Reﬁned blow-up results for nonlinear fourth order differential equations

We study a class of nonlinear fourth order differential equations which arise as models of suspension bridges. When it comes to power-like nonlinearities, it is known that solutions may blow up in ﬁnite time, if the initial data satisfy some positivity assumption. We extend this result to more general nonlinearities allowing exponential growth and to a wider class of initial data. We also give some hints on how to prevent blow-up.


Introduction
In this paper we are interested in finite space blow up of solutions of the ordinary differential equation where T ∈ R and f is a locally Lipschitz function. This equation arises in several contexts. We refer to the monograph by Peletier-Troy [15] and to [1,3,4,7,11,16] and references therein for several different applications.
In this paper we investigate further an ideal model obeying to (1) which was introduced in [6]. Consider an infinite beam subject to the restoring forces of a large number of nonlinear two-sided springs as in Figure 1. If the beam had finite length, it could model the roadway of a suspension bridge and the springs could model the hangers. In recent years, the nonlinear structural behavior of suspension bridges has been uncovered, see e.g. [5,6,14,17], and this suggests to consider semilinear versions of (1). Let u = u(x) denote the vertical displacement of the beam in position x. Assume that, besides the nonlinear restoring force g = g(u) due to the springs, there is a uniform downwards load p(x) ≡ p > 0 acting on the beam, for instance, its weight per unit length. The equation modeling an infinite beam having flexural rigidity EI, constant tension T ≥ 0, and subject to both a downwards load p (its weight) and to the restoring action g = g(u) due to some elastic springs reads EI u (x) − T u (x) = p − g(u(x)) (x ∈ R) .
For physical reasons, one has that u → g(u) is increasing. Let u p > 0 be the unique solution of g(u p ) = p. Put f (w) := g(w + u p ) − p so that f is also increasing and f (0) = 0. Put w(x) = u(x) − u p ; then w solves the equation which is precisely (1). The first purpose of the present paper is to refine previous blow-up results for the solutions of (1) when the nonlinearity f satisfies f ∈ Lip loc (R) , f (w) w > 0 for every w ∈ R \ {0}. (2) To start, let us recall the following statements proved in [2]: Proposition 1. Let T ∈ R and assume that f satisfies (2).
(i) If a local solution w of (1) blows up at some finite R ∈ R, then then any local solution of (1) exists for all x ∈ R.
If both the conditions in (4) are satisfied then global existence follows from classical theory of ODE's; but (4) merely requires that f is "one-sided at most linear" so that statement (ii) is far from being trivial and, as shown in [8], it does not hold for differential equations of order at most 3. On the other hand, Proposition 1 (i) states that, under the sole assumption (2), the only way that finite space blow up can occur is with "wide and thinning oscillations" of the solution w; again, in [8] we showed that this kind of blow up is a phenomenon typical of (at least) fourth order problems such as (1) since it does not occur in related lower order equations. Note that assumption (4) includes, in particular, the cases where f is either concave or convex.
Subsequently, it was proved in [10] (see also [8,9] for related work) that finite space blow up indeed occurs when f has a superlinear power-like behavior: If w = w(x) is a local solution of (1) in a neighborhood of x = 0 which satisfies then w blows up in finite space for x > 0, that is, there exists R < +∞ such that (3) holds.
A detailed description of the oscillations of the solution w was also given in [10] and slightly more general nonlinearities f were considered: one could replace (5) with the assumption that f is bounded from above and below by the same power. But the problem whether blow up occurs under more general assumptions on f was left open. In the present paper we prove a stronger version of Proposition 2, for more general superlinearities f satisfying (2) and for initial data possibly violating (6). Moreover, we give some hints on how to prevent blow up and to obtain global solutions.

Main results
We assume that f satisfies the regularity conditions and the growth conditions where F (w) := w 0 f (s) ds is an antiderivative of f . We also assume that f is increasing and that A typical function f that satisfies all these assumptions is the function f which appears in (5). A further example is obtained by defining f (w) by (5) for all w ≤ 0 and by f (w) = β(e γw − 1) for all w ≥ 0, where β, γ > 0 are arbitrary. In particular, f (w) is allowed to grow exponentially as w → +∞.
In order to study equation (1), we shall first introduce the energy function Then, if w solves (1), there holds for some C ∈ R. Besides the energy (10), we also introduce the auxiliary function so that If T ≥ 0 and (2) holds, by (13) we infer that Note that (7)-(8) strengthen (2), while (6) reduces to H(0) > 0.
Theorem 3. Let T > 0. Assume that f is increasing and satisfies (7)-(8)- (9). Assume that w = w(x) is a local solution of (1) in a neighborhood of x = 0 which satisfies Then, w blows up in finite space for x > 0, that is, there exists R < +∞ such that (3) holds.
Theorem 3 can easily be extended to the case T = 0, if one also assumes that wf (w) ≥ Cw 2 . Under this additional assumption, one may replace our Lemma 6 below by [8,Lemma 9] and then the rest of our proof applies verbatim. However, we shall merely focus on the case T > 0 for simplicity.
If assumption (15) is violated, then Theorem 3 becomes false and the trivial solution w(x) ≡ 0 serves as a counterexample. One is thus led to investigate what happens when H(0) ≤ 0 = E(0) but w ≡ 0. In this respect, we list a few simple conditions which are satisfied by all nontrivial solutions. Theorem 4. Let T > 0. Assume that f is increasing and satisfies (7)-(8)- (9). Assume that w = w(x) is a local nontrivial solution of (1) in a neighborhood of x = 0 which may be continued on [0, +∞). Then The condition E(0) = 0 defines a 3D manifold in the phase space R 4 but since the stable manifold of {0} is a 2D manifold, see [2, Proposition 20], one has probability 0 to find a global solution, even if E(0) = 0. Condition (ii) suggests to seek initial data with "very negative" H(0) and one is led to conjecture that the blow up place might be monotonically linked to H(0). In Section 3 we numerically show that this is not the case and that different conjectures should be made. A problem left open in our paper is precisely to determine a global nontrivial solution of (1).
In the whole, our numerical results suggest the following  Conjecture 5. Consider the equation with the initial conditions Although E(0) = 0 for all θ, for no values of θ may the solution be extended to [0, +∞). The blow up place R(θ) does not depend monotonically on the value of H(0). For any θ we have 3.064 ≤ R(θ) ≤ 3.857836.

Preliminary lemmas
In this section we prove some lemmas which will enable us to reach the proofs of our main results. We point out that in some of the following statements the function f is not required to satisfy assumptions (7)-(8)- (9) but only weaker assumptions. However, all the results hold under assumptions (7)-(8)- (9). We first prove some qualitative properties of the solution of (1).
Lemma 6. Let T > 0 and assume that f ∈ Lip loc (R) satisfies (8). Let w be a solution of (1) defined on some maximal interval [0, R). Then, for the function H defined in (13), the following alternative holds: Proof. If R < +∞, Proposition 1 states that there exists a sequence of local maxima m j such that m j R and w(m j ) → ∞ as j → ∞. Hence, (11) shows that (14). We have so proved that if H(x) remains bounded, then R = +∞.
For the remaining statements, we refer to [2,Theorem 8]. In the case where R < ∞ (statement (ii)) one should invoke once more Proposition 1. 2 Next, we turn our attention to geometric properties of the solution, such as monotonicity and concavity. The next two statements are also obtained by exploiting the features of the energy functions. In particular, the next result may not hold if the assumption H(m) > 0 is violated, see [10].
We now prove very precise geometric properties of w in these intervals; what follows can be extended to intervals where w is negative.
By (14) we infer that 0 < H(0) < w (z 1 )w (z 1 ) = H(z 1 ) < H(z 2 ) = w (z 2 )w (z 2 ) which, together with the just proved Item (i), shows that w (z 2 ) < 0 < w (z 1 ). This implies the existence of a first point r ∈ (z 1 , z 2 ) such that w (r) = 0 and w changes sign at r. Since Lemma 7 ensures that w < 0 on (m, z 2 ), we have r ∈ (z 1 , m] and we need only show that r is unique. If not, there exists a second point σ ∈ (r, m] such that w (σ) = 0 and w changes sign at σ. Since w changes from positive to negative at the point r, we must have w (r) ≤ 0. Similarly, we have w (σ) ≥ 0. Hence, Moreover, since w (x) < 0 for x ∈ (r, σ), we have 0 < w (σ) < w (r) and, in turn, with strict inequalities if T > 0. Finally, recalling that (2) implies the monotonicity of F in [0, ∞), since w (x) > 0 for x ∈ (r, σ), we have F (w(r)) < F (w(σ)). Combined with (23) and (24), this gives in contradiction with (11). 2 We now introduce two further energy functions. Let w = w(x) be a local solution of (1) and let Lemma 9. Assume f is increasing with f (0) = 0 and satisfies (7). Let T ≥ 0 and let w = w(x) be a nontrivial local solution of (1). Then Φ and Ψ are strictly convex functions. Moreover, if w admits a local maximum at some m such that w(m) > 0 and H(m) > 0, then Φ and Ψ are strictly increasing for x ≥ m.
Proof. Since f is increasing, our assumption (7) implies that By differentiating and by using (1) we obtain By (26) and recalling that T ≥ 0, we obviously have Φ (x) > 0 for almost all x, except for at most some isolated x where Φ (x) = 0 or where Φ is not defined (when w(x) = 0). If the local maximum m exists, the assumptions w(m) > 0 and H(m) > 0 imply that w (m) < 0, that is, Φ (m) ≥ 0 and hence Φ (x) > 0 for all x > m. This proves the statements for Φ. Since E(x) = Ψ(x) − Φ(x), by (11) we obtain Ψ (x) = Φ (x) and Ψ (x) = Φ (x), which prove the statements also for Ψ. 2 Finally, we prove a crucial result regarding solutions which are eventually of one sign. If T ≤ 0, we recall from [2,Theorem 4] that no such solutions exist, if one assumes (2) and Proposition 10. Let T ≤ 0 and let f satisfy (2) and (27). If w is a nontrivial global solution of (1), then w(x) changes sign infinitely many times as x → +∞ and as x → −∞.
It is also known [2] that, under the sole assumption (2), this phenomenon may not occur when T > 0. In our next result, we show that solutions which are eventually of one sign must necessarily decay to zero. To see that such solutions may actually exist, we note that w = αe −x satisfies (1) with T = 2 and f (w) = w for any α = 0, while f (w) = w satisfies (7) and (8) near w = 0.
Lemma 11. Let T > 0 and assume that f satisfies (7) and (8). If w is a solution of (1) on [0, +∞) and w is eventually of one sign, then its energy (10) is equal to zero and we have Proof. Assume that w is eventually positive, the other case being similar. We claim that w is eventually of one sign as well. If not, consider an interval (a, b) where w(x) ≥ 0, w (x) < 0 and w (a) = w (b) = 0. Multiplying (1) by w and integrating, we then obtain Since the left hand side is the sum of non-positive terms, we get w ≡ 0 in (a, b); this makes w constant in (a, b) and hence w ≡ 0 in (a, b), being w (a) = w (b) = 0, a contradiction.
If the conclusions of Lemma 6(i) hold, then w(x) → 0 as x → ∞, so Since w(x) attains a limit as x → ∞, we must then have Using our definition (13), we conclude that for large enough x, so the expression in brackets is bounded from above as x → ∞. This implies Recalling (29) and also (8), we conclude that In view of (1), this implies w (x) → −∞ as x → ∞, which is contrary to (29). 2 We conclude this section by remarking that the first part of (8) implies that F (w) ≥ c|w| 2+δ for all w ∈ R. Therefore, we have 5 Proof of Theorem 3 Step 1. Organization of the proof. Denote by [0, R) the maximal interval of continuation of the local solution w = w(x). In order to prove that R < +∞, we need some delicate estimates, see Steps 2-3-4-5 below. Once these estimates are obtained, in Step 6 we prove that R < +∞. However, before doing this, we need to remark some preliminary facts, regardless of whether R is finite or infinite.
Recall that we are assuming either H(0) > 0 or E(0) = 0. In the former case, we have G(x), H(x) → ∞ as x → R by Lemma 6(ii), so we may assume that G(0), H(0) > 0 by using a translation, if necessary. In the latter case, the conclusions of Lemma 6(i) cannot hold. If they did hold, then (28) would also hold by [2, Proposition 1] and the conserved energy (10) would be zero, contrary to assumption. In particular, the conclusions of Lemma 6(ii) hold, so we may still assume that G(0), H(0) > 0 by using a translation, if necessary. We may thus invoke Lemma 8 in what follows.
From Proposition 1 in the case R < +∞ and from Lemma 11 in the case R = +∞, we infer that there exists an increasing sequence {z j } j∈N such that: (i) z j R as j → ∞; (ii) w(z j ) = 0 and w has constant sign in (z j , z j+1 ) for all j ∈ N. Furthermore, in each interval (z j , z j+1 ) where w(x) > 0 the following facts occur: (iii) 0 < w (z j ) < −w (z j+1 ) and there exists a unique m j ∈ (z j , z j+1 ) such that w (m j ) = 0; (iv) w (z j+1 ) < 0 < w (z j ), there exists a unique r j ∈ (z j , z j+1 ) where w changes sign, and r j ≤ m j . Similar facts as (iii)-(iv) (with obvious changes) occur in intervals (z j , z j+1 ) where w(x) < 0. So, for all j let m j ∈ (z j , z j+1 ) be the point where |w(x)| attains its maximum on [z j , z j+1 ] and let M j = w(m j ). If R < +∞, by Proposition 1 we infer that lim sup j→∞ |M j | = +∞ .
In view of (8) and (11) we then infer Hence, recalling the definition of G in (12) and noticing that w(m h )w (m h ) < 0, we get By (14) we then deduce that lim x R G(x) = +∞ without extracting subsequences. In particular, we get that on the whole sequence {m j } of maxima of |w(x)|. Using again (11) we obtain that which, replaced into (33), proves that lim whenever R < +∞. If R = +∞, in what follows we assume that the local solution w = w(x) can be continued as x → +∞ so that w is defined (at least) on [0, +∞). By Lemma 11 we know that w(x) changes sign infinitely many times as x → +∞. Note that by (12) and Lemma 6(ii), we have again (33), whereas by (11), we have again (34). Hence, we obtain (35) also in the case R = +∞. We now prove some estimates related to the points found in Lemma 8. For the sake of simplicity, we denote by (z j , z j+1 ) an interval where w(x) > 0 and by (z j−1 , z j ) an interval where w(x) < 0; moreover, we put M j = w(m j ) > 0 and M j−1 = w(m j−1 ) < 0. Clearly the estimates below can be reversed on intervals where w has the opposite sign.
Step 2. We prove that lim Assume for contradiction that the claim is false so that lim sup j→∞ (r j − z j ) > 0. Then there exists a > 0 and a subsequence (still denoted in the same way) such that (r j −z j ) ≥ a for all j. By Lemma 6 we know that w (z j ) 2 = G(z j ) → +∞ as j → ∞. In turn, by Lemma 8, we know that w (x) → +∞ for all x ∈ [z j , r j ] as j → ∞. Finally, this proves that Let h(x) := (x − z j ) 3 (r j − x) 4 . By (37) and assumption (8), we infer that h(z j +σ)f (w(z j +σ))−T h (z j +σ)w(z j +σ)+h (z j +σ)w(z j +σ) → +∞ ∀σ ∈ (0, r j −z j ) (38) as j → ∞. Multiply (1) by h(x) and integrate over [z j , r j ]. Since h, h , h vanish in {z j , r j } and h (r j ) = 0, four integrations by parts yield This contradicts (38) unless (36) holds. This proves the claim of Step 2.
Step 3. We prove that there exists C 1 > 0 such that if j is sufficiently large, then In what follows c denotes a positive constant which depends on δ and which may vary from line to line, and also within the same formula. Let h(x) := sin 4 π x − z j r j − z j so that Similarly, two integrations by parts yield By solving this biquadratic algebraic inequality we infer that (52) holds for some C 3 > 0.

Proof of Theorem 4
If E(0) = 0, then blow up occurs by Theorem 3. The same is true by translation, if H(x) > 0 for some x ≥ 0. In particular, we have E(0) = 0 and H(x) ≤ 0 for all x ≥ 0. Using Lemma 6(i), we conclude that both w(x) and H(x) approach zero as x → ∞. Since H(x) is increasing by (14), the result follows.