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Positive solutions for a nonlinear Kirchhoff type beam equation
Applied Mathematics Letters 18 (2005) 479–482 www.elsevier.com/locate/aml
Positive solutions for a nonlinear Kirchhoff type beam equation✩ To Fu Ma Department of Mathematics, State University of Maringá, 87020-900 Maringá, PR, Brazil Received 13 February 2004; accepted 1 March 2004
Abstract The existence of two solutions, a positive and a negative, for a nonlinear fourth order equation with nonlinear boundary conditions is considered. The problem models the bending equilibrium of extensible beams on nonlinear elastic foundations. © 2005 Elsevier Ltd. All rights reserved. MSC: 34B18; 74K10 Keywords: Beam equation; Multiplicity; Variational method
1. Introduction In this note we show the existence of multiple solutions for a class of fourth order equation of the type iv 0 < x < 1, u (x) − m(u 22 ) u (x) = f (x, u(x)), u (0) = u (1) = 0, (1.1) u (0) − m(u 22 ) u (0) = −g(u(0)), u (1) − m(u 22 ) u (1) = g(u(1)), 1 where m ∈ C(R+ ), f ∈ C([0, 1] × R), g ∈ C(R) are real functions, and u 22 = 0 |u (s)|2 ds. ✩ This work was done while the author was visiting the Department of Mathematical Sciences, Florida Institute of Technology.
The hospitality of the Department and of its head, Professor V. Lakshmikantham, is gratefully acknowledged. Sponsored by CAPES/Brazil, Grant No BEX-0706-03-5. E-mail address: [email protected]. 0893-9659/$ - see front matter © 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2004.03.013
480
T.F. Ma / Applied Mathematics Letters 18 (2005) 479–482
This kind of problem arises in the theory of bending extensible elastic beams on nonlinear elastic foundations. The solution u = u(x) represents a thin extensible elastic beam which is resting on two elastic bearings located at its ends, given by the function g. The function f acts as a force exerted on the beam by the foundation. Then the problem models the bending equilibrium of the system. We refer the reader to [1,6,8,10] for a discussion about Kirchhoff type strings and beams. Fourth order problems with nonlinear boundary conditions involving third order derivatives were considered by several authors, e.g., [2–5,7–9]. Among them, positive solutions were only discussed in [9], for a related problem with m = 0. Our objective is to establish an existence result for positive solutions of (1.1), with m = 0, which is essentially different from the ones in [9]. It is worth noting that in the applications the function m is required to be a line with positive slope. Our main result allows m to have this property. Theorem 1. Suppose that m ≥ 0 and that f, g satisfy the sign conditions f (x, s)s ≥ 0 and g(s)s > 0. Suppose also that there are λ, µ > 0 such that f (x, u) g(u) ≤ f∞ < λ > µ, (1.2) and lim inf lim sup |u|→∞ u u |u|→∞ with min{1, µ} − λθ > 0,
(1.3)
where θ > 0 is the embedding constant for → solution. If in addition, there are α, β > 0 such that f (x, u) g(u) lim inf and lim sup ≥ f0 > α < β, u→0 u u u→0 H 2 (0, 1)
L 2 (0, 1).
Then problem (1.1) has at least one (1.4)
with
α < 0, 2 then problem (1.1) has at least two nonzero solutions, one positive and one negative. β−
(1.5)
2. Positive solutions Our argument for obtaining positive solutions is based on the following lemma. Lemma 1. Suppose that g(s)s ≥ 0. Then if f ≥ 0, any solution u of the problem (1.1) is non-negative. In addition, if u = 0, then u > 0. On the other hand, if f ≤ 0 then any nonzero solution of problem (1.1) is negative. Proof. Assume f ≥ 0. By integration we obtain from (1.1) that 1 1 2 g(u(0)) = m |u (s)| ds (u(1) − u(0)) + (1 − x) f (x, u(x)) dx,
0 1
g(u(1)) = m 0
(2.1)
0
|u (s)| ds (u(0) − u(1)) + 2
1
x f (x, u(x)) dx.
(2.2)
0
So, if u(0) ≤ u(1), one gets from (2.1) that g(u(0)) ≥ 0. From the sign assumption on g one has u(0) ≥ 0 and therefore u(1) ≥ 0. If u(1) ≤ u(0) then from (2.2) we also infer that u(0) ≥ 0, u(1) ≥ 0.
T.F. Ma / Applied Mathematics Letters 18 (2005) 479–482
Now, let us put
1
v =u −m
481
|u (s)| ds u. 2
(2.3)
0
Then v = f ≥ 0, and since u (0) = u (1) = 0, we also get v(0) ≤ 0 and v(1) ≤ 0, so that the maximum principle implies that v ≤ 0. Therefore 1 u − m |u (s)|2 ds u ≤ 0, 0 < x < 1, u(0) ≥ 0, u(1) ≥ 0, 0
and the result follows again from the maximum principle. The second part of the lemma, concerned with f ≤ 0, is proved in an analogous way.
As discussed in [8], (classical) solutions of problem (1.1) are critical points of the C 1 functional Ψ : H 2 (0, 1) → R defined by 1 1 2 1 2 Ψ (u) = u 2 + M(u 2 ) − F (x, u(x)) dx + G(u(0)) + G(u(1)), 2 2 0 where
M(t) =
t
F (x, t) =
m(s) ds, 0
t
f (x, s) ds
and
G(t) =
0
In addition,
Ψ (u), ϕ =
1
+ m(u 22 )
t
g(s) ds. 0
1
u (x)ϕ (x) dx u (x)ϕ (x) dx 0 0 1 f (x, u(x))ϕ(x)dx + g(u(0))ϕ(0) + g(u(1))ϕ(1), − 0
for all u, ϕ ∈
H 2 (0, 1).
In H 2 (0, 1) we consider the equivalent norm
u2H 2 = u 22 + |u(0)|2 + |u(1)|2 . In order to find critical points for Ψ we show that it is bounded from below and satisfies (PS), the Palais–Smale compactness condition. We recall that (u n ) is a Palais–Smale sequence if Ψ (u n ) is bounded and Ψ (u n ) → 0. If any Palais–Smale sequence has a convergent subsequence, then one simply says that Ψ satisfies (PS). Lemma 2 ([8]). Every bounded Palais–Smale sequence of Ψ has a convergent subsequence. Proof. Let u n be a bounded Palais–Smale sequence of Ψ . Then from the compact embedding of H 2 (0, 1) in H 1 (0, 1) and C[0, 1], going to a subsequence if necessary, there exists u such that (u n ) converges to u weakly in H 2 (0, 1) and strongly in both H 1 (0, 1) and C[0, 1]. Thus, 1 1 m(u n 22 ) u n (u n − u) dx + f (x, u n )(u n − u) dx → 0 0
0
and g(u n (0))(u n (0) − u(0)) + g(u n (1))(u n (1) − u(1)) → 0.
482
T.F. Ma / Applied Mathematics Letters 18 (2005) 479–482
Since (u n ) is a Palais–Smale sequence, we know that Ψ (u n ), u n − u → 0, and then from the above limits 1 u n (u n − u) dx → 0. 0
From this we infer that u n − u H 2 → 0.
u Proof of Theorem 1. Let us truncate f by putting f (x, u) = f (x, u + ) and F (x, u) = 0 f (x, s) ds. We seek critical points of Ψ + , the corresponding functional with F replaced by F. Note that we do not truncate g. The assumptions (1.2) give, for some C > 0, 1 λ µ Ψ + (u) ≥ u 22 − u22 + (|u(0)|22 + |u(1)|2 ) − C 2 2 2 1 λθ u2H 2 − C → ∞, ≥ min{1, µ}u2H 2 − 2 2 as u H 2 → ∞, because (1.3) holds. Then any (PS) sequence of Ψ + is bounded and by Lemma 2 (PS) holds. Hence Ψ + has a global minimum, say u 1 , which is a solution of problem (1.1) with f replaced by f . Now noting that f (x, u 1 ) ≥ 0, we see from Lemma 1 that u 1 ≥ 0, and therefore f (x, u 1 ) = f (x, u 1 ), showing that u 1 is in fact a solution of problem (1.1). However, this solution could be zero. We are going to show that u 1 is nonzero. In fact, for t > 0 small, one has from (1.4) and (1.5) that 1 α Ψ + (t) = − F (x, t) dx + 2G(t) ≤ − t 2 + βt 2 < 0. 2 0 Since Ψ + (0) = 0, it follows that u 1 = 0. From Lemma 1 we see that u 1 > 0 is a solution of problem (1.1). In an analogous way, replacing f (x, u) by f˜(x, u) = f (x, u − (x)), we conclude that problem (1.1) also has a negative solution u 2 . References [1] A. Arosio, A geometrical nonlinear correction to the Timoshenko beam equation, Nonlinear Anal. 47 (2001) 729–740. [2] M. Feˇckan, Free vibrations of beams on bearings with nonlinear elastic responses, J. Differential Equations 154 (1999) 55–72. [3] E. Feireisl, Non-zero time periodic solutions to an equation of Petrovsky type with nonlinear boundary conditions: slow oscillations of beams on elastic bearings, Ann. Scuola Norm. Sup. Pisa Cl. Sci. 20 (1993) 133–146. [4] M.R. Grossinho, T.F. Ma, Symmetric equilibria for a beam with a nonlinear elastic foundation, Portugal. Math. 51 (1994) 375–393. [5] M.R. Grossinho, St.A. Tersian, The dual variational principle and equilibria for a beam resting on a discontinuous nonlinear elastic foundation, Nonlinear Anal. 41 (2000) 417–431. [6] G. Kirchhoff, Vorlessunger über Mathematiche Physik: Mechanik, Teubner, Leipzig, 1876. [7] T.F. Ma, J. da Silva, Iterative solutions for a beam equation with nonlinear boundary conditions of third order, Appl. Math. Comput. 159 (2004) 11–18. [8] T.F. Ma, Existence results and numerical solutions for a beam equation with nonlinear boundary conditions, Appl. Numer. Math. 47 (2003) 189–196. [9] T.F. Ma, Positive solutions for a beam equation on a nonlinear elastic foundation, Math. Comput. Modelling 39 (2004) 1195–1201. [10] S. Woinowsky-Krieger, The effect of axial force on the vibration of hinged bars, J. Appl. Mech. 17 (1950) 35–36.
Positive solutions for a nonlinear Kirchhoff type beam equation✩ To Fu Ma Department of Mathematics, State University of Maringá, 87020-900 Maringá, PR, Brazil Received 13 February 2004; accepted 1 March 2004
Abstract The existence of two solutions, a positive and a negative, for a nonlinear fourth order equation with nonlinear boundary conditions is considered. The problem models the bending equilibrium of extensible beams on nonlinear elastic foundations. © 2005 Elsevier Ltd. All rights reserved. MSC: 34B18; 74K10 Keywords: Beam equation; Multiplicity; Variational method
1. Introduction In this note we show the existence of multiple solutions for a class of fourth order equation of the type iv 0 < x < 1, u (x) − m(u 22 ) u (x) = f (x, u(x)), u (0) = u (1) = 0, (1.1) u (0) − m(u 22 ) u (0) = −g(u(0)), u (1) − m(u 22 ) u (1) = g(u(1)), 1 where m ∈ C(R+ ), f ∈ C([0, 1] × R), g ∈ C(R) are real functions, and u 22 = 0 |u (s)|2 ds. ✩ This work was done while the author was visiting the Department of Mathematical Sciences, Florida Institute of Technology.
The hospitality of the Department and of its head, Professor V. Lakshmikantham, is gratefully acknowledged. Sponsored by CAPES/Brazil, Grant No BEX-0706-03-5. E-mail address: [email protected]. 0893-9659/$ - see front matter © 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2004.03.013
480
T.F. Ma / Applied Mathematics Letters 18 (2005) 479–482
This kind of problem arises in the theory of bending extensible elastic beams on nonlinear elastic foundations. The solution u = u(x) represents a thin extensible elastic beam which is resting on two elastic bearings located at its ends, given by the function g. The function f acts as a force exerted on the beam by the foundation. Then the problem models the bending equilibrium of the system. We refer the reader to [1,6,8,10] for a discussion about Kirchhoff type strings and beams. Fourth order problems with nonlinear boundary conditions involving third order derivatives were considered by several authors, e.g., [2–5,7–9]. Among them, positive solutions were only discussed in [9], for a related problem with m = 0. Our objective is to establish an existence result for positive solutions of (1.1), with m = 0, which is essentially different from the ones in [9]. It is worth noting that in the applications the function m is required to be a line with positive slope. Our main result allows m to have this property. Theorem 1. Suppose that m ≥ 0 and that f, g satisfy the sign conditions f (x, s)s ≥ 0 and g(s)s > 0. Suppose also that there are λ, µ > 0 such that f (x, u) g(u) ≤ f∞ < λ > µ, (1.2) and lim inf lim sup |u|→∞ u u |u|→∞ with min{1, µ} − λθ > 0,
(1.3)
where θ > 0 is the embedding constant for → solution. If in addition, there are α, β > 0 such that f (x, u) g(u) lim inf and lim sup ≥ f0 > α < β, u→0 u u u→0 H 2 (0, 1)
L 2 (0, 1).
Then problem (1.1) has at least one (1.4)
with
α < 0, 2 then problem (1.1) has at least two nonzero solutions, one positive and one negative. β−
(1.5)
2. Positive solutions Our argument for obtaining positive solutions is based on the following lemma. Lemma 1. Suppose that g(s)s ≥ 0. Then if f ≥ 0, any solution u of the problem (1.1) is non-negative. In addition, if u = 0, then u > 0. On the other hand, if f ≤ 0 then any nonzero solution of problem (1.1) is negative. Proof. Assume f ≥ 0. By integration we obtain from (1.1) that 1 1 2 g(u(0)) = m |u (s)| ds (u(1) − u(0)) + (1 − x) f (x, u(x)) dx,
0 1
g(u(1)) = m 0
(2.1)
0
|u (s)| ds (u(0) − u(1)) + 2
1
x f (x, u(x)) dx.
(2.2)
0
So, if u(0) ≤ u(1), one gets from (2.1) that g(u(0)) ≥ 0. From the sign assumption on g one has u(0) ≥ 0 and therefore u(1) ≥ 0. If u(1) ≤ u(0) then from (2.2) we also infer that u(0) ≥ 0, u(1) ≥ 0.
T.F. Ma / Applied Mathematics Letters 18 (2005) 479–482
Now, let us put
1
v =u −m
481
|u (s)| ds u. 2
(2.3)
0
Then v = f ≥ 0, and since u (0) = u (1) = 0, we also get v(0) ≤ 0 and v(1) ≤ 0, so that the maximum principle implies that v ≤ 0. Therefore 1 u − m |u (s)|2 ds u ≤ 0, 0 < x < 1, u(0) ≥ 0, u(1) ≥ 0, 0
and the result follows again from the maximum principle. The second part of the lemma, concerned with f ≤ 0, is proved in an analogous way.
As discussed in [8], (classical) solutions of problem (1.1) are critical points of the C 1 functional Ψ : H 2 (0, 1) → R defined by 1 1 2 1 2 Ψ (u) = u 2 + M(u 2 ) − F (x, u(x)) dx + G(u(0)) + G(u(1)), 2 2 0 where
M(t) =
t
F (x, t) =
m(s) ds, 0
t
f (x, s) ds
and
G(t) =
0
In addition,
Ψ (u), ϕ =
1
+ m(u 22 )
t
g(s) ds. 0
1
u (x)ϕ (x) dx u (x)ϕ (x) dx 0 0 1 f (x, u(x))ϕ(x)dx + g(u(0))ϕ(0) + g(u(1))ϕ(1), − 0
for all u, ϕ ∈
H 2 (0, 1).
In H 2 (0, 1) we consider the equivalent norm
u2H 2 = u 22 + |u(0)|2 + |u(1)|2 . In order to find critical points for Ψ we show that it is bounded from below and satisfies (PS), the Palais–Smale compactness condition. We recall that (u n ) is a Palais–Smale sequence if Ψ (u n ) is bounded and Ψ (u n ) → 0. If any Palais–Smale sequence has a convergent subsequence, then one simply says that Ψ satisfies (PS). Lemma 2 ([8]). Every bounded Palais–Smale sequence of Ψ has a convergent subsequence. Proof. Let u n be a bounded Palais–Smale sequence of Ψ . Then from the compact embedding of H 2 (0, 1) in H 1 (0, 1) and C[0, 1], going to a subsequence if necessary, there exists u such that (u n ) converges to u weakly in H 2 (0, 1) and strongly in both H 1 (0, 1) and C[0, 1]. Thus, 1 1 m(u n 22 ) u n (u n − u) dx + f (x, u n )(u n − u) dx → 0 0
0
and g(u n (0))(u n (0) − u(0)) + g(u n (1))(u n (1) − u(1)) → 0.
482
T.F. Ma / Applied Mathematics Letters 18 (2005) 479–482
Since (u n ) is a Palais–Smale sequence, we know that Ψ (u n ), u n − u → 0, and then from the above limits 1 u n (u n − u) dx → 0. 0
From this we infer that u n − u H 2 → 0.
u Proof of Theorem 1. Let us truncate f by putting f (x, u) = f (x, u + ) and F (x, u) = 0 f (x, s) ds. We seek critical points of Ψ + , the corresponding functional with F replaced by F. Note that we do not truncate g. The assumptions (1.2) give, for some C > 0, 1 λ µ Ψ + (u) ≥ u 22 − u22 + (|u(0)|22 + |u(1)|2 ) − C 2 2 2 1 λθ u2H 2 − C → ∞, ≥ min{1, µ}u2H 2 − 2 2 as u H 2 → ∞, because (1.3) holds. Then any (PS) sequence of Ψ + is bounded and by Lemma 2 (PS) holds. Hence Ψ + has a global minimum, say u 1 , which is a solution of problem (1.1) with f replaced by f . Now noting that f (x, u 1 ) ≥ 0, we see from Lemma 1 that u 1 ≥ 0, and therefore f (x, u 1 ) = f (x, u 1 ), showing that u 1 is in fact a solution of problem (1.1). However, this solution could be zero. We are going to show that u 1 is nonzero. In fact, for t > 0 small, one has from (1.4) and (1.5) that 1 α Ψ + (t) = − F (x, t) dx + 2G(t) ≤ − t 2 + βt 2 < 0. 2 0 Since Ψ + (0) = 0, it follows that u 1 = 0. From Lemma 1 we see that u 1 > 0 is a solution of problem (1.1). In an analogous way, replacing f (x, u) by f˜(x, u) = f (x, u − (x)), we conclude that problem (1.1) also has a negative solution u 2 . References [1] A. Arosio, A geometrical nonlinear correction to the Timoshenko beam equation, Nonlinear Anal. 47 (2001) 729–740. [2] M. Feˇckan, Free vibrations of beams on bearings with nonlinear elastic responses, J. Differential Equations 154 (1999) 55–72. [3] E. Feireisl, Non-zero time periodic solutions to an equation of Petrovsky type with nonlinear boundary conditions: slow oscillations of beams on elastic bearings, Ann. Scuola Norm. Sup. Pisa Cl. Sci. 20 (1993) 133–146. [4] M.R. Grossinho, T.F. Ma, Symmetric equilibria for a beam with a nonlinear elastic foundation, Portugal. Math. 51 (1994) 375–393. [5] M.R. Grossinho, St.A. Tersian, The dual variational principle and equilibria for a beam resting on a discontinuous nonlinear elastic foundation, Nonlinear Anal. 41 (2000) 417–431. [6] G. Kirchhoff, Vorlessunger über Mathematiche Physik: Mechanik, Teubner, Leipzig, 1876. [7] T.F. Ma, J. da Silva, Iterative solutions for a beam equation with nonlinear boundary conditions of third order, Appl. Math. Comput. 159 (2004) 11–18. [8] T.F. Ma, Existence results and numerical solutions for a beam equation with nonlinear boundary conditions, Appl. Numer. Math. 47 (2003) 189–196. [9] T.F. Ma, Positive solutions for a beam equation on a nonlinear elastic foundation, Math. Comput. Modelling 39 (2004) 1195–1201. [10] S. Woinowsky-Krieger, The effect of axial force on the vibration of hinged bars, J. Appl. Mech. 17 (1950) 35–36.