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MAT 280 Laplacian Eigenfunctions Theory Applications and Computations Lectures 910 Laplacian Eigenvalue Problems for General Domains II Computation of Eigenvalues Lecturer Naoki Saito Scribe Patrick SatarzadehAllen Xue May 1 2007 The basic references for this lecture are the texts by Strauss 2 Sec 113 and Courant and Hilbert 3 Sec V11 V12 For more advanced treatments see 4 Chap 4 and 5 Chap 1 The Dirichlet Laplacian eigenvalues in particular A1 the smallest is quite im portant in many applications To compute them we will use the Rayleigh quotient For any trial function w 6 039 with w i 0 we have HVwHZ A1 S sz If w ltp1 we get exactly A1 but we do not know ltp1 at this point So we should be satis ed with a moderately clever choice of w that might provide a relatively good approximation Example 01 In Lecture 8 we already discussed the approximate minimum of MP 1 iv2 dm m min 0 we Cg0l 1 weC2Q wzoo LUZ Cl 0 We choose the trial function w 30zl 7 which gives an estimate 5 1 10 where the true eigenvalue is A1 71392 x 98696 In this case we have done well but in general it is too dif cult to come up with a good trial function to produce a good approximation of A1 or Aj 1 RayleighRitz Approximation RRA Let w1 wn 6 039 w i 0 be arbitrary trial functions Let A ajk B bjk where ajk bjk are de ned as 177 ltij7Vwkgt ij Vwk dw Q 1 A bjk ltwjwkgt LUj LUk dw n for j k l n Thus A and B are n gtlt n symmetric matrices Then e roots of the polynomial equation detA 7 AB are approxi ation to the rst 71 eigenvalues A1 A Before presenting an informal proof of this let us consider the following example Example 11 Consider the radial vibrations of a circular membrane 9 z y z2 y2 lt 1 Then the Dirichlet Laplacian DL problem is iAu Au which can be written in the polar coordinates r 6 as frurrru 0lt7quotltl7 u0atrl Here we apply the fact that radial function depends only on the parameter r ie Way Wquot Then the Rayleigh quotient is given by 1 lelZdw 27T Tude 0 2 Q 1 w2d 27T Tude 0 Now we ask what trial functions should we use The trial functions are required to satisfy the boundary conditions 100 nite7 11170 0 101 A simple choice of a pair of them is l 7 r2 and l 7 r22 In this case we can compute the entries for matrices A and B in 1 as follows A7473 33 37m 2 Hence detA 7 AB det 473731254 w2lt2gtltgtltgt2 Solving the characteristic equation detA 7 AB 0 gives us the estimates A 64 i 8 34 A 64 8x34 A1 2 578417 2 l 2 368825 3 On the otherhand the true eigenvalues are given by A1 jg1 5783 A2 jg 304705 4 where j0k denote the kth zero of the Bessel function of the rst kind of order 0 It becomes clear that our estimate for the rst eigenvalue is good but the second has signi cant error To improve we need to use either better trial functions or possibly use three trial functions 11 Informal derivation of the RayleighRitz Approximation RRA Let mi1 C 039 be arbitrary trial functions and linearly independent As an approximation to the true minimum problem MP see Lecture 8 let us impose an additional condition for w wac chwkw ck some constant 5 k1 So we only seek w of the form 5 Therefore in general 5 1 2 A1 because we impose more constraints in MPn If we are incredibly smart so that wac would be an eigenfunction then we have iAw Aw and ltVwVw7gt Vwij due 9 2 iAwwjdac6iwwjdac 6 n 395V Mwywjk where we used Green s rst identity in Therefore recalling the de nitions of oak and bjk we get ltchVwk ijgt A ltchwk wjgt chajk AchbJk 7 k k k k So we can write them in a matrix form Ac ABC7 c 31 0 which leads to detA 7 AB 0 If we are not so smart which is usually the case we can still use this determinant as our approximation method This leads to the Minimwc Principle See 5 Sec 13 for more details 2 Minimax Principle In reality what we really want is an exact formula to compute the eigenvalues instead of an approximation Let A g A 3 g A be the roots of detA 7 AB 0 From numerical linear algebra see Sec 21 it is easy to see that 5 max max ltAc7cgt 0 cTBc 7 0 ltBccgt Thus we have A max Zj k ajijCk 0 EM bjkejek Wangwt vltzkckwkgtgt max 8 C7 ltZj cgwaZk Ckwkgt Hle 2 max 2 wespan iquot M W H with w E spanw1 nun Hence we have from the MPn An s A Thus this allows us to take the minimum on the RHS to get An min A 9 where the minimization is taken over all possible choices of 101 wn C 039 Theorem 21 Minimax Principle let tugEL1 C 039 be an arbitrary set of n trial functions De ne A by 8 Then the nth eigenvalue is A An mm An where min is taken over all possible choices of the n trial functions 1071 C 039 Before presenting the proof consider the following intermission which plays a role in the proof 21 Intermission Rayleigh Quotient amp Linear Algebra This intermission is taken from 1 Lect 27 Let A E Rmxm be symmetric and z E R Then we know A has real eigenvalues A1 Am and real orthogonal eigenvectors cpl cpm De ne T A at Ann 10 rm x ltgt Clearly if ac cpj then rltpj Aj Of course we do not know 4p a priori So a motivating question is Given ac what scalar a acts most like an eigenvalue for ac The answer to this question leads to the following least square problem minHAw 7awH2 11 046R The answer to this least square problem is o The reason is as follows Hm7w ltm7wmm7wm wTA 7 aITA 7 aIw wTAZw 7 2awTAw aszw 9 2 T T By 67HAac 7 awHZ 0 we get 2am ac 7 290 A90 0 So a wTAw a wTw 12 is the solution of the least square problem 11 Thus if ac is close to one of the eigenvectors then the a in 12 should be close to the corresponding eigenvalue Also notice that Vrw Aw 7 13 So if ac cpj then Vrltpj 0 Conversely if Vrac 0 for ac 7 0 then ac is an eigenvector of A and Mac is the corresponding eigenvalue Thus the key point is that cpj are the stationary points of Further we can show that we 7 ma 0ltHw W112 as w a 90 14 This property is the basis for the method of power iterations See 1 Lecture 27 and 6 Sec 82 22 Proof of Theorem 21 Proof First let s x 101 wn C 039 Then choose 0102 07 not all zeros such that wltzgt zckwkltzgt L wwl k1 ie ltwltpkgt 20 ltwjltpkgt 07 k 1 n 71 j1 It is clear now that we have n unknowns and n 7 1 equations Therefore it is possible to nd 071 not all zero By the minimum principle we have llellZ n 7 15 11sz On the other hand the maximum in 8 is taken over all possible 07 So 2 Hva g A 16 1le Therefore we have 2 n S Hva S A 17 Hw HZ Now 17 must be true for all possible 101 wn C 039 which allows us to take minimum on the RHS ie A g min A 18 where the minimum is taken over all possible wj Yet we still need to show that An min A is attainable Choose wj e77 forj 1 71 where 1 With this choice we have HWZjCM HZ A max 19 00 H 27 01 th Using Green s rst identity G1 results in ltVltPj7VltPkgt PAW kgt A jk Therefore we have Aez An 02 max 7 2 g A A 20 0 2739 Ci 2739 0739 so that A 3 An Therefore An min A is attainable D References 1 L N TREFETHEN AND D BAU Ill Numerical Linear Algebra SIAM 1997 2 WA STRAUSSPartialIquotJ7 39 39 l 39 An J 39 John Wi ley amp Sons 1992 3 R COURANT AND D HILBERT Methods of Mathematical Physics Wiley lnterscience 1953 4 E B DAVIES Spectral Theory and Di erential Operators Cambridge UniV Press 1995 5 A HENROT Extremum Problems for Eigenvalues of Elliptic Operators Birkh iuser 2006 6 G H GOLUB AND C F VAN LOAN Matrix Computations 3rd Ed Johns Hopkins UniV Press 1996 MAT 280 Laplacian Eigenfunctions Theory Applications and Computations Lecture 4 Diffusions and Vibrations in 2D and 3D 1 Basics Lecturer Naoki Saito Scribe Brendan FarrellAllen Xue April 10 2007 The basic reference for this lecture is 1 Sec10l 1 Wave Equation and Heat Equation Consider a bounded domain 9 C Rd wave equation utt CZAU in Q d23 heat equation 1 U kAu in Q with one of the three boundary conditions BC on 39 u 0 D 3U a E 0 N u E au 7 0 R with initial conditions IC We 0 f 06 Ma 0 996 u0 D 3U 5 0 N 2 61 au 0 R 3V 149670 x 3 uw70 WC 0 The abbreviations for the boundary conditions used here are Dirichlet D Neu mann N Robin R For the Robin BC a is a constant We use the method of separation of variables and set uac t Ttvac which leads to the following equations T A1 From wave equation 7 7 7A CZT 1 4 T A1 From heat equation 7 7 7A kT 1 Later in this lecture we will show that A 2 0 for at least either D N or R in 2 is satis ed Regardless of whether we consider the heat or the wave equation we reach 7A1 A1 in Q where 1 satis es either D N or R Lots of mathematics are involved to prove that the set of A satisfying 5 is dis crete ie A1 A2 and there exist the corresponding eigenfunctions ltp1 ltp2 that are mutually orthogonal We ll cover those math later but at this point we assume the existence of A1 A2 and ltp1 ltp2 Once we have the eigenpairs An ltpnff1 we can write the solutions for 1 as wave equation uac t 21 An cosxnct B sinxAnctltpnac heat equation uac t 21 Ane Aquotktltpnac 5 6 where An and B are appropriate constants Preliminary some important formulas used in the following sections QVfdwanude 1 is normal vector and d5 is a surface measure on 69 o Divergence Theorem 0 Green s rst identity G1 For um E 02 uAvdwVuVvdw ugdS n 9 an 5 2 o Green s second identity G2 For um E 02 31 3U QuAv 7 vAu dw An 7 11 d5 0 The de nition of the directional derivative along 1 5 A quy 7 2 Orthogonality of the Eigenfunctions De ne the inner product ltfggt fxgxdx whereQ 6 Rd dac dzldzg dzd 9 Consider two functions u 1 E 02 with W Q U 69 02 condition can be weakened we have uAv 7 A101 VUV1 7 Vu zl Then integrate both sides in Q uAv 7 Amy dw V UVU 7 Vu zl dw n n 2 V UVU 7 Vuv d5 8 an 17 62 u 7 7 7 d5 39 u 3V Day 7 where a is derived by divergence theorem and b is from the de nition 7 Now we can show that any um 6 029 satisfying either D N or R also satisfy ltu7Avgt Auw Proof Equation 8 is equivalent to 31 3U ltu7Avgt7ltAu711gt 7 An 7 11 d5 3 If u and 1 satisfy D or N it is obvious that the above is equal to 0 If u and 1 satisfy R we get 1127 7 0 u7ao 7 117au 0 D Therefore each of these three classical BC s is symmetric Suppose both u 1 are real eigenfunctions satisfying 7Au Alu 7A1 A21 and satisfying either D N or R Then A1 A2 are reals and if A1 7 A2 then um 0 Proof A1ltuu Alum lt7Au7u lt rAugt u7A1u ATltu7ugt7 which implies 0 Since 7 0 A1 T1 lt A1 6 R Similarly A1ltuo 7 A2ltuo Alum 7 u Ago lt7Auo 7 u 7A0 u7Ao 7 u 7A0 0 which implies A1 7 A2uo 0 Since A1 7 A2 um 0 D u We summarize the information above with the following theorem Theorem 21 In the eigenvalue problem 5 we have the following facts 0 all the eigenvalues are real 0 the eigenfunctions can be chosen to be realvalued o the eigenfunctions corresponding to distinct eigenvalues are necessarily or thogonal 0 all the eigenfunctions can be chosen to be orthogonal iiei orthonormal 4 3 Multiplicity of the Eigenvalues De nition 31 An eigenvalue A has multiplicity m if it has m linearly indepen dent eigenfunctions The eigenspace EA is a linear space spanned by the set of eigenfunctions corresponding to A So in this case dimE m Notice if dimE m and EA spanw1 wm but ltwiwjgt 7 617 then we can use the Gram Schmidt orthogonalization method to get EA spanltp1 ltpm with ltltPi7ltpjgt 6H 4 Generalized Fourier Series Because of the Theorem 21 we have for f 6 L262 fw Emma u ow n1 This is a generalization of the Fourier series and we can discuss the decay of fn etc Theorem 41 As in 5 let Ak be the Dirichletlaplacian eigenvalues let Vk be the Neumannlaplacian eigenvalues and let pk be the Robinlaplacian eigenval ues where k E N Then kgt07 ukZO andpk207ifa20 Proof Let u and 1 are corresponding eigenfunctions Use Green s rst identity G1 a uAvdwVuVvdw uidlS7 n 9 an 5 6 uAudwVu2dw uiudS7 n 9 an 5 For the boundary condition D Sew u V 2d u7AudwVu2dw0 AWZO Q n fnu due 5 But Vulz 7 0 Since if so u const then u E 0 which con icts with the fact that u is eigenfunction Therefore A gt 0 For the boundary condition N V 2d V M 2 0 f9 u dw Here Vulz 0 is acceptable ie u E const 7 0 Then V 2 0 where V 0 corresponds to the eigenfunction ltp0z E const 7 0 For the boundary condition R we have iplulzdwqul2dw u7audS n 9 an afanlul2d5 f9 qulZdw gt 0 i P 7 7 fnlulzdw ifaZO 5 Completeness of 903nm in the LQsense See 2 Sec33 34 and 3 Chapter 4 for the elementary discussion on the com pleteness of a set of basis in L262 For all f 6 L262 we have N O aSN OO n1 7 L2 Equivalently 00 f Z fmpn in the L2 sense n1 This is important because if it were not the case we could not represent arbitrary L262 function in terms of ltpnn6N We will discuss more about the complete ness later Example Diffusion in a 3D cube LetQQzyz0ltxlt7r 0ltylt7r 0ltzlt7r DE utkAu inQ BC u0 on69 IC ufw wEQt0 Then by the separation of variables let uac t Tt as before we have 7A1 A1 in Q with 1 0 on 6Q Because the sides of Q are parallel to the axes one can do the separation of vari ables again Wag72 X96YyZ2 X Y Y 7 7 7 7A T X T Y T Y BC s are also separated as X0 X7r Y0 Y7r Z0 Z7r 0 Therefore we can solve vzy 2 for lmn 6 N3 vyz sinlx sinmy sinnz lmnw whose orthonormal version is 32 sinlz sinmy sinnz Then A lmn l2 m2 i 712 Finally we get the solution uw t Z Almne l2m2 2kt sinlx sinmy sinnz lmn where AWN 3ltf ilmngt Here different values for l m n can result in the same eigenvalue For example A 27 The valid values for lmn are 5 1 1 15 1 1 15 and 333 In other words the multiplicity of A 27 is four 7 MAT 280 Laplacian Eigenfunctions Theory Applications and Computations Lecture 7 Nodal Sets Lecturer Naoki Saito Scribe Bradley MarchandAllen Xue April 20 2007 Consider the eigenfunctions Mac of the Laplacian 7A2 A1 in o 1 with Dirichlet Neumann or Robin boundary conditions We de ne the nodal set N as follows De nition 11 The nodal set N is the set of points in 9 such that the eigenfunc tions of l are zero Thus N pc 6mm 0 Note that since 9 is open no point on the boundary of Q is in N ie 69 N Nodal sets are important because they allow us to visualize the sets where Mac gt 0 and Mac lt 0 They mark a natural division of the domain into regions It is this property that interested a German physicist named Ernst Chladni 6 Chladni s most well known work was developing techniques for demonstrating the various modes of vibration in a surface By drawing a bow over a piece of metal lightly covered in sand Chladni was able to visually show the nodal sets for varying frequencies Similar techniques are particularly important in making musical in struments since the symmetries in the nodal lines can be interpreted as a measure of quality To gain a better understanding of nodal sets let us consider a few examples 1 Figure 1 First four eigenfunctions for 1D Dirichlet Laplacian eigenvalue problem in the domain of S2 07 E Red points are Nodal nodes Example 12 ID Vibrating String Earlier we showed that the eigenfunctions for the 1D Dirichlet Laplacian eigen value problem in the domain of S2 0 6 are 2233 2 sin gpnv n E N 2 From the eigenfunctions we can see that M an Nodalsetforgpn x kl2n l n Note that k 0 and k n are not in N since they are on the boundary The nodes in the nodal set can also be given a physical interpretation Let us consider a wave equation with the Dirichlet boundary condition We already know that ua3 t A cos at B sin ctva 3 is a solution for the wave equation no initial condition is xed at this point Here 22213 is an eigenfunction of the Dirichlet Laplacian Then for all a E N um t E 0 for all t Z 0 Thus all the nodes in the nodal set are stationary points for all time This has physical implications if we consider a guitar string If the guitar player holds his nger on the middle of the string then the only tones that have nodes at 1 g survive see Figure 2 which means 2 Figure 2 If a guitar string is held in the middle only eigenfunctions with nodal points in the middle are possible Thereby limiting the possible sounds that the string can produce TMTC that the frequencies 7 with even n survive and that the frequencies with odd n cannot happen Example 13 The Square in 2D Let us consider the eigenfunctions of Dirichlet Laplacian on a 2D square ie S2 51y 0 lt 1 lt 7T 0 lt y lt 77 Previously we found that the eigenfunctions and eigenvalues are 9933 y 2 3283 ml my with n m E N 4 Since we know what the eigenvalues and functions are we can tabulate them in order of increasing eigenvalues For eigenvalues with multiple eigenfunctions ie eigenvalues with multiplicity we consider a linear combination of the eigenfunc ons For example there are two corresponding eigenfunctions 9012 and 9021 for A 5 Thus we write 9033 y A9012z y B9021 13 y as the eigenfunction for A 5 Table 3 contains a few of the tabulated eigenvalues and functions Note that by ordering the eigenvalues in this fashion we are able to indeX the eigenvalues and eigenfunctions with one indeX instead of two See the Table 3 There are some interesting questions we would like to consider First how many n M96711 2 Asinxsiny 5 8 Asin 2x siny B sinrsin 2y Asin 2sin 2y 10 Asin3xsiny Bsinxsin3y Figure 3 Table of eigenvalues and eigenfunctions for DirichletLaplacian problem on the 2D square whose side length is 7139 where A B are appropriate coef cients ways can a given integer A be written as the sum of two squares of integers Sec ondly we want to ask what do the nodal sets corresponding to different eigen values look like The rst question is answered by Number Theory Speci cally we know the following Theorem 14 A positive integer n can be written as the sum of two squares if and only if no prime congruent to 3 modulo 4 appears an odd number of times in the factorization of 71 into primes See 2 Chap 72 for details Let us investigate a bit about the second question First we consider eigenvalues of single multiplicity Since the eigenvalue is of single multiplicity we know that ltpz y A sin nz sin my for some n m in N Thus we can easily see that the nodal set will be the set x z 7rp U y y nq where p q E N such that 1 lt p lt n divides n and 1 lt q lt m divides m Those interested in the number of divisors of n and m can once again look to Number Theory for the answer 2 Chap 3 In the case where A 2 we get the lines z 7r2 and y 7r2 For eigenvalues of double multiplicity the nodal sets can be much more exotic since we have a linear combination of two eigenfunctions When A or B is zero we have the solutions that were discribed above For different values of A and B we get different looking nodal sets Images of nodal sets for A 10 can be seen in Figure 4 More images of nodal sets on the square can be seen in 3 Sec 104 Trefethen 5 Chap 11 provides insight into the programming involved in solving certain differential equations numerically 9031 9031 9013 9031 9013 9013 9031 139013 9031 139013 Figure 4 Some nodal sets for A 10 of the Dirichlet Laplacian on the square Example 15 The 3D Ball Recall that the eigenfunctions for the DirichletLaplacian on a 3D ball ie Q 334 z 932 32 Z2 lt a2 are J 12 A477quot where we use the spherical coordinates which were introduced in Lecture 6 WW 7quot 6 gb Pgmcos 6 A cos mgb B sin mgb 5 The nodal set for mm is a union of the following kinds of surfaces i Spheres inside 9 corresponding to the zeros of the Bessel functions ii Vertical planes ie gb const and iii Horizontal planes ie 6 const which in fact contains j l spheres m vertical planes and E m horizontal planes So how many regions can the nodal set of the DirichletLaplacian divide a gen eral domain 9 into assuming 9 is connected The following theorem limits the possibilities Theorem 16 Courant Nodal Domain Theorem i The rst eigenfunction p1 corresponding to the smallest eigenvalue A1 cannot have any nodes ii For n 2 2 MW corresponding to the nth eigenvalue counting multiplicity divides the domain 9 into at least 2 and at most 71 pieces Discussions on nodal sets and the Courant Nodal Domain Theorem can be found in 1 Vol 1 Sec V5 V16 and 3 Sec 104 We will prove this later in this course when we deal with the general eigenvalue problem using the calculus of variations At this point it is easy to show that 490 must divide 9 into at least 2 pieces if n 2 2 if we assume i is correct Proof We know that ltp1 is perpendicular to on for n 2 2 So 1ww due 0 o and from i we know that ltp1ac gt 0 or ltp1ac lt 0 in 9 Thus 7490 must change its sign in Q So there exist zeros of 490 in Q by the continuity of on These zeros form a nodal set D REMARK In 1950 Szego conjectured a similar theorem to the Courant Nodal Domain The orem for the biharmonic eigenvalue problem Azu Au in Q 6 u0 ona Szego s conjecture was the following Conjecture 17 Szego 1950 If 9 C R2 is a nice domain ie 69 is an analytic curve then p1 for 6 does not change its sign However surprisingly the conjecture is not even true for the rst eigenfunction For the details and its history including Szego s conjecture see 4 MAT 280 Laplacian Eigenfunctions Theory Applications and Computations Lecture 2 Sturm Liouville Theorem PreHistory of the Laplacian Eigenvalue Problem in Rd Lecturer Naoki Saito Scribe Adam DobrinAllen Xue April 3 2007 0 Review Vibrations of A One Dimensional String In Lecture 1 for the problem of Vibration of 1D string depending on the type of boundary condition BC we consider the following utt czum forz 6 06 andt gt 0 u0 t u t 0 Dirichlet BC or um0t umw 0 Neumann BC fort 2 0 1 um 0 f967 M967 0 996 fOM E 075 To solve for uz t we assume that the solution is independent in time and space That is we can write our solution as uz t X where X and Tt do not depend on each other After separating T part we had 7X AX for z 6 06 and A 2 0 X0 XZ 0 Dirichlet BC 2 or X 0 X Z 0 Neumann BC This is a 1 D version of the Laplacian eigenvalue problem given a general shape 9 C Rd iAuu inQCRd u f on 69 Dirichlet BC 3 6 l g on 69 Neumann BC 3V REMARK In 1D problem 2 we get 0139 Z 2 Dirichlet BC 12 Neumann BC Notice that Q 06 with Z Therefore the eigenvalues re ect the geometric information of Q in this 1D case the volume of Q the length of In 1D this line of work culminated in the work of Sturm and Liouville 1836 37 which also accounts nonuniform strings I Sturm and Liouville s work Given 9 a b De ne 17 Lamb 2 fsupportedonab fz2wzdxltoo 4 with wx gt 0 and w e Oab equipped with the weighted inner product for all f g 6 Li a b as A b 7 mm fltzgtgltzgtwltzgtdz De ne L a b a L a b such that for f E Limb if W pf 5 where r E Ola b r gt 0 on ab andp E Oab is real valued We consider the Regular Sturm Liouville Problem RSLP fAwf0 81f82f0 6 where ij Otjfa ojf a j b gf UJ forj l 2 with constants 04739 04 6739 and We say that the boundary conditions in 6 are selfadjoint relative to if Mfg i f 0 for all fg satisfying ij ng 0 j 12 For any f g belonging to a certain subspace of Limb and satisfying the self adjoint boundary conditions we can easily show that the differential operator de ned in 5 satis es lt f ggtw ltf ggtw Such an is called a selfadjoint operator Note that for the vibration of the one dimensional string with homogeneous Dirich let boundary conditions r E l p E 0 a1 101L 0 61 0 Bi 0 a2 0 02062land 0 11 The SturmLiouville Theorem For every RSLP 6 the following hold All eigenvalues are real D Eigenfunctions corresponding to distinct eigenvalues are orthogonal with respect to lt w U The eigenspace for any eigenvalue A is at most 2 dimensional and these two eigenfunctions can be chosen to be orthogonal 4 15 the n ih eigenfunction corresponding to the n ih distinct eigenvalue has n 7 l zeros in a b L11 For any f E 02ab satisfying 81f 82f 0 but not necessarily f Aw f 0 the series 21 ltf ngtw 15 converges uniformly to f 6 For any f E Limb ity fllimam 21 l ltf ngtw 2 Parseval s equal In higher dimensions we shall not delve into generalities of spatially varying coef cients such as pz in RSLP 6 We shall stick with the simple Laplacian eigenvalue problem in Rd d gt 1 2 PreHistory of the Laplacian Eigenvalue Prob lem in Rd 21 The Lorentz Conjecture from 2 In late October of 1910 a Dutch physicist H A Lorentz delivered a series of ve Wolfskehl lectures via a donation of Mr Wolfskehl who intended to pay the prize for a person who solved The Fermat Conjecture titled Old and New Prob lems of Physics at Gottingen University in Germany Referring to a Cambridge physicist J H Jeans s work in radiation theory Lorentz said In an enclosure with a perfectly re ecting surface there can form standing electromagnetic waves analogous to tones over an organ pipe we shall con ne our attention to very high overtones Jeans asks for the energy in the frequency interval dV To this end he cal culates the number of overtones which lie between frequencies V and V dV and multiplies this number by the energy which belongs to the frequency V and which according to a theorem of statistical me chanics is the same for all frequencies It is here that there arises the mathematical problem to prove that the number of su ciently high overtones which lie between V and V dV is independent of the shape of the enclosure and is simply propor tional to its volume For many shapes for which calculations can be carried out this theorem has been veri ed in a Leiden dissertation There is no doubt that it holds in general even for multiply connected regions Similar theorems for other vibrating structures like mem branes air masses etc should also hold If we express the Lorentz conjecture in a vibrating membrane it becomes of the 4 following form N A ZAWltA 1 An Dirichlet Laplacian eigenvalues 3 An lt A 7 DA 27139 A mathematician D Hilbert was attending these lectures and predicted as follows This theorem would not be proved in my life time But in fact Hermann Weyl a graduate student at that time was also attending these lectures Weyl proved this conjecture four months later in February of 1911 N 22 Weyl s work Let Q C Rd with volume of Q f9 dac lt 00 Consider the vibration problem utt Au in Q uact 0 on 69 8 Mam mo utw0 9w in 9 Using the separation of variables uac t X we reach to T AX 7 7 7 7 XT 7 AXT T X A replacing X by u we get the Dirichlet Laplacian eigenvalue problem iAu Au in Q u 0 on 69 9 62 62 62 t t 39 39 39 t If u i 0 satis es 9 then it is called a eigenfunction and the corresponding A LetL7AwhereA VV is called the eigenvalue De ne EA u E DL Lu Au the eigenspace corresponding to A with dim EA multiplicity of A In this problem A consists of countably nite number of eigenvalues with nite multiplicity ie we can order them as Mgugmgugam 5 Let Leak Awk k 12 And let f E 00 ie f 6 06 and f 0 on 69 where Q Q U 69 In fact it is ok to assume f 6 L262 Then f f ltpkgtltpk This is called an eigenfunction expansion of f be cause wk ltplgt 6M and pkheN is an orthonormal basis ONB for short of L262 here ltf79gt f9 fw9w dw Expanding the initial conditions into the eigenbasis we get M8 M8 90 ltf7ltPkgtltPk7 990 lt97ltPkgtltpk w H H w H H Then we get the solution to the vibration problem ultmgt ltf m m lt37 sin minim So the key was the Laplacian eigenvalue problem 9 Weyl s Theorem k i Ak as k a 00 10 where Cd 2m dl Equivalently NM when 11 where N k E Nl Ak A This equivalence is clear since NOW k dZ so k N Odl lAk Weyl s Conjecture 1 iod llamHA 0Ad 12 NO Odl ld 4 where l ld volume in Rd and wow volume in Rd l or the area in Rd This conjecture has not been completely solved yet and started the eld known as spectral geometry 23 Can we hear the shape of a drum In 1966 Mark Kac Rockefeller Univ asked Can we hear the shape of a drum 2 In other words how much can we know about the shape geometric infor mation of Q from the Laplacian eigenvalues Anf1 Kac proceeded to show that for all bounded Q C R2 00 7A 69 fl n 77 t 2 t 0 13 Ze 4m 8 7mm gtas1 gt n1 also 7A 191 1691 1 7 77 7 1 t 0 8 4m 8H 3 Hmas l 7 14 n1 if Q has r holes and Q and holes are polygons In 1967 McKean and Singer generalized Kac s result to Q C Rd For more about this work and the related work up to 1987 see 3 References 1 G B FOLLAND Fourier Analysis and Its Applications Sec 35 36 as well as other chapters BrooksCole 1992 2 M KAC Can one hear the shape of a drum Amer Math Monthly vol 73 no 4 part 2 pp 1 23 1966 3 M H PROTTER Can one hear the shape of a drum Revisited SIAM Review vol 29 no 2 pp 185 197 1987 4 W A STRAUSS PartialD39Jj 39 39Fl 39 An J 39 Chap 10 11 John Wiley amp Sons 1992 5 K YOSIDA lectures on Di erential and Integral Equations Chap 2 Wiley Interscience 1960 Republished by Dover in 1991 6 N YOUNG An Introduction to Hilbert Space Chap 9 10 11 Cambridge Univ Press 1988 MAT 280 Laplacian Eigenfunctions Theory Applications and Computations Lectures 89 Laplacian Eigenvalue Problems for General Domains I Eigenvalues as Minima of the Potential Energy Lecturer Naoki Saito Scribe Milad SabetimaniAllen Xue April 24 amp 26 2007 In this lecture we will consider Laplacian eigenvalue problems for general do mains Since the explicit formulas only exist for special domains eg rectangles disks balls etc what can we say about An ltpn for a domain 9 of general shape The basic references for this lecture are the texts by Strauss 6 Sec 111 112 and Courant and Hilbert 1 Sec V11 For the details and the survey up to the recent results consult 3 1 The Eigenvalues as the Minima of the Potential Energy Consider the following Dirichlet Laplacian DL Problem where Q is an open domain with general shape lt co and 69 is piecewise smoooth iAu Au in Q 1 u0 ona In this lecture we list 0ltA1A2An7 where each eigenvalue is repeated according to its multiplicity Now consider the following minimization problem MP m min Hva2 MP wecgm HwHZ wzo HVwHZ Md 2 Rayleigh quotient And illeHZ lelZ dw is the potential energy or roughness of ww in 9 is called the where 039 w 6 029 lw 0 on 69 The term Notice that if is solution for MP then so is a where a is an arbitrary nonzero constant Theorem 11 Let v 2 A1m min w and w6039 w 20 p1 arg min vaw wecgm lelz wzo then 7Altp1 Alon In other words the rst eigenvalue is the minimum of the potential energy and the rst eigenfunction is the ground state state of the lowest energy 3 Proof From now on we will call a function from 039 a trial function Let u be the solution of MP with minimum value m 2 0 Then for any trial function w 6 039 we have 2 2 m f9 qul dw lt f9 lel dw39 f9 lulz d3 7 f9 lwlz d3 2 Let 1 6 039 be any other trial function such that wac 8U where 8 is any real constant Then de ne A f9 lVu 81 2 due 8 f9 luevl2 dw which has a minimum at 8 0 ie fO 0 Expanding f 8 in 8 yields f9 qul2 28Vu V1 EZ VU Z dw 8 f9 u 26m 82112 dw Using the quotient rule for differentiation and substituting 8 0 we obtain that 0 f0 2Vu V1 dim qulZ dim 2m dim A simple algebraic manipulation produces leVulZ dw VuV1dw 2 uvdwmuvdw 9 f9 d9 9 9 Also by Green s rst identity G11 we may write VuVwa0Audw valdS0 n 9 an 5 The last equality follows from the fact that 1 6 039 ie vl an 0 Therefore Aumuvdw0 n This is true for any 1 6 039 Therefore we must have Au mu 0 ie m and u are the eigenpair for the Dirichlet Laplacian problem 1 We still need to show m is actually the smallest eigenvalue ie m A1 u ltp1 1See Lecture 4 for some details Let 7Avj Ajvj where Aj is any eigenvalue of the Dirichlet Laplacian problem 1 Then by de nition m lt 9 V jlz dw 7 f9 vjAvj dw 7 f9 x dw f9 1172 dw f9 1172 dw 7 f9 1172 dw A72 The rst equality follows from G1 So m Aj W and m is an eigenvalue of 1 So m A1 andu ea In this proof we apply the idea of calculus of variations Classical but excellent general references on calculus of variations are 1 Chap IV 4 Part II and 5 Chap II Finally an excellent treatment of calculus of variations related to PDEs is 8 Chap 8 r W dx Example 12 Find m min 01 w 6 030 1 wecgm f0 u2 dx wzo Answer m 7T2 since the solution of this MP is wz sin 7m ltp1 and A1 7T2 Interesting to Note We can easily pick a function w 6 030 1 to get an ap proximate value of m For example we can choose wz am1 7 a is an arbitrary constant Let us compare the true solution and this w Here we choose the functions with unit L2 norm See Figure 1 xEsimm 1 sin 7Tx 2 da 7T2 986967 01 2 z17 7 x 2 da 10 2 The Other Eigenvalues Theorem 21 Minimum Principle for the nth Eigenvalue Suppose that Aj avg g are already known Then A min WW2 Mm wecgm Hsz w jgto jeh W71 AH sm 7m Figure 1 Plots of MP solution sin 7m and a trial function 30zl 7 assuming that the minimum exists Furthermore the minimizing function is ie the nth eigenfunction Note that this theorem implies MA 3 An Vn 2 2 Proof By assumption there exists that is a solution to MPn Let m be the minimum value of MP So ulm 0 and u l ltp1 ltpn1 As in the proof of the preVious theorem let wac evw where w and 1 satisfy the conditions for MP Then exactly as before we have Au mm 1 dw 07 3 for anyz 6 039 withv l21ltpn1 Now consider forj l n 7 l QAu mm pj dw 2 Qu Am mltpj dw W 7 MWj due lt4 2 0 Q where a is derived by Green s second identity G21 b is from the fact that Altpj 7AM and c is derived by the fact u 1 ltpj Now let hac be an arbitrary trial function and set lth7ltpkgt ltltPk7ltl7kl we have 7 jetsam ck lt5 k1 Thenoltpj Oforj1n71 Since hltpj 6 039 W E 1 n 7 1 3 is valid foro de ned in 5 From 3 and 4 n 1 Au mlu lt1 ch kgt dw Au mm h dw 07 Vh 6 039 9 k1 n This implies that iAu mm Similarly to the previous theorem with induction we can show that m An u on D Remark 22 The existence of the minima MP and MP is a delicate mathe matical issue that we have avoided which led to the theory of Sobolev spaces In fact there are domains D with rough boundaries for which MP does not have any solution at all For further information see 7 8 Chap 5 9 Chap 6 and 10 Chap 7 8 Also 11 is the paper that put the end to the confusion of the two different de nitions of the Sobolev spaces References 1 R COURANT D HILBERT Methods of Mathematical Physics Vol I Wiley lnterscience 1953 2 LC EVANS Partial Di erential Equations AMS 1998 3 A HENROT Extremum Problems for Eigenvalues of Elliptic Operators Frontiers in Mathematics Birkhauser 2006 1See Lecture 4 for some details 4 S G MIKHLIN Mathematical Physics An Advanced Course North Holland Publ 1970 5 V I SMIRNOV A Course of Higher Mathematics Vol IV Pergamon PressAddison Wesley 1964 6 W A STRAUSS Partial Iquot 39 F 39 An J 39 John Wi ley amp Sons 1992 7 R A ADAMS AND J J F FOURNIER Sobolev Spaces 2nd Ed Elsevier Science Ltd Oxford UK 2003 8 L C EVANS Partial Di erential Equations AMS 1998 9 G B FOLLAND Introduction to Partial Di erential Equations Princeton Univ Press 1995 10 E H Loss AND M Loss Analysis 2nd Ed AMS 2001 11 N G MEYERS AND J SERRIN HW Proc Nat Acad Sci vol 51 pp 1055 1056 1964 MAT 280 Laplacian Eigenfunctions Theory Applications and Computations Lectures 12l3 Laplacian Eigenvalue Problems for General Domains IV Asymptotics of the Eigenvalues Lecturer Naoki Saito Scribe Ernest WoeiAllen Xue May 8 amp 10 2007 The basic references for this lecture are 1 Sec 116 2 Sec V12 and 3 Sec 112 1 Asymptotics of the Eigenvalues Our main purpose here is to show An T 00 and Vn T 00 as n a 00 and also to show how the eigenvalues go to in nity Theorem 11 Weyl Consider thefollowingDirichlet Laplacian D L problem iAu Au in 97 u 0 on 397 where Q C R2 9 is open and lt 00 Then 1 An 7 47139 SEQ n 19quot 1 For general domains in higher dimensions the following are also true 32 2 n 6 o ForQ CR3 lim 7 l TLHOO n AZ 0 d o For 9 C Rd lim 7 7d where Cd 47fd2P 7 1 Waco n 2 Before discussing a rough proof let us list a couple of examples Example 12 1D interval Given 9 0 Z for the DL problem then may and 12 7T 7T 1 Li 753071 e 9 Notice that for NeumannLaplacian N L and RobinLaplacian RL problems we have the same results Example 13 2D rectangle of sides a and b Let Q E R2 l 0 lt z lt 10 lt y lt b for the DL problem then W 2 77m 2 AnlmTgtZm12 2 Because these are naturally indexed by Z m it is dif cult to see the relationship between 1 and 2 So we ll introduce the socalled enumeration function NA n ENlAng A egif1 3 A2 3 S An S thenNn n In this particular case N A is the number of Z m E N2 satisfying 2 m2 A a2 b2 7 7T2 7 Figure l Lattice points contained inside the ellipse in the rst quadrant The z 2 y A ellipse equation 7 7 7 a b 7T2 see Figure 1 For each 7 m we can associate a square of area 1 which implies area of the ellipse a b l Aab lt if 7 NO 7 4 7T 7T 7T 4 47139 Now for large A the difference between N A and is proportional to the length of the perimeter So lt Aab 7c SNOx for some 0 gt 0 By setting A An we get n b An b a foxAngngi 47139 47139 where c is independent of 71 Now we have Then One can get the same thing for the NL case To proceed further we still need several other theorems Theorem 14 Maximin Principle Fix it E N with n 2 2 Fixn 7 l arbitrary trialfunctions yl 7 E 03 91 for the D L problem yl 7yn1 6 029 for the N L problem De ne 2 A Vw nak n 7 wow lle wyjgt0 j1n71 w70 2 A HVwH w602lt9gt HwHZ ltwgtyjgt0 j1n71 2110 Then 3 Ann 2177600261 m m 21760261 Proof We will prove this theorem for DL case rst The NL case can be proved similarly Note that yl 7 are xed at the moment Given the eigenfunc tions ltpj 21 of DL problem let w 204le j1 12 uses f 6 6139 and 6 PCQ azj From the assumption we have 107 0 for h l7 7n 7 1 Assume also Will 1f0rj 1771 Hence 0 must satisfy n TL 0 ltZci i7ykgt 01 ltltPj7ykgt 7 k 17717 1 11 j1 Since there are n 7 1 equations and n unknowns we can choose 01 7 on so that not all of them are zeroes In particular 21 c 31 0 Then by the de nition of Am we have 2 2 cm lt7Altpj7 W Z Z CjCkWk j k k vaF j ATL S 2 llwll ch k ltltPj7ltPkgt chjck6jgtk j k j k 2 AjCj M u lt An N H M l O NJQ where a is derived by Green s rst identity This inequality holds for each choice of yL7 711731 Hence we have max Am 3 An 3 y1gtmyn71chQ To show 3 is in fact equal we only need to nd a special choice of yL7 7 yn1 that attains equality in 3 So let yj ltpj j l7 7n 7 1 By the minimum principle MP and the de nition of Am we know that for this choice of yj p we have max Am An D Theorem 15 Vj S Ag 1727 Note that this is dz erent from the Friedlander Theorem that claims Vj1 S A forj 17 27 whose proofis much more dif cult Proof By the minimum principle we have HVwHZ HVwHZ n 1 lmw7 wwmlmw w0 w0 Now 03 C 029 so the search space for the Neumann case is larger There fore A1 2 V1 Now let n 2 2 For the same reason we have A 2 V7 This holds for each set of n 7 1 trial functions So by the M aximz39n principle V V 1 Remark 16 Any additional constraint will increase the value of the maximin Example 17 1D String Let 9 06 We have 2 7 1 2 in 1 2 7 H72 Theorem 18 If9 C 9 then An9 2 An9 Proof For simplicity let s write An An97 A An9 Let w 6 039 be an arbitrary trial function in 9 De ne w 6 039 such that ww ww ifw E 97 0 if ac E 9 9 So every trial function in 9 corresponds to a trial function in 9 but not conversely ie 3 trial functions for 9 that do not satisfy the Dirichlet boundary condition for 9 So compared to the trial function for 9 the trial function for 9 have the extra constraint of vanishing on 39 So by Remark 16 we get An 2 ALL Here we avoided to show w 6 039 but for the details see 2 Sec V11 D For the Neumann case there eXists a counterexample see 4 Sec 132 as fol lows Figure 2 w w Example 19 Consider a 2D rectangle of sides a and b with a gt b See Figure l 9 04 a a Figure 3 Neumann Case Counter Example Let 9 i 0 lt z lt 17 0 lt y lt b and Q be the inscribed thin rectangle as shown in Figure 3 Clearly Q C 9 We already know the Neumann eigenvalues and eigenfunctions for a rectangle V 2 92 g WW may 7T W 12 whereZm07 7 7 Clearly V1 V09 07 1 E c Since a gt b the second smallest eigenvalue and its corresponding eigenfunction are 7 7T 2 7139 V2 V10 gt 7 7792 77010 C 39 COS ltgt a a For V3 we have several possibilities depending on the relationship between a and b Here are just two examples i Ifg gt ie b lt a lt 2b we have 7139 2 7139 V3 V01 gt 7 7793 77001 C 39 COS 9 7 b b ii Ifg lt ie a gt 2b we have 27139 2 27139 V3 V20 7 7793 1020 C 39 COS The point is that the second smallest eigenvalue V2 of a 2D rectangle only depends on the longer side ofthe rectangle in this case 1 Now the longer side of Q is equal to xa 7 04 b 7 B By choosing appro priate a gt 07 B gt 0 we can have a 7 04 b 7 B gt a In other words we can have V2 lt V even ifQ lt 9 2 Subdomains The next step toward the proof of An 7 007 17quot 7 46quot as n 7 00 is to divide 9 into a nite number of subdomains 91 79m by introducing smooth boundary surfaces partitions F1 F2 See Figure 4 Let 0 lt A1 3 A2 3 be the eigenvalues for 9 Let 0 lt X1 3 X2 3 be the collection of all the eigenvalues Aj 2k1ltkltm JEN in the ascending order By the Maximin principle each X can be obtained as 7 Vw 2 An max E mpgn71 wecgm wiyiy But each yj j 17 771 7 l are supported on only one of the subdomains 91 79m So An S X by Remark 16 8 As for the Neumann case again list all the eigenvalues of the subdomains as 0 1 32 3m lt 3m 3 Now in the Maximinprmciple the trial functions y17 7M4 for n do not have to vanish at 39 and F73 j 1727 So there eXist less constraints than in the Dirichlet case for 9 hence we have n S An Summarizing all the results so far we have Theorem 21 Vn S An S n7 3n S n S n Now let a 571 U 72 U U m where 97 are all all rectangles see Figure 5 Figure 4 DiVision of 9 into a nite number of subdomains 917 927 7 9m with smooth boundary surfaces F17 F27 Let MX neNXnX 9 Then by counting integer lattice points in each rectangle 97 and 9 we have imMg LZLH A700 A 7 7T 47 Since MOW n we get Hm my Figure 5 9 represented as a collection of rectangles Similarily we can get 1 n 47139 1m 7 7 L700 71 By the sandwich theorem we have 1 An 47139 1m 7 7 L700 71 10 Remark 22 For a more general domain it can be approximated by unions of rectangles Using the similar arguments as before it is possible to prove An 47139 lim 7 2D version of Weyl s asymptotic formula TLHOO n Figure 6 Example of an approximation of Q by the union of a uniform squares For the details see 2 Sec VI 2 and 3 Sec 112 References 1 W A STRAUSS Partial Dijfkrential Equations An Introduction John Wi ley amp Sons 1992 2 R COURANT D HILBERT Methods of Mathematical Physics Vol I WileyInterscience 1953 3 P R GARABEDIAN Partial Dijfkrential Equations AMS Chelsea Publish ing 1964 4 A HENROT Extremum Problems for Eigenvalues of Elliptic Operators Frontiers in Mathematics Birkh user 2006

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