Elliptic Partial Differential Equations. Leonardo Abbrescia

Transcription

1 Elliptic Partial Differential Equations Leonardo Abbrescia

2 Advised by Daniela De Silva, PhD and Ovidiu Savin, PhD.

3 Contents Introduction and Acknowledgments Variational Methods and Sobolev Embedding Theorems 3. Laplacian A non-linear PDE and Method of Continuity Non-linear variations on Manifolds Regularity Theory Harnack Inequality for Divergence Equations 3 3. Regularity Estimates Harnack Inequality Structural Inequalities Moser Iteration Applications of the Harnack Inequality Harnack Inequality for Non-Divergence Equations ABP Estimate Measure theory and Harnack Inequality Curved C,α Domains Estimate for Laplacian General Elliptic Operators

4 Introduction and Acknowledgments Second order elliptic partial differential equations are fundamentally modeled by Laplace s equation u =. This thesis begins with trying to prove existence of a solution u that solves u = f using variational methods. In doing so, we introduce the theory of Sobolev spaces and their embeddings into L p and C k,α. We then move on to applying our techniques to a non-linear elliptic equation on a compact Riemannian manifold. We introduce the method of continuity along the way to provide another way of solving the equation. We move onto proving Schauder estimates for general elliptic equations in divergence form: i (a ij j u) + c(x)u = f with various assumptions on a, c, and f. We conclude our study of equations in divergence form by proving the Harnack Inequality using Moser iteration. Personally I would have liked to have proved the Harnack inequality in my own flavor, but due to lack of time, I had to follow very closely the proof given in []. The second half of the thesis revolves around equations in non-divergence form: a ij u ij =. As a disclaimer, I wrote the second half separately from the first and so my notations change heavily. We first start with the proof of the ABP maximum principle which is used heavily in, not only the proof for the Harnack inequality for non-divergence equations, but for the section on curved C,α domains. In that section I go over a completely new way of getting regularity estimates: approximation by polynomials. We first show how this can be used for u = f and then for general elliptic operators. We conclude the paper by introducing Krylov s regularity results for flat domains and generalize it to curved domains whose boundaries look locally like the graph of a C,α function. As one more disclaimer, I was not able to prove every single detail in this book due to lack of time. I leave tiny bits and pieces as exercises for the reader, but the overwhelming majority is proved in rigorous detail. The writing of this paper was a long and arduous process. It grew out of many discussions with Professor Daniela De Silva and Professor Ovidiu Savin. I am very grateful for their insights and very very patient guidance. I would not have been able to set my path as a mathematician without them. I d like to give thanks to Professor Michael Weinstein for being an exceptional guidance counselor and instructor whom I ve learned a lot from about singular integrals, methods of characteristics, and graduate schools. I d like to give special thanks to Professor Duong Phong, whose Analysis II class helped refine the details of this thesis.

5 Variational Methods and Sobolev Embedding Theorems. Laplacian We begin with the simplest problem. Let R n be a bounded domain, and f a function on. We wish to find a u such that { u = f in u = on The way we will find the solution to this problem is by finding the minimum for a specific functional. This is the idea: let g(x) be a function on R such that G(x) on with G (x) = g(x) and we want to find an x with g(x ) =. Then one way to approach this problem is to find an x such that G(x) G(x ) x = G (x ) = = g(x ) =. By comparisons, g(x ) = would be the equivalent of u =. So now we try to find the equivalent of G. Define the functional I(u) = Du + fu. This functional is going to be defined for u W, (). Now lets go over some definitions. Definition.. Let B be a normed Banach space. Then its completion B := { {u k } B, u k is cauchy }. Example.. Q = R. Example.3. {C (), u p < } = L p () (modulo the equivalence that f g if f g on a set of measure zero). Example.4. W, () = {u k C, u k u l, Du k Du l }. Also, W, () = {u L (), L () u = lim k u k, Du k Du l, Du k v L.}. Now we go back to our question: does there exist a u W, () such that I(u) I(u ) for any u W, ()? We will begin to show this in two steps. Our first is to show that our functional is bounded below. We will show that C > such that I(u) C for any u W, (). This will at least give us a starting point to find a minimum because we no longer have the ambiguity of I(u) exploding to. Next, assuming this, if I(u) did have a minimum, then min I(u) <. Now pick a minimizing sequence {u j } such that I(u j ) min I(u). The reason why this can be picked will be shown later. We wish to show that u j u and I(u ) = min I(u). Claim. We claim that C > such that I(u) C u W, (). First recall the definition of I: I(u) = Du dx + fudx. Then applying Hölder s inequality and Cauchy s inequality, we have I(u) Du dx f u ( ɛ Du dx u + ) ɛ f. Now note that we would be done with our claim if we show that Du ɛ u. This is saying that the gradient controls our function u. But we are in luck because we choose u to have compact support, so it is zero on the boundary. Then this implies that the gradient not only approximates values near u, but tells us what they are. 3

6 Lemma.5. There exists ɛ = ɛ() such that ɛ u 4 Du. Proof. First note that this will imply I(u) Now we move onto a claim: ( 4 Du ) ɛ f dx. Claim. Let be convex, and bounded in R n. Then I will show that u C (), x, then (diam )n Du(y) u(x) u(y) dy dy. n x y n Proof of Claim. Let x, y be arbitrary, and denote r = x y. Let ω = (y x)/ x y be the unit vector in this direction. Then we have that u(y) u(x) = = r r u(x) u(y) u(x) u(y) dy i= r d u(x + tω) dt dt n r u x i (x + tω)ω i dt Du(x + tω) ω dt Du(x + tω) dt dy. We transform the RHS of this integral inequality using the polar transformation centered at x: y (r, ω). Let l ω denote the distance from x to. lω [ r ] u(x) u(y) dy Du(x + tω) dt r n dr dσ(ω) = S n lω [ r S n t lω l n ω S n (diam )n n (diam )n = n (diam )n = n (diam )n = n ] r n dr Du(x + tω) dt dσ(ω) Du(x + tω) dt dσ(ω) n lω Du(x + tω) dt dσ(ω) S n lω S n Du(x + tω) t n dt dσ(ω) t n Du(x + tω) dy t n Du(y) dy. x y n With the proof of the claim out of the way, we can prove the following corollary: Corollary.6. Let u C (). Then for any x R n, we have Du u(x) c n dy. x y n R n 4

7 Proof of Corollary. We begin by introducing a little bit of notation: ū (y) := u(y) dy. Then we have for = B R () u(x) ū BR () = B R () B R () B R () = c n B R () B R () B R () (R) n n (u(x) u(y)) dy u(x) u(y) dy B R () Du(y) dy. x y n Du(y) dy x y n Notice that we will be done with our corollary if we show that ū BR () as R. But this is where we use the fact that u has compact support: ū BR () = B R () Du(y) x y n dy = Du(y) dy as R. R x y n n Now that we are done with this corollary, we will quote a lemma that will be proved later: Lemma.7 (Estimates for Integral Operator). Assume u(x) K(x, y) v(y) dy. Then for any p, we have u(x) p dx A v(y) p dy where { A = max sup x K(x, y) dy, sup y } K(x, y) dx.. Applying this lemma to Corollary.6 gives us u p C Du p for u C () (and in fact for u W, () by approximations). We are finally done with our Lemma.5. Now that we are done with the first part of the problem, we go back to the infimum question. Assume we know that inf u W, () I(u) > and pick a minimizing sequence, i.e. u k W, () such that I(u k ) inf I(u). By approximations, we may assume u k C (). The question now is do the u k s converge? To do this we have to go through a few things: Claim. I claim that {u k } is a bounded sequence, i.e. C > that is independent of k such that u k C, Du C. Proof of Claim. The fact that u k is a minimizing sequence of I(u) implies that k, C I(u k ) 4 Du ɛ f. We can bring the f term to the other side and get C Du. Poincaré s inequality then implies that C 3 u. 5

8 Ok now we have that {u k } is a bounded sequence. Recall that in a finite dimensional vector space, boundedness implies pre-compactness. However, our functional space W, () is infinite dimensional, so we need to find a weaker substitute called weak compactness. Definition.8. Let B be a Banach space and B be its dual space (space of bounded linear functionals), i.e., l B is linear and l, u C u for any u B. Let {u k } B. Then we say that u k u weakly if l B, l, u k l, u. It is easy to see that if u k u in the usual sense, then u k u weakly. Let l B so we have l, u k u C u k u. The converse is not true. For an easy example, let u k be an orthonormal basis in an infinite dimensional Hilbert Space. Then u k u l =, so we obviously do not have convergence. On the other hand, from Parseval s formula, we have u k l, u k = l. Then this implies that l, u k l = u k but u k. Now we go to a result from analysis: Theorem.9 (Bamach-Alaoglu). Let B be a reflexive separable Banach space. sequence u k B, there exists a subsequence u kl B such that u kl u weakly. Then for any bounded We will apply this to our problem. We have that I(u k ) inf I(u) and u k + Du C. By passing through our subsequence, we have a u W, () such that u k u and Du k Du. Now the question that we have to answer is if u is the minimum that we seek after. However, we can t say that u k u, Du k Du implies I(u k ) I(u ). The problem with this is that I(u) is not continuous with respect to weak convergence. However, it is lower semi-continuous! i.e., u k u weakly in L implies that u lim inf k u k. Here is the proof of this: Proof. u = uū = lim k uū k lim inf k u u k. In particular, I(u ) = Du + f, u lim inf k = lim inf I(u k). k However, since the following inequality is automatic, and we have that I(u ) = inf I(u). ( Du k + f, u k inf I(u) I(u ) lim inf I(u k), k We used the Banach-Alaoglu Theorem: if B is a reflective and separable Banach space and {u j } is a bounded sequence, then there exists a weakly precompact sequence. The missing steps we have are to show the proof of this, and completely show that u C Du for u C (). The proof of Theorem.9 will be left as a black box. We move onto the proof of Lemma.7. ) 6

9 Proof of Lemma.7. Our main tool will be Hölder s inequality. Choose p and p such that /p + /p =. Then we have u(x) K(x, y) p K(x, y) p v(y) dy [ ([ u(x) p dx ] [ ] p K(x, y) dy K(x, y) v(y) p p dy ] p [ ] ) p K(x, y) dy K(x, y) v(y) p dy dx [ sup x ] p ( ) p K(x, y) dy K(x, y) v(y) p dy dx. Since v(y) p dy is a constant in terms of x, we can take it out of the integral of the RHS. Then switching the order of integration and applying the same bounding trick we have [ u(x) p dx sup x u p p A p p + v p p u p A v p. ] p [ p K(x, y) dy sup y ] K(x, y) dx v(y) p dy Notice that this lemma would be pointless if A = because then we learn nothing know with u p. We are in luck because we can actually deduce that A is finite in our case! The reason for this is because x y n yields a singularity of dimension strictly less than n when taking the supremum over the y s, and so it integrable. Finally since x is over a set of compact support, it is bounded by a constant. Hence for u C () = u p C Du p. Now we are finally done with our result from earlier. Now we propose a question: how do we sharpen our estimates? A better way to visualize this question is by noticing that the Kernel is integral for any power less than n. We actually used the worst power in our previous proof. The answer to our question comes from the following inequalities: Theorem. (Sobolev Inequality). Let R n and u C (). Then for any p < n we have u np n p C n,p Du p. Theorem. (Trudinger Ineqality). For p = n, there exists constants K, C > such that ( ) n u(x) n exp C. K Du n Theorem. (Morrey s Inequality). For any p > n, we have where α = n/p and u Cα () C Du p u(x) u(y) u C α () = sup u + sup x y x y α. Lets first think of why the first inequality is a better estimate than the Poincaré inequality that we have. It is better because np/(n p) > p and we know that the L p norms grow bigger as p increases. This means that we have essentially closed the gap between u and Du that we got from Lemma.7. Ok now lets try to 7

10 prove this. The proof is going to follow the same theme from Lemma.7 s proof. From Hölder s inequality we have u(x) K(x, y) α ( K(x, y) α v(y) β) v(y) β dy ( ) /a [ K(x, y) αa dy ] /c ( K(x, y) ( α)c v(y) ( β)c dy ) /b v(y) βb dy where /a + /b + /c =. We need to choose our parameters wisely so that we have our desired estimates. One obvious constraint to put is βb = p and ( β)c = p because we want v(y) on the RHS to have powers of p. Raising everything to the power q and integrating gives us (( u(x) q dx (( = ) q/a [ K(x, y) αa dy ) q/a [ K(x, y) αa dy ] q/c ( K(x, y) ( α)c v(y) p dy ] q/c K(x, y) ( α)c v(y) p dy v βq p ) ) βq v(y) p dy dx In order to make our calculations a little bit easier take q = c. Then we can do the following: (( ) q/a [ ] ) u(x) q dx K(x, y) αa dy K(x, y) ( α)c v(y) p dy v βq p dx ( sup x ( sup x ) q/a [ K(x, y) αa dy q/a [ K(x, y) dy) αa sup y ) dx. ] K(x, y) ( α)c v(y) p dy v βq p dx ] K(x, y) ( α)c dx v p+βq p Recall that so far < α < is arbitrary. Choose it so that αa = ( α)c. Now lets play around with these parameters. Recall that we chose q = c and ( β)c = p = β = p/q. We can then plug this into βb = p = b = p/β to get b = p β = p p = pq q q p. Now that we have parameters b and c, we can plug this into /a + /b + /c = to get the parameter a. After skipping some steps we see that /a = /p. Finally recall that we have αa = ( α)c. plugging in our value for a and c and solving for α gives us α = q(p ) p + pq q, αa = pq p + pq q. When applied to our gradient estimates, this means we need the integral K(x, y) αa dy to be finite. Following our explanation after our proof of Lemma.7 tells us that we need the integral to be finite. Comparing the powers would require ( ) pq p+pq q dy x y n pq (n ) p + pq q < n. 8

11 After playing around with this inequality we get q < pn n p. What suffices to show the full proof of Theorem. is that our coefficient in front of Du p must depend only on n and p. Additionally, setting K(x, y) = x y n+ in our generalization of Lemma.7 would mean that A would only be finite when ( sup x x y n ) pq pq+p q dx <. One can see that our kernel will be integrable (n ) pq+p q < n p q < n. We now prove the following general lemma where u no longer has compact support: Lemma.3. Assume p q < n. If u satisfies u(x) u(y) dy then (diam )n n ( + q u u q c ) + q p p n n + q p Proof. It suffices to show, after dropping some constants, { sup x ( pq Du(y) dy x y n (diam ) n Du n + p. p ) pq } + q pq+p q ( p + q dx x y n ) + q p p n + q n p + q. () p The way to do this is from this simple fact: µ <, dy x y µn c n µ µ. The way to prove this fact is to show that it is true for balls and then for general through rearrangement inequalities. Then () easily follows. We have finally finished the proof of the Sobolev inequality. Observe now that p q < np n q < n p. We can ask the question if this holds for q np n p? We d think that the answer is no! Before we do that we give an alternative proof for the Sobolev Inequality: Theorem.4. Let u C () and p < n. Then u np n p C s Du p. Proof. Consider first p = and u. Notice that we can write u(x) in the following way: This makes sense because u(x) = χ {u>t} = χ {u>t} dt. { u > t u t and so the RHS becomes u dt. Since p = our goal is to show that u n C Du n. Then we see u n n χ {u>t} ( ) n n dt. 9

12 But the inside of this inequality is χ {u>t} ( ) n n = ( ( ) n ) n n n χ {u>t} ( ) n dx = n ( dx {u>t} ) n n = (Vol{u > t}) n n Notice that this has dimension of surface area because we can interpret Vol as having n dimensions, then we take the n th root of it leaving one spacial dimension, and then we raise it to the n + th dimension again. Then we can use the following isoperimetric inequality: (Vol{u > t}) n n C S Area( {u > t}). This gives us now the inequality u n n C s Area( {u > t})dt. Now we will apply iterated integrals using the coarea formula: let u be a real valued function that isn t constant. Then dx = Du dσ tdt () is the coarea formula. What we are doing is integrating the {u = t} level set with the dσ t measure and then integrating with respect to dt. Applying this gives us u n n C s Area( {u > t})dt ( ) = C s dσ t dt u=t = C s Du dx. We are done with the proof because for p we notice that u L p u p L. With this being said we turn back to the Trudinger inequality: for p = n, there exists constants K n, C n > such that ( exp K n (diam ) n u(x) u Du n ) n n dx Cn. Notice that this is stronger than what the Sobolev inequality says for p < n because exponentials are always greater than polynomials, and so this is stronger than u u p <. However it is weaker than u u C because this deals with the supremum, which is the best bound we can have. Proof of the Trudinger Inequality. We will recall the integral inequality ( + q u u q c ) + q p p n n + q Du p (3) p and drop terms involving because in the end they re just constants. We expand the exponential as a power series to get ( ) n n u(x) u exp dx = K n Du n k! k= = k= ( u(x) u K n Du n ( u u kn n k! K n Du n ) kn n dx ) kn n (4)

13 Now we go back to (3) and plug in p = n to get the following inequality: u u q Du n c n ( ( q + q )) + q n n c n q n. This was verified on paper by playing little tricks. Anyway, we plug this into (4) with q = kn n : ( ) n n ( u(x) u exp dx K n Du n k! k= = k! = k= k= Then something called Stirling s formula makes this converge. k k k! K n ( K n kn n ( kn c n n ( ) k kn c n n ) k n. K n n n (n ) ) ) kn n n Recall that we have shown this basic inequality before: Consider p, q and assume p q and p q < n. Then ( + q u u q c ) + q p p (diam ) n n n + q Du p n + p p for convex and u satisfying one of our integral inequalities. Then certainly we have u u c n ( + q p n + q p ) + q p (diam ) n Du n + p (5) p We have used this to show the Sobolev Inequality and Trudinger inequality. Now we prove Morrey s inequality. Theorem.5 (Morrey s Inequality). Let p > n, α = n p, = Sn. Then Proof. Recall that u C α u C α C Du p. u(x) u(y) = u + sup x y x y α where < α <. It the suffices to show that each term is bounded by Du p i.e. u C Du p and [u] C α C Du p. Clearly the first part follows from (5) with q =. Now we prove the second part. Fix x, y, x y and let δ = x y. Now define := B δ (x) B δ (y). Clearly this is convex. Then we have the following inequalities u(x) u(y) u(x) u + u u(y) n ( (diam ) Du(z) dz + x z n B δ(x) ) Du(z) dz y z n Now notice that diam( ) δ and B δ (x). Finally since all the terms inside the integral are positive then we can write ( ) Du(z) Du(z) u(x) u(y) C dz + dz. x z n B δ (y) y z n

14 Now at this point we don t have to reinvent the wheel so we see now that our function satisfies the correct requirements of (5) so we can write And we are done with the proof. u(x) u(y) C Du diam(b δ(x)) n B δ (x) n + p C Du p (δ) n (ω n δ n ) n p C Du p δ n p n δ n C Du p δ α Theorem.6 (Sobolev Embedding Theorem). Let u W k,p (). Then u L () u W k,p () for /p < k/n. Proof. Lets recall Morrey s inequality. This says that if /p > /n, then u(x) u(y) u + sup x y x y α = u C α C Du p for u C and α = n/p. Let k =. Then the theorem follows from Morrey s inequality because /p < /n and u is obviously bounded by u C α. Now assume the theorem holds for k and B, consider D β u where β k. Set v = D β u. Then by definition we have Dv p u W k+,p. Since by we are assuming p < k+ n pick some ɛ > such that p = k+ n ɛ. Now recall the Sobolev inequality: v q Dv p for p q < n p n q. We will choose a q satisfying (for some ɛ < ɛ) q = p n + ɛ = k + n n + ɛ ɛ = k n + ɛ ɛ < k n. Now we have v q u W k+,p for q < k n and by definition u W u k,q W k+,p for q < k n. And so our induction hypothesis tells us u u W k,q u W k+,p. As a consequence we have then that if u W k,p () then u C () if /p < k/n. The way to see this is the following: Let {u j } C () and u j u with respect to W (). This of course exists because k,p W k,p () = {C () W k,p () < }. Then we apply the Sobolev embedding theorem to u j u m. Then we have that u j u m C C u j u m W k,p (). This implies that u j converges uniformly and so lim u j is continuous. In fact, u j u uniformly in the usual sense because uniform convergence implies that u j u in L p. However, since W k,p () convergence is stronger, we have u j u in the usual sense. Lets summarize what we ve done. Let R n, f L (). Define ( ) I(u) = Du + fu dx for u W, (). Then we showed that u W, () with I(u ) = inf I(u). We will observe that the minimum of a functional I(u) is going to be a generalized solution of the Euler-Lagrange equation for I(u). The basis on which we will set our ground on is that if x = min x f(x) = f (x ) =.

15 Let ϕ C () and consider for t << the function R t A(t) = I(u + tϕ). Then we see that t = is going to be a minimum for A(t) = = da dt = d ( ) t= dt t= D(u + tϕ) + f (u + tϕ) dx = d ( ) dt t= ( Du + tdu Dϕ + t Dϕ ) + f (u + tϕ) dx = (Du D ϕ + fϕ) dx n = u ϕ + fϕ dx. x j x j j= Now if we assume temporarily that u C () then we are allowed to integrate by parts and get ( = ) u x + f ϕdx. j Since this is true for any ϕ then we have u + f =, which is the Laplace equation that we wanted to solve from the first page.. A non-linear PDE and Method of Continuity.. Non-linear variations on Manifolds Before we can extend our current groundwork to general manifolds, we need to expand our theory to more general boundary values. Let R n+ + = {x R n+ x n+ } and let R n+ +. Assume u C (). Then we will attempt to show u(, ) Lp (R n ) C n+ u Lp (R n+ + ). (6) Pick d > diam. Since u has compact support we have d u(x, ) = u(x, ) u(x, d) = u (x, x n+ )dx n+ x n+ ( d ) p /p ( ) /q u d (x, x n+ ) x n+ dx n+ q dx n+ d p u(x, ) p C u (x, x n+ ) x n+ dx n+ d p u(x, ) p dx C u (x, x n+ ) x n+ dx n+ dx. Taking the pth root gives us the desired claim. We will now attempt to show that this inequality will allow us to define u for an arbitrary u W,p (). Take u j C () where u j u in W,p. We know this sequence converges because W,p is the completion of C with respect to the p norm. Then by our inequality (6), we can see (u j u k ) p C u j u k W,p by definition and so u j is cauchy. But since L p is complete we can then formally define u := lim u j 3

16 where the limit is taken over the L p norm. Recall that we still had being some some simple semi-circle in R n+ +. Lets extend the notion of boundary values in more general with C. Let be a small subset of such that. Let v C ( ). We will define (using norms) v. Let y. Since the boundary is smooth we can map into an upper half sphere as before with y x. Of course, we can go backwards. So then we can say v(y) = v(y(x)) =: u(x) and note that via our definitions, v = u(x, ) and from our previous observations we have u(, ) L p (R n ) C u x n+ L p (R n+ ) = C v(y(x)) x n+ n+ C v y l l= Lp (R n+ ) C v W,p (R n+ ). L p (R n+ ) Noticed that (7) we did something very fishy that I will now justify. We did the change of variables from integrating with respect to the x coordinates to integrating with respect to the y coordinates. The problem with this is that the integrals might not be bounded in the correct way. Recall dy = det y l x j dx and c det yl x C. And so we have for a general function f j f( ) p L = f(y) p dy p y = (f y)(x) P det y l x j dx (7) and so we have and (7) is valid. c f y p L p x f p L p y C f y p L p x Claim. The following inequality holds for any v (not just those supported in a boundary neighborhood): v L p ( ) C v W,p (). Proof. Note that cdx dσ Cdx and we are able to apply the same argument as above (i.e. norms are equivalent under change of variable). So now the problem is to deal with the full. Since is compact we cover it = N α= α and pick a partition of unity χ α C ( α ) and χ =. Then v Lp N χ ( ) α v LP ( ) α= N χ α v W,p () α= N ( χα v L p () + D(χ α v) L ()) p α= C v W,p () Where we have bounded χ α v p v and expanded the second term using the Leibniz rule. 4

17 With this done we can now extend our previous work to more general boundary conditions. Say you wanted to solve u = f with u = g. Then if g L p ( ) then choose a G W,p () whose restriction is equal to g and then consider v = u G and the problem v = u G and v. We will begin by analyzing a Non-Linear PDE. Let M be a compact Riemannian manifold of dimension. Let ds = g ij dx i dx j i,j= be the Riemannian metric. Let v C (M) and let R be a given negative constant. We want to solve u + λe u+v R =. (8) Notice that the exponential of u makes this a very non-linear equation. Let g := det g ij. Then we will define the laplacian as follows u = g = = j,k= j,k= j,k= k ( gg ij j u ) g ( gg jk k j u ) + first order terms g jk k j u + first order terms. Example.7. In Euclidean space, we have ds = (dx i ) = g ij = g ij = δ ij and so u = u x j + first order terms. In order to solve this, we would have to consider the following functional: I(u) = Du gdx + R u gdx subject to the constraint V M M M e u+v gdx =, V = M gdx and attempt to mimic our work for u = f (show that the function is bounded from below and find a minimum). Note that Du = j u and Du = j u k u then we need the metric to contract the indices so we introduce g jk. Note that we can assume v gdx = (9) V because we can shift everything about constants. We will prove boundedness from below for our functional, but first lets compare with R n and I(u) = Du dx + fudx for any u W, () and f L (). The way we proved this was by showing that u dx C Du dx, 5

18 but can we do this on a general compact manifold? No! In general we have the following Poincaré inequality u ū L (M) C Du L (M), ū = u gdx. V Now we can write our functional in the following way I(u) = Du gdx + R (u ū) gdx } M {{ M } bounded as before In order to deal with this difficulty we exploit our constraint = e u+v gdx V M ( = log e u+v ) gdx V M M + Rū gdx. } M {{ } can blow up and note that V gdx has total measure. Now recall consider Jensen s Inequality: If dµ = then ( ) log(f)dµ log fdµ. We will also use the fact that in the following way ( ) a + b log log a + log b ( = log e u+v ) gdx log(e u+v ) gdx V M V M = (u + v) gdx = u gdx V M V M where we have used (9) in the last step. So now we have that our constraint implies ū and so we do indeed have our bound I(u) = Du gdx + R u gdx. Now as before lets pick a minimizing sequence u j C (M) such that I(u j ) min I(u) and each u j satisfies the constraint. Clearly Du L (M) C and so u j u j ū j + ū j C Du j + R I(u j) C. This means that our minimizing sequence is bounded and so we can apply the Banach-Alaoglu Theorem to show that u W, (M) such that Du j Du weakly and u j u weakly. Now the question is if u is the minimum or not? We do have I(u ) lim inf Du j gdx + Ru j gdx = lim inf I(u j ) = lim j I(u j). The real question is if u satisfies the constraint i.e. we wish to show e u +v gdx = () V M The bottom-line question here is if we can pass through the limit. We need to strengthen our converging hypothesis as much as possible (i.e. from weak convergence to point-wise convergence). In fact, we claim that for the minimizing sequence we have, Du j Du weakly but u j u in L. For this we will need 6

19 Lemma.8 (Rellich s Lemma). Let M be a compact manifold and {u j } W,p (M) with u j W p (M) C with C independent of j, then there exists u W,p (M) and a subsequent {u jk } such that u jk u in L p (M). Note. Note that this also holds for W,p (R n ) if supp u j k R n j. Now recall a theorem from measure theory: Theorem.9. If u j u in L p for p <, then there exists u jk u point wise a.e. Note that all together, we can take a subsequence of our subsequence to find a sequence Du j C, u j u almost everywhere. We will see that in two dimensions these properties imply that e uj+v e u +v in L. Since these values are always positive, the L norm is just convergence of the integral, which is what we needed. To see this, Taking the L norm will give us e uj+v e u +v gdx M e uj+v e u +v = [ [ M M = = u u j L d [ e tu +( t)uj+v] dt dt (u u j )e tu +( t)uj+v dt. u u j e tu +( t)uj+v ] gdx dt u u j ] / [ gdx [ M M e (tu +( t)uj+v) gdx] / dt e (tu +( t)uj+v) gdx] / dt and now note that we need the second integral to be uniformly bounded for the RHS of the inequality to go to zero. We claim that e wj gdx C (independent of j) if w j L C and Du j L C. In this case we have w j = (tu +( t)u j +v) satisfying the condition because Du j L C and Du L lim inf Du j L C and similarly the L norm of u j and u is bounded. Now lets see why this claim is true. Recall the Trudinger inequality: u C (B) implies that ( ) n u(x) n exp C K Du L n and so for n = we have Now note that we can write u(x) = ( ) u(x) exp C. K Du L u(x) K Du L ( ) u(x) + K Du L K Du L (K Du L ). Since the exponential function is increasing we can write [ ( ) ] u(x) exp u(x) exp + K Du L (K Du L ) exp u(x) e (K Du L ) exp ( ) u(x) K Du L e (K Du L ) C 7

20 And this concludes the proof. A great exercise is as follows. Let M be a compact Riemannian n-manifold and show that e w gdx C exp (C Du n L + n w n L n). As a summary, we have shown that for a sequence satisfying I(u j ) min I(u) and V eu j+v gdx = then u j u (as explained above) and V eu +v gdx = so we have inf I(u) I(u ) inf I(u). Now we claim that u satisfies our partial differential equation in the Euler-Langrange sense. Fix ϕ C (M) and consider u + tϕ + c t. We add the constant c t so that this function still satisfies the constraint. To see what c t has to be note = e u +tϕ+ct+v gdx V M ( e ct = e u +tϕ+v ) gdx V M ( c t = log e u +tϕ+v ) gdx V and thus I(u + tϕ + c t ) I(u ) for any t and so we leave it as an exercise to show in detail d dt I(u + tϕ + c t ) =. t= Lets recap our problem. Our functional was I(u) = Du gdx + R M M M u gdx subject to the constraint e u+v gdx = () V M where v is some smooth function and V is the volume over this compact manifold. We obtained a u subject to the same constraint that satisfies u + λe u +v R =. Recall that this general laplacian has the form ( u = g i g j u ). Where does this come from? If we assume u t is smooth then we can check d Du t gdx = d ( ) g ij i u t j u t gdx dt M dt = d ( ) ( u t i g ij ) j u t gdx dt = d ( ) u t u t gdx dt So we see that this laplacian is the perfect analogue of the one taken on R n. We want to derive the generalized Euler-Lagrange equation for our PDE, so let ϕ C (M) and u t = u + tϕ + c t, where c t is picked so that u t also satisfies the constraint (). Lets figure out what λ has to be in order to for our equation to be solved correctly. Integrate on both sides of our PDE to get = u g + λ e u +v g R gdx M M M ( = i gg ij ) j u + λ e u +v g R gdx g M = λv RV M M 8

21 so we see that λ has to equal R. The reason the first term vanishes is because we are integrating an exact form over a compact manifold. So then we have that our PDE is u + Re u+v R = () and u satisfies () in the general sense if ϕ c we have = ( u + Re u+v R ) ϕ g and (3) is the generalized Euler-Lagrange equation... Regularity Theory = ( g ij i j ϕ + Re u+v Rϕ ) gdx (3) Lets finally start some regularity theory. Let g ij be a smooth Riemannian metric and assume Λ g ij λ >. If u W, loc satisfies u = f in the weak sense then. If f W k,p and < p < the u W k+,p and for any ( u W k+,p () C, f W k,p ( ) + u W k,p ( )). If f C k,α and < p < the u C k+,α and for any ( u C k+,α () C, f C k,p ( ) + u W k,p ( )) Claim. It follows easily that if u satisfies (3) in the generalized sense, then u C and actually satisfies () in the standard sense. Proof. Indeed we can let Re u+v + R = f and f L = W, by the Trudinger s inequality (it tells us that this exponential is bounded by the L norm of the weak derivative) then by regularity we will get u W,. Now recall that Morrey s inequality tells us that if p < k n then W k,p is embedded in a Hölder space, but since p = n = we have u C α. But then since u = f this implies f C α and we can apply regularity to get u C,α. We can iterate this and find of course that u C. We will prove these regularity statements by the method of continuity and a priori estimates. Consider u + Re u+v R = on (M, g ij (x)) where v is a smooth function. Note that u + Re u R = admits a solution u so we see that the difficulty comes from the v term. Let t [, ] and introduce the family of equations u + Re u+tv R = (4) And consider the set I = {t [, ] (4) admits a solution u t }. Note that I because I is a solution. Then we obviously want to show I = [, ], so we need to show that I is open and closed. Lets discuss this very briefly (to be made concise later). Say we want to show I is open. We will do this by an analogue of the implicit function theorem. Recall that it says given f(x, y ) = and f y (x, y ) then ɛ > such that for x x < ɛ, then!y such that f(x, y) =. Now let f(t, u) = u + Re u+tv R. Then we want to solve f(x, u) = where we know f(t, u ) =. So our goal is going to be an implicit function theorem for Banach spaces and we need to check that f u (t, u ) in a way we will define later. Now lets briefly discuss how we will show that I is closed. Take t j I such that (4) will admit a solution called u j, and assume t j T. Then closeness of I is equivalent to showing that (4) will admit a solution for T. It will suffice to have a subsequence converge in C. 9

22 Proof. Lets start by showing that I is closed. Let t j T with t j I and let u j be the corresponding solution of (5). Observe that if suffices to show that C independent of j so that u j C 3 C where in general u C3 () = α u C (). α 3 The reason why this will help is that this would imply that β, then β u j is an equicontinuous family. Then the Arzela-Ascolati Theorem tells us that if we have an equicontinuous family, then by going through a subsequence β u j we have uniform convergence to D α u T where u T C. However we don t necessarily have that T I because we need u T to be smooth and so far it is only C. This is fixed however by our regularity observations. Before we can apply our regularity conditions we must show that our equation is uniformly elliptic. Recall that a second order PDE is said to be uniformly elliptic if the leading coefficient satisfies λ ξ a α ξ α Λ ξ. α = This we need to show that our Laplacian is elliptic. u = g i ( g j u) = g gg ij i j u + first order terms. So the symbol (the middle term of the ellipticity inequality requirement) of our Laplacian is going to be σ (x, ξ) = g ij ξ i ξ j and since g is positive definite we definitely have the ellipticity requirement. So then we can apply our regularity theorems by viewing u = f C C,α. So now we have to prove the a priori estimate u j C 3 C. We will use the maximum principle. Since there are so many different formulations of maximum principles, it is a good idea to simply examine what happens near a maximum. Let u C satisfy u + R tw+u R = (by denoting u t = u) and then I claim that u C C where C is independent of t. Let x such that u(x ) is a maximum. Then u(x ) and since R < we have u = Re tw+u R R Re tw+u R R e tw+u e tw+u tw + u u tw w C. But since x is a maximum we have that x, u(x) u(x ) w C and applying the same argument for the minimum we have u C w C. In order to get higher derivatives we write u = f C and so for f L p p < we have u W,p C α for some α when n < kp. Now lets show that I is open. Let t I i.e. there exists u which is a smooth solution of (4). Let (u, t) F (u, t) = u + Re tw+u R. Then we want to show that δ > so that t t < δ implies u t C F (u t, t) =. The main tool will be the Implicit Function Theorem for Banach Spaces, which goes as follows.

23 Let B and B be Banach spaces. Let F C and consider B R (u, t) F (u, t) B. Assume that F (u, t ) =. Let F u (u, t ) be the derivative of F at (t, u ) viewed as a linear operator B B. Note that if h B C F u (u, t )h h B (5) B then F u (u, t ) is injective and surjective. So assume that (5) holds. Then the Implicit Function Theorem for Banach Spaces says that δ >, V that is a neighborhood of u such that!u t V with F (u t, t) =. Before we can even apply this theorem, we need to mae sense of derivatives in terms of Banach spaces. Let B u F (u) B. Then F is differentiable at u if L : B B that is a bounded linear operator satisfying F (t, u + h) = F (t, u) + Lh + E(t, u, h) with E(u, h) B lim =. h h B In order to apply our theorem we need so specify our Banach spaces. We will want to have (t, u) R C,α F (u, t) C,α. What we will then need to check are the assumptions of the implicit function theorem. Let t, u R C,α satisfy the conditions of the IFT. Lets determine L. Consider the expression F (t, u + h): F (t, u + h) = (u + h) + Re tw+u+h R = F (t, u ) + h + Re tw+u+h Re tw+u = F (t, u ) + h + Re tw+u ( e h ) = F (t, u ) + ( + Re t+u) h + Re tw+u (e h h). So we have figured out our L. Now we leave it as an exercise to show that Re tw+u (e h h) C,a C h C,α. The main tool for this is the integral form of Taylor s Remainder Theorem, which starts as follows h ( t) d dt [f (u + th)] dt = f (u + th)dt + h( t)f (u + th) However we can write the LHS as h and so we have the final part as = ( t) d dt [f (u + th)] dt = d dt f(u + th)dt hf (u) = f(u + h) f(u) hf (u). ( t)f (u + th)dt t f(u + h) = f(u) + f (u)h + h ( t)f (u + th)dt. Thus we have that L = F u (t, u ) : C,α h h+re tw+u h C,α, which is clearly a bounded operator. It is trickier to show that it is injective and bijective. Assume = Lh = h + Re tw+u h. Let x me a maximum (again this means h ). Then we have Re tw+u h = h(x )

24 and since the coefficients are both strictly positive, it means that h(x ). Since this is a maximum it means that x, h(x) h(x ). Applying the same process to the minimum we get that h(x) and so h. This implies that the kernel of L is zero, and so it is injective. In order to finally complete the proof we will show that L is onto i.e. f C,α, we want to show h C,α so that Lh = f. We will show this by variational methods again. Set I(h) = [ gg ij i h j h Re tw+u h + fh ] gdx for some h W, (M). We leave it as an exercise to show that I(h) attains its minimum for some h. One way to do this is by showing that I(h) Dh + ɛ h ɛ f and using our tricks. Assuming the exercise, we invoke the black box to make h smooth. Recall that it says if f C k,α then h C k+,α C ( Lh C k,α + h C k,α). We now improve the black box by saying that if ker L =, then h C k+,α C Lh C k,α. (6) Now we will prove this little lemma. We will use weak compactness. If {u l } C k,α with u l C k,α C with C independent of l, then either k < k, β or k = k with < β < α then there exists a convergent subsequence in C k,β by the routine application of the Arzela-Ascolti Theorem. Now assume (7) does not hold. Then for any N, h N C k+,α such that h N C k+,α > N Lh n C k,α. Set h N h N = h N C k+,α and notice that h N C k+,α = and so this implies that L h C k,α < N. By the weak compactness, going through a subsequence, we can assume that h N h in C k,α. Applying the black box gives h N h ( M C k+,α C L h N h M C k,α + h N h ) M C k,α which implies that h N h in C k+,α. Thus L h N Lh in C k,α and so h ker L. Since this contradicts the fact that ker L =. h C k+,α = lim h N C k+,α =

25 3 Harnack Inequality for Divergence Equations 3. Regularity Estimates Suppose that u W, () solves i (a ij j u) + cu = f in the generalized sense, where a ij is uniformly elliptic i.e. < λ a ij Λ. The question we want to answer is: when is u regular i.e. u C α, W k,, C, etc? Lets look at the simplest case where a ij is constant and c = f =. In this case u solving a ij i j u = in the generalized sense. This means that v W, (), a ij i u j v =. (7) Then we will show that u C () and for α = k, < r < R, we have D α u C λ,λ,k (R r) k u. (8) B R (x ) Note that this inequality is very powerful because we have that the derivative is being bounded by the function, where w usually have it the other way around. Proof. We apply (7) with v = χ u with χ, χ on B r (x ) and χ C (B R (x )). Additionally assume Dχ(x ) R r. Applying (7) gives us χ a ij i u j u = a ij (χ j χ)u i u and note that ellipticity gives us a ij u i v j (a ij u i u j ) (a ij v i v j ). Putting absolute values gives us ( ) ( ) χ a ij i u j u a ij i u j uχ a ij i χ j χ u a ij i u j uχ + 4 a ij i χ j χ u a ij i u j uχ 4 a ij i χ j χ u where we have used the standard fact that ab ɛ a + ɛ b where we chose ɛ to give us the right coefficients. Then from the ellipticity condition we have λ Du a ij i u j uχ 4 a ij i χ j χ u 6 Λ Dχ u 8Λ (R r) u B R (x ) which proves the inequality for k =. To prove α = k Z, we proceed by induction. Assume u C. Then D α u C and satisfies the same equation a ij i j (D α ) =. Applying the previous case with k = gives (by induction) D(D α u) C ( B r+r r(x ) r ) D α u B R r (x ) C ( R+r r ) ( ) R R+r k u B R (x ) C = u. (R r) (k+) B R (x ) 3

26 As one can expect, for the non-smooth case, we use mollifiers. Take η C ( x < ) with R n η = and define η ɛ = ɛ η( x n ɛ ). Then define u ɛ (x) = u(x y)η ɛ (y)dy = u(y)η ɛ (x y)dy which is well defined for dist(x, ) > ɛ. By an exercise we leave to the reader, note that if u L p () for < p <, then for any K and ɛ < ɛ K, then u ɛ u in L p (K). Using this, and by a single exercise that shows a ij i u ɛ j v = a ij i u j v ɛ we can conclude that if u satisfies a ij i j u = in the generalized sense then so does u ɛ. Thus we have that u ɛ C and a ij i j u ɛ = and thus we can apply our estimate (8) to find D α u ɛ (R r) k u ɛ. B R (x ) Lets try to generalize this. We will do this with the following theorem. Theorem 3.. Let u W, () be a weak solution to in R n. Assume that i) < λ a ij Λ i (a ij j u) + c(x)u = f (9) ii) a ij are continuous with modulus of continuity τ i.e. a ij (x) a ij (y) τ( x y ) iii) c L n, f L q where n < q < n. Then for any B R (x), u C α (B R (x)) with α = n q and < α < with the estimate u Cα (B) C n,λ,λ,τ, c p ( f Lq () + u W, ()). In order to prove this we will need a few lemmas. They are as follows. Lemma 3.. Assume u W, (B) and Then u C α and u C α (B) C(M + u L ). u ū x dx M r n+α. Proof. First note that what we are doing makes some sort of sense. In the assumption, we have the difference between u and its average at x being squared, so that explains the α. Then since we are integrating over the ball, we have the factor of r n. Let < r < R and look at the difference between averages ū r (x ) ū R (x ) ū r (x ) u(x) + u(x) ū R ū r (x ) ū R (x ) ( ū r u(x) + u(x) ū R ). 4

27 We now integrate over the ball of radius r and using the fact that the LHS is a constant and the assumption of the lemma we have ( ) r n ū r (x ) ū R (x ) ū r u(x) + u(x) ū R ( ) M r n+α + u(x) ū R B R (x ) CM ( r n+α + R n+α) ( ) n ) R ū r (x ) ū R (x ) CM (r α + R α r ( ) n ) R CM ( + R α r ( ( ) n ) R ū r (x ) ū R (x ) CM + R α. () r Let r = l L, R = l L and plug this into () to see that ( ( l ) ) L n ū r ū R CM + l ( l L) α L Now use () to see that = CM( + n )( l L) α = CM( l L) α. () ū m L ū l L m ū i L ū i i=l CM m ( i L) α i=l CM( l L) α () where we have used the fact that we had a telescoping sequence and that we have a geometric sequence that we have bounded by the highest term. This implies that {ū l L} is a cauchy sequence and so we can define u := lim l ū l L(x ). We will now show that this is independent of L. Take L < L. Then () implies that ( ) L n ) ū l L(x ) ū l L ( CM + ( l L ) α L and taking l shows that u is independent of L. From all of our hard work, we can say that u = u almost everywhere by the Lebesgue Differentiation Theorem which says that if u L, then lim r ū r (x ) = u(x ) for almost every x. Now taking m in () we are able to get ū m L(x ) ū l L(x ) CM( l L) α u (x ) ū l L(x ) CM( l L) α u (x ) ū r (x ) CMr α. (3) 5

28 Letting r = implies that u (x ) L (B ) C(M + u L (B )) Now in order to complete this theorem we need to estimate [u ] C α. Let x, y such that B r (x), B r (y) and B r (x) B r (y). Denote δ = x y and let z be the point midway between x and y. By convexity we can see that B δ (z) B r (x) B r (y). Then we write u (x) u (y) u (x) ū δ (x) + u (y) ū δ (y) + ū δ (x) u(z) + ū δ (y) u(z) u (x) u (y) C ( u (x) ū δ (x) + u (y) ū δ (y) + ū δ (x) u(z) + ū δ (y) u(z) ) CM δ α + C ū δ (x) u(z) + C ū δ (y) u(z) where we have used (3) on the first two terms. Integrating the inequality with respect to z and using the assumptions of the lemma yields the required results. Lemma 3.3. Assume that Then u C α. Du M r n +α. Proof. The details of the proof will be left as an exercise. Here is a sketch of it. Recall the Poincaré inequality: u ū S λ S Du. Then you need to show that S S u ū r (x ) Cr Du where c is independent of r. The way to do this is to apply the Poincaré Inequality with r =. Then consider the rescaling ũ = u(rx). After doing this, apply the assumption of the lemma and you ll find yourself in the position of Lemma 3.. Lets recall our goals. We were proving Schauder estimates to get regularity. Assume u W, is a weak solution of i (a ij j u) = i.e. a ij j u i v = v W,. Then we will show that u C α where < λ a ij Λ and a ij (x) a ij (y) τ( x y ) with τ as R. We will be using the following key estimate for < r < R [( r ) n ] Du C + τ(r) Du. (4) R B R (x ) Proof. We first prove the case where we have a constant coefficient i.e. τ. Then we will show that ( r ) n Du C Du (5) R B R (x ) and use a the Lemma of De Giorgi (seen later). By rescaling v(x) = u(rx) what we need to show becomes ( r ) n R Dv C Dv R R B (x) B r R (x ) B s(x ) Dv Cs n B (x ) Dv. 6

29 This will follow from previous work. Take s small (s < ). Then By the Sobolev Embedding Theorem and by the first theorem proved in this section we are able to say Dv sup Dv s n B s(x ) B s α k Cs n B 34 (x ) B (x ) Dv D α (Dv) s n and we are done showing (5), which corresponds to the case where a ij are constants. Since this case will follow when we prove the Lemma of De Giorgi, we move on to the non-constant case. We will prove the non-constant case as a perturbation of the equation a ij (x ) i j w =, which we have already finished. To carry this out, consider the following Dirichlet problem { aij (x ) i j w = weak sense w u W, (B R (x )). Set v = u w and u = v + w. Then [ ] Du Dv + Dw [ ( r ) n ] Dv + c Dw R B R (x ) [ ( r ) n ] Dv + c Du + Dv R B R (x ) [ ( r ) n ] C Du + Dv R B R (x ) where we have used the fact that w solves our PDE with constant coefficients, saw that the integral over r is than the integral over R, and saw max{, rn R } =. Now we need to control the integral of Dv, so we n use the fact that v = u w, where u solves i (a ij j u) = and w solves a ij (x ) i j w = in the weak sense. Now since v W,, we are able to use it as a test function in the definition of weak derivatives to get a ij (x )D j vd i v = a ij (x )(D j u + D j w)d i v = a ij (x )D j ud i v = (a ij (x ) a ij (x))d j ud i v + a ij (x)d j ud i v = (a ij (x ) a ij (x))d j ud i v. We now use the fact that a ij is elliptic and use its modulus of continuity to see λ Dv τ(r) Du Dv B R (x ) B R (x ) τ(r) Dv Cτ(R) B R (x ) B R (x ) B R (x ) Du + Dv Du (6) 7

30 which we can plug into (6) to see that we have finally proved (4) for the case of non-constant a ij. We still need to relate everything to get that u C α. Now we finally state the Lemma of De Girogi: Lemma 3.4 (Lemma of De Giorgi). Let ϕ(r) with ϕ(r) as r, and A, B. Assume that < r < R and α > β > such that [( r ) α ] ϕ(r) A + ɛ ϕ(r) + BR β. (7) R Then < β < γ < α, ɛ such that ɛ < ɛ implies [( r ) ] γ ϕ(r) C ϕ(r) + Br β. (8) R We will apply the Lemma with ϕ(r) = Du because (4) implies that ϕ(r) satisfies (7) with B = and α = n. Then by the Lemma of De Giorgi we have ( r ) γ ϕ(r) C ϕ(r) R and by taking R = we imply that ϕ(r) Cr γ after absorbing ϕ() into the constant. We are now in the case of Lemma 3.3 because we are free to choose any γ < α = n and u ū r (x ) dx Cr γ+. Hence u C α. Recall that we consider weak solutions of i (a ij j u) + c(x)u = f where < λ a ij Λ and a ij (x) a ij (y) τ( x y ) where τ as R, c L n and f L q for n < q < n. We want to show that u Cα for α = n q. Proof. We already got the case for c = f =, where for < r < R the main tool was showing { ( r ) n } Du C Du + τ(r) Dv. (9) R B R (x ) B R (x ) Recall that we defined v = u w where w solves { aij (x ) i j w = on B R (x ) u w W, (B R (x )) in the generalized sense. By the same argument as before, (9) can also be established for non-trivial lower ordered terms. Lets now estimate Dv in (9): a ij (x )D i vd j v = a ij (x )(D i u D i w)d j v = a ij (x )D i ud j v = a ij (x)d i ud j v + (a ij (x ) a ij (x))d i ud j v = c(x)uv + fv + (a ij (x ) a ij (x))d i ud j v. (3) 8

31 These calculations show two new terms that were not there before in the proof of the previous theorem. We handle these terms by bounding their derivatives. Recall the Sobolev inequality: and use the Holder s and the Sobolev inequality B R (x ) v L n n C Dv L ( c(x)uv u L u L ) / cv B R (x ) [ ( )? ( v n n ) ]? / c where we have to figure out which powers to use. We need Lesbegue conjugates p + q p = n n = p = n n, q = n. This implies ( cv ) n ( ) v n n /n n ( c ) n = v c L n n L n = and so we let Putting this together yields c(x)uv u L v L n n c L n u L Dv L c L n ɛ Dv L + ɛ u L c L n and now the first term gets absorbed in (9). The second term in (3) can be bounded as follows fv f B L n v n+ L n n R f L n n+ Dv L ɛ Dv L + ɛ f. L n+ n Once again the Dv term gets absorbed in (9). Putting everything together gives us the following estimate { (( r ) n ) } Du C + τ(r) Du + C u L R (B R ) c L n (B R ) + C f. (3) B R (x ) L n+ n Since we want to show u C α, we look for things of the form Du Mr n+ α. Now recall the Lemma of De Giorgi: ϕ(r) as r, ϕ(r). Assume that < r < R, β < α and (( r ) α ) ϕ(r) A + ɛ ϕ(r) + BR β. R Then < β < γ < α, ɛ > such that ɛ < ɛ implies ( r ) γ ϕ(r) C ϕ(r) + Br β. R 9