Part IV, Chapter 19
Main results on well-posedness
The starting point of this chapter is the model problem derived in §18.3. Our
goal is to specify conditions under which this problem is well-posed. Two
important results are presented: the Lax–Milgram Lemma and the Banach–
Nečas–Babuška Theorem. The former provides a sufficient condition for wellposedness, whereas the latter, relying on slightly more sophisticated assumptions, provides necessary and sufficient conditions.
19.1 Mathematical setting
We consider the following model problem:
Find u ∈ V such that
a(u, w) = ℓ(w), ∀w ∈ W,
(19.1)
The mathematical setting for (19.1) is as follows:
(i) The spaces V and W are Banach spaces over the field K = R or C,
equipped with norms denoted k·kV and k·kW , respectively. In many applications, V and W are Hilbert spaces.
(ii) In the real case, a is a bounded bilinear form on V ×W , and we write
a ∈ B(V, W ). The boundedness of a means that
kakB(V,W ) :=
sup
(v,w)∈V ×W
a(v, w)
kvkV kwkW
(19.2)
is finite. It is henceforth implicitly understood that suprema are always
taken over nonzero arguments. In the complex case, a is sesquilinear (we
still write a ∈ B(V, W )) and ℜ(a(v, w)) replaces a(v, w) in (19.2).
(iii) In the real case, ℓ is a bounded linear form on W , and we write ℓ ∈ W ′ =
L(W ; R). The boundedness of ℓ means that
♥♥(JLG) Dec 29,
2016
248
Chapter 19. Main results on well-posedness
kℓkW ′ := sup
w∈W
ℓ(w)
kwkW
(19.3)
is finite. In the complex case, ℓ is antilinear (we still write ℓ ∈ W ′ ) and
ℜ(ℓ(w)) replaces ℓ(w) in (19.3).
Remark 19.1 (Use of absolute value and modulus). In the real case,
using the absolute value of the numerator in the definitions of kakB(V,W ) and
kℓkW ′ gives the same result (consider ±w in the supremum). In the complex
case, using the modulus instead of the real part gives the same result (replace
w by ξw where ξ is any complex number such that |ξ| = 1).
⊓
⊔
We use the following definition of well-posedness by Hadamard [277]:
Definition 19.2 (Well-posedness). Problem (19.1) is said to be well-posed
if it admits one and only one solution for all ℓ ∈ W ′ , and there is c, uniform
with respect to ℓ, such that the a priori estimate kukV ≤ c kℓkW ′ holds.
A key idea to address the well-posedness of (19.1) is to introduce the
Banach operator A ∈ L(V ; W ′ ) that is naturally associated with the bilinear
form a on V ×W by setting
hAv, wiW ′ ,W = a(v, w),
∀(v, w) ∈ V × W ′ .
(19.4)
One can verify that the operator A is linear and bounded since
kAvkW ′ = sup
w∈W
hAv, wiW ′ ,W
≤ kakB(V,W ) kvkV ,
kwkW
for all v ∈ V . The problem (19.1) can be reformulated as follows: Find u ∈ V
such that Au = ℓ in W ′ . Hence, proving the existence and uniqueness of the
solution to (19.1) amounts to proving that the Banach operator A is bijective.
19.2 Lax–Milgram Lemma
The Lax–Milgram Lemma is applicable only if the solution and test spaces are
the same. Thus, we set W = V , and the model problem (19.1) becomes
Find u ∈ V such that
(19.5)
a(u, w) = ℓ(w), ∀w ∈ V.
Lemma 19.3 (Lax–Milgram). (i) Let V be a Hilbert space over R, let a ∈
B(V, V ), and let ℓ ∈ V ′ . Assume that the following coercivity property hods:
There is a real number α > 0 and ξ = ±1 such that
ξa(v, v) ≥ αkvk2V ,
∀v ∈ V.
(19.6)
Part IV. Galerkin Approximation
249
Then, the problem (19.5) is well-posed with the a priori estimate kukV ≤
1
′
α kℓkV . (ii) If V is a Hilbert space over C, the same conclusion holds with the
following coercivity property: There is a α ∈ R, α > 0 and ξ ∈ C with |ξ| = 1
such that
ℜ (ξa(v, v)) ≥ αkvk2V ,
∀v ∈ V.
(19.7)
Proof. Although this lemma is a consequence of the BNB Theorem 19.10 (see
Remark 19.12 below), we present a direct proof in the real case with ξ = 1
for completeness. Let A : V → V ′ be the bounded linear operator defined
by (19.4). We need to prove that A is bijective. Let A∗ : V → V ′ be the
adjoint operator of A such that hA∗ v, wiV ′ ,V = hAw, viV ′ ,V for all v, w ∈ V .
We prove that A and A∗ are injective and that the range of A is closed (the
last two conditions imply that A is surjective since im(A) = (ker(A∗ ))⊥ ; see
Lemma A.48(iii)). The injectivity of A∗ results from the fact that A∗ v = 0
implies
0 = hA∗ v, viV ′ ,V = hAv, viV ′ ,V = a(v, v) ≥ αkvk2V ,
so that v = 0. Furthermore, we observe that
αkvkV ≤
ξa(v, v)
a(v, ξw)
a(v, w)
≤ sup
= sup
= kAvkV ′ .
kvkV
w∈V kwkV
w∈V kwkV
(19.8)
This bound shows that A is injective and that its range is closed (if (Ayn )n∈N
is a Cauchy-sequence in im(A), (yn )n∈N is then a Cauchy-sequence in V , hence
converges to some y ∈ V , and Ayn → Ay since A is bounded). Finally, the a
ℓ(u)
′
priori estimate results from αkukV ≤ a(u,u)
⊓
⊔
kukV = kukV ≤ kℓkV .
Example 19.4 (Laplacian). Consider the weak formulation (18.8) of the
Poisson problem with homogeneous Dirichlet condition. The functional setting
1
is V = W =
R bilinear form is
R H0 (D) equipped with the norm k·kH 1 (D) , the
a(v, w) := D ∇v·∇w dx, and the linear form is ℓ(w) := D f w dx. By the
Cauchy–Schwarz inequality, the forms a and ℓ are bounded on V ×V and V ,
respectively. Moreover, a is V -coercive since, for all v ∈ V ,
a(v, v) = k∇vk2L2 (D) ≥
2
min(1, CP,D
)
1
1 2
kvk2L2 (D) ≥
k∇vk2L2 (D) + CP,D
kvk2H 1 (D) ,
2
2
2
owing to the Poincaré inequality (B.77) (with p = 2), so that ξ = 1 here.
Hence, by the Lax–Milgram Lemma, the problem (18.8) is well-posed.
⊓
⊔
Example 19.5 (Complex case). Consider the PDE iu − ∆u = f in
D with f ∈ L2 (D; C) and homogeneous Dirichlet condition. Recalling Remark 18.7, the functional setting is V = W = H01R(D; C) equipped
with the
R
∇v·∇w
dx
+
dx,
ivw
norm k·kH 1 (D;C) , the sesquilinear form
is
a(v,
w)
=
Dπ
D
R
and the antilinear form is ℓ(w) = D f w dx. Then, taking ξ = e−i 4 , one read√
ily verifies that the coercivity condition (19.7) holds with α = 22 .
⊓
⊔
250
Chapter 19. Main results on well-posedness
Remark 19.6 (Hilbert spaces). The Lax–Milgram Lemma holds in Hilbert
spaces only (i.e., not in more general Banach spaces) since coercivity is essentially an Hilbertian property; see Exercise 19.5.
⊓
⊔
Remark 19.7 (Coercivity with modulus). The coercivity property is
sometimes stated as follows in the literature: There is α ∈ R, α > 0, such that
|a(v, v)| ≥ αkvk2V ,
∀v ∈ V.
(19.9)
This condition is equivalent to the coercivity property stated in Lemma 19.3;
for a proof of this fact, see Exercise 19.1 in the real case and Brezis [97, p. 366]
in the complex case.
⊓
⊔
Whenever V is a real Hilbert space and the bilinear form a is symmetric
and coercive with ξ = 1, the problem (19.5) can be interpreted as a minimization problem (or a maximization problem if ξ = −1). In this context, (19.5)
is called a variational formulation.
Proposition 19.8 (Variational formulation). Let V be a real Hilbert
space, let a ∈ B(V, V ) be a coercive bilinear form with ξ = 1, and let ℓ ∈ V ′ .
Assume that the bilinear form a is symmetric, i.e.,
a(v, w) = a(w, v),
∀v, w ∈ V.
(19.10)
Then, introducing the (so-called) energy functional E : V → R such that
E(v) =
1
a(v, v) − ℓ(v),
2
(19.11)
u solves (19.5) if and only if u minimizes E over V .
Proof. The proof relies on the following identity: for all u, w ∈ V and t ∈ R,
E(u + tw) = E(u) + t(a(u, w) − ℓ(w)) +
t2
2 a(w, w),
(19.12)
which results from the symmetry of a. Assume that u solves (19.5). Then,
since a(w, w) ≥ 0 owing to the coercivity of a (i.e., ξ = 1), (19.12) implies
that u minimizes E over V . Conversely, assume that u minimizes E over V . The
right-hand side of (19.12) is a second-order polynomial in t that is minimal
at t = 0. Hence, the derivative of this polynomial vanishes at t = 0, which
amounts to a(u, w) − ℓ(w) = 0. Since w is arbitrary in V , u solves (19.5). ⊓
⊔
Remark 19.9 (Link with Riesz–Fréchet). Given a symmetric coercive
bilinear form a on a Hilbert space V , ((·, ·))V := a(·, ·) defines an inner product in V , and the induced norm is equivalent to k·kV owing to the coercivity
and boundedness of a. Solving problem (19.5) amounts to finding the representative u ∈ V of the linear form ℓ ∈ V ′ , i.e., ((u, w))V = ℓ(w) for all w ∈ V .
This problem is well-posed by the Riesz–Fréchet Theorem A.44.
⊓
⊔
Part IV. Galerkin Approximation
251
19.3 Banach–Nečas–Babuška (BNB) Theorem
The BNB Theorem plays a fundamental role in this book. We use this terminology since, to our knowledge, the BNB Theorem was stated by Nečas
in 1962 [364] and Babuška in 1972 in the context of finite element methods
[33, p. 112]. From a functional analysis point of view, the BNB Theorem is
a rephrasing of two fundamental results by Banach: the Closed Range Theorem and the Open Mapping Theorem. We present two settings for the BNB
Theorem depending on whether test functions in the model problem belong
to a reflexive Banach space or to the dual of a Banach space. Recall from
Definition A.25 that a Banach space W is said to be reflexive if the canonical
isometry JW : W → W ′′ is an isomorphism; this is always the case if W is a
Hilbert space.
19.3.1 Test functions in reflexive Banach space
Theorem 19.10 (Banach–Nečas–Babuška (BNB)). Let V be a Banach
space and let W be a reflexive Banach space, both over R. Let a ∈ B(V, W )
and let ℓ ∈ W ′ . Then, the problem (19.1) is well-posed if and only if:
(bnb1)
(bnb2)
inf sup
v∈V w∈W
∀w ∈ W,
a(v, w)
=: α > 0,
kvkV kwkW
[ ∀v ∈ V, a(v, w) = 0 ] =⇒ [ w = 0 ].
1
′
α kℓkW
Moreover, the a priori estimate kukV ≤
equivalent to the following inf-sup condition:
∃α > 0,
(19.13a)
αkvkV ≤ sup
w∈W
(19.13b)
holds. Condition (bnb1) is
a(v, w)
,
kwkW
∀v ∈ V.
(19.14)
In the complex case, a(v, w) must be replaced by ℜ(a(v, w)) in (bnb1).
Proof. We give the proof in the real case; the complex case is treated similarly. We consider the operator A ∈ L(V ; W ′ ) defined by (19.4) and use
Theorem A.59(iii) (with W ′ in lieu of W ), i.e., A is bijective if and only if
condition (A.31) holds and A∗ : W ′′ → V ′ is injective. Condition (A.31) states
that kAvkW ′ ≥ αkvkV for all v ∈ V , which is equivalent to (bnb1) since
kAvkW ′ = sup
w∈W
hAv, wiW ′ ,W
a(v, w)
= sup
.
kwkW
w∈W kwkW
Furthermore, the injectivity of A∗ amounts to [ A∗ w′′ = 0 ] =⇒ [ w′′ = 0 ] for
all w′′ ∈ W ′′ . Since W is reflexive, this condition can be rewritten as
[ A∗ (JW w) = 0 ] =⇒ [ w = 0 ],
∀w ∈ W,
which is equivalent to (bnb2) since a(v, w) = hAv, wiW ′ ,W = hJW w, AviW ′′ ,W ′ =
hA∗ (JW w), viV ′ V . Finally, the a priori estimate results from the inequalities
ℓ(w)
′
⊓
⊔
αkukV ≤ supw∈W a(u,w)
kwkW = supw∈W kwkW = kℓkW .
252
Chapter 19. Main results on well-posedness
Remark 19.11 (Proving (bnb1)). Finding, for all v ∈ V , a “partner”
wv ∈ W such that kwv kW ≤ ω1 kvkV and a(v, wv ) ≥ ω2 kvk2V , with positive
numbers ω1 , ω2 independent of v, proves the inf-sup condition (bnb1) with
2
α= ω
ω1 . Observe that the coercivity property amounts to saying that wv = ξv
is a good partner.
⊓
⊔
Remark 19.12 (Link with Lax–Milgram). Let V be a real Hilbert space
and let a ∈ L(V ×V ; R) be a coercive bilinear form. Then, the proof of the Lax–
Milgram Lemma shows that a satisfies conditions (bnb1) and (bnb2) (with
W = V ). The proof can be readily adapted to the complex case. The converse
statement is wrong: conditions (bnb1) and (bnb2) do not imply coercivity.
Hence, (19.6) is not necessary for well-posedness in general, while (bnb1)(bnb2) are necessary and sufficient. However, it turns out that coercivity is
both necessary and sufficient for well-posedness when the bilinear form a is
symmetric and positive-semidefinite; see Exercise 19.6.
⊓
⊔
Remark 19.13 (T -coercivity). Let V and W be two Hilbert spaces. Then,
(bnb1)-(bnb2) are equivalent to the existence of a bijective operator T ∈
L(V ; W ), a positive number η and ξ = ±1 such that
ξa(v, T v) ≥ ηkvk2V ,
∀v ∈ V,
see Exercise 19.9. This property is called T -coercivity in Bonnet-Ben Dhia
et al. [67, 68]. Note that T -coercivity is a notion relevant in Hilbert spaces
only, since it relies on the coercivity of the operator T ∗ A ∈ L(V ; V ′ ), see
Remark 19.6. The advantage of this notion over coercivity is the possibility
of treating different trial and test spaces and using a test function different
from v ∈ V to estimate kvk2V .
⊓
⊔
19.3.2 Test functions in dual Banach space(♦)
The requirement on the reflexivity of the space W in the BNB Theorem 19.10
can be removed if the model problem is reformulated so that the test functions
act on the problem data instead of the data acting on the test functions.
Instead of (19.1), consider the following model problem:
Find u ∈ V such that
(19.15)
a(u, w′ ) = hw′ , f iW ′ ,W , ∀w′ ∈ W ′ .
In (19.15), the data is f ∈ W and the test functions belong to W ′ , whereas
in (19.1), the data is ℓ ∈ W ′ and the test functions belong to W . The functional
setting of problem (19.15) turns out to be more natural when considering
problems involving first-order PDEs; see §18.2.2. (Note that it sounds more
natural that tests functions act on data than the other way around.)
Theorem 19.14 (Banach–Nečas–Babuška (BNB)). Let V and W be two
Banach spaces over R. Let a ∈ B(V, W ′ ) and let f ∈ W . Then, the problem
(19.15) is well-posed if and only if:
Part IV. Galerkin Approximation
(bnb1’)
(bnb2’)
inf sup
v∈V w′ ∈W ′
′
′
∀w ∈ W ,
253
a(v, w′ )
:= α > 0,
kvkV kw′ kW ′
(19.16)
[ ∀v ∈ V, a(v, w′ ) = 0 ] =⇒ [ w′ = 0 ].
(19.17)
1
α kf kW
Moreover, the a priori estimate kukV ≤
holds. In the complex case,
the real part of the numerator is taken in (bnb1’).
Proof. We give the proof in the real case; the complex case is treated similarly.
Let A : V → W be defined by hw′ , AviW ′ ,W = a(v, w′ ) for all (v, w′ ) ∈ V ×W ′ .
The operator A is linear and bounded owing to the boundedness of a on
V × W ′ . Problem (19.15) can be reformulated as follows: Find u ∈ V such
that Au = f in W . Hence, proving the existence and uniqueness of the solution
to (19.15) amounts to proving that the Banach operator A is bijective. To this
purpose, we observe that conditions (bnb1’) and (bnb2’) are just statement
(iii) using (A.32) in Theorem A.59 (with the spaces V and W ). Finally, the a
priori estimate is proved as above.
⊓
⊔
19.3.3 Example 1: Darcy’s equations
Consider the weak formulation (18.12). This formulation fits the setting of
our model problem (19.1) with the functional spaces V = H(div; D)×H01 (D)
and W = L2 (D)×L2 (D), the bilinear form
Z
a(v, w) :=
{σ·τ + ∇p·τ + (∇·σ)q} dx,
(19.18)
D
with v := (σ, p) and w := (τ , q), and the linear form ℓ(w) :=
Proposition 19.15. Problem (18.12) is well-posed.
R
D
f q dx.
Proof. We equip the Hilbert spaces V and W with the norms kvkV =
{kσk2H(div;D) + kpk2H 1 (D) }1/2 and kwkW = {kτ k2L2 (D) + kqk2L2 (D) }1/2 with
v = (σ, p) and w = (τ , q), respectively. Since the bilinear form a and linear form ℓ are obviously bounded, it remains to check conditions (bnb1) and
(bnb2).
,q))
(1) Proof of (bnb1). Let (σ, p) ∈ V and define S := sup(τ ,q)∈W a((σ,p),(τ
k(τ ,q)kW .
Since V ⊂ W , we can take (σ, p) as the test function. Since p vanishes at the
boundary, a((σ, p), (σ, p)) = kσk2L2 (D) , whence we infer that
kσk2L2 (D) =
a((σ, p), (σ, p))
k(σ, p)kV ≤ S k(σ, p)kV .
k(σ, p)kV
Moreover, taking (∇p, ∇·σ) as the test function leads to
R
n
o 21
{∇p·τ + (∇·σ)q} dx
2
2
D
= sup
k∇pkL2 (D) + k∇·σkL2 (D)
k(τ , q)kW
(τ ,q)∈W
R
σ·τ dx
a((σ, p), (τ , q))
D
+ sup
.
≤ sup
k(τ , q)kW
(τ ,q)∈W k(τ , q)kW
(τ ,q)∈W
254
Chapter 19. Main results on well-posedness
Hence, {k∇pk2L2 (D) +k∇·σk2L2 (D) }1/2 ≤ S+kσkL2 (D) . Squaring this inequality
and combining it with the above bound on kσkL2 (D) , we infer that
k∇pk2L2 (D) + kσk2H(div;D) ≤ 2S2 + 3S k(σ, p)kV .
Owing to Poincaré’s inequality, we infer that CP,D kpkL2 (D) ≤ k∇pkL2 (D) .
2
) yields γk(σ, p)k2V ≤ 2S2 + 3Sk(σ, p)kV .
Hence, letting γ := 21 min(1, CP,D
Using Young’s inequality to absorb the last term on the right-hand side into
′
2
2
′
the left-hand side, we infer a bound of the
√ ′form γ k(σ, p)kV ≤ S , γ > 0,
which is just condition (bnb1) with α = γ .
(2) Proof of (bnb2). Let (τ , q) ∈ W be such that Ra((σ, p), (τ , q)) = 0 for
all (σ, p) ∈ V . This means on the one hand that D ∇p·τ dx =R 0 for all
p ∈ H01 (D), so that ∇·τ = 0. On the other hand, we obtain that D {σ·τ +
(∇·σ)q} dx = 0 for all σ ∈ H(div; D). Taking σ ∈ [C0∞ (D)]d , we infer that
q ∈ H 1 (D) and ∇q = τ ; taking now σ ∈ H(div; D) yields
Z
1
.
{σ·∇q + (∇·σ)q} dx = hσ·n, qi − 12
0=
2
H
D
(∂D),H (∂D)
R
Observing that τ ∈ H(div; D), we infer that 0 = D {τ ·τ + (∇·τ )q} dx =
kτ k2L2 (D) since ∇·τ = 0; as a result, τ = 0. Finally, ∇q = 0 so that q is
R
constant in D; since D (∇·σ)q dx = 0 for all σ ∈ H(div; D), q is identically
zero in D.
⊓
⊔
19.3.4 Example 2: first-order PDE(♦)
Consider the weak formulation (18.19). This formulation fits the setting of
the model problem (19.15) with test functions in the dual of a Banach space.
More precisely, we set V := {v ∈ W 1,1 (D) | v(0) = 0}, W := L1 (D), and
W ′ = L∞ (D) with D = (0, 1). The bilinear form is given by (here, we denote
d
derivatives by dt
and reserve the primes to duality)
a(v, w′ ) =
Z
1
0
dv ′
w dt,
dt
and the right-hand side by hw′ , f iW ′ ,W =
∀(v, w′ ) ∈ V ×W ′ ,
R1
0
(19.19)
w′ f dt with f ∈ W .
Proposition 19.16. Problem (18.19) is well-posed.
Proof. We equip the Banach spaces V and W with the norms kvkV =
′
′
1
∞
kvkL1 (D) + k dv
dt kL (D) and kw kW = kw kL (D) , and we verify the conditions
(bnb1’) and (bnb2’) from Theorem 19.14.
−1
(1) Proof of (bnb1’). Let v ∈ V and set D± = ( dv
({t ∈ D | ± dv
dt )
dt (t) > 0}).
′
Taking wv = χD+ − χD− where χS denotes the characteristic function of a
measurable set S, we infer that
Part IV. Galerkin Approximation
255
R 1 dv ′
Z 1
dv w dt
a(v, w′ )
a(v, wv′ )
0 dt v
dt = dv 1
sup
≥
=
=
dt
dt L (D) .
′
′
′
′
′
′
′
kwv kW
kwv kL∞ (D)
w ∈W kw kW
0
Invoking the Poincaré–Friedrichs inequality on V (see (B.79) with p = 1 and
the bounded linear form f (v) = v(0)) yields (bnb1’).
R1
′
(2) Proof of (bnb2’). Let w′ ∈ W ′ be such that 0 dv
dt w dt = 0 for all v ∈
′
V . Taking v in C0∞ (D), we infer that dw
dt = 0 in the distribution sense.
Lemma B.50 implies that w′ is a constant. Choosing v(t) = t as a test function
R1
leads to 0 w′ dt = 0 and, as a consequence, w′ = 0.
⊓
⊔
Exercises
Exercise 19.1 (Coercivity with modulus). Let V be a real Hilbert space
and let a ∈ B(V, V ). Show that a satisfies (19.6) if and only if it satisfies (19.9).
(Hint: assume for nonzero elements v, w ∈ V that a(v, v) < 0 and a(w, w) > 0,
and consider the second-order polynomial a(v + λw, v + λw) for all λ ∈ R.)
Exercise 19.2 (Reflexivity). Let V, W be two Banach spaces such that
there is an isomorphism A ∈ L(V ; W ). Assume that V is reflexive. Prove
that W is reflexive. (Hint: consider the map A∗∗ ◦ JV ◦ A−1 .)
Exercise 19.3 (Orthogonal projection). Let V be a real Hilbert space
equipped with the inner product (·, ·)V and induced norm k·kV . Let U be a
nonempty, closed, and convex subset of V . Let f ∈ V .
(i) Show that there is a unique u in U such that kf −ukV = minv∈U kf −vkV .
(Hint: Recall that 14 (a − b)2 = 21 (c − a)2 + 12 (c − b)2 − (c − 12 (a + b))2 and
show that a minimizing sequence is a Cauchy sequence.)
(ii) Show that u ∈ U is the solution to the above minimization problem if
and only if (f − u, v − u)V ≤ 0 for all v ∈ U . (Hint: proceed as in the
proof of Proposition 19.8.)
(iii) Assuming that U is a (non-trivial) subspace of V , prove that the unique
minimizer is characterized by (f − u, v)V = 0 for all v ∈ U , and prove
that the mapping ΠU : V ∋ f 7→ u ∈ U is linear and kΠU kL(V ;U ) = 1.
(iv) Let a be a bounded, symmetric, and V -coercive bilinear form. Let ℓ ∈ V ′ .
Set E(v) = 12 a(v, v) − ℓ(v). Show that there is a unique u ∈ V such that
E(u) = minv∈U E(v) and that u is a minimizing solution if and only if
a(u, v − u) ≥ ℓ(v − u) for all v ∈ U .
Exercise 19.4 (Fixed-point argument). The goal of this exercise is to
derive another direct proof of the Lax–Milgram Lemma. Let A ∈ L(V ; V ) be
defined by (Av, w)V = a(v, w) for all v, w ∈ V . Let L be the representative in
V of the linear form ℓ ∈ V ′ . Let λ > 0. Consider the map Tλ : V → V such
that Tλ (v) = v − λ(Av − L) for all v ∈ V . Prove that if λ is small enough,
256
Chapter 19. Main results on well-posedness
kTλ (v) − Tλ (w)kV ≤ ρλ kv − wkV for all v, w ∈ V with ρλ ∈ (0, 1). Infer
that (19.5) is well-posed. (Hint: use Banach’s Fixed-Point Theorem.)
Exercise 19.5 (Coercivity implies Hilbert space). Let V be a Banach
space. Prove that V can be equipped with a Hilbert structure with the same
topology if and only if there is a coercive operator in L(V ; V ′ ). (Hint: Think
of ξ(hAv, wiV ′ ,V + hAw, viV ′ ,V ) in the real case and extend the idea to the
complex case.)
Exercise 19.6 (Coercivity as necessary condition). Let V be a reflexive Banach space and let A ∈ L(V ; V ′ ) be a monotone self-adjoint operator; see §A.4.4. Prove that A is bijective if and only if A is coercive (with
ξ = 1). (Hint: Prove that, if A is monotone and self-adjoint, hAv, wiV ′ ,V ≤
1
1
hAv, viV2 ′ ,V hAw, wiV2 ′ ,V for all v, w ∈ V (in the real case); then, use this inequality in the inf-sup condition. Adapt the argument to the complex case.)
Exercise 19.7 (Darcy). Prove that problem (18.14) is well-posed. (Hint:
adapt the proof of Proposition 19.15.)
Exercise 19.8 (First-order PDE). Prove that problem (18.20) is wellposed. (Hint: adapt the proof of Proposition 19.16.)
Exercise 19.9 (T -coercivity). Let V and W be two Hilbert spaces. Prove
that (bnb1)-(bnb2) are equivalent to the existence of a bijective operator
T ∈ L(V ; W ) and a number η > 0 such that a(v, T v) ≥ ηkvk2V for all v ∈ V .
(Hint: to construct T , consider A∗ and use the Riesz–Fréchet Theorem.)
Exercise 19.10 (Sign-changing diffusion). Let the domain D in Rd be
partitioned into two disjoint subdomains D1 and D2 . Set Σ := ∂D1 ∩ ∂D2 .
Let κ1 , κ2 be two real numbers such that κ1 > 0 and κ2 < 0. Set κ(x) =
κ1 1D1 (x)+κ2 1D2 (x) for all x ∈ D. Let V := H01 (D) be equipped with
R the norm
k∇vkL2 (D) . The goal is to show that the bilinear form a(u, w) := D κ∇u·∇w
satisfies conditions (bnb1)-(bnb2) on V ×V ; see Chesnel and Ciarlet [136] for
further insight. Set Vm := {v|Dm | v ∈ V }, for all m ∈ {1, 2}, equipped with
the norm k∇vm kL2 (Dm ) , and let γ0,m be the traces of functions in Vm on Σ.
(i) Let S1 ∈ L(V1 ; V2 ) be such that γ0,2 (S1 v1 ) = γ0,1 (v1 ). Let T : V → V
be such that, for all v ∈ V , T v(x) = v(x) if x ∈ D1 and T v(x) =
−v(x) + 2S1 (v|D1 )(x) if x ∈ D2 . Prove that T ∈ L(V ) and that T is an
isomorphism. (Hint: verify that T ◦ T = IV , the identity in V .)
(ii) Assume that |κκ21 | > kS1 k2L(V1 ;V2 ) . Prove that conditions (bnb1)-(bnb2)
are satisfied. (Hint: use T -coercivity from Remark 19.13.)
(iii) Let D1 = (−a, 0) × (0, 1) and D2 = (0, b) × (0, 1) with a > b > 0. Show
that if |κκ12 | 6∈ [1, ab ], (bnb1)-(bnb2) are satisfied. (Hint: consider the map
S1 ∈ L(V1 ; V2 ) such that S1 v1 (x, y) = v1 (− ab x, y) for all v1 ∈ V1 , and the
map S2 ∈ L(V2 ; V1 ) such that S2 v2 (x, y) = v2 (−x, y) if x ∈ (−b, 0) and
S2 v2 (x, y) = 0 otherwise, for all v2 ∈ V2 .)
Part IV. Galerkin Approximation
257
Solution to exercises
Exercise 19.1 (Coercivity with modulus). It is clear that (19.6) implies (19.9). Conversely, assume that the bilinear form satisfies (19.9). Reasoning by contradiction, assume that there are nonzero elements v, w ∈ V
such that a(v, v) < 0 and a(w, w) > 0. Then, the second-order polynomial
a(v + λw, v + λw) for all λ ∈ R has at least one root λ∗ ∈ R. Property (19.9)
implies that v + λ∗ w = 0. Hence, a(v, v) = λ2∗ a(w, w) > 0 which contradicts
a(v, v) < 0.
Exercise 19.2 (Reflexivity). Let us prove that A∗∗ ◦ JV ◦ A−1 = JW . Let
w ∈ W . We observe that
hA∗∗ (JV (A−1 (w))), w′ iW ′′ ,W ′ = hJV (A−1 (w), A∗ (w′ ))iV ′′ ,V ′
= hA∗ (w′ ), A−1 (w)iV ′ ,V
= hw′ , AA−1 (w)iW ′ ,W
= hw′ , wiW ′ ,W = hJW (w), w′ iW ′′ ,W ′ ,
for all w′ ∈ W ′ . Hence, A∗∗ (JV (A−1 (w))) = JW (w) in W ′ , which proves
that A∗∗ ◦ JV ◦ A−1 = JW . Since A is an isomorphism, A∗∗ is also an isomorphism (see Remark A.60). Moreover, JV is an isomorphism since V is
reflexive. Hence, JW is an isomorphism, which proves the reflexivity of W .
Exercise 19.3 (Orthogonal projection).
(i) Let (un )n∈N be a minimizing sequence in U . Then, a direct calculation
shows that
1
1
1
kun − um k2V = kf − un k2V + kf − um k2 − kf − 12 (un + um )k2V
4
2
2
1
1
2
≤ kf − un kV + kf − um k2 ,
2
2
which shows that (un )n∈N is a Cauchy sequence in V . Since V is a Hilbert
space and U is closed, its limit, say u, is in U , and by construction, u is
a minimizer. Uniqueness follows from the above formula (consider for un
and um two minimizers).
(ii) We proceed as in the proof of Proposition 19.8 with E(v) = kf − vk2V .
Let u be such that E(u) = minv∈U E(v). Formula (19.12) becomes
E(u + tw) = E(u) − 2t(f − u, w)V + t2 kwk2V .
For all v ∈ U and all t ∈ [0, 1], take w = v − u and observe that u +
tw ∈ U by convexity of U . Hence, E(u + tw) ≥ E(u) for all t ∈ [0, 1].
Since p(t) := E(u + tw) is a polynomial in t and p(t) ≥ p(0) for all
t ∈ [0, 1], we infer that its derivative at t = 0 is non-negative, yielding
−(f − u, v − u)V = p′ (0) ≥ 0 for all v ∈ U . Conversely, if u ∈ U satisfies
this property, evaluating E(u + tw) with w = v − u at t = 1 yields
E(v) ≥ E(u) for all v ∈ U .
258
Chapter 19. Main results on well-posedness
(iii) When U is a subspace of V , (v − u) spans U , so that the characterization
becomes (f − u, v)V ≤ 0 for all v ∈ U , and since the same inequality
holds for −v, we infer that (f − u, v)V = 0 for all v ∈ U . The linearity of
the map ΠU results from this characterization, which also yields
kΠU f k2V = kf k2V − kf − ΠU (f )k2V ,
for all f ∈ V , showing that kΠU kL(V ;U ) ≤ 1. Equality is attained on the
elements of U (unless U = {0} so that kΠU kL(V ;U ) = 0).
(iv) All of the above can be extended by replacing the old functional kf −vk2V
by the new functional E(v). In particular,
α
1
1
1
kun −um k2V ≤ a(un −um , un −um ) = E(un )+ E(um )−E( 12 (un +um )).
8
8
2
2
Exercise 19.4 (Fixed-point argument). We observe that
kTλ (v) − Tλ (w)k2V = k(v − w) − λA(v − w)k2V
= kv − wk2V − 2λa(v − w, v − w) + λ2 kA(v − w)k2V
≤ (1 − 2λα + λ2 kak2V ×V )kv − wk2V .
Taking λ = kakα2
yields the assertion. The above bound shows that the map
V ×V
Tλ is continuous and contracting. Owing to Banach’s Fixed-Point Theorem,
there exists a unique u ∈ V such that Tλ (u) = u, which is equivalent to u
solving (19.5).
Exercise 19.5 (Coercivity implies Hilbert space). Setting ((v, w))V :=
ξ(hAv, wiV ′ ,V + hAw, viV ′ ,V ) defines an inner product in V ; the only nontrivial property is the positive-definiteness which follows from the coercivity of
A. The induced norm is equivalent to k·kV since, by coercivity and continuity
of A,
2αkvk2V ≤ ((v, v))V ≤ 2kAkL(V ;V ′ ) kvk2V .
In the complex case, one sets ((v, w))V := ξhAv, wiV ′ ,V + ξhAw, viV ′ ,V .
Exercise 19.6 (Coercivity as necessary condition). Assume that A is
monotone and self-adjoint. Let v, w ∈ V . For all t ∈ R, we infer that
0 ≤ hA(v + tw), (v + tw)iV ′ ,V = hAv, vi + 2thAv, wiV ′ ,V + t2 hAw, wi,
where we have used the monotonicity and self-adjointness of A. The above
right-hand side is a second-order polynomial in t taking only non-negative
1/2
1/2
values. We infer that hAv, wiV ′ ,V ≤ hAv, viV ′ ,V hAw, wiV ′ ,V . Assume now
that A is bijective. Then, condition (bnb1) implies that there is α > 0 such
that
hAv, wiW ′ ,W
,
∀v ∈ V.
αkvkV ≤ sup
kwkW
w∈W
Part IV. Galerkin Approximation
259
Using the above bound on hAv, wiV ′ ,V yields
1
αkvkV ≤
1
hAv, viV2 ′ ,V
sup
w∈W
2
hAw, wiW
′ ,W
1
1
2
≤ hAv, viV2 ′ ,V kAkL(V
;V ′ ) .
kwkW
2
Hence, coercivity holds with parameter kAk α ′ and ξ = 1. In the complex
L(V ;V )
case, the above is adapted by considering the real part of duality brackets.
Exercise 19.7 (Darcy). We verify conditions (bnb1)-(bnb2) for the bilinear
form
Z
a((σ, p), (τ , q)) =
(σ·τ + ∇p·τ + σ·∇q) dx,
D
for the functional spaces V = W := L2 (D)×H01 (D) equipped with the norm
k(σ, p)kV = {kσk2L2 (D) + k∇pk2L2 (D) }1/2 (owing to the Poincaré inequality for
p). Set
a((σ, p), (τ , q))
.
S := sup
k(τ , q)kV
(τ ,q)∈V
Then,
kσk2L2 (D) = a((σ, p), (σ, −p)) ≤ S k(σ, p)kV ,
k∇pk2L2 (D) = a((σ, p), (∇p, −p)) ≤ 2S k(σ, p)kV ,
so that (bnb1) holds. Let now (τ , q) ∈ V be such that a((σ, p), (τ , q)) = 0
for all (σ, p) ∈ V . Testing with (σ, p) = (τ , −q) yields kτ k2L2 (D) = 0 so that
τ = 0, while testing with (σ, p) = (∇q, −q) yields k∇qk2L2 (D) = 0 so that
q = 0. Hence, (bnb2) holds.
Exercise 19.8 (First-order PDE). Let D = (0, 1). We verify conditions
(bnb1)-(bnb2) for the bilinear form
Z 1
dv
a(v, w) =
w dt,
0 dt
and the functional spaces V := {v ∈ H 1 (D) | v(0) = 0} and W = L2 (D)
2
equipped with the norms kvkV = k dv
dt kL (D) (owing to the Poincaré inequality)
and kwkW = kwkL2 (D) , respectively. We first observe that
R dv
w dt
a(v, w)
dv
D dt
= sup
,
kvkV = k dt kL2 (D) = sup
2
kwk
2
w∈W kwkW
L (D)
w∈L (D)
R1
so that (bnb1) holds. Let now w ∈ W be such that 0 dv
dt w dt = 0 for all
v ∈ V . Taking first v ∈ C0∞ (D), we infer that dw
=
0
in
the distribution
dt
sense, so that w is constant in D. Taking v = t then shows that w = 0, so
that (bnb2) holds.
260
Chapter 19. Main results on well-posedness
Exercise 19.9 (T -coercivity). Assume the existence of a bijective operator
T ∈ L(V ; W ) and a number η > 0 such that a(v, T v) ≥ ηkvk2V for all v ∈ V .
η
since, for all v ∈ V with v 6= 0,
Then, (bnb1) holds with α = kT kL(V
;W )
ηkvk2V ≤ a(v, T v) ≤
a(v, w)
a(v, T v)
kT vkW ≤ sup
kT kL(V ;W ) kvkV .
kT vkW
w∈W kwkW
(bnb2) holds since, considering w ∈ W such that a(v, w) = 0 for all v ∈ V ,
the bijectivity of T implies that there is vw ∈ V with T vw = w and taking
vw in the above statement leads to 0 = a(vw , w) = a(vw , T vw ) ≥ ηkvw k2V ,
whence vw = 0 so that w = 0. Conversely, assume (bnb1)-(bnb2). Then,
A ∈ L(V ; W ′ ) is an isomorphism, and the same holds for A∗ ∈ L(W ; V ′ )
(see Remark A.60). Set T = (A∗ )−1 JV →V ′ where JV →V ′ is the isomorphism
resulting from the Riesz–Fréchet Theorem. Then, T ∈ L(V ; W ), and T is an
isomorphism. Moreover, for all v ∈ V ,
a(v, T v) = hAv, T viW ′ ,W = hAv, (A∗ )−1 JV →V ′ viW ′ ,W
= hJV →V ′ v, viV ′ ,V = kvk2V .
Exercise 19.10 (Sign-changing diffusion).
(i) It is clear that the map T is linear and bounded. Moreover, for all v ∈ V ,
we observe that (T ◦ T )(v)(x) = v(x) for all x ∈ D1 , while
(T ◦ T )(v)(x) = −T (v)(x) + 2S1 ((T v)|D1 )(x)
= −(−v(x) + 2S1 (v|D1 )(x)) + 2S1 (v|D1 )(x) = v(x),
for all x ∈ D2 . Hence, T ◦ T = IV , and this implies that T is bijective.
(ii) To prove T -coercivity, we observe that, for all v ∈ V ,
Z
Z
Z
a(v, T v) =
κ1 |∇v|2 dx −
κ2 |∇v|2 dx + 2
κ2 ∇v·∇(S1 v|D1 ) dx
D1
D2
D2
Z
= κ1 kv|D1 k2V1 − κ2 kv|D2 k2V2 + 2
κ2 ∇v·∇(S1 v|D1 ) dx
≥ (κ1 − η
−1
D2
2
|κ2 |kS1 kL(V1 ;V2 ) )kv|D1 k2V1
− κ2 (1 − η)kv|D2 k2V2 ,
where η > 0 can be chosen as small as needed. Under the assumption
that |κκ12 | > kS1 k2L(V1 ;V2 ) , it is possible to choose η ∈ (0, 1) such that both
real numbers (κ1 − η −1 |κ2 |kS1 k2L(V1 ;V2 ) ) and −κ2 (1 − η) are positive.
This yields T -coercivity, and, therefore, conditions (bnb1)-(bnb2) are
satisfied.
(iii) The map S1 is in L(V1 ; V2 ), and one verifies that kS1 k2L(V1 ;V2 ) = ab . Hence,
(bnb1)-(bnb2) are satisfied if |κκ21 | > ab . Reasoning similarly with the
operator S2 shows that T -coercivity holds provided |κκ21 | < kS2 k2L(V2 ;V1 ) ,
and one verifies that kS2 k2L(V2 ;V1 ) = 1.