Lectures on

Weak Solutions of Elliptic Boundary Value Problems

S. KesavanThe Institute of Mathematical Sciences,

CIT Campus, Taramani,Chennai - 600 113.

e mail: kesh@imsc.res.in

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1 Some abstract variational problems

Most partial differential equations in engineering and physics arise out ofvariational principles. We usually have a class of admissible solutions (say,displacements) and an energy functional associated to these admissible func-tions; we seek to minimize the energy to identify the solution of the prob-lem. The corresponding Euleri-Lagrange equation gives the partial differ-ential equation we started with. In the next section, we will see severalexamples of this general situation. In this section, we will discuss some ab-stract variational problems which will form the basis of our study of ellipticboundary value problems.

One of the classical results in functional analysis is the minimization ofthe norm (or distance) in a Hilbert space.

Theorem 1.1 Let H be a real Hilbert space whose norm and inner productare denoted ‖.‖ and (., .) respectively. Let K ⊂ H be a closed convex subset.Let x ∈ H. Then there exists a unique y ∈ K such that

‖x− y‖ = minz∈K‖x− z‖. (1.1)

Further, y can be characterized as the unique vector in K such that, for allz ∈ K, we have

(x− y, z − y) ≤ 0. (1.2)

Proof: Let d = infz∈K ‖x − z‖ ≥ 0. Let yn be a minimizing sequence inK, i.e. yn ∈ K and ‖x − yn‖ → d. Then it is clear that yn is a boundedsequence in H and so, since H is reflexive, we can extract a weakly convergentsubsequence ynk. Let ynk y weakly in H. Then, since K is closed andconvex, it is also weakly closed and so y ∈ K. Also, since the norm is aweakly lower semi-continuous functional, we have

‖x− y‖ ≤ lim infk→∞

‖x− ynk‖ = d.

Since y ∈ K, we also have ‖x− y‖ ≥ d. Thus we have proved the existenceof a vector y ∈ K satisfying (1.1). If y′ ∈ K also satisfied (1.1), then by theconvexity of K, we have 1

2(y+ y′) ∈ K, and by the parallelogram identity we

have ∥∥∥∥y + y′

2− x∥∥∥∥2

=1

2‖y − x‖2 +

1

2‖y′ − x‖2 −

∥∥∥∥y − y′2

∥∥∥∥2

< d2

which contradicts the definition of d. This proves the uniqueness of y.We will now prove the characterization of y via (1.2). Let x ∈ H and

y ∈ K satisfy (1.1). Let z ∈ K. If 0 < t < 1, then, by the convexity of K, itfollows that tz + (1− t)y ∈ K. Then, by virtue of (1.1), we have

‖x− y‖ ≤ ‖x− (tz + (1− t)y)‖ = ‖(x− y)− t(z − y)‖.

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Thus,‖x− y‖2 ≤ ‖x− y‖2 − 2t(x− y, z − y) + t2‖z − y‖2

which yields

(x− y, z − y) ≤ t

2‖z − y‖2

and (1.2) follows on letting t → 0. Conversely, if x ∈ H and y ∈ K satisfy(1.2), then, for z ∈ K, we have

‖x− y‖2 − ‖x− z‖2 = 2(x− y, z − y)− ‖z − y‖2 ≤ 0

from which it follows that x and y satisfy (1.1).

Remark 1.1 If H = R2 with the usual euclidean inner product and if K isa closed convex subset of the plane, the relation (1.2) can be geometricallyinterpreted as follows: the lines joining the point y with x and with any pointz ∈ K will always contain an obtuse angle. See Figure 1.

K

y = PK(x)x

z

Figure 1

Notation Let H be a real Hilbert space and let K ⊂ H be a closed convexsubset. Let x ∈ H. We will denote the unique vector y ∈ K satisfying(1.1) or, equivalently (1.2), whose existence is guaranteed by the precedingtheorem, by PKx. The mapping PK : H → K is, in general, a nonlinear mapand is usually called the projection of H onto K.

Corollary 1.1 Let H be a Hilbert space and let K ⊂ H be a closed convexsubset. Then, for x1 and x2 in H, we have

‖PKx1 − PKx2‖ ≤ ‖x1 − x2‖. (1.3)

Proof: It follows from (1.2) that

(x1 − PKx1, PKx2 − PKx1) ≤ 0(x2 − PKx2, PKx1 − PKx2) ≤ 0.

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Adding these two relations, we get

((x1 − x2)− (PKx1 − PKx2), PKx2 − PKx1) ≤ 0

which leads to

‖PKx2 − PKx1‖2 ≤ (x2 − x1, PKx2 − PKx1) ≤ ‖x2 − x1‖ ‖Pkx2 − PKx1‖

by the Chauchy-Schwarz inequality, from which (1.3) follows immediately.

Definition 1.1 Let H be a real Hilbert space. and let a : H ×H → R be abilinear form. The bilinear form is said to be continuous if there exists aconstant M > 0 such that for all u and v in H, we have

|a(u, v)| ≤ M‖u‖ ‖v‖.

It is said to be H-elliptic if there exists a constant α > 0 such that for everyv ∈ H, we have

a(v, v) ≥ α‖v‖2.

The bilinear form is said to be symmetric if for every u and v in H, wehave

a(u, v) = a(v, u).

Example 1.1 Let H = RN with the usual euclidean inner product. Thenevery N × N matrix with real entries defines a continuous bilinear form. IfA = (aij), 1 ≤ i, j ≤ N , and if we have u = (u1, · · · , uN) and v = (v1, · · · , vN),then the bilinear form is defined via the relation

a(u, v)def=

N∑i,j=1

aijujvi = vTAu

(if we consider u and v as column vectors and if vT denotes the transpose ofv). By the Cauchy-Schwarz inequality, we have

|a(u, v)| = |vTAu| = |(v, Au)|≤ ‖v‖ ‖Au‖ ≤ ‖A‖ ‖u‖ ‖v‖.

If A is a symmetric and positive definite matrix, then the bilinear form issymmetric and H-elliptic since we know that

N∑ij=1

aijvivj ≥ α‖v‖2

where α > 0 is the smallest eigenvalue of the matrix A.

We will see several examples of elliptic bilinear forms in the next section.

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Theorem 1.2 Let H be a real Hilbert space and let a : H × H → R be acontinuous, symmetric and H-elliptic bilinear form. Let K ⊂ H be a closedconvex set. Let f ∈ H. Then, there exists a unique u ∈ K such that

a(u, v − u) ≥ (f, v − u), for every v ∈ K. (1.4)

Further, u ∈ K can be characterized as the minimizer of J(.) over K, where

J(v) =1

2a(v, v)− (f, v), v ∈ H.

Proof: Define, for u, v ∈ H,

< u, v >def= a(u, v).

Then, by the bilinearity, symmetry and ellipticity of a(., .), this defines aninner product on H. Let the corresponding norm be denoted by ‖.‖a so that,for all v ∈ H, we have

‖v‖2a = a(v, v).

The continuity and ellipticity of the bilinear form yield

α‖v‖2 ≤ ‖v‖2a ≤ M‖v‖2

for every v ∈ H and so the two norms are equivalent and H is a Hilbertspace with the new norm as well. Now, by the Riesz representation theorem,there exists f ∈ H such that for every v ∈ H, we have

a(f , v) = < f, v > = (f, v)

since v 7→ (f, v) defines a continuous linear functional on H with the norm‖.‖a. Then

12‖v − f‖2

a = 12a(v − f , v − f)

= 12a(v, v)− a(v, f) + 1

2a(f , f)

= 12a(v, v)− (f, v) + 1

2‖f‖2

a

= J(v) + 12‖f‖2

a.

Since 12‖f‖2

a is a constant, minimizing J overK is the same as minimizing ‖v−f‖a over K which yields, thanks to the characterization (1.2), the existenceof a unique u ∈ K such that

< f − u, v − u > ≤ 0

for all v ∈ K, which is the same as (1.4).

We now show that we can relax the condition of symmetry of the bilinearform and still have a unique solution to (1.4). Of course, we can no longeridentify the solution as the minimizer of a functional of the form J .

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Theorem 1.3 (Stampacchia) Let H be a real Hilbert space and let K ⊂ Hbe a closed convex set. Let a(., .) be a continuous and H-elliptic bilinear formon H. Given f ∈ H, there exists a unique u ∈ K such that (1.4) is satisfiedfor every v ∈ K.

Proof: Let w ∈ H be a fixed element. By the continuity of a(., .), the mapv 7→ a(w, v) is a continuous linear functional and so there exists (by the Rieszrepresentation theorem) an element, denoted Aw, in H such that

(Aw, v) = a(w, v)

for every v ∈ H. The bilinearity of a(., .) implies the linearity of the mapw 7→ Aw and, the continuity and H-ellipticity of a(., .) imply that

‖Aw‖ ≤ M‖w‖,(Aw,w) ≥ α‖w‖2

(1.5)

for all w ∈ H. Thus A is a bounded linear operator on H. Now, (1.4) canbe writtten as

(Au, v − u) ≥ (f, v − u), for all v ∈ K. (1.6)

Let ρ > 0 be a positive constant, to be chosen presently. Then (1.6) isequivalent to finding u ∈ K such that

(ρf − ρAu+ u− u, v − u) ≤ 0

for all v ∈ K. In other words (cf. (1.2)) we seek u ∈ K such that

u = PK(ρf − ρAu+ u).

Thus we are looking for a fixed point of the map F : H → H (whose rangelies in K) defined by

F (v)def= PK(ρf − ρAv + v).

Now, if v, w ∈ H, we have by Corollary 1.1, that

‖F (v)− F (w)‖ ≤ ‖(v − w)− ρA(v − w)‖.

Thus, by (1.5), we get

‖F (v)− F (w)‖2 ≤ ‖v − w‖2 − 2ρ(A(v − w), v − w) + ρ2‖A(v − w)‖2

≤ (1− 2ρα + ρ2M2)‖v − w‖2.

If we now choose ρ such that 0 < ρ < 2αM2 , then

1− 2ρα + ρ2M2 < 1

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and so F is a contraction mapping and by the contraction mapping theorem,F has a unique fixed point u which must belong to K. This completes theproof.

If K = V , a closed subspace of H, then it is obviously convex. In thatcase, for any w ∈ V , set v = w + u in (1.4). Thus, we get, for every w ∈ V ,

a(u,w) ≥ (f, w).

Since V is a subspace, this inequality also holds when −w replaces w and sowe have

a(u,w) = (f, w), for every w ∈ V. (1.7)

Notice that if we set w = u, then

α‖u‖2 ≤ a(u, u) = (f, u) ≤ ‖f‖ ‖u‖,

which yields

‖u‖ ≤ 1

α‖f‖. (1.8)

Thus, the mapping f 7→ u is a bounded linear operator from H into V . Wehave thus proved the following result.

Theorem 1.4 (Lax-Milgram lemma) Let H be a Hilbert space and let V ⊂ Hbe a closed subspace. Let a(., .) be a continuous and H-elliptic bilinear formon H. Let f ∈ H. Then, there exists a unique u ∈ V such that (1.7) holds.In particular, this is true when V = H itself. The mapping G : H → Vdefined by Gf = u is a bounded linear operator and (1.8) holds. In addition,if a(., .) is symmetric, then u is the minimizer of the functional

J(v) =1

2a(v, v)− (f, v)

over V.

Remark 1.2 In particular, if a(u, v) = (u, v), the inner product on H, itfollows that PV satisfies the relation

(PV f, w) = (f, w)

for every w ∈ V . In this case it is now clear that the projection PV is a linearoperator and is called the orthogonal projection of H onto V .

Remark 1.3 It is a simple exercise, based on the Riesz representation the-orem, to see that in Theorems 1.2-1.4, we can have, on the right-hand sideof (1.4), ϕ(v−u) where ϕ ∈ H ′, the dual of H, in place of (f, v−u), f ∈ H.

Remark 1.4 When the bilinear form a(., .) is symmetric, and when K = H,the global minimum of the corresponding functional J defined in the Lax-Milgram lemma (Theorem 1.4) is attained at u ∈ V which satisfies (1.7).

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These can be considered as the Euler-Lagrange equations of the unconstrainedoptimization problem. When K is a closed convex proper subset, then wehave a constrained optimization problem and we cannot expect the solutionto satisfy equations but only the inequalities (1.4). These are called varia-tional inequalities.

We will conclude this section with yet another abstract variational prob-lem.

If Σ and V are Hilbert spaces, then a bilinear form b : Σ× V → R is saidthe be continuous if there exists a constant M > 0 such that

|b(σ, v)| ≤ M‖σ‖Σ‖v‖V

for every σ ∈ Σ and for every v ∈ V .

Theorem 1.5 (Babuska-Brezzi) Let Σ and V be real Hilbert spaces. Leta : Σ× Σ→ R and b : Σ× V → R be continuous bilinear forms. Let

Z = σ ∈ Σ | b(σ, v) = 0 for every v ∈ V .

Assume that a(., .) is Z-elliptic, i.e. there exists a constant α > 0 such thatfor every σ ∈ Z, we have

a(σ, σ) ≥ α‖σ‖2Σ.

Assume further that the bilinear form b(., .) satisfies the Babuska-Brezzi con-dition (also called the inf-sup condition): there exists β > 0 such that forevery v ∈ V , we have

supτ∈Στ 6=0

b(τ, v)

‖τ‖Σ

≥ β‖v‖V . (1.9)

Let κ ∈ Σ and let ` ∈ V . Then, there exists a unique pair (σ, u) ∈ Σ × Vsuch that

a(σ, τ) + b(τ, u) = (κ, τ)Σ for every τ ∈ Σ,b(σ, v) = (`, v)V for every v ∈ V.

(1.10)

where (., .)Σ and (., .)V denote the inner products in Σ and V respectively.

Proof: As in the case of Theorem 1.3, we can define continuous linear oper-ators A : Σ → Σ and B : Σ → V such that for every σ, τ ∈ Σ and for everyv ∈ V , we have

(Aσ, τ)Σ = a(σ, τ)(Bσ, v)V = b(σ, v).

Then (1.10) is equivalent to solving the system of equations:

Aσ +B∗u = κ,Bσ = `,

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where B∗ : V → Σ is the adjoint of B.By virtue of (1.9), we deduce that for every v ∈ V we have

‖B∗v‖Σ ≥ β‖v‖V .

Thus B∗ has closed range and is injective. Since we are dealing with contin-uous linear operators, it follows that B is surjective. Let us choose σ1 ∈ Σsuch that Bσ1 = `. Now, Z = ker(B) is a closed subspace of Σ and so, bythe Z-ellipticity of a(., .), it follows from the Lax-Milgram lemma (Theorem3.1.4) that there exists a unique σ0 ∈ Z such that

a(σ0, τ) = (κ, τ)Σ − a(σ1, τ) (1.11)

for every τ ∈ Z. Since Z = ker(B), it follows that Bσ0 = 0 and so, ifwe set σ = σ1 + σ0, we still have Bσ = `. It also follows from (1.11) thatκ− Aσ ∈ Z⊥, the orthogonal complement of Z in Σ. But we know that

Range(B∗) = (ker(B))⊥ = Z⊥.

Thus, there exists u ∈ V such that B∗u = κ−Aσ and so (σ, u) solves (1.10).To prove the uniqueness, let, if possible (σ, u) be another solution pair.

Set σ′ = σ − σ and u′ = u− u. Then

a(σ′, τ) + b(τ, u′) = 0, for every τ ∈ Σ,b(σ′, v) = 0, for every v ∈ V.

(1.12)

Thus, σ′ ∈ Z and setting τ = σ′ in the first equation of (1.12), we geta(σ′, σ′) = 0 which yields σ′ = 0. It follows from this that B∗u′ = 0 whichimplies that u′ = 0. This proves the uniqueness of the solution and completesthe proof.

Remark 1.5 As mentioned in Remark 1.3, it is easy to see in the case of(1.10) the right-hand sides (κ, τ)Σ and (`, v)V can be replaced by continu-ous linear functionals ϕ(τ) and f(v) respectively. In this case we will haveA : Σ→ Σ′, B : Σ→ V ′ and B∗ : V → Σ′, where Σ′ and V ′ are the duals ofΣ and V respectively.

Remark 1.6 If the bilinear form a(., .) is symmetric, then we can considerthe constrained optimization problem: find σ ∈ Σ such that Bσ = ` and

J(σ) = minBτ=`

J(τ)

where

J(τ) =1

2a(τ, τ)− (κ, τ)Σ.

Then we introduce the Lagrangian

L(τ, v)def=

[1

2a(τ, τ)− (κ, τ)Σ

]+ [b(τ, v)− (`, v)V ]

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and look for a saddle point. The corresponding set of equations satisfied bythe saddle point will be preciosely the set (1.10).

Another way to look at (1.10) is from the point of view of elasticity. Thefirst equation in (1.10) can be viewed as a constitutive law (for instance astress-strain relationship like Hooke’s law) and the second equation will cor-respond to the equlibrium equation or the equation of balance of forces.

In the next section, we will see numerous applications of the theoremsof Lax-Milgram, Babuska-Brezzi and Stampacchia. While the Lax-Milgramformulation is the starting point of the so called direct finite element meth-ods, the Babuska-Brezzi theorem will provide the basis for the mixed finoiteelement methods.

2 Examples of elliptic boundary value prob-

lems

In this section, we will present several examples of elliptic boundary valueproblems. In each case, we will state what we mean by a weak solution ofthe problem and study its existence and uniqueness using the results of theprevious section.

Throughout this section, unless otherwise specified, Ω will denote a boundedopen set in RN and Γ = ∂Ω will denote its boundary.

2.1 The Dirichlet problem for second order elliptic operators

Consider the following problem:

−∆u = f in Ω,u = 0 on Γ,

(2.1)

where f : Ω → R is a given function. A classical solution of (2.1) whenf ∈ C(Ω), is a function u ∈ C2(Ω) which satisfies this equation pointwise. Ifu is a classical solution, let us multiply the differential equation in (2.1) byϕ ∈ D(Ω) and integrate over Ω to get

−∫

Ω

(∆u)ϕ dx =

∫Ω

fϕ dx.

Applying Green’s theorem (integration by parts) and using the fact that ϕvanishes on the boundary, we get∫

Ω

∇u.∇ϕ dx =

∫Ω

fϕ dx.

Since u ∈ C2(Ω) and u = 0 on Γ, we have that u ∈ H10 (Ω) and also f ∈ L2(Ω).

Further, D(Ω) is dense in H10 (Ω) and both sides of the above equation are

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continuous in ϕ with respect to the H10 (Ω)-topology. Thus, it follows by

density that ∫Ω

∇u.∇v dx =

∫Ω

fv dx, for every v ∈ H10 (Ω). (2.2)

Notice that in (2.2) we do not need any information on the second derivativesof u. Hence we say that if u ∈ H1

0 (Ω) satisfies (2.2), then it is a weak solu-tion of (2.1). We have just seen that every classical solution is automaticallya weak solution. We now show that a weak solution always exists uniquelyfor certain classes of functions f .

Theorem 2.1 Let Ω ⊂ RN be a bounded open set and let Γ = ∂Ω. Letf ∈ L2(Ω). Then there exists a unique weak solution u ∈ H1

0 (Ω) to (2.1).Further, u can be characterized as the minimizer in H1

0 (Ω) of the functionalJ : H1

0 (Ω)→ R defined by

J(v) =1

2

∫Ω

|∇v|2 dx−∫

Ω

fv dx.

Proof: Set V = H10 (Ω) and let

a(u, v) =

∫Ω

∇u.∇v dx.

Since, by Poincare’s inequality, this is an inner product on H10 (Ω) whose norm

|.|1,Ω is equivalent to the usual norm ‖.‖1,Ω, this is a symmetric, continuousand V -elliptic bilinear form on V . Further v 7→

∫Ωfv dx is a continuous

linear functional on v since, again, by Poincare’s inequality, we have∣∣∣∣∫Ω

fv dx

∣∣∣∣ ≤ |f |0,Ω|v|0,Ω ≤ C|f |0,Ω|v|1,Ω.

The result now follows from the Lax-Milgram lemma (Theorem 1.4).

Remark 2.1. The same result is valid when f ∈ H−1(Ω), the dual of H10 (Ω).

We only need to replace∫

Ωfv dx by < f, v >, the duality pairing between

H−1(Ω) and H10 (Ω), i.e. the action of the linear functional f on v, in (2.2).

If ϕ ∈ D(Ω), we get from (2.2) and Green’s theorem that

−∫

Ω

(∆u)ϕ dx =

∫Ω

fϕ dx.

Thus −∆u = f in the sense of distributions. Further, if we know a priorithat a weak solution u belongs to C2(Ω) and if f ∈ C(Ω), then u has tobe a classical solution of (2.1). For, in that case we have u = 0 on Γ andsince D(Ω) is dense in L2(Ω), we deduce that as functions in L2(Ω) we have

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−∆u = f and so they are equal almost everywhere; but since they arecontinuous functions, they are therefore equal pointwise everywhere.

The question, therefore, is: when is a weak solution smooth enough? Thisis answered by a regularity theorem. We merely state for the record that iff ∈ L2(Ω) and if Ω is a ‘reasonably smooth domain’, then the weak solutionu ∈ H1

0 (Ω) is actually in H2(Ω) ∩H10 (Ω). If f ∈ Hm(Ω) and if Ω is of class

Cm+2, then u ∈ Hm+2(Ω) ∩H10 (Ω). Thus, combining such information with

the Sobolev imbedding theorems, we can decide whether a weak solution isclassical or not.

We now turn to the inhomogeneous Dirichlet problem. Let f : Ω → Rand g : Γ→ R be given functions. Consider the problem:

−∆u = f in Ω,u = g on Γ.

(2.3)

As before, would like to look for a weak solution inside H1(Ω). If u ∈ H1(Ω),

then its trace belongs to H12 (Γ). Thus, we assume that g ∈ H 1

2 (Γ) and thatf ∈ L2(Ω). Then, by the trace theorem, there exists u ∈ H1(Ω) such that

u|Γ = γ0(u) = g.

Now define

K = u+H10 (Ω) = v ∈ H1(Ω) | v − u ∈ H1

0 (Ω).

Thus, K is a closed affine subspace of H1(Ω) and we seek u ∈ K. We definea weak solution of (2.3) as an element u ∈ K such that∫

Ω

∇u.∇v dx =

∫Ω

fv dx for every v ∈ H10 (Ω). (2.4)

We leave it to the reader to check that every classical solution is weak andthat every weak solution in C2(Ω), with f ∈ C(Ω) and g ∈ C(Γ), is classical.Choosing v = ϕ ∈ D(Ω), we see that a weak solution satisfies −∆u = f inthe sense of distributions and we have imposed the condition that u = g on Γin the sense of trace. As for the existence of a weak solution u, set u = u+w,where w ∈ H1

0 (Ω). Then, we seek w ∈ H10 (Ω) such that∫

Ω

∇w.∇v dx =

∫Ω

fv dx−∫

Ω

∇u.∇v dx (2.5)

for every v ∈ H10 (Ω). Now the map

v 7→∫

Ω

fv dx−∫

Ω

∇u.∇v dx

defines a continuous linear functional on H10 (Ω) and so can be written as

< F, v >, with F ∈ H−1(Ω). Thus, by the Lax-Milgram theorem or by

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Theorem 2.1, we deduce the existence of a unique w satisfying the aboveequation which leads to the existence of a weak solution u to (2.3).

We now prove the uniqueness of the weak solution (this is not obviousfrom the uniqueness of w, since u is not uniquely defined!). If u1 and u2 aretwo weak solutions to (2.3), then u1 − u2 ∈ H1

0 (Ω) and∫Ω

∇(u1 − u2).∇v dx = 0

for every v ∈ H10 (Ω). Choosing v = u1− u2, we deduce that |u1− u2|1,Ω = 0,

which, by Poincare’s inequality, implies that u1 = u2.We finally show that u depends continuously on the data f and g. Since

the trace map γ0 is surjective, it has a continuous right inverse. Thus thereexists a constant C > 0 and for every g ∈ H

12 (Γ), there exists u ∈ H1(Ω)

such that γ0(u) = g and

‖u‖1,Ω ≤ C‖g‖ 12,Γ.

Setting v = w in (2.5), we get

|w|21,Ω ≤ |f |0,Ω|w|0,Ω + |u|1,Ω|w|1,Ω ≤ (C1|f |0,Ω + ‖u‖1,Ω)|w|1,Ω

using Poincare’s inequality. Thus it follows that

|w|1,Ω ≤ C1|f |0,Ω + ‖u‖1,Ω ≤ C1|f |0,Ω + C|g| 12,Γ.

Again, by Poincare’s inequality, ‖w‖1,Ω ≤ C2|w|1,Ω and so since u = u + w,we deduce that there exists a constant C > 0, depending only on Ω, suchthat

‖u‖1,Ω ≤ C(|f |0,Ω + |g| 12,Γ)

which proves the continuous dependence of u on the data f and g.

Remark 2.2 Existence, uniqueness and continuous dependence on the dataare the criteria for the well-posedness of a problem in the sense of Hadamard.

We can generalize the preceding considerations to cover the case of asecond order elliptic operator. Let aij ∈ C1(Ω), 1 ≤ i, j ≤ N be functionssatisfying the ellipticity condition

N∑i,j=1

aij(x)ξiξj ≥ α|ξ|2 (2.6)

for all x ∈ Ω and for all ξ = (ξ1, · · · , ξN) ∈ RN , where α > 0 is a constantindependent of x and ξ. Let a0 ∈ C(Ω). Consider the problem:

−∑N

i,j=1∂∂xi

(aij

∂u∂xj

)+ a0u = f in Ω,

u = 0 on Γ.

(2.7)

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The differential operator in the above equation is said to be a uniformlyelliptic operator in divergence form.. If f ∈ L2(Ω), a weak solution isdefined as u ∈ H1

0 (Ω) such that∫Ω

N∑i,j=1

aij∂u

∂xj

∂v

∂xidx+

∫Ω

a0uv dx =

∫Ω

fv dx (2.8)

for every v ∈ H10 (Ω). Again, if f ∈ C(Ω), it is easy to check that every

classical solution is a weak solution and also that a smooth weak solution isclassical. If a0(x) ≥ 0 for all x ∈ Ω, then the bilinear form

a(u, v) =

∫Ω

N∑i,j=1

aij∂u

∂xj

∂v

∂xidx+

∫Ω

a0uv dx

is H10 (Ω)-elliptic by virtue of the ellipticity condition (2.6) and Poincare’s

inequality. Thus, by the Lax-Milgram lemma (Theorem 1.4), there exists aunique weak solution in this case. In addition if aij = aji for all 1 ≤ i, j ≤ N ,then a(., .) is symmetric and so the weak solution u minimizes the functionalJ on H1

0 (Ω), where

J(v) =1

2

∫Ω

N∑i,j=1

aij∂v

∂xj

∂v

∂xidx+

1

2

∫Ω

a0v2 dx−

∫Ω

fv dx.

Notice that to define and prove the existence of a weak solution via (2.8), itsuffices that aij, a0 are all in L∞(Ω) and f ∈ H−1(Ω).

More generally, we can consider the following second order elliptic bound-ary value problem:

−∑N

i,j=1∂∂xi

(aij

∂u∂xj

)+∑N

i=1 ai∂u∂xi

+ a0u = f in Ω,

u = 0 on Γ,

(2.9)

Where the aij, 1 ≤ i, j ≤ N satisfy the ellipticity condition (2.6) and ai ∈C(Ω) for 0 ≤ i ≤ N . A weak solution is a function u ∈ H1

0 (Ω) such that

a(u, v) =

∫Ω

fv dx (2.10)

for every v ∈ H10 (Ω), where

a(u, v) =

∫Ω

N∑i,j=1

aij∂u

∂xj

∂v

∂xidx+

∫Ω

N∑i=1

ai∂u

∂xiv dx+

∫Ω

a0uv dx.

This bilinear form is not symmetric in general. If it is H10 (Ω)-elliptic, then

a unique weak solution exists by the Lax-Milgram lemma (Theorem 1.4). Ingeneral, we have the following result.

15

Theorem 2.2 Assume that aij ∈ L∞(Ω) for 1 ≤ i, j ≤ N and satisfy theellipticity condition (2.6). Assume that ai ∈ L∞(Ω) for 0 ≤ i ≤ N . Letf ∈ L2(Ω). When f = 0, the set of weak solutions of (2.9) is a finitedimensional subspace of H1

0 (Ω). Assume that the dimension is d. Then,there exists a d-dimensional subspace F ⊂ L2(Ω) such that (2.10) has asolution if, and only if, f ∈ F⊥, the orthogonal complement of F in L2(Ω).

Proof: Choose λ > 0 such that

a0(x) + λ ≥ γ > 0

for all x ∈ Ω. Let |ai|0,∞,Ω ≤ β for 0 ≤ i ≤ N . Then, for any v ∈ H10 (Ω), we

have

a(v, v) + λ∫

Ωv2 dx ≥ α|v|21,Ω − β|v|1,Ω|v|0,Ω + γ|v|20,Ω

= α|v|21,Ω +

(γ

12 |v|0,Ω − βγ−

12

2|v|1,Ω

)2

− β2

4γ|v|21,Ω

≥(α− β2

4γ

)|v|21,Ω.

We can choose λ large enough so that α− β2

4γ> 0. Then the bilinear form

a(u, v) + λ

∫Ω

uv dx

will be H10 (Ω)-elliptic and given f ∈ L2(Ω), there will be a unique solution

udef= Gf ∈ H1

0 (Ω) such that

a(u, v) + λ

∫Ω

uv dx =

∫Ω

fv

for every v ∈ H10 (Ω). It is clear that the map f 7→ u = Gf is continuous

from L2(Ω) into H10 (Ω) (cf. Theorem 1.4) and since H1

0 (Ω) is compactlyimbedded in L2(Ω) by the Rellich-Kondrosov theorem, it follows that G is acompact operator of L2(Ω) into itself. Now, if u is a weak solution of (2.9),then clearly

u = G(f + λu).

If we set v = f + λu, then

v − λGv = f.

Now, G is compact and λ > 0. So I − λG is invertible unless λ−1 is aneigenvalue of G. Thus, if λ−1 is not an eigenvalue, then a solution existsfor all f (and d = 0). If λ−1 is an eigenvalue, then it has finite geometricmultiplicity, since G is compact. So, if f = 0, the set of solutions is thecorresponding eigenspace, which has finite dimension, say, d. Now, when

16

f 6= 0, by the Fredholm alternative (cf. Kesavan [4]), solutions exist if, andonly if, f satisfies the compatibility condition viz. f ∈ ker(I − λG∗)⊥ andker(I − λG∗) has the same dimension d as ker(I − λG).

Remark 2.3 If we can show that d = 0, then we saw that a weak solutionexists uniquely for all f ∈ L2(Ω). It can be shown, for instance, by meansof a maximum principle that if a0 ≥ 0, then whatever may be ai ∈ L∞(Ω)for 1 ≤ i ≤ N , we have d = 0. Thus without further hypotheses on theai, 1 ≤ i ≤ N ,we have the existence and uniqueness of a weak solution to(2.9) when a0 ≥ 0 (cf. Gilbarg and Trudinger [3]).

2.2 The Neumann problem

Let ν denote the unit outer normal on the boundary Γ of a boundedopen set Ω ⊂ RN . We denote by ∂u

∂νthe exterior normal derivative on Γ of

a smooth function u defined on Ω. We have ∂u∂ν

= ∇u.ν. If f : Ω → R is agiven function, then consider the following boundary value problem:

−∆u+ u = f in Ω,∂u∂ν

= 0 on Γ.

(2.11)

If u is a classical (i.e. smooth enough) solution of (2.11), then, u ∈ H1(Ω)and, by Green’s formula, we have for every v ∈ H1(Ω),∫

Ω

∇u.∇v dx+

∫Ω

uv =

∫Ω

fv dx. (2.12)

If f ∈ L2(Ω), then we define a weak solution of (2.11) as a function u ∈H1(Ω) satisfying (2.12) for every v ∈ H1(Ω). If we set

a(u, v) =

∫Ω

∇u.∇v dx+

∫Ω

uv dx,

then this is just the inner propduct in H1(Ω) and so we trivially have thesymmetry, continuity and H1(Ω)-ellipticity of the bilinear form and the ex-istence of a unique weak solution follows immediately from the Lax-Milgramlemma (Theorem 1.4). Further, u will minimize the functional

J(v) =1

2

∫Ω

|∇v|2 dx+1

2

∫Ω

|u|2 dx−∫

Ω

fv dx

over all of H1(Ω).Now let u be a weak solution which also belongs to H2(Ω). Then, by

retracing the passage from (2.11) to (2.12), we get∫Ω

(−∆u)v dx+

∫Ω

uv dx+

∫Γ

∂u

∂νv dσ =

∫Ω

fv dx,

17

where ∂u∂ν

is nothing but the trace γ1(u) which is well defined since u ∈ H2(Ω).If we choose v ∈ D(Ω), then the integral on Γ will vanish and we deduce fromthis that −∆u + u = f in the sense of distributions. Since D(Ω) is dense inL2(Ω), and since all the integrals over Ω are continuous with respect to v inthe L2(Ω)-topology, we also deduce that this equality holds in the sense offunctions in L2(Ω). Consequently the above relation be rewritten as∫

Ω

(−∆u+ u− f)v dx+

∫γ

∂u

∂νv dσ = 0

for all v ∈ H1(Ω) and this now reduces to∫Γ

∂u

∂νv dσ = 0

for all v ∈ H1(Ω). But v|Γ = γ0(v) ∈ H 12 (Γ). Since γ0 is surjective onto this

space which is dense in L2(Γ), we deduce from this that

γ1(u) =∂u

∂ν= 0

as an element of L2(Γ). Thus, if u ∈ H2(Ω) ⊂ H1(Ω) is a weak solution, itsatisfies the differential equation of (2.11) in the sense of distributions andthe boundary condition in the sense of trace. If, further, u ∈ C2(Ω) (andf ∈ C(Ω)), then u will be a classical solution of (2.11).

Remark 2.4 We must emphasize here the important difference between theDirichlet and Neumann problems. In case of the former, the boundary con-dition u = 0 (or u = g) had to be imposed a priori in the space of admissiblefunctions where we seek a (weak) solution. On the other hand, for the Neu-mann problem, we imposed no such condition on the function space and itturns out that the boundary condition emerges as a natural consequence ofthe weak formulation. Dirichlet type boundary conditions are referred to asessential boundary conditions and have to be taken into account while for-mulating the problem in the weak sense while Neumann type boundary con-ditions are called natural boundary conditions and take care of themselves.

We can also study the Neumann problem for a general second order uni-formly elliptic operator as in (2.7) or (2.9). In this case, the natural boundarycondition will take the form

N∑i,j=1

aij∂u

∂xjνi = 0.

The quantity on the left-hand side of the above equation is called the conor-mal derivative associated to the differential operator and if A = (aij) is thematrix of the coefficients of the highest order terms, this is denoted ∂u

∂νA.

18

Let us now look at the inhomogeneous Neumann problem. Let f : Ω→ Rand g : Γ→ R be given functions. Then we look for a function u such that

−∆u+ u = f in Ω,∂u∂ν

= g on Γ.

(2.13)

We will look for a weak solution in H1(Ω) and so we need that g ∈ H− 12 (Γ).

If < ., . >Γ denotes the duality pairing between H−12 (γ) and H

12 (Γ), then a

weak solution of the inhomogeneous Neumann problem (2.13) is a functionu ∈ H1(Ω) such that∫

Ω

∇u.∇v dx+

∫Ω

uv dx =

∫Ω

fv dx+ < g, v >Γ, (2.14)

for every v ∈ H1(Ω). If g ∈ L2(Γ), then we can replace < g, v >Γ by∫

Γgv dσ.

Again, it is easy to see, from the Lax-Milgram lemma, that a unique solutionalways exists to (2.14). If u ∈ H2(Ω) is a weak solution, then as beforewe can deduce that −∆u + u = f in the sense of distributions and thatγ1(u) = ∂u

∂ν= g. For a weak solution to belong to H2(Ω), we must obviously

have f ∈ L2(Ω) and g ∈ H 12 (Γ).

If we consider the Neumann problem without the lower order term, viz.

−∆u = f in Ω,∂u∂ν

= g on Γ,

then its weak formulation would be to look for u ∈ H1(Ω) such that∫Ω

∇u.∇v dx =

∫Ω

fv dx+

∫Γ

gv dσ,

assuming g ∈ L2(Γ). If f ∈ L2(Ω), then a weak solution will exist if, andonly if, f and g satisfy the following compatibility condition:∫

Ω

f dx+

∫γ

g dσ = 0.

That it is necessary comes from integrating both sides of the differentialequation over Ω and using Green’s theorem. Conversely, from the theory ofthe Fredholm alternative (similar to the argument in the proof of Theorem2.2), it follows that the above condition is sufficeint as well (since we canshow that the solution space when f = 0 and g = 0 is one dimensional andconsists of constant functions) and in this case we have an infinity of weaksolutions, since adding a constant function to an arbitrary solution producesanother one. Thus, there exists a unique solution which is orthogonal to thesubspace of constant functions in L2(Ω), i.e. satisfying∫

Ω

u dx = 0.

19

In the same vein, we can consider weak formulations for other types ofboundary value problems for second order elliptic operators. For instance,we can consider a Robin condition which is of the form ∂u

∂ν+αu = 0 on Γ for

some constant α > 0. We can also consider an oblique derivative problem i.e.with α1

∂u∂ν

+ α2∂u∂τ

= 0 on Γ, where α1 and α2 are constants and ∂u∂τ

denotesthe tangential derivative along the boundary when Ω ⊂ R2. Some of thesewill figure in the exercises at the end of this chapter. Another problem is themixed problem of the following form:

−∆u = f in Ω,u = 0 on Γ1,∂u∂ν

= 0 on Γ2,

where Γ = Γ1 ∪ Γ2 with Γ1 ∩ Γ2 = ∅. If the surface measure of Γ1 is strictlypositive, then Poincare’s inequality is still available for the space

V = v ∈ H1(Ω) | v|Γ1 = 0.

Thus we can apply the Lax-Milgram lemma to the problem: find u ∈ V suchthat for every v ∈ V , ∫

Ω

∇u.∇v dx =

∫Ω

fv dx.

This is the weak formulation of the mixed problem stated above (check!).Notice that the essential condition v = 0 on Γ1 was imposed on the space Vwhile nothing was done about the natural condition ∂u

∂ν= 0 on Γ2.

2.3 The biharmonic operator

The biharmonic operator is given by ∆2, where, as usual, ∆ is the Laplaceoperator. This is a differential operator of the fourth order. The Dirichletproblem for the biharmonic operator is the following:

∆2u = f in Ω,u = ∂u

∂ν= 0 on Γ.

(2.15)

Let f ∈ L2(Ω). If ϕ ∈ D(Ω), then multiplying both sides of the differentialequation by ϕ and repeatedly using Green’s formula, we get∫

Ω

∆u∆ϕ dx =

∫Ω

fϕ dx. (2.16)

For (2.16) to make sense, it is sufficient that u ∈ H2(Ω) and in that caseboth the traces u|Γ = γ0(u) and ∂u

∂ν|Γ = γ1(u) are both well defined. Since we

require them both to vanish, we have by the trace theorem that u ∈ H20 (Ω).

Further, D(Ω) is dense in H20 (Ω) and both sides of (2.16) are continuous with

respect to ϕ in the H20 (Ω)-topology. Thus (2.16) holds for all ϕ ∈ H2

0 (Ω).

20

Thus a weak solution of (2.15) is a function u ∈ H20 (Ω) which satisfies

(2.16) for all ϕ ∈ H20 (Ω).

Consider the bilinear form on H20 (Ω) given by

a(u, v) =

∫Ω

∆u∆v dx.

This is clearly continuous, since by the Cauchy-Schwarz inequality, we have

|a(u, v)| ≤ |∆u|0,Ω|∆v|0,Ω ≤ C‖u‖2,Ω‖v‖2,Ω.

Further, we know that u 7→ |∆u|0,Ω defines a norm on H20 (Ω) equivalent to

the usual norm (Exercise!). So

a(v, v) = |∆u|20,Ω ≥ α‖u‖22,Ω

and so a(., .) is H20 (Ω)-elliptic. Hence, by the Lax-Milgram lemma, there

exists a unique weak solution which also minimizes the functional

J(v) =1

2

∫Ω

|∆v|2 dx−∫

Ω

fv dx

over all of H20 (Ω). Again, we have continuous dependence on the data since

α‖u‖22,Ω ≤

∫Ω

|∆u|2 dx =

∫Ω

fu dx ≤ |f |0,Ω|u|0,Ω ≤ |f |0,Ω‖u‖2,Ω

whence it follows that

‖u‖2,Ω ≤1

α|f |0,Ω.

As usual, if u is a weak solution, we have ∆2u = f in the sense of distri-butions. If u ∈ H4(Ω) ∩ H2

0 (Ω), then ∆2u = f as L2(Ω) functions. If, inaddition, we know that u ∈ C4(Ω), then f ∈ C(Ω) and u will be a classicalsolution.

Remark 2.5 The weak formulation also makes sense if f ∈ H−2(Ω), in whichcase we replace

∫Ωfϕ by < f, ϕ >, the duality bracket between H−2(Ω) and

H20 (Ω).

Remark 2.6 While the Dirichlet problem for the Laplace operator could beused to model the behaviour of a membrane fixed along the boundary andacted upon by a vertical force when Ω ⊂ R2. In the same way, the Dirichletproblem for the biharmonic operator describes the bending of a thin elas-tic plate which is clamped along the boundary and acted upon by a verticalforce. The plate itself is a three-dimensional body and is approximated byits middle surface which occupies the region Ω ⊂ R2. If we wish to considera plate which is simply supported on the boundary and is fixed along it, thenthe boundary condition u = 0 will be retained but the condition ∂u

∂ν= 0

21

will be replaced by another boundary condition(which will correspond to thenotion of a natural boundary condition for the biharmonic operator).

Let us now describe another weak formulation of the Dirichlet problemfor the biharmonic operator when f ∈ L2(Ω). Set σ = −∆u ∈ L2(Ω). Thenwe can write ∫

Ω

στ dx+

∫Ω

(∆u)τ dx = 0 (2.17)

for every τ ∈ L2(Ω). Then (2.16) becomes

−∫

Ω

σ∆v dx =

∫Ω

fv dx (2.18)

for every v ∈ H20 (Ω). If we set Σ = L2(Ω) and V = H2

0 (Ω) and define

a(σ, τ) =∫

Ωστ dx for every σ, τ ∈ Σ,

b(τ, v) =∫

Ω(∆v)τ dx for every τ ∈ Σ and v ∈ V,

then the system (2.17)-(2.18) is as in the Babuska-Brezzi theorem (Theorem1.5). Indeed, the bilinear form a(., .) is L2(Ω)-elliptic and so trivially Z-elliptic for any closed subspace Z of Σ. Also,

supτ∈Στ 6=0

b(τ, v)

|τ |0,Ω≥∫

Ω∆v∆v dx

|∆v|0,Ω= |∆v|0,Ω ≥ α‖v‖2,Ω

for every v ∈ V and so the Babuska-Brezzi condition is also verified. Thusthere exists a unique pair (σ, u) ∈ Σ × V satisfying (2.17)-(2.18). Clearly,then u ∈ H2

0 (Ω) and σ = −∆u and so u satisfies (2.16).

This formulation has no intrinsic value as we seem only to have increasedthe number of unknowns. Let us now describe another formulation of thesame problem. Let Σ = H1(Ω) and let V = H1

0 (Ω). Define

a(σ, τ) =∫

Ωστ dx for all σ, τ ∈ Σ,

b(σ, v) = −∫

Ω∇τ.∇v dx for all τ ∈ Σ, v ∈ V .

Consider the problem: find (σ, u) ∈ Σ× V such that

a(σ, τ) + b(τ, u) = 0 for every τ ∈ Σ,

−b(σ, v) =∫

Ωfv dx for every v ∈ V .

(2.19)

In this case also b(., .) satisfies the Babuska-Brezzi condition. Indeed,

supτ∈Στ 6=0

b(τ, v)

‖τ‖1,Ω

≥∫

Ω∇v.∇v dx‖v‖1,Ω

=|v|21,Ω‖v‖1,Ω

≥ β‖v‖1,Ω

22

by Poincare’s inequality. Unfortunately, the bilinear form a(., .) is not Z-elliptic since we cannot expect the L2(Ω)-norm to majorize the H1(Ω)-norm.Thus, we cannot directly apply Theorem 1.5 to prove the existence of asolution. However, it is easy to check that if a solution exists, then it isunique. We now prove the following result.

Theorem 2.3 Assume that the weak solution u ∈ H20 (Ω) of (2.16) satisfies

the regularity condition u ∈ H3(Ω). Then (−∆u, u) = (σ, u) ∈ Σ× V is theunique solution of (2.19).

Proof: Clearly (σ, u) = (−∆u, u) satisfies (2.17)-(2.18). But if τ ∈ H1(Ω),then

b(τ, u) =

∫Ω

(∆u)τ dx = −∫

Ω

∇u.∇τ dx = b(τ, u).

Also, we have

−∫

Ω

σ∆v dx =

∫Ω

fv dx

for every v ∈ H20 (Ω) which yields∫

Ω

∇σ.∇v dx =

∫Ω

fv dx.

In particular, this is true for every v ∈ D(Ω). But D(Ω) is dense in H10 (Ω)

and both sides of the above relation are continuous with respect to v for theH1

0 (Ω)- topology. So the above relation also holds for all v ∈ H10 (Ω). Thus,

σ = σ = −∆u and u = u satiusfy (2.19). This completes the proof.

Remark 3.2.7 This latter formulation is very useful from the computationalpoint of view, as we shall see later. It is due to Ciarlet and Raviart [1]. Itdepends on the regularity result that a weak solution of (2.15) belongs toH3(Ω). If Ω is of class C∞, we in fact have that u ∈ H4(Ω) when f ∈ L2(Ω).If Ω is a polygon, or if it is a Lipschitz domain, then we do have u ∈ H3(Ω)by a result of Kondrat’ev [5].

3 Eigenvalue problems

Let Ω ⊂ RN be a bounded open set with boundary Γ. In this section we willconsider the eigenvalue problem: find λ ∈ R and function(s) u 6≡ 0 such that

−∆u = λu in Ω,u = 0 on Γ.

(3.1)

Given a solution pair (λ, u) of the above problem, we say that λ is an eigen-value and u is an eigenfunction of the Laplace operator (with homogeneousDirichlet boundary conditions). Notice that if u1 and u2 are eigenfunctionsfor an eigenvalue λ, then so is αu1 + βu2 for any constants α and β. Thus,

23

the set of all eigenfunctions corresponding to an eigenvalue, together with thenull function, forms a vector space and this space is called the eigenspacecorresponding to the eigenvalue λ.

Remark 3.1 In a manner analogous to (3.1), we can pose eigenvalue prob-lems for other (homogeneous) boundary conditions. We can also pose theproblem for other elliptic operators. The results in this section give a flavourof the kind of properties that eigenvalues and eigenfunctions of elliptic opera-tors generally enjoy. Of course, depending on the operator and the boundaryconditions, the validity of the various properties will differ.

Theorem 3.1 There exists an orthonormal basis wn of L2(Ω) and a se-quence of positive real numbers λn, with λn → ∞ as n → ∞, such thatwn ∈ H1

0 (Ω) ∩ C∞(Ω),

0 < λ1 ≤ λ2 ≤ · · · ≤ λn ≤ λn+1 ≤ · · · ,

and−∆wn = λnwn in Ω.

Further, the dimension of the eigenspace of each λn is finite.

Proof: Given f ∈ L2(Ω), define Gf ∈ H10 (Ω) as the weak solution of the

problem: −∆u = f , in Ω and u = 0 on Γ. Thus, for every v ∈ H10 (Ω), we

have ∫Ω

∇(Gf).∇v dx =

∫Ω

fv dx.

Then G : L2(Ω) → H10 (Ω) is a continuous linear map. Since Ω is bounded,

H10 (Ω) is compactly imbedded in L2(Ω) by the Rellich-Kondrosov theorem

and so after composition by this inclusion, we can consider G as a compactmap from L2(Ω) into itself (with range contained in H1

0 (Ω)). This map isself-adjoint since, for arbitrary f and g in L2(Ω), we have∫

Ω

(Gf)g dx =

∫Ω

∇(Gf).∇(Gg) dx =

∫Ω

f(Gg) dx.

Further, ∫Ω

(Gf)f dx = =

∫Ω

|∇(Gf)|2 dx = |Gf |21,Ω

which is strictly positive if f 6= 0.It now follows from the theory of compact self-adjoint and positive definite

operators on a separable Hilbert space that there exists an orthonormal basiswn of eigenfunctions and a family µn of eigenvalues, with finite geometricmultiplicity, decreasing to zero as n → ∞ such that Gwn = µnwn (see, forinstance, Kesavan [4]). Thus, clearly wn ∈ H1

0 (Ω). Since G is positivedefinite, we have µn 6= 0 for all n. Setting λn = µ−1

n , we get

wn = G(λnwn)

24

and so for every v ∈ H10 (Ω), we have∫Ω

∇wn.∇v dx = λn

∫Ω

wnv dx

which is the weak formulation of (3.1). In particular, wn = 0 on Γ andsatisfies −∆wn = λnwn in the sense of distributions.

Finally, let x ∈ Ω and let r > 0 be such that the ball B(x; r) ⊂ Ω.By an interior regularity result, we get that wn ∈ H2(B(x; r)) since wn ∈L2(B(x; r)) and −∆wn = λnwn in this ball. This, in turn, now implies thatwn ∈ H4(B(x; r)) and so on. Thus wn ∈ Hk(B(x; r)) for all positive integersk and, by the Sobolev imbedding theorem, it follows that wn ∈ C∞(B(x; r)).Since x ∈ Ω was arbitrarily chosen, it follows that wn ∈ C∞(Ω). This com-pletes the proof.

Remark 3.2 If Ω is of class C∞, then wn ∈ C∞(Ω) for all positive integersn.

Remark 3.3 If H10 (Ω) is provided with the inner-product

(u, v) =

∫Ω

∇u.∇v dx,

then λ−12

n wn is an orthonormal basis for H10 (Ω). Indeed,

1√λkλm

∫Ω

∇wk.∇wm dx =

(λkλm

) 12∫

Ω

wkwm dx = δkm.

Further, if u ∈ H10 (Ω) is such that (u,wm) = 0 for all m, then

0 =

∫Ω

∇u.∇wm dx = λm

∫Ω

uwm dx

and so∫

Ωuwm dx = 0 for all m. Since wm is an orthonormal basis for

L2(Ω), it follows that u = 0. This proves the claim.It is now easy to see that the Fourier expansion of a function u ∈ H1

0 (Ω)is given by

u =∞∑n=1

(∫Ω

uwn dx

)wn

in both the spaces L2(Ω) and H10 (Ω).

Example 3.1 let N = 1 and let Ω = (0, 1). Then the problem (3.1) readsas:

−u′′ = λu in (0, 1),u(0) = u(1) = 0.

25

Multiplying the differential equation by u and integrating by parts, using theboundary condition, we get∫ 1

0

|u′|2 dx = λ

∫ 1

0

|u|2 dx.

Thus, it follows that λ ≥ 0. But if λ = 0, u′ = 0 and so u is a constant andmust vanish since it vanishes at the end points. Thus λ > 0. Now, a generalsolution of the differential equation u′′ + λu = 0, for positive λ is given by

u(x) = A cos√λx+B sin

√λx.

Since u(0) = 0, we get A = 0. Thus B 6= 0 and since u(1) = 0, it followsthat

√λ is an integer multiple of π. Thus the eigenvalues are n2π2∞n=1 and

the corresponding eigenfunctions are constant multiples of wn(x) = sinnπx.We know from the theory of Fourier series that

√2 sinnπx∞n=1 forms an

orthonormal basis for L2(0, 1).

Example 3.2 Let N = 2 and let Ω = (0, 1)× (0, 1). It is easy to see that ifΛnm = (n2 +m2)π2 and if Wnm = sinnπx sinmπy, then

−∆Wnm = ΛnmWnm

in Ω and that Wnm vanishes on the boundary of Ω. Thus Λnm are all eigenval-ues with corresponding eigenfunctions Wnm. Are there any other eigenvalues?It can be shown that 2Wnm∞n,m=1 is an orthonormal basis for L2(Ω) andso there can be no further eigenfunctions. Thus these are all the eigenvaluesand eigenfunctions in this case.

Remark 3.4 When we numbered the eigenvalues, we wrote the sign ‘≤’between them. In the case of Example 3.1, all these eigenvalues were dis-tinct. However, in the case of Example 3.2, we have λ1 = Λ11 = 2π2, whileλ2 = λ3 = Λ12 = Λ21 = 5π2. In the latter case, the eigenspace is two-dimensional and is spanned by the functions sin πx sin 2πy and sin 2πx sinπy.Thus when we number the eigenvalues in increasing order, we repeat the sameeigenvalue as many times as the dimension of the eigenspace, and this num-ber is called its geometric multiplicity.

We now give a variational characterization of the eigenvalues and eigen-functions obtained in Theorem 3.1. We define, for v ∈ H1

0 (Ω), v 6= 0, theRayleigh quotient R(v) by

R(v)def=

∫Ω∇v.∇v dx∫Ωv2 dx

.

We denote by Vm the space spanned by the firstm eigenfunctions w1, · · · , wm.By the orthonormality of the eigenfunctions, they are linearly independentand so dim(Vm) = m.

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Theorem 3.2 Let m be an aribtrary positive integer. Then λm = R(wm).Further,

λm = maxv∈Vmv 6=0

R(v) (3.2)

= minv⊥Vm−1v 6=0

R(v) (3.3)

= minW⊂H1

0(Ω)

dim(W )=m

maxv∈Wv 6=0

R(v). (3.4)

In particular,λ1 = min

v∈H10(Ω)

v 6=0

R(v) (3.5)

Proof: If (λ, u) is an eigen pair, then for every v ∈ V , we have∫Ω

∇u.∇v dx = λ

∫Ω

uv dx. (3.6)

It is now obvious from this that λm = R(wm) for any positive integer m. Letv ∈ Vm. Then

v =m∑k=1

αkwk,

where αk =∫

Ωvwk dx. Using the orthonormality of the wk, we get

R(v) =

∑mk=1 λkα

2k∑m

k=1 α2k

≤ λm

since the λk are increasing with k. Thus the maximum value of R(v) as vvaries over Vm is less than, or equal to, λm. Since wm ∈ Vm and R(wm) = λm,we get (3.2).

Now, let v ⊥ Vm−1 (this orthogonality holds in L2(Ω) as well as in H10 (Ω)

simultaneously). Then

v =∞∑k=m

αkwk = lim`→∞

∑k=m

αkwk.

If we set v` =∑`

k=m αkwk, then v` → v as `→∞ both in H10 (Ω) and L2(Ω).

Thus, it follows that R(v`)→ R(v). Now,

R(v`) =

∑`k=m λkα

2k∑`

k=m α2k

≥ λm

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and so R(v) ≥ λm as well. Again wm ⊥ Vm−1 and this establishes (3.3). Inparticular m = 1 gives (3.5).

Finally, let W be an m-dimensional subspace of H10 (Ω). Then obviously

W ∩ V ⊥m−1 6= 0 (why?). Thus, there exists w 6= 0, w ∈ W ∩ V ⊥m−1. Then by(3.3), R(w) ≥ λm and so

maxv∈Wv 6=0

R(v) ≥ R(w) ≥ λm.

Again (3.4) follows since dim(Vm) = m and on this subspace, the maximumvalue of the Rayleigh quotient is indeed λm. This completes the proof.

Remark 3.5 The characterization (3.4) is intrinsic in the sense that it doesnot depend on the choice of eigenfunctions.

We will now prove an important property of the eigenpair (λ1, w1).

Lemma 3.1 Let w ∈ H10 (Ω), w 6= 0 be such that R(w) = λ1. Then w is an

eigenfunction corresponding to λ1.

Proof: Let v ∈ H10 (Ω) be arbitrarily chosen. Let t > 0. Then w+tv ∈ H1

0 (Ω)and by virtue of (3.5) we have that R(w+ tv) ≥ λ1 = R(w). By normalizing,we can assume that |w|20,Ω = 1. Then∫

Ω∇(w + tv).∇(w + tv) dx∫

Ω(w + tv)2 dx

≥∫

Ω

∇w.∇w dx = λ1.

Cross multiplying and simplifying, we get

t2∫

Ω

∇v.∇v dx+ 2t

∫Ω

∇w.∇v dx ≥ λ1

(2t

∫Ω

wv dx+ t2∫

Ω

v2 dx

).

Dividing throughout by 2t and letting t→ 0, we get∫Ω

∇w.∇v dx ≥ λ1

∫Ω

wv dx.

We can replace v by −v and get the reverse inequality as well and so wsatisfies (3.6) and this completes the proof.

Theorem 3.3 Let Ω ⊂ RN be an open and connected set. Then λ1 is asimple eigenvalue (i.e. dim(V1) = 1) and any eigenfunction corresponding toit does not change sign in Ω. In particular, we can always choose w1 to bestrictly positive in Ω.

Proof: Let w be any eigenfunction corresponding to λ1. Then we know thatw+ and w− also belong to H1

0 (Ω). Setting v = w+ and v = w− in the weakform ∫

Ω

∇w.∇v dx = λ1

∫Ω

wv dx,

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we easily deduce that

R(w+) = R(w−) = λ1.

Consequently, by the preceding lemma, both w+ and w− will be eigenfunc-tions, if they are non-zero and hence will also be in C∞(Ω). But since−∆w+ = λ1w

+ ≥ 0, it follows from the strong maximum principle thatw+ ≡ 0 or w+ > 0 in Ω. The same conclusion holds for w− as well. But bothw+ and w− cannot be simultaneously non-zero. Thus w = w+ or w = w−

and so w cannot change sign in Ω. This then implies that we cannot havetwo mutually orthogonal eigenfunctions for λ1 and so λ1 is simple and wecan always choose an eigenfunction which is strictly positive in Ω. This com-pletes the proof.

Remark 3.6 This theorem can also be proved as a consequence of the Krein-Rutman theorem. In finite dimensional spaces, we have an analogous resultknown as the Perron-Frobenius theorem which states that for a non-negativeirreducible matrix, the spectral radius is a simple eigenvalue and that thereexists a strictly positive eigenvector. Here, the operatorG defined in Theorem3.1 can be shown to have analogous properties in infinite dimensions and soits spectral radius µ1 is a simple eigenvalue with eigenfucntions of constantsign. The advantage of the above proof is that it easily extends to any ellipticeigenvalue problem of the form: find (λ, u) ∈ R× (H1

0 (Ω)\0) such that forall v ∈ H1

0 (Ω),∫Ω

N∑i,j=1

aij∂u

∂xj

∂v

∂xidx+

∫Ω

a0uv dx = λ

∫Ω

uv dx

with the matrix (aij) being symmetric and satisfying the ellipticity condition(2.6) and a0 ≥ 0 so that the corresponding Rayleigh quotient characteriza-tion and the strong maximum principle work.

Remark 3.7 To apply the strong maximum principle we also need that anyeigenfunction u satisfies u ∈ C(Ω). This is true even for Lipschitz domainswhen N ≤ 3 since u ∈ H2(Ω) by the regularity theorem which is imbeddedin C(Ω) when N ≤ 3 by the Sobolev imbedding theorem. Otherwise, we needto assume more smoothness of the boundary so that we can successively ap-ply the regularity theorem to ultimately get u ∈ C(Ω) by Sobolev inclusion.

Remark 3.8 Just as the stationary problem −∆u = f in Ω and u = 0 on Γmodels the displacement of a membrane fixed along Γ and acted upon by aforce f , the eigenvalue problem describes the vibration of a membrane fixedalong Γ, when Ω ⊂ R2. The harmonics of a membrane fixed along Γ aregiven by wm(x) and

√λmt.

Theorem 3.4 (Monotonicity of the spectrum with respect to the domain)Let Ω1 and Ω2 be two bounded domains in RN such that Ω1 ⊂ Ω2. Let

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λn(Ωi), i = 1, 2, denote the sequence of eigenvalues of the Laplace opera-tor with homogeneous Dirichlet boundary conditions, numbered in increasingorder of magnitude, for the domains Ωi, i = 1, 2. Then, for every positiveinteger n, we have

λn(Ω1) ≥ λn(Ω2).

Proof: Observe that if u ∈ H10 (Ω1), then its extension by zero outside Ω1,

denoted u, is in H1(RN) and its restriction to Ω2 is in H10 (Ω2). Further∫

Ω1

|∇u|2 dx =

∫Ω2

|∇u|2 dx and

∫Ω1

u2 dx =

∫Ω2

u2 dx.

The result is now an immediate consequence of the min-max characterizationof eigenvalues given by (3.4).

We saw that any eigenfunction corresponding to the first eigenvalue λ1

has to be of constant sign. Consequently, since eigenfunctions of differenteigenvalues are orthogonal to each other in L2(Ω), it follows that any eigen-function of an eigenvalue λn, where n ≥ 2, must change sign in Ω. A nodalline of an eigenfunction is a curve in Ω, other than Γ, along which an eigen-function vanishes. A nodal domain is a connected open subset of Ω wherethe eigenfunction has a constant sign. Thus, an eigenfunction corrspondingto λ1 has exactly one nodal domain, viz. Ω itself and has no nodal lines.

Proposition 3.1 Let λk, k ≥ 2 be an eigenvalue such that λk < λk+1. Thenan eigenfunction u of λk has at most k nodal domains.

Proof: Let u be an eigenfunction for λk with ` nodal domains denotedΩi, 1 ≤ i ≤ `. In each Ωi, the function u satisfies −∆u = λku and it vanisheson the boundary of this domain. Define, for 1 ≤ i ≤ `,

ui =

u|Ωi in Ωi,0, outside Ωi.

Then ui ∈ H10 (Ω) and clearly

∫Ωuiuj dx = 0 whenever i 6= j. Thus the

dimension of the space V = spanu1, · · · , u` is `. Now, since −∆ui = λkui

in Ωi and ui = 0 on ∂Ωi, we have∫Ωi

|∇ui|2 dx = λk

∫Ωi

(ui)2 dx

and so ∫Ω

|∇ui|2 dx = λk

∫Ω

(ui)2 dx

for each 1 ≤ i ≤ `. If v =∑`

k=1 αkuk is an arbitrary element of V , then it

is immediate to verify that R(v) = λk as well. Thus, since dim(V ) = `, itfollows from (3.4) that λ` ≤ λk. Since λk+1 > λk, it follows that ` ≤ k, which

30

completes the proof.

Remark 3.9 A slight refinement of the above proof, involving subtler prop-erties of solutions to elliptic equations, implies in fact that for every positiveinteger k, an eigenfunction corresponding to the eigenvalue λk has at most knodal domains. Thus the additional hypothesis λk < λk+1 is not necessary.This is the famous Courant’s nodal line theorem and a proof can be found inCourant and Hilbert [2], Volume I. In particular, an eigenvalue correspond-ing to λ2 must have at least 2 nodal domains since it is orthogonal to anyeigenfunction of λ1 (which will be of constant sign) and, on the other hand, itwill have at most 2 nodal domains, by Courant’s theorem. Thus it will haveexactly two nodal domains. This is not true in general for an eigenfunctioncorresponding to λk when k ≥ 3. For example, in the unit square in R2, wehave λ2 = λ3 and so eigenfunctions of λ3 will have only 2 nodal domains andnever 3. Thus Courant’s theorem is optimal.

We thus see that the min-max characterization of eigenvalues (3.4) isvery powerful and leads to several interesting properties of eigenvalues andeigenfunctions.

References

[1] Ciarlet, P. G. and Raviart, P. A. A mixed finite element method forthe biharmonic equation, pp. 125-145 in Mathematical Aspects of FiniteElements in Partial Differential Equations, C. de Boor (ed.), AcademicPress, 1974.

[2] Courant, R. and Hilbert, D. Methods of Mathematical Physics (Vols.I and II), Interscience, 1962.

[3] Gilbarg, D. and Trudinger, N. Elliptic Partial Differential Equationsof Second Order, Springer-Verlag, 1977.

[4] Kesavan, S. Functional Analysis, TRIM Series, 52, Hindustan BookAgency, 2009.

[5] Kondrat’ev, V. A. Boundary value problems for elliptic equations indomains with conical or angular points, Trudy Markov Mat. Obsc., 16,pp. 209-292, 1967.

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