It is proved that if a smooth function , such that where UN is the normal derivative of u on S, has a closed smooth surface S of zeros, then the function has also a closed smooth surface Se of zeros. Here u is a smooth function and is a sufficiently small number.
zero surfaces, perturbation theory
Let be a bounded domain containing inside a connected closed C3-smooth surface S, which is the set of zeros of a function , so that Consider the scattering problem:
(1)
Let N = Ns be the unit normal to S, such that where UN is the normal derivative of u on S. Let , where is sufficiently small. Assume that
(2)
The purpose of this paper is to prove Theorem 1.
Theorem 1. Under the above assumptions there exists a smooth closed surface Se such that Ue = 0 on Se.
In Section 2 Theorem 1 is proved.
Although there are many various results on perturbation theory, see [2], [3], the result formulated in Theorem 1 is new.
Proof of Theorem 1
Consider the following equation for t:
(3)
where N = N(s) is the normal to S at the point s and t is a parameter. Using the Taylor's formula and relation (1), one gets from (3)
where is the Lagrange remainder in the Taylor's formula and
Since the functions u and belong to C3(D), the function has a bounded derivative with respect to t uniformly with respect to
Consider equation (4) as an equation for t = t(s) in the space C(S). Rewrite (4) as
Let us check that the operator B satisfies the contraction mapping theorem in the set
where is a small number, and
First, one should check that B maps M into itself. One has
We have chosen N so that This is possible because equation (1) implies that is orthogonal to S at the point Assumption (2) implies that for sufficiently small ∈ one has
Sinceis continuously differentiable, one has
Thus, if (13) holds then B maps M into itself.
Let us check that B is a contraction mapping on M. One has
is sufficiently small. Indeed,
if δ is suciently small. Here C4 is a constant.
Thus, B is a contraction on M. By the contraction mapping principle, equation (6) is uniquely solvable for t. Its solution t = t(s) allows one to construct the zero surface Se of the function Ue by the equation r = s + t (s) N , where r = r (s) is the radius vector of the points on Se.
Theorem 1 is proved.
Remark 1. Condition (2) is a sufficient condition for the validity of Theorem 1. Although this condition is not necessary, if it does not hold one can construct counterexamples to the conclusion of Theorem 1. For example, assume that
Then the function does not have zeros in i 3.
Remark 2. In scattering theory the following question is of interest: assume that u (x) is an entire function of exponential type, Assume that u = 0 on S, where S is a closed smooth connected surface in i 3.
Is there another closed smooth connected surface of zeros of an entire function Ue of exponential type,
We will not use Theorem 1 since assumption (2) may not hold, but sketch an argument, based on the fact that S in the above question is the intersection of an analytic set with i 3, see, for example, [1] for the definition and properties of analytic sets. The functions u and ue in Remark 2 solve the differential equation
The function uN may vanish on S at most on the closed set which is of the surface measure zero (by the uniqueness of the solution to the Cauchy problem for equation (16)). For every point the argument given in the proof of Theorem 1 yields the existence of t(s), the unique solution to (6). Since S is real analytic the set Se , defined in the proof of Theorem 1, is analytic and is a part of the analytic set defined by the equation ue = 0. In our problem S is a bounded closed real analytic surface. The set Se can be continued analytically to an analytic set which intersects the real space i 3 over a real analytic surface Se. It is still an open problem to prove (or disprove) that the analytic continuation of the set Se intersects i 3 over a bounded closed real analytic surface
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