Abstract
Let \((M,g)\) be a (complete) Riemannian surface, and let \(\Omega \subset M\) be an open subset whose closure is homeomorphic to a disk. We prove that if \(\partial \Omega \) is smooth and it satisfies a strong concavity assumption, then there are at least two distinct orthogonal geodesics in \(\overline{\Omega }=\Omega \bigcup \partial \Omega \). Using the results given in Giambò et al. (Adv Differ Eq 10:931–960, 2005), we then obtain a proof of the existence of two distinct brake orbits for a class of Hamiltonian systems. In our proof we shall use recent deformation results proved in Giambò et al. (Nonlinear Anal Ser A Theory Methods Appl 73:290–337, 2010).
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Notes
One can choose \(\phi \) such that \(\vert \phi (q)\vert ={\mathrm {dist}}(q,\partial \Omega )\) for all \(q\) in a (closed) neighborhood of \(\partial \Omega \).
Observe that, with our definition of \(\phi \), then \(\nabla \phi \) is a normal vector to \(\partial \Omega \) pointing outwards from \(\Omega \).
By geometrically distinct curves we mean curves having distinct images as subsets of \(\overline{\Omega }\).
The map \(\widetilde{H}^1(\mathbb S^1\times \mathbb S^1)\rightarrow \widetilde{H}^1(\Delta ^1)\) is induced by the diagonal inclusion of \(\mathbb S^1\) into \(\mathbb S^1\times \mathbb S^1\). It takes both generators \(\pi _1^*(\omega )\) and \(\pi _2^*(\omega )\) of \(H^1(\mathbb S^1\times \mathbb S^1)\) to the generator \(\omega \) of \(H^1(\mathbb S^1)\cong H^1(\Delta ^1)\).
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Communicated by P. Rabinowitz.
Appendix A. An estimate on the relative category
Appendix A. An estimate on the relative category
Let \(n\ge 1\) be an integer; \(\mathbb S^n\) is the \(n\)-dimensional sphere, and \(\Delta ^n\subset \mathbb S^n\times \mathbb S^n\) is the diagonal. We want to estimate the relative Lusternik–Schnirelman category of the pair \((\mathbb S^n\times \mathbb S^n,\Delta ^n)\), and to this aim we will prove an estimate on the relative cuplength of the pair.
For a topological space \(X\) and an integer \(k\ge 0\), we will denote by \(H^k(X)\) and \(\widetilde{H}^k(X)\) respectively the \(k\)th singular cohomology and the \(k\)th reduced singular cohomology group of \(X\). For a topological pair \((X,Y)\), \(H^k(X,Y)\) is the \(k\)th relative singular cohomology group of the pair; in particular, \(H^k(X,\emptyset )=H^k(X)\). Given \(\alpha \in H^p(X,Y)\) and \(\beta \in H^q(X,Z)\), \(\alpha \cup \beta \in H^{p+q}(X,Y\bigcup Z)\) will denote the cup product of \(\alpha \) and \(\beta \); recall that \(\alpha \cup \beta =(-1)^{pq}\beta \cup \alpha \).
The notion of relative cuplength, here recalled, will be also used.
Definition 10.1
The number \({\mathrm {cuplength}}(X,Y)\) is the largest positive integer \(k\) for which there exists \(\alpha _0\in H^{q_0}(X,Y)\) (\(q_0\ge 0\)) and \(\alpha _i\in H^{q_i}(X)\), \(i=1,\ldots ,k\) such that
and
where \(\cup \) denotes the cup product.
As for the absolute Lusternik–Schirelmann category, we have the following estimate of relative category by means of relative cuplenght, cf. e.g. [3, 4]
Proposition 10.2
\({\mathrm {cat}}_{\mathbb S^n \times \mathbb S^n,\Delta ^n}(\mathbb S^n \times \mathbb S^n) \ge {\mathrm {cuplength}}(\mathbb S^n\times \mathbb S^n,\Delta ^n) + 1\).
Therefore, to prove that \(cat_{\mathbb S^n \times \mathbb S^n,\Delta ^n}(\mathbb S^n \times \mathbb S^n) \ge 2\) it will be sufficient to prove the following
Proposition 10.3
For all \(n\ge 1\), \({\mathrm {cuplength}}(\mathbb S^n\times \mathbb S^n,\Delta ^n)\ge 1\).
Proof
The statement is equivalent to proving the existence of \(p\ge 0\), \(q\ge 1\), \(\alpha \in H^p(\mathbb S^n\times \mathbb S^n,\Delta ^n)\) and \(\beta \in H^q(\mathbb S^n\times \mathbb S^n)\) such that \(\alpha \cup \beta \ne 0\). This will follow immediately from the Lemma below. \(\square \)
Lemma 10.4
For \(n\ge 1\), the group \(H^{2n}(\mathbb S^n\times \mathbb S^n, \Delta ^n)\) is isomorphic to \(\mathbb {Z}\), and the map \(H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n) \times H^n(\mathbb S^n\times \mathbb S^n)\ni (\alpha ,\beta )\mapsto \alpha \cup \beta \in H^{2n}(\mathbb S^n\times \mathbb S^n, \Delta ^n)\) is surjective.
Proof
It is well known that \(H^k(\mathbb S^n)\cong \mathbb {Z}\) for \(k=0,n\), and \(H^k(\mathbb S^n)=0\) if \(k\ne 0,n\). It follows \(H^n(\mathbb {S}^n\times \mathbb {S}^n)\cong \bigoplus _{k=0}^nH^k(\mathbb S^n)\otimes H^{n-k}(\mathbb S^n)\cong \mathbb {Z}\oplus \mathbb {Z}\). If \(\omega \) is a generator of \(H^n(\mathbb S^n)\), then the two generators of \(H^n(\mathbb S^n\times \mathbb S^n)\cong \mathbb {Z}\oplus \mathbb {Z}\) are \(\pi _1^*(\omega )\) and \(\pi _2^*(\omega )\), where \(\pi _1,\pi _2:\mathbb S^n\times \mathbb S^n\rightarrow \mathbb S^n\) are the projections.
For the computation of \(H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n)\), we use the long exact sequence of the pair \((\mathbb S^n\times \mathbb S^n,\Delta ^n)\) in reduced cohomology:
Since \(\Delta ^n\) is homeomorphic to \(\mathbb S^n\), then \(\widetilde{H}^{n-1}(\Delta ^n)=0\). Thus, the group \(H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n)\) can be identified with the subgroup of \(\widetilde{H}^n(\mathbb S^n\times \mathbb S^n)\) given by the kernel of the map \(\mathfrak i^*:\widetilde{H}^n(\mathbb S^n\times \mathbb S^n)\rightarrow \widetilde{H}^n(\Delta ^n)\). This map takes each of the two generators \(\pi _i^*(\omega )\), \(i=1,2\), to \(\omega \) (here we identify \(\Delta ^n\) with \(\mathbb S^n\)), so that \(H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n)\) is the subgroup of \(\widetilde{H}^n(\mathbb S^n\times \mathbb S^n)\) generated by \(\pi _1^*(\omega )-\pi _2^*(\omega )\), which is isomorphic to \(\mathbb {Z}\).
Finally, let us compute \(H^{2n}(\mathbb S^n\times \mathbb S^n,\Delta )\) using again the long exact sequence of the pair \((\mathbb S^n\times \mathbb S^n,\Delta ^n)\) in reduced cohomology:
Clearly, \(\widetilde{H}^{2n}(\Delta ^n)=0\), and if \(n>1\), also \(\widetilde{H}^{2n-1}(\Delta ^n)=0\). When \(n=1\), then \(\widetilde{H}^{2n-1}(\Delta ^n)=\widetilde{H}^1(\Delta ^1)\cong \mathbb {Z}\), however the map \(\widetilde{H}^1(\Delta ^1)\rightarrow \widetilde{H}^2(\mathbb S^1\times \mathbb S^1)\) is identically zero, because the previous map of the exact sequence \(\widetilde{H}^1(\mathbb S^1\times \mathbb S^1)\rightarrow \widetilde{H}^1(\Delta ^1)\) is clearly surjective.Footnote 4 In both cases, \(n=1\) or \(n>1\), we obtain \(H^{2n}(\mathbb S^n\times \mathbb S^n,\Delta ^n)\cong \widetilde{H}^{2n}(\mathbb S^n\times \mathbb S^n)\cong \mathbb {Z}\). A generator of \(\widetilde{H}^{2n}(\mathbb S^n\times \mathbb S^n)\) is \(\pi _1^*(\omega )\cup \pi _2^*(\omega )\).
In conclusion, using the above identifications, the map \(H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n) \times H^n(\mathbb S^n\times \mathbb S^n)\ni (\alpha ,\beta )\mapsto \alpha \cup \beta \in H^{2n}(\mathbb S^n\times \mathbb S^n, \Delta ^n)\) reads as the bilinear map \(\mathbb {Z}\times (\mathbb {Z}\oplus \mathbb {Z})\rightarrow \mathbb {Z}\) that takes \(\big (1,(1,0)\big )\) to \((-1)^{n+1}\) and \(\big (1,(0,1)\big )\) to \(1\). This is clearly surjective. \(\square \)
From Propositions 10.2 and 10.3 we get:
Corollary 10.5
For all \(n\ge 1\), \({\mathrm {cat}}_{\mathbb S^n \times \mathbb S^n}(\mathbb S^n \times \mathbb S^n,\Delta ^n)\ge 2\).
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Giambò, R., Giannoni, F. & Piccione, P. Multiple brake orbits in \(m\)-dimensional disks. Calc. Var. 54, 2553–2580 (2015). https://doi.org/10.1007/s00526-015-0875-5
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DOI: https://doi.org/10.1007/s00526-015-0875-5