M ath 5 2 7 Fall 2 0 0 9 L ecture 1 9 N ov. 1 6, 2 0 0 9 ) S ecod- Order Elliptic Equatios: Weak S olutios 1. Defiitios. I this ad the followig two lectures we will study the boudary value problem Here or L u = i, j L u = a i j x) u x i ) L u = f i ; u = 0 o. 1 ) x j + i= 1 a i j x) u x i x j + i = 1 b i x) u x i + c x) u A x) D u) + b x) D u + c x) u, 2) b i x) u x i + c x) u A x) : D 2 u + b x) D u + c x) u. 3) I the first case it s said the equatio is i divergece form, i the secod odivergece form. Remark 1. If a i j C 1 the the two forms are basically equivalet. Remark 2. For the equatio to be elliptic, we eed to assume A x) symmetric, ad positive defiite. Remark 3. No-zero boudary values ca easily be icorporated. I the followig we will see that the right settig for the weak solutio is u H 1 ). I this case, if u = g 0 o the boudary, we ca exted g to a H 1 fuctio G i 1, ad the the equatio for v = u G satisfies v = 0 o. I the followig, we require a i j, b i, c L, that is uiformly bouded. Remark 4. Note that uder such assumptio, the divergece ad odivergece forms are ot equivalet aymore. der such assumptio, the followig defiitio is meaigful. Deote by B[, ] the biliear form [ B[ u, v] a i j u x i u x j + ] b i u x i v + c u v dx 4) i= 1 for u, v H 1 0 ). Defiitio 5. We say that u H 0 1 ) is a weak solutio of the problem if for all v H 0 1 ). L u = f i ; u = 0 o. 5) B[ u, v] = f, v) f v dx 6) Remark 6. Note that the boudary coditio u = 0 does ot appear i the equatio B[ u, v] = f, v). It s istead guarateed by the requiremet u H 0 1 ). 2 We try to establish a complete theory of such elliptic equatios, ad hope that we ca obtai results similar to that of the Poisso equatio, which is the special case a i j = δ i j, b i = 0, c = 0. I particular, we would like to be able to show that f H k u H k + 2. It turs out that to fulfill this, we eed to make oe further assumptio, that is the operator L is uiformly elliptic. 1. Some regularity is required for g. More precisely we eed g H 1 / 2 ). Of course, sice g should be the trace of u, it must satisfy such requiremet. u 2. Suppose we try to form a weak solutio formulatio for the Neuma problem L u = f, = g, what fuctio space should we take?
Defiitio 7. We say the partial differetial operator L is uiformly elliptic if there exists a costat θ > 0 such that a i j 2 x) ξ i ξ j θ ξ 7) for a. e. x ad all ξ R. 2. Existece ad uiqueess of weak solutio. I our geeral case here, it is ot possible to fid a explicit formula as we did for the Poisso equatio. Therefore we eed to show existece implicitly. Oe way is through the Lax-Milgram theorem. We assume f L 2 ). Theorem 8. Lax-Milgram Theorem) Let H be a real Hilbert space, with orm ad ier product, ). Let, deote the pairig of H with its dual. Let be a biliear mappig. Let f: H R be a bouded liear fuctioal o H. With the above settig, if ther are costats α, β > 0 such that B[, ] : H H R 8) ad B[ u, v] α u v boudedess) 9) β u 2 B[ u, u] Coercivity) 1 0) the there exists a uique elemet u H such that for all v H. B[ u, v] = f, v 1 1 ) Remark 9. I our problem, H is the space H 1 0 ), whose ier product is u, v) u v + D u D v dx 1 2) Its dual is the space H 1 which cotais L 2, ad for f L 2 H 1, the pairig with ay v H 1 0 is give by f, v f v dx. 1 3) For the characterizatio of H 1, see Evas 5. 9. 1. Proof. First we defie a operator à through à u, v = B[ u, v] 1 4) for all v H. As B[ u, v] α u v, we coclude that à u H. Now apply the Riesz represetatio theorem, we ca defie aother operator A: H H such that A u, v) = Ãu, v 1 5) for all v H. It is easy to check that A is a bouded liear operator. Apply the Riesz represetatio theorem agai, we ca fid f H such that ) f, v = f, v 1 6) for all v. Now what we eed to show becomes Existece: For all f H, we ca fid u such that A u, v) = words existece is equivalet to that A: H H is oto. ) f, v or equivaletly Au = f. I other
iqueess: For ay f H there is at most oe u such that A u = f. I other words, the mappig A is oe-to-oe. It is easy to check that uiqueess follows immediately from coercivity of B. Now we show that A is oto, that is R A) = H. Assume the cotrary, that is R A) is a geuie closed subspace of H. The sice H is a Hilbert space, we ca fid a ozero v H such that v R A). Now compute Cotradictio! 0 = Av, v) = B v, v) β v 2 v = 0 1 7) Now we are ready to show the existece ad uiqueess of our problem. Note that the boudedess of the coefficiets a i j, b i, c guaratees the boudedess of B. However coercivity is ot always satisfied. Theorem 1 0. There is a umber γ 0 such that for each µ γ ad each fuctio f L 2 ), there exists a uique weak solutio u H 0 1 ) of the boudary-value problem L u + µ u = f i ; u = 0 o. 1 8) Proof. See Evas pp. 300 301. Remark 1 1. From the proof we ca see that for the Poisso equatio, γ ca be take to be 0. 3 3. Regularity. Theorem 1 2. Iterior H 2 -regularity) Assume a i j C 1 ), b i, c L ), ad f L 2 ). Suppose further that u H 1 ) is a weak solutio of the elliptic PDE The ad for each ope subset V, we have the estimate The costat C depedig oly o V, ad the coefficiets of L. Remark 1 3. There are several poits worth oticig. L u = f i. 1 9) 2 u H loc ) 20) u H 2 V) C f L 2 + u L 2). 21 ) 1. As a i j is assumed to be i C 1, it does t matter whether L is i divergece or odivergece form. Note that this assumptio is ideed ecessary i the proof. 2. We does ot require u H 1 0 ). Thus the importat poit here is that, eve if u oly satisfies the 2 weak form of the equatio locally, the it is H loc there. 3. Iterior refers to the fact that our estimate u H 2 V) C f L 2 + u L 2). 22) caot reach the boudary. As we will see i the proof, the costat C teds to ifiity as V gets larger. Proof. First we illustrate the idea through Poisso equatio, whose weak formulatio is D u D v = f v. 23) Now if we ca set v = u, the after itegratio by parts we have u) 2 = f u 2 f 2 + 1 u) 2 u) 2 4 2 f 2. 24) 3. I fact we ca take γ to be egative by takig advatage of the Poicare iequalities.
Furthermore, if we ca set v = u, the weak formulatio gives D u 2 = f u f 2 + u 2. 25) The above gives u) 2 + D u 2 + ) u 2 C f 2 + u 2 which leads to our desired result. Of course the major problem i the above proof is that we caot take v = u ad v = u. The way to fix this is to cut-off. Fix ay V, ad choose a ope set W such that V W. The we ca costruct a smooth fuctio ζ such that ζ 1 o V; ζ 0 o R W; 0 ζ 1. 27) Now the idea is, istead of usig u, we use ζu). This test fuctio vaishes at the boudary, which is ecessary for it to be i H 1 0. However, sice u H 1 oly, we caot take two derivatives. So fially we have to take the followig more tricky versio: where the differece quotiet Now we have D k h u x) a i j u x i v x j dx = 26) v D k h ζ 2 D k h u ) 28) u x + h e k ) u x) h = =. 29) [ a i j h u x i D k ζ 2 D h k u ) ] x j h D ) k a i j u x i ζ 2 D h k u ) x j a i j x + h e k ) D h ) k u x i ζ 2 D h k u ) x j + D h k a ) i j u x i ζ 2 D h k u ) x j = ζ 2 a i j x + h e k ) D h ) k u h x i Dk u ) x j + + + 2 ζ ζ x j a i j x + h e k ) D k h u x i ) Dk h u ) 2 ζ ζ x j Dk h a i j) u x i Dk h u ) ζ 2 D k h a i j) u x i Dk h u ) x j. 30) Now the first term ca be estimated as ζ 2 a i j x + h e k ) D h ) k u h x i Dk u ) θ ζ 2 h 2 Dk D u x θ j D h 2 k D u. 31 ) If we ca fid a uiform i h) upper boud for this term, we are doe. To do this, we eed to give upper bouds for the other three terms. The key observatio is that, each of them is bouded either by C D 2 u D u 32) or by C V D u 2. 33)
Same is true for all the terms comig from b D u, c u ad f. Thus we obtai u H 2 V ) C ) f L 2 ) + u H 1 ). 34) To obtai the desired result, we eed to estimate u H 1 ) by f L 2 ad u L 2. 4 fortuately this is ot possible. O the other had fortuately the above argumet works ot oly for but also for ay Ũ satisfyig V Ũ. Thus we have u H 2 V ) C f ) L 2 + u ) Ũ H 1 Ũ ). 35) To estiamte u ) H 1, we take v = ζ 2 u with ζ 1 o Ũ istead of V, ad go through estimates similar Ũ but simpler) to what we have doe above. Remark 1 4. There is o difficulty extedig this result to higher order, basically gettig f H m m+ 2 ), as log as we are willig to assume a i j, b i, c C m + 1 ). See Evas pp. 31 4 31 6. H loc Now what happes at the boudary? Theorem 1 5. Boudary H 2 regularity) Assume a i j C 1 Ū ), b i, c L ) ad f L 2 ). Suppose that u H 0 1 ) is a weak solutio of the elliptic boudary-value problem L u = f i ; u = 0 o. 36) Assume fially is C 2. The u H 2 ) ad we have the estimate u H 2 ) C ) f L 2 ) + u L 2 ). 37) The costat C depedig oly o ad the coefficiets of L. Remark 1 6. Note the differeces! 1. a i j C 1 Ū ) istead of C 1 ). 2. Assumptio o the regularity of is ecessary. 3. u eeds to solve the boudary value problem, ot just satisfy the equatio. Proof. We oly sketch the mai ideas. For details see Evas pp. 31 7 322. First cosider the case where the boudary is x = 0. Now we ca take the cuf-off fuctio ζ such that ζ 1 o B 0, 1 /2) but 0 outside B 0, 1 ). Now it is easy to check that v D k h ζ 2 D k h u ) 38) u ca serve as a test fuctio that is, i H 1 0 ) as log as k etries i D 2 u except u x x.. This way we ca obtai bouds for all To estimate u x x, write the equatio i odivergece form, ad move all terms to the RHS except for a u x x. Now we have u H 2 ) C f L 2 + u H 1 ). 39) This time o trick is eeded to estimate u H 1, sice u H 0 1 allows us to use Poicare iequality. Fially, i the geeral case, we first do a partitio of uity ad the apply a chage of variables to straighte the boudary locally. It turs out that, uder the assumptio C 2, the ew equatio still satisfies the coditios i the theorem. 4. If we oly wat a boud for u H 1 ), o work eeds to be doe as u is assumed to be i H 1.
Remark 1 7. Oe ca also obtai higher order versios of boudary regularity uder the assumptio that See Evas pp. 323 326. a i j, b i, c C m + 1 Ū ), C m + 2. 40)