Constant sign solution for a simply supported beam equation

Electronic Journal of Qualitative Theory of Differential Equations 2017, No. 59, 1–17; doi: 10.14232/ejqtde.2017.1.59 http://www.math.u-szeged.hu/ejqtde/ Constant sign solution for a simply supported beam equation Alberto Cabada and Lorena SaavedraB Instituto de Matemáticas, Facultade de Matemáticas, Universidade de Santiago de Compostela Santiago de Compostela, Galicia, Spain Received 28 July 2017, appeared 18 August 2017 Communicated by Gabriele Bonanno Abstract. The aim of this paper is to ensure the existence of constant sign solutions for the fourth order boundary value problem: (u(4)(t)−p u00(t) + c(t)u(t) = h(t)(≥0),t∈I≡[a,b], u(a) = u00(a) = u(b) = u00(b) = 0 , where c,h∈C(I)and p≥0. This problem models the behavior of a suspension bridge assuming that the vertical displacement is small enough. By using variational methods, we weaken the previously known sufficient conditions on cto ensure that the obtained solution is of constant sign. Keywords: differential equations, boundary value problems, constant sign solutions, variational approach. 2010 Mathematics Subject Classification: 34B05, 34B60, 47J30. 1 Introduction The study of different fourth order differential equations coupled with the simply supported beam boundary conditions has been wider treated along the literature. For instance, in [9] and [10] there are obtained sufficient conditions that ensure that the problem: (u(4)(t) + c(t)u(t) = h(t)(≥0),t∈[0, 1], u(a) = u(b) = u00(a) = u00(b) = 0 , (1.1) has a unique solution, which has constant sign on the interval [0, 1]. Both papers improve previous results obtained in [4] and [16]. In [7] the strongly inverse positive (negative) character of the operator u(4)(t) + p1(t)u(3)(t) + p2(t)u00(t) + M u(t), BCorresponding author. Email: lorena.saav[email protected] 2A. Cabada and L. Saavedra coupled with the same simply supported beam boundary conditions as in (1.1), where p1∈ C3(I)and p2∈C2(I), is determined by the spectrum of the operator with suitable related boundary conditions. As it has been shown in [11], the study of this kind of problems is very important, since they are used to model different kind of bridges. In addition, in [12, Chapter 2] the author describes several models for suspension bridges. For instance, in [12, Section 2.6.3], he considers a hinged beam (which represents the roadway) subject to non-linear forces along two-sided springs (the hangers of the suspension bridge). In such a case, the one dimension mathematical model for the vertical displacement of the roadway is given by: (E I u(4)(t)−T u00(t) + g(u(t)) = q(t),t∈I, u(a) = u(b) = u00(a) = u00(b) = 0 , (1.2) where aand bare the extremes of the studied bridge, Eand Iare two positive constants given by the material of the beam (the Young’s module and the moment of inertia), T≥0 is the constant strength tension, q(t)is a downwards distributed load which acts on the beam and gis the restoring force. This model involves a non-linear part given by the function g. But in order to study it, it is very important to know first its the linear part. Indeed, in some cases, for instance if the vertical displacement is small enough, we can consider gas a linear function in the way g(u(t)) = k u(t), for a constant k∈R. A more general problem consists on considering this restoring force as a non-autonomous function g(t,u(t)) = f(t)u(t), where f is a continuous function on I. In particular the fact that the displacement of the bridge occurs in the same direction as the external force is fundamental in order to ensure the stability of the considered structure. In [4,13] the existence of one or multiple positive solutions of some suitable non-linear problems are considered. The used tools are strongly involved with the constant sign of the related Green’s function. In this paper we study the existence of constant sign solutions of the following fourth-order problem: T[p,c]u(t)≡u(4)(t)−p u00(t) + c(t)u(t) = h(t),t∈I, (1.3) coupled with the boundary conditions: u(a) = u(b) = u00(a) = u00(b) = 0 . (1.4) Remark 1.1. Realize that if in (1.2) we consider the non-autonomous function g(t,u(t)) = f(t)u(t)and divide by E I, then problem (1.3)–(1.4) is a particularization of (1.2) with p= T E I ≥0, c(t) = f(t) E I and h(t) = q(t) E I ≥0 (because qis a downwards load). Let us denote the correspondent space of definition as follows: X=nu∈C4(I)|u(a) = u(b) = u00(a) = u00(b) = 0o. (1.5) Realize that problem (1.1), studied in [9,10], is a particular case of (1.3)–(1.4) with p=0. We want to recall that despite the problem studied in [7] is more general than (1.3)–(1.4), here we weaken the sufficient conditions on cto ensure the existence of constant sign solutions. In that reference, it is imposed that the continuous function cremains between two values which we obtain by means of spectral theory. With the results which we will prove below, we allow cto pass throw these values in some sense. Constant sign solution for a simply supported beam equation 3 In the next section, for convenience of the reader we introduce some previous results which we use along the paper. Then in Section 3, before showing the main existence results, we formulate the variational approach of problem (1.3)–(1.4) and we obtain different previous results which will be used along the paper. In Section 4, we will obtain sufficient conditions to ensure that problem (1.3)–(1.4) has a unique solution. In fact, Section 4is devoted to prove sufficient conditions which guarantee that the operator T[p,c]is either strongly inverse positive in Xor strongly inverse negative in X. In Section 5, we obtain different conditions for functions h>0 and cthat ensure that the unique solution of the problem (1.3)–(1.4) is either positive or negative. Finally, in Section 6, we show an example where we apply our results. 2 Preliminaries In this section, we introduce several tools and results which will be used along the paper. We consider a general nth-order linear operator: Ln[M]u(t)≡u(n)(t) + p1(t)u(n−1)(t) + ···+pn−1(t)u0(t) + (pn(t) + M)u(t), (2.1) with t∈Iand pk∈Cn−k(I),k=1, . . . , n. Definition 2.1. The nth-order linear differential equation: Ln[M]u(t) = 0, t∈I, (2.2) is said to be disconjugate on Iif every non trivial solution has less than nzeros at I, multiple zeros being counted according to their multiplicity. We introduce a definition to our particular problem (1.3) in the space X. Definition 2.2. The operator T[p,c]is said to be strongly inverse positive (strongly inverse negative) in X, if every function u∈Xsuch that T[p,c]u0 in I, satisfies u>0 (u<0) on (a,b)and, moreover u0(a)>0 and u0(b)<0 (u0(a)<0 and u0(b)>0). Let us denote gp,cthe related Green’s function to operator T[p,c]in X. Next results, collected in [7], show a relationship between the Green’s function’s sign and the previous definition. Theorem 2.3. Green’s function related to operator T[M]in X is positive a.e. on (a,b)×(a,b)and, moreover, ∂ ∂tgp,c(t,s)|t=a>0and ∂ ∂tgp,c(t,s)|t=b<0a.e. on (a,b), if, and only if, operator T[M]is strongly inverse positive in X. Theorem 2.4. Green’s function related to operator T[M]in X is negative a.e. on (a,b)×(a,b)and, moreover, ∂ ∂tgp,c(t,s)|t=a<0and ∂ ∂tgp,c(t,s)|t=b>0a.e. on (a,b), if, and only if, operator T[M]is strongly inverse negative in X. • Let λp 1>0 be the least positive eigenvalue of T[p, 0]in X. • Let λp 2<0 be the maximum between: 4A. Cabada and L. Saavedra λp 20<0, the biggest negative eigenvalue of T[p, 0]in X1=nu∈C4(I)|u(a) = u(b) = u0(b) = u00(b) = 0o, λp 200 <0, the biggest negative eigenvalue of T[p, 0]in X3=nu∈C4(I)|u(a) = u0(a) = u00(a) = u(b) = 0o. • Let λp 3>0 be the minimum between: λp 30>0, the least positive eigenvalue of T[p, 0]in U=nu∈C4(I)|u(a) = u0(a) = u(b) = u00(b) = 0o, λp 300 >0, the least positive eigenvalue of T[p, 0]in V=nu∈C4(I)|u(a) = u00(a) = u(b) = u0(b) = 0o. Remark 2.5. In [5] we prove that the second order linear differential equation u00(t) + m u(t) = 0 , is disconjugate on Iif, and only if, m∈−∞,π b−a2. In particular: if p≥0, then u00(t)− p u(t) = 0 is a disconjugate equation in every real interval I. Hence, under this disconjugacy condition, in [7] it is proved the existence of λp 1>0, λp 20<0, λp 200 <0, λp 30>0 and λp 300 >0. Thus, the previous eigenvalues are well-defined. As a consequence of [7, Theorem 6.1] we can state the following result. Corollary 2.6. Consider the operator T[p,c]u(t)≡u(4)(t)−p u00(t) + c(t)u(t), where p ∈Rand p≥0. Then, • If −λp 1<c(t)≤ −λp 2for every t ∈I, then T[p,c]is strongly inverse positive in X. • If −λp 3≤c(t)<−λp 1for every t ∈I, then T[p,c]is strongly inverse negative in X. Moreover, in [7], there are obtained the values of λp 1,λp 2and λp 3. In particular, we have: The eigenvalues of the operator T[p, 0]in Xare given by λp 1(k) = k4π b−a4 +k2pπ b−a2 , (2.3) where k∈ {1, 2, 3, . . . }. Obviously, the least positive eigenvalue is given by λp 1≡λp 1(1) = π b−a4+pπ b−a2. Moreover, we denote as λp 1(2) = 16 π b−a4+4pπ b−a2the second positive eigenvalue of T[p, 0]in X. It is clear that if we denote λas an eigenvalue of T[p, 0]and its associated eigenfunction as u∈X1, then function v(t):=u(1−t)is an eigenfunction associated to λin X3. As a consequence, the eigenvalues of T[p, 0]on the spaces X1and X3are the same. So, in the previous definitions λp 2=λp 20=λp 200. Constant sign solution for a simply supported beam equation 5 One can verify that such eigenvalues are given as −λ, where λis a positive solution of tan b−a 2q2√λ−p q2√λ−p = tanh b−a 2q2√λ+p q2√λ+p , (2.4) in particular, λp 2is the opposite of the least positive solution of this equation. Similarly, the eigenvalues of T[p, 0]in Uand Vare the same and we can conclude that λp 3=λp 30=λp 300. In particular, the eigenvalues are given as the positive solutions of the following equality: tan (b−a)q√p2+4λ−p √2! qpp2+4λ−p = tanh (b−a)q√p2+4λ+p √2! qpp2+4λ+p , (2.5) and λp 3is the least positive solution of this equation. 3 Variational approach In this section we obtain the variational approach of problem (1.3)–(1.4) and some results which will be used in our main results. First, we consider the Hilbert space H:=H2(I)∩H1 0(I), where: H2(I) = {u∈L2(I)|u0,u00 ∈L2(I)}, and H1 0(I) = {u∈L2(I)|u0∈L2(I),u(a) = u(b) = 0}. We say that u∈His a weak solution of (1.3)–(1.4) if it satisfies: Zb au00(t)v00(t)dt +pZb au0(t)v0(t)dt +Zb ac(t)u(t)v(t)dt =Zb ah(t)v(t)dt ,∀v∈H. (3.1) For a function f∈C(I). Let us denote: fm:=min t∈If(t),fm:=max t∈If(t)and f±(t) = max {0, ±f(t)},t∈I. If p=0 and a=0, b=1, we have the following result, see [17,18]. Proposition 3.1. Let c(t)6=−k4π4for any k ∈Nand all t ∈[0, 1]. Let p =0, a =0and b =1, then the problem (1.3)–(1.4)has a unique solution u ∈X. Moreover, if −π4<cm<0, then kukC([0,1]) ≤π 2(π4+cm)khkC([0,1]) . Now, we enunciate an equivalent result to this proposition, which refers to our case. Proposition 3.2. Let c(t)6=−k4π b−a4−k2pπ b−a2for any k ∈ {1, 2, 3, . . . }and all t ∈I. Then the problem (1.3)–(1.4)has a unique solution u ∈X. Moreover, if −π b−a4−pπ b−a2<cm<0, then kukC([0,1]) ≤π 2π b−a4+pπ b−a2+cmkhkC([0,1]) . 6A. Cabada and L. Saavedra Proof. If c(t)6=−k4π b−a4−p k2π b−a2for any k∈ {1, 2, 3, . . . }and t∈I, it means that, since c∈C(I), either there exist k∈ {1, 2, 3 . . . }such that c(t)∈ −(k+1)4π b−a4 −p(k+1)2π b−a2 ,−k4π b−a4 −p k2π b−a2! or that cm>−π b−a4−pπ b−a2, i.e. there is no any eigenvalue of T[0, p]between cmand cm. As a consequence, the existence of a unique solution of problem (1.3)–(1.4) is ensured. Now, let us see the boundedness. We have the two following Wirtinger inequalities for every u∈H, (see [14,17]): kukL2(I)≤b−a π u0 L2(I)≤b−a π2  u00 L2(I), (3.2) and, kukC(I)≤√b−a 2 u0 L2(I). (3.3) Now, multiplying equation (1.3) by the unique solution u∈Xand integrating, we have: Zb au(4)(t)u(t)dt −pZb au00(t)u(t)dt +Zb ac(t)u2(t)dt =Zb ah(t)u(t)dt , which is equivalent to: Zb au002(t)dt +pZb au02(t)dt =Zb ah(t)u(t)dt −Zb ac(t)u2(t)dt . Now, taking into account the inequalities (3.2), the Hölder inequality and that cm≤0 we have:  u00  2 L2(I)+p u0  2 L2(I)≥π b−a2  u0  2 L2(I)+p u0  2 L2(I), and, Zb ah(t)u(t)dt −Zb ac(t)u2(t)dt ≤khkC(I)Zb a|u(t)|dt −cmkuk2 L2(I) ≤khkC(I)√b−akukL2(I)−cmkuk2 L2(I) ≤khkC(I)√b−ab−a π u0 L2(I)−cmb−a π2  u0  2 L2(I). So, combining the last two inequalities we arrive to: π b−a2 +p+cmb−a π2! u0 L2(I)≤khkC(I)√b−ab−a π, which is equivalent to:  u0 L2(I)≤π √b−akhkC(I) π b−a4+pπ b−a2+cm, and this combined with the inequality (3.3) gives our result. Constant sign solution for a simply supported beam equation 7 Remark 3.3. We note that previous inequality includes Proposition 3.1 as a particular case. For an arbitrary nonnegative continuous function r(t)≥0 in I, we define the scalar product: (u,v) = Zb au00(t)v00(t)dt +pZb au0(t)v0(t)dt +Zb ar(t)u(t)v(t)dt ,u,v∈H, (3.4) and kuk= (u,u)1/2 its associated norm. We have the following inequality: |u(t)−u(s)|=Zt su0(r)dr≤√t−s u0 L2(I)≤√t−sb−a π u00 L2(I)≤√t−sb−a πkuk. Thus, we can affirm that the embedding of Hinto C(I)is compact. Let f(t)and h(t)be continuous functions on I, following the arguments shown in [9], using the Riesz Representation Theorem we can define Sf:H→Hand h∗∈Hsuch that: (Sfu,v) = Zb af(t)u(t)v(t)dt ,(h∗,v) = Zb ah(t)v(t)dt ,u,v∈H. (3.5) Now, let us introduce some results which make a relation between this norm and the norms k·kC(I)and k·kL2(I). Such a result generalizes [9, Lemma 7]. Lemma 3.4. Let u ∈H, r ∈C(I), r ≥0in I and k·k be the norm associated to the scalar product (3.4). Then: kukC(I)≤1 √δ1kuk, and, kukL2(I)≤kuk qπ b−a4+pπ b−a2+mint∈I{r(t)} , where: δ1=max 4p b−a,4π2 (b−a)3. (3.6) Proof. Using the inequalities (3.2)–(3.3), we have that the two following inequalities are satisfied: pkuk2 C(I)≤b−a 4pZb a(u0(t))2dt ≤b−a 4Zb a(u00(t))2dt +pZb a(u0(t))2dt +Zb ar(t)u2(t)dt =b−a 4kuk2, kuk2 C(I)≤b−a 4Zb a(u0(t))2dt ≤(b−a)3 4π2Zb a(u00(t))2dt ≤(b−a)3 4π2Zb a(u00(t))2dt +pZb a(u0(t))2dt +Zb ar(t)u2(t)dt =(b−a)3 4π2kuk2. 8A. Cabada and L. Saavedra So, if p6=0, kukC(I)≤min (sb−a 4p,p(b−a)3 2π)kuk=1 √δ1kuk, moreover, if p=0, kukC(I)≤p(b−a)3 2πkuk=1 √δ1kuk. On another hand, kuk2 L2(I)=π b−a4+pπ b−a2+mint∈I{r(t)} π b−a4+pπ b−a2+mint∈I{r(t)}Zb au2(t)dt ≤Rb a(u00(t))2dt +pRb a(u0(t))2dt +Rb ar(t)u2(t)dt π b−a4+pπ b−a2+mint∈I{r(t)} =kuk2 π b−a4+pπ b−a2+mint∈I{r(t)}. From classical arguments, see [1], we obtain the following result, where we see that a weak solution of (3.1) in Hunder suitable conditions is indeed a classical solution of (1.3)–(1.4) in X. Proposition 3.5. Let c,h∈C(I). If u ∈H is a weak solution of (3.1), then u is a classical solution of (1.3)–(1.4)in X. Next result improves [9, Lemma 8]. Lemma 3.6. Let Sf:H→H be the operator previously defined in (3.5). Then:  Sf ≤1 δ1Zb a|f(t)|dt . Proof. Using Lemma 3.4 we can deduce the following inequalities which prove the result:  Sf =sup kuk=1 Sfu =sup kuk=1 sup kvk=1Zb af(t)u(t)v(t)dt≤sup kuk=1 sup kvk=1Zb a|f(t)||u(t)||v(t)|dt ≤sup kuk=1kukC(I)sup kvk=1kvkC(I)Zb a|f(t)|dt ≤1 δ1Zb a|f(t)|dt . Repeating the previous argument, we have:  Sf(un−um) ≤1 √δ1Zb a|f(t)|dt kun−umkC(I). Thus, from the compact embedding of Hinto C(I), we can affirm that Sf:H→His a compact operator. The proof of next result is analogous to [9, Lemma 9]. Lemma 3.7. Let h∗∈H be previously defined in (3.5). Then: kh∗k≤sb−a π b−a4+pπ b−a2+mint∈I{r(t)}khkC(I). Constant sign solution for a simply supported beam equation 9 4 Strongly inverse positive (negative) character of T[p,c]in X This section is devoted to prove maximum and anti-maximum principles for the problem (1.3)–(1.4). These results generalize those obtained in [9,10] for p=0. The proofs follow similar arguments to the ones given in such articles. We point out that on them there is no reference to spectral theory. Moreover, we also generalize Corollary 2.6, in the sense that we allow cto pass throw the given eigenvalues. The first result ensures the existence of a unique solution of the problem under certain hypotheses and gives sufficient conditions to affirm that the operator (1.3) is strongly inverse positive in X. Theorem 4.1. Let c ,h∈C(I)be such that Zb ac−(t)dt <δ1, where δ1has been defined in (3.6). Then problem (1.3)–(1.4)has a unique classical solution u ∈X and there exists R >0(depending on c and p) such that kukC(I)≤RkhkC(I). Moreover, if c(t)≤ −λp 2, for every t ∈I, then T[p,c]is strongly inverse positive in X. Proof. First, we decompose c(t) = c+(t)−c−(t). And, we write the problem (1.3)–(1.4) as follows u(4)(t)−p u00(t) + c+(t)u(t) = c−(t)u(t) + h(t),t∈I, u(a) = u(b) = u00(a) = u00(b) = 0 . 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