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Stanford MATH 220 - Lecture Notes

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Comments on Rayleigh-Ritz Approximation and MinimaxPrincipleLet {w1, . . . , wn} be n linearly independent trial functions. Let A = (ajk), B = (bjk)where ajk= h∇wj, ∇wki and bjk= hwj, wki. Consider the equationdet(A − λB) = 0.Assume λ∗1, . . . , λ∗nare the n real roots of this equation. We know that there exists a set ofvectors {vj} which form a basis for Rnsuch thatAvj= λ∗jBvjj = 1, . . . , nand the {vj} are mutually orthogonal with respect to B; that is,Bvj· vi= 0 i 6= j.LetXi≡ span{v1, . . . vi}.Lemma 1. For {w1, . . . , wn} a set of linearly independent trial functions and A, B, {vj}defined as above, the ithroot of the equation det(A − λB) = 0 satisfiesλ∗i= maxc∈Xi,c6=0Ac · cBc · cProof. First, we note thatλ∗i=Avi· viBvi· vi≤ maxc∈XiAc · cBc · c.Next, let c ∈ Xi. Then c =Pij=1cjvj. Therefore,Ac · cBc · c=APij=1cjvj·Pij=1cjvjBPij=1cjvj·Pij=1cjvj=Pij=1cjλ∗jBvj·Pij=1cjvjPij=1c2jBvj· vj=Pij=1c2jλ∗jBvj· vjPij=1c2jBvj· vj≤ λ∗i.Therefore, taking the maximum of both sides over all possible c ∈ Xi, we get the desiredresult. ¤1Remark. Using the fact thatAc · cBc · c=||∇(Pnj=1cjwj)||2L2||Pnj=1cjwj||2L2,we haveλ∗i= maxc∈Xi,c6=0(||∇(Pnj=1cjwj)||2L2||Pnj=1cjwj||2L2).Lemma 2. The ithDirichlet eigenvalue is given byλi= min maxc∈Xi,c6=0(||∇(Pnj=1cjwj)||2L2||Pnj=1cjwj||2L2)where the minimum is taken over all possible sets of linearly independent functions {w1, . . . , wn}.Remark: From the above remark, we see thatλi= min λ∗i(w1, . . . , wn)where the minimum is taken over all possible sets of linearly independent trial functions.Proof. Fix { w1, . . . , wn}. Choose a linear combination w =Pnj=1cjwjsuch that• c = (c1, . . . , cn) ∈ Xi• w is orthogonal to the first i − 1 eigenfunctions (denoted ui).The first condition impliesc · Bvj= 0 j = i + 1, . . . , n.The second condition implieshw, uji = 0 j = 1, . . . , i − 1.In particular, we have n − 1 equations for our n unknowns c1, . . . , cn. Therefore, such afunction exists.By the Minimum Principle for the itheigenvalue, we haveλi≤||∇w||2L2||w||2L2≤ maxc∈Xi||∇(Pnj=1cjwj)||2L2||Pnj=1cjwj||2L2.Taking the minimum of both sides over all possible sets of linearly independent trialfunctions, we conclude thatλi≤ min maxc∈Xi,c6=0||∇(Pnj=1cjwj)||2L2||Pnj=1cjwj||2L2.2Next, let {w1, . . . , wn} be the first n eigenfunctions. Thenmaxc∈Xi||∇(Pnj=1cjwj)||2L2||Pnj=1cjwj||2L2=||∇wi||2L2||wi||2L2= λi.Therefore,min maxc∈Xi||∇(Pnj=1cjwj)||2L2||Pnj=1cjwj||2L2≤ λi.Consequently, we get the desired result.

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