Higher regularity for parabolic equations and maximum
principles
1
Differentiability of weak solutions
As for the elliptic case, we can provide an improved regularity for weak solutions. We now
aim to show that if the coefficients are more regular than u is more regular.
Idea: We establish this by roughly speaking replacing derivatives of the formal calculation
by difference quotients, and carefully deriving estimates for these quantities.
Theorem 1.1. Let u be a weak solution of
X
X
ut (x, t) −
Di (aij (x, t)Dj u(x, t)) +
bi Di u + c(x, t)u = f (x, t)
ij
i
where f ∈ L2 (ΩT ), ai,j , bi , c ∈ L∞ (ΩT ), aij ∈ C 1 (ΩT ),
||aij + bi + c||L∞ (ΩT ) ≤ K1 ,
and
||Daij ||L∞ (ΩT ) ≤ K2 ,
X
λ|ξ|2 ≤
aij (x, t)ξi ξj ≤ Λ|ξ|2
ij
00
Then for any Ω ⊂⊂ Ω and 0 < ρ < T , we have that D2 u and ut are in L2 (Ω00 × (2ρ, T − 2ρ))
and
(1.1)
kD2 ukL2 (Ω00 ×(2ρ,T −2ρ)) ≤ C kukL2 (ΩT ) + kDukL2 (ΩT ) + kf kL2 (ΩT ) ,
∂u
k kL2 (Ω00 ×(2ρ,T −2ρ)) ≤ C kukL2 (ΩT ) + kDukL2 (ΩT ) + kf kL2 (ΩT )
(1.2)
∂t
where C(K1 , n, Ω00 , ρ, λ, K2 ).
1
P
Proof. Let Ω00 ⊂⊂ Ω0 ⊂⊂ Ω. We set f¯ := − i bi Di u − cu + f so that f¯ ∈ L2 (ΩT ).
Let
u(x + hek , t) − u(x, t)
h
be the difference quotient in space and let uε be the time mollification of u. Let η be a smooth
function supported in Ω0 × (ρ, T − ρ), η ∈ (0, 1) and such that η = 1 on Ω00 × (2ρ, T − 2ρ).
Dkh u(x, t) :=
We take the test function ϕ = Dk−h (η 2 Dkh uε ). We have that
Z
−
ΩT
Z
n
X
(aij Dj u)ε Di ϕdx dt =
Z
∂ϕ
dx dt +
uε
∂t
ΩT
ΩT
ij
We denote
Z
A := −
uε
ΩT
(f¯)ε ϕ dx dt.
∂Dk−h (η 2 Dkh uε )
dx dt.
∂t
We have by using IPP in time
Z
∂uε −h 2 h
Dk (η Dk uε ) dx dt
ΩT ∂t
Z
∂uε 2 h
)η Dk uε dx dt
=−
Dkh (
∂t
ΩT
Z
∂Dkh uε 2 h
=−
η Dk uε dx dt
∂t
ΩT
Z
1 ∂|Dkh uε |2 2
η dx dt.
=−
∂t
ΩT 2
A=
Using again IPP in time, we get
Z
Z
1 h 2 ∂η 2
|Dkh uε |2 dx dt.
|Dk uε |
dx dt ≤ C||∂t η||L∞
A=
2
∂t
ΩT
ΩT
We denote
Z
B=−
ΩT
n
X
(aij Dj u)ε Di (Dk−h (η 2 Dkh uε )) dx dt.
ij
2
We have by IPP in space
Z
B=
ΩT
Z
=
ΩT
Z
=
ΩT
Z
+
ΩT
Z
+
ΩT
Z
+
ΩT
n
X
Dkh ((aij Dj u)ε )Di (η 2 Dkh uε ) dx dt
ij
n
X
(Dkh aij Dj u)ε + (aij (x + hek , t)Dkh Dj u)ε (η 2 Di Dkh uε + 2ηDi ηDkh uε ) dx dt
ij
n
X
(Dkh aij Dj u)ε (η 2 Di Dkh uε ) dx dt
ij
n
X
(Dkh aij Dj u)ε (2ηDi ηDkh uε ) dx dt
ij
n
X
ij
n
X
(aij (x + hek )Dkh Dj u)ε (η 2 Di Dkh uε ) dx dt
(aij (x + hek )Dkh Dj u)ε (2ηDi ηDkh uε ) dx dt
ij
= B1 + B2 + B3 + B4 .
Denote
Z
C :=
−(f¯)ε Dk−h (η 2 Dkh uε ) dx dt.
ΩT
Using Hölder’s inequality, we have
Z
Z
δ
2
¯
C ≤ C(δ)
(f )ε dx dt +
|Dk−h (η 2 Dkh uε )|2 dx dt
C
ZΩT
ZΩT
δ
(f¯)2ε dx dt +
|D(η 2 Dkh uε )|2 dx dt
≤ C(δ)
C
ZΩT
ZΩT
δ
≤ C(δ)
(f¯)2ε dx dt +
|2ηDηDkh uε + η 2 DDkh uε |2 dx dt
C
ZΩT
Z ΩTZ
≤ C(δ)
(f¯)2ε dx dt + δ
η 2 |DDkh uε |2 dx dt
ΩT ΩT
Z ΩT
+C
η 2 |Dkh uε |2 dx dt.
ΩT
3
Here we used that
Z
Z
h
2
h
2
|2ηDηDk uε + η DDk uε | dx dt =
|2ηDηDkh uε + η 2 Dkh Duε |2 dx dt
ΩT
Z
Z
2
2
2
h
≤C
η |Dη| |Dk uε | + C
η 2 |DDkh uε |2 dx dt
ΩT
ZΩT
Z
≤C
η 2 |Dkh uε |2 + C
η 2 |DDkh uε |2 dx dt.
ΩT
ΩT
ΩT
We have that B3 = A + C − B1 − B2 − B4 .
From the uniform ellipticity of aij , we have that for ε small
Z
2
λ
η 2 |Dkh Duε |2 dx dt ≤ B3 .
3 ΩT
Now by using Hölder’s inequality we have that
Z
η 2 |DDkh uε |2 dx dt
ΩT
Z
Z
2
2
2
h
2
+ C(||aij ||C1 + ||Dη||L∞ )
η |Dk uε | dx dt +
−B1 − B2 − B4 ≤ (δ1 + δ2 )
ΩT
Taking δ1 + δ2 + δ3 ≤
Z
η
λ
2
η |Duε | dx dt
2
2
ΩT
λ
we get that
2
|Dkh Duε |2 dx dt
≤
C(||aij ||2C1
+
||Dη||2L∞ )
ΩT
Z
2
η
|Dkh uε |2 dx dt
+C
2
+
ΩT
Z
η |Duε | dx dt
Z
2
ΩT
(f¯)2ε dx dt.
(1.3)
ΩT
Recall that
Z
Z
X
2
2
¯
(f )ε dx dt ≤ C(1 + ||c||L∞ +
||bi ||L∞ )
ΩT
(u2ε
2
+ |Duε | +
fε2 ) dx dt
.
ΩT
i
Passing to the limit h → 0 and ε → 0 in (1.3), we get that D2 u in L2 (Ω00 × (2ρ, T − 2ρ) and
we get the desired estimate.
∂u
∈ L2loc (ΩT ). We know that
We now show that
∂t
Z
Z X
Z
n
∂ϕ
−
uε
dx dt +
(aij Dj u)ε Di ϕdx dt =
(f¯)ε ϕ dx dt.
∂t
ΩT ij
ΩT
ΩT
4
u(x, t + h) − u(x, t)
is the quotient differWe take ϕ(x, t) = η 2 T h (uε ) where T h u(x, t) =
h
ence in time, η is the same smooth function as before and uε is the regularization in time.
Integrating by part in time, we get
Z X
Z
Z
n
∂η 2 T h (uε )
2 h
dx dt +
−
(aij Dj u)ε Di (η T (uε ))dx dt =
uε
(f¯)ε η 2 T h (uε ) dx dt
∂t
ΩT ij
ΩT
ΩT
Z
Z X
Z
n
∂uε 2 h
(aij Dj u)ε Di (η 2 T h (uε )) dx dt =
η T (uε ) dx dt +
(f¯)ε η 2 T h (uε ) dx dt
=
∂t
ΩT ij
ΩT
ΩT
It follows that (here we use IPP in space)
Z
Z
Z X
n
∂uε 2 h
2 h
(aij Dj u)ε Di (η T (uε ))dx dt +
η T (uε ) dx dt = −
(f¯)ε η 2 T h (uε ) dx dt
∂t
ΩT
ΩT ij
ΩT
Z X
Z
n
Di ((aij Dj u)ε )η 2 T h (uε ) dx dt +
(f¯)ε η 2 T h (uε )dx dt
=
ΩT
≤
1
4
+
Hence
Z
ΩT
1
∂uε 2 h
η T (uε ) dx dt −
∂t
2
ΩT
ij
Z
η 2 (T h (uε ))2 dx dt + C
n
X
(Di ((aij Dj u)ε )2 dx dt
Z
ΩT
ΩT
Z
Z
1
4
Z
η 2 (T h (uε ))2 dx dt + C
ΩT
2
ij
(f¯)2ε η 2 dx dt.
ΩT
h
Z
2
(f¯)2ε η 2 dx dt
η (T (uε )) dx dt ≤ C
ΩT
ΩT
n
X
Z
+C
ΩT
(Di ((aij Dj u)ε )2 , dx dt. (1.4)
ij
Using the previous inequality for D2 u, and the fact that aij ∈ C 1 (ΩT ) we have that for
ε small
Z X
Z X
n
n
2
(Di ((aij Dj u)ε ) dx dt =
[(Di (aij )Dj u)ε + (aij Di Dj u)ε ]2 dx dt
ΩT
ΩT
ij
ij
Z
≤
||Daij +
ΩT
aij ||2L∞
n
X
[(Dj u)2ε + (Di Dj u)2ε ] dx dt
ij
≤ C kukL2 (ΩT ) + kDukL2 (ΩT ) + kf kL2 (ΩT )
Passing to the limit h → 0, ε → 0 in (1.4), we get that
Z
∂u
| |2 η 2 dx dt ≤ C kukL2 (ΩT ) + kDukL2 (ΩT ) + kf kL2 (ΩT )
ΩT ∂t
5
Remark 1.2. Using the energy estimate, we can replace the L2 norm of Du by the L2 norm
of u. in the right side of the inequalities of the Theorem. In addition, it follows by induction
that if aij have bounded spatial derivatives of order less or equal to some positive integer
k and if f, c, bi have bounded spatial derivatives of order less or equal to k − 1, then the
weak solution has spacial derivatives of order less or equal to k + 1 in L2loc and ut has k − 1
derivatives in L2loc and a similar argument gives higher regularity in time if the coefficients
has higher regularity with respect to time. In particular, if the coefficients are C ∞ then the
weak solution u is locally C ∞ .
Remark 1.3. For operators with smooth coefficients on smooth domains with smooth data f ,
one can obtain regularity results for weak solutions by deriving energy estimates for higherorder derivatives of the approximate Galerkin solutions um and taking the limit as m → ∞.
A repeated application of this procedure, and the Sobolev theorem, implies, from the Sobolev
embedding theorem, that the weak solutions constructed via the Galerkin method are smooth,
classical solutions. For a discussion of this regularity theory, see the book of Evans.
Lemma 1.4 (Weak maximum principle). Let ∂p ΩT = (Ω × {0}) ∪ (∂Ω × (0, T )). Let u be
weak subsolution to
n
X
∂t u −
Dj (aij Di u) = 0
ij
with aij bounded and λ|ξ|2 ≤
Pn
i,j
aij ξi ξj ≤ Λ|ξ|2 . Then
sup u ≤ sup u.
ΩT
∂p ΩT
Proof. Let M := sup u. We have that
∂p ΩT
Z
−
ΩT
∂ϕ
dx dt +
uε
∂t
Z
ΩT
n
X
(aij Dj u)ε Di ϕ dx dt ≤ 0.
ij
Taking ϕ = (uε − k)+ (χh (t))ε where k > M . ϕ is an admissible test function (see exercice 1
set 8) since ∂t ϕ ∈ L2 (ΩT ) and ϕ ∈ L2 (0, T ; W01,2 (Ω)) and vanishes at t = 0 and t = T for ε
small. We have
Z
Z X
n
∂((uε − k)+ (χh (t))ε
−
uε
dx dt +
(aij Di u)ε Di (((uε − k)+ (χh (t))ε ) dx dt ≤ 0.
∂t
ΩT i,j
ΩT
Using that
Z
ΩT
∂[((uε − k)+ (χh (t))ε ]
dx dt =
uε
∂t
Z
ΩT
Z
+
ΩT
6
∂(χh (t))ε
dx dt
∂t
1 ∂(uε − k)2+
(χh (t))ε dx dt
2
∂t
(uε − k)2+
and integrating by parts we get that
Z
Z
[∂((uε − k)+ (χh (t))ε ]
1
(χh (t))ε
uε
dx dt =
(uε − k)2+
dx dt
∂t
2 ΩT
∂t
ΩT
Z
χh (t)
ε→0 1
=
dx dt
(u − k)2+
2 ΩT
∂t
Z
1
h→0
= −
(u − k)2+ (x, T ) dx.
2 ΩT
We obtain by first letting ε → 0 and then h → 0 that,
1
0≥
2
Z
(u −
k)2+ (x, T ) dx
n
X
aij Dj (u − k)+ Di (u − k)+ dx dt
Z
+
ΩT
ΩT
Z
i,j
Z
1
(u − k)2+ (x, T ) dx + λ
|D(u − k)+ |2 dx dt
2 ΩT
Z
ZΩT
1
λ
≥
(u − k)2+ (x, T ) dx +
(u − k)2+ dx dt.
2 ΩT
µ ΩT
≥
It follows that (u−k)+ ≡ 0 and the result follows since k is an arbitrary constant k > M .
Lemma 1.5 (Weak maximum principle for c ≥ 0). Let c ∈ L∞ (ΩT ) with c ≥ 0. Let u be
weak subsolution to
∂t u − ∆u + c(x, t)u = 0.
Then,
sup u ≤ sup u+ .
ΩT
∂p ΩT
Proof. We use that u+ = max(u, 0) is a subsolution to ∂t u − ∆u ≤ 0 and the fact that
sup u ≤ sup u+ .
ΩT
ΩT
Remark 1.6. Unlike the elliptic case, the condition c ≥ 0 is not needed. Indeed, we can
consider w = ue−Kt which satisfies
wt − ∆w + (c(x, t) + K)w = 0
and taking K > ||c||L∞ we are left with an equation with c̃ = c + K ≥ 0.
Lemma 1.7 (Strong Maximum principle). Let u be a continuous weak subsolution to ut −
∆u = 0 in ΩT with Ω being a bounded and connected domain. Suppose that u attains its
maximum on ΩT at (x0 , t0 ) ∈ ΩT , then u is constant in Ωt0 .
7
We can state as follows this maximum principle: if u attains a maximum (a nonnegative
maximum if (c 6= 0) at some interior point, then u is constant at all earlier times
Proof. Strong comparison principle for weak solutions follows from Harnack type arguments.
Let M := max u and consider the function v := M − u. Then v is a positive weak solution
ΩT
of vt − ∆v = 0. Let x0 ∈ Ω0 ⊂⊂ Ω and x0 ∈ Ω00 ⊂⊂ Ω0 . Here we change the cubes used
in the Harnack inequality but the proof of Harnack holds for any domains Ω1 × (t1 , t2 ) and
Ω2 × [t3 , t4 ) with 0 < t1 < t2 < t3 < t4 < T and Ω2 ⊂⊂ Ω1 ⊂⊂ Ω: for any positive
supersolution of ut − ∆u = 0 there exists a constant C(n) > 0 such that
Z
u ≤ C(n) ess inf u.
Ω2 ×[t3 ,t4 )
Ω1 ×(t1 ,t2 )
Applying the previous Harnack inequality, for 0 < ε < h < t0 , we get that
Z
(M − u) dx dt ≤ C(n) 00ess inf (M − u) ≤ v(x0 , t0 ) = M − M = 0.
0≤
Ω0 ×(ε,h)
Ω ×[t0 ,t0 +ε)
It follows that M − u ≡ 0 in Ω0 × (ε, h) for any ε > 0, h < t0 and x0 ∈ Ω0 ⊂⊂ Ω. Using the
connectednees of Ω it follows that u ≡ M in Ω × (0, t0 ).
8