On the identification H = H 0 in the Lions theorem
and a related error
J. Simon
CNRS, Laboratoire J.-A. Dieudonné, Université de Nice, France
Keywords: Navier-Stokes equations, heat equation, Lions theorem, error
Abstract
We show that the common identification of H with its dual space
in the famous functional frame V ⊂ H = H 0 ⊂ V 0 is incompatible with
the distributional frame for some standard pdes and we recall an error
related to this unnecessary identification which is repeated from years.
1
Introduction
The general frame for solving linear parabolic pdes with the Lions theorem is V ⊂ H ⊂ D0 (Ω). To remain in distribution spaces, the common
identification H = H 0 ⊂ V 0 requires that V 0 ⊂ D0 (Ω).
We show that this condition is satisfied for particular pdes, such as the
heat equation with the Dirichlet condition, but not for some standard pdes,
such as the heat equation with the Neumann condition or the Navier–Stokes
equations.
We show that the identification H = H 0 does not improves results if
V 0 ⊂ D0 (Ω) and else is a source of error, such as the resolution of the
Navier–Stokes equations in a distribution sense with f ∈ L2 (0, T ; V0 ) which
is repeated from years.
2
Functional frame for the Lions theorem
The Lions theorem
Let us recall this basic tool for the resolution of parabolic linear equations.
Theorem 1 Let V and H be two Hilbert spaces such that
V⊂
H,
→
V is dense in H,
1
(1)
T > 0, a ∈ L∞ (0, T ; L2 (V × V )) such that there exists two real numbers
α > 0 and β ≥ 0 such that, for all v ∈ V ,
a(v, v) ≥ αkvk2V − βkvk2H ,
`1 ∈ L1 (0, T ; H 0 ), `2 ∈ L2 (0, T ; V 0 ) and u0 ∈ H.
Then there exists a unique solution u ∈ L2 (0, T ; V ) ∩ C([0, T ]; H) of

 d (u, v) + a(u, v) = h` , vi 0
∀v ∈ V,
1
H
H ×H + h`2 , viV 0 ×V ,
dt
(2)

u(0) = u0 . u
t
This result was given in [2], th. 2.1 p. 219, hyp. (1.16) p. 218 and th. 3.1
p. 221. Another proof is in [7], th. 1.
The notation ⊂
stands for imbedding with continuous identity, H 0 is the
→
dual space of H and L2 (V × V ) is the set of bilinear continuous forms on V .
Distributions frame
When the Lions theorem is used to solve a scalar pde in an open set Ω ⊂ Rd ,
one has
V⊂
H⊂
D0 (Ω).
→
→
A temptation
From (1), we get H 0 ⊂
V 0 (identifying f ∈ H 0 with its restriction f |V ).
→
The Hilbert space H being isomorfic with its dual space by the Riesz–
Fréchet theorem, it could be identified with H 0 (identifying f ∈ H with
If ∈ H 0 , defined by hIf, vi = (f, v)H ). It would provide the unified frame
V⊂
H=
H0 ⊂
V 0.
↔
→
→
In order to remain in distribution spaces, this identification would require
V0⊂
D0 (Ω).
→
(3)
This is for example necessary to solve an equation of type ∂u/∂t − ∆u = f
with f ∈ L2 (0, T ; V 0 ) (unless partial derivatives be extended to another
space than the distributions one).
Remind that (3) is satisfied if
D(Ω) ⊂
V,
→
D(Ω) is dense in V .
(4)
Conversely, to get a satisfying meaning, (3) requires, by the following Proposition 2:
V ∩ D(Ω) is dense in D(Ω),
D(Ω) ∩ V is dense in V .
(5)
Remark. It is not clear wether (3) requires (4) or not, but it does not matter.
2
3
Necessary condition
Let us give a general condition for a dual space to be included into another
one, which contains (5) and an analogous vectorial condition which will be
used in Section 6 for the Navier–Stokes equations.
Proposition 2 Let E and V be two separated topological vector spaces
being vector subspaces of a same vector space. To be valid with unique
eventual continuous extensions, the imbedding
V 0 ⊂ E0
requires that V ∩ E is dense in E and E ∩ V is dense in V . u
t
Proof. The imbedding V 0 ⊂ E 0 requires that:
To every fV ∈ V 0 corresponds a unique fE ∈ E 0
such that fE = fV in E ∩ V ,
(6)
which requires that V ∩ E is dense in E. Indeed, should it be nondense, by
(a corollary of) the Hahn–Banach theorem, there would exist g ∈ E 0 such
that g 6= 0 and g|E∩V = 0. Extending g by 0 in V \ E, we would obtain f
defined in E ∪ V such that
f |E = g ∈ E 0 ,
f |E 6= 0,
f |V = 0 ∈ V 0 .
Then the function 0 in E ∪V would satisfy 0|V = 0 = f |V and 0|E = 0 6= f |E
which would contradict (6).
The inclusion V 0 ⊂ E 0 also requires that:
No fE ∈ E 0 can correspond to two distinct fV ∈ V 0 ,
(7)
which requires that E ∩ V is dense in V . Indeed, else, exchanging E and V
in the above proof, we would get f defined in E ∪ V such that 0|E = f |E
and 0|V 6= f |V , with f |V ∈ V 0 , which would contradict (7). u
t
4
Heat equation with the Dirichlet condition
Problem
We seek for u = u(t, x) such that
∂u
− ∆u = f,
∂t
Given u0 ∈ L2 (Ω) and
u|∂Ω = 0,
u|t=0 = u0 .
f ∈ L2 (0, T ; H −1 (Ω)),
(8)
a solution u ∈ L2 (0, T ; H01 (Ω)) ∩ C(0, T ; L2 (Ω)) is obviously obtained using
the Lions theorem with
V = H01 (Ω),
H = L2 (Ω).
3
Compatibility with the identification H = H 0
The compatibility condition (4) is satisfied here, that is, D(Ω) ⊂
H 1 (Ω)
→ 0
1
1
and D(Ω) is dense in H0 (Ω) since, by definition, H0 (Ω) is the closure of
D(Ω) in H 1 (Ω). Therefore the identification of H with H 0 , that is, the
identification of L2 (Ω) with its dual space, is compatible here with
the other identifications and imbeddings in distribution spaces; it leads to
H01 (Ω) ⊂
L2 (Ω) =
(L2 (Ω))0 ⊂
(H01 (Ω))0 ⊂
D0 (Ω).
↔
→
→
→
(9)
With these identifications, the heat equation can be solved with
f ∈ L2 (0, T ; (H01 (Ω))0 ).
(10)
Although possible, this identification does not provide better results since
then (10) is equivalent to (8). This is a redundancy if H −1 (Ω) denotes
the image of (H01 (Ω))0 by these identifications; it is a theorem if, as we
recommend, it denotes the space of sums of derivatives of order ≤ 1 of
elements of L2 (Ω); for more details, see Section 8.
5
Heat equation with the Neumann condition
Problem
We seek now for u such that
∂u
− ∆u = f,
∂t
∂u = 0,
∂n ∂Ω
u|t=0 = u0 .
Given u0 ∈ L2 (Ω) and
f ∈ L1 (0, T ; L2 (Ω)),
a solution u ∈ L2 (0, T ; H 1 (Ω)) ∩ C(0, T ; L2 (Ω)) is obtained using the Lions
theorem with now
V = H 1 (Ω), H = L2 (Ω).
Remark.
The Neumann condition is satisfied here in the following weak form
R
∇
·
(v∇u)
= 0 for all v ∈ H 1 (Ω) which is meaningful in W −1,∞ (0, T )
Ω
because ∇ · (v∇u) = v∆u + ∇v · ∇u ∈ W −1,∞ (0, T ; L1 (Ω)).
Incompatibility with the identification H = H 0
The compatibility condition (5) is not satisfied here since D(Ω) is not
dense in H 1 (Ω). Therefore, by Proposition 2,
(H 1 (Ω))0 6⊂ D0 (Ω).
So the identification of H with H 0 , that is again the identification of L2 (Ω)
with its dual space, is dangerous here since (L2 (Ω))0 is identified with
a subspace of (H 1 (Ω))0 which is not a distribution space.
4
Danger of the identification
Would we identify L2 (Ω) =
(L2 (Ω))0 ⊂
(H 1 (Ω))0 , it would be temptating to
↔
→
generalize the above existence result by considering f valued in the larger
space (H 1 (Ω))0 , namely f ∈ L2 (0, T ; (H 1 (Ω))0 ).
The Lions theorem would still give a solution u unfortunately it could
not be a distributional solution of the heat equation for this f since f =
∂u/∂t − ∆u ∈ D0 (0, T ; D0 (Ω)) would imply (H 1 (Ω))0 ⊂ D0 (Ω).
Although nobody felt in this trap, notice that first rank mathematicians
felt in the similar one for the Navier–Stokes equations, see Section 7.
The dual of H 1 (Ω)
Given R a continuous inverse of the trace operator, the decomposition v =
(v − R(v|∂Ω )) + R(v|∂Ω ) gives H 1 (Ω) = H01 (Ω) ⊕ R(H 1/2 (∂Ω)). By duality,
this provides
(H 1 (Ω))0 ' (H01 (Ω))0 × (R(H 1/2 (∂Ω)))0 .
With the identification (H01 (Ω))0 =
H −1 (Ω) (which is always compatible with
↔
the distributions frame), this shows that (H 1 (Ω))0 is strictly “greater” than
H −1 (Ω) and that an element of the first space cannot be equal to an element
of the second (as y ∈ R2 cannot be equal to x ∈ R).
6
The linearized Navier–Stokes equations
Problem
We seek now for u = (u1 , u2 , u3 )(t, x) and p = p(t, x) such that
∂u
− ν∆u + ∇p = f, ∇ · u = 0, u|∂Ω = 0,
∂t
Given u0 ∈ H and
f ∈ L2 (0, T ; (H −1 (Ω))3 )
u|t=0 = u0 .
a velocity pressure solution u ∈ L2 (0, T ; (H01 (Ω))3 ) ∩ C(0, T ; (L2 (Ω))3 ), p ∈
W −1,∞ (0, T ; (L2loc (Ω))3 ) is classically obtained by the Lions theorem with V
and H and by the de Rham theorem. Remind that V and H are the closure,
respectively in (H 1 (Ω))3 and (L2 (Ω))3 , of {v ∈ (D0 (Ω))3 : ∇ · v = 0}.
Incompatibility with the identification H = H0
Since u has three scalar components, the compatibility condition (5) is replaced here by an analogous vectorial condition which is not satisfied.
Indeed V, and even H, are not dense in (D(Ω))3 since ∇ · v = 0 for all v in
H, and therefore for all v in its closure in (D(Ω))3 . Therefore, by Proposition 2,
H0 6⊂ (D0 (Ω))3 , V0 6⊂ (D0 (Ω))3 .
(11)
5
Therefore the identification of H with H0 is dangerous since this last
is not a distribution space.
Proposition 3 A distributional solution (u, p) of the (linearized) Navier–
Stokes equations can neither be obtained for f ∈ L2 (0, T ; H0 ) nor for f ∈
L2 (0, T ; V0 ). u
t
Proof. Let c ∈ V0 and f be the constant distribution with value c. Would a
distributional solution exist, it should satisfy
f=
∂u
− ∆u + ∇p ∈ D0 (0, T ; (D0 (Ω))3 )
∂t
and then c ∈ (D0 (Ω))3 which would contradict (11). In the nonlinear case,
the additional term (u · ∇)u would not change this result. u
t
Another proof of Proposition 3 is given in [6], cor. 3 p. 229.
The dual spaces of H and V
Decomposing (L2 (Ω))3 = H ⊕ H⊥ , we get by duality
((L2 (Ω))3 )0 ' H0 × (H⊥ )0 .
With the identification of L2 (Ω) with its dual space, this shows that H0
is strictly “smaller” than (L2 (Ω))3 and that their respective elements can
never be equal.
Similarly, ((H01 (Ω))3 )0 ' V0 ×(V⊥ )0 and an element of V0 cannot be equal
to an element of (H −1 (Ω))3 .
7
Some inaccuracies
Incorrect results
Various authors wrote than the Navier–Stokes equations, linearized or not,
stationary or not (this plays no part at this point) have a distributional
solution (u, p) with f ∈ L2 (0, T ; V0 ), or f ∈ V0 . Let us cite in particular:
• J.-L. Lions [4], Property (6.34) p. 69, in 1969.
• R. Temam [8], chap. III, § 1, Proposition 1.1 p. 266, in 1977.
• R. Dautray and J.-L. Lions [1], t. 3, chap. XIX, § 2, Proposition 1
p. 843, in 1984.
• F. Ortegon [5] lines following Theorem 1 p. 60, in 2001.
6
These results are not correct in view of Proposition 3. We emphasize the
point because these works being deservedly widely cited, the error carries
on spreading.
The error follows from the confusion of two distinct duality products and
from the use of the de Rham theorem without satisfying its hypothesis. Let
us give some details for better clarity: In [8], the author writes (Equation
(1.90), p. 267)
“ (u(t) − u0 , v) + ν((U (t), v)) = hF (t), vi,
∀t, ∀v ∈ V
(1.90)
or
hu(t) − u0 − ν∆U (t) − F (t), vi,
∀v ∈ V ”
and concludes by the de Rham theorem that there exists P (t) such that
u(t) − u0 − ν∆U (t) − F (t) = −∇P (t).
With the notations of [8], (1.90) means that
Z
Z
(u(t) − u0 ) · v + ν
∇U (t) · ∇v = hF (t), viV0 ×V
Ω
Ω
which is correct. It follows that
hu(t) − u0 − ν∆U (t), vi(H −1 (Ω))3 ×(H01 (Ω))3 = hF (t), viV0 ×V
but these two duality products cannot be identified as it is done in the
equation following (1.90). Moreover the use of the de Rham theorem here
would require u(t) − u0 − ν∆U (t) − F (t) to belong to (D0 (Ω))3 , which is not
the case because V0 6⊂ (D0 (Ω))3 .
Another point of view
Given f ∈ L2 (0, T ; V0 ), there exists f ∈ L2 (0, T ; (H −1 (Ω))3 ) such that, for
all v ∈ V,
(12)
hf , vi(H −1 (Ω))3 ×(H01 (Ω))3 = hf, viV0 ×V
(in D0 (0, T )). Then, as seen in the above statement of the problem, there
exists u ∈ L2 (0, T ; V) and p ∈ W −1,∞ (0, T ; L2loc (Ω)) such that
∂u
− ∆u + ∇p = f .
∂t
(13)
For a given f , u is unique although there exists infinitely many such f .
Moreover f can be choosed such that p be “almost” arbitrary. Indeed
if π is arbitrary in L2 (0, T ; L2loc (Ω)) and (u, p) is solution for the extension
f , then (u, p + π) is as well solution for the other extension f + ∇π of f
because, for all v ∈ V,
Z
h∇π, vi(H −1 (Ω))3 ×(H01 (Ω))3 = − π∇ · v = 0
Ω
7
since v|∂Ω = 0 and ∇ · v = 0.
This shows that, would the Navier–Stokes equations, linearized or not,
be meaningful (but this never holds!) with f ∈ L2 (0, T ; V0 ), they would not
provide any information on the pressure p.
What can be obtained with f ∈ L2 (0, T ; V0 )?
a) Given f ∈ L2 (0, T ; V0 ) and u0 ∈ H, the Lions theorem provides a unique
solution u ∈ L2 (0, T ; V) ∩ C(0, T ; H) such that u(0) = u0 of the variational
equation, in D0 (0, T ),
Z
Z
d
∇u · ∇v = hf, viV0 ×V , ∀v ∈ V.
u·v+ν
dt Ω
Ω
It is for example proved by J.-L. Lions [4], thm. 6.1 p. 69 (in the nonlinear
case), or R. Temam [8], ec. (1.31) p. 253 and proof of thm. 1.1 p. 254.
b) Obviously u is the unique solution of the equivalent equation, in D0 (0, T ),
D ∂u
∂t
− ν∆u, v
E
(H −1 (Ω))3 ×(H01 (Ω))3
= hf, viV0 ×V ,
∀v ∈ V.
c) This u is also the unique solution of the equivalent equation, in D0 (0, T ; V0 ),
dIu
+ Au = f
dt
R
where A and I areR defined from V into V0 by hAu, viV0 ×V = ν Ω ∇u · ∇v
and hIu, viV0 ×V = Ω u · v for all v ∈ V, see for example [8], eq. (1.33) p. 254.
d) Moreover, there exists infinitely many distributions f satisfying (12) and,
for each of them, a distribution p such that (u, p) satisfies (13).
e) But it is not possible to get a distribution p such that (u, p) be solution
of the Navier–Stokes equations with this f .
8
Complements
Definition of H −1
Various authors define H −1 (Ω) as “the dual space of H01 (Ω)”, which is an
abbreviation for “the image of this dual space by the identifications leading
to (9)”. Without identifying L2 (Ω) with its dual space, it writes
n
H −1 (Ω) := v ∈ D0 (Ω) :
o
|hf, viD0 (Ω)×D(Ω) |
<∞ .
kvkH 1 (Ω)
v∈D(Ω), v6=0
sup
This definition posseses three inconvenients:
8
(14)
• It does not describes H −1 (Ω) in terms of partial derivatives.
• It does not extend to W −1,1 (Ω) (W −1,p is the “dual space” of W01,p
only if p > 1).
0
• It does not extend to H −1 (Ω; E) if E is not reflexive or, at least, a
dual space.
These inconvenients disappear with the following definition,
H −1 (Ω) := {v ∈ D0 (Ω) : v = v0 + ∂1 v1 + · · · + ∂d vd , vi ∈ L2 (Ω), ∀i},
that is, the space of “derivatives of order ≤ 1 of elements of L2 ”. This
definition is also classic and equivalent to (14), see for example [3].
An abstract equation
The variational equation (2) given by the Lions theorem is equivalent to the
strong equation
dIu
+ Au = `1 + `2
dt
where I is the Riesz–Fréchet isometry from H onto H 0 , A ∈ L(V ; V 0 ) is
defined by hAw, viV 0 ×V := a(w, v) for all w and v in V , and H 0 ⊂
V 0.
→
0
The operator A is not a partial differential one if V is not a distribution
space.
Generalization of the Lions theorem
Existence in theorem 1, but not uniqueness, may be extended to a form
a ∈ L2 (0, T ; L2 (V × V )) provided that continuity in H of u is replaced by
weak continuity, see [7], th. 2.
9
Conclusion
Which frame choose?
The inclusion V ⊂
H makes up the general frame to solve the parabolic linear
→
pdes, including the above three, with the Lions theorem.
On the contrary, the identification H =
H 0 coupled with H 0 ⊂
V 0 is com↔
→
patible with the imbedding in distribution spaces solely for very particular
pdes. It is unnecessary for pdes with which it is compatible and a source of
error for the other ones.
9
Which model choose?
Heat equation with Dirichlet condition is the ideal model of application of
the Lions theorem since it is the most simple parabolic pde of physics.
On the contrary, it is exceptional for its compatibility with the identification H =
H 0 . Therefore, with this identification, the “model” turns into a
↔
“trap model”.
Thanks
The author is indebted with Enrique Fernández Cara, Paolo Galdi and
Jérôme Lemoine for their comments which allow me to improve this work.
References
[1] R. Dautray, J.-L. Lions, Analyse mathématique et calcul numérique
pour les sciences et les techniques, 8 volumes, Masson, 1984.
[2] Lions, J.-L. Quelques remarques sur les équations différentielles opérationnelles du 1er ordre, Rend. Sem. Mat. Padova, 33 (1963), 213–225.
[3] J.-L. Lions. Problèmes aux limites dans les équations aux dérivées partielles, Presses Univ. Montreal, 1965 (= Œuvres choisies, I, Équations
aux dérivées partielles, Interpolation, EDP Sciences, 2003, 431–576).
[4] J.-L. Lions. Quelques méthodes de résolution des problèmes aux limites non linéaires, Dunod & Gauthier-Villars, 1969.
[5] F. Ortegón Gallego. Algunos modelos de la mecánica de fluidos. Bol.
Soc. Esp. Mat. Apl., 17 (2001), 51–81.
[6] J. Simon. On the existence of the pressure for solutions of the variational Navier–Stokes equations. J. Math. Fluid Mech., 1 (1999), 225–
234.
[7] J. Simon. Una generalización del teorema de Lions–Tartar, Bol. Soc.
Esp. Mat. Apl., 40 (2007), 43–69.
[8] R. Temam. Navier–Stokes equations, North-Holland, 1977.
10