AN ERROR ESTIMATE FOR THE FINITE DIFFERENCE
APPROXIMATION TO DEGENERATE
CONVECTION-DIFFUSION EQUATIONS IN SEVERAL SPACE
DIMENSIONS.
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Abstract. We consider first-order finite difference schemes for a nonlinear degenerate convection-diffusion equation in several space dimensions, and prove
an L1loc error estimate. Precisely, we prove that the L1loc difference between
the approximate solution given by the semidiscrete scheme and the unique
entropy solution converges at a rate O(∆x1/(10+d) ) where ∆x is the spatial
mesh size and d is the spatial dimension.
1. Introduction
We consider the nonlinear, possibly strongly degenerate convection-diffusion
problem
∂t u + ∇ · f (u) = ∆A(u), (x, t) ∈ ΠT ,
(1.1)
u(x, 0) = u0 (x),
x ∈ Rd ,
where ΠT = Rd ×(0, T ) for some fixed final time T > 0, u(x, t) is the scalar unknown
function that is sought. The initial function u0 : Rd → R is a given integrable and
bounded function, while the flux function f : R → Rd and the diffusion function
A : R → R are given functions satisfying:
(i) f (0) = 0 and f ∈ Liploc (R; Rd ).
(ii) A is continuously differentiable with A(0) = 0, A0 ≥ 0 and A ∈ Liploc (R).
By strongly degenerate it is ment that we allow A0 (u) = 0 for all u in some interval
[α, β] ⊂ R. The class of equations under consideration therefore contains the heat
equation, the porous medium equation, the two phase flow equation and conservation laws. In the nondegenerate (uniformly parabolic) case of problem (1.1) it is
well known that it admits a unique classical (smooth) solution. However strongly
degenerate equations must be considered in the weak sense and will in general possess discontinuous (shock wave) solutions. As in the case of conservation laws it is
necessary to introduce the notion of entropy solution in order for (1.1) to be well
posed.
2. Preliminary material
η
Set A (u) := A(u)+ηu for any fixed η > 0, and consider the uniformly parabolic
problem
(
uηt + ∇ · f (uη ) = ∆Aη (uη ), (x, t) ∈ ΠT ,
(2.1)
uη (x, 0) = u0 (x),
x ∈ R.
It is well known that (2.1) admits a unique classical (smooth) solution.
We collect some relevant (standard) a priori estimates in the next three lemmas.
Key words and phrases. Degenerate convection-diffusion equations, finite difference schemes,
rate of convergence.
1
2
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Lemma 2.1. Suppose u0 ∈ L∞ (Rd ) ∩ L1 (Rd ) ∩ BV (Rd ), and let uη be the unique
classical solution of (2.1). Then for any t > 0,
kuη (·, t)k 1 d ≤ u0 1 d ,
L (R )
η
L (R )
0
ku (·, t)kL∞ (Rd ) ≤ ku kL∞ (Rd ) ,
0
|uη (·, t)|
d ≤ u
d .
BV (R )
BV (R )
For a proof of the previous and next lemmas, see for example [31].
Lemma 2.2. Suppose u0 ∈ L∞ (Rd ) ∩ L1 (Rd ) and ∇ · (f (u0 ) − ∇A(u0 )) ∈ L1 (Rd ).
Let uη be the unique classical solution of (2.1). Then for any t1 , t2 > 0,
kuη (·, t2 ) − uη (·, t1 )k 1
≤ ∇ · (f (u0 ) − ∇A(u0 )) 1 d |t2 − t1 | .
L (R)
L (R )
η
These results above imply that the family {u }η>0 is relatively compact in
C([0, T ]; L1loc (Rd )). If u = limη→0 uη , then
kuη − ukL1 (ΠT ) ≤ Cη 1/2 ,
for some constant C which does not depend on η, see [14]. Moreover, u is an entropy
solution according to the following definition.
Definition 2.1. An entropy solution of (1.1) is a measurable function u = u(x, t)
satisfying:
(D.1) u ∈ L1 (ΠT ) ∩ L∞ (ΠT ) ∩ C((0, T ); L1 (Rd )).
(D.2) A(u) ∈ L2 ([0, T ]; H 1 (Rd )).
(D.3) For all constants c ∈ R and all non-negative test functions φ ∈ C0∞ (Rd ×
[0, T ), the following entropy inequality holds:
ZZ
|u − c|∂t φ + sign(u − c)(f (u) − f (c)) · ∇ϕ + |A(u) − A(c)|∆ϕ dtdx
ΠT
Z
+
|u0 − c|ϕ(x, 0) dx ≥ 0
Rd
The uniqueness of entropy solutions follows from [4].
3. Difference scheme
Let us first consider the case d = 2. Consider a uniform cartesian grid {Ci,j }(i,j)
where
Ci,j = [xi−1/2 , xi+1/2 ) × [yj−1/2 , yj+1/2 )
where xi = i∆x and yj = j∆y. Suppose that u is a solution to (1.1). Then
ZZ
Z
Z
∂
u(x, y, t) dxdy =
f (u) · n ds −
∇A(u) · n ds
∂t
Ci,j
∂Ci,j
∂Ci,j
where n denotes the normal vector. Along each edge this vector is constant so we
obtain
ZZ
Z yj+1/2
Z yj+1/2
∂
u(x, y, t) dxdy = −
f 1 (u(xi+1/2 , y, t)) dy +
f 1 (u(xi−1/2 , y, t)) dy
∂t
Ci,j
yj−1/2
yj−1/2
Z xi+1/2
Z xi+1/2
−
f 2 (u(x, yj+1/2 , t)) dx +
f 2 (u(x, yj−1/2 , t)) dx
xi−1/2
yj+1/2
xi−1/2
Z yj+1/2
Z
∂x A(u(xi+1/2 , y, t)) dy −
+
yj−1/2
Z xi+1/2
∂x A(u(xi−1/2 , y, t)) dy
yj−1/2
Z xi+1/2
∂y A(u(x, yj+1/2 , t)) dx −
+
xi−1/2
∂y A(u(x, yj−1/2 , t)) dx.
xi−1/2
AN ERROR ESTIMATE
3
Now denote by ui,j the average value of u on each cell. That is
ZZ
1
ui,j =
u(x, y, t) dxdy.
∆x∆y
Ci,j
In order to obtain a scheme from the above expression we approximate the terms
by expressions in the average values {ui,j }. That is, suppose we can find
Z yj+1/2
1
1
Fi±1/2,j
≈
f k (u(xi±1/2 , y, t)) dy,
∆y yj−1/2
Z xi+1/2
1
2
Fi,j±1/2 ≈
f 2 (u(x, yj±1/2 , t)) dx
∆x xi−1/2
where F k is an expression in the average values {ui,j }. Furthermore we approximate
A(ui+1,j ) − A(ui,j )
∆x
where y ∈ [yj−1/2 , yj+1/2 ]. The other differentialts are treated in the same way.
Hence
∂
1 1
1 2
1
1
ui,j ≈ −
Fi+1/2,j − Fi−1/2,j
−
Fi,j+1/2 − Fi,j−1/2
∂t
∆x
∆y
1
[A(ui+1,j − 2A(ui,j ) + A(ui−1,j )]
+
∆x2
1
+
[A(ui,j+1 − 2A(ui,j ) + A(ui,j−1 )] .
∆y 2
∂x A(u(xi+1/2 , y, t)) ≈
Concerning the definition of F k , k = 1, 2 we make the following definition.
Definition 3.1. (Numerical flux) We call a function F ∈ C 1 (R2 ) a numerical flux
for f given that F (u, u) = f (u) for u ∈ R. If
∂
∂
F (u, v) ≥ 0 and
F (u, v) ≤ 0
∂u
∂v
holds for all u, v ∈ R we call it monotone.
We will also assume F to be Lipschitz continuous in each variable. That is, there
exists a constant K such that for real numbers u, v and w
|F (u, w) − F (u, v)| ≤ K|w − v| and |F (u, v) − F (w, v)| ≤ K|u − w|.
1
Let F and F 2 be numerical fluxes for f 1 and f 2 respectively. Then we may
1
take Fi+1/2,j
:= F (ui,j , ui+1,j ) and so on. The above ideas are easily extended to
arbitrary dimension d. For simplicity let xi = i∆x and take
Cα = [xα1 −1/2 , xα1 +1/2 ) × · · · × [xαd −1/2 , xαd +1/2 )
for each α in Zd . Let ek be the element taking the value one in the k-th entry and
zero else. Then we define
σα±ek − σα
Dk± (σα ) = ±
k = 1, . . . , d.
∆x
for any real valued multisequence {σα }α∈Zd . The semidiscrete approximation of
problem (1.1) is the solution of the scheme
(
Pd
Pd
∂t uα + i=1 Di− F i (uα , uα+ei ) = i=1 Di− Di+ A(uα ), α ∈ Zd , t ∈ (0, T ),
uα (0) = u0α ,
α ∈ Zd .
(3.1)
A natural choice would be
Z
1
0
uα :=
u0 (x)dx.
(3.2)
∆xd Cα
4
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
In order to get bounds independent of ∆x we let
kσk1 = ∆xd
X
|σα | and |σ|BV =
α
d
XX
α
|σα+ei − σα |.
i=1
If these are bounded we say that σ = {σα } is in `1 and of bounded variation. Let
u(t) = {uα (t)}α∈Zd and u0 = {uα (0)}α∈Zd and define the operator A : `1 (Zd ) →
`1 (Zd ) by
d
X
(A(u))α =
Di− F i (uα , uα+ei ) − Di+ A(uα ) .
i=1
Then (3.1) may be considered as the Cauchyproblem
(
du
dt + A(u) = 0, t > 0,
u(0) = u0 .
(3.3)
This problem has a unique continuously differentiable solution since A is Lipschitz
continuous for each ∆x > 0. The solution does define a strongly continuous semigroup S(t) on `1 . We want to show that this semigroup is `1 contractive. This
follows by the theory presented in [9]. A is accretive if
X
sign(uα − vα )(A(u) − A(v))α ≥ 0.
(3.4)
α
1
d
for any u and v in ` (Z ) ([26, 28]). Now
(A(u) − A(v))α =
d
X
Di− F i (uα , uα+ei ) − F i (vα , vα+ei ) − Di+ (A(uα ) − A(vα )) .
i=1
denote the partial derivatives of F i with respect to the first and
and
Let
second variable respectively. Since F i is continuously differentiable there exist for
each (α, i) som number τα,i such that
Fvi
Fui
F i (uα , uα+ei ) − F i (vα , uα+ei ) = Fui (τα,i , uα+ei )(uα − vα )
and similarily a number θα,i such that
F i (vα , uα+ei ) − F i (vα , vα+ei ) = Fvi (vα , θα,i )(uα+ei − vα+ei ).
Let wα = uα − vα then
F i (uα , uα+ei ) − F i (vα , vα+ei )
= F i (uα , uα+ei ) − F i (vα , uα+ei ) + F i (vα , uα+ei ) − F i (vα , vα+ei )
= Fui (τα,i , uα+ei )wα + Fvi (vα , θα,i )wα+ei
Let A0 = a. Then there exist some ξα such that
A(uα ) − A(vα ) = a(ξα )wα
Using these expressions we obtain
X
sign(uα − vα )(A(u) − A(v))α
α
=
d
XX
α
−
i=1
d
XX
α
sign(wα )Di− Fui (τα,i , uα+ei )wα + Fvi (vα , θα,i )wα+ei
i=1
sign(wα )Di− Di+ (a(ξα )wα ) := T1 − T2 .
(3.5)
AN ERROR ESTIMATE
5
Consider T1 first.
1
Di− Fui (τα,i , uα+ei )wα + Fvi (vα , θα,i )wα+ei =
F i (τα,i , uα+ei )wα
∆x u
− Fui (τα−ei ,i , uα )wα−ei + Fvi (vα , θα,i )wα+ei − Fvi (vα−ei , θα−ei ,i )wα .
Hence
d 1 XX i
T1 =
F (τα,i , uα+ei )|wα | − Fui (τα−ei ,i , uα )sign(wα )wα−ei
∆x α i=1 u
i
i
+ Fv (vα , θα,i )sign(wα )wα+ei − Fv (vα−ei , θα−ei ,i )|wα |
d X
1 X X i
Fu (τα,i , uα+ei )|wα | −
Fui (τα,i , uα+ei )sign(wα+ei )wα
=
∆x i=1 α
α
X
X
i
i
+
Fv (vα , θα,i )sign(wα )wα+ei −
Fv (vα , θα,i )|wα+ei |
α
α
It follows since F i is monotone for each 1 ≤ i ≤ d that T1 ≥ 0. Considering T2 we
have
d
1 XX
T2 =
a(ξα+ei )sign(wα )wα+ei − 2a(ξα )|wα | + a(ξα−ei )sign(wα )wα−ei
∆x2 i=1 α
from wich it follows that T2 ≤ 0.
Lemma 3.1. Suppose F i is monotone for each 1 ≤ i ≤ d. Then there exists a
unique solution u = {uα } to (3.1) on [0, T ] with the properties:
(a) ku(t)k1 ≤ ku0 k1 .
(b) For every α ∈ Zd and t ∈ (0, T )
inf {u0β } ≤ uα (t) ≤ sup{u0β }.
β
β
(c) |u(t)|BV ≤ |u0 |BV .
(d) If v = {vα } is a solution of the same problem with initial data v 0 then
ku(t) − v(t)k1 ≤ ku0 − v 0 k1 .
Proof. Parts (a),(c) and (d) follows since S(t) is a contraction semigroup. Part (b)
follows from [5].
Lemma 3.2. Suppose F i is monotone for each 1 ≤ i ≤ d. If u is a solution to
(3.1) and A(u0 ) ∈ `1 (Zd ), then for each h > 0
ku(t + h) − u(t)k`1 ≤ kA(u0 )k`1 h.
Proof. Suppose that ku0 (t)k ≤ C. Then
Z t+h
Z
ku(t + h) − u(t)k = k
u0 (s) dsk ≤
t
t+h
ku0 (s)k ds ≤ Ch
t
and so Lipschitz continuity would follow. We claim that
∂ 0
ku (t)k ≤ 0.
∂t
(3.6)
6
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Now
∂ 0
∂
ku (t)k = kA(u(t))k
∂t
∂t "
#
X
∂
d
=
∆x
sign(A(u(t))α )A(u(t))α
∂t
α
X
= ∆xd
sign(A(u(t))α )∂t A(u(t))α
α
Now,
d
∂t A(u(t))α =
=
∂ X − i
Di F (uα (t), uα+ei (t)) − Di+ A(uα (t))
∂t i=1
d
X
Di− Fui (uα (t), uα+ei (t))u0α (t) + Fvi (uα (t), uα+ei (t))u0α+ei (t)
i=1
−
d
X
Di− Di+ a(uα (t))u0α (t)
i=1
=−
d
X
Di− Fui (uα (t), uα+ei (t))A(u(t))α + Fvi (uα (t), uα+ei (t))A(u(t))α+ei
i=1
+
d
X
Di− Di+ a(uα (t))A(u(t))α .
i=1
Considering the similarity between this computation and (3.5) it is seen that (3.6)
holds. It follows that ku0 (t)k ≤ kA(u0 )k and so the result follows.
3.1. The numerical entropy flux. It turns out that we need some more conditions on F i , 1 ≤ i ≤ d than just demanding it to be monotone. Lemma 3.3 provides
us with a sufficient condition.
Definition 3.2. Given an entropy, entropy flux pair (ψ, q) and a numerical flux F .
Suppose that Q ∈ C 1 (R2 ) satisfies
Q(u, u) = q(u),
∂
∂
Q(v, w) = ψ 0 (v) F (v, w),
∂v
∂v
∂
∂
Q(v, w) = ψ 0 (w)
F (v, w),
∂w
∂w
then we call Q a numerical entropy flux.
A natural question would now be for what type of numerical fluxes, if any, does
such a function exists.
Lemma 3.3. The numerical flux F has a numerical entropy flux Q, independent
of the chosen entropy, entropy flux pair, if there exist F1 , F2 ∈ C 1 (R) such that
F (u, v) = F1 (u) + F2 (v),
F10 (u)
+
F20 (u)
0
= f (u)
for all u, v ∈ R.
Proof. Let (ψ, q) be an entropy, entropy flux pair. Then q has the form
Z u
q(u) =
ψ 0 (z)f 0 (z)dz + C
c
(3.7)
(3.8)
AN ERROR ESTIMATE
for some constant C. Define Q by
Z
Z u
ψ 0 (z)F10 (z)dz +
Q(u, v) =
7
v
ψ 0 (z)F20 (z)dz + C.
(3.9)
c
c
It is easily verified that Q is a numerical entropy flux for F .
Note that if Q is supposed to have symmetric partial derivatives, then (3.7)
and (3.8) are necessary conditions. Fortunately there exist some numerical flux
functions which satisfies lemma 3.3.
Example 3.1. (The Engquist-Osher flux) Let
0
f+
(s) = max(f 0 (s), 0)
0
f−
(s) = min(f 0 (s), 0).
and
Then, in the terminology of lemma 3.3, let F (u, v) = F1 (u) + F1 (v) where
Z u
Z v
0
0
F1 (u) = f (0) +
f+ (s)ds
and
F2 (v) =
f−
(s)ds.
0
0
It is easily seen to satisfy the criteria given in lemma 3.3 and it is also clearly
monotone.
Example 3.2. Let a, b ∈ R and define
F1 (u) = af (u) + bu and F2 (v) = (1 − a)f (v) − bv.
Note that F (u, v) = F1 (u) + F2 (v) is monotone if
a inf {f 0 (x)} ≥ −b and
x
(1 − a) sup{f 0 (x)} ≤ b
x
This example includes both the upwind scheme and the Lax-Friedrichs scheme.
From a more general point of view we may consider any flux splitting. That is
f (u) = f + (u) + f − (u) where (f + (u))0 ≥ 0 and (f − (u))0 ≤ 0 for all u ∈ R. Then
the numerical flux F defined by
F (u, v) = f + (u) + f − (v)
satisfies the assumptions of lemma 3.3. Note also that any convex combination of
numerical flux functions which satisfies the hypothesis of lemma 3.3 itself satisfies
the lemma.
If lemma 3.3 holds we have a representation of Q given by (3.9). It follows that
Z v
Q(u, v) = q(u) +
ψ 0 (z)F20 (z)dz.
u
Note that we may obtain another representation depending on F1 by splitting up
the first integral.
Lemma 3.4. Let Q be a numerical entropy flux associated with the entropy, entropy
flux pair (ψ, q) and the monotone numerical flux F . Then
ψ 0 (u)(F (u, w) − F (v, u)) ≥ Q(u, w) − Q(v, u)
for all u, v, w ∈ R.
Proof. Let u be fixed. Define p(v, w) = p1 (w) + p2 (v) where
p1 (w) = −ψ 0 (u)F (u, w) + Q(u, w) + ψ 0 (u)f (u) − q(u),
p2 (v) = ψ 0 (u)F (v, u) − Q(v, u) − ψ 0 (u)f (u) + q(u).
Then we have
p(v, w) = −ψ 0 (u)(F (u, w) − F (v, u)) + (Q(u, w) − Q(v, u))
8
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
and so the lemma is proved if we can show that p(v, w) ≤ 0 for all v, w ∈ R. Let
us differentiate p1 and p2 .
∂
∂
p01 (w) = −ψ 0 (u)
F (u, w) + ψ 0 (w)
F (u, w)
∂w
∂w
∂
= ψ 00 (ξ1 )
F (u, w)(w − u)
∂w
for some ξ1 ∈ int(u, w). Similarly
∂
∂
p02 (v) = ψ 0 (u) F (v, u) − ψ 0 (v) F (v, u)
∂v
∂v
∂
= ψ 00 (ξ2 ) F (v, u)(u − v)
∂v
for some ξ2 ∈ int(u, v). Since F is monotone and ψ is convex we may infer that if
z ∈ R then p0i (z)(z − u) ≤ 0 for both i = 1 and i = 2. It remains to observe that
p1 (u) = p2 (u) = 0 and so
Z z
pi (z) =
p0i (ξ)dξ ≤ 0, for i = 1, 2.
u
Hence p(v, w) ≤ 0.
4. Error estimate
be the solution to (3.1) with A replaced by Aη . We associate with
Let
it the piecewise constant function uηα : Rd × [0, T ] → R defined by
{uηα }α∈Zd
uηα (x, t) = uηα (t),
for x ∈ Cα .
(4.1)
To derive the error estimate we need many of the uniform bounds from section 2
and 3. For these assumptions to hold we make the following assumptions on the
initial data u0 :
(i) u0 , u0α ∈ L1 (Rd ) ∩ L∞ (Rd ) ∩ BV (Rd ).
(ii) ∇A(u0 ) ∈ (BV (Rd ))d and kA(u0α )k`1 is bounded independently of ∆x.
Note that if u0 satisfies the above assumptions and u0α is given by (3.2) then u0α
also satisfies the above. We may now state the theorem.
Theorem 4.1. Let u be the entropy solution to (1.1) and uηα the piecewise constant
approximation defined above with η = (∆x)2/(10+d) . Suppose u0 , u0α satisfies (i) an
(ii). Then for each R > 0 there exist a constant C independent of ∆x such that for
all t ∈ (0, T )
Z
|uηα (x, t) − u(x, t)| dx ≤ C (∆x)1/(10+d) + ku0α − u0 kL1 (Rd ) .
B(0,R)
The constant C may depend on u0 , u0α , T, f and A.
The next lemma is a substitution for the chainrule when working with difference
quotients and piecewise constant functions.
Lemma 4.1. Let u : Zd → R. Suppose A : R → R is strictly increasing and continuously differentiable. Then for each (α, i) ∈ Zd × {1, . . . , d} there exist sequences
{τα,i } and {θα,i } such that both τα,i and θα,i are in int(uα , uα+ei ) and
Di+ signε (A(uα ) − A(c)) = sign0ε (A(τα,i ) − A(c))Di+ A(uα ),
Di+ |A(uα ) − A(c)|ε = signε (A(θα,i ) − A(c))Di+ A(uα ).
Let u : Rd → R be differentiable and set u = u(y) where y = (y1 , . . . , yd ). Then for
1≤i≤d
sign0ε (A(τα,i ) − A(u))A(u)yi = −(signε (A(θα,i ) − A(u)))yi .
(4.2)
AN ERROR ESTIMATE
9
Proof. Consider the first equality. By the mean value theorem there exists a real
number ξ ∈ int(A(uα ) − A(c), A(uα+ei ) − A(c)) such that
sign0ε (ξ) (A(uα+ei ) − A(uα )) = signε (A(uα+ei ) − A(c)) − signε (A(uα ) − A(c)).
Hence
sign0ε (ξ)Di+ A(uα ) = Di+ (signε (A(uα ) − A(c)))
By the intermediate value theorem there exist a τα,i ∈ int(uα , uα+ei ) such that
A(τα,i ) − A(c) = ξ. The existence of θα,i is proved the same way. To prove (4.2)
note that
sign0ε (A(τα,i ) − A(u))A(u)yi Di+ A(uα ) = [Di+ signε (A(uα ) − A(u))]A(u)yi
= −Di+ (|A(u − α) − A(u)|ε )yi
= −(signε (A(θα,i ) − A(u)))yi Di+ A(uα ).
Hence if Di+ A(uα ) 6= 0 (4.2) follows. Next, assume that Di+ A(uα ) = 0, then since
A is injective we have uα = uα+ei . It follows that τα,i = θα,i = uα and equation
(4.2) is simply the normal chain rule.
η
η
Let τα,i
and θα,i
satisfy the lemma for uα = uηα and u = uη . As in (4.1) we
η
η
η
associate with τα,i
and θα,i
the piecewise constant functions τ ηα,i and θα,i respectively. In what follows we will only be working with the regularized problem and
so to ease notation we supress the η. Furthermore we will only use the piecewise
constant functions so we simply write uα when we really mean uηα and so on.
4.0.1. Rewriting the continuous equation. Define an entropy, entropy flux pair (ψε , qε )
by
Z u
ψε (u, c) =
signε (A(z) − A(c)) dz,
Z cu
qεi (u, c) =
ψε0 (z, c)(f i (z))0 dz,
1 ≤ i ≤ d,
c
where ψε0 is the derivative with respect to the first varible. Let ϕ be a non negative
function in the space
D(Π2T )0 := ϕ ∈ D(Π2T ) : ϕ|t=0 = ϕ|t=T ≡ 0, ϕ|s=0 = ϕ|s=T ≡ 0
where D(Π2T ) denotes the space of compactly supported smooth functions. Let
u = u(y, s) solve equation (2.1). Multiply the equation by ψε0 (u, c)ϕ and integrate
in both space and time to get
Z
Z
ψε (u, c)s ϕ+ψε0 (u, c)divy (f (u)−f (c))ϕ dyds =
(ψε0 (u, c)ϕ)∆y A(u) dyds.
ΠT
ΠT
ψε0 (u, c)
Note that
= signε (A(u) − A(c)). Integration by parts and the chain rule
gives
Z
ψε (u, c)ϕs − divy (qε (u, c))ϕ dyds
ΠT
Z
=
signε (A(u) − A(c))∇y A(u) · ∇y ϕ + sign0ε (A(u) − A(c))|∇y A(u)|2 ϕ dyds.
ΠT
(4.3)
Using the chain rule and integration by parts we get
Z
Z
signε (A(u) − A(c))∇y (A(u)) · ∇y ϕ dyds = −
ΠT
|A(u) − A(c)|ε ∆y ϕ dyds.
ΠT
(4.4)
10
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Let c = uα (x, t) in (4.3) and (4.4). Then integrate in x and t to obtain
ZZ
ψε (u, uα )ϕs − divy (qε (u, uα ))ϕ dX
Π2T
ZZ
−|A(u) − A(uα )|ε ∆y ϕ + sign0ε (A(u) − A(uα ))|∇y A(u)|2 ϕ dX.
=
Π2T
Adding and subtracting we get
ZZ
|u − uα |ϕs − divy (qε (u, uα ))ϕ dX
Π2T
=
X ZZ
Π2T
i
sign0ε (A(τα,i ) − A(u))((A(u)yi )2 )ϕ∆xi dX
ZZ
−
|A(u) − A(uα )|ε (∆y ϕ) dX
Π2T
ZZ
(|u − uα | − ψε (u, uα ))ϕs dX
+
Π2T
+
X ZZ
i
sign0ε (A(uα ) − A(u))ϕ − sign0ε (A(τα,i ) − A(u))ϕ∆xi (A(u)yi )2 dX.
Π2T
(4.5)
where ϕ∆xi (x, t, y, s) := ϕ(x + ∆xei , t, y, s).
4.0.2. Rewriting the semidiscrete equation. Let us try to rewrite the semidiscrete
equation (3.1) in a similar way. Multiply by ψε0 (uα , c)ϕ and integrate in both time
and space to obtain
Z
ψε (uα , c)t ϕ +
X
ΠT
ψε0 (uα , c)Di− F i (uα , uα+ei )ϕ dxdt
i
!
Z
signε (A(uα ) − A(c))
=
ΠT
X
Di− Di+ A(uα )
ϕ dxdt. (4.6)
i
Using integration by parts and Leibniz rule for difference quotients we get
Z
X
ψε (uα , c)ϕt −
ψε0 (uα , c)Di− F i (uα , uα+ei )ϕ dxdt
ΠT
=
i
XZ
i
+
Di+ signε (A(uα ) − A(c))Di+ A(uα )ϕ∆xi dxdt
(4.7)
ΠT
XZ
i
signε (A(uα ) − A(c))Di+ A(uα )Di+ ϕ dxdt.
ΠT
By lemma 4.1 we obtain
Z
Di+ signε (A(uα ) − A(c))Di+ A(uα )ϕ∆xi dxdt
ΠT
Z
=
sign0ε (A(τα,i ) − A(c))[Di+ A(uα )]2 ϕ∆xi dxdt.
ΠT
(4.8)
AN ERROR ESTIMATE
11
Concerning the second term on the right of (4.7) we add and subtract to apply
lemma 4.1 again. Using integration by parts we obtain
Z
signε (A(uα ) − A(c))Di+ A(uα )Di+ ϕ dxdt
ΠT
Z
signε (A(θj ) − A(c))Di+ A(uα )Di+ ϕ dxdt
=
ΠT
Z
[signε (A(uα ) − A(c)) − signε (A(θα,i ) − A(c))] Di+ A(uα )Di+ ϕ dxdt
+
ΠT
Z
|A(uα ) − A(c)|ε Di− Di+ ϕ dxdt
=−
ΠT
Z
[signε (A(uα ) − A(c)) − signε (A(θα,i ) − A(c))] Di+ A(uα )Di+ ϕ dxdt.
+
ΠT
(4.9)
Let c = u(y, s) in (4.6), (4.7), (4.8) and (4.9). Integrate in y and s and apply the
equalities (4.8) and (4.9).
ZZ
ψε (uα , u)ϕt −
Π2T
=
X
ψε0 (uα , u)Di− F i (uα , uα+ei )ϕ dX
i
X ZZ
Π2T
i
sign0ε (A(τα,i ) − A(u))(Di+ A(uα ))2 ϕ∆xi dX
ZZ
−
|A(uα ) − A(u)|ε
Π2T
+
X
Di− Di+ ϕ dX
i
X ZZ
Π2T
i
[signε (A(uα ) − A(u)) − signε (A(θα,i ) − A(u))] Di+ A(uα )Di+ ϕ dX.
(4.10)
4.0.3. Adding up the equations. Let us first note that we may write zero according
to the following computations. Recall lemma 4.1.
ZZ
0=
Π2T
=
Div+ (signε (A(uα ) − A(u))∇y A(u)ϕ) dX
X ZZ
i
Π2T
Di+ (signε (A(uα ) − A(u)))A(u)yi ϕ∆xi
+ signε (A(uα ) − A(u))A(u)yi Di+ ϕ dX
=
X ZZ
i
Π2T
Di+ (signε (A(uα ) − A(u)))A(u)yi ϕ∆xi
− (|A(uα ) − A(u)|ε )yi Di+ ϕ dX
=
X ZZ
i
Π2T
sign0ε (A(τα,i ) − A(u))Di+ A(uα )A(u)yi ϕ∆xi
+ |A(uα ) − A(u)|ε Di+ ϕyi dX.
(4.11)
12
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Adding equations (4.5), (4.10) and subtract two times (4.11) we get:
ZZ
|uα − u|(ϕt + ϕs )
Π2T
−
X
(ψε0 (uα , c)Di− F i (uα , uα+ei ) + qε (u, c)yi )ϕ dX
(4.12)
i
=
X ZZ
Π2T
i
sign0ε (A(τα,i ) − A(u)) Di+ A(uα ) − A(u)yi
ZZ
−
|A(uα ) − A(u)|ε
Π2T
X
2
ϕ∆xi dX
(Di− Di+ ϕ + 2Di+ ϕyi + ϕyi yi ) dX
(4.13)
(4.14)
i
ZZ
+
(|uα − u| − ψε (uα , u))ϕt dX
(4.15)
(|u − uα | − ψε (u, uα ))ϕs dX
(4.16)
Π2T
ZZ
+
Π2T
+
X ZZ
Π2T
i
signε (A(uα ) − A(u))
− signε (A(θα,i ) − A(u)) Di+ A(uα )Di+ ϕ dX
+
X ZZ
i
Π2T
(4.17)
sign0ε (A(uα ) − A(u))ϕ
− sign0ε (A(τα,i ) − A(u))ϕ∆xi (A(u)yi )2 dX.
(4.18)
4.1. Obtaining the inequality. Following lemma 3.3 we may define the numerical
entropy flux Qc,i
ε (uα , uα+ei ) by
Z uα+e
i
c,i
i
Qε (uα , uα+ei ) = qε (uα , c) +
ψε0 (z, c)(F2i )0 (z)dz.
(4.19)
uα
By lemma 3.4
ψε0 (uα , c)Di− F i (uα , uα+ei ) ≥ Di− Qc,i
ε (uα , uα+ei ).
The term (4.13) is positive and so
ZZ
X
i
|uα − u|(ϕt + ϕs ) −
(Di− Qu,i
ε (uα , uα+ei ) + qε (u, uα )yi )ϕ dX
Π2T
i
ZZ
+
Π2T
X
|A(uα ) − A(u)|ε
(Di− Di+ ϕ + 2Di+ ϕyi + ϕyi yi ) dX ≥ <1 + <2 , (4.20)
i
where
<1 := (4.15) + (4.16), and <2 := (4.17) + (4.18).
Integration by parts and (4.19) yields
ZZ
i
−
Di− Qu,i
ε (uα , uα+ei )ϕ + qε (u, uα )yi ϕ dX
Π2T
ZZ
=
Π2T
+
i
Qu,i
ε (uα , uα+ei )Di ϕ + qε (u, uα )ϕyi dX
ZZ
=
Π2T
qεi (uα , u)Di+ ϕ + qεi (u, uα )ϕyi dX
Z
ZZ
uα+ei
+
Π2T
uα
ψε0 (z, u)(F2i )0 (z) dz Di+ ϕ dX.
AN ERROR ESTIMATE
13
Next we need an expression for qεi (uα , u)Di+ ϕ + qεi (u, uα )ϕyi . Let us rewrite qεi (u, c)
using integration by parts.
Z u
∂
qεi (u, c) =
signε (A(z) − A(c)) (f i (z) − f i (c)) dz
∂z
c
= signε (A(u) − A(c))(f i (u) − f i (c))
Z u
∂
−
(signε (A(z) − A(c)))(f i (z) − f i (c)) dz.
c ∂z
Observe that signε (A(uα ) − A(u)) = −signε (A(u) − A(uα )) and so
qεi (uα , u)Di+ ϕ + qεi (u, uα )ϕyi
= signε (A(uα ) − A(u))(f i (uα ) − f i (u))(Di+ ϕ + ϕyi )
Z uα
∂
i
i
−
(signε (A(z) − A(u)))(f (z) − f (u)) dz Di+ ϕ
∂z
u
Z u
∂
i
i
(signε (A(z) − A(uα )))(f (z) − f (uα )) dz ϕyi .
−
uα ∂z
(4.21)
Let
γ1i
ZZ
uα
Z
=
Π2T
ZZ
γ2i =
Z
Π2T
ZZ
γ3i =
u
u
∂
(signε (A(z) − A(u)))(f i (z) − f i (u))dz Di+ ϕ dX,
∂z
Π2T
∂
(signε (A(z) − A(uα )))(f i (z) − f i (uα ))dz ϕyi dX,
uα ∂z
Z uα+e
i
ψε0 (z, u)(F2i )0 (z)dz Di+ ϕ dX.
uα
Pd
Let γj = i=1 γji for j = 1, 2 and 3. We obtain from 4.20 the inequality
ZZ
|uα − u|(ϕt + ϕs ) dX
Π2T
+
X ZZ
i
+
signε (A(uα ) − A(u))(f i (uα ) − f i (u))(Di+ ϕ + ϕyi ) dX
Π2T
X ZZ
i
Π2T
|A(uα ) − A(u)|ε (Di− Di+ ϕ + 2Di+ ϕyi + ϕyi yi ) dX
≥ γ1 + γ2 − γ3 + <1 + <2 . (4.22)
The next step is to define ϕ. In fact ϕ is going to be in the space
S (Π2T )0 := ϕ ∈ S (Π2T ) : ϕ|t=0 = ϕ|t=T ≡ 0, ϕ|s=0 = ϕ|s=T ≡ 0
where S (Π2T ) is the space of rapidly decreasing functions (See [27]). To make
sure that equation (4.22) applies to these functions we make the following simple
observation.
Lemma 4.2. The space D(Π2T )0 is dense in S (Π2T )0 .
Proof. The proof of theorem 7.10 given in [27] clearly adapts to this situation. By the above observation equation (4.22) applies to ϕ ∈ S (Π2T )s,t=0 given that
the terms involved represents tempered distributions. Let ρ ∈ C0∞ (R) satisfy
Z
supp(ρ) ⊂ [−1, 1], ρ(σ) ≥ 0,
ρ(σ)dσ = 1.
R
14
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
For x ∈ Rd , t ∈ R and r, r0 , α0 > 0, let ωr (x) = 1r ρ( xr1 ) · · · 1r ρ( xrd ), ρα0 (ξ) = α10 ρ( αξ0 )
and ρr0 (t) = r10 ρ( rt0 ). Let ν < τ be two numbers in (0, T ). For any α0 > 0 define
Z t
α0
ρα0 (ξ)dξ.
ψ(ν,τ ) (t) = Hα0 (t − ν) − Hα0 (t − τ ), Hα0 (t) =
−∞
0
Now let R > R and define
1
0
φ̂(x) =
exp(−(|x| − R0 )) = eR e−|x|
|x| ≤ R0
elsewhere.
Then let 0 < α be so small that supp(ωα ) ⊂ B(0, R0 − R) and define φ = φ̂ ∗ ωα .
Note that this implies φ(x) = 1 for x ∈ B(0, R). Let us first make an observation
concerning the derivatives of φ̂ outside B(0, R).
Lemma 4.3. For each multi-index α there exist numbers k, z0 , . . . , zk , n0 , . . . , nk
and multiindices β0 , . . . , βk such that
Dα e−|x| =
k
X
j=0
zj
xβj −|x|
e
|x|nj
where |βj | ≤ nj for each 0 ≤ j ≤ k.
Proof. The proof is by induction. This is obviously true for |α| = 0. Suppose the
result is true for α, we want to prove that it is true for α + ei . Since differentiation
is linear it suffices to consider each term separatly. Hence suppose that |β| ≤ n.
Then
β
∂
xβ+ei
xβ+ei
x −|x|
xβ−ei −|x|
e
= βi
e
− n n+2 e−|x| + n+1 e−|x|
n
n
∂xi |x|
|x|
|x|
|x|
and the result follows.
As a consequence we obtain the following result.
Lemma 4.4. For each multi-index α there exist a constant Cα such that
|Dα φ(x)| ≤ Cα φ(x).
Furthermore φ is in S (Rd ).
Proof. First consider x in the compact B(0, R0 ). By continuity Dα φ is bounded on
B(0, R0 ). Since φ > 0 it is uniformly bounded from below by some δ > 0 on B(0, R0 ).
Hence there exists a constant such that |Dα φ(x)| ≤ Cφ(x) for all x ∈ B(0, R0 ). Next
consider x ∈ Rd \ B(0, R0 ). Then by the above lemma Dα φ̂(x) = f (x)φ̂(x) for some
f ∈ Cb (Rd \ B(0, R)). Hence
|Dα φ(x)| = |(Dα φ̂ ∗ ωα )(x)|
Z
=|
Dα φ̂(x − y)ωα (y)dy|
B(0,R0 −R)
Z
=|
f (x − y)φ̂(x − y)ωα (y)dy|
B(0,R0 −R)
Z
≤C
φ̂(x − y)ωα (y)dy = Cφ(x).
B(0,R0 −R)
It now follows that φ is in Sd since (1 + |x|2 )N φ(x) is bounded for each N =
0, 1, . . . .
AN ERROR ESTIMATE
15
Let
α0
Ψ(x, t) = ψ(ν,τ
) (t)φ(x)
and
ϕ(x, t, y, s) = Ψ(x, t)ωr (x − y)ρr0 (t − s).
To make sure that ϕ is in D(Π2T )0 , let 0 < r0 < min(ν, T − τ ) and 0 < α0 <
min(ν − r0 , T − τ − r0 ). Observe that ϕ has some very important properties:
ϕt + ϕs = Ψt ωr ρr0 ,
ϕxi + ϕyi = Ψxi ωr ρr0 ,
ϕxi xi + 2ϕxi yi + ϕyi yi = Ψxi xi ωr ρr0 .
for 1 ≤ i ≤ d. In equation (4.22) these expressions appear with difference quotients
instead of x-derivatives. It should then be expected that these equalities turns into
good approximations as long as ∆x tends relatively fast to zero compared to r.
This will be seen in what follows. Consider the first term on the left in (4.22).
ZZ
|uα − u|(ϕt + ϕs ) dX
Π2T
ZZ
ZZ
=
Π2T
|uα − u|ρα0 (t − ν)φωr ρr0 dX −
Π2T
|uα − u|ρα0 (t − τ )φωr ρr0 dX
Hence
ZZ
Π2T
|uα − u|ρα0 (t − ν)φωr ρr0 dX
+
X ZZ
Π2T
i
+
X ZZ
i
Π2T
signε (A(uα ) − A(u))(f i (uα ) − f i (u))(Di+ ϕ + ϕyi ) dX
|A(uα ) − A(u)|ε (Di− Di+ ϕ + 2Di+ ϕyi + ϕyi yi ) dX + γ3
ZZ
≥
Π2T
|uj − u|ρα0 (t − τ )φωr ρr0 dX + γ1 + γ2 + <1 + <2 . (4.23)
4.2. Estimates. The subject of this section is to find bounds on the “unwanted”
terms in (4.23). In these computations we let C denote a generic constant. By
constant it is meant that it does not depend on the small variables but it might
depend on T and R. Similarly we let Γ = Γ(∆x, η, α0 , r, r0 ) denote a generic
function with the property that it is locally bounded, positive and increasing in each
variable. Note that given Γ1 and Γ2 we can always pick Γ = max{Γ1 , Γ2 }. That is,
taking the maximum of two increasing functions we obtain an increasing function.
Thus we may work with this class of functions in a similar way as with constants.
The following simple computation should be kept in mind while simplifying the
expressions below. Given two positive functions f and g we have
Γ1 f + Γ2 g ≤ Γf + Γg = Γ(g + f ).
Now consider the second term on the left in (4.23). Since
Di + ϕ + ϕyi = Ψxi ωr ρr0 + (Di+ ϕ − ϕxi )
16
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
we get
ZZ
Π2T
signε (A(uα ) − A(u))(f i (uα ) − f i (u))(Di+ ϕ + ϕyi ) dX
ZZ
=
Π2T
signε (A(uα ) − A(u))(f i (uα ) − f i (u))(Ψxi ωr ρr0 ) dX
ZZ
+
Π2T
signε (A(uα ) − A(u))(f i (uα ) − f i (u))(Di+ ϕ − ϕxi ) dX
=: β1i + β2i .
Now, using that f i ∈ Liploc and (4.4) we get
ZZ
i
β1 ≤ C
|uα − u|ϕ dX.
Π2T
Next, we want to find a bound on β2i , but first we write down some standard
computations as an easy reference.
Lemma 4.5. Let Dik =
∂k
.
∂xk
i
Then

k
X
|Dik ϕ(x, t, y, s)| ≤ CΨ(x, t)χ|xi −yi |≤r 
j=0
r


1  Y
ρr (xj − yj ) ρr0 (t − s).
k+1
j6=i
Let ϕσ (x, t, y, s) = ϕ(x + σi , t, y, s), and suppose |σ| ≤ ∆x. Then


Y
1
|Dik ϕσ (x, t, y, s)| ≤ Γ(r) k+1 χ|xi −yi |≤r+∆x  ρr (xj − yj ) ρr0 (t − s).
r
j6=i
Considering the difference quotient applied to ωr we have


0
Y
kρ k∞
|Di+ ωr (x − y)| ≤
χ|xi −yi |≤r+∆x  ρr (xj − yj ) .
r2
j6=i
Proof. Recall that ρr (σ) = 1r ρ
σ
r
. Note that
Dik ωr (x) = Dik (ρr (x1 ) · · · ρr (xd )) =
1
rk+1
ρ(k)
x i
r


Y
 ρr (xj ) .
j6=i
Since supp(ρ) ⊂ [−1, 1] we have


x (k)
Y
kρ
k
1
∞
j 
|Dik ωr (x)| ≤
χ|xi |≤r 
ρ
rk+1
r
r
j6=i
By lemma 4.4 and Leibniz rule
|Dik ϕ|
=|
k X
k
j=0
j
Dik−j ΨDij ωr ρr0 | ≤ CΨ
k
X
j=0
|Dij ωr |ρr0 .
AN ERROR ESTIMATE
17
Combining this with the above estimate we obtain the first statement. Consider
the second statement. By the first statement we have
|Dik ϕ(x+σi , t, y, s)|

k
X

≤ CΨ(x + σi , t)χ|xi +σ−yi |≤r
j=0

k
X

≤ Cχ|xi −yi |≤r+∆x
j=0
r


1  Y
ρr (xj − yj ) ρr0 (t − s)
k+1
j6=i

r

1  Y
ρr (xj − yj ) ρr0 (t − s)
k+1
j6=i


Y
1
≤ Γ(r) k+1 χ|xi −yi |≤r+∆x  ρr (xj − yj ) ρr0 (t − s).
r
j6=i
Since Ψ ≤ 1 and if |xi − yi | ≥ r + ∆x, then
|xi + σ − yi | ≥ |xi − yi | − |σ| ≥ r + ∆x − ∆x = r,
so it follows that χ|xi +σ−yi |≤r ≤ χ|xi −yi |≤r+∆x .
Di+ ωr (x) = Di+ (ρr (x1 ) · · · ρr (xd )) =
ρr (xi + ∆x) − ρr (xi )
∆x


Y

ρr (xj ) .
j6=i
If |xi | ≥ r + ∆x then both ρr (xi + ∆x) = 0 and ρr (xi ) = 0 so supp(D+ (ρr )) ⊂
[−r − ∆x, r + ∆x]. By the mean value inequality and the fact kρ0r k∞ = kρ0 k∞ r−2
we get
kρ0 k∞
∆x.
|ρr (xi + ∆x) − ρr (xi )| ≤
r2
The last statement follows.
In some of the following estimates it will be convenient to consider uα as sequence in the i’th variable depending on the remaining variables (x̂i , t) where
x̂i = (x1 , . . . , xi−1 , xi+1 , . . . , xd ). Recall that xj = j∆x and Ij = (xj−1/2 , xj+1/2 ].
There exist a map j : R → Z defined by x ∈ Ij(x) . Given a sequence {aj } = a we
get a piecewice constant function by a(x) = aj(x) or equivalently
X
a(x) =
aj χIj (x).
j
Conversly, if we are given any function a : R → R we define a sequence (ä)j = a(xj ).
Note that for any sequence a, ä = a, and if a is a piecewice constant function on
our grid, then ä(x) = a(x). This follows since
X
ä(x) =
(ä)j χIj (x) = a(x)
(4.24)
j
where we used that a is piecewice constant in the last equality. These operators
commute with composition. Suppose f : R → R, then f (a) = f (a) since f (a(x)) =
f (aj(x) ) = (f (a))j(x) = f (a)(x) or equivalently


X
X
f (a(x)) = f 
aj χIj (x) =
f (aj )χIj (x) = f (a)(x)
(4.25)
j
j
since the sum contains only one term for each x. We imediatly get
Z
Z
X
X
f (a(x))dx =
f (aj ) χIj (x)dx = ∆x
f (aj ).
R
j
R
j
(4.26)
18
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Estimate 1.
|β2i | ≤ Γ(r, η)
∆x
rη
∆x
1+
r
Proof. Let us start by the equality
1
∆x
(Di+ ϕ − ϕxi )(x, t, y, s) =
∆x
Z
(∆x − σ)ϕxi xi (x + σi , t, y, s) dσ
(4.27)
0
where σi denotes the vector that is σ in the i’th entry and zero else. To see this let
f (σ) := ϕ(x + σi , t, y, s). Then using the fundamental theorem and integration by
parts we obtain
Z ∆x
Z ∆x
0
0
f (∆x) − f (0) =
f (σ) dσ = ∆xf (0) −
(σ − ∆x)f 00 (σ) dσ.
0
0
Then use the chain rule to obtain f 0 (σ) = ϕxi (x + σi , t, y, s) and f 00 (σ) = ϕxi xi (x +
σi , t, y, s). Hence
Z Z Z ∆x
1
signε (A(uα ) − A(u))(f i (uα ) − f i (u))ϕσxi xi (∆x − σ)dσ dX
β2i =
2
∆x
ΠT 0
where ϕσi (x, t, y, s) = ϕ(x + σi , t, y, s). The idea is to move one derivative over from
the testfunctions, and then use the BV bound to obtain an estimate. Let
uj (x̂i , t) := (u¨α )j (x̂i , t) = uα (x1 , . . . , xi−1 , xj , xi+1 , . . . , xd , t).
From (4.24) we have
uα (x, t) =
X
uj (x̂i , t)χIj (xi ).
j
By (4.25) we may write
signε (A(uα ) − A(u))(f i (uα ) − f i (u))(x, t, y, s)
X
=
signε (A(uj ) − A(u))(f i (uj ) − f i (u)) (x̂i , t, y, s)χIj (xi )
j
=:
X
Θj (x̂i , t, y, s)χIj (xi )
j
Using summation by parts we get
Z Z ∆x
1
signε (A(uα ) − A(u))(f i (uα ) − f i (u))ϕσxi xi (∆x − σ)dσdxi
∆x R 0
Z ∆x X
Z
1
=
Θj
χIj (xi )ϕσxi xi (∆x − σ)dxi dσ
∆x 0
R
j
Z ∆x X
1
=
Θj (ρj − ρj−1 ) (∆x − σ)dσ
∆x 0
j
Z ∆x X
=
Θj (D− ρj )(∆x − σ)dσ
0
=−
j
X
j
+
Z
∆x
ρj (∆x − σ)dσ,
D Θj
0
where ρj = ρj (σ, x̂i , t, y, s) = ϕxi (x1 , . . . , xi−1 , xj+1/2 + σ, xi+1 , . . . , xd , t, y, s). By
lemma 4.5 we have
Y
1+r
|ϕxi (x + σi , t, y, s)| ≤ C 2 χ|xi −yi |≤r+∆x
(ρr (xk − yk )) ρr0 (t − s).
r
k6=i
AN ERROR ESTIMATE
19
Hence
Z ∆x
Y
1+r
|
ρj (∆x − σ)dσ| ≤ C(∆x)2 2 χ|xj+1/2 −y|≤r+∆x
(ρr (xk − yk )) ρr0 (t − s).
r
0
k6=i
Now
D + Θj =
=
1
∆x
Z
Z
1
∆x
uj+1
uj+1
uj
∂ signε (A(z) − A(u))(f i (z) − f i (u)) dz
∂z
sign0ε (A(z)−A(u))A0 (z)(f i (z)−f i (u))+signε (A(z)−A(u))(f i )0 (z) dz
uj
=: T1 + T2 .
Let us estimate each term separatly. By the mean value theorem
|A(z) − A(u)| ≥ η|z − u|.
Let K = kf i kLip , then
|f (z) − f (u)| ≤ K|z − u| ≤
K
|(A(z) − A(u)|.
η
Hence
|T1 | = |
1
∆x
uj+1
Z
sign0ε (A(z) − A(u))A0 (z)(f i (z) − f i (u)) dz|
uj
≤
Z
K
|
η∆x
uj+1
sign0ε (A(z) − A(u))A0 (z)|A(z) − A(u)| dz|.
uj
Let σ(z) = A(z) − A(u). Then we obtain
|T1 | ≤
K
|
η∆x
Z
σ(uj+1 )
sign0ε (σ)|σ| dσ| ≤ C
σ(uj )
K +
|D A(uj )|,
η
since sign0ε (σ)|σ| ≤ Cε ε and σ(uj+1 ) − σ(uj ) = A(uj+1 ) − A(uj ). Consider T2 . Since
both these functions are bounded we have
Z uj+1
1
|T2 | = |
signε (A(z) − A(u))(f i )0 (z) dz| ≤ C|D+ uj |.
∆x uj
Using the above computations and estimates we obtain
|
1
∆x
Z Z
R
0
∆x
signε (A(uα ) − A(u))(f i (uα ) − f i (u))ϕσxi xi (∆x − σ)dσdxi |
=|
X
j
+
Z
∆x
ρj (∆x − σ)dσ|
D Θj
0


X K
≤C(∆x)2 
|D+ A(uj )| + |D+ (uj )| 
η
j


Y
1+r
×  2 χ|xi −yi |≤r+∆x
(ρr (xk − yk )) ρr0 (t − s)
r
k6=i
20
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
It follows by (4.26) and since uα is of bounded variation that
Z TZ
Z X
K +
1+r
|β2i | ≤ C(∆x)2 2 (r + ∆x)
···
|D A(uj )| + |D+ (uj )| dx̂i dt
r
η
0
R
R j
Z
K +
1+r
|Di A(uα )| + |Di+ (uα )| dxdt
= C∆x 2 (r + ∆x)
r
ΠT η
1
1+r
≤ C∆x 2 (r + ∆x)
+1
r
η
The estimate follows.
Next, let us consider the term (4.14). Note that
Di− Di+ ϕ + 2Di+ ϕyi + ϕyi yi = Ψxi xi ωr ρr0 + (Di− Di+ ϕ − ϕxi xi ) + 2(Di+ ϕ − ϕxi )yi
Hence we rewrite (4.14).
ZZ
|A(uα ) − A(u)|ε (Di− Di+ ϕ + 2Di+ ϕyi + ϕyi yi ) dX
(4.28)
Π2T
ZZ
=
Π2T
|A(uα ) − A(u)|ε Ψxi xi ωr ρr0 dX
ZZ
+
Π2T
(4.29)
|A(uα ) − A(u)|ε (Di− Di+ ϕ − ϕxi xi ) dX
(4.30)
|A(uα ) − A(u)|ε (Di+ ϕ − ϕxi )yi dX
(4.31)
ZZ
+2
Π2T
=:ζ1i + ζ2i + ζ3i
(4.32)
By lemma 4.4 there exist a constant C such that |Ψxi xi | ≤ CΨ. Since A ∈ Liploc
we conclude that
ZZ
i
|uα − u|ϕ dX.
|ζ1 | ≤ C
Π2T
Estimate 2.
|
d
X
ζ2i + ζ3i | ≤ Γ(r)
i=1
∆x (∆x)2
+
r2
r3
∆x
.
1+
r
Proof. Consider the term (4.30). We use the same strategy as with the β2i terms.
Put f (σ) := ϕ(x + σi , t, y, s). Then for z ∈ R we have
Z
1
1
1 z
f (z) − f (0) = zf 0 (0) + z 2 f (2) (0) + z 3 f (3) (0) −
(σ − z)3 f (4) (σ) dσ.
2
6
6 0
Let z be ∆x and −∆x in the above equation. Then add the two equations to get
f (∆x) − 2f (0) + f (−∆x) − (∆x)2 f (2) (0)
Z
Z
1 ∆x
1 0
=−
(σ − ∆x)3 f (4) (σ) dσ +
(σ + ∆x)3 f (4) (σ) dσ
6 0
6 −∆x
It follows that
Di+ Di− ϕ(x) − ϕxi xi (x) = −
1
6(∆x)2
Z
∆x
(σ − ∆x)3
0
1
+
6(∆x)2
Z
∂4
ϕ(x + σi , t, y, s) dσ
∂x4i
0
−∆x
(σ + ∆x)3
∂4
ϕ(x + σi , t, y, s) dσ.
∂x4i
AN ERROR ESTIMATE
21
Splitting (4.30) according to this equality we get
ZZ
|A(uα ) − A(u)|ε (Di− Di+ ϕ − ϕxi xi ) dX
Π2T
=−
+
1
6(∆x)2
ZZ
1
6(∆x)2
ZZ
∆x
Z
Π2T
|A(uα ) − A(u)|ε (σ − ∆x)3
0
Z
Π2T
∂4
ϕ(x + σi , t, y, s) dσ dX
∂x4i
0
|A(uα ) − A(u)|ε (σ + ∆x)3
−∆x
∂4
ϕ(x + σi , t, y, s) dσ dX
∂x4i
=:T1 + T2 .
Let uj (x̂i , t) be defined as in estimate 1. By (4.25) we have
X
|A(uα ) − A(u)|ε (x, t, y, s) =
|A(uj ) − A(u)|ε (x̂i , t, y, s)χIj (xi )
j
=:
X
Θj (x̂i , t, y, s)χIj (xi ).
j
Let
∂4
ϕ(x
∂x4i
Z
−
0
∆x
+ σi , t, y, s) =
∂4
ϕσ (x, t, y, s).
∂x4i
Now consider T1 .
∂4 σ
ϕ dxi dσ
∂x4i
R
Z ∆x
Z
X
∂4
=−
|A(uj ) − A(u)|ε (x̂i , t, y, s)
(σ − ∆x)3 χIj (xi ) 4 ϕσ dxi dσ
∂xi
0
R
j
Z ∆x
X
= − ∆x
(σ − ∆x)3
Θj D− ρj dσ
Z
|A(uα ) − A(u)|ε (σ − ∆x)3
0
=∆x
X
j
D + Θj
Z
∆x
(σ − ∆x)3 ρj dσ
0
j
where
ρj (σ, x̂i , t, y, s) =
∂3
ϕ(x1 , . . . , xj+1/2 + σ, . . . , xd , t, y, s).
∂x3i
Now we use lemma 4.5 to estimate this term. By 4.26 we get the following estimate
Z Z Z ∆x
1
∂4
|T1 | =|
|A(uα ) − A(u)|ε (σ − ∆x)3 4 ϕ(x + σi , t, y, s) dσ dX|
2
6(∆x)
∂xi
Π2T 0
Z Z TZ
Z X
Z ∆x
1
=|
···
D+ Θj
(σ − ∆x)3 ρj dσ dx̂i dt dyds|
6(∆x) ΠT 0 R
R j
0
!
Z
Z ∆x
r + ∆x
+
3
≤Γ(r)
|D A(uα )|
(σ − ∆x) dσ dxdt
(∆x)2 r4 ΠT i
0
(∆x)2
∆x
2 r + ∆x
≤Γ(r)(∆x)
= Γ(r) 3
1+
r4
r
r
The same estimate apply to T2 . Let us consider (4.31). Integration by parts gives
ZZ
ζ3i =2
|A(uα ) − A(u)|ε (Di+ ϕ − ϕxi )yi dX
Π2T
ZZ
=−2
Π2T
signε (A(uα ) − A(u))A(u)yi (Di+ ϕ − ϕxi ) dX
22
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
By (4.27) and lemma 4.5 we obtain
Z Z Z ∆x
1
i
|ζ3 | =2
|
signε (A(uα ) − A(u))A(u)yi (∆x − σ)ϕσxi xi dσ dX|
2
∆x
ΠT 0
Z
Z ∆x
r + ∆x
(∆x − σ) dσ dyds
|A(u)
|
≤Γ(r)
yi
(∆x)r3 ΠT
0
∆x
∆x
≤Γ(r) 2 1 +
r
r
The result follows.
Estimate 3.
|γ1 | + |γ2 | ≤ Γ(r)
ε
ηr
∆x
1+
r
Proof. We start with the terms γ1i . Consider the integral
Z uα
sign0ε (A(z) − A(u))(f i (z) − f i (u))A0 (z)dz.
I=
u
By the mean value theorem |A(z) − A(u)| ≥ η|z − u|. Let K = kf i kLip , then
|f i (z) − f i (u)| ≤ K|z − u| ≤
K
|(A(z) − A(u)|.
η
Observe that sign(uα − u)(A(z) − A(u)) = |A(z) − A(u)| for all z ∈ int(uα , u). Let
σ(z) = |A(z) − A(u)|. Then σ 0 (z) = sign(uα − u)A0 (z). Recalling that sign0ε is
symmetric we can make a change of variables.
Z uα
K
|I| ≤ |
sign0ε (A(z) − A(u))|A(z) − A(u)|A0 (z)dz|
η u
Z σ(uα )
K
≤ |
sign0ε (σ)(A(z) − A(u))dσ|
η 0
Z
K ε
2
ε
0
≤
signε (σ)σdσ = K
1−
.
η 0
η
π
To finish the estimate we need a bound on kDi+ ϕkL1 (Π2T ) . Note that
Di+ ϕ = Di+ Ψωr ρr0 + Ψ∆xi Di+ ωr ρr0 ,
(4.33)
kDi+ φkL1 (Rd )
and so since
is bounded independently of ∆x we get by lemma 4.5
Z
ZZ
r + ∆x
r + ∆x
∆xi
Ψ
dxdt
≤
C
1
+
.
(4.34)
|Di+ ϕ| dX ≤ C +
r2
r2
ΠT
Π2T
Therefore
|γ1i |
ε
≤ Γ(r)
ηr
∆x
1+
.
r
|γ2i | is estimated in a similar way. Just note that ϕyi = −Ψ(ωr )xi ρr0 so by lemma
4.5 we get
ZZ
Z
C
C
|ϕyi | dX ≤
Ψdxdt ≤ .
r ΠT
r
Π2T
Estimate 4.
∆x
|γ3 | ≤ Γ(r)
r
∆x
1+
.
r
AN ERROR ESTIMATE
23
Proof. Since F i is Lipschitz in each variable (F2i )0 (z) is bounded. Hence
Z uα+e
i
signε (A(z) − A(u))(F2i )0 (z)dz| ≤ C∆x|Di+ uα |.
|
uα
kDi+ uα (·, t)kL1 (Rd )
Now,
is bounded independently of ∆x. We use (4.33) and 4.5
to obtain the following inequality:
ZZ
|Di+ uα ||Di+ ϕ| dX
Π2T
ZZ
≤
Π2T
|Di+ uα ||Di+ Ψ|ωr ρr0 dX + C
Hence
r + ∆x
r2
Z
|Di+ uα |Ψ∆xi dxdt.
ΠT
r + ∆x
.
|γ3 | ≤ C∆x 1 +
r2
Let us now consider now the terms in <1 .
Estimate 5.
|<1 | ≤ Γ(r0 , r)
ε
ηr0
Proof. Note that |A(z) − A(u)| ≥ η|z − u| and
signε (A(z) − A(u)) = sign(uα − u)signε (|A(z) − A(u)|)
for all z ∈ int(u, uα ). Let ζ(z) =
Z
||uα − u| − ψε (uα , u)| =
πη
2ε |z
uα
− u|. Then we have
sign(uα − u) − signε (A(z) − A(u))dz
Z uα
= sign(uα − u)
1 − signε (|A(z) − A(u)|)dz
Zuuα
≤ sign(uα − u)
1 − signε (η|z − u|)dz
u
u
2ε
=
πη
ε
≤
η
Z
ζ(uα )
(1 − sin(ζ))χζ≤π/2 dζ
0
Hence in order to estimate the terms (4.15) and (4.16) it suffices to find an estimate
on kϕt kL1 (Π2T ) and kϕs kL1 (Π2T ) respectively. First differentiate to obtain
ZZ
Z
ZZ
|ϕt | dX ≤
|Ψt |dxdt +
Ψωr |(ρr0 )t | dX.
Π2T
Now
Π2T
ΠT
ZZ
Π2T
Ψωr |(ρr0 )t | dX ≤ C
1
r0
Z
Ψ dxdt ≤
ΠT
C
,
r0
and
Z
Z
|(ψ α0 )t |φ dxdt
|Ψt |dxdt =
ΠT
ΠT
T
Z
Z
φ(x)(ρα0 (t − ν) + ρα0 (t − τ )) dxdt ≤ C.
=
0
Rd
It follows that
kϕt kL1 (Π2T ) ≤
C
(1 + r0 ).
r0
24
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Hence
ε
r0 η
we note that |ϕs | = Ψωr |(ρr0 )t | and so
ε
|(4.16)| ≤ C
.
r0 η
|(4.15)| ≤ Γ(r0 )
Concerning kϕs kL1 (Π2T )
Estimate 6.
|<2 | ≤ Γ(r, ε, r0 )
∆x
ε2 rd r0
1+
∆x
r
Proof. To estimate this term note that |sign0ε (z)| ≤
inequality we obtain
C
ε.
So by the mean value
∆x +
|Di A(uα )|
ε
since θα,i ∈ int(uα , uα+ei ). Since kDi+ A(uα )kL2 (Rd ) is bounded independently of
∆x and η we are left with a similar computation as in estimate 4. Hence
r + ∆x
∆x
∆x
∆x
1+
≤ Γ(r)
1+
|(4.17)| ≤ C
ε
r2
εr
r
|signε (A(uα ) − A(u)) − signε (A(θα,i ) − A(u))| ≤ C
Let us first split (4.18) according to the following equality:
sign0ε (A(uα ) − A(u))ϕ − sign0ε (A(τα,i ) − A(u))ϕ∆xi
= sign0ε (A(uα ) − A(u))(ϕ − ϕ∆xi )
+ sign0ε (A(uα ) − A(u)) − sign0ε (A(τα,i ) − A(u)) ϕ∆xi . (4.35)
The first term on the right gives rise to the term
ZZ
sign0ε (A(uα ) − A(u))(ϕ − ϕ∆xi )(A(u)yi )2 dX
Π2T
ZZ
= −∆x
Π2T
sign0ε (A(uα ) − A(u))(A(u)yi )2 Di+ ϕ dX.
We now use that kA(u)yi kL2 (ΠT ) to obtain
ZZ
sign0ε (A(uα ) − A(u))(ϕ − ϕ∆xi )(A(u)yi )2 dX|
|
Π2T
≤
C
ε
1+
r + ∆x
r2
≤ Γ(r)
∆x
εr
1+
∆x
r
We now consider the second term on the right of (4.35). Now, since kϕk∞ ≤ C rd1r0
d ZZ
X
sign0ε (A(uα ) − A(u)) − sign0ε (A(τα,i ) − A(u)) (A(u)yi )2 ϕ∆xi dX
i=1
Π2T
d
∆x X
≤C 2
ε i=1
ZZ
Π2T
|Di+ A(uα )|(A(u)yi )2 ϕ∆xi dX
Z
d Z
∆x X
+
2
|Di A(uα )| dxdt
(A(u)yi ) dyds
≤C 2 d
ε r r0 i=1
ΠT
ΠT
Hence we obtain
∆x
|(4.18)| ≤ Γ(r)
εr
∆x
1+
r
+C
∆x
ε2 rd r0
.
AN ERROR ESTIMATE
25
Let us turn back to inequality (4.23). Let Ξ(∆x, r, η, ε, r0 ) denote the remaining
terms.
ZZ
ZZ
Ĉ
|uα − u|ϕ dX +
|uα − u|ρα0 (t − ν)φωr ρr0 dX + Ξ(∆x, r, η, ε, r0 )
Π2T
Π2T
ZZ
≥
Π2T
|uα − u|ρα0 (t − τ )φωr ρr0 dX.
Letting α0 tend to zero we get
Z τ
Ĉ
κ(t)dt + κ(ν) + Ξ(∆x, r, η, ε, r0 ) ≥ κ(τ ).
ν
where
Z
ZZ
|uα (x, t) − u(y, s)|φ(x)ωr (x − y)ρr0 (t − s) dydsdx.
κ(t) :=
Rd
ΠT
By Grönwall’s inequality
κ(τ ) ≤ (κ(ν) + Ξ(∆x, r, η, ε, r0 )) 1 + Ĉ(τ − ν)eĈ(τ −ν) .
(4.36)
The last estimate depends on how well u0α approximates u0 .
Estimate 7.
κ(ν) ≤ ku0α − u0 kL1 (Rd ) + C(ν + r + r0 )
Proof. First observe that
|uα (x, ν) − u(y, s)| ≤ |uα (x, ν) − u0α (x)| + |u0α (x) − u0 (y)| + |u0 (y) − u(y, s)|.
It follows since kφk∞ ≤ 1 that
Z ZZ
κ(ν) =
|uα (x, ν) − u(y, s)|φ(x)ωr (x − y)ρr0 (ν − s) dydsdx
Rd
ΠT
Z
≤
|uα (x, ν) − u0α (x)| dx
d
R
Z Z
+
|u0α (x) − u0 (y)|ωr (x − y) dydx
Rd Rd
ZZ
+
|u0 (y) − u(y, s)|ρr0 (ν − s) dyds
ΠT
=: T1 + T2 + T3 .
By the L1 -Lipschitz continuity in time |T1 | ≤ Cν. Now
Z
Z Z
T2 ≤
|u0α (x) − u0 (x)| dx +
|u0 (x) − u0 (y)|ωr (x − y) dydx.
Rd
Rd
Rd
Furthermore, if we let y = x + h we obtain
Z Z
Z Z
|u0 (x) − u0 (y)|ωr (x − y) dydx =
|u0 (x + h) − u0 (x)|ωr (−h) dh dx
Rd Rd
Rd
Rd
Z Z
0
0
=
|u (x + h) − u (x)| dx ωr (−h) dh
d
Rd
ZR
≤
|u0 |BV |h|ωr (−h) dh
Rd
≤ |u0 |BV r
26
K. H. KARLSEN, N. H. RISEBRO, AND E. B. STORRØSTEN
Also
ZZ
|u0 (y) − u(y, ν)| dy +
T3 =
ΠT
ZZ
|u(y, ν) − u(y, s)|ρr0 (ν − s) dyds
ΠT
≤ Cν + |u0 |BV r0 .
Estimate 8.
Z
|uα (x, τ ) − u(x, τ )| dx ≤ κ(τ ) + |u0 |BV r + Cr0
B(0,R)
Proof. First observe that
|uα (x, τ ) − u(x, τ )| ≤ |uα (x, τ ) − u(y, s)| + |u(y, s) − u(x, τ )|
and so, since
Z
|uα (x, τ ) − u(x, τ )|φ(x) dx
Rd
Z Z
=
|uα (x, τ ) − u(x, τ )|φ(x)ωr (x − y)ρr0 (τ − s) dydsdx
Rd
ΠT
we have
Z
B(0,R)
|uα (x, τ ) − u(x, τ )| dx
Z Z
|uα (x, τ ) − u(x, τ )|φ(x)ωr (x − y)ρr0 (τ − s) dydsdx
≤
Rd ΠT
Z Z
≤ κ(τ ) +
|u(y, s) − u(x, τ )|φ(x)ωr (x − y)ρr0 (τ − s) dydsdx
Rd
Since
Z
Rd
Z
ΠT
|u(y, s) − u(x, τ )|φ(x)ωr (x − y)ρr0 (τ − s) dydsdx ≤ |u0 |BV r + Cr0
ΠT
the result follows.
Recall that we wrote u = uη and uα = uηα to simplify notation. By estimate 7
and 8 we turn (4.36) into
Z
|uηα (x, τ ) − uη (x, τ )| dx
B(0,R)
≤ C(r + r0 ) + ku0α − u0 kL1 (Rd ) 1 + Ĉ(τ − ν)eĈ(τ −ν)
+ (C(ν + r + r0 ) + Ξ(∆x, r, η, ε, r0 )) 1 + Ĉ(τ − ν)eĈ(τ −ν)
Let u denote the entropy solution to the degenerate problem. By (??)
Z
|uηα (x, τ ) − u(x, τ )| dx
B(0,R)
√
η) + ku0α − u0 kL1 (Rd ) 1 + Ĉ(τ − ν)eĈ(τ −ν)
+ (C(ν + r + r0 ) + Ξ(∆x, r, η, ε, r0 )) 1 + Ĉ(τ − ν)eĈ(τ −ν)
≤ C(r + r0 +
Combining estimate 1,2,3,4,5 and 6 we see that
2
∆x
ε
ε
∆x
∆x
|Ξ(∆x, r, η, ε, r0 )| ≤ Γ(r, ε, r0 , η)
+
+
+ 2 d
1+
r2
ηr ηr0
ε r r0
r
AN ERROR ESTIMATE
27
√
Let r = r0 , η = r, ε = r4 , ν = 2r and ∆x = r10+d . Then, by the above there
exists constants C1 and C2 such that
Z
|uηα (x, τ ) − u(x, τ )| dx ≤ C1 (∆x)1/(10+d) + C2 ku0α − u0 kL1 (Rd ) ,
B(0,R)
where C2 is only dependent on T, kf kLip and kAkLip .
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(Kenneth Hvistendahl Karlsen)
Center of Mathematics for Applications (CMA)
University of Oslo
P.O. Box 1053, Blindern
N–0316 Oslo, Norway
E-mail address: [email protected]
(Nils Henrik Risebro)
Center of Mathematics for Applications (CMA)
University of Oslo
P.O. Box 1053, Blindern
N–0316 Oslo, Norway
E-mail address: [email protected]
(Erlend Briseid Storrøsten)
Center of Mathematics for Applications (CMA)
University of Oslo
P.O. Box 1053, Blindern
N–0316 Oslo, Norway
E-mail address: [email protected]