AN H-SYSTEM FOR A REVOLUTION SURFACE WITHOUT BOUNDARY

We study the existence of solutions an H-system for a revolution surface without boundary for H depending on the radius f . Under suitable conditions we prove that the existence of a solution is equivalent to the solvability of a scalar equation N(a) = L/√2, where N : ⊂ R+ → R is a function depending on H . Moreover, using the method of upper and lower solutions we prove existence results for some particular examples. In particular, applying a diagonal argument we prove the existence of unbounded surfaces with prescribed H .


Introduction
The prescribed mean curvature equation for a vector function X : Ω → R 3 is given by the following nonlinear system of partial differential equations: (1.1) Here Ω ⊂ R 2 is a bounded domain, ∧ denotes the exterior product in R 3 and H : R 3 → R is a given function.It is well known that if X is isothermal, namely then H is the mean curvature of the surface parameterized by X (see, e.g., [8]).Equation (1.1) is also known in the literature as an H-system.The parametric Plateau and Dirichlet problems for (1.1) have been extensively studied by different authors (see [3][4][5][8][9][10]). Nonparametric and more general quasilinear equations are considered in [1,2,6,7].
2 An H-system for a revolution surface without boundary We will consider the particular case of a revolution surface X(u,v) = f (u)cosv, f (u)sinv,g(u) (1.3) with f ,g ∈ C 2 (I) ∩ C(I) such that f > 0 over the open interval I ⊂ R. Then (1.1) reads where H : R + × R → R is given.
It is easy to see that any solution of (1.4) verifies the equality Hence, the isothermal condition (1.2) holds if and only if c = 0. We will study (1.4) for a compact surface without boundary.Without loss of generality we may assume that I = (0,L), and hence the problem reads In particular, when H depends only on the radius f , from the equality we easily reduce problem (1.6) to a single equation: indeed, if H(t) = t 0 sH(s)ds, it holds that g (t) = 2 H( f (t)), and g (L) = 2 H( f (L)).Thus, solving (1.6) is equivalent to obtain a positive solution of the problem with H : R + → R. We remark that if H > 0 then g > 0, and if f is a positive solution of (1.8) the parametrization X given in (1.3) defines a regular revolution surface.For example, this holds when H is positive.We will also consider the case L = +∞, namely the problem P. Amster et al. 3 where r > 0 is a constant.Note that if f is a positive solution of (1.9) then g (+∞) = 2 H(r).Thus, if H(r) > 0 it follows that the surface parameterized by X is unbounded in the direction z → +∞ of the upper halfspace R 2 × R + .The paper is organized as follows.In Section 2 we prove that under suitable conditions the existence of a positive solution of (1.8) is equivalent to the solvability of the scalar equation N(a) = L/ √ 2, where N is defined by (1.11)Moreover, we prove existence and uniqueness of solutions for some particular examples.
In Section 3 we apply the method of upper and lower solutions and a diagonal argument in order to prove the existence of solutions of problem (1.9).

A scalar equation for (1.8)
In this section we study the existence of positive solutions of (1.8).Let us first note that if φ is defined as in (1.11), the problem may be written as Then we have the following theorem.

Then (2.1) admits at most one positive solution f with a
Proof.Let f be a positive solution of (2.1) with a = f C([0,1]) , and fix x 0 ∈ (0,L) such that f (x 0 ) = a.Multiplying the equation by f it follows by integration that , and hence f (x) = a.We conclude that x 0 is the only critical point of f .Thus, (2.4) 4 An H-system for a revolution surface without boundary This implies that (2.5) In particular, x 0 = L − x 0 , and then x 0 = L/2.Furthermore, for x = 0 we obtain and extend it by symmetry for x < L/2.It is immediate to verify that f is a positive solution of problem (2.1).Moreover, from the above computations it is clear that if f is a positive solution with f C([0,1]) = a, then f = f .Remark 2.2.The proof of existence of a solution in the previous theorem holds for any Remark 2.3.If H is bounded in a neighborhood of 0, then φ (0 + ) < 0 and hence 0 ∈ Ꮽ.

Upper and lower solutions and unbounded revolution surfaces
In this section we apply the method of upper and lower solutions in order to solve a nonhomogeneus Dirichlet problem associated to (1.4).In particular, applying a diagonal argument we prove the existence of solutions of (1.9).We recall that (α,β) ∈ (C 2 ([0,+∞))) 2 is an ordered couple of a lower and an upper solution of the problem if α ≤ β and For simplicity we will assume that H is continuously differentiable.
In particular, if (3.2) holds we may take β ≡ r as an upper solution.
Theorem 3.2.Let (α,β) be an ordered couple of a lower and an upper solution of (1.9), let N > 0 and let c N be any constant with α(N) ≤ c N ≤ β(N).Then the Dirichlet problem 6 An H-system for a revolution surface without boundary This choice of λ implies that the function ξ(x) := −4H(x)x H(x) − λx is non-increasing.We will construct a sequence { f n } given recursively by f 0 = α and f n+1 the unique solution of the linear problem We claim that { f n } is non-decreasing, with α ≤ f n ≤ β.Indeed, as by the comparison principle we deduce that f 1 ≥ α.Now assume that f n ≥ f n−1 then and we deduce that f n+1 ≥ f n .On the other hand, f 0 = α ≤ β, and if f n ≤ β we have that As using again the comparison principle we deduce that f n+1 ≤ β.
It follows that { f n } converges pointwise to some function f .By the standard a priori bounds and using the fact that α ≤ f n ≤ β for each n we have that for some constant C. Thus, if we suppose that f n f uniformly, taking a subsequence we may assume that f n − f C([0,N]) ≥ ε for some ε > 0. By the Sobolev imbedding H 2 (0,N) C 1 ([0,N]), taking a subsequence we may assume that f n converges to some function g = f P. Amster et al. 7 for the C 1 -norm, a contradiction.Hence f n → f uniformly, and It follows that f is a solution of the problem.Remark 3.3.In the previous proof, it is easy to see that the convergence is more accurate for smaller values of λ.Indeed, if λ ≥ λ, with λ as before, the corresponding sequence { f n } given recursively by f 0 = α and f n+1 the unique solution of the linear problem is non-decreasing and converges to a solution of the problem.We claim that f n ≤ f n for every n: indeed, this is trivial for n = 0, and if the claim is true for n we have that Using the inductive hypothesis and the fact that { f k } is nondecreasing, it follows that f n+1 ≥ f n+1 .
To conclude this remark, note that { f n } and { f n } converge to the same solution.Indeed, it suffices to replace β by β = lim n→∞ f n (3.14) in the proof of Theorem 3.2.As β ≤ β, the definition of { f n } coincides with the previous one, and f n ≤ β for every n.
Proof.For any N ∈ N, by the previous theorem we may choose a solution f N of (3.3) with for any N ≥ M. For M = 1 we may take a subsequence, still denoted { f N }, which converges uniformly in [0,1] to some function f 1 .Repeating the procedure we may assume that f N converges uniformly in [0,M] to a function f M .Then f : [0,+∞) → [0,+∞) given by f (x) = f N (x) if x ≤ N solves (1.9).Indeed, it is clear that f is well defined, and that f (0 and the proof follows. Example 3.5.Assume that (3.2) holds, and that ( H 2 ) ≤ 0 on [0,r].Then (α,β) given by is an ordered couple of a lower and an upper solution of (1.9).Indeed, From the previous theorem we deduce that (1.9) admits at least one positive solution between α and β.
3.1.Some numerical experiments.The method described in the proof of Theorem 3.2 can be implemented as a numerical method to compute the solution in an effective way.At each step of the iterative procedure, we have to solve a linear differential equation, with Dirichlet boundary conditions.With that purpose, we use the standard finite difference method: We split the interval [0,N] into k small sub-intervals of length h = k/N, and we denote by f i n the approximate value of f n (x i ).Then, we approximate the linear problem (3.6) by the linear system of equations subject to the boundary conditions Let us recall that the energy introduced in (2.3) is constant, for any solution of the problem, therefore we can use the discrete quantity as a test for the accuracy of the method.We stop the iteration when this quantity is close enough to a constant, for the desired precision ε 0 , that is, when

Lower solution
Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 E(h) H system: iterative method (precision = 0.1) We have implemented this numerical scheme using GNU Octave for different choices of H.In Figure 3.1, we present the case H(x) = x, N = 1, λ = 10 and ε 0 = 0.1.