Volume 8 (2007), Issue 1, Article 2, 13 pp.
A SOBOLEV-TYPE INEQUALITY WITH APPLICATIONS
RAMÓN G. PLAZA
D EPARTAMENTO DE M ATEMATICAS Y M ECANICA
IIMAS-UNAM
A PDO . P OSTAL 20-726
C.P. 01000 M EXICO DF, M EXICO
plaza@mym.iimas.unam.mx
Received 02 August, 2006; accepted 09 February, 2007
Communicated by C. Bandle
A BSTRACT. In this note, a Sobolev-type inequality is proved. Applications to obtaining linear
decay rates for perturbations of viscous shocks are also discussed.
Key words and phrases: Sobolev-type inequality, Linear decay rates, Viscous shocks.
2000 Mathematics Subject Classification. 26D10, 35L67.
1. T HE I NEQUALITY
The purpose of this contribution is to prove the following.
Theorem 1.1. Let ψ be a real-valued smooth localized function with non-zero integral,
Z
(1.1)
ψ(x) dx = M 6= 0,
R
satisfying
Z
(1.2)
|xi ∂ j ψ(x)| dx ≤ C,
for all i, j ≥ 0.
R
Then there exists a uniform constant C∗ > 0 such that
1/2
1/2
sup |u(x)| ≤ C∗ kukL2 kux − αψkL2 ,
(1.3)
x
1
for all u ∈ H (R) and all α ∈ R.
Clearly, this result is an extension of the classical Sobolev inequality
kuk2∞ ≤ 2kukL2 kux kL2 .
Assuming ψ satisfies (1.1) and (1.2), inequality (1.3) is valid for any u ∈ H 1 (R) and all α ∈ R;
here the constant C∗ > 0 is independent of u and α, but depends on ψ. This result may be
useful while studying the asymptotic behavior of solutions to evolution equations that decay to
This research was partially supported by the University of Leipzig and the Max Planck Society, Germany. This is gratefully acknowledged.
209-06
2
R AMÓN G. P LAZA
a manifold spanned by a certain function ψ (see Section 2 below). It is somewhat surprising
that the result holds for all α ∈ R. The crucial fact is that the antiderivative of ψ cannot be in
L2 , thanks to hypothesis (1.1). In this fashion we avoid the case ux ∈ span{ψ}.
We would like to establish (1.3) by extremal functions. Since the solution to the minimization problem associated with (1.3) may not exist, our approach will consist of studying
a parametrized family of inequalities for which we can explicitly compute extremal functions
for each parameter value.
Theorem 1.2. Under the assumptions of Theorem 1.1, there exists a constant c∗ > 0 such that
c∗ ≤ ρ−1 kuk2L2 + ρkux − αψk2L2 ,
(1.4)
for all ρ > 0, α ∈ R, and u in a dense subset of H 1 (R) with u(0) = 1. Moreover, c∗ is also
uniform under translation ψ̃(·) = ψ(· + y), where y ∈ R (even though hypothesis (1.2) is not
uniform by translation).
Proposition 1.3. Theorem 1.2 implies Theorem 1.1.
Proof. It suffices to show that
(1.5)
1/2
1/2
|u(0)| ≤ C∗ kukL2 kux − αψkL2 ,
with uniform C∗ > 0, also by translation. Indeed, we can always take, for any y ∈ R, ũ(x) :=
u(x + y), ψ̃(x) := ψ(x + y), yielding
1/2
1/2
|u(y)| = |ũ(0)| ≤ C∗ kũkL2 kũx − αψ̃kL2
1/2
1/2
= C∗ ku(· + y)kL2 kux (· + y) − αψ(· + y)kL2
1/2
1/2
= C∗ kukL2 kux − αψkL2 ,
∀y ∈ R,
by uniformity of C∗ and by translation invariance of Lp norms. This shows (1.3).
Now assume Theorem 1.2 holds. If u(0) = 0 then (1.5) holds trivially. In the case u(0) 6= 0,
consider ũ = u/u(0), α̃ = α/u(0) and apply (1.4),
c∗ u(0)2 ≤ ρ−1 kuk2L2 + ρkux − αψk2L2 .
Minimizing over ρ yields ρ = kukL2 /kux − αψkL2 , so that
c∗ u(0)2 ≤ 2kukL2 kux − αψkL2 .
p
This proves (1.5) with C∗ = 2/c∗ .
Therefore, we are left to prove Theorem 1.2.
1.1. Proof of Theorem 1.2. Without loss of generality assume that
kψkL2 = 1.
(1.6)
Since u ∈ H 1 , we may use the Fourier transform, and the constraint u(0) = 1 becomes
Z
(1.7)
û(ξ) dξ = 1,
R
up to a constant involving π. Note that the expression on the right of (1.4) defines a family of
functionals parametrized by ρ > 0,
Z
Z
ρ
−1
2
(1.8)
J [u] := ρ
|û(ξ)| dξ + ρ |iξ û(ξ) − αψ̂(ξ)|2 dξ.
R
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S OBOLEV- TYPE INEQUALITY
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We shall see by direct computation that the minimizer u exists and is unique (given by a simple
formula) for each ρ and α. Denote û = v + iw, ψ̂ = η + iθ (real and imaginary parts). Then
each functional (1.8) can be written as
Z
Z
ρ
−1
2
2
(1.9) J [u] = ρ
(v + w ) dξ + ρ (ξ 2 (v 2 + w2 ) + 2αξ(wη − vθ) + α2 (η 2 + θ2 )) dξ.
R
R
The constraint (1.7) splits into
minimization problem
R
v dξ = 1 and
R
w dξ = 0. Hence, we have the following
min J ρ [(v, w)]
u∈H 1 (R)
subject to
Z
I1 [(v, w)] =
v dξ − 1 = 0,
ZR
I2 [(v, w)] =
w dξ = 0,
R
for each ρ > 0 and α ∈ R. The Lagrange multiplier conditions
1
D
J [(v, w)]
2 (h1 ,0)
1
D
J [(v, w)]
2 (0,h2 )
= µD(h1 ,0) I1 [(v, w)],
= νD(0,h2 ) I2 [(v, w)],
yield
Z
−1
Z
2
(ρ v + ρξ v − ραθξ)h1 dξ = µ
h1 dξ,
R
Z
R
−1
Z
2
(ρ w + ρξ w + ραηξ)h2 dξ = ν
R
h2 dξ,
R
for some (µ, ν) ∈ R2 and for all test functions (h1 , h2 ). Therefore
ρ−1 v + ρξ 2 v − ραξθ = µ,
ρ−1 w + ρξ 2 w + ραξη = ν.
Denote λ = µ + iν. Multiply the second equation by i, and solve for v and w to obtain
ρλ − iαρ2 ξ ψ̂(ξ)
.
1 + ρ2 ξ 2
Equation (1.10) is, in fact, the expression for the minimizer. Whence, we can compute the
minimum value of J ρ for each ρ > 0, in terms of λ and α. Substituting (1.10) one obtains
(after some computations),
(1.10)
û =
ρ(|λ|2 + α2 |ψ̂|2 )
.
1 + ρ2 ξ 2
Hence we easily find that the minimum value of J ρ is given by
Z
Z
ρdξ
|ψ̂(ξ)|2
ρ
2
2
Jmin = |λ|
+
α
ρ
dξ
2 2
2 2
R 1+ρ ξ
R 1+ρ ξ
(1.11)
= π|λ|2 + α2 Γ(ρ),
ρ−1 |û|2 + ρ|iξ û − αψ̂|2 =
where
Z
(1.12)
Γ(ρ) := ρ
R
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
|ψ̂(ξ)|2
dξ.
1 + ρ2 ξ 2
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R AMÓN G. P LAZA
Now we find the Lagrange multiplier λ in terms of α using the constraint (1.7), which implies
Z
Z
ρdξ
iξ ψ̂(ξ)
2
1=λ
− αρ
dξ = λπ + αΘ(ρ),
2
2
2 2
R 1+ρ ξ
R 1+ρ ξ
where
(1.13)
2
Z
Θ(ρ) := −ρ
R
iξ ψ̂(ξ)
dξ.
1 + ρ2 ξ 2
Solving for λ we find,
(1.14)
λ=
1
(1 − αΘ(ρ)).
π
Observe that since ψ is real, then ψ̂(ξ) = ψ̂(−ξ) and therefore Θ(ρ) ∈ R for all ρ > 0. This
readily implies that λ ∈ R and, upon substitution in (1.11), that
(1.15)
ρ
Jmin
=
1
(1 − αΘ(ρ))2 + α2 Γ(ρ).
π
The latter expression is a real quadratic polynomial in α ∈ R. Minimizing over α we get
(1.16)
α=
Θ(ρ)
∈ R.
πΓ(ρ) + Θ(ρ)2
Thus, we can substitute (1.16) in (1.15), obtaining in this fashion the lower bound
Γ(ρ)
> 0.
πΓ(ρ) + Θ(ρ)2
R
Remark 1.4. The choice (1.16) corresponds to taking α = iξ ûψ̂dξ ∈ R, as the reader may
easily verify using (1.10). Intuitively, the most we can do with α in (1.8) is to remove the
ψ̂-component of û. In other words, if we minimize kux − αψkL2 over α we obtain α =
R 2
R
R
¯
ux ψ dx
ψ dx = iξ ûψ̂dξ (recall kψkL2 = 1). We can substitute its value in the expression of the minimizer to compute the lower bound I(ρ).
ρ
Jmin
≥ I(ρ) :=
We do not need to show that (1.10) is the actual minimizer. The variational formulation
simply helps us to compute a lower bound for the functional in terms of ρ. Next, we study the
behavior of Θ(ρ) and Γ(ρ) for all ρ > 0. We are particularly interested in what happens for
large ρ. In addition, we have to prove that the lower bound is uniform in y ∈ R if we substitute
ψ(·) by ψ(· + y), a property that was required in the proof of Proposition 1.3.
Lemma 1.5. There holds
(i) Γ(ρ) ∈ R+ for all ρ > 0 and it is invariant under translation ψ(·) → ψ(· + y) for any
y ∈ R,
(ii) C −1 ρ ≤ Γ(ρ) ≤ Cρ for ρ ∼ 0+ , and some C > 0,
(iii) Γ(ρ) → πM 2 as ρ → +∞,
(iv) Θ(ρ) ≤ Cρ2 for ρ ∼ 0+ , and
(v) Θ(ρ) is uniformly bounded under translation ψ(·) → ψ(· + y) with y ∈ R, as ρ → +∞.
\
Proof. (i) is obvious, as |ψ(·
+ y)(ξ)| = |eiξy ψ̂(ξ)| = |ψ̂(ξ)|; also by (1.1), it is clear that
Γ(ρ) > 0, for all ρ > 0.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
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S OBOLEV- TYPE INEQUALITY
5
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(ii) follows directly from Γ(ρ) ≤ ρ |ψ̂|2 dξ = ρ for all ρ > 0, because of (1.6), and from
noticing that
Z
|ψ̂(ζ/ρ)|2
Γ(ρ) =
dζ
2
R ζ +1
Z
Z
=
+
|ζ|≤1
|ζ|≥1
Z
Z
1
ρ
2
≥
|ψ̂(ζ/ρ)| dζ =
|ψ̂(ξ)|2 dξ.
2 |ζ|≤1
2 |ξ|≤1/ρ
Since kψkL2 = 1, we have for ρ sufficiently small,
Z
1
|ψ̂(ξ)|2 dξ ≥ ,
2
|ξ|≤1/ρ
and thus Γ(ρ) ≥ 14 ρ = C −1 ρ for ρ ∼ 0+ .
(iii) to prove (iii), notice that |ψ̂| is bounded, ψ̂(ζ/ρ) → ψ̂(0) as ρ → +∞ pointwise, and
(ζ 2 + 1)−1 is integrable; therefore we clearly have
Z
Z
|ψ̂(ζ/ρ)|2
|ψ̂(0)|2
Γ(ρ) =
dζ
−→
dζ = π|ψ̂(0)|2 = πM 2 > 0,
2+1
2
ζ
1
+
ζ
R
R
as ρ → +∞.
(iv) follows directly from hypothesis (1.2), as
Z
Z
|ξ ψ̂(ξ)|
2
2
|Θ(ρ)| ≤ ρ
dξ ≤ ρ
|ξ ψ̂(ξ)| dξ ≤ Cρ2 .
2
2
R 1+ρ ξ
R
Note that this estimate is valid also by translation, even though ψ(· + y) may not satisfy (1.2).
(v) in order to prove (v), we first assume that ψ itself satisfies (1.1) and (1.2). Split the integral
into two parts,
Z
Z
iξ ψ̂(ξ)
iξ ψ̂(ξ)
Θ(ρ) = −
dξ −
dξ := I1 + I2 .
2
2
2
2
|ξ|≤1 ξ + 1/ρ
|ξ|≥1 ξ + 1/ρ
I2 is clearly bounded as ρ → +∞ by hypothesis (1.2),
Z
Z
|ξ ψ̂(ξ)|
|I2 | ≤
dξ ≤
|ξ ψ̂(ξ)| dξ ≤ C.
2
2
|ξ|≥1 ξ + 1/ρ
R
Denote
(
φ(ξ) :=
1
(ψ̂(ξ)
ξ
dψ̂
(0)
dξ
− ψ̂(0)) for ξ 6= 0,
for ξ = 0.
φ is continuous. Then, I1 can be further decomposed into
Z
Z
iξ dξ
iξ 2 φ(ξ)
I1 = −ψ̂(0)
dξ
−
dξ.
2
2
2
2
|ξ|≤1 ξ + 1/ρ
|ξ|≤1 ξ + 1/ρ
The first integral is clearly zero for all ρ > 0, and the second is clearly bounded as
Z
Z
ξ 2 |φ(ξ)|
dξ ≤
|φ(ξ)| dξ ≤ C.
2
2
|ξ|≤1 ξ + 1/ρ
|ξ|≤1
Therefore, Θ(ρ) is bounded as ρ → +∞.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
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R AMÓN G. P LAZA
Now, let us suppose that ψ(·) = ψ0 (· + y) for some fixed y ∈ R, y 6= 0, where ψ0 satisfies
(1.1) and (1.2). Then clearly ψ̂(ξ) = eiξy ψ̂0 (ξ) and
Z
iξeiξy ψ̂0 (ξ)
Θ(ρ) = −
dξ.
2
2
R ξ + 1/ρ
Assume that y > 0 (the case y < 0 is analogous); then consider the function
izeizy ψ̂0 (z)
g(z) = 2
,
z + 1/ρ2
for z in Im z > 0, and take the upper contour
C = [−R, R] ∪ {z = Reiθ ; θ ∈ [0, π]},
for some R > 0 large. Then g(z) is analytic inside C except at the simple pole z = i/ρ. (When
y < 0 one takes the lower contour that encloses the pole at z = −i/ρ.) By complex integration
of g along C in the counterclockwise direction, and by the residue theorem, one gets
Z
g(z) dz = 2πi Resz=i/ρ g(z) = −πe−y/ρ ψ̂0 (i/ρ).
C
Therefore it is easy to see that the value Θ(ρ) is uniformly bounded in y ∈ R as
|Θ(ρ)| ≤ π|ψ̂0 (i/ρ)| → π|M | > 0
when ρ → +∞. This completes the proof of the lemma.
Remark 1.6. If we consider the solution uρ to
−uxx +
(1.17)
1
u = ψx ,
ρ2
then, after taking Fourier transform, one finds
ûρ (ξ) =
iξ ψ̂(ξ)
,
+ 1/ρ2
ξ2
R
so that uρ (0) = ûρ dξ = −Θ(ρ). The claim that uρ (0) is bounded as ρ → +∞ is plausible
because in the limit (formally) we have −uρxx = ψx or uρx = −ψ. Since ψ is integrable, uρ
should be bounded. The bound Θ(ρ) ∼ e−|y|/ρ represents the (slow) exponential decay of the
Green’s function solution to (1.17).
In Lemma 1.5, we have shown that Θ(ρ) and Γ(ρ) are uniformly bounded for ρ large and
in y ∈ R. The same applies to I(ρ). For ρ near 0, since both tend to zero as ρ → 0+ , by
L’Hôpital’s rule we get
lim+ I(ρ) = lim+
ρ→0
ρ→0
dΓ
dρ
π dΓ
dρ
+
2Θ dΘ
dρ
= π −1 > 0,
because (ii) implies (dΓ/dρ)|ρ=0+ ≥ C −1 > 0, and dΘ/dρ is bounded as ρ → 0+ by (iv).
Therefore, the constant I(ρ) is uniformly bounded from above and below for all ρ > 0, in
ρ
particular for ρ → +∞. This implies the uniform boundedness from below of Jmin
and of
ρ
J [u] for all u in the constrained class of functions considered in Theorem 1.2. Furthermore,
the lower bound is uniform by translation as well. This completes the proof.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
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S OBOLEV- TYPE INEQUALITY
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Remark 1.7. The corresponding Fourier L1 estimate
1/2
1/2
kûkL1 ≤ CkûkL2 kiξ û − αψ̂kL2 ,
(from which the result can be directly deduced), does not hold. Here it is a counterexample: let
ψ be a nonnegative function with compact support and let Ψ be its antiderivative. Set
u(x) := Ψ(x) − Ψ(x/L),
where L > 0 is large. Then there is R > 0 such that u vanishes outside |x| ≤ RL. Henceforth
kukL2 ≤ CL for some C > 0. Moreover, we also have ux − ψ = ψ(x)/L, and consequently
kux −ψkL2 ≤ C/L. This implies that the product kukL2 kux −ψkL2 remains uniformly bounded
in L. Now, the Fourier transform of u is
i
ψ̂(Lξ) − ψ̂(ξ) .
û(ξ) = Ψ̂(ξ) − LΨ̂(Lξ) =
ξ
Since ψ̂ has compact support, it vanishes outside |ξ| ≤ R̃, for some R̃ > 0. Now, |ψ̂(0)| =
M > 0 implies that |ψ̂(ξ)| > 0 near ξ = 0, and we can choose L sufficiently large such that
|ψ̂(ξ)| ≥ c0 for R̃/L ≤ |ξ| ≤ δ∗ /2, where δ∗ = sup {δ > 0 ; |ψ̂(ξ)| > 0 for 0 ≤ |ξ| < δ}, and
c0 is independent of L. Therefore
|û(ξ)| =
|ψ̂(ξ)|
c0
≥
,
|ξ|
|ξ|
for all R̃/L ≤ |ξ| ≤ δ∗ /2, and the L1 norm of û behaves like
Z
dξ
kûkL1 ≥ c0
∼ c0 ln L → +∞,
R̃/L≤|ξ|≤δ∗ /2 |ξ|
as L → +∞.
2. A PPLICATIONS TO V ISCOUS S HOCK WAVES
To illustrate an application of uniform inequality (1.3), consider a scalar conservation law
with second order viscosity,
(2.1)
ut + f (u)x = uxx ,
where (x, t) ∈ R × [ 0, +∞), f is smooth, and f 00 ≥ a > 0 (convex mode). Assume the
triple (u− , u+ , s) (with u+ < u− ) is a classical shock front [5] satisfying the Rankine-Hugoniot
jump condition1 −s[u] + [f (u)] = 0, and Lax entropy condition f 0 (u+ ) < s < f 0 (u− ). A
shock profile [1] is a traveling wave solution to (2.1) of form u(x, t) = ū(x − st), where ū
satisfies ū00 = f (ū)x − sū0 , with 0 = d/dz, z = x − st, and ū → u± as z → ±∞. Without
loss of generality we can assume s = 0 by normalizing f (see e. g. [3]), so that f (u± ) = 0,
f 0 (u+ ) < 0 < f 0 (u− ) and the profile equation becomes
ūx = f (ū).
(2.2)
Such a profile solution exists, and under the assumptions, it is both monotone ūx < 0 and
exponentially decaying up to two derivatives
|∂xj (ū(x) − u± )| . e−c|x| ,
for all 0 ≤ j ≤ 2 and some constant c > 0 (see [7, 8, 1] and the references therein)2 .
We will show that the following consequence of Theorem 1.1 is useful to obtain decay rates
for solutions to the linearized equations for the perturbed problem.
1
2
Here [g] denotes the jump g(u+ ) − g(u− ) for any g.
In the sequel “.” means “≤” modulo a harmless positive constant.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
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R AMÓN G. P LAZA
Lemma 2.1. Let ū be the shock profile solution to (2.2). Then
kuk2L∞ . kukL2 kux − αūx kL2 ,
(2.3)
for all u ∈ H 1 (R) and all α ∈ R.
Proof. Follows immediately from Theorem 1.1 with ψ = ūx , which satisfies hypotheses (1.1)
and (1.2), as ūx is exponentially decaying and has non-zero integral [u] 6= 0.
Consider a solution to (2.1) of the form u+ ū, u being a perturbation; linearizing the resulting
equation around the profile we obtain
ut = Lu := uxx − (f 0 (ū)u)x ,
(2.4)
where L is a densely defined linear operator in, say, L2 . In [4], Goodman introduced the flux
transform F : W 1,p → Lp , where Fu := uxx − f 0 (ū)ux as a way to cure the negative sign
of f 00 (ū)ūx < 0. That is, if u solves (2.4) then clearly its flux variable v := Fu satisfies the
“integrated” equation [2],
ut = Lu := uxx − f 0 (ū)ux ,
(2.5)
which leads to better energy estimates. Another feature of the flux transform formulation is the
following inequality (see [4] for details, or [6] – Chapter 4, Proposition 4.6 – for the proof).
Lemma 2.2 (Poincaré-type inequality). There exists a constant C > 0 such that for all 1 ≤ p ≤
+∞ and u ∈ Lp ,
ku − δūx kLp ≤ CkFukLp ,
(2.6)
where δ is given by
1
δ=
Z
(2.7)
and Z =
R
R
Z
uūx dx,
R
ū2x dx > 0 is a constant.
Here we illustrate an application of the uniform estimate (2.3) to obtain sharp decay rates for
solutions to the linearized perturbation equation, using the flux formulation due to Goodman.
Proposition 2.3 (Goodman [4]). For all global solutions to ut = Lu, with suitable initial
conditions, there holds
ku(t) − δ(t)ūx kL∞ . t−1/2 ku(0)kW 1,1 ,
(2.8)
where δ(t) is given by (2.7).
Remark 2.4. This is a linear stability result with a sharp decay rate (the power t−1/2 is that of
the heat equation, and therefore, optimal). Notice also that δ(t) depends on t, corresponding (at
least at this linear level) to an instantaneous projection onto the manifold spanned by ū. The
need of a uniform inequality for all δ ∈ R such as (2.3) is thus clear. For a very comprehensive
discussion on (nonlinear) “wave tracking” and stronger results, see Zumbrun [9].
Remark 2.5. The formal adjoint of the integrated operator is given by
L∗ u := uxx + (f 0 (ū)u)x .
Note that if v and w are solutions to vt = Lv and wt = −L∗ w, respectively, then
Z
Z
d
v(t)w(t) dx = (wLv − vL∗ w) dx = 0,
dt R
R
and hence
Z
Z
v(t)w(t) dx =
v(0)w(0) dx, for all t ≥ 0.
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S OBOLEV- TYPE INEQUALITY
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In the sequel, we will gloss over many details, such as global existence of the solutions to
the linear equations, or the correct assumptions for initial conditions in suitable spaces (which
are standard and can be found elsewhere [2, 9]), and concentrate on filling out the details of the
proof of Proposition 2.3 sketched in [4].
2.1. Energy Estimates. We start with the basic energy estimate.
Lemma 2.6. Let v be a solution to either vt = Lv or vt = L∗ v. Then for all t ≥ s ≥ 0 we have
the basic energy estimate
Z
1
1d
2
2
(2.9)
kv(t)kL2 ≤ −kvx (t)kL2 −
f 00 (ū)|ūx |v(t)2 dx < 0,
2 dt
2 R
and,
kv(t)k2L2 ≤ kv(s)k2L2 ,
(2.10)
t
Z
(2.11)
s
Z tZ
f 00 (ū)|ūx |v(τ )2 dxdτ ≤ kv(s)k2L2 .
(2.12)
s
1
kvx (τ )k2L2 dτ ≤ kv(s)k2L2 ,
2
R
Proof. Follows by standard arguments. Multiply vt = Lv by v and integrate by parts once to
get (2.9). Likewise, multiply vt = L∗ v by v and integrate by parts twice to arrive at the same
estimate. The negative sign in (2.9) is a consequence of compressivity of the wave f 00 (ū)ūx < 0.
Estimates (2.10) – (2.12) follow directly from (2.9).
Next, we establish decay rates for vt and w, and solutions to vt = Lv and wt = L∗ w.
Lemma 2.7. Let v be a solution to vt = Lv. Then the following decay rate holds
kvt (t)kL2 . t−1/2 kv(0)kL2 .
(2.13)
Proof. First observe that vt = Fvx , and therefore vtt = (Fvx )t = Fvtx = Lvt , that is, vt
solves the integrated equation as well, and hence, the estimates (2.9) – (2.12) hold for vt also.
In particular, the L2 norm of vt is decreasing. To show (2.13) it suffices to prove
kvt (t)k2L2 . kvx (s)k2L2 ,
(2.14)
for all t > s + 1, s ≥ 0. Integrate (2.14) in s ∈ [0, t − 1] and use (2.11) to obtain
Z t
2
−1
kvt (t)kL2 . (t − 1)
kvx (s)k2L2 ds . (t − 1)−1 kv(0)k2L2 . t−1 kv(0)k2L2 ,
0
for all t ≥ 2, yielding (2.13). To show (2.14) differentiate vt = Lu with respect to x, multiply
by vx and integrate by parts to obtain
Z
1d
1
2
2
(2.15)
kvx (t)kL2 = −kvxx (t)kL2 −
f 0 (ū)x vx2 dx ≤ M kvx (t)k2L2 ,
2 dt
2 R
where M := sup |f 0 (ū)x |. By Gronwall’s inequality
(2.16)
kvx (T + t)k2L2 ≤ eM t kvx (T )k2L2 ,
for all t, T ≥ 0. Integrating (2.15) in t ∈ [s, T ],
Z T
Z Z
1 T
2
2
2
(2.17)
kvx (T )kL2 ≤ kvx (s)kL2 −
kvxx (τ )kL2 dτ −
f 0 (ū)x vx (τ )2 dxdτ.
2
s
s
R
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
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10
R AMÓN G. P LAZA
Estimate the last integral using (2.16), to obtain
Z T Z
Z T −s
0
2
f (ū)x vx dxdτ ≤ M
eM τ kvx (τ )k2L2 dτ ≤ eM (T −s) kvx (s)k2L2 .
s
0
R
Upon substitution in (2.17),
Z T
kvxx (τ )k2L2 dτ ≤ 21 (1 + eM (T −s) )kvx (s)k2L2 .
s
Likewise, from (2.16) it is easy to show that
Z TZ
m M (T −s)
e
kvx (s)k2L2 ,
|f 0 (ū)|vx (τ )2 dxdτ ≤
M
s
R
m Mt
0
where m := sup |f (ū)|. Denoting µ(t) := max 12 (1 + eM t ), N
e
, we see that both
Z T
Z TZ
2
kvxx (τ )kL2 dτ, and
|f 0 (ū)|vx (τ )2 dxdτ,
s
s
−s)kvx (s)k2L2 .
are bounded by µ(T
(2.10) for vt we obtain
(T −
s)kvt (T )k2L2
R
2
Since the L norm of vt is decreasing, integrating inequality
Z
T
kvt (τ )k2L2 dτ
≤
s
T
Z
k(Lv)(τ )k2L2 dτ
=
s
Z
T
.
kvxx (τ )k2L2
Z
T
Z
dτ +
s
s
|f 0 (ū)|vx (τ )2 dxdτ
R
. µ(T − s)kvx (s)k2L2 .
Choose T − s ≡ 1 to finally arrive at
kvt (t)k2L2 ≤ kvt (1 + s)k2L2 . µ(1)kvx (s)k2L2 ,
for all t > 1 + s, establishing (2.14). This proves the lemma.
Lemma 2.8. Let w be a solution to wt = L∗ w. Then the following decay rate holds
kw(t)kL∞ . t−1/4 kw(0)kL2 .
(2.18)
Proof. Recall that (2.9) – (2.12) hold for w. In particular, by convexity f 00 ≥ a > 0 and (2.12),
we have
Z tZ
(2.19)
|ūx |w(τ )2 dxdτ ≤ a−1 kw(0)k2L2 ,
0
R
for all t ≥ 0. Differentiate wt = L∗ w with respect to x, multiply by wx and integrate by parts
to obtain, for all t ≥ 0,
Z
Z
1d
3
1
2
2
00
2
kwx (t)kL2 = −kwxx (t)kL2 −
f (ū)|ūx |wx (t) dx −
f 0 (ū)xxx w(t)2 dx.
2 dt
2 R
2 R
The firstR two terms on the right hand side have the right sign for decay. We must control the
term − f 0 (ū)xxx w2 dx. For that purpose, use the equation for w and the profile equation to
compute
Z
Z
Z
1d
2
2
|ūx |w(t) dx = − |ūx |wx (t) dx − f 00 (ū)|ūx |2 w(t)2 dx.
2 dt R
R
R
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S OBOLEV- TYPE INEQUALITY
11
This provides the cancellation we need, as the decreasing L2 norm we seek will be that of wx
plus a multiple of |ūx |1/2 w. First note by the smoothness of f and convexity that there exists
A > 0 such that
|f 0 (ū)xxx | ≤ A|ūx |2 .
This implies
Z
d 1
1 −1
2
2
kwx (t)kL2 + Aa
|ūx |w(t) dx
dt 2
2
R
Z
Z
3
1
2
00
2
= −kwxx (t)kL2 −
f (ū)|ūx |wx (t) dx −
f 0 (ū)xxx w(t)2 dx+
2 R
2 R
Z
Z
−1
2
−1
− Aa
|ūx |wx (t) dx − Aa
f 00 (ū)|ūx |2 w(t)2 dx
R
R
Z
A
≤ J(t) −
|ūx |2 w(t)2 dx,
2 R
where
J(t) :=
−kwxx (t)k2L2
3
−
2
for all t ≥ 0. Denoting Ā = Aa
Z
−1
00
2
f (ū)|ūx |wx (t) dx − Aa
−1
R
Z
|ūx |wx (t)2 dx ≤ 0,
R
and defining
R(t) :=
kwx (t)k2L2
Z
+ Ā
|ūx |w(t)2 dx,
R
we have thus shown that R(t) is the decaying norm we were looking for, as dR/dt ≤ 0. Integrating R(t) ≤ R(τ ) according to custom with respect to τ ∈ [0, t], for fixed t ≥ 0, and using
(2.11) and (2.19), one can estimate
Z t
tR(t) ≤
R(τ ) dτ ≤ 21 kw(0)k2L2 + Āa−1 kw(0)k2L2 . kw(0)k2L2 .
0
Therefore,
kwx (t)kL2 . t−1/2 kw(0)kL2 ,
for all t > 0 large. By the classical Sobolev inequality and (2.10) we obtain
1/2
1/2
kw(t)kL∞ . kwx (t)kL2 kw(t)kL2 . t−1/4 kw(0)kL2 ,
as claimed.
2.2. Proof of Proposition 2.3. If u solves ut = Lu, then its flux transform v = Fu is a solution
to vt = Lv. Apply the uniform Sobolev-type inequality (2.3) to v, substituting1 α by
Z
1
δ̃(t) =
vx (t)ūx dx,
Z R
R
(with Z = R |ūx |2 dx), and the Poincaré-type inequality (2.6) (with p = 2), to obtain
kv(t)k2L∞ . kv(t)kL2 kvx − δ̃(t)ūx kL2
. kv(t)kL2 k(Fvx )(t)kL2 = kv(t)kL2 kvt (t)kL2 .
Then, using the estimate (2.13), we arrive at
(2.20)
1
kv(t)k2L∞ . (t − s)−1/2 kv(s)k2L2 ,
Here the uniformity of inequality (2.3) in α ∈ R plays a crucial role.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
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12
R AMÓN G. P LAZA
for all t ≥ s + 2. For fixed T > 0 define the linear functional A : L2 → R as
Z
Ag :=
v(T )g dx,
R
2
for all g ∈ L , with norm
Z
kAk = sup v(T )g dx .
kgkL2 =1
R
For every g ∈ L with kgkL2 = 1, we can always solve the equation wt = −L∗ w = −wxx −
(f 0 (ū)w)x on t ∈ [0, T ] “backwards” in time, with w(T ) = g. Thus, by Remark 2.5
Z
Z
Z
v(T )g dx = v(T )w(T ) dx = v(0)w(0) dx ≤ kv(0)kL1 kw(0)kL∞ ,
2
R
R
R
for all T > 0. Making the change of variables w̃(x, t) = w(x, T − t) we readily see that w̃
satisfies w̃t = L∗ w̃ with w̃(0) = g, and we can use estimate (2.18), yielding
kw(0)kL∞ = kw̃(T )kL∞ . T −1/4 kgkL2 .
Thus,
kv(T )kL2
Z
= sup v(T )g dx ≤ kv(0)kL1 kw(0)kL∞ . T −1/4 kv(0)kL1 ,
kgkL2 =1
R
for all T > 0. Choose s = t/2 in (2.20), and apply the last estimate with T = t/2, to get
(2.21)
kv(t)kL∞ . (t/2)−1/4 kv(t/2)kL2 . t−1/2 kv(0)kL1 ,
which corresponds to the optimal decay rate for solutions to the integrated equation.
To prove the decay rate (2.8) for the original solution to the unintegrated equation ut = Lu,
apply the Poincaré-type inequality again (now with p = ∞) together with (2.21),
ku(t) − δ(t)ūx kL∞ . kv(t)kL∞ . t−1/2 kv(0)kL1 . t−1/2 ku(0)kW 1,1 .
This completes the proof.
ACKNOWLEDGEMENTS
The motivation to prove inequality (1.3) originated during my research on viscous shock
waves towards my doctoral dissertation [6], written under the direction of Prof. Jonathan Goodman. I thank him for many illuminating discussions, useful observations, and his encouragement. I am also grateful to Prof. Stefan Müller for suggesting the counterexample in Remark
1.7.
R EFERENCES
[1] I.M. GELFAND, Some problems in the theory of quasi-linear equations, Amer. Math. Soc.
Transl., 29(2) (1963), 295–381.
[2] J. GOODMAN, Nonlinear asymptotic stability of viscous shock profiles for conservation laws,
Arch. Rational Mech. Anal., 95 (1986), 325–344.
[3] J. GOODMAN, Stability of viscous scalar shock fronts in several dimensions, Trans. Amer. Math.
Soc., 311(2) (1989), 683–695.
[4] J. GOODMAN, Remarks on the stability of viscous shock waves, in Viscous Profiles and Numerical
Methods for Shock Waves, M. Shearer, ed., SIAM, Philadelphia, PA, 1991, 66–72.
[5] P.D. LAX, Hyperbolic systems of conservation laws II, Comm. Pure Appl. Math., 10 (1957), 537–
566.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
http://jipam.vu.edu.au/
S OBOLEV- TYPE INEQUALITY
13
[6] R.G. PLAZA, On the Stability of Shock Profiles, PhD thesis, New York University, 2003.
[7] D. SERRE, Systems of Conservation Laws 1: Hyperbolicity, entropies, shock waves, Cambridge
University Press, 1999.
[8] J. SMOLLER, Shock Waves and Reaction-Diffusion Equations, Springer-Verlag, New York, Second ed., 1994.
[9] K. ZUMBRUN, Refined wave-tracking and nonlinear stability of viscous Lax shocks, Methods
Appl. Anal., 7(4) (2000), 747–768.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 2, 13 pp.
http://jipam.vu.edu.au/