An Obstacle Problem for Elastic Graphs

An obstacle problem for elastic graphs

Anna Dall’Acqua Klaus Deckelnick

Abstract We consider an obstacle problem for elastic curves given by graphs with ﬁxed ends. We give conditions on the obstacles that assures existence of solutions and discuss their regularity. At the end we present a numerical method to obtain approximate solutions.

Keywords: Obstacle problem, elastic energy, regularity, fourth order.

MSC(2010): 49J05, 49J40, 49N60, 53A04

1 Introduction

For a regular curve γ, Bernoulli-Euler’s elastic energy is given by Z κ2(s) ds γ where κ is the curvature of γ and s denotes the arclength parameter. It is well known that this energy is used in applications in order to model the bending energy of an elastic rod but it also appears in image processing, see for example [18] and [4]. In this paper we focus on curves that can be written as the graph of a function u : [0, 1] → R, in which case u00(x) d u0(x) κ(x) = = . (1) 3 p 0 2 (1 + u0(x)2) 2 dx 1 + u (x)

The elastic energy of the graph is then given by

Z 1 Z 1 00 2 2p 0 2 u (x) W (u) = κ(x) 1 + u (x) dx = 5 dx 0 0 (1 + u0(x)2) 2 and we are interested in minimising the energy W subject to the constraints that the curve is ﬁxed at its ends and lies above a given function ψ : [0, 1] → R. Deﬁning the convex set

M := {u ∈ W 2,2(0, 1) | u(0) = u(1) = 0, u(x) ≥ ψ(x), x ∈ [0, 1]} , we are thus led to the obstacle problem

min{W (v) | v ∈ M}. (P)

Concerning the obstacle itself we shall assume in what follows that

ψ ∈ C0([0, 1]) with ψ(0), ψ(1) < 0 and max ψ(x) > 0. (2) x∈[0,1]

1 If a mimimum exists then this describes the optimal shape of an elastic curve ﬁxed at its ends and constrained to lie above the obstacle ψ. Note that in (P) the length of the curve is neither prescribed nor penalised, so that this model might not cover certain applications in materials science. Nevertheless we believe that the setting we study gives rise to an interesting task in the calculus of variations and leave the problem of including a length penalisation for future work. A more general obstacle problem which also involves the elastic energy has recently been studied in [17], where the author considers for > 0 the functional

Z b Z b 2 2p 0 2 p 0 2 W(u) = κ(x) 1 + u (x) dx + 1 + u (x) dx (3) a a Z p − (1 − α) 1 + u0(x)2 dx. {u=ψ}

Here, the third integral represents an adhesion energy, in which the function α :[a, b] → (0, 1) is given. In [17] the elastic energy is treated as a perturbation and the Γ–limit of the functionals W as → 0 is examined. In a more recent work [16] the same author addresses the minimisation problem for an energy functional similar to the one given above on general curves and in a periodic setting. He also gives conditions on the obstacle and on the parameters appearing in the energy functional that assures that a minimiser admits a graph representation. The presence of the term penalising the length of the curve makes the analysis different from ours. While we are not aware of other works concerning the obstacle problem for the elastic energy with prescribed boundary (other than in the periodic setting named above), there are contributions concerning confinement problems. The papers [10] and [7] consider closed elastic curves that are confined to lie in a bounded and Lipschitz open set Ω so that the boundary of Ω acts as an obstacle and prove existence and regularity properties of minimisers. Let us return to the minimisation problem (P). For small |u0| it would be viable to replace W (u) by the quadratic energy

Z 1 E(u) = (u00(x))2 dx 0 leading to the one–dimensional biharmonic obstacle problem. It is well–known that E is strictly convex on M and hence has a unique minimum which is characterised by the biharmonic variational inequality. In contrast, the solution of (P) is more involved: The functional W is no longer convex and we can ensure W 2,2–coercivity only on subsets that are uniformly bounded in W 1,∞. Hence, in order to prove the existence of a minimum of W it will be crucial to control the ﬁrst order derivative. We shall pursue two approaches. In the ﬁrst we will consider for δ ∈ (0, 1] the following penalised functional

Z 1 δ 0 6 Wδ(u) = W (u) + |u (x)| dx on M. 6 0 In this case one easily gets control on the W 1,∞- norm of a minimising sequence and hence the existence of a minimiser uδ for any δ ∈ (0, 1]. Under additional assumptions on the obstacle we are able to prove a uniform bound on the W 2,2 ∩ W 1,∞- norm of the family (uδ)δ∈(0,1]. By taking the limit δ & 0 we get to a minimiser of the original problem.

2 In the second approach we work directly with the original functional and, under suitable assumptions -in addition to (2)- on the obstacles, are able to gain unifom bounds on the W 1,∞–norm of a minimising sequence. We will also study shortly the minimisation problem in the class of symmetric functions (u(x) = u(1 − x), x ∈ [0, 1]). Here we are able to show that if kψk∞ is too large, then the minimisation problem has no solution. Let us next suppose that u ∈ M is a (local) minimum of W . In order to examine the regularity properties of u we study the associated variational inequality. Since M is convex, the functions u + t(v − u) belong to M for all v ∈ M and t ∈ [0, 1] and hence due to the minimality of u

d W (u + t(v − u)) ≥ 0 . dt t=0 This leads to the variational inequality

W 0(u)(v − u) ≥ 0 ∀v ∈ M, (4) where W 0(u)(ϕ) denotes the ﬁrst variation of W at u in direction ϕ given by

Z 1 00 00 Z 1 00 2 0 0 0 u (x)ϕ (x) u (x) u (x)ϕ (x) W (u)(ϕ) = 2 5 dx − 5 7 dx 0 (1 + u0(x)2) 2 0 (1 + u0(x)2) 2

Z 1 ϕ00(x) Z 1 u0(x) = 2 κ(x) dx − 5 κ(x)2 ϕ0(x) dx. (5) 0 2 p 0 2 0 1 + u (x) 0 1 + u (x)

Note that in view of the lack of convexity of W , (4) is only a necessary but not a suﬃcient condition for a minimum of W . We shall prove that a solution u of (4) belongs to C∞ on {u > ψ} and satisﬁes on (0, 1) ∩ {u > ψ} ! 1 d κ0(x) 1 + κ3(x) = 0 (6) p1 + (u0(x))2 dx p1 + (u0(x))2 2

(see Proposition 3.2) together with the boundary conditions

u(0) = u(1) = 0, κ(0) = κ(1) = 0, (7)

(see Corollary 3.3) so that (4) gives rise to a variational inequality involving a highly nonlinear differential operator of fourth order. In addition we shall prove that any concave solution u of (4) has a point of contact with the obstacle (see Corollary 3.4) and we study the regularity at this free boundary. We prove that u is twice continuously differentiable on [0, 1], while the third derivative is of bounded variation on this interval (see Theorem 5.1). Our paper is organized as follows. In Section 2 we show that it is enough to carry out the minimisation of W in the class of concave functions and derive an upper bound on infv∈M W (v). Section 3 is devoted to the above-mentioned approach for the existence result via penalisation and also to the regularity of the solution of (4) above the obstacle, (6) and (7). In Section 4 we derive the existence result via energy-bounds and we consider also the symmetric case. Section 5 is devoted to the global regularity of solutions of (4) and in the final section we use a numerical method in order to obtain approximate solutions of (4) and present results of test calculations.

3 2 Preliminaries

We begin by showing that, in the minimisation, it is enough to consider concave functions. As we shall occasionally consider symmetric functions we also introduce

Msym := {u ∈ M | u(x) = u(1 − x), x ∈ [0, 1]}. (8)

Lemma 2.1. For every u ∈ M there exists a concave function v ∈ M such that W (v) ≤ W (u) and, if u is not concave, then W (v) < W (u). If u ∈ Msym, then also v ∈ Msym. Moreover 0 0 kv kLp ≤ ku kLp for p ∈ [2, ∞] . (9) Proof. The idea is to look for the concave envelope of u, which in view of [19, (3)] is the solution of the following obstacle problem. Minimise the Dirichlet energy

Z 1 (η0(x))2 dx , 0 on the convex set K given by

1,2 K := {η ∈ W0 (0, 1) : η ≥ u a.e. in (0, 1)} .

By [15, Theorem 2.1, Sec. II.2] the solution v ∈ K exists and is unique. In order to further examine the solution v we follow [9, Lemma 4.1]. With the arguments in [15, Sec. IV.2] we 2,2 2,2 1,2 see that v is the weak limit in W (0, 1) for ε & 0 of vε ∈ W (0, 1) ∩ W0 (0, 1) that solve

00 00 + −vε = (−u ) θε(vε − u) in (0, 1) , vε(0) = vε(1) = 0 , with  1 for t ≤ 0 ,  t θε(t) = 1 − ε for 0 ≤ t ≤ ε ,  0 for t ≥ ε.

2,2 1,2 00 00 2 In particular, v ∈ W ∩ W0 (0, 1) so that v ∈ M. Furthermore, since vε * v in L (0, 1) we infer that v00 ≤ 0 a.e. in (0, 1) conﬁrming that v is concave. In order to compare the energies, let I := {x ∈ (0, 1) : u(x) = v(x)} be the coincidence set. Then we have v0 = u0 a.e. in I and v00 = 0 a.e. on (0, 1) \ I. We compute

Z v00(x)2 Z v00(x)2 W (v) = 5 dx = 5 dx I (1 + v0(x)2) 2 I (1 + u0(x)2) 2 Z 00 2 Z 00 + 2 vε (x) ((−u ) (x)) ≤ lim inf 5 dx ≤ 5 dx ε→0 I (1 + u0(x)2) 2 I (1 + u0(x)2) 2 Z u00(x)2 Z u00(x)2 ≤ 5 dx + 5 dx = W (u). I (1 + u0(x)2) 2 (0,1)\I (1 + u0(x)2) 2

Furthermore, if W (v) = W (u), then we have u00 ≤ 0 a.e. in I and u00 = 0 a.e. in (0, 1) \ I, which implies that u is concave. From the construction one also sees that if u is symmetric then also v is symmetric.

4 We ﬁnally show (9). Let again I be the coincidence set. For p ∈ [2, ∞) using the convexity of x 7→ |x|p we compute

Z 1 Z Z |v0(x)|p dx = |u0(x)|p dx + |v0(x)|p dx 0 I [0,1]\I Z Z ≤ |u0(x)|p dx + |u0(x)|p − p|v0(x)|p−2v0(x)(u0(x) − v0(x)) dx I [0,1]\I Z 1 Z = |u0(x)|p dx − p |v0(x)|p−2v0(x)(u0(x) − v0(x)) dx . (10) 0 [0,1]\I

Now, the second integral on the right hand side of (10) is equal to zero. To see this, we may consider separately the connnected components of [0, 1] \ I since there are at most countably many. Let A = (a1, a2) be a connected component of [0, 1] \ I. Since on A the derivative v0 is constant we see that

Z Z a2 0 p−2 0 0 0 0 p−2 0 0 0 |v (x)| v (x)(u (x) − v (x)) dx = |v (a1)| v (a1) (u (x) − v (x)) dx = 0 , A a1 using that u(ai) = v(ai), i = 1, 2 since ai ∈ I. Hence (9) follows from (10) ﬁrst for p ∈ [2, ∞) and then for p = ∞ by a limit process.

Remark 2.2. We deduce from Lemma 2.1 that any minimiser of W in M is necessarily concave.

Let us next recall some results for the unconstrained case. Suppose that u ∈ C4(0, 1) solves (6) in (0, 1). We introduce the auxiliary function

u00(x) v : [0, 1] → R, v(x) = 5 , (1 + (u0(x))2) 4 which already appears in the work of Euler [11]. In view of Proposition 3.2 (b) below with δ = 0 we obtain that v satisﬁes the following second order ordinary diﬀerential equation

0 ! 0 d v (x) κ(x)u (x) 0 − 3 + 1 v (x) = 0 , dx (1 + (u0(x))2) 4 (1 + (u0(x))2) 4

(this is (21) with δ = 0). If u is additionally symmetric then so is v and, using the maximum principle, one deduces that the function v is constant. This allows to derive an explicit formula for u. To do so, consider the function Z s c0 c0 1 G : R → (− , ),G(s) := 5 dτ , (11) 2 2 0 (1 + τ 2) 4 with Z 1 √ Γ(3/4) c0 := dτ = π ' 2.396280469.... . (12) 2 5 R (1 + τ ) 4 Γ(5/4) This function is bijective and strictly increasing with G0(s) > 0, hence invertible. We have

5 Lemma 2.3 ([8], Lemma 4). For c ∈ (0, c0) the functions 2 2 [0, 1] 3 x 7→ uc(x) := − (13) p4 −1 c 2 p4 −1 c 2 c 1 + (G ( 2 − cx)) c 1 + (G ( 2 )) are solutions of (6) on (0, 1) and satisfy uc(0) = uc(1) = 0. Furthermore, we have

00 0 −1 c uc (x) uc(x) = G ( − cx), 5 = −c, x ∈ [0, 1]. (14) 2 0 2 (1 + uc(x) ) 4

2 In particular, W (uc) = c .

In addition, the family of functions (uc)c∈(0,c0) is increasing in c and converges uniformly as c ↑ c0 to the function U0 : [0, 1] → R given by  2  , x ∈ (0, 1), p4 −1 c0 2 U0(x) := c0 1 + (G ( 2 − c0x)) (15)  0, x = 0, 1.

2 Although U0 has unbounded slope at x = 0 and x = 1 it is smooth as a curve in R with 2 corresponding elastic energy given by c0. We now use the functions uc in order to derive an upper bound on inf{W (v): v ∈ M}. Observe that the bound is independent of the obstacle.

Lemma 2.4 (Upper bound on the minimal energy). For any ψ satisfying (2) there holds

2 inf{W (v): v ∈ M} ≤ inf{W (v): v ∈ Msym} ≤ c0 .

1 Proof. To begin, note that (2) implies the existence of 0 < x¯ < 2 such that ψ(x) < 0 for x ∈ [0, x¯] ∪ [1 − x,¯ 1]. (16)

1 For 0 < δ < 2 we make the ansatz

 −1 c  G ( 2 )x , 0 ≤ x ≤ δ   x−δ −1 c vδ(x) = (1 − 2δ)uc 1−2δ + G ( 2 )δ , δ < x < 1 − δ (17)   −1 c G ( 2 )(1 − x) , 1 − δ ≤ x ≤ 1.

0 0 −1 c By Lemma 2.3 we have uc(0) = uc(1) = 0 and uc(0) = −uc(1) = G ( 2 ) so that vδ ∈ 2,2 1,2 W ∩ W0 (0, 1). Furthermore, vδ is symmetric and satisﬁes

Z 1−δ 2 r 1 x − δ 0 x − δ 2 W (vδ) = 2 κ[uc] 1 + uc dx (1 − 2δ) δ 1 − 2δ 1 − 2δ Z 1 1 2p 0 2 1 1 2 = κ[uc](y) 1 + uc(y) dy = W (uc) = c . 1 − 2δ 0 1 − 2δ 1 − 2δ Now, given > 0 we ﬁrst choose n o δ = min x,¯ 2 2(c0 + )

6 −1 c and then c < c0 in such a way that G ( 2 )δ ≥ maxx∈[0,1] ψ(x). With these choices one easily veriﬁes that vδ(x) ≥ ψ(x), x ∈ [0, 1] so that vδ ∈ Msym and hence

1 2 1 2 2 inf W (v) ≤ inf W (v) ≤ W (vδ) = c ≤ c0 ≤ c0 + . v∈M v∈Msym 1 − 2δ 1 − 2δ Since > 0 is arbitrary, the result follows.

The following lemma contains useful compactness and lower semicontinuity properties of the functional W provided that a uniform bound on the ﬁrst order derivatives is available.

Lemma 2.5. Suppose that (uk)k∈N ⊂ M is a sequence, which satisﬁes 0 W (uk) ≤ C, max |uk(x)| ≤ C (18) x∈[0,1] uniformly in k ∈ N. Then there exists a subsequence (ukj )j∈N and u ∈ M such that 00 00 2 1 uk * u in L (0, 1), ukj → u in C ([0, 1]) , j → ∞ and W (u) ≤ lim inf W (ukj ). j j→∞

Proof. In view of our assumptions (18) the sequence (uk)k∈N is uniformly bounded in 2,2 1,2 2,2 1,2 W (0, 1) ∩ W0 (0, 1) and hence there exists u ∈ W (0, 1) ∩ W0 (0, 1) and a subse- 2,2 quence (ukj )j∈N, such that ukj * u in W (0, 1), j → ∞. Using a well–known embedding 1 result we obtain that ukj → u in C ([0, 1]), j → ∞. Clearly u ∈ M. Moreover

1 00 1 00 00 Z u (x) Z uk (x) u (x) W (u) = lim u00 (x) dx = lim j dx kj 5 0 5 5 j→∞ 0 (1 + u0(x)2) 2 j→∞ 0 (1 + u (x)2) 4 (1 + u0(x)2) 4 kj 1 1 ≤ lim inf W (uk ) 2 W (u) 2 , j→∞ j completing the proof of the lemma.

3 Existence via penalisation

Let us deﬁne for δ ∈ (0, 1] the functional on M Z 1 δ 0 6 Wδ(u) := W (u) + |u (x)| dx, u ∈ M. 6 0 Lemma 3.1 (Existence of the minimiser for the penalised problem). For every δ ∈ (0, 1] there exists a concave function uδ ∈ M such that

Wδ(uδ) = inf Wδ(u) . u∈M

Proof. Let (uk)k∈N ⊂ M be a sequence with Wδ(uk) & infu∈M Wδ(u), k → ∞. In view of Lemma 2.1 we may assume that the functions uk are concave. Then we have for k ∈ N 1 1 00 Z Z |u (x)| 5 00 k 0 2 4 |uk(x)|dx = 5 (1 + uk(x) ) dx 0 2 4 0 0 (1 + uk(x) ) ! 1 1 Z 1 00 2 2 Z 1 2 |uk(x)| 0 5 ≤ C 5 dx (1 + |uk(x)| )dx 0 2 2 0 (1 + uk(x) ) 0 ≤ C(δ) Wδ(uk) + 1 ≤ C˜(δ).

7 0 Since uk(0) = uk(1) = 0 there exists ξk ∈ (0, 1) with uk(ξk) = 0 and hence Z 1 0 00 ˜ |uk(x)| ≤ |uk(y)|dy ≤ C(δ), x ∈ [0, 1], k ∈ N. 0 Therefore we may deduce from Lemma 2.5 that there exists a subsequence, that we still denote by (uk)k∈N, and uδ ∈ M such that 00 00 2 1 uk * uδ in L (0, 1), uk → uδ in C ([0, 1]) , k → ∞ and W (uδ) ≤ lim inf W (uk). k→∞

R 1 0 6 R 1 0 6 In particular, uδ is concave. Since 0 uk(x) dx → 0 uδ(x) dx we deduce that uδ is a minimum of Wδ.

Clearly, the minimiser uδ constructed in Lemma 3.1 satisﬁes Z 1 0 0 5 0 0 W (uδ)(v − uδ) + δ uδ(x) (v (x) − uδ(x))dx ≥ 0 ∀v ∈ M. (19) 0 The following result establishes the smoothness of a solution of (19) away from the obstacle. It is convenient to include at this stage the case δ = 0, so that we infer the subsequent properties also for the functional W (even though the existence of a solution of (19) for δ = 0 hasn’t yet been established). For convenience we temporarily drop the index δ from the solution uδ. Proposition 3.2 (Regularity and Euler-Lagrange equation above the obstacle). Suppose that u ∈ M solves (19) for some δ ∈ [0, 1] and that u(x) > ψ(x) for all x ∈ E := (x1, x2) ⊂ (0, 1). Then: (a) u ∈ C∞(E¯) and u satisﬁes on E ! 1 d κ0(x) 1 5 + κ3(x) − δ(u0(x))4u00(x) = 0 . (20) p1 + (u0(x))2 dx p1 + (u0(x))2 2 2

00 1 u (x) 0 2 4 (b) The function v(x) := κ(x)(1 + u (x) ) = 5 with κ = κ[u] satisﬁes on E (1 + u0(x)2) 4

0 ! 0 d v (x) κ(x)u (x) 0 5 0 4 00 p 0 2 − 3 + 1 v (x) = − δ(u (x)) u (x) 1 + (u (x)) . (21) dx (1 + (u0(x))2) 4 (1 + (u0(x))2) 4 2

∞ Proof. (a) Extending ϕ ∈ C0 (E) trivially to [0, 1] it follows from our assumptions that u + εϕ ∈ M for |ε| ≤ 0. Hence by (5) Z 1 0 = W 0(u)(ϕ) + δ u0(x)5 ϕ0(x)dx (22) 0 Z x2 ϕ00(x) Z x2 u0(x) Z 1 = 2 κ(x) dx − 5 κ2(x) ϕ0(x) dx + δ u0(x)5 ϕ0(x)dx 0 2 p 0 2 x1 1 + u (x) x1 1 + u (x) 0

∞ 2,2 for any ϕ ∈ C0 (E). By density the above relation holds for any ϕ ∈ W0 (E). ∞ In what follows we adopt ideas from [6, Proof of Thm. 3.9]. For η ∈ C0 (E) let Z x Z y 2 3 ϕ(x) := η(s) ds dy + α(x − x1) + β(x − x1) , x ∈ E, x1 x1

8 with 1 Z x2 3 Z x2 Z y α = η(y) dy − 2 η(s) ds dy , x2 − x1 x1 (x2 − x1) x1 x1 α 1 Z x2 Z y β = − − 3 η(s) ds dy . x2 − x1 (x2 − x1) x1 x1 2,2 Then ϕ ∈ W0 (E) and |α|, |β|, kϕkC1 ≤ CkηkL1 , with C depending only on x1, x2. Inserting ϕ into (22) we obtain

Z x2 η(x) ∞ κ(x) dx ≤ Ckηk 1 ∀η ∈ C (E) 0 2 L 0 x1 1 + u (x) 2,2 0 with a constant C depending only on the W -norm of u, δ and x1, x2. Since u is bounded, the inequality above shows that κ ∈ L∞(E) and hence u ∈ W 2,∞(E) = C1,1(E¯). 1,∞ ∞ Let us next show that κ ∈ W (E). For η ∈ C0 (E) consider

Z x Z x2 3 2 2 3 ϕ(x) := η(y) dy + η(y) dy − 2 (x − x1) + 3 (x − x1) . x1 x1 (x2 − x1) (x2 − x1)

2,2 0 ∞ Then, ϕ ∈ W0 (E) with kϕkC0 ≤ CkηkL1 and kϕ kL1 ≤ CkηkL1 . Using the L -bound on the curvature, we ﬁnd with this choice of ϕ in (22)

Z x2 0 η (x) ∞ κ(x) dx ≤ Ckηk 1 ∀η ∈ C (E), 0 2 L 0 x1 1 + u (x) from which we deduce that κ ∈ W 1,∞(E) and hence u ∈ W 3,∞(E) = C2,1(E¯). Integrating by parts in (22) and using (1) we ﬁnd that

Z x2 ϕ0(x) Z x2 u0(x) Z 1 −2 κ0(x) dx − κ2(x) ϕ0(x) dx + δ u0(x)5 ϕ0(x)dx = 0 0 2 p 0 2 x1 1 + u (x) x1 1 + u (x) 0 ∞ for all ϕ ∈ C0 (E), so that there exists a constant c ∈ R such that κ0(x) 1 u0(x) δ + κ2(x) − u0(x)5 = c a.e. in E. 1 + u0(x)2 2 p1 + u0(x)2 2

From this relation we deduce that κ ∈ W 2,∞(E) and hence that u ∈ W 4,∞(E) = C3,1(E¯). Diﬀerentiating the above equation then yields (20). The smoothness of u follows with the help of a bootstrap argument. (b) In the case δ = 0 this has been established in [8, Lem.3] and we repeat here only the main steps of the computation. Since v0(x) κ0(x) 1 = + κ(x)2u0(x) 3 p 0 2 (1 + u0(x)2) 4 1 + u (x) 2 one sees that ! ! d v0(x) d κ0(x) 1 = + κ(x)2u00(x) + κ(x)κ0(x)u0(x). 3 p 0 2 dx (1 + u0(x)2) 4 dx 1 + u (x) 2 Combining the above relations with (20) the identity (21) follows.

9 Corollary 3.3. For any δ ∈ [0, 1] every solution u ∈ M of (19) belongs to C∞ in a neighbourhood of x = 0 and x = 1 and satisﬁes κ(0) = κ(1) = 0.

Proof. Since u(0) > ψ(0), there isx ˆ > 0 such that u(x) > ψ(x), x ∈ [0, xˆ]. Choose ϕ ∈ C2([0, 1]) such that ϕ(0) = 0, ϕ0(0) 6= 0, supp(ϕ) ⊂ [0, xˆ) and u ± ϕ ∈ M. Then by (19) u satisﬁes Z 1 0 = W 0(u)(ϕ) + δ u0(x)5 ϕ0(x)dx . (23) 0 Applying Proposition 3.2 with E = (0, xˆ) we obtain that u ∈ C∞([0, xˆ]) and that (20) holds in (0, xˆ). Integrating by parts in (23) we derive

κϕ0 xˆ κ(0)ϕ0(0) 0 = 2 02 = −2 0 2 . 1 + u 0 1 + u (0) It follows that κ(0) = 0 and in the same way we obtain κ(1) = 0.

Corollary 3.4 (Touching the obstacle). Suppose that u ∈ M is concave and solves (19) for some δ ∈ [0, 1]. Then there exists x0 ∈ (0, 1) such that u(x0) = ψ(x0). Proof. Assume by contradiction that u(x) > ψ(x) for all x ∈ [0, 1]. We deduce from Proposition 3.2 (a) that u ∈ C∞([0, 1]) and that u solves (20) in (0, 1). Let us denote by v the function deﬁned in Proposition 3.2 (b). Then in view of (21) v satisﬁes on (0, 1)

0 ! 0 d v (x) κ(x)u (x) 0 5 0 4 00 p 0 2 − 3 + 1 v (x) = − δ(u (x)) u (x) 1 + (u (x)) ≥ 0 , dx (1 + (u0(x))2) 4 (1 + (u0(x))2) 4 2 since u is concave. The maximum principle and Corollary 3.3 yield

0 2 1 0 2 1 v(x) ≥ min(v(0), v(1)) = min(κ(0)(1 + u (0) ) 4 , κ(1)(1 + u (1) ) 4 ) = 0, x ∈ [0, 1].

On the other hand, the concavity of u implies that v(x) ≤ 0, x ∈ [0, 1], so that v ≡ 0 on [0, 1]. This shows that u00 ≡ 0 on [0, 1] and therefore u ≡ 0 since u(0) = u(1) = 0. But then, ψ(x) ≤ u(x) = 0, x ∈ [0, 1] contradicting (2).

Let us return to the problem of minimising W over M. Our aim is to obtain a minimum 0 by considering the above sequence (uδ)0<δ≤1 and proving a bound on maxx∈[0,1] |uδ(x)| which is uniform in δ. This will be achieved under an additional condition on the obstacle ψ. In order to formulate this condition we assume that ψ ∈ C2(0, 1) and deﬁne the set

A := {x ∈ (0, 1) : ψ(x) ≥ 0 and ψ00(x) ≤ 0} .

Theorem 3.5. Suppose that ψ ∈ C2(0, 1) and that

( 00 00 ) n 0 ψ (x)x o n 0 ψ (x)(1 − x) o 1 max max G(ψ (x)) − 5 , max G(−ψ (x)) − 5 < c0. x∈A (1 + ψ0(x)2) 4 x∈A (1 + ψ0(x)2) 4 2 (24) Then infv∈M W (v) is attained.

10 Proof. Let us ﬁx δ ∈ (0, 1] and denote by uδ ∈ M the minimum of Wδ constructed in Lemma 3.1. Since uδ is concave we have that

0 0 0 max |uδ(x)| = max{uδ(0), −uδ(1)}. (25) x∈[0,1]

In view of Corollary 3.4 there exists x0 ∈ (0, 1) with uδ(x0) = ψ(x0) so that we may deﬁne

x1 := inf{x ∈ (0, 1) : uδ(x) = ψ(x)} > 0.

∞ Hence uδ > ψ in (0, x1) and Proposition 3.2 (a) implies that uδ ∈ C ([0, x1]). Proceeding 0 2 1 as in the proof of Corollary 3.4 we infer that the function vδ(x) = κδ(x)(1 + (uδ(x)) ) 4 satisﬁes 00 uδ (x1 − 0) vδ(x) ≥ min(vδ(0), vδ(x1 − 0)) = vδ(x1 − 0) = 5 , x ∈ (0, x1), (26) 0 2 4 (1 + uδ(x1) )

0 2 1 since vδ(0) = κδ(0)(1 + uδ(0) ) 4 = 0 by Corollary 3.3 and vδ(x1 − 0) ≤ 0 by the concavity of uδ. Here, vδ(x1 − 0) denotes the leftsided limit of vδ at x1. Using the fact that uδ − ψ has a local minimum at x1 and is twice continuously diﬀerentiable on [0, x1] we have that 0 0 00 00 uδ(x1) = ψ (x1) and ψ (x1) ≤ uδ (x1 − 0) ≤ 0, so that in particular x1 ∈ A. As a result, we infer from (26)

00 00 00 d 0 uδ (x) uδ (x1 − 0) ψ (x1) G(uδ(x)) = 5 = vδ(x) ≥ 5 ≥ 5 , x ∈ (0, x1), dx 0 2 4 0 2 4 0 2 4 (1 + uδ(x) ) (1 + uδ(x1) ) (1 + ψ (x1) ) from which we obtain after integration

00 0 0 ψ (x1) G(uδ(x1)) − G(uδ(0)) ≥ x1 5 . 0 2 (1 + ψ (x1) ) 4

0 0 Rearranging and recalling that uδ(x1) = ψ (x1) we end up with

00 00 0 0 ψ (x1)x1 n 0 ψ (x)x o G(uδ(0)) ≤ G(ψ (x1)) − 5 ≤ max G(ψ (x)) − 5 0 2 x∈A 0 2 (1 + ψ (x1) ) 4 (1 + ψ (x) ) 4 since x1 ∈ A. In a similar way we obtain

00 0 n 0 ψ (x)(1 − x) o G(−uδ(1)) ≤ max G(−ψ (x)) − 5 . x∈A (1 + ψ0(x)2) 4

Abbreviating

( 00 00 ) n 0 ψ (x)x o n 0 ψ (x)(1 − x) o β = max max G(ψ (x)) − 5 , max G(−ψ (x)) − 5 (27) x∈A (1 + ψ0(x)2) 4 x∈A (1 + ψ0(x)2) 4

1 we infer with the help of (25) and the fact that β < 2 c0 that 0 −1 max |uδ(x)| ≤ G (β) (28) x∈[0,1] uniformly in δ ∈ (0, 1].

11 In order to complete the proof we choose a sequence (δk)k∈N such that δk ∈ (0, 1], limk→∞ δk =

0 and set uk := uδk . Using (28) we infer from Lemma 2.5, that there exists a subsequence, again denoted by (uk)k∈N, and u ∈ M such that for arbitrary v ∈ M

W (u) ≤ lim inf W (uk) ≤ lim inf Wδ (uk) ≤ lim inf Wδ (v) = W (v), k→∞ k→∞ k k→∞ k

since uk is a minimum of Wδk . Hence, W (u) = infv∈M W (v). Note that the expression that appears in condition (24) is quite natural since we have in view of (14) for c ∈ (0, c0)

00 0 uc (x)x c G(uc(x)) − 5 = , ∀x ∈ [0, 1]. (29) 0 2 2 (1 + uc(x) ) 4 We use this observation in order to derive a necessary condition for (24).

2 1 Lemma 3.6. Suppose that ψ ∈ C (0, 1) and that β in (27) satisﬁes 0 ≤ β < 2 c0. Then ψ(x) ≤ u2β(x), x ∈ [0, 1].

Proof. Assume that minx∈[0,1](u2β − ψ)(x) < 0. Due to continuity there exists a c ∈ (2β, c0) such that minx∈[0,1](uc − ψ)(x) < 0. Choose x0 ∈ [0, 1] such that (uc − ψ)(x0) = 0 0 00 minx∈[0,1](uc − ψ)(x). Then (uc − ψ)(x0) < 0, x0 ∈ (0, 1), ψ (x0) = uc(x0) and ψ (x0) ≤ 00 uc (x0) < 0. Since ψ(x0) > uc(x0) we deduce that x0 ∈ A which combined with (29) yields

00 00 c 0 uc (x0)x0 0 ψ (x0)x0 = G(uc(x0)) − 5 ≤ G(ψ (x0)) − 5 ≤ β, 2 0 2 0 2 (1 + uc(x0) ) 4 (1 + ψ (x0) ) 4 a contradiction. Hence minx∈[0,1](u2β − ψ)(x) ≥ 0 and the result follows.

4 Existence via a bound on the functional

Our starting point is the following observation. If u ∈ M then by the deﬁnition of the function G we ﬁnd Z 1 00 0 0 0 0 u (x) p G(u (0)) + G(−u (1)) = G(u (0)) − G(u (1)) = − 5 dx ≤ W (u) , 0 (1 + u0(x)2) 4 and hence G(max{u0(0), −u0(1)}) + G(min{u0(0), −u0(1)}) ≤ pW (u) . (30) 0 If we want to exploit (30) in order to derive an upper bound on maxx∈[0,1] |u (x)| it is useful to obtain some control on min{u0(0), −u0(1)}. To do so, we introduce the quantity

n nψ(x) o n ψ(x) oo α := min max : x ∈ (0, 1) , max : x ∈ (0, 1) . x 1 − x

c0 Theorem 4.1. If the obstacle ψ satisﬁes ψ(x) < ucˆ(x), x ∈ [0, 1] with cˆ := 2 + G(α) then inf W (u) is attained. u∈M

12 Proof. Let (uk)k∈N be a minimising sequence in M. Lemma 2.1 allows us to assume that all elements of the minimising sequence are concave. Due to the assumption on ψ and by a continuity argument we infer that there exists δ > 0 (depending only on ψ) such that ψ(x) ≤ ucˆ−δ(x), x ∈ [0, 1]. Hence ucˆ−δ ∈ M and we may assume that

2 W (uk) ≤ W (ucˆ−δ) = (ˆc − δ) for any k ∈ N .

0 0 In order to make use of (30) we now derive a lower bound for min{uk(0), −uk(1)}. Fix k ∈ N. For any x ∈ (0, 1) we have ψ(x) u (x) u (x) − u (0) ≤ k = k k = u0 (ξ) ≤ u0 (0) x x x − 0 k k for some ξ ∈ (0, x) in view of the concavity of uk. Hence ψ(x) u0 (0) ≥ max : x ∈ (0, 1) ≥ α k x and similarly we derive

ψ(x) −u0 (1) ≥ max : x ∈ (0, 1) ≥ α. k 1 − x Combining these bounds with the monotonicity of G we deduce that

0 0 G(min{uk(0), −uk(1)}) ≥ G(α), k ∈ N.

Inserting this estimate and the bound on W (uk) into (30) and recalling the deﬁnition ofc ˆ we get c G(max{u0 (0), −u0 (1)}) ≤ cˆ − δ − G(α) = 0 − δ . k k 2 It follows that for all k ∈ N

0 0 0 −1 c0 max |uk(x)| = max{uk(0), −uk(1)} ≤ G ( − δ) = C x∈[0,1] 2 and the existence of a minimiser follows from Lemma 2.5 using that (uk)k∈N is a minimising sequence.

4.1 The symmetric case

The above arguments simplify considerably if we look for minimisers in the class Msym deﬁned in (8).

2 Lemma 4.2. Suppose that W (˜u) < c0 for some u˜ ∈ Msym or that ψ(x) < U0(x), x ∈ [0, 1]. Then inf W (v) is attained. v∈Msym

2 Proof. Suppose ﬁrst that W (˜u) < c0 for someu ˜ ∈ Msym. Let (uk)k∈N ⊂ Msym be a minimising sequence for W , whose elements can be chosen to be concave functions in view 0 0 0 of Lemma 2.1. In particular, maxx∈[0,1] |uk(x)| = uk(0) = −uk(1), so that (30) simpliﬁes to

0 1p G( max |uk(x)|) ≤ W (uk), k ∈ N. (31) x∈[0,1] 2

13 2 Since by our assumption inf W (v) < c0 we may assume that there exists δ > 0 such v∈Msym 2 that W (uk) ≤ (c0 − δ) for all k ∈ N. Inserting this bound into (31) we infer that

0 −1 c0 − δ max |uk(x)| ≤ G ( ), k ∈ N, x∈[0,1] 2 and the existence of a minimiser follows by applying Lemma 2.5 and using that (uk)k∈N is a minimising sequence. Next, if ψ(x) < U0(x), x ∈ [0, 1], then by a continuity argument there exists c ∈ (0, c0) such 2 2 that ψ(x) ≤ uc(x), x ∈ [0, 1]. Hence, uc ∈ Msym with W (uc) = c < c0 and we are back to the previous case.

We conjecture that the minimum obtained in Lemma 4.2 also minimises W over M provided that the obstacle ψ is symmetric. While we are currently unable to prove a corresponding result, it is not diﬃcult to see that a solution u ∈ Msym of the variational inequality

0 W (u)(v − u) ≥ 0 ∀v ∈ Msym (32) also solves (4). To show this, let us write an arbitrary function v ∈ M as v = v1 + v2, 1 where v1(x) = 2 (v(x) + v(1 − x)) and v2 = v − v1. Clearly, v1 ∈ Msym and v2 satisﬁes 0 v2(1 − x) = −v2(x). In view of (5) we see that W (u)(v2) = 0 so that by (32)

0 0 W (u)(v − u) = W (u)(v1 − u) ≥ 0, ∀v ∈ M, and hence u is a solution of (4). Our ﬁnal result shows that the minimum is not attained if the obstacle is too large. More precisely we have:

1 8 Lemma 4.3. Let ψ be symmetric with ψ( 2 ) > c2 . Then inf W (v) is not attained. 0 v∈Msym

Proof. Suppose that there exists u ∈ Msym such that W (u) = inf W (v). Then, in view v∈Msym 2 1 of Lemma 2.4, W (u) ≤ c0. We calculate for 0 ≤ t < 2

Z t 00 Z 1 00 0 0 1 u (s) 2 |u (s)| G(u (t)) = G(u ( )) + 5 ds ≤ 5 ds 2 1 0 2 4 0 2 4 2 (1 + u (s) ) t (1 + u (s) ) 1 1 1 Z 00 2 1 1 1 2 |u (s)| 2 1 1 2 2 2 ≤ − t 5 ds ≤ − t W (u) 2 0 (1 + u0(s)2) 2 2 2 1 c0 ≤ 1 − 2t 2 . 2

0 c0 1 In particular we have G(u (t)) < 2 , 0 < t < 2 so that we deduce

0 −1 0 −1 1 c0 1 u (t) = G (G(u (t)) ≤ G (1 − 2t) 2 , 0 < t < . 2 2

1 Integrating the above inequality over (0, 2 ), performing several changes of variable as well

14 as integration by parts we obtain Z 1 Z c0 1 2 1 c0 4 2 −1 2 −1 u( ) ≤ G (1 − 2t) dt = 2 τG (τ)dτ 2 0 2 c0 0 Z ∞ Z ∞ 4 s 8 d 1 = 2 G(s) 5 ds = − 2 G(s) 1 ds c0 0 (1 + s2) 4 c0 0 ds (1 + s2) 4 " #s=∞ 8 Z ∞ 1 8 G(s) = 2 3 ds − 2 1 c 0 2 2 c 2 4 0 (1 + s ) 0 (1 + s ) s=0 8 s s=∞ 8 1 = √ = < ψ( ), 2 2 2 c0 1 + s s=0 c0 2 which is a contradiction.

5 Global regularity

Suppose that u ∈ M is a solution of the variational inequality

W 0(u)(v − u) ≥ 0 ∀v ∈ M. (33)

We infer from Proposition 3.2 and Corollary 3.3 that u is C∞ on {u > ψ} and satisfies (6) on (0, 1) ∩ {u > ψ} as well as (7). Furthermore, by Corollary 3.4 there exists x0 ∈ (0, 1) such that u(x0) = ψ(x0). In what follows we address the global regularity of a solution u of (33). Our aim is to show that u has continuous second derivatives and that the third derivative is of bounded variation on (0, 1). This result was obtained by Cimatti in [5] for a solution of the biharmonic variational inequality on a interval and we follow his approach. Due to the nonlinearity an extra step is needed in this case to control the second derivative of the solution. In higher dimensions regularity results on the solution of the linear fourth n order obstacle problem on a bounded domain Ω ⊂ R have been established by Frehse in [12], [13] and by Caffarelli and Friedman in [2] (see also [3]). Frehse has shown in [12] that 3,2 2,∞ u ∈ Wloc (Ω) and in [13] that u ∈ Wloc (Ω). Caffarelli and Friedman in [2] studied the regularity up to the boundary and in particular prove that in dimension two the solution is twice continuously differentiable. Theorem 5.1. Let u ∈ M be a solution of (33). Then u ∈ C2([0, 1]) and u000 ∈ BV (0, 1). Proof. Since u ∈ M is a solution of (33), the map

Z 1 ϕ00(x) Z 1 u0(x) C∞(0, 1) 3 ϕ 7→ 2 κ(x) dx − 5 κ(x)2 ϕ0(x) dx , 0 0 2 p 0 2 0 1 + u (x) 0 1 + u (x) is a non-negative distribution. Hence by [21, Lem. 37.2] there exists a non-negative Radon measure µ such that Z 1 ϕ00(x) Z 1 u0(x) Z 1 2 κ(x) dx − 5 κ(x)2 ϕ0(x) dx = ϕ(x) µ(dx) , 0 2 p 0 2 0 1 + u (x) 0 1 + u (x) 0

∞ for all ϕ ∈ C0 (0, 1). With Proposition 3.2 we see that the measure µ has support on the coincidence set {x ∈ (0, 1) : u(x) = ψ(x)}. Let m be the distribution function of µ, i.e.

m(x) = µ(0, x) for all x ∈ (0, 1) .

15 Then m is a function of bounded variation, more precisely it is monotonically increasing and right-continuous ([1, Thm. 2.30]), and

Z 1 Z 1 0 ∞ ϕ(x) µ(dx) = − ϕ (x)m(x)dx for all ϕ ∈ C0 (0, 1), 0 0 see [20, Chap.III, Thm 14.1]. Since the relation

Z 1 ϕ00(x) Z 1 u0(x) Z 1 2 κ(x) dx = 5 κ(x)2 ϕ0(x) dx − ϕ0(x)m(x) dx , (34) 0 2 p 0 2 0 1 + u (x) 0 1 + u (x) 0

∞ is then valid for all ϕ ∈ C0 (0, 1), by a density argument we see that it is satisﬁed for all 2,2 ϕ ∈ W0 (0, 1). Proceeding as in the proof of Proposition 3.2 (a) one proves ﬁrst that κ ∈ L∞(0, 1) and 1,∞ then that κ ∈ W (0, 1). Notice that in this case we have x1 = 0 and x2 = 0 so that the special test functions ϕ simplify considerably. The presence of the extra term

Z 1 ϕ0(x)m(x) dx 0 does not create diﬃculties since the function m is bounded. From the gained regularity for κ it follows that u ∈ W 3,∞(0, 1) and, in particular, that κ and the second derivative of u are continuous. So we may integrate once by parts in (34) and obtain ! Z 1 κ0(x) u0(x) 2 + κ(x)2 − m(x) ϕ0(x) dx = 0 , 0 2 p 0 2 0 1 + u (x) 1 + u (x)

2,2 for all ϕ ∈ W0 (0, 1). Hence there exists c ∈ R such that κ0(x) u0(x) 2 + κ(x)2 − m(x) = c 1 + u0(x)2 p1 + u0(x)2 and u000 ∈ BV (0, 1).

6 Numerical approximation

In this section we use a numerical method to calculate approximate solutions of the variational inequality (4). In [14] a numerical method for an obstacle problem with the energy functional W given in (3) is developed. The author introduces an approximation of W which acts on the space of continuous, piecewise linear functions, while the obstacle constraint is incorporated via a penalty term. The resulting discrete functionals are shown to Γ-converge to the functional W in a setting where a uniform bound on the ﬁrst derivatives is available. Our approach slightly diﬀers from the one presented in [14] in that we discretize the Euler-Lagrange equation for the following penalised functional 2,2 1,2 Wγ : W (0, 1) ∩ W0 (0, 1) → R with Z 1 Z 1 2p 0 2 γ 2 Wγ(u) := κ(x) 1 + u (x) dx + (u(x) − ψ(x))− dx, γ 1, 0 2 0

16 2,2 1,2 where a− := min{0, a}. For the ﬁrst variation of Wγ in direction ϕ ∈ W (0, 1)∩W0 (0, 1) we derive as in (5)

Z 1 ϕ00(x) Z 1 u0(x) W 0 (u)(ϕ) = 2 κ(x) dx − 5 κ(x)2 ϕ0(x) dx γ 0 2 p 0 2 0 1 + u (x) 0 1 + u (x) Z 1 +γ (u(x) − ψ(x))− ϕ(x) dx 0 Z 1 κ0(x)ϕ0(x) Z 1 u0(x) = −2 dx − κ(x)2 ϕ0(x) dx 0 2 p 0 2 0 1 + u (x) 0 1 + u (x) κϕ0 1 Z 1 +2 02 + γ (u(x) − ψ(x))− ϕ(x) dx 1 + u 0 0 where we have integrated by parts and used the relation (1). Introducing κ as a second dependent variable, critical points of Wγ are therefore given by solutions (u, κ) of the system

Z 1 κ0(x)ϕ0(x) Z 1 u0(x) −2 dx − κ(x)2 ϕ0(x) dx 0 2 p 0 2 0 1 + u (x) 0 1 + u (x) Z 1 1,2 +γ (u(x) − ψ(x))− ϕ(x) dx = 0 ∀ϕ ∈ W0 (0, 1) (35) 0 Z 1 Z 1 u0(x) κ(x)η(x) dx + η0(x) dx = 0 ∀η ∈ W 1,2(0, 1) (36) p 0 2 0 0 0 1 + u (x) together with the boundary conditions

u(0) = u(1) = κ(0) = κ(1) = 0. (37)

Note that (36) is simply (1) written in variational form. We use a finite element method 1 in order to solve (35)–(37) numerically. Let xj = jh, j = 0,...,N + 1 with h = N+1 and define the finite element space

0 1,2 Vh := {ϕh ∈ C ([0, 1]) | ϕh|[xj−1,xj ] ∈ P1, 1 ≤ j ≤ N + 1, ϕh(0) = ϕh(1) = 0} ⊂ W0 (0, 1), where P1 denotes the space of polynomials of degree ≤ 1. The scheme now reads: Find (uh, κh) ∈ Vh × Vh such that

Z 1 κ0 (x)ϕ0 (x) Z 1 u0 (x) −2 h h dx − κ (x)2 h ϕ0 (x) dx 1 + u0 (x)2 h p 0 2 h 0 h 0 1 + uh(x) Z 1 +γ (uh(x) − ψ(x))− ϕh(x) dx = 0 ∀ϕh ∈ Vh (38) 0 Z 1 Z 1 u0 (x) κ (x)η (x) dx + h η0 (x) dx = 0 ∀η ∈ V . (39) h h p 0 2 h h h 0 0 1 + uh(x)

Note that the functions φ1, . . . , φN ∈ Vh with φj(xk) = δjk, j = 1, . . . , N, k = 0,...,N + 1 form a basis of Vh. Expanding

N N X X uh = ujφj, κh = κjφj j=1 j=1

17 and using ϕh = φj, ηh = φj in (38) and (39) respectively we ﬁnd that (38), (39) is equivalent 2N 2N to a system F (u, κ) = 0, where u = (u1, . . . , uN ), κ = (κ1, . . . , κN ) and F : R → R . In practice this system was solved iteratively with the help of a descent method which reduced the value of Wγ in each step. In our calculations we obtained good results when coupling penalisation parameter and grid size according to γ = 2h−3. In our ﬁrst two examples we have chosen the following non-symmetric obstacles 1 π ψ (x) := sin(2πx − ) and ψ (x) = −22(x − .97)(x − .4)(x − .5)(x − .02) (40) 1 2 4 2 and the corresponding solutions are shown in Figure 1. In our second test we consider the symmetric situation and we choose the symmetric obstacle

ψ(x) := αx(1 − x) − 0.1, (41)

1 1 with α > 0.4 in order to guarantee that maxx∈[0,1] ψ(x) > 0. Note that ψ( 2 ) < U0( 2 ) for α < 8 + 0.4 = α∗ ≈ 3.75. Comparing the numerical plots of the functions U and ψ we c0 0 ∗ see that indeed ψ < U0 if α < α (see Figure 3). In this range of the parameter α, Lemma 4.2 would ensure the existence of a minimum in the class of symmetric functions. Figure 2 show the corresponding solutions for α = 2 and α = 3.7 respectively. In the latter ﬁgure one observes that the solution develops large derivatives near the boundary and the algorithm is not able to calculate solutions closer to α∗. This suggests that the minimisation problem might not have a solution in M if the condition ψ(x) < U0(x), x ∈ [0, 1] is violated.

0.5

0.4 0.4

0.3

0.2 0.2

0.1

0 0

−0.1

−0.2 -0.2

−0.3

−0.4 -0.4

−0.5 -0.6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 1: Solutions for the obstacles ψ1 (left) and ψ2 (right) given in (40).

References

[1] Giuseppe Buttazzo, Mariano Giaquinta, and Stefan Hildebrandt. One-dimensional variational problems, volume 15 of Oxford Lecture Series in Mathematics and its Ap- plications. The Clarendon Press Oxford University Press, New York, 1998. An introduction. [2] Luis A. Caffarelli and Avner Friedman. The obstacle problem for the biharmonic operator. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, 6(1):151– 184, 1979. [3] Luis A. Caffarelli, Avner Friedman, and Alessandro Torelli. The two-obstacle problem for the biharmonic operator. Pacific J. Math., (2):325–335, 1982.

18 0.5 1

0.45 0.9

0.4 0.8

0.35 0.7

0.3 0.6

0.25 0.5

0.2 0.4

0.15 0.3

0.1 0.2

0.05 0.1

0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 2: Solutions for ψ in (41) with α = 2 (left) and α = 3.7 (right).

0.8

0.6

0.4 Out[19]=

0.2

0.2 0.4 0.6 0.8 1.0

Figure 3: Plot verifying the condition ψ(x) < U0(x), x ∈ [0, 1] for ψ in (41) and α = 3.7.

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Anna Dall’Acqua,, UniversitätUlm, Helmholtzstraße 18, 89081 Ulm, Germany [email protected] Klaus Deckelnick,, Otto-von-Guericke-UniversitätMagdeburg, Universitätsplatz2, 39016 Magdeburg, Ger- many [email protected]