Topics In Math
Topics In Math MAT 280
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MAT 280 Laplacian Eigenfunctions Theory Applications and Computations Lecture 11 Laplacian Eigenvalue Problems for General Domains III Completeness of a Set of Eigenfunctions and the Justi cation of the Separation of Variables Lecturer Naoki Saito Scribe Alexander SheynisAllen Xue May 3 2007 1 Completeness of a Set of Eigenfunctions 11 The Neumann Boundary Condition In this lecture we begin examining a generalized look at the Laplacian Eigenvalue Problem particularly related to generalized domains Our goal here is to establish the notion of the completeness of a set of eigenfunctions which we then use to justify separation of variables a tool we so far have taken for granted Basic references for this lecture are 1 Sec 113 115 2 Chap 11 and 3 Sec 33 For more advanced treatments see 4 Sec 65 In the earlier lectures we used N V or AgN for the Neumann Laplacian eigenvalues 77 or ltpj for the Neumann Laplacian eigenfunctions For simplicity let us use Vi y for Neumann BC and Aj ltpj for Dirichlet BC 1 And we number V7 in ascending order 0 V1 lt V2 3 V3 3 also recall that w1ac const Theorem 11 For the Neumann Condition we de ne a trial function as any ww 6 029 such that ww i 0 Then similarly to the Dirichlet case MP MP n RRA see Lecture 9 and the minimax principle are all valid Note w 6 029 means that there is no constraint at 69 In particular w may not satisfy the Neumann condition nor the Dirichlet condition As such the Neu mann condition is also referred to as the free condition and the Dirichlet condition is referred to as the xed condition 7 WW2 m7 Ullll wec2ltngt le w0 Let u 6 029 attain this minimum Set w u 81 V 1 6 029 The similar procedure leads to as a result of variational calculus Proof De ne ew v71 muvdw 0 vi 6 029 1 9 By Green s rst identity 3U l ltgt Au mu1 dw 71 d5 2 9 an 5 i Let s choose Mac no where 0 on 69 Then the right hand side of 2 equals 0 which implies that Aumu 0 inside 9 A arbitrary 02 function in 9 11090 ii Since Au mu 0 in Q for M 1 6 029 in 2 we have 20 geek 00n6 2 39 12 Completeness in the L2 sense The notion of completeness for eigenfunctions is similar to that of a complete basis which many of us have seen in an analysis class or any class that signif icantly dealt with Fourier series in which a set of eigenfunctions is complete if any arbitrary function in the space of concern can be exactly represented as a lin ear combination of the eigenfunctions here the exact in terms of the norm of that space To this end we begin with the following theorem to illustrate this point Theorem 12 Both the DirichletLaplacian DL and the Neumannlaplacian NL eigenfunctions are complete in the L2 sense ie V f E L2Q Hf Zg CMPn Z OasNaoo whereenM39 2 W lt3gt Hf T 2771 dnlbn a 0 as N a 00 where dn or ltf7ltpngt 0 V71 6 N gt f E 07 ae and zbn 0 V71 6 N gt fEO7 lleu Remark This is important since 724 are not useful if they are not com plete In other words if 3f 6 L262 such that Hf 7 21 enon gt 0 then on spans only a subspace of L262 Proof The proof depends on the following two facts i The existence of the minima of the Rayleigh quotient and ii An T 00 m T 00 as n a 00 will be proved in the lecture of eigenvalue asymptotics We ll prove 3 for all f 6 039 for DL and for all f 6 029 for NL respectively For an arbitrary f 6 L262 see examples in 4 Sec 65 and 2 Chap 11 Let s start with the DL case Given an arbitrary function f 6 039 set A N WW We 7 Z cmmw 6 039 n1 3 By the orthogonality for j l n N ltTN7ltP1gt ltf ch n7lt jgt ltf7ltPjgt 0739 ltltPj7ltPjgt 0 n1 So TN satis es the conditions of MPN1 Thus we have HVwH2 lt HWsz2 wec mxwzo HwHZ HTNHZ r 3971 N N1 Now we have HerHZ W0 7 21 Wt dw sz 7 220an W7 ZZCanVw Wan due 9 71 m 71 4 By Green s rst identity we have VfVltpndwifAltpndw hde u m n 9 an 5 and similarly Vltpn Vltpn 5mnAnHltPnH2 Q So 4 will give us N HWNHZ lVledw 7 Zcixnlmuz n1 Since An gt 0 we have HWNHZ lt lVledw HVsz 9 Therefore HWNHZ HTNHZ W lt Hm N1 S N1 gt Hr As we will show later ANH T 00 as N a 00 we have that HTNHZ l 0as N a 00 The Neumann case can be derived similarly E 2 Justi cation of the Separation of Variables Via Com pleteness Prior to all this we heuristically used the separation of variables technique to solve certain partial differential equations With the idea of completeness we can justify its use if we have that onheN and zb neN are a complete orthogonal basis 21 Separation of the Time Variable To justify the separation of the time variable we examine the heat equation for a more practical treatment of the justi cation One can adjust this argument to validate the separation of the time variable in a more general context but for our purposes it is not necessary The heat conduction is modeled by U kAu in 9 u 0 on 69 5 uw0 fw in Q A similar model can be formed for the Neumann and Robin conditions As we know the separation of variables leads to uav7 t ZAne Awtmn n1 But we will prove this by assuming certain differentiability on u and u 6 H3 ie uHZ HVuHZ lt 00 instead of the separation of variables For the details see 4 Chap 7 Proof Let uac t be a solution to 5 For each It u t 6 L262 and satis es the Dirichlet condition Since onheN is complete in L262 we can expand u t as ultw7 t Emma 6 Note that cnt is unknown at this point Plugging 6 into 5 term by term differentiation of the series is assumed derivatives taken in this manner fall into the weak derivative sense we get 2 Chtltpnw k 2 GntAltPnw Z kAnCnOtypnCB Thanks to the completeness of ltpnn6N we can deduce out ikAncnOt Hence we have cnt Ane k w An arbitrary const 7 D 22 Separation of the Spatial Variables For simplicity we will only justify separation of the spatial variables for the 2D rectangle Q 1 gtlt 2 C R2 with 1 2 C R as shown in Figure 1 12 011312 Figure 1 Q C R2 with 112 C R n 62 62 Wlth A L 67112 6m 6yy let iaw p OIWiW and iayw y mpg with appropriate boundary conditions for the D N or R cases Theorem 21 The set of products izltpfnymmeN2 is a complete set of eigenfunctions for 7A in Q with the given boundary conditions Proof We have A iltp3ny famine and W96 fagMM an Hence 0 8m7 iWW lDhmmew are eigenpairs and we clearly see that Wi m duo const Sm6mm We know that 6N is complete in L2 1739 j 1 2 so the question is whether this 2D eigenvalue problem has eigenfunctions other than the product form evil Suppose u is such an eigenfunction non product form so that u satis es iAu Au u ag 0 i Suppose A an mhmmeNz Then u L ltp1ltp n Via the Fundamental Theo n rem of orthogonality of Laplacian eigenfunctions see Lecture 4 Thus we have 0 ltu7ltpiltp3ngt 1ux7y id gammy By the completeness of nym6N in 1212 we must have fIl uz dz 0 for y E 2 ae Similarly by the completeness of izn6N in L2Il uzy 0 ae So such a u is not an eigenfunction which means that we must have A 6 an mhmmeNg ie A an 6m for some mm 6 N2 It is possible that an 6m may have multipilicity greater than 1 So let oiofnhmmg be the corresponding eigen functions where A N2 Then set ltu7 i 3ngt A i 7amp7 79 Z Cnm ilt gt nlt9gtv Wlth Cnm Hltplltp2 Hz39 nm6A Now rzy 1 oiofnhmmd by construction So rzy E 0 ae Via a similar argument as before Thus um 2 Castaway nm6A Therefore oiofnhmmew is complete in L262 1 References 1 W A STRAUSS Partial D39Jj quot l 39 An J 39 BrooksCole Publishing Company 1992 2 R YOUNG An Introduction to Hilbert Space Cambridge UniV Press 1988 3 G B FOLLAND Fourier Analysis and Its Applications BrooksCole Pub lishing Company 1992 4 L C EVANS Partial Di erential Equations AMS 1998
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