Positivity for the clamped plate equation under high tension

In this article, we consider positivity issues for the clamped plate equation with high tension γ>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma >0$$\end{document}. This equation is given by Δ2u-γΔu=f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta ^2u - \gamma \Delta u=f$$\end{document} under clamped boundary conditions. Here, we show that given a positive f, i.e. upwards pushing, we find a γ0>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma _0>0$$\end{document} such that for all γ≥γ0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma \ge \gamma _0$$\end{document} the bending u is indeed positive. This γ0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma _0$$\end{document} only depends on the domain and the ratio of the L1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^1$$\end{document} and L∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^\infty $$\end{document} norm of f. In contrast to a recent result by Cassani and Tarsia, our approach is valid in all dimensions.


Introduction
The Boggio-Hadamard conjecture states, that for a given convex, open, bounded set Ω ⊆ ℝ n , an f ≥ 0 and outer normal of Ω , a solution u to is nonnegative, i.e. u ≥ 0 (cf. [26,27]). Problem (1.1) models the bending u of a clamped plate Ω under a force f. Hence the problem can be restated as: The conjecture was substantiated by Boggio's explicit formula [6] (see also [15,Lemma 2.27] or [28]) for the Greens function of problem (1.1) on the unit disc, because this function is positive, and furthermore by Almansi's result to calculate the Greens function for certain domains by the Greens function of the unit ball (see [3]). (1.1) Does upward pushing yield upward bending? Several other domains than the disc have been found on which such a positivity preserving property holds (see the references below). Hadamard himself claimed in [27] that such a property for all limaçons is true, which turned out to be wrong in general, though some of them still possess this property (see [15, Fig. 1.2]). Remarkable is that such limaçons are not convex. In [20], Grunau and Robert showed that positivity preserving is preserved under small regular perturbations of the domain in dimensions n ≥ 3.
The conformal invariance of the problem was also successfully used to construct domains with a positivity preserving property by, e.g. Dall'Acqua and Sweers in [10] (see also the references therein for more information on such domains).
On the other hand, several counterexamples have been found by now. The first one was by Duffin on an infinite strip [12] and shortly after Garabedian [14] showed that on an elongated ellipse the Greens function changes sign. By now even for uniform forces, i.e. f ≡ 1 , counterexamples have been found by Grunau and Sweers in [23] and [24]. We refer to [15,Sect. 1.2] for a comprehensive historical overview to this problem.
Instead of examining (1.1) for positivity, Cassani and Tarsia in [9] examined positivity issues for with > 0 big enough. The basic motivation is that for big enough, the influence of Δu ( −Δu = f satisfies positivity preserving via the maximum principle) becomes stronger than that of Δ 2 u . In more technical detail Cassani and Tarsia conjectured the existence of a 0 = 0 (f , Ω) ≥ 0 , such that u ≥ 0 for all ≥ 0 and provided a proof for dimensions n = 2, 3 , smooth, bounded Ω and positive f ∈ L 2 (Ω) . In this article we provide a different approach, which is valid for all dimensions, see Theorem 1.1.
In dimension n = 1 , this positivity preserving property is true for all > 0 independent of f. This was shown by Grunau in [19] Proposition 1.
The parameter is usually called tension, if it is positive. Several results concerning (1.2) have been achieved, which are usually concerned with vibrations of the plate, i.e. eigenvalue problems. Bickley gave some explicit calculations for the spectrum in the unit disc in [5] already in 1933. Hence the existing literature for these eigenvalue problems is quite vast and is still developing, see, e.g. [8,4] and the references therein.
Other modifications for (1.1) concerning positivity issues are, e.g. changing the boundary conditions to so called Steklov conditions. This has been examined by, e.g. Bucur and Gazzola in [7].
Different elliptic differential operators of higher order, their respective fundamental solutions and their sign close to a singularity have also been examined by Grunau, Romani and Sweers in [22] in a more systematic approach to understand better the loss of positivity preserving.
Instead of (1.2), we examine the following boundary value problem for positivity preserving. This is obviously equivalent, but (1.3) yields the advantage, that the singularity of the equation is more prominent and hence yields easier access to necessary estimates. Here and > 0 , and the solution u ∈ W 4,p (Ω) for all 1 < p < ∞ (see, e.g. [15,Corollary 2.21] and the references therein for existence, regularity and uniqueness to (1.3)).
Please note that we do not have any restrictions on the dimension, i.e. n ∈ ℕ arbitrary. Furthermore, our method yields that 0 does not depend on f directly, only on .
The strategy of the proof is as follows: The limiting problem of (1.3) is which admits a maximum principle, and establishes positivity of u on any Ω � ⊂⊂ Ω for small. We proceed by contradiction and assume Theorem 1.1 is false. Hence for every > 0 , we find a nonnegative f ∈ L ∞ (Ω) satisfying (1.4), such that u is not positive in Ω.
Then, we examine a blow-up of our solutions u , which is weighted by the supremum of the modulus of the Laplacian at the boundary, i.e. sup Ω 2 |Δu | . After a careful analysis (see Sects. 2 and 3), we can show that this blow-up converges in a suitable sense to a solution of Δ 2 u − Δu = 0 on the half-space with Dirichlet boundary conditions (see Sect. 4). With a uniqueness result shown in Appendix A, we explicitly calculate this limit and obtain positivity of the Laplacian of u on the boundary for small. This is crucial, as in the presence of Dirichlet boundary conditions in (1.3) the Laplacian is the second normal derivative of u on the boundary, and therefore positivity of the Laplacian on the boundary gives positivity of u close to the boundary, see Sect. 4.
Similar strategies of examining a blow-up to the half space and using explicit formulas have been employed by Grunau and Robert in [20] and Grunau, Robert and Sweers in [21] to show lower bounds for the Greens function of a polyharmonic operator. This method was later refined by Pulst in his PhD-thesis [30] to also obtain such estimates, if non-constant lowerorder terms are present. If variable coefficients in the principal part of the operator are given by a power of a second-order elliptic linear operator, such estimates were found by the same method by Grunau in [18].
Our blow-up strategy needs careful estimates for the singular problem (1.3). Estimates for these kinds of problems have a long history, see, e.g. [13,17,25] and [29]. We are not aware of any specific estimates, which would help in our specific situation. For this reason and for the sake of completeness, we derive them here.

Preliminary estimates
We proceed by contradiction and assume we find nonnegative f ∈ L ∞ (Ω) for ↓ 0 satisfying (1.4), such that the solution u of (1.3) is sign-changing. As (1.3) is homogeneous of degree one, we may assume by scaling By the Banach-Alaoglu Theorem, we get after passing to subsequence and relabelling In particular, we have f ≢ 0.
The limiting problem of (1.3) is thought to be the second-order boundary-value problem We also consider These two problems admit some important estimates: Proof Both problems admit by standard elliptic theory, see [16] Theorem 9.15, unique solutions u respectively u 0, ∈ W 2,p (Ω) ↪ C 1, (Ω) for all 1 < p < ∞ and 1 − (n∕p) > > 0 with In particular, the set of all u, u 0, for any f , f with (2.2) and (2.1) is compact in C 1 (Ω) , and in particular and weakly in W 2,p (Ω) for all 1 < p < ∞.
Equation (2.3) is indeed the limiting problem of (1.3) in the sense of the following proposition.
Proof Multiplying (2.10) by v and integrating by parts, we get Replacing 2 Δ 2 u by Δv in the second term with (2.10), we continue Passing to a subsequence, we get Multiplying (2.10) by some ∈ C ∞ 0 (Ω) and passing to a subsequence, we get and v ∈ W 1,2 0 (Ω) is harmonic in Ω , hence v = 0 and u → u weakly in W 1,2 0 (Ω) . Returning to (2.12), we improve now to which is (2.11). ◻ The following proposition shows that the Laplacian cannot be bounded throughout Ω. Choosing ∈ C ∞ 0 (Ω) with 0 ≤ ≤ 1 and ≡ 1 in Ω � , we get Letting ↗ Ω , we get from (2.2) that and the proposition follows.

The Laplacian on the boundary
In this section, we investigate the values of the Laplacian on the boundary and put With subscripts ±, we denote the positive, respectively, negative part, i.e.

Furthermore, we set
The quantity M will be crucial throughout the exposition. Our goal is to show that it has the same asymptotic as itself, i.e. we find constants c 0 , C > 0 such that for > 0 small. A first step in this direction is Proposition 3.3, which will later be improved to our desired result in Proposition 4.2 and (4.19).
With the maximum principle, we get the following estimates. Since v ≥ − 2 f − + ,+ on Ω 0 , as v = 0 on Ω by above, we get from the meanvalue estimate for superharmonic functions or by Alexandroff's maximum principle, as v ∈ W 2,n (Ω) , see [16] Theorem 9.1, that v ≥ − 2 f − + ,+ in Ω 0 , hence Ω 0 = � , and the left estimate in (3.5) follows. The right estimate is obtained by symmetry observing that f ≥ 0.
Next for x ∈ Ω with u (x) = min Ω u and assuming that this minimum is negative, we see Then, we get with (3.4) and (2.9) that which is (3.6). ◻ Using the fourth-order equation, we get estimates for the Laplacian.

Blow-up
In this section, we consider a blow-up of our solutions u by translating and rescaling with x 0, ∈ ℝ n . We will have to choose x 0, differently in different steps in the proof of Theorem 1.1. In, e.g. the proof of Claim 1 in the proof of Theorem 1.1, we choose x 0, ∈ Ω such that max x∈ Ω | 2 Δu (x)| is attained, while in the proof of Claim 2, we choose x 0, ∈ Ω to attain − . This will not cause a problem, because the constants yielded by these claims do not depend on . We put , we see u ,ũ ∈ W 2,2 loc (ℝ n ). We also have to stretch ũ to get a nontrivial limit, and it turns out that reaching bounded values of the Laplacian of ũ on the boundary is the right measure for stretching.  3) and x 0, ∈ Ω , we get for any subsequence with after rotating Ω such that  for small depending on Ω and , which yields the first convergence in (4.6) after passing to a subsequence.
Proceeding from (4.3), we get from fourth-order L p −estimates, see [1,2] Sect. 10, after flattening the boundary of Ω , as Ω ∈ C 4 , with (4.4) and (4.10) that and small depending on Ω and . After passing to a subsequence, we obtain with (4.5) the second convergence in (4.6).
Actually by fourth-order higher-order L p −estimates, see [1,2] Sect. 10, we get that the blow-up ũ ∞ is smooth on ℝ n + . Now we are able to give a lower bound for M which improves the asymptotic in Proposition 3.3.  Proof From (2.7) and, as u 0, = 0 on Ω by (2.4), we get by Taylor's expansion for any x ∈ ℝ n + and any 0 < < 1 that As u 0, = 0 on Ω by (2.4), we get with (2.8), (4.5) and Ω ∈ C 4 after passing to a subse- which is (4.14).
By our investigation of half space solutions in Appendix A, Proposition A.3 applied to ũ ∞ andũ 0,∞ with Proposition 4.1 (4.7) and Proposition 4.3 (4.14) determines ũ ∞ uniquely as the one-dimensional solution and immediately yields the following Proposition.
Proof If on contrary ∕M → 0 for a subsequence → 0 , then we get from Proposition 4.3 and (2.7) that = 0 , hence with Proposition 4.4 (4.16) after passing to this subsequence we have that ũ ∞ ≡ 0.
On the other hand, choosing x 0, ∈ Ω in such a way that we get from the convergence in Proposition 4.1 (4.6) that hence Δũ ∞ (0) ≠ 0 andũ ∞ ≢ 0 . This is a contradiction, and the claim follows.

Claim 3
for some C = C(Ω, ) < ∞ and small depending on Ω and .

Claim 5
for small depending on Ω and .

Uniqueness of a bi-Laplace equation on the half space
In this section, we show the following uniqueness theorem.
The proof is based on an energy type estimate and on the one-dimensional case. We start with the one-dimensional case and show the following lemma: Then, v = 0.
Proof Since the differential equation is ordinary and linear, the solution space of the equation itself is of dimension 4. By inserting the following functions, we see that they constitute a basis of the solution space The following functions are therefore a basis of the solution space including the initial conditions at t = 0: Hence v has to be of the form Inserting the exponential function for cosh and sinh yields Since the solution is bounded, we have A = −B . Again the boundedness then yields B = 0 , to rule out linear growth. Hence v = 0. .
For our next step, we introduce a bit of notation for half spaces Next we show an energy type estimate.

Lemma A.2 Let v satisfy
Then, there exists a constant C = C(n) > 0 , such that for all R > 1 we have and for any k ≥ 1 i.e. is a cut-off function for the ball B R (0) . Then, v 4 and its first derivative are zero on B + 2R . Therefore, partial integration and the differential equation itself yield By the previous identity and Young's inequality with an > 0 we get Now choosing small enough, we can absorb the terms with as a prefactor into the left hand side Let x = (y, t) ∈ ℝ n + , such that y ∈ ℝ n−1 and t ≥ 0 . By k y v , we denote any partial derivative of v of order k only after horizontal directions, i.e. indices in {1, … , n − 1} . Then, for every k ∈ ℕ the function k y v ∶ ℝ n + → ℝ still solves the differential equation, satisfies the Dirichlet boundary conditions and by (A.2) is again bounded. Now we can iteratively apply Lemma A.2 for k ≥ ∈ ℕ and obtain By choosing k = , (A.2) yields If k > 2n , this yields for R → ∞: |∇ k−j y v| 2 + | k−j y v| 2 dL n .
Hence k y v = 0 . This implies Therefore k−1 y v(y, t) is independent of y ∈ ℝ n−1 and t ↦ k−1 y v(y, t) satisfies the assumptions of Lemma A.1. Hence, we also have Especially we have for all i = 1, … , n − 1 . Hence again k−2 y v(y, t) is independent of y and therefore again satisfies the assumptions of Lemma A.1. Iterating this final process yields which is the desired conclusion. ◻

Remark:
The proof above works, because both Δ 2 and −Δ are monotone operators which are added correctly. If we would destroy this monotonicity by, e.g. examining Δ 2 + Δ , Theorem A.1 is not true anymore. For example, a bounded nontrivial solution to Δ 2 v + Δv = 0 , is (y, t) ↦ 1 − cos t . and see v ∈ C ∞ loc (ℝ n + ) and Then, the uniqueness in Proposition A.1 gives v ≡ 0 , which is (A.4), and by direct calculation which is (A.5). For > 0 , we get by the strict convexity of the exponential function which is (A.6).
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