ELLIPTIC EQUATIONS

ELLIPTIC EQUATIONS 3 Theorem 1.5 (Riesz). Let Hbe a separable Hilbert space, and let T be an element of its dual H0, i.e., a linear application T: H!R...

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ELLIPTIC EQUATIONS LUIGI ORSINA

Contents 1. Existence with regular data in the linear case 1.1. Minimization in Banach spaces 1.2. Hilbert spaces 1.3. Sobolev spaces 1.4. Weak solutions for elliptic equations 1.5. A nonlinear equation 2. Regularity results 2.1. Examples 2.2. Stampacchia’s theorems 3. Existence via duality for measure data 3.1. Measures 3.2. Duality solutions for L1 data 3.3. Duality solutions for measure data 3.4. Regularity of duality solutions 4. Existence via approximation for measure data 5. Nonuniqueness for distributional solutions References

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1. Existence with regular data in the linear case Before stating and proving the existence theorem for linear elliptic equations, we need some tools. 1.1. Minimization in Banach spaces. Let E be a Banach space, and let J : E → R be a functional. Definition 1.1. A functional J : E → R is said to be weakly lower semicontinuous if un * u



J(u) ≤ lim inf J(un ). n→+∞

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Definition 1.2. A functional J : E → R is said to be coercive if lim

kukE →+∞

J(u) = +∞.

Example 1.3. If E = R, the function J(x) = x2 is an example of a (weakly) lower semicontinuous and coercive functional. Another example is J(u) = kukE . Theorem 1.4. Let E be a reflexive Banach space, and let J : E → R be a coercive and weakly lower semicontinuous functional (not identically equal to +∞). Then J has a minimum on E. Proof. Let m = inf J(v) < +∞, v∈E

and let {vn } in E be a minimizing sequence, i.e., vn is such that lim J(vn ) = m.

n→+∞

We begin by proving that {vn } is bounded. Indeed, if it were not, there would be a subsequence {vnk } such that lim kvnk k = +∞.

k→+∞

Since J is coercive, we will have m = lim J(vn ) = lim J(vnk ) = +∞, n→+∞

k→+∞

which is false. Therefore, {vn } is bounded in E and so, being E reflexive, there exists a subsequence {vnk } and an element v of E such that vnk weakly converges to v as k diverges. Since J is weakly lower semicontinuous, we have m ≤ J(v) ≤ lim inf J(vnk ) = lim J(vn ) = m, n→+∞

k→+∞

so that v is a minimum of J.



1.2. Hilbert spaces. 1.2.1. Linear forms and dual space. We recall that a Hilbert space H is a vector space where a scalar product (·|·) is defined, which is complete with respect to the distance induced by the scalar product by the formula p d(x, y) = (x − y|x − y). Examples of Hilbert spaces are R (with (x|y) = x y), RN (with the “standard” scalar product), `2 , and L2 (Ω) with Z (f |g) = f g. Ω

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Theorem 1.5 (Riesz). Let H be a separable Hilbert space, and let T be an element of its dual H 0 , i.e., a linear application T : H → R such that there exists C ≥ 0 such that |hT, xi| ≤ Ckxk,

(1.1)

∀x ∈ H.

Then there exists a unique y in H such that hT, xi = (y|x),

∀x ∈ H.

Proof. Denote by {eh } a complete orthonormal system in H, i.e. a sequence of vectors of H such that (eh |ek ) = δhk , and such that, for every x in H, one has +∞ X x= (x|eh )eh . h=1

It is then well known that there exists a bijective isometry F from H to `2 , defined by F(x) = {(x|eh )}. We claim that {hT, eh i} belongs to `2 . Indeed, if n X yn = hT, eh ieh , h=1

we have, by linearity and by (1.1),

so that

n X (hT, eh i)2 = hT, yn i ≤ Ckyn k = C

n X

h=1

h=1 n X

! 12 (hT, eh i)2

,

(hT, eh i)2 ≤ C 2 ,

h=1

which yields (letting n tend to infinity) that {hT, eh ii} belongs to `2 . Therefore, one has, again by linearity and by (1.1), +∞ X hT, xi = (x|eh )hT, eh i,

∀x ∈ H.

h=1

Let now y be the vector of H defined by y=

+∞ X

hT, eh ieh .

h=1

Then, since hT, eh i = (y|eh ), one has hT, xi =

+∞ X h=1

(x|eh )(y|eh ),

∀x ∈ H,

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and the right hand side is nothing but the scalar product in `2 of F(x) and F(y). Since F is an isometry, we then have hT, xi = (y|x),

∀x ∈ H,

as desired. Uniqueness follows from the fact that (y|x) = (z|x) for every x in H implies y = z (just take x = y − z).  Corollary 1.6. The map T 7→ y is a bijective linear isometry between H 0 and H. Proof. Since hT + S, xi = hT, xi + hS, xi, and hλT, xi = λhT, xi, it is clear that the map T 7→ y is linear. In order to prove that it is an isometry, we have |hT, xi| = |(y|x)| ≤ kykkxk, which implies kT k ≤ kyk. Furthermore kyk2 = (y|y) = hT, yi ≤ kT kkyk, so that kyk ≤ kT k. The map is clearly injective, and it is surjective since the application x 7→ (y|x) is linear and continuous on H (by Cauchy-Schwartz inequality).  1.2.2. Bilinear forms. An application a : H × H → R such that a(λx + µy, z) = λa(x, z) + µa(y, z), and a(z, λx + µy) = λa(z, x) + µa(z, x), for every x and y in H, and for every λ and µ in R, is called bilinear form. A bilinear form is said to be continuous if there exists β ≥ 0 such that |a(x, y)| ≤ βkxkkyk, ∀x, y ∈ H, and is said to be coercive if there exists α > 0 such that a(x, x) ≥ αkxk2 ,

∀x ∈ H.

An example of bilinear form on H is the scalar product, which is both continuous (with β = 1, thanks to the Cauchy-Schwartz inequality), and coercive (with α = 1, by definition of the norm in H). Theorem 1.7. Let a : H × H → R be a continuous bilinear form. Then there exists a linear and continuous map A : H → H such that a(x, y) = (A(x)|y),

∀x, y ∈ H.

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Proof. Since a is linear in the second argument and continuous, for every fixed x in H the map y 7→ a(x, y) is linear and continuous, so that it belongs to H 0 . By Riesz theorem, there exists a unique vector A(x) in H such that a(x, y) = (A(x)|y),

∀x, y ∈ H.

Since a is linear in the first argument, the map x 7→ A(x) is linear. Furthermore, by the continuity of a, kA(x)k2 = (A(x)|A(x)) = a(x, A(x)) ≤ βkxkkA(x)k, so that kA(x)k ≤ βkxk, and the map is continuous.



1.2.3. Banach-Caccioppoli and Lax-Milgram theorems. Theorem 1.8 (Banach-Caccioppoli). Let (X, d) be a complete metric space, and let S : X → X be a contraction mapping, i.e., a continuous application such that there exists θ in [0, 1) such that d(S(x), S(y)) ≤ θ d(x, y),

∀x, y ∈ X.

Then there exists a unique x in X such that S(x) = x. Proof. Let x0 in X be fixed, and define x1 = S(x0 ), x2 = S(x1 ), and, in general, xn = S(xn−1 ). We then have, since S is a contraction mapping, d(xn+1 , xn ) = d(S(xn ), S(xn−1 )) ≤ θ d(xn , xn−1 ), and iterating we obtain d(xn+1 , xn ) ≤ θn d(x1 , x0 ). Therefore, by the triangular inequality, d(xn , xm ) ≤

n−1 X h=m

d(xh+1 , xh ) ≤

n−1 X

θh d(x1 , x0 ) =

h=m

θm − θn . 1−θ

h

Since {θ } is a Cauchy sequence in R (being convergent to zero), it then follows that {xn } is a Cauchy sequence in (X, d), which is complete. Therefore, there exists x in X such that xn converges to x. Since S is continuous, on one hand S(xn ) converges to S(x), and on the other hand S(xn ) = xn+1 converges to x so that x is a fixed point for S. If there exist x and y such that S(x) = x and S(y) = y, then, since S is a contraction mapping, d(x, y) = d(S(x), S(y)) ≤ θ d(x, y), which implies (since θ < 1) d(x, y) = 0 and so x = y.



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Theorem 1.9 (Lax-Milgram). Let a : H × H → R be a continuous and coercive bilinear form, and let T be an element of H 0 . Then there exists a unique x in H such that (1.2)

a(x, z) = hT, zi,

∀z ∈ H.

Proof. Using the Riesz theorem and Theorem 1.7, solving the equation (1.2) is equivalent to find x such that a(x, z) = (A(x)|z) = (y|z) = hT, zi,

∀z ∈ H,

i.e., to solve the equation A(x) = y. Given λ > 0, this equation is equivalent to x = x − λA(x) + λy, which is a fixed point problem for the function S(x) = x − λA(x) + λy. Since, being A linear, one has S(x1 ) − S(x2 ) = x1 − x2 − λA(x1 ) + λA(x2 ) = x1 − x2 − λA(x1 − x2 ), in order to prove that S is a contraction mapping, it is enough to prove that there exists λ > 0 such that kx − λA(x)k ≤ θkxk, for some θ < 1 and for every x in H. We have kx − λA(x)k2 = kxk2 + λ2 kA(x)k2 − 2λ(A(x)|x). Recalling Theorem 1.7 and the definition of A, we have kA(x)k2 ≤ β 2 kxk2 ,

(A(x)|x) = a(x, x) ≥ αkxk2 ,

so that kx − λA(x)k2 ≤ (1 + λ2 β 2 − 2λα)kxk2 . , we have θ2 = 1 + λ2 β 2 − 2λα < 1, so that S is a If 0 < λ < 2α β2 contraction mapping.  Remark 1.10. Not every function which has a fixed point is a contraction; for example, the identity map (which has infinitely many fixed points), and the function f : [0, 1] → [0, 1] defined by f (x) = x2 (so that f (0) = 0, and f (1) = 1), are not contractions. It is therefore useful to have other fixed point theorems, under different assumptions on the map S. This is the case of Schauder’s theorem. Theorem 1.11 (Schauder). Let K be a convex, closed, bounded subset of a Banach space, and let S : K → K be a continuous function such that S(K) is compact. Then S has at least a fixed point.

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1.3. Sobolev spaces. The Banach spaces where we will look for solutions are space of functions in Lebesgue spaces “with derivatives in Lebesgue spaces” (whatever this means). Warning: This section is a short summary of the results contained in Chapter IX of the book by H. Brezis (see [2]): we refer to it for further results and proofs. 1.3.1. Definition of Sobolev spaces. Let Ω be a bounded, open subset of RN , N ≥ 1, and let u be a function in L1 (Ω). We say that u has a weak (or distributional) derivative in the direction xi if there exists a function v in L1 (Ω) such that Z Z ∂ϕ u v ϕ, ∀ϕ ∈ C01 (Ω). =− ∂x i Ω Ω ∂u as the function v. If u In this case we define the weak derivative ∂x i has weak derivatives in every direction, we define its (weak, or distributional) gradient as the vector   ∂u ∂u ∇u = . ,..., ∂x1 ∂xN

If p ≥ 1, we define the Sobolev space W 1,p (Ω) as  W 1,p (Ω) = u ∈ Lp (Ω) : ∇u ∈ (Lp (Ω))N . The Sobolev space W 1,p (Ω) becomes a Banach space under the norm kukW 1,p (Ω) = kukLp (Ω) + k∇uk(Lp (Ω))N , and W 1,2 (Ω) is a Hilbert space under the scalar product Z Z (u|v)W 1,2 (Ω) = uv + ∇u · ∇v. Ω

Ω 1,2

For historical reasons the space W (Ω) is usually denoted by H 1 (Ω): we will use this notation from now on. Since we will be dealing with elliptic problems with zero boundary conditions, we need to define functions which somehow are “zero” on the boundary of Ω. Since ∂Ω has zero Lebesgue measure, and functions in W 1,p (Ω) are only defined up to almost everywhere equivalence, there is no “direct” way of defining the boundary value a function u in some Sobolev space. We then give the following definition. Definition 1.12. We define W01,p (Ω) as the closure of C01 (Ω) in the norm of W 1,p (Ω). If p = 2, we will denote W01,2 (Ω) by H01 (Ω), which is a Hilbert space. From now on we will mainly deal with W01,p (Ω).

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1.3.2. Properties of Sobolev spaces. Since a function in W01,p (Ω) is “zero at the boundary” it is possible to control the norm of u in Lp (Ω) with the norm of its gradient in the same space. This is known as Poincar´e inequality. Theorem 1.13 (Poincar´e inequality). Let p ≥ 1; then there exists a constant C, only depending on Ω, N and p, such that (1.3)

kukLp (Ω) ≤ C k∇uk(Lp (Ω))N ,

∀u ∈ W01,p (Ω).

Proof. We only give an idea of the proof in dimension 1. Let u belong to C01 ((0, 1)). Then Z x Z x 0 u (t) dt = u0 (t) dt, ∀x ∈ (0, 1). u(x) = u(0) + 0

0

Thus, by H¨older inequality Z x p Z p p 0 0 p |u(x)| = u (t) dt ≤ x 0

x

0

0

p

Z

1

|u (t)| ≤

|u0 (t)|p .

0

C01 ((0, 1))

functions. The Integrating this inequality yields the result for 1,p result for functions in W0 (Ω) then follows by a density argument.  As a consequence of Poincar´e inequality, we can define on W01,p (Ω) the equivalent norm built after the norm of ∇u in (Lp (Ω))N . From now on, we define kukW 1,p (Ω) = k∇uk(Lp (Ω))N . 0

Even though functions in W01,p (Ω) should only belong to Lp (Ω), the assumptions made on the gradient allow to improve the summability of functions belonging to Sobolev spaces. This is what is stated in the following “embedding” theorem. p Theorem 1.14. Let 1 ≤ p < N , and let p∗ = NN−p (p∗ is called the Sobolev embedding exponent). Then there exists a constant Sp (depending only on N and p) such that

(1.4)

kukLp∗ (Ω) ≤ Sp kukW 1,p (Ω) , 0

∀u ∈ W01,p (Ω).

Remark 1.15. The fact that p∗ is the correct exponent can be easily recovered by a scaling argument. Indeed, if u belongs to W01,p (RN ), then u(λx) belongs to the same space. But then Z Z 1 q |u(λx)| dx = N |u(y)|q dy, λ RN RN and Z Z 1 p |∇u(λx)| dx = N −p |∇u(y)|p dy. λ N N R R

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Therefore, if (1.4) holds for some constant C (independent on λ) and some exponent q, one should have N N −p = , q p which implies q =

Np N −p

= p∗ . ∗

By (1.4), the embedding of W01,p (Ω) in Lp (Ω) is continuous. We recall that a map T : X → Y (with X and Y Banach spaces) is said to be compact if the closure of T (B) is compact in Y for every bounded set B in X. To obtain compactness of the embedding of W01,p (Ω) in Lebesgue spaces, we cannot consider exponents up to p∗ . Theorem 1.16. Let 1 ≤ p < N , and let 1 ≤ q < p∗ . Then the embedding of W01,p (Ω) into Lq (Ω) is compact. ∗

Remark 1.17. The fact that the embedding of W01,p (Ω) into Lp (Ω) is not compact is at the basis for several nonexistence results for equations like −∆u = uq if q is “too large”. But this is another story. . . An important role will be played by the dual of a Sobolev space. We have the following representation theorem. 0

Theorem 1.18. Let p > 1, and let T be an element of (W01,p (Ω)) . 0 Then there exists F in (Lp (Ω))N such that Z hT, ui = F · ∇u, ∀u ∈ W01,p (Ω). Ω 0

The dual of W01,p (Ω) will be denoted by W −1,p (Ω), while the dual of H01 (Ω) is H −1 (Ω). Remark 1.19. The space H01 (Ω) is a Hilbert space. Therefore, by Theorem 1.5, it is isometrically equivalent to its dual H −1 (Ω). Furthermore, by Poincar´e inequality, H01 (Ω) is embedded into L2 (Ω), which is itself a Hilbert space. Since the embedding is continuous and dense, we also have that the the dual of L2 (Ω) (which is L2 (Ω)) is embedded into H −1 (Ω). We therefore have H01 (Ω) ⊂ L2 (Ω) ≡ (L2 (Ω))0 ⊂ (H01 (Ω))0 = H −1 (Ω). If we identify both L2 (Ω) and its dual, and H01 (Ω) and its dual, we obtain a contradiction (since H01 (Ω) and L2 (Ω) are different spaces). Therefore, we have to choose which identification to make: which will be that L2 (Ω) is equivalent to its dual.

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Remark 1.20. Since, by Sobolev embedding, W01,p (Ω) is continuously ∗ ∗ embedded in Lp (Ω), we have by duality that (Lp (Ω))0 is continuously 0 embedded in W −1,p (Ω). If we define p∗ = (p∗ )0 =

Np , Np − N + p

we then have 0

Lp∗ (Ω) ⊂ W −1,p (Ω). If p = 2, we have 2∗ =

2N , N +2

and the embedding of L2∗ (Ω) into H −1 (Ω).

The final result on Sobolev spaces will be about composition with regular functions. Theorem 1.21 (Stampacchia). Let G : R → R be a lipschitz continuous functions such that G(0) = 0. If u belongs to W01,p (Ω), then G(u) belongs to W01,p (Ω) as well, and (1.5)

∇G(u) = G0 (u) ∇u,

almost everywhere in Ω.

Remark 1.22. Recall that a lipschitz continuous function is only almost everywhere differentiable, so that the right-hand side of (1.5) may not be defined. We have however two possible cases: if k is a value such that G0 (k) does not exist, either the set {u = k} has zero measure (and so, since identity (1.5) only holds almost everywhere, this value does not give any problems), or the set {u = k} has positive measure. In this latter case, however, we have both ∇u = 0 and ∇G(u) = 0 almost everywhere, so that (1.5) still holds. Let k > 0; in what follows, we will often use composition of functions in Sobolev spaces with the lipschitz continuous functions (1.6)

Tk (s) = max(−k, min(s, k)), k −k k −k

and (1.7)

Gk (s) = s − Tk (s) = (|s| − k)+ sgn(s).

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−k k

By Theorem 1.21, we have ∇Tk (u) = ∇u χ{|u|≤k} ,

∇Gk (u) = ∇u χ{|u|≥k} ,

almost everywhere in Ω. 1.4. Weak solutions for elliptic equations. We have now all the tools needed to deal with elliptic equations. 2

1.4.1. Definition of weak solution. Let A : Ω → RN be a matrix-valued measurable function such that there exist 0 < α ≤ β such that A(x)ξ · ξ ≥ α|ξ|2 ,

(1.8)

|A(x)| ≤ β,

for almost every x in Ω, and for every ξ in RN . We will consider the following uniformly elliptic equation with Dirichlet boundary conditions ( −div(A(x) ∇u) = f in Ω, (1.9) u=0 on ∂Ω, where f is a function defined on Ω which satisfies suitable assumptions. If the matrix A is the identity matrix, problem (1.9) becomes ( −∆u = f in Ω, u=0 on ∂Ω, i.e., the Dirichlet problem for the laplacian operator. 1.4.2. Classical solutions and weak solutions. Suppose that the matrix A and the functions u and f are sufficiently smooth so that one can “classically” compute −div(A(x)∇u). If ϕ is a function in C01 (Ω), we can then multiply the equation in (1.9) by ϕ and integrate on Ω. Since −div(A(x)∇u) ϕ = −div(A(x)∇u ϕ) + A(x)∇u · ∇ϕ, we get Z

Z A(x)∇u · ∇ϕ −



Z div(A(x)∇u ϕ) =



f ϕ. Ω

By Gauss-Green formula, we have (if ν is the exterior normal to Ω) Z Z div(A(x)∇u ϕ) = A(x)∇u · ν ϕ = 0, Ω

∂Ω

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since ϕ has compact support in Ω. Therefore, if u is a classical solution of (1.9), we have Z Z A(x)∇u · ∇v = f v, ∀v ∈ C01 (Ω). Ω



We now remark that there is no need for A, u, ϕ and f to be smooth in order for the above identity to be well defined. It is indeed enough that A is a bounded matrix, that u and ϕ belong to H01 (Ω), and that f is in L2 (Ω) (or in L2∗ (Ω), thanks to Sobolev embedding, see Remark 1.20). We therefore give the following definition. Definition 1.23. Let f be a function in L2∗ (Ω). A function u in H01 (Ω) is a weak solution of (1.9) if Z Z f v, ∀v ∈ H01 (Ω). A(x)∇u · ∇v = (1.10) Ω



If u is a weak solution of (1.9), and u is sufficiently smooth in order to perform the same calculations as above “going backwards”, then it can be proved that u is a “classical” solution of (1.9). The study of the assumptions on f and A such that a weak solution is also a classical solution goes beyond the purpose of this text. 1.4.3. Existence of solutions (using Lax-Milgram). Theorem 1.24. Let f be a function in L2∗ (Ω). Then there exists a unique solution u of (1.9) in the sense of (1.10). Proof. We will use Lax-Milgram theorem. Indeed, if we define the bilinear form a : H01 (Ω) × H01 (Ω) → R by Z a(u, v) = A(x)∇u · ∇v, Ω

and the linear and continuos (thanks to Sobolev embedding) functional T : H01 (Ω) → R by Z hT, vi = f v, Ω

solving problem (1.9) in the sense of (1.10) amounts to finding u in H01 (Ω) such that a(u, v) = hT, vi,

∀v ∈ H01 (Ω),

which is exactly the result given by Lax-Milgram theorem. In order to apply the theorem, we have to check that a is continuous and coercive

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(the fact that it is bilinear being evident). We have, by (1.8), and by H¨older inequality, Z |A(x)||∇u||∇v| ≤ β kukH 1 (Ω) kvkH 1 (Ω) , |a(u, v)| ≤ 0



0

so that a is continuous. Furthermore, again by (1.8), we have Z Z a(u, u) = A(x)∇u · ∇u ≥ α |∇u|2 = α kuk2H 1 (Ω) , Ω



so that a is also coercive.

0



1.4.4. Existence of solutions (using minimization). If the matrix A satisfies (1.8) and is symmetrical, existence and uniqueness of solutions for (1.9) can be proved using minimization of a suitable functional. Theorem 1.25. Let f be a function in L2∗ (Ω), and let J : H01 (Ω) → R be defined by Z Z 1 J(v) = A(x)∇v · ∇v − f v, ∀v ∈ H01 (Ω). 2 Ω Ω Then J has a unique minimum u in H01 (Ω), which is the solution of (1.9) in the sense of (1.10). Proof. We begin by proving that J is coercive and weakly lower semicontinuous on H01 (Ω), so that a minimum will exist by Theorem 1.4. Recalling (1.8) and using H¨older and Sobolev inequalities, we have Z α J(v) ≥ |∇v|2 − kf kL2∗ (Ω) kvkL2∗ (Ω) 2 Ω α ≥ kvk2H 1 (Ω) − S2 kf kL2∗ (Ω) kvkH 1 (Ω) , 0 0 2 and the right hand side diverges as the norm of u in H01 (Ω) diverges, so that J is coercive. Let now {vn } be a sequence of functions which is weakly convergent to some v in H01 (Ω). Since f belongs to L2∗ (Ω), ∗ and vn converges weakly to v in L2 (Ω), we have Z Z lim f vn = f v, n→+∞





so that the weak lower semicontinuity of J is equivalent to the weak lower semicontinuity of Z K(v) = A(x)∇v · ∇v. Ω

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By (1.8) we have Z K(v − vn ) =

A(x) ∇(v − vn ) · ∇(v − vn ) ≥ 0, Ω

which, together with the symmetry of A, implies Z Z Z (1.11) 2 A(x)∇v · ∇vn − A(x)∇v · ∇v ≤ A(x)∇vn · ∇vn . Ω





Since ∇vn converges weakly to ∇v in (L2 (Ω))N , and since A(x)∇v is fixed in the same space, we have Z Z lim A(x)∇v · ∇vn = A(x)∇v · ∇v, n→+∞





so that taking the inferior limit in both sides of (1.11) implies Z Z A(x)∇vn · ∇vn = lim inf K(vn ), A(x)∇v · ∇v ≤ lim inf K(v) = n→+∞



n→+∞



which means that K is weakly lower semicontinuous on H01 (Ω), as desired. Let now u be a minimum of J on H01 (Ω). We are going to prove that it is unique. Indeed, if u and v are both minima of J, one has u + v  u + v  J(u) ≤ J , J(v) ≤ J , 2 2 that is, u + v  J(u) + J(v) ≤ 2J , 2 which becomes (after cancelling equal terms and multiplying by 4) Z Z Z 2 A(x)∇u · ∇u + 2 A(x)∇u · ∇u = A(x)∇(u + v) · ∇(u + v). Ω





Using the fact that A is symmetric, expanding the right hand side, and cancelling equal terms, we arrive at Z Z Z A(x)∇u · ∇u − 2 A(x)∇u · ∇v + A(x)∇v · ∇v ≤ 0, Ω





which can be rewritten as Z A(x)∇(u − v) · ∇(u − v) ≤ 0. Ω

Using (1.8) we therefore have α ku − vk2H 1 (Ω) ≤ 0, 0

which implies u = v, as desired.

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We are now going to prove that the minimum u is a solution of (1.9) in the sense of (1.10). Given v in H01 (Ω) and t in R, we have J(u) ≤ J(u + tv), that is Z Z Z Z 1 1 A(x)∇u·∇u− f u ≤ A(x)∇(u+tv)·∇(u+tv)− f (u+tv). 2 Ω 2 Ω Ω Ω Expanding the right hand side, cancelling equal terms, and using the fact that A is symmetric, we obtain Z Z Z t2 t A(x)∇u · ∇v + A(x)∇v · ∇v − t f v ≥ 0. 2 Ω Ω Ω If t > 0, dividing by t and then letting t tend to zero implies Z Z f v ≥ 0, A(x)∇u · ∇v − Ω



while if t < 0, dividing by t and then letting t tend to zero implies the reverse inequality. It then follows that Z Z f v, ∀v ∈ H01 (Ω), A(x)∇u · ∇v = Ω



and so u solves (1.9) (in the sense of (1.10)). In order to prove that such a solution is unique, we are going to prove that if u solves (1.9), then u is a minimum of J. Indeed, choosing u − v as test function in (1.10), we have Z Z Z A(x)∇u · ∇u − A(x)∇u · ∇v = f (u − v). Ω





This implies Z Z Z 1 1 J(u) + A(x)∇u · ∇u − A(x)∇u · ∇v = J(v) − A(x)∇v · ∇v, 2 Ω 2 Ω Ω which implies J(u) ≤ J(v) since Z Z Z 1 1 A(x)∇u · ∇u − A(x)∇u · ∇v + A(x)∇v · ∇v 2 Ω 2 Ω Ω is nonnegative by (1.8) since it is equal to Z 1 A(x)∇(u − v) · ∇(u − v). 2 Ω 

16

LUIGI ORSINA

1.5. A nonlinear equation. Let now a : Ω × R → R be a continuous function such that α ≤ a(x, s) ≤ β ,

(1.12)

∀x ∈ Ω , ∀s ∈ R ,

with 0 < α ≤ β in R. Given a function f in L2∗ (Ω), we ask ourselves whether there exists a weak solution of the equation ( −div(a(x, u) ∇u) = f in Ω, (1.13) u=0 on ∂Ω, that is, a function u in H01 (Ω) such that Z Z a(x, u) ∇u · ∇v = f v, Ω

∀v ∈ H01 (Ω) .



Remark that, since a(x, u) belongs to L∞ (Ω) being a bounded, the integral on the left hand side is well defined. Of course, since the equation is nonlinear, Lax-Milgram theorem cannot be applied, so that we will need to follow another approach. The first idea is to perform a change of variable if a does not depend on the variable x. Indeed, in this case, we define Z s A(s) = a(t) dt , 0 0

and v = A(u). Then ∇v = A (u) ∇u = a(u) ∇u, so that u solves (1.13) if and only if v is a solution in H01 (Ω) of −∆v = f . Since this latter equation has a unique solution v, and since A is invertible (being strictly increasing as its derivative a(s) is positive), then u = A−1 (v) is the solution of (1.13). Unfortunately, such a trick does not work if the function a depends also on the variable x. In this case, even if we can define Z u(x) v(x) = a(x, t) dt , 0

we have Z ∇v = a(x, u) ∇u +

u(x)

∇a(x, t) dt , 0

so that an extra term appears (which requires a to be C 1 ). Luckily, Lax-Milgram theorem (or a change of variable) is not our only tool. We can for example consider the following functional: Z Z 1 2 J(v) = a(x, v) |∇v| − f v , v ∈ H01 (Ω) . 2 Ω Ω Using the assumptions on a, it is easy to see that J is both coercive and weakly lower semicontinuous in H01 (Ω) (see the proof of Theorem

ELLIPTIC EQUATIONS

17

1.25), so that there exists at least a minimum u of J on H01 (Ω) by Weierstrass’ theorem. Since u is a minimum, then J(u) ≤ J(u + tv) for every t in R, and for every v in H01 (Ω). Starting from this inequality, we arrive to Z 1 0≤ [a(x, u + tv) − a(x, u)]|∇u|2 2 Ω Z Z Z t2 2 +t a(x, u + tv)∇u · ∇v + a(x, u + tv)|∇v| − t f v . 2 Ω Ω Ω Dividing by t > 0, and letting t tend to zero, we will obtain (should every passage be correct, which is not) Z Z Z 1 0 2 a (x, u) |∇u| v + a(x, u) ∇u · ∇v − f v, 0≤ 2 Ω Ω Ω and the reverse inequality dividing by t < 0 and letting t tend to zero. In other words, u in H01 (Ω) would be such that Z Z Z 1 0 2 a(x, u) ∇u · ∇v + a (x, u) |∇u| v = f v , ∀v ∈ H01 (Ω) . 2 Ω Ω Ω This identity, however, has at least two problems: first of all, the function a is only continuous, so that a0 (x, s) may not exist. This, however, can be solved: just suppose that a has a continuous derivative (remark that this assumption is not needed to prove that J has a minimum). Furthermore, even if a has a continuous derivative, the term Z a0 (x, u) |∇u|2 v Ω

is not necessarily well defined for every v in H01 (Ω); indeed, the term |∇u|2 only belongs to L1 (Ω), while H01 (Ω) functions are not necessarily bounded (this only happens if the space dimension is one). Therefore, we have to restrict the class of test functions we consider in the inequality J(u) ≤ J(u + tv): we have to consider functions v in H01 (Ω) ∩ L∞ (Ω); once again, this is not enough: we also need that a0 (x, s) is bounded (note that the assumption “a bounded” does not imply “a0 bounded”. . .). Thus, under all of these assumptions, any minimum u of J is such that Z Z Z 1 0 2 a (x, u) |∇u| v = f v, a(x, u) ∇u · ∇v + 2 Ω Ω Ω

18

LUIGI ORSINA

for every v in H01 (Ω) ∩ L∞ (Ω), so that it is a weak solution of the equation ( −div(a(x, u) ∇u) + 12 a0 (x, u) |∇u|2 = f in Ω, u=0 su ∂Ω. But this is not equation (1.13)! Our “functional” idea is wrong — since the derivative of a product is not the product of derivatives. . . It is now time for the right way of proving the existence of solutions for (1.13). Let v in L2 (Ω) be fixed. Then since a is bounded and strictly positive, there exists a unique solution u in H01 (Ω) of ( −div(a(x, v) ∇u) = f in Ω, (1.14) u=0 su ∂Ω. Thus, the function S : L2 (Ω) → H01 (Ω) defined by S(v) = u is well defined. Furthermore, since H01 (Ω) is embedded in L2 (Ω), S is a function from L2 (Ω) into itself. Thus, a solution of (1.13) is a fixed point for S. We are going to prove the existence of a fixed point using Schauder’s theorem 1.11. To apply Schauder’s theorem, we begin by observing that there exists R > 0 such that kS(v)kL2 (Ω) ≤ R for every v in L2 (Ω). Indeed, choosing u = S(v) as test function in the weak formulation of (1.14), and using the fact that a(x, s) ≥ α > 0, we have Z Z Z 2 2 (1.15) α |∇u| ≤ a(x, v) |∇u| = f u ≤ kf kL2 (Ω) kukL2 (Ω) . Ω





Recalling Poincar´e inequality, we get kuk2L2 (Ω) ≤ C kf kL2 (Ω) kukL2 (Ω) , which implies the result choosing R = C kf kL2 (Ω) . Thus, the ball BR of L2 (Ω) centered at the origin, and with radius R, is convex, closed and bounded in L2 (Ω), and S maps BR into itself. Let us prove that S is continuous. Let {vn } be a sequence which strongly converges to v in L2 (Ω), and let un = S(vn ) be the solutions of (1.14). Using (1.15) we get, recalling that the norm of un in L2 (Ω) is smaller than R, Z α |∇un |2 ≤ R kf kL2 (Ω) , Ω

and so un is bounded in H01 (Ω). Thus, up to subsequences, vn converges to v almost everywhere, and un tends to u weakly in H01 (Ω) and strongly

ELLIPTIC EQUATIONS

19

in L2 (Ω). Therefore, we can pass to the limit in the identities Z Z a(x, vn ) ∇un · ∇z = f z , ∀z ∈ H01 (Ω) , Ω



to prove that u is a solution of (1.14) (with a(x, v)), so that u = S(v) (by uniqueness). Since the limit u does not depend on the extracted subsequences, then un = S(vn ) converges to u = S(v) in L2 (Ω), and so S is continuous. The compactness of S(BR ) is easy to prove, since we have proved that S(BR ) is bounded in H01 (Ω): by Rellich-Kondrachov theorem, the closure of S(BR ) is compact in L2 (Ω), as desired. Therefore, by Schauder’s theorem, there exists at least a solution u of (1.13). 2. Regularity results Warning to the reader: from now, unless explicitly stated, N ≥ 3. Thanks to the results of the previous section, we have existence of solutions for linear elliptic equations with data f in L2∗ (Ω). The solu∗ tion u belongs to H01 (Ω) and (thanks to Sobolev embedding) to L2 (Ω). One then wonders whether an increase on the regularity of f will yield more regular solutions. 2.1. Examples. We are going to study a model case, in which the solution of (1.9) can be explictly calculated. This example will give us a hint on what happens in the general case. Example 2.1. Let Ω = B 1 (0), let N ≥ 3, let α < N , and define 2

f (x) =

1 . |x|α (− log(|x|))

It is well known that f belongs to Lp (Ω), with p = study the regularity of the solution u of ( −∆u = f in Ω, u=0 on ∂Ω,

N . α

We are going to

taking advantage of the fact that the solution will be radially symmetric. Recalling the formula for the laplacian in radial coordinates, we have 1 1 − N −1 (ρN −1 u0 (ρ))0 = α . ρ ρ (− log(ρ))

20

LUIGI ORSINA

Multiplying by ρN −1 and integrating between 0 and ρ, we obtain

ρ

N −1

Z

0

ρ

tN −1−α dt. log(t)

u (ρ) = 0

Dividing by ρN −1 and integrating between ing that u( 12 ) = 0) Z u(ρ) = −

1 2

1

Z

sN −1

ρ

0

s

1 2

and ρ we then get (recall-

 tN −1−α dt ds. log(t)

We are integrating on the set E = {(s, t) ∈ R2 : ρ ≤ s ≤ 12 , 0 ≤ t ≤ s}, t 1/2 ρ

E ρ

1/2

s

which, after exchanging t with s, becomes E = {(t, s) ∈ R2 : 0 ≤ t ≤ 1 , max(ρ, t) ≤ s ≤ 21 }, 2 s 1/2 E ρ ρ

1/2

t

Exchanging the integration order, we then have ! Z 1 2 tN −1−α ds u(ρ) = − dt N −1 log(t) 0 max(ρ,t) s # Z 1 N −1−α " 2−N 2 t 1 1 = − (max(ρ, t))2−N dt N − 2 0 log(t) 2 Z 1 N −1−α Z 1 N −1−α 2 t 2 t 2N −2 1 (max(ρ, t))2−N = dt − dt. N − 2 0 log(t) N −2 0 log(t) Z

1 2

ELLIPTIC EQUATIONS

21

Since α < N , the first integral is bounded, so that it is enough to study the behaviour near zero of the function Z 1 N −1−α 2 t (max(ρ, t))2−N dt v(ρ) = log(t) 0 Z ρ N −1−α Z 1 1−α 2 t t = ρ2−N dt + dt log(t) log(t) 0 ρ = ρ2−N w(ρ) + z(ρ). It is easy to see (using the de l’Hopital rule), that if α 6= 2 ρN −α w(ρ) ≈ , log(ρ)

ρ2−α and z(ρ) ≈ , log(ρ)

as ρ tends to zero, so that, if α 6= 2, u(ρ) ≈

ρ2−α , log(ρ)

as ρ tends to zero. This implies that u belongs to L∞ (Ω) if α < 2, N , if 2 < α < N . Recalling that while it is in Lm (Ω), with m = α−2 N p f belongs to L (Ω) with p = α , we therefore have that u belongs to L∞ (Ω) if f belongs to Lp (Ω), and p > N2 , while it is in Lm (Ω), with p , if f belongs to Lp (Ω), with 1 < p < N2 . m = NN−2p If α = 2, then w(ρ) ≈

ρN −α , log(ρ)

and z(ρ) ≈ log(− log(ρ)),

so that u is in every Lm (Ω), but not in L∞ (Ω), if f belongs to Lp (Ω) with p = N2 . In this case (which we will not study in the following), it can be proved that e|u| belongs to L1 (Ω). Observe that if α = N2+2 , so that f belongs to L2∗ (Ω), we get that ∗ u belongs to L2 (Ω), which is exactly the results we already knew by Sobolev embedding. Also remark that the above example gives informations also if f does not belong to L2∗ (Ω) (i.e., if N2+2 < α < N ), an assumption under which we do not have any existence results (yet!). If we want to take α = N , we need to change the definition of f . We fix β > 1 and define f (x) =

|x|N

1 , (− log(|x|))β

22

LUIGI ORSINA

which is a function belonging to L1 (Ω). Performing the same calculations as above, we obtain Z 1 2 1 dt u(ρ) = , N −1 β−1 ρ t (− log(t))β−1 so that u(ρ) ≈

ρN −2

1 , (− log(ρ))β−1

as ρ tends to zero. Observe that in this case f belongs to L1 (Ω) for ·1 = NN−2 if and every β > 1, but u belongs to Lm (Ω), with m = NN−2·1 only if β > 2 − N2 . If 1 < β ≤ 2 − N2 , the solution u belongs “only” to Lm (Ω), for every m < NN−2 . We leave to the interested reader the study of the case N = 2. 2.2. Stampacchia’s theorems. The regularity results we are going to prove now show that the previous example is not just an example. We begin with a real analysis lemma. Lemma 2.2 (Stampacchia). Let ψ : R+ → R+ be a nonincreasing function such that ψ(h) ≤

(2.16)

M ψ(k)δ , (h − k)γ

∀h > k > 0,

where M > 0, δ > 1 and γ > 0. Then ψ(d) = 0, where δγ

dγ = M ψ(0)δ−1 2 δ−1 . Proof. Let n in N and define dn = d(1 − 2−n ). We claim that nγ

ψ(dn ) ≤ ψ(0) 2− δ−1 .

(2.17)

Indeed, (2.17) is clearly true if n = 0; if we suppose that it is true for some n, then, by (2.16), ψ(dn+1 ) ≤

(n+1)γ nγδ M ψ(dn )δ − δ−1 δ − δ−1 (n+1)γ −γ ≤ M ψ(0) 2 2 d = ψ(0) 2 , (dn+1 − dn )γ

which is (2.17) written for n + 1. Since (2.17) holds for every n, and since ψ is non increasing, we have nγ

0 ≤ ψ(d) ≤ lim inf ψ(dn ) ≤ lim ψ(0)δ−1 2− δ−1 = 0, n→+∞

n→+∞

as desired.  The first result (due to Guido Stampacchia, see [6]), deals with bounded solutions for (1.9).

ELLIPTIC EQUATIONS

23

Theorem 2.3 (Stampacchia). Let f belong to Lp (Ω), with p > N2 . Then the solution u of (1.9) belongs to L∞ (Ω), and there exists a constant C, only depending on N , Ω, p and α, such that kukL∞ (Ω) ≤ C kf kLp (Ω) .

(2.18)

Proof. Let k > 0 and choose v = Gk (u) as test function in (1.9) (Gk (s) has been defined in (1.7)). Defining Ak = {x ∈ Ω : |u(x)| ≥ k} one then has, since ∇v = ∇u χAk by Theorem 1.21, and using (1.8) Z Z Z Z 2 α |∇Gk (u)| ≤ A(x)∇u · ∇u χAk = f Gk (u) = f Gk (u). Ak





Ak

Using Sobolev inequality (in the left hand side), and H¨older inequality (in the right hand side), one has Z  21 Z Z  22∗  21∗ ∗ α 2∗ 2∗ 2∗ |f | |Gk (u)| ≤ |Gk (u)| . S22 Ak Ak Ak Simplifying equal terms, we thus have  22∗  2 2∗ Z Z ∗ S ∗ 2 |f |2∗ . |Gk (u)|2 ≤ α Ak Ak Recalling that f belongs to Lp (Ω), and that p > 2∗ since p > have (again by H¨older inequality)  2 ∗ Z S2 kf kLp (Ω) 2 2∗ 2∗ 2∗ |Gk (u)| ≤ m(Ak ) 2∗ − p . α Ak

N , 2

we

We now take h > k, so that Ah ⊆ Ak , and Gk (u) ≥ h − k on Ah . Thus, ∗  2 S2 kf kLp (Ω) 2 2∗ 2∗ 2∗ (h − k) m(Ah ) ≤ m(Ak ) 2∗ − p , α which implies  m(Ah ) ≤

S22 kf kLp (Ω)

2∗

α

2∗

2∗

m(Ak ) 2∗ − p . (h − k)2∗

We define now ψ(k) = m(Ak ), so that ψ(h) ≤ where  M=

S22 kf kLp (Ω) α

M ψ(k)δ , (h − k)γ

2∗ ,

δ=

2∗ 2∗ − , 2∗ p

γ = 2∗ .

24

LUIGI ORSINA

The assumption p > N2 implies δ > 1, so that applying Lemma 2.2, we have that ψ(d) = 0, where ∗

d2 = C(Ω, N, p) M. Since m(Ad ) = 0, we have |u| ≤ d almost everywhere, which implies kukL∞ (Ω) ≤ d = C(N, Ω, p, α) kf kLp (Ω) , as desired.



Remark 2.4. Observe that, in order to prove the previous theorem, we did not use two of the properties of the equation: that the matrix A is bounded from above (we only used its ellipticity) and, above all, the fact that the equation was linear: in other words, the proof above also holds for every uniformly elliptic operator (for example, for the equation studied in §5 of the previous section). The second results deals with the case of unbounded solutions. Theorem 2.5 (Stampacchia). Let f belong to Lp (Ω), with 2∗ ≤ p < N . Then the solution u of (1.9) belongs to Lm (Ω), with m = p∗∗ = 2 Np , and there exists a constant C, only depending on N , Ω, p and N −2p α, such that kukLp∗∗ (Ω) ≤ C kf kLp (Ω) .

(2.19)

Proof. We begin by observing that if p = 2∗ , then p∗∗ = 2∗ , so that the result is true in this limit case by the Sobolev embedding. Therefore, we only have to deal with the case p > 2∗ . The original proof of Stampacchia used a linear interpolation theorem; i.e., it is typical of a linear framework. We are going to give another proof, following [1], which makes use of a technique that can be applied also in a nonlinear context. Let k > 0 be fixed, let γ > 1 and choose v = |Tk (u)|2γ−2 Tk (u) as test function in (1.9) (Tk (s) has been defined in (1.6)). We obtain, by Theorem 1.21, Z Z 2γ−2 (2γ − 1) A(x)∇u · ∇Tk (u) |Tk (u)| = f |Tk (u)|2γ−2 Tk (u). Ω



Using (1.8), and observing that ∇u = ∇Tk (u) where ∇Tk (u) 6= 0, we then have Z Z 2 2γ−2 α (2γ − 1) |∇Tk (u)| |Tk (u)| ≤ |f | |Tk (u)|2γ−1 . Ω



ELLIPTIC EQUATIONS

25

Since, again by Theorem 1.21, |∇Tk (u)|2 |Tk (u)|2γ−2 = γ12 |∇|Tk (u)|γ |2 , we have Z Z α (2γ − 1) γ 2 |∇|Tk (u)| | ≤ |f | |Tk (u)|2γ−1 . 2 γ Ω Ω Using Sobolev inequality (in the left hand side), and H¨older inequality (in the right hand one), we obtain α (2γ − 1) S22 γ 2

Z

γ2∗

 22∗

|Tk (u)|

Z ≤ kf kLp (Ω)



(2γ−1)p0

|Tk (u)|

 10 p

.

Ω ∗∗

We now choose γ so that γ2∗ = (2γ − 1)p0 , that is γ = p2∗ (as it is easily seen). With this choice, γ > 1 if and only if p > 2∗ (which is true). Since p < N2 , we also have 22∗ > p10 , and so Z

p∗∗

 22∗ − 10 p

|Tk (u)|

≤ C(N, Ω, p, α) kf kLp (Ω) .



Observing that

2 2∗



1 p0

=

1 , p∗∗

we have therefore proved that

kTk (u)kLp∗∗ (Ω) ≤ C(N, Ω, p, α) kf kLp (Ω) ,

∀k > 0.

Letting k tend to infinity, and using Fatou lemma (or the monotone convergence theorem), we obtain the result.  Remark 2.6. The results of theorems 2.3 and 2.5 are somehow “natural” if we make a mistake. . . Indeed, let u be the solution of −∆u = f , with f in Lp (Ω). Then, if we read the equation, we have that u has two derivatives in Lp (Ω), so that it belongs to W02,p (Ω). By Sobolev ∗ embedding, u then belongs to W01,p (Ω) and, again by Sobolev embed∗∗ ding, to Lp (Ω) (or to L∞ (Ω) if p > N2 ). The “mistake” here is to deduce from the fact that the sum of (some) derivatives of u belongs to Lp (Ω), the fact that all derivatives are in the same space. Surprisingly, it turns out that, in the case of the laplacian, the fact that −∆u belongs to Lp (Ω) actually implies that u is in W02,p (Ω) (this is the so-called Calderun-Zygmund theory), so that the “mistake” is not an actual one. . . Summarizing the results of this section, we have the following picture.

26

LUIGI ORSINA

H01 (Ω)

?

H01 (Ω)

∗∗

1

Lp (Ω)

L∞ (Ω)

Theorem 2.5

Theorem 2.3

2N N +2

N 2

p

We will deal with the “?” part in the forthcoming section (actually, in all the forthcoming sections). 3. Existence via duality for measure data We are now going to deal with existence results for data which do not belong to L2∗ (Ω) (i.e., they are not in H −1 (Ω)), so that neither LaxMilgram theorem nor minimization techniques can be applied. Before going on, we need some definitions. 3.1. Measures. We recall that a nonnegative measure on Ω is a set function µ : B(Ω) → [0, +∞] defined on the σ-algebra B(Ω) of Borel sets of Ω (i.e., the smallest σ-algebra containing the open sets) such that µ(∅) = 0 and such that +∞ [

µ



En =

n=1

+∞ X

µ(En ),

n=1

for every sequence {En } of disjoint sets in B(Ω). This latter property is called σ-additivity. A σ-additive measure µ is also σ-subadditive, i.e., one has +∞ +∞ [  X En ≤ µ(En ), µ n=1

n=1

for every sequence {En } of sets in B(Ω). A nonnegative measure µ is also monotone, i.e., one has that A⊆B

implies µ(A) ≤ µ(B).

A measure µ is said to be regular if for every E in B(Ω) and for every ε > 0 there exist an open set Aε , and a closed set Cε , such that Cε ⊆ E ⊆ Aε ,

µ(Aε \ Cε ) < ε.

A measure µ is said to be bounded if µ(Ω) < +∞. The set of nonnegative, regular, bounded measures on Ω will be denoted by M+ (Ω). We

ELLIPTIC EQUATIONS

27

define the set of bounded Radon measures on Ω as M(Ω) = {µ1 − µ2 , µi ∈ M+ (Ω)}. Given a measure µ in M(Ω), there exists a unique pair (µ+ , µ− ) in M+ (Ω) × M+ (Ω) such that µ = µ+ − µ− , and such there exist E + and E − in B(Ω), disjoint sets, such that µ± (E) = µ(E ∩ E ± ),

∀E ∈ B(Ω).

The measures µ+ and µ− are the positive and negative parts of the measure µ. Given a measure µ in M(Ω), the measure |µ| = µ+ + µ− is said to be the total variation of the measure µ. If we define kµkM(Ω) = |µ|(Ω), the vector space M(Ω) becomes a Banach space, which turns out to be the dual of C 0 (Ω). A bounded Radon measure µ is said to be concentrated on a Borel set E if µ(B) = µ(B ∩ E) for every Borel set B. In this case, we will write µ E. For example, we have µ± = µ E ± , with E ± as above. Given two Radon measures µ and ν, we say that µ is absolutely continuous with respect to ν if ν(E) = 0 implies µ(E) = 0. In this case we will write µ << ν. Two Radon measures µ and ν are said to be orthogonal if there exists a set E such that µ(E) = 0, and ν = ν E. In this case, we will write µ ⊥ ν. For example, given a Radon measure µ, we have µ+ ⊥ µ− . Theorem 3.1. Let ν be a nonnegative Radon measure. Given a Radon measure µ, there exists a unique pair (µ0 , µ1 ) of Radon measures such that µ = µ0 + µ1 , µ0 << ν, µ1 ⊥ ν. Proof. Suppose that µ is nonnegative, and define A = {µ(E) : E ∈ B(Ω), ν(E) = 0}. Let α = sup A, and let En be a maximizing sequence, i.e., a sequence of Borel sets such that lim µ(En ) = α,

n→+∞

ν(En ) = 0.

If we define E as the union of the En , clearly ν(E) = 0 (since ν is σsubadditive), and µ(E) = α (since µ(E) ≥ µ(En ) for every n). Define now µ1 = µ E, µ0 = µ − µ1 .

28

LUIGI ORSINA

Clearly, µ1 ⊥ ν (since ν(E) = 0, and since µ1 is concentrated on E by definition). On the other hand, if ν(B) = 0, then µ0 (B) = 0; and indeed, if it were µ0 (B) > 0 for some B 6= E, then 0 < µ0 (B) = µ(B) − µ(B ∩ E) = µ(B \ E), so that B ∪ E will be such that ν(B ∪ E) = 0, and µ(B ∪ E) = µ(E) + µ(B \ E) = α + µ(B \ E) > α, thus contradicting the definition of α. As for uniqueness, if µ = µ0 + µ1 = µ00 + µ01 , then µ0 − µ00 = µ01 − µ1 . If ν(B) = 0, we will have (µ1 − µ01 )(B) = 0. Since µ1 − µ01 is also orthogonal with respect to ν, this implies that (µ1 − µ01 )(E) = 0 for every Borel set E, so that µ1 = µ01 , hence µ0 = µ00 . If the measure µ has a sign, it is enough to apply the result to µ+ and µ− .  Examples of bounded Radon measures are the Lebesgue measure LN concentrated on a bounded set of RN , or the measure defined by ( 1 if x0 ∈ E, δx0 (E) = 0 if x0 6∈ E, which is called the Dirac’s delta concentrated at x0 . We clearly have δx0 ⊥ LN . Another example of Radon measure is the measure defined by Z f (x) dx,

µ(E) = E

with f a function in L1 (Ω). In this case µ << LN , and Z Z ± ± µ (E) = f (x) dx, |µ|(E) = |f (x)| dx. E

E

1

Therefore, L (Ω) ⊂ M(Ω). For sequences of measures, we have two notions of convergence: the weak∗ : Z Z ϕ dµn → ϕ dµ, ∀ϕ ∈ C00 (Ω), Ω



and the narrow convergence: Z Z ϕ dµn → ϕ dµ, Ω

∀ϕ ∈ Cb0 (Ω).



For positive measures, narrow convergence is equivalent to weak∗ convergence and convergence of the “masses” (i.e., µn (Ω) converges to µ(Ω)). If xn is a sequence in Ω which converges to a point x0 on ∂Ω, then δxn converges to zero for the weak∗ convergence (since the

ELLIPTIC EQUATIONS

29

measure δx0 is indeed the zero measure in Ω), but not for the narrow convergence. Measures can be approximated (in either convergence) by sequences of bounded functions. Before dealing with existence results for elliptic equations with measure data, we will begin with a particular case. 3.2. Duality solutions for L1 data. Let f and g be two functions in L∞ (Ω), and let u and v be the solutions of ( ( −div(A∗ (x) ∇v) = g in Ω, −div(A(x) ∇u) = f in Ω, v=0 on ∂Ω. u=0 on ∂Ω, where A∗ is the transposed matrix of A (note that A∗ satisfies (1.8) with the same constants as A). Since both u and v belong to H01 (Ω), u can be chosen as test function in the formulation of weak solution for v, and vice versa. One obtains Z Z Z Z ∗ ug = A (x)∇v · ∇u = A(x)∇u · ∇v = f v. Ω







In other words, one has Z

Z

f v,

ug = Ω



for every f and g in L∞ (Ω), where u and v solve the corresponding problems with data f and g respectively. Clearly, both u and v belong to L∞ (Ω) by Theorem 2.3, but we remark that the two integrals are well-defined also if f only belongs to L1 (Ω), and u only belongs to L1 (Ω) (always maintaining the assumption that g — and so v — is a bounded function). This fact inspired to Guido Stampacchia the following definition of solution for (1.9) if the datum is in L1 (Ω). Definition 3.2. Let f belong to L1 (Ω). A function u in L1 (Ω) is a duality solution of (1.8) with datum f if one has Z Z ug = f v, Ω





for every g in L (Ω), where v is the solution of ( −div(A∗ (x) ∇v) = g in Ω, v=0 on ∂Ω. Theorem 3.3 (Stampacchia). Let f belong to L1 (Ω). Then there exists a unique duality solution of (1.8) with datum f . Furthermore, u belongs to Lq (Ω), for every q < NN−2 .

30

LUIGI ORSINA

Proof. Let p >

N 2

and define the linear functional T : Lp (Ω) → R by Z hT, gi = f v. Ω

By Theorem 2.3, the functional is well-defined; furthermore, since (2.18) holds, there exists C > 0 such that Z |hT, gi| ≤ |f | |v| ≤ kf kL1 (Ω) kvkL∞ (Ω) ≤ C kf kL1 (Ω) kgkLp (Ω) , Ω

so that T is continuous on Lp (Ω). By Riesz representation Theorem 0 for Lp spaces, there exists a unique function up in Lp (Ω) such that Z hT, gi = up g, ∀g ∈ Lp (Ω). Ω ∞

p

Since L (Ω) ⊂ L (Ω), we have Z Z up g = hT, gi = f v, Ω

∀g ∈ L∞ (Ω),



so that up is a duality solution of (1.9), as desired. We claim that up does not depend on p; indeed, if for example p > q > N2 , we have Z Z Z up g = fv= uq g, ∀g ∈ L∞ (Ω), Ω





so that up = uq in L1 (Ω) (and so they are almost everywhere the same function). Therefore, there exists a unique function u which is a 0 duality solution of (1.9), and it belongs to Lp (Ω) for every p > N2 ; i.e., u belongs to Lq (Ω) for every q < NN−2 , as desired.  N q Remark that the fact that u belongs to L (Ω) for every q < N −2 is consistent with the results of the last part of Example 2.1 (the case α = N ). 3.3. Duality solutions for measure data. The case of L1 (Ω) data is only a particular one, since L1 (Ω) is a subset of M(Ω). However, recalling that M(Ω) is the dual of C 0 (Ω), the proof of Theorem 3.3 could be performed in exactly the same way if one knew that the solution of (1.9) were not only bounded, but also continuous on Ω if the datum is in Lp (Ω) with p > N2 . This is exactly the case if the boundary of Ω is sufficiently regular. Theorem 3.4 (De Giorgi). Let Ω be of class C 1 , and let f be in Lp (Ω), with p > N2 . Then the solution u of (1.9) with datum f belongs to C 0 (Ω), and there exists a constant Cp such that kukC 0 (Ω) ≤ Cp kf kLp (Ω) .

ELLIPTIC EQUATIONS

31

Thanks to the previous result, we thus have the following existence result. Theorem 3.5. Let µ be a measure in M(Ω). Then there exists a unique duality solution of (1.8) with datum µ, i.e., a function u in L1 (Ω) such that Z Z ug = v dµ, ∀g ∈ L∞ (Ω), Ω



where v is the solution of (1.9) with datum g and matrix A∗ . Furthermore, u belongs to Lq (Ω), for every q < NN−2 . 3.4. Regularity of duality solutions. If the datum f belongs to Lp (Ω), with 1 < p < 2∗ , then the duality solution of (1.9) is more regular. Theorem 3.6. Let f belong to Lp (Ω), 1 < p < 2∗ . Then the duality ∗∗ p . solution u of (1.8) belongs to Lp (Ω), p∗∗ = NN−2p Np Proof. Let q = N p−N , and define T : Lq (Ω) → R as in the proof of +2p Theorem 3.3. We then have Z |f | |v| ≤ kf kLp (Ω) kvkLp0 (Ω) . |hT, gi| ≤ Ω

By Theorem 2.5, the norm of v in Lr (Ω) is controlled by a constant times the norm of g in Ls (Ω), with r = s∗∗ . Taking r = p0 , this gives s = q; hence, |hT, gi| ≤ C kf kLp (Ω) kgkLq (Ω) , 0

so that the function u which represents T belongs to Lq (Ω); since we p have q 0 = NN−2p , the result is proved.  p∗∗ Once again, the fact that u belongs to L (Ω) is consistent with the results of Example 2.1 (the case N2+2 < α < N ). The picture at the end of Section 2 can now be improved as follows.

H01 (Ω)

H01 (Ω)

?

?

L N −2 −ε (Ω)

Lp (Ω)

Lp (Ω)

L∞ (Ω)

Theorem 3.3

Theorem 3.6

Theorem 2.5

Theorem 2.3

∗∗

N

1

∗∗

2N N +2

N 2

p

32

LUIGI ORSINA

4. Existence via approximation for measure data The result of Theorem 3.5 is somewhat unsatisfactory: even though it proves that there exists a unique solution by duality of (1.9) if the datum belongs to M(Ω), it only states that the solution belongs to some Lebesgue space, and does not say anything about the gradient of such a solution. In order to prove gradient estimates on the duality solution we have to proceed in a different way. Theorem 4.1. Let µ belong to M(Ω). Then the unique duality solution of (1.8) with datum f belongs to W01,q (Ω), for every q < NN−1 . Proof. Let fn be a sequence of L∞ (Ω) functions which converges to µ in M(Ω), with the property that kfn kL1 (Ω) ≤ kµkM(Ω) , and let un be the unique solution in H01 (Ω) of ( −div(A(x) ∇un ) = fn in Ω, un = 0 on ∂Ω. Let k > 0 and choose v = Tk (un ) as test function of the weak formulation for un . We obtain, recalling that ∇un = ∇Tk (un ) where ∇Tk (un ) 6= 0, and using (1.8), Z Z Z 2 fn Tk (un ) ≤ k kµkM(Ω) , A(x)∇un · ∇Tk (un ) = |∇Tk (un )| ≤ α Ω





where in the last passage we have used that |Tk (un )| ≤ k. Using Sobolev embedding in the left hand side, we have Z  22∗ α 2∗ |Tk (un )| ≤ k kµkM(Ω) . S22 Ω Observing that |Tk (un )| = k on the set An,k = {x ∈ Ω : |un (x)| ≥ k}, we have 2 α 2 k (m(An,k )) 2∗ ≤ k kµkM(Ω) , 2 S2 which implies  kµk  N M(Ω) N −2 m(An,k ) ≤ C , k with C depending only on N and α. Now we fix λ > 0, and we have {|∇un | ≥ λ} = {|∇un | ≥ λ, |un | < k} ∪ {|∇un | ≥ λ, |un | ≥ k}, so that {|∇un | ≥ λ} ⊂ {|∇un | ≥ λ, |un | < k} ∪ An,k .

ELLIPTIC EQUATIONS

33

Since Z k kµkM(Ω) 1 m({|∇un | ≥ λ, |un | < k}) ≤ 2 |∇Tk (un )|2 ≤ , λ Ω λ2 we have  N  kµk k kµkM(Ω) M(Ω) N −2 , m({|∇un | ≥ λ}) ≤ + C λ2 k 1

N −2

N −1 for every k > 0. If we choose k = λ N −1 kµkM(Ω) , the above inequality becomes  kµk  N M(Ω) N −1 . m({|∇un | ≥ λ}) ≤ C λ Let q < NN−1 be fixed, and let t > 0. Then Z Z Z q q |∇un | = |∇un | + |∇un |q Ω {|∇un |
= tq m(Ω) +

t

C(q − 1) kµkM(Ω) . N N − q t N −1 −q N −1

Choosing t = kµkM(Ω) , we obtain Z |∇un |q ≤ Cq kµkqM(Ω) , (4.20) Ω

so that un is bounded in W01,q (Ω), with q < NN−1 . Note that Cq diverges as q tends to NN−1 . Therefore, up to subsequences, un converges to some function uq weakly in W01,q (Ω) and strongly in L1 (Ω). Since un , being a weak solution, is such that Z Z un g = fn v, ∀g ∈ L∞ (Ω), ∀n ∈ N, Ω



we can pass to the limit as n tends to infinity to have Z Z uq g = v dµ, ∀g ∈ L∞ (Ω), Ω



so that uq (which belongs to W01,q (Ω) for some q < NN−1 ) is the duality solution of (1.9) with datum µ. This fact is true for every q < NN−1 , so that uq does not depend on q. It then follows that the duality solution u of (1.9) belongs to W01,q (Ω) for every q < NN−1 . 

34

LUIGI ORSINA

Remark 4.2. If µ = f is a function in L1 (Ω), and fn converges to f strongly in L1 (Ω), we have that fn is a Cauchy sequence in L1 (Ω). Thus, if we repeat the proof of the previous theorem working with un − um , using the linearity of the operator, and “keeping track” of fn − fm , we find that (4.20) becomes Z |∇un − um |q ≤ Cq kfn − fm kqL1 (Ω) , Ω

for every q < NN−1 . Since {fn } is a Cauchy sequence in L1 (Ω), it then follows that un is a Cauchy sequence in W01,q (Ω), for every q < NN−1 . This implies that un strongly converges to the solution u in W01,q (Ω), for every q < NN−1 , so that (up to subsequences) ∇un converges to ∇u almost everywhere in Ω. Remark 4.3. If µ = f is a function in L1 (Ω), and we repeat the proof of the previous theorem working with un − vn , where vn is the solution of (1.9) with a datum gn which converges to f in L1 (Ω), we find as before that Z |∇(un − vn )|q ≤ C kfn − gn kqL1 (Ω) , (4.21) Ω

for every q < NN−1 . Since {fn − gn } tends to zero in L1 (Ω), it then follows that un − vn tends to zero in W01,q (Ω), for every q < NN−1 . In other words, the solution u found by approximation does not depend on the sequence we choose to approximate the datum f . We already knew this fact (since every approximating sequence converges to the duality solution which is unique), but this different proof may be useful if, for example, the differential operator is not linear, but allows to prove (4.21) in some way, so that the concept of duality solution is not available. If the datum f is “more regular”, one expects solutions with an increased regularity. We already know, from Theorem 3.6, that the summability of u increases with the summability of f , but what happens to the gradient? Recall that if the datum f is “regular” (i.e., if it belongs to L2∗ (Ω)), the summability of u increases with that of f , but the gradient of u always belongs to (L2 (Ω))N . Surprisingly, this is not the case for “bad” solutions, as the following theorem shows. Theorem 4.4. Let f be a function in Lm (Ω), 1 < m < 2∗ . Then the ∗ m duality solution of (1.9) belongs to W01,m (Ω), m∗ = NN−m .

ELLIPTIC EQUATIONS

35

Proof. Let fn = Tn (f ), and let un be the unique solution of ( −div(A(x) ∇un ) = fn in Ω, un = 0 on ∂Ω. Since we already know that un will converge to the duality solution of (1.9), it is clear that in order to prove the result it will be enough to ∗ prove an a priori estimate on un in W01,m (Ω). In order to do that, we fix h > 0 and choose ϕh (un ) = T1 (Gh (un )) as test function in the weak formulation for un . If we define Bh = {x ∈ Ω : h ≤ |un | ≤ h + 1}, and Ah = {x ∈ Ω : |un | ≥ h} (for the sake of simplicity, we omit the dependence on n on the sets), we obtain, recalling (1.8), Z Z Z Z 2 α |∇un | ≤ A(x)∇un · ∇ϕh (un ) = fn ϕh (un ) ≤ |f |. Bh





Ak

Let now 0 < λ < 1; we can then write +∞

Z Ω

Z +∞ X 1 |∇un |2 ≤ |∇un |2 λ λ (1 + |u |) (1 + h) n Bh Bh h=0 Z +∞ +∞ Z X X 1 1 |f | = |f | α(1 + h)λ Ah α(1 + h)λ k=h Bk h=0 h=0 k +∞ Z X X 1 |f | = α(1 + h)λ h=0 k=0 Bk Z +∞ Z X 1−λ |f |(1 + |un |)1−λ |f | (1 + k) ≤C ≤C

X |∇un |2 = (1 + |u|)λ h=0 +∞ X ≤

Z

k=0



Bk

Z

(1−λ)m0

≤ C kf kLm (Ω)

(1 + |un |)



1 m0

.



Let now q > 1 be fixed. Then, by Sobolev and H¨older inequality, 1 Sqq

Z

q∗

|un | Ω

 qq∗

q |∇un |q |∇un | = (1 + |un |)λ 2 λ 2q Ω (1 + |un |) ΩZ  2q Z 1− 2q 2 λq |∇un | ≤ (1 + |un |) 2−q λ Ω (1 + |u|) Ω Z  q0 2m (1−λ)m0 ≤ C kf kLm (Ω) (1 + |un |) Ω Z 1− 2q λq × (1 + |un |) 2−q .

Z



q



Z

36

LUIGI ORSINA

We now choose λ and q in such a way that λq (1 − λ)m0 = q ∗ = . 2−q This implies N (2 − q) Nm λ= , q = m∗ = . N −q N −m It is easy to see that 1 < m < 2∗ implies 0 < λ < 1, as desired. We thus have q Z  qq∗ Z 1− 2m Z q∗ q∗ q |un | (1 + |un |) ≤C |∇un | ≤ C kf kLm (Ω) . Ω



Ω q q∗

q 2m

Since > 1− is true (being equivalent to m < N2 ), we obtain ∗ from the first and third term that un is bounded in Lq (Ω) (which ∗∗ is again Lm (Ω), see Theorem 2.5) by a constant depending (among other quantities) on the norm of f in Lm (Ω). Once un is bounded, the boundedness of |∇un | in Lq (Ω) (with q = m∗ ) then follows comparing the second and the third term.  We can now draw the complete picture.

1, NN −ε −1

W0



W01,p (Ω)

(Ω)

H01 (Ω)

∗∗

N

H01 (Ω)

∗∗

L N −2 −ε (Ω)

Lp (Ω)

Lp (Ω)

L∞ (Ω)

Theorem 4.1

Theorem 4.4

Theorem 2.5

Theorem 2.3

1

2N N +2

N 2

p

5. Nonuniqueness for distributional solutions If the datum µ is a measure, we have proved in Theorem 4.1 that the sequence un of approximating solutions is bounded in W01,q (Ω), for every q < NN−1 . Therefore, and up to subsequences, un weakly converges to the solution u in W01,q (Ω), for every q < NN−1 . Choosing a C01 (Ω) test function ϕ in the formulation (1.10) for un , we obtain Z Z A(x)∇un · ∇ϕ = fn ϕ, Ω



which, passing to the limit, yields Z Z A(x)∇u · ∇ϕ = ϕ dµ ∀ϕ ∈ C01 (Ω), Ω



ELLIPTIC EQUATIONS

37

so that u is a solution in the sense of distributions of (1.9). Since the definition of solution in the sense of distributions can always be given (even when the notion of duality solution is unavailable due for example to the operator being nonlinear), one may wonder whether there is a way of proving uniqueness of distributional solutions (not passing through duality solutions). The following example is due to J. Serrin (see [5]). Let ε > 0 and Aε (x) be the symmetric matrix defined by xi xj aεij (x) = δij + (aε − 1) 2 . |x| If aε =

N −1 , ε(N −2+ε)

then the function wε (x) = x1 |x|1−N −ε

is a solution in the sense of distributions of − div(Aε (x) ∇wε ) = 0,

(5.22)

in RN \ {0}.

Indeed, if we rewrite w(x) = x1 |x|α and aij (x) = δij + β

xi xj , |x|2

simple (but tedious) calculations imply wx1 (x) = |x|α + αx21 |x|α−2 ,

wxi (x) = αx1 xi |x|α−2 ,

so that N X

aij (x) wxi (x) = δ1j |x|α + (αβ + α + β)x1 xj |x|α−2 .

i=1

Therefore, (A(x) ∇w)x1 = αx1 |x|α−2 + (αβ + α + β)[2x1 |x|α−2 + (α − 2)x31 |x|α−4 ], and (A(x) ∇w)xj = (αβ + α + β)[x1 |x|α−2 + (α − 2)x1 x2j |x|α−4 ], so that div(A(x)∇w) = x1 |x|α−2 [α + (N − 1 + α)(αβ + α + β)]. Given 0 < ε < 1, if we choose α = 1 − N − ε, and β = have α + (N − 1 + α)(αβ + α + β) = 0,

N −1 ε(N −2+ε)

+ 1, we

38

LUIGI ORSINA

so that w is a solution of (5.22) if x 6= 0. To prove that wε is a solution in the sense of distributions in the whole RN , let ϕ be a function in C01 (Ω), and observe that since |Aε (x)∇wε | belongs to L1 (Ω), we have Z Z ε ε A (x)∇w · ∇ϕ = lim+ Aε (x)∇wε · ∇ϕ. r→0

RN

RN \Br (0)

Using Gauss-Green formula, and recalling that wε is a solution of the equation outside the origin, we have Z Z ε ε A (x)∇w · ∇ϕ = − lim+ ϕ Aε (x)∇wε · ν dσ, r→0

RN

∂Br (0)

where ν is the exterior normal to Br (0), i.e., ν = xr . By a direct computation, x Aε (x)∇wε · = Qx1 |r|α−1 , r N −1 with Q = 1 + αβ + α + β = − ε . Therefore, recalling the value of α, and rescaling to the unit sphere, Z Z N −1 1 ε ε − ϕ A (x)∇w · ν dσ = ϕ(ry)x1 dσ. ε rε ∂B1 (0) ∂Br (0) Using again the Gauss-Green formula, we have Z Z e1 · ∇ϕ(rx) dx, ϕ(ry)x1 dσ = r ∂B1 (0)

B1 (0)

where e1 = (1, 0, . . . , 0). Therefore, since 0 < ε < 1, we have Z Z ε ε 1−ε lim+ ϕ A (x)∇w · ν dσ = lim+ r e1 · ∇ϕ(rx) dx = 0, r→0

∂Br (0)

r→0

B1 (0)

so that wε is a solution in the sense of distributions of −div(Aε ∇wε ) = 0 in the whole RN . Let now Ω = B1 (0) be the unit ball, and let vε be the unique solution of ( −div(Aε (x) ∇v ε ) = div(Aε (x) ∇x1 ) in Ω, vε = 0 on ∂Ω, which exists since div(Aε (x) ∇x1 ) is a regular function belonging to H −1 (Ω) (as can be easily seen). Therefore, the function z ε = v ε + x1 is the unique solution in H 1 (Ω) of the problem ( −div(Aε (x) ∇z ε ) = 0 in Ω, z ε = x1 on ∂Ω,

ELLIPTIC EQUATIONS

39

so that the function uε = wε − z ε is a solution in the sense of distributions of ( −div(Aε (x) ∇uε ) = 0 in Ω, uε = 0 on ∂Ω, which is not identically zero since z ε belongs to H 1 (Ω), while wε belongs N to W01,q (Ω) for every q < qε = N −1+ε . Hence, the problem ( −div(Aε (x) ∇u) = f in Ω, u=0 on ∂Ω, has infinitely many solutions in the sense of distributions, which can be written as u = u + t uε , t in R, where u is the duality solution. One may observe that the solution found by approximation belongs to W01,q (Ω) for every q < NN−1 , while the solution of the above example belongs to W01,q (Ω) for some q < NN−1 , and that we are not allowed to take ε = 0 since in this case aε diverges. Thus one may hope that there is still uniqueness of the solution obtained by approximation. However, it is possible to modify Serrin’s example in dimension N ≥ 3 (see [4]) to find a nonzero solution in the sense of distributions for ( −div(B ε (x) ∇u) = 0 in Ω, u=0 on ∂Ω, which belongs to W01,q (Ω), for every q < NN−1 . Here   x21 x1 x2 (aε − 1) x2 +x2 0   1 + (aε − 1) x21 +x22 1 2     x22 , B ε (x) =  (aε − 1) x21 x22 1 + (aε − 1) x2 +x2 0  x1 +x2   1 2   0 0 I N −2 where I is the identity , and aε is as above, with ε fixed p matrix in R ε 2 2 ε−1 so that w (x) = x1 ( x1 + x2 ) belongs to W 1,q (R2 ) for every q < 2. On the other hand, in dimension N = 2 there is a unique solution in the sense of distributions belonging to W01,q (Ω), for every q < 2. The proof of this fact uses Meyers’ regularity theorem for linear equations with regular data.

Theorem 5.1 (Meyers). Let A be a matrix which satisfies (1.8). Then there exists p > 2 (p depends on the ratio αβ and becomes larger as αβ tends to 1) such that if u is a solution of (1.9) with datum f belonging to L∞ (Ω), then u belongs to W01,p (Ω).

40

LUIGI ORSINA

Theorem 5.2. Let N = 2. Then there exists a unique solution in the sense of distributions of (1.9) such that u belongs to W01,q (Ω), for every q < 2. Proof. Since the equation is linear, it is enough to prove that if u is such that Z A(x) ∇u · ∇ϕ = 0, ∀ϕ ∈ C01 (Ω), Ω

then u = 0. Since u belongs to W01,q (Ω), for every q < 2, it is enough to prove that Z A(x) ∇u · ∇ϕ = 0, ∀ϕ ∈ W01,p (Ω), Ω

for some p > 2, implies u = 0. Let B be a subset of Ω, and let vB be the solution of ( −div(A∗ (x) ∇vB ) = χB in Ω, v=0 on ∂Ω. By Meyers’ theorem, vB belongs to W01,p (Ω), for some p > 2. Hence Z A(x) ∇u · ∇vB = 0, Ω

while, choosing u as test function in the weak formulation for vB (which can be done using a density argument and the regularity of ∇vB ), we have Z Z A∗ (x) ∇vB · ∇u =



u. B

Therefore, Z u = 0,

∀B ⊆ Ω,

B

and this implies u ≡ 0.

 References

[1] L. Boccardo, D. Giachetti, Some remarks on the regularity of solutions of strongly nonlinear problems, and applications, Ricerche Mat., 34 (1985), 309– 323. [2] H. Brezis, Analyse fonctionnelle, Masson, Paris, 1987. [3] M.M. Porzio, A uniqueness result for monotone elliptic problems, C. R. Math. Acad. Sci. Paris, 337 (2003), 313–316. [4] A. Prignet, Remarks on existence and uniqueness of solutions of elliptic problems with right-hand side measures, Rend. Mat., 15 (1995), 321–337. [5] J. Serrin, Pathological solutions of elliptic differential equations, Ann. Scuola Norm. Sup. Pisa Cl. Sci., 18 (1964), 385–387.

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41

[6] G. Stampacchia, Le probl`eme de Dirichlet pour les ´equations elliptiques du second ordre ` a coefficients discontinus, Ann. Inst. Fourier (Grenoble), 15 (1965), 189–258.