−p ≥ 0, λ, t ∈ [0, 1], we arrive tomp ≤ λm(p− 1)(x′(t))p−2 ≤λ f (t, x(t), x′(t))(p− 1)(x′(t))p−2 ≤λM(p− 1)|x′(t)|p−2 ≤ Mpfor all λ, t ∈ [0, 1], from where, keeping in mind that x′(t) > 0 on [0, 1], we getmp ≤ λ f (t, x(t), x′(t))(p− 1)|x′(t)|p−2 ≤ Mp for all λ, t ∈ [0, 1],which yields the required bounds for x′′(t).Now, introduce setsC1+[0, 1] = {x ∈ C1[0, 1] : x(t) > 0 on [0, 1], x(1) = φp(B)}and, in case that H holds,V = {x ∈ C1[0, 1] : A− σ ≤ x ≤ L + σ, F1 − σ ≤ x′ ≤ L1 + σ}.Introduce also the map Λλ : V → C1+[0, 1] defined byΛλx = λ∫ t1f (s, x(s), x′(s))ds + φp(B) for λ ∈ [0, 1].Lemma 2.5. Let H hold andf (t, x, y) ∈ C([0, 1]× [A− σ, L + σ]× [F1 − σ, L1 + σ]). (2.4)Then Λλ, λ ∈ [0, 1], is well defined and continuous.6 P. Kelevedjiev and S. BojerikovProof. Clearly, because of (2.4), (Λλx)′(t) = λ f (t, x(t), x′(t)), x ∈ V, is continuous on [0, 1] foreach λ ∈ [0, 1]. Next, observe that for each x ∈ V we haveλ f (t, x(t), x′(t)) ≤ λM ≤ M for λ, t ∈ [0, 1].Integrating this inequality from 1 to t, t ∈ [0, 1), we getλ∫ t1f (s, x(s), x′(s))ds ≥ M(t− 1), t ∈ [0, 1],from where it followsλ∫ t1f (s, x(s), x′(s))ds ≥ −M, t ∈ [0, 1],and−M + φp(B) ≤ (Λλx)(t), t ∈ [0, 1].By (1.5) and Proposition 2.3, we have0 < (F1 − σ)p−1 < (B−M)p−1 ≤ −M + Bp−1 = −M + φp(B)and then,0 < (F1 − σ)p−1 < (Λλx)(t), t ∈ [0, 1].Obviously, (Λλx)(1) = φp(B). Finally, (2.4) implies that the map Λλ, λ ∈ [0, 1], is continuouson V.Further, introduce the setsC2BC[0, 1] = {x ∈ C2[0, 1] : x(0) = A, x′(1) = B},K = {x ∈ C2BC[0, 1] : x′(t) > 0 on [0, 1]}and the map Φp : K → C1+[0, 1] defined by Φpx = φp(x′).Lemma 2.6. The map Φp is well defined and continuous.Proof. For each x ∈ K we have x′(t) > 0, t ∈ [0, 1]. Then,(Φpx)(t) = x′(t)|x′(t)|p−2 = x′(t)p−1 > 0 on [0, 1] (2.5)and, obviously, (Φpx)′(t) = (p− 1)(x′(t))p−2x′′(t) is continuous on [0, 1]. Also, (Φpx)(1) =x′(1)|x′(1)|p−2 = φp(B). So, Φpx ∈ C1+[0, 1]. The continuity of Φp follows from x′ ∈ C[0, 1]and (2.5).It is well known that the inverse function of φp(s) is φq(s) = s|s|q−2, q−1 + p−1 = 1, p > 1.Using it, we introduce the map Φq : C1+[0, 1]→ K, defined by(Φqy)(t) =∫ t0φq(y(s))ds + A, t ∈ [0, 1].But, for y ∈ C1+[0, 1] we have y(t) > 0 on [0, 1] and so(Φqy)(t) =∫ t0(y(s))1p−1 ds + A, t ∈ [0, 1].On the solvability of a BVP for p-Laplacian differential equations 7Lemma 2.7. The map Φq : C1+[0, 1]→ K is well defined, the inverse map of Φp and continuous.Proof. For each fixed y ∈ C1+[0, 1] we get a unique x(t) = (Φqy)(t) =∫ t0 (y(s))1p−1 ds + A. Infact, to establish the veracity of the first two assertions, we have to show that x ∈ K or, whatis the same, to show that x is a unique C2[0, 1]-solution to the BVPx′|x′|p−2 = y, t ∈ [0, 1], x(0) = A, x′(1) = B (2.6)with x′(t) > 0 on [0, 1].The last follows immediately from x′(t) = (y(t))1p−1 on [0, 1]. Then, x′|x′|p−2 = (x′(t))p−1 =y(t) for t ∈ [0, 1]. Besides, x′(1) = (y(1)) 1p−1 = (φp(B))1p−1 = B and x(0) = A. Now, thecontinuity of y′(t) and y(t) > 0 on [0, 1] imply thatx′′(t) =1p− 1 (y(t))2−pp−1 y′(t)exists and is continuous on [0, 1]. Thus, x(t) is a solution to (2.6) and is in C2[0, 1].To complete the proof we just have to observe that the continuity of Φq follows from thecontinuity of y1/(p−1)(t) on [0, 1].3 Proof of main resultProof of Theorem 1.1. We will prove the assertion for an arbitrary fixed p > 2. Introduce the setU = {x ∈ K : A− σ < x < L + σ, F1 − σ < x′ < L1 + σ, mp − σ < x′′(t) < Mp + σ}and consider the homotopyHλ : U × [0, 1]→ Kdefined by Hλ(x) := ΦqΛλ j, where j : U → C1[0, 1] is the embedding jx = x. To show that allassumptions of Theorem 2.1 are fulfilled observe firstly that U is an open subset of K, and Kis a convex subset of the Banach space C2[0, 1]. For the fixed points of Hλ, λ ∈ [0, 1], we haveΦqΛλ j(x) = xandΦpx = Λλ j(x),which is the operator form of the family{φp(x′) = λ∫ t1 f (s, x(s), x′(s))ds + φp(B), t ∈ (0, 1),x(0) = A, x′(1) = B.(3.1)Thus, the fixed points of Hλ coincide with the C2[0, 1]-solutions of (3.1). But, it is obvious thateach C2[0, 1]-solution of (3.1) is a C2[0, 1]-solution of (2.1). So, all conclusions of Lemma 2.4are valid in particular and for the C2[0, 1]-solutions of (3.1) which allow us to conclude thatthe C2[0, 1]-solutions of (3.1) do not belong to ∂U and so the homotopy is fixed point freeon ∂U. On the other hand, it is well known that j is completely continuous, that is, it mapseach bounded set to a compact one. Thus, j(U) is a compact set. Besides, it is clear thatj(U) ⊂ V. Then, according to Lemma 2.5, Λλ(j(U)) ⊆ C1+[0, 1] is compact. Finally, the set8 P. Kelevedjiev and S. BojerikovΦq(Λλ(j(U)) ⊂ K is compact, by Lemma 2.7. So, the homotopy is compact. Now, since forx ∈ U we have Λ0 j(x) = φp(B) = Bp−1, the map H0 maps each x ∈ U to the unique solutionl = Bt + A ∈ K to the BVPx′ = B, t ∈ (0, 1),x(0) = A, x′(1) = B,i.e., it is a constant map and so is essential, by Theorem 2.2. So, we can apply Theorem 2.1. Itinfers that the map H1(x) has a fixed point in U. It is easy to see that it is a C2[0, 1]-solution ofthe BVPs of families (3.1) and (2.1) obtained for λ = 1 and, what is the same, of (1.1), (1.2).An elementary consequence of the just proved theorem is the following.Corollary 3.1. Let A ≥ 0, H and (1.5) hold, and f (t, x, y) be continuous for (t, x, y) ∈ [0, 1]×[A − σ, L + σ] × [F1 − σ, L1 + σ]. Then for each p > 2 BVP (1.1), (1.2) has at least one strictlyincreasing solution in C2[0, 1] with positive values on (0, 1].We illustrate this result by the following example.Example 3.2. Consider the BVP(φp(x′))′ =(2x′ − 1)(x′ − 10)√x + 1+ 100, t ∈ (0, 1),x(0) = 2, x′(1) = 5,where p > 2 is fixed.It is easy to check that H holds for F2 = 1, F1 = 2.1, L1 = 11.9, L2 = 13 and σ = 0.1;moreover, we can take L = 14, m = −0.5 and M = 0.5. The function f (t, x, y) = (2y−1)(y−10)√x+1+100is continuous for (t, x, y) ∈ [0, 1] × [2, 14] × [2.1, 11.9]. Thus, we can apply Corollary 3.1 toconclude that this BVP has a positive strictly increasing solution in C2[0, 1].References[1] R. P. Agarwal, H. Lü, D. O’Regan, An upper and lower solution method for the one-dimensional singular p-Laplacian, Mem. Differential Equations Math. Phys. 28(2003), 13–31.MR1986715[2] R. P. Agarwal, H. Lü, D. O’Regan, Positive solutions for the singular p-Laplace equation,Houston J. Math. 31(2005), 1207–1220. MR2175432[3] R. P. Agarwal, D. Cao, H. Lü, D. O’Regan, Existence and multiplicity of positive solu-tions for singular semipositone p-Laplacian equations, Canad. J. Math. 58(2006), 449–475.MR2223452[4] A. 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On the solvability of a boundary value problem for p-Laplacian differential equations

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