Gis´ele Ruiz Goldstein, Jerome A. Goldstein ∗, Rosa Maria Mininni, Silvia Romanelli†
March 26, 2018
Abstract
The two related one space dimensional singular linear parabolic equations (1), (2) studied by H. Brezis et al. [2] have different scaling properties. These scaling properties lead to new variants of the Hardy and Caffarelli-Kohn-Nirenberg inequalities. These results are proved, and they imply some non wellposedness results when the constant in the singular potential term is large enough.
Key words. Scaling, Caffarelli-Kohn-Nirenberg inequality, Hardy inequality, (C0)
semi-group.
AMS subject classifications. 35K65, 26D10, 47D06.
1
Introduction
Hardy’s inequality and its important generalization by Caffarelli, Kohn and Nirenberg (CKN) [3] are used (among other things) in connection with establishing wellposedness and illposedness for certain singular parabolic equations. For simplicity we deal only with one space dimensional equations. The Hardy and CKN inequalities for functions on (0, ∞) can be proved (among other ways) by scaling techniques. Here we look at these issues from a reverse perspective. We start with the two equations (1), (2) below that possess scaling properties and from them derive an associated Hardy-type inequality. We use these inequalities to learn new facts about (1) and (2).
In a very interesting 1971 paper, H. Brezis et al. [2] considered the equations ut= 1 2uxx+ γ xux, (1) vt= x vxx+ α vx. (2) ∗
Department of Mathematical Sciences, University of Memphis, 373 Dunn Hall, Memphis, TN 38152-3240, USA (ggoldste@memphis.edu, jgoldste@memphis.edu)
†
Department of Mathematics, University of Bari Aldo Moro, Via E. Orabona 4, 70125 Bari, Italy (rosamaria.mininni@uniba.it, silvia.romanelli@uniba.it)
Here x ≥ 0, t ≥ 0, γ > −1/2, α = γ + 1/2 > 0; the initial conditions are u(x, 0) = f (x), v(x, 0) = g(x) = f (√x); and the boundary conditions at the origin are
ux(0, t) = 0,
√
x v(x, t) → 0 as x → 0. These problems are equivalent under the change of variable
v(x, t) = u(√x, t/2).
The compatibility condition f0(0) = 0 is also assumed. These problems also arose in the range condition calculations of [7].
An interesting aspect of these problems is the scaling property: (1) scales like λ2, while (2) scales like λ. This means the following. For functions on (0, ∞) let (U (λ) f )(x) = f (λ x). Then U (λ)−1 = U (1/λ) and it is easily seen that, for D = d/dx,
U (λ)−1DnU (λ) = λnDn, U (λ)−1xαU (λ) = λ−αxα, U (λ)−1xαDnU (λ) = λn−αxαDn.
Here n ∈ N, α ∈ R, and xα is the operator of multiplication of a function of x by xα. Rewrite (1), (2) as
ut= A u, vt= B v; (3)
then
U (λ)−1A U (λ) = λ2A, U (λ)−1B U (λ) = λB,
which we summarize by saying “A scales like λ2” and “B scales like λ ”. The same assertions hold for A + c/x2 and B + c/x where c is a constant.
Hardy’s inequality in L2(0, ∞) says Z ∞ 0 |u0(x)|2dx ≥ 1 4 Z ∞ 0 |u(x)|2 x2 dx (4)
for u ∈ Cc∞(0, ∞), with 1/4 being the optimal constant. In terms of quadratic forms, this says
h−D2u, ui ≥ 1 4 1 x2u, u ,
and both −D2 and 1/x2 scale like λ2. The scaling feature is preserved if the Lebesgue measure dx is replaced by xβdx for any β ∈ R, with a β dependent optimal constant, and the resulting version of Hardy’s inequality becomes the one-dimensional CKN inequality. For completeness, we will include the known scaling proof of the CKN inequality in Section
2. In particular, there are many proofs of Hardy’s inequality, but scaling is of primary concern here.
To return to (1) and (2), the above discussion suggests that there could be a Hardy or CKN inequality relating A u and B u (see (3)) with c/x2 and c/x in L2(0, ∞) and Lp(0, ∞) as well. This is true and is the main result of this paper. Using this we will discuss some non wellposedness results for the associated initial value problems when c exceeds the critical Hardy (or CKN) constant.
2
The CKN inequality in one dimension
Proposition 1. Let 1 < p < ∞, β ∈ R, u ∈ Cc1(0, ∞). Then
Z ∞ 0 |u0(x)|pxβdx ≥ 1 + β − p p pZ ∞ 0 |u(x)|p xp x βdx. (5)
The constant is best possible and is not attained if u 6= 0.
Hardy’s inequality corresponds to β = 0, and this reduces to (4) when β = 0 and p = 2. Here is the scaling proof of Proposition 1 (cf. [6]).
Proof. Take u to be real valued, u ∈ Cc1(0, ∞). Then
|u(x)|p = Z 1 0 d dλ(|u(λ x)| p) dλ (6) = Z 1 0
p |u(λ x)|p−1sign0(u(λx)) u0(λx) x dλ,
where sign0(z) =
z
|z| or 0 according as z 6= 0 or z = 0. Let y = λx, and assume 1 + β − p < 0. Thus Z ∞ 0 |u(x)|pxβ−pdx = Z ∞ 0 Z 1 0
p |u(λ x)|p−1sign0(u(λx)) u0(λx) xβ−p+1dλ dx
= Z ∞
0
Z 1
0
p |u(y)|p−1sign0(u(y)) u0(y) yβ−p+1λ−(β−p+2)dλ dy
and
Z 1
0
λ−(β−p+2)dλ = 1 p − β − 1.
If 1 + β − p > 0, replace R01. . . dλ by −R1∞. . . dλ in (6). The result is, for 1 + β − p 6= 0,
|1 + β − p| Z ∞ 0 |u(x)|p |x|p x βdx =Z ∞ 0
p |u(y)|p−1sign0(u(y)) u0(y) y1−pyβdy
≤ p Z ∞ 0 |u(y)|p yp yβdy 1/p0Z ∞ 0 |u0(y)|pyβdy 1/p , (7)
which is finite by H¨older’s inequality when 1 p+ 1 p0 = 1. Then (7) implies 1 + β − p p pZ ∞ 0 |u(y)|p yp y βdy ≤ Z ∞ 0 |u0(y)|pyβdy,
the desired inequality. Note that in (7) the two terms, |u(y)| y
p
and |u0(y)|p, differ by a small amount (y) > 0 for y near zero. Furthermore equality in (7) is attained only when R |f g| dµ = ||f ||p||g||p, which requires g to be in a one-dimensional space determined by
f when f 6= 0.
A careful analysis leads to conclusion that the constant is best possible but there are no nonzero extremals. The extension from real u to complex u is done in the usual way.
Note that (5) is in fact interesting even when β = p − 1, since the constant 0 on the right hand side is best possible. That is, for every > 0 there is a nonzero u such that
Z ∞ 0 |u0(x)|pxβdx < Z ∞ 0 |u(x)|p xp x βdx.
These considerations lead to some dissipativity results for A and B, which we state here as our main theorem.
Theorem 1. Let H = L2(0, ∞). • (a) OnH, for c ∈ R, let
Ac:= 1 2D 2+γ xD + c x2 = A0+ c x2 with D(Ac) = {u ∈ H2(0, ∞) : u0(0) = 0, u0 x ∈H, c x2u ∈H}
and γ > −1/2. Then A0+ ωI is m-dissipative iff ω ≤ γ −
1
4 and Acis m-dissipative on H iff c ≤ 1
8 − γ
2. Thus, Ac generates a (C0) quasicontractive semigroup for c ≤ 1 8− γ 2. • (b) On H, for c ∈ R, let Bc:= xD2+ αD + c x = B0+ c x with D(Bc) = {u ∈ Hloc2 (0, ∞) : xu00, u0, c xu ∈H}.
Then B0 is m-dissipative on H for all α > 0. But for α > 0, c > 0, Bc is not
3
The Hardy inequalities for A and B
We begin with the proof of Theorem 1. Proof. We work inH = L2(0, ∞). DenoteD
0= Cc2(0, ∞). Let D = d/dx, L = 2γ
x D with
D(D) = D(L) = D0. Then D ∗
= −D and ˜A = D2+ L is twice A0 = 12D2+ 12L. Note
that A and, respectively, B defined in (3) coincide with A0 and, respectively, with B0. We
have hLu, ui = 2γ Z ∞ 0 u0(x) x u(x) dx = −2γ Z ∞ 0 u(x) u 0(x) x − u(x) x2 ! dx = 2γ Z ∞ 0 |u(x)|2 x2 dx − hu, Lui, whence RehLu, ui = γ Z ∞ 0 |u(x)|2 x2 dx. So
Reh ˜Au, ui = hD2u, ui + RehLu, ui = − Z ∞ 0 |u0(x)|2dx + γ Z ∞ 0 |u(x)|2 x2 dx ≤ (γ − 1 4) Z ∞ 0 |u(x)|2 x2 dx
by Hardy’s inequality (4), and the constant (γ − 1
4) is best possible (and not attained). In other words, RehA0u, ui ≤ ( γ 2 − 1 8) Z ∞ 0 |u(x)|2 x2 dx, and Reh−A0u, ui ≥ ( 1 8 − γ 2) Z ∞ 0 |u(x)|2 x2 dx, (8)
is the sharp L2 Hardy’s inequality for −A0. Since γ > −1/2, we have
1 8− γ 2 < 1 8 + 1 4 = 3 8. Hardy’s inequality in the form
Reh−A0u, ui ≥ c Z ∞ 0 |u(x)|2 x2 dx fails to hold if c > 1 8 − γ 2.
Now consider the operator B0 = x D2+ α D. For u ∈ Cc2(0, ∞), hB0u, ui = hx u00+ α u0, ui = Z ∞ 0 x u00(x) u(x) dx + α Z ∞ 0 u0(x) u(x) dx, RehB0u, ui = − Z ∞ 0 |u0(x)|2x dx + (α − 1) Z ∞ 0 d dx |u(x)|2 2 dx = − Z ∞ 0 |u0(x)|2x dx ≤ 0 (9)
and (9) is the CKN inequality (5) with β = 1, p = 2. The constant on the right hand side of (5) is 0, but it is best possible. Thus for any > 0 there is a 0 6= u ∈ Cc2(0, ∞) such
that Reh−B0u, ui > − Z ∞ 0 |u(x)|2 x dx. (10)
The assertions involving dissipativity onH for A0and B0now follow. For the m-dissipative
assertion involving A0 and B0, the range conditions follow from the proofs in [2] and [7],
which show that the range of λI − (A0+ ωI) and λI − B0 for large λ contain C0[0, ∞) ∩
L2(0, ∞) which is dense in H. The semigroup generation assertions follow from the Hille-Yosida theorem [9].
Now we consider c ∈ R and the perturbed operators Ac:= 1 2D 2+ 1 2L + c x2, Bc:= xD 2+ αD + c x. Observe that U (λ)−1AcU (λ) = λ2Ac, U (λ)−1BcU (λ) = λ Bc.
The new Hardy’s inequality (8) implies
h−Acu, ui ≥ 0
for all u ∈ Cc2(0, ∞) iff c ≤ 1 8−
γ
2, and it fails when c > 1 8 −
γ 2.
The assertions about Ac and Bc now follow using the range condition calculations in
[7]. That Acand Bcgenerate (C0) quasicontractive semigroups for c ≤ 18−γ2 follows from
the Hille - Yosida theorem [8]. This completes the proof of Theorem 1.
Our next comments concern various notions of wellposedness of the Cauchy problems (1) and (2) in a broad context.
4
The Non-quasidissipative Case
The recent research monograph by Cialdea and Maz’ya [4] studies ∂u∂t = Lu where L is a second order uniformly elliptic operator with complex coefficients and various bound-ary conditions. These operators generate (C0) semigroups on Lp spaces in a very general
context. But for those operators only quasicontractive semigroups are considered, that is, semigroups T satisfying kT (t)k ≤ M eωt for M = 1 rather than M ≥ 1. This often
leads to cleaner theorems involving necessary and sufficient conditions rather than merely sufficient conditions. In cases where one can construct a (C0) semigroup which is not
qua-sicontractive, one is often working with analytic semigroups. Here is a concrete nontrivial example.
The parabolic problem
ut= Gu = ∆u +
c x2u
in Lp(RN) where G = Gp with a suitable domain in Lp(RN) is quasidissipative iff G is
m-dissipative iff c < (N −22 )2(pp40) where 1p + p10 = 1. (See [1].) By a result in [9] the
semigroup etGp has an analytic extension into the (maximal) sector
Σ(θp) = {z ∈ C : Re z > 0, |arg(z)| < θp =
π
2 − arctan
|p − 2|
2√p − 1}, 1 < p < ∞. For 0 < α < θp with p 6= 2, e−iαGp and eiαGp both generate (C0) semigroups which are
not quasicontractive, see [5]. Suppose µ0 ∈ σp(Ac), that is, Acϕ = µ0ϕ for some µ0 ∈ C
and some 0 6= ϕ ∈H. Then
Acψ = λ2µ0ψ for ψ = U (λ) ϕ.
Then all positive multiples of µ0 are eigenvalues of Ac.
If Reµ0> 0, then {λµ0 : λ > 0} ⊂ σp(Ac) and so Accannot generate a (C0) semigroup.
Indeed, if λ > 0 and Acϕ = λµ0ϕ, then et Acϕ = et λµ0ϕ and so the estimate
k et Ac k≤ M eω t
for t ≥ 0 cannot hold for fixed constants M, ω. The inequality
hAcu, ui > 0 will hold for some real unit vector u provided c > 1
8 − γ
2, and we will have hAcu, ui ≥ ε k u k2
for some ε > 0 provided −12 < γ ≤ 14. Then necessarily c > 0. By scaling, write
uλ(x) = λ1/2u(λ x);
uλ will be a unit vector and
hAcuλ, uλi ≥ λ2ε k uλk2
for all λ > 0. Consequently Ac cannot be quasidissipative and so cannot generate a
Now we address Bc. The CKN inequality for Bcfails for all c > 0. The above arguments
for Ac adapt readily to Bc and show that Bc cannot generate a (C0) quasicontractive
semigroup on L2(0, ∞) for any c > 0. Moreover, Bc cannot generate a (C0) semigroup on
L2(0, ∞) if Bc has an eigenvalue with positive real part.
To summarize, working in the context of quasicontractive semigroups rather than gen-eral (C0) semigroups restricts the generality but has certain other advantages. We discuss
Acand Bcin this context. Ac(resp. Bc) fails to be quasidissipative when c >
1 8− γ 2 (resp. c > 0).
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