Composite Membranes. 8th August 2006

Transcription

1 Composite Membranes Aldo Lopes UFMG, Russell E. Howes, Brigham Young University Cynthia Shepherd, California State University - Northridge Cynthia.Shepherd.87@csun.edu 8th August

2 Contents 1 Introduction 3 2 The irichlet Problem Spectral Theorem Variations 7 4 Variations With Fixed Boundary 10 5 Critical Eigenvalues 12 2

3 1 Introduction The study of vibrating membranes is an aspect of mathematical physics that is primarily concerned with the geometries of the membrane corresponding to its frequencies. Membranes consisting of two materials of different densities are called composite membranes. We are interested in determining the shape of membranes with minimal eigenvalues; in particular the first eigenvalue, or fundamental frequency, of such membranes. In this paper 1 we will be studying a membrane divided into two subsets, a lower density and a higher density c. The boundary between and c is a hypersurface and will be called. We will first look at the associated irichlet boundary PE problem. We will review the Spectral Theorem, which is an important result related to our work and then consider variations that fix the boundary of and preserves the volume of and show that such variations do exist under our given specifications. The main purpose of this paper is to determine conditions on and its corresponding first eigenfunction, u, for the critical points of the smallest eigenvalue, λ. We will also prove that for any critical point of λ, is a level set of u. 1 Partially supported by NSF grant OISE:

4 2 The irichlet Problem The composite membrane eigenvalue problem for a bounded domain R n is the irichlet boundary PE problem: ( ),α, { u + αχ u = λu in u = 0 in u is a normalised eigenfunction of the above problem, is the region of previously defined, and we are interested in minimising λ 0, the lowest eigenvalue or lowest possible value of λ for a given. We will only worry about the eigenfunction u associated with λ 0, and we want to find what configurations of u and will make λ 0 = 0. χ is the characteristic function: { 0 if x χ (x) = c (= ) 1 if x Let R n be an open, connected and bounded set. We define N as the boundary of, i.e., = N. Here N is smooth and a C hypersurface. In fact, N = f 1 (a), where a is a regular value of the function f : U R with N U. F : ( ɛ, ɛ) R n a C function is the variation of. If is necessary, we can consider the extension of F : F : ( ɛ, ɛ) U R n a C function and U R n is an open set. We define now the follwing function that for each time: F t = F (t, ) : R n F t ( ) = t Restricting the time if necessary, one always has that F t : R n is a diffeomorphism onto its image: F t : t For each time we consider N = and N t = t also a C hypersurface. And =. F t : N N t is also a diffeomorphism. Let s start by defining some functions. We previously defined F : ( ɛ, ɛ). efinition 1. For some fixed t ( ɛ, ɛ), we define F t : as F t = F (t, x). 4

5 Figure 1: The membrane Let s place some constraints on our F t : For any t 1, t 2, (t 1 + t 2 ) ( ɛ, ɛ): F t1 F t2 = F t1 +t 2 This structure makes things nice for us it makes the F t s associative, and gives us for each F t an inverse, so that F t F t = F 0 = F. We will refer to F 0 as F from here on out. efining F t leads to a natural definition of t and thus χ t, and prompts us to redefine the first eigenfunction: efinition 2. For some fixed t ( ɛ, ɛ), we define u t : R as u t = u(t, x). We will newly define u = u 0. A few observations: We want to rewrite χ t as a deformation of χ. We notice that if x = F t (y), y = F t (x) = Ft 1 (x) and χ (y) = χ t F t (y) χ (F t ) 1 (x) = χ t (x) = χ F t 5

6 2.1 Spectral Theorem Usually one also has a boundary condition coming from the original problem. We will consider the case where the membrane is fixed along,i.e., F t (x) = 0 for all x and all t R. This gives rise to the same condition for u, the irichlet boundary condition. Thus, one is bound to study the possible solutions of the PE problem { u = λu in u = 0 in Analogous to the linear algbra concepts, a nontrivial u satisfyng such conditions will be called an eigenfunction of the Laplacian and the associated real number λ an eigenvalue. Recall that given two functions φ and ψ in wich go to zero at its boundary, their L 2 - inner product is given by φ, ψ = ϕψdx dy One then has the classical Theorem 2.1. (Spectral Theorem for the Laplacian with the irichlet boundary condition). There is an unbounded discrete set of (positive) eigenvalues (0 <)λ 1 < λ 2 λ 3... and associated set of eigenfunctions u i, which may be taken mutually orthogonal and with norm equal to 1, with respect to the L 2 inner product in, which is complete in the sense that any smooth function v : R such that v = 0 at may be written as the Fourier series v = v, u i u i i=1 Remark. In the case of one space variable, i.e., of a vibrating string, the u is are given by sine functions, the original Fourier series situation. Observe that the (orthonormal) system of eigenfunctions plays the usual role of an orthonormal basis in finite dimensional linear algebra, when we use orthogonal projection to write a vector in that basis (in that case one has a finite sum of projected vectors). It is also called a Hilbert basis for the space os functions in (with the irichlet boundary condition). The numbers λ i give the frequency of the basic solutions (called harmonics) for the vibrating membrane. For this reason, λ 1, the first eigenvalue, is also called the fundamental frequency or pitch of the membrane 6

7 3 Variations We say that F preserves the volume if vol( t ) = vol() t ( ɛ, ɛ) Since bounds, there exists ν : R n, such that: 1. ν(p) = 1 2. ν points to exterior of 3. ν(p) T p Where T p is the tangent plane at p to. ν is the exterior unit normal vector field along. Now, let F : ( ɛ, ɛ) and p. We will denote by X(p) = F (0, p) Rn t the variation vector of F. X(p) is the initial velocity of variation at points of. X : N R n is a C vector field on N. We have a formula for the volume of t : v(t) = dx t Now we have the following important lemma. Lemma 3.1. v (0) = divx dx = X, ν dx Proof. We will show two different proofs. (1) As in the formula above, we have: v (t) = d ( ) dx = (χ t ) dx = dt t (χ t F t ) dx The function χ is not smooth neither continuous. Then we consider a smooth function g δ : R. dg δ : R n R is a linear application. Also we have: lim δ 0 g δ = χ 7

8 Now we can compute this integral (g δ F t ) dx Remember that X = F is the variation vector. Then we have: t t g δ (F ( t, x)) dx = dg δ (x) F (0, x).( 1) dx t=0 t = g δ (x), X(x) dx = X(g δ )(x) dx But we have And div(g δ X) = g δ, X + g δ divx div(g δ X) = g δ X, ν = 0 because X is zero along the boundary. As we have lim δ 0 g δ = χ, then the integral converges too. It means (χ F t ) dx = (χ )divxdx = X, ν ds (2) Using change of variables we have: v(t) = Where JF t = det(df t ). Then v (t) = d dy = d dt t dt JF t dy JF t dx = d dt JF t dx Now, we will work only with d dt JF t. df t is the Jacobian matrix for the function F t. F t : R n R n F t (x 1, x 2,..., x n ) = (F 1 t x, F 2 t x,..., F n t x) Where x = (x 1, x 2,..., x n ) and t ( ε, ε). df t = (F 1 t ) 1 (F 1 t ) 2 (F 1 t ) n (F 2 t ) 1 (F 2 t ) 2 (F 2 t ) n. (Ft n ) 1.. (Ft n ) 2 (Ft n ) n 8

9 Where (F i t ) j = F i t x j and i, j = 1,..., n. If we look at the determinant of df t by its columns, we get: det(df t ) = det[(f t ) 1 (F t ) 2 (F t ) n ] Now, we can differentiate with respect to t: d dt det(df t) = det[(f t ) 1 (F t ) 2 (F t ) n ] + + det[(f t ) 1 (F t ) 2 (F t ) n] Where (F t ) k is the derivative of the k-th column with respect to t. We can take a look at one of the terms of the sum above: (Ft 1 ) 1 (Ft 1 ) k (Ft 1 ) n det[(f t ) 1 (Ft 1 ) k (F t ) n ] = det (F 2 t ) 1 (F 2 t ) k (F 2 t ) n. (Ft n ) 1.. (Ft n ) k (Ft n ) n We can compute the determinant of the matrix above as a sum as below, where P is the set of all permutations of the index i and σ is its permutation. ε σ (Ft 1 ) σ(1) (Ft 2 ) σ(2) (Ft k ) σ(k) (Ft n ) σ(n) σ P Where ε σ = ±1 is the signal of the permutation.we need this computation only when t = 0. In this time we have for i, j = 1,..., n: { (F i 0 ) j = 0 i j (F i 0) i = 1 This means that F 0 = Id. So, it is clear that: d dt det df t t=o = (F 1 0 ) + (F 2 0 ) + + (F n 0 ) It is easy to see that ( ),,, x 1 x 2 x n ( df 1 0 dt, df 0 2 dt,, df ) 0 n = F 0 dt The end of this proof it is made by the Gauss-Green Theorem. If we called F0 i = X i for i = 1,..., n and div X = X X n: v (0) = div X dx = X, ν ds 9

10 4 Variations With Fixed Boundary Lemma 4.1. Let f : R be a piecewise smooth function such that fds = 0 then there exists a volume-preserving normal variation whose variation vector is fn. If, in addition, f 0 on, the variation can be chosen so that it fixes the boundary. Proof. Let g = 0 on and gdm 0. We can let g = 1 on. Now, in order to extend tf + tg over all of, Take any point q. We will define d(, q) to be the length of the normal vector to passing through q (touching at point p; thus we also define a function p(q) mapping to.) We also define d(p, ) for any point p as the length of the vector normal to at p whose endpoint is on (we will say that the length is negative if q is inside.) We will also define the bump function, B : [0, 1] R so that B(x) = 1 if x 1, B(x) = 0 if x 2, and B(x) is 3 3 C on [0, 1]. We then define our extension function, ( ) d(, q) φ(q) = B d(p(q), ) It is easily shown that φ(q) is uniformly 1 on, 0 on, and C on. Now, set x(t, t) = x 0 + φ(q)(tf + tg)n, g : R is a piecewise smooth function. The volume function is V (t, t) = x, N ds Since we are considering the case where the volume is constant, we have Then we have V t V (t, t) = V (0, 0) = (0,0) dm 0 Now we want to apply the Implicit Function Theorem. 10

11 We have V (t, t) = V (0, 0) so, let G : R R R G := V (t, t) V (0, 0) continuously differentiable on an interval containing (0, 0) and G(0, 0) = 0. The Jacobian is [ ] V V dv = t t So, our M = [ ] V t det(m) t=0= V = gdm 0 by definition. t t=0 Then, by the Implicit Function Theorem, there is an open set ( ε, ε) R containing t = 0 such that tɛ( ε, ε)! ϕ(t)ɛ( ε, ε) such that ϕ(t) = t and G(t, ϕ(t)) = 0 V (t, ϕ(t)) = V (0, 0). 11

12 5 Critical Eigenvalues Theorem 5.1. Let R n be bounded, measurable and with a smooth boundary,u,f as defined above. Then λ (0) = α u 2 f ds Proof. Observe that u 0 + αχ u 0 = λu 0 u t + αχ t u t = λu t We multiply the first equation by u t and the second by u, then add them and integrate both side over to obtain u t u 0 u 0 u t + α (χ t χ ) u 0 u t = (λ(t) λ) u 0 u t The first integrand on the left-hand side of the equation is equal to the divergence of u u t u t u. By the divergence theorem, this integral has value 0 since both u and u t are uniformly 0 on. Our equation reduces to α (χ t χ ) u 0 u t = (λ(t) λ) u 0 u t We take the derivative of both sides with respect to t, and then terms cancel out when we evaluate at t = 0: ( ) d ( ) α χ dt t u0 u t + χ t u 0 u t χ u 0 u t = = λ (t) u 0 u t + λ(t) u 0 u t λ 0 u 0 u t Evaluate at t = 0: α d ( ) χ dt t t=0 u2 = λ (0) u 2 = λ (0) since the L 2 -norm of u is 1. The question remains, how can we take the derivative of a non-differentiable function? Well, we can approximate χ by a C function, g ɛ, by defining g ɛ to be equal to χ on all points p where d(p, ) ɛ and C on all points (on the ring around, g slopes from 0 to 12

13 1). Clearly as ɛ 0, g ɛ χ, but g ɛ is still a C function which we can differentiate. α d dt (χ t ) t=0 u 2 = α d dt (χ F t ) t=0 u 2 α d ( ) u 2 g ɛ F t dt t=0 = α u 2 X (g ɛ ) = α u 2 X, g ɛ where X is the vector field associated with F. Now, by the properties of divergence, = div(g ɛ u 2 X) = u 2 X, g ɛ + g ɛ div(u 2 X) g ɛ div(u 2 X) = g ɛ u 2 X, µ since u 0 on. Thus λ (0) = α div(g ɛ u 2 X) α χ u 2 divx = α u 2 X, g ɛ = u 2 X, g ɛ u 2 X, g ɛ u 2 X, g ɛ = α g ɛ div(u 2 X) u 2 divx = α u 2 X, ν by the ivergence Theorem. We previously stated that f = X, ν. is made up of two parts, and. On, u 0, so the previous result reduces to λ (0) = α u 2 fds which is a nice simple expression for λ (0). efinition 3. (, ) is a critical point if for all variations f where fds = 0 we have λ (0) = 0. Proposition 5.2. (, ) is a critical point for our problem if and only if = u 1 (c) where c is a constant. 13

14 Proof. ( ) Assume, in order to prove the contra-positive, that u is not constant over. u 2 is not constant over. Then define ū 2 = u2 ds ds Now define positive, smooth functions φ, ψ such that φ = 0 when u 2 ū 2 0 and, ψ = 0 when u 2 ū 2 0 Then let f = (φ + ψ)(u 2 ū 2 ) (φ + ψ)(u 2 ū 2 ) = 0 Consider u2 fds u 2 fds = u 2 (φ+ψ)(u 2 ū 2 ) = (φ+ψ)(u 2 ū 2 ) 2 + ū 2 (φ+ψ)(u 2 ū 2 ) Then since ū 2 is constant, we can factor it out and that term goes to zero by our definition of f, leaving u 2 fds = (φ + ψ)(u 2 ū 2 ) 2 0 ( ) Trivial, since if u = c constant, λ (0) is clearly 0. References [1] Pedrosa, H.L. Renato, Some Results Regarding Symmetry and Symmetry-breaking Proprieties of Optimal Composite Membranes, Progress in Nonlinear ifferential Equations and Their Applications, Vol 66, (2005) [2] Barbosa, Joäo Lucas and do Carmo, Manfredo Stability of Hypersurfaces with Constant Mean Curvature, Mathematische Zeitschrift 185, (1984) 14

15 This research was conducted in the IRES program at UNICAMP, Brazil, funded by NSF and CNPq, and was advised by Renato Pedrosa of the UNICAMP Mathematics epartment. The authors would like to thank the epartment of Mathematics for their hospitality. 15