Volume 9 (2008), Issue 2, Article 33, 13 pp.
ASYMPTOTIC BEHAVIOR FOR DISCRETIZATIONS OF A SEMILINEAR
PARABOLIC EQUATION WITH A NONLINEAR BOUNDARY CONDITION
NABONGO DIABATE AND THÉODORE K. BONI
U NIVERSITÉ D ’A BOBO -A DJAMÉ
UFR-SFA, D ÉPARTEMENT DE M ATHÉMATIQUES ET I NFORMATIQUES
16 BP 372 A BIDJAN 16, (C ÔTE D ’I VOIRE ).
nabongo_diabate@yahoo.fr
I NSTITUT NATIONAL P OLYTECHNIQUE H OUPHOUËT-B OIGNY DE YAMOUSSOUKRO
BP 1093 YAMOUSSOUKRO , (C ÔTE D ’I VOIRE ).
Received 16 October, 2007; accepted 17 March, 2008
Communicated by C. Bandle
A BSTRACT. This paper concerns the study of the numerical approximation for the following
initial-boundary value problem:

u = uxx − a|u|p−1 u, 0 < x < 1, t > 0,

 t
ux (0, t) = 0 ux (1, t) + b|u(1, t)|q−1 u(1, t) = 0, t > 0,
(P)


u(x, 0) = u0 (x) > 0, 0 ≤ x ≤ 1,
where a > 0, b > 0 and q > p > 1. We show that the solution of a semidiscrete form of
(P ) goes to zero as t goes to infinity and give its asymptotic behavior. Using some nonstandard
schemes, we also prove some estimates of solutions for discrete forms of (P ). Finally, we give
some numerical experiments to illustrate our analysis.
Key words and phrases: Semidiscretizations, Semilinear parabolic equation, Asymptotic behavior, Convergence.
2000 Mathematics Subject Classification. 35B40, 35B50, 35K60, 65M06.
1. I NTRODUCTION
Consider the following initial-boundary value problem:
(1.1)
(1.2)
(1.3)
ut = uxx − a|u|p−1 u,
0 < x < 1, t > 0,
ux (0, t) = 0 ux (1, t) + b|u(1, t)|q−1 u(1, t) = 0,
u(x, 0) = u0 (x) > 0,
t > 0,
0 ≤ x ≤ 1,
where a > 0, b > 0, q > p > 1, u0 ∈ C 1 ([0, 1]), u00 (0) = 0 and u00 (1) + b|u0 (1)|q−1 u0 (1) = 0.
314-07
2
NABONGO D IABATE AND T HÉODORE K. B ONI
The theoretical study of the asymptotic behavior of solutions for semilinear parabolic equations has been the subject of investigation for many authors (see [2], [4] and the references cited
therein). In particular, in [4], when b = 0, the authors have shown that the solution u of (1.1) –
(1.3) goes to zero as t tends to infinity and satisfies the following :
1
for t ∈ [0, +∞),
(1.4)
0 ≤ ku(x, t)k∞ ≤
1
(ku0 (x)k∞ + a(p − 1)t) p−1
1
lim t p−1 ku(x, t)k∞ = C0 ,
(1.5)
t→∞
1
p−1
1
. The same results have been obtained in [2] in the case where b > 0
where C0 = a(p−1)
and q > p > 1.
In this paper we are interested in the numerical study of (1.1) – (1.3). At first, using a
semidiscrete form of (1.1) – (1.3), we prove similar results for the semidiscrete solution. We also
construct two nonstandard schemes and show that these schemes allow the discrete solutions to
obey an estimation as in (1.4). Previously, authors have used numerical methods to study the
phenomenon of blow-up and the one of extinction (see [1] and [3]). This paper is organized as
follows. In the next section, we prove some results about the discrete maximum principle. In
the third section, we take a semidiscrete form of (1.1) – (1.3), and show that the semidiscrete
solution goes to zero as t tends to infinity and give its asymptotic behavior. In the fourth section,
we show that the semidiscrete scheme of the third section converges. In Section 5, we construct
two nonstandard schemes and obtain some estimates as in (1.4). Finally, in the last section, we
give some numerical results.
2. S EMIDISCRETIZATIONS S CHEME
In this section, we give some lemmas which will be used later. Let I be a positive integer,
and define the grid xi = ih, 0 ≤ i ≤ I, where h = 1/I. We approximate the solution u of
the problem (1.1) – (1.3) by the solution Uh (t) = (U0 (t), U1 (t), . . . , UI (t))T of the semidiscrete
equations
d
Ui (t) = δ 2 Ui (t) − a|Ui (t)|p−1 Ui (t), 0 ≤ i ≤ I − 1, t > 0,
(2.1)
dt
(2.2)
d
2b
UI (t) = δ 2 UI (t) − a|UI (t)|p−1 UI (t) − |UI (t)|q−1 UI (t),
dt
h
Ui (0) = Ui0 > 0,
(2.3)
t > 0,
0 ≤ i ≤ I,
where
Ui+1 (t) − 2Ui (t) + Ui−1 (t)
, 1 ≤ i ≤ I − 1,
h2
2U1 (t) − 2U0 (t)
2UI−1 (t) − 2UI (t)
δ 2 U0 (t) =
, δ 2 UI (t) =
.
2
h
h2
The following lemma is a semidiscrete form of the maximum principle.
δ 2 Ui (t) =
Lemma 2.1. Let ah (t) ∈ C 0 ([0, T ], RI+1 ) and let Vh (t) ∈ C 1 ([0, T ], RI+1 ) such that
d
(2.4)
Vi (t) − δ 2 Vi (t) + ai (t)Vi (t) ≥ 0, 0 ≤ i ≤ I, t ∈ (0, T ),
dt
(2.5)
Vi (0) ≥ 0,
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
0 ≤ i ≤ I.
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D ISCRETIZATIONS OF A S EMILINEAR PARABOLIC E QUATION
3
Then we have Vi (t) ≥ 0 for 0 ≤ i ≤ I, t ∈ (0, T ).
Proof. Let T0 < T and let m = min0≤i≤I,0≤t≤T0 Vi (t). Since for i ∈ {0, . . . , I}, Vi (t) is a
continuous function, there exists t0 ∈ [0, T0 ] such that m = Vi0 (t0 ) for a certain i0 ∈ {0, . . . , I}.
It is not hard to see that
Vi (t0 ) − Vi0 (t0 − k)
dVi0 (t0 )
(2.6)
= lim 0
≤ 0,
k→0
dt
k
δ 2 Vi0 (t0 ) =
(2.7)
(2.8)
δ 2 Vi0 (t0 ) =
V1 (t0 ) − V0 (t0 )
≥ 0 if
h2
i0 = 0,
Vi0 +1 (t0 ) − 2Vi0 (t0 ) + Vi0 −1 (t0 )
≥ 0 if 1 ≤ i0 ≤ I − 1,
h2
VI−1 (t0 ) − VI (t0 )
≥ 0 if i0 = I.
h2
Define the vector Zh (t) = eλt Vh (t) where λ is large enough such that ai0 (t0 ) − λ > 0. A
straightforward computation reveals:
dZi0 (t0 )
(2.10)
− δ 2 Zi0 (t0 ) + (ai0 (t0 ) − λ)Zi0 (t0 ) ≥ 0.
dt
(2.9)
δ 2 Vi0 (t0 ) =
dZ (t )
0 0
We observe from (2.6) – (2.9) that idt
≤ 0 and δ 2 Zi0 (t0 ) ≥ 0. Using (2.10), we arrive at
(ai0 (t) − λ)Zi0 (t0 ) ≥ 0, which implies that Zi0 (t0 ) ≥ 0. Therefore, Vi0 (t0 ) = m ≥ 0 and we
have the desired result.
Another form of the maximum principle is the following comparison lemma.
Lemma 2.2. Let Vh (t), Uh (t) ∈ C 1 ([0, ∞), RI+1 ) and f ∈ C 0 (R × R, R) such that for t ∈
(0, ∞),
(2.11)
(2.12)
dUi (t)
dVi (t)
− δ 2 Vi (t) + f (Vi (t), t) <
− δ 2 Ui (t) + f (Ui (t), t),
dt
dt
Vi (0) < Ui (0),
0 ≤ i ≤ I,
0 ≤ i ≤ I.
Then we have Vi (t) < Ui (t), 0 ≤ i ≤ I, t ∈ (0, ∞).
Proof. Define the vector Zh (t) = Uh (t) − Vh (t). Let t0 be the first t > 0 such that Zi (t) > 0
for t ∈ [0, t0 ), i = 0, . . . , I, but Zi0 (t0 ) = 0 for a certain i0 ∈ {0, . . . , I}. We observe that
dZi0 (t0 )
Zi (t0 ) − Zi0 (t0 − k)
= lim 0
≤ 0.
k→0
dt
k
 Z (t )−2Z (t )+Z (t )
i +1 0
i0 0
i0 −1 0

≥ 0 if 1 ≤ i0 ≤ I − 1,
 0
h2



2Z1 (t0 )−2Z0 (t0 )
δ 2 Zi0 (t0 ) =
≥0
if i0 = 0,
h2




 2ZI−1 (t0 )−2ZI (t0 )
≥0
if i0 = I,
h2
which implies:
dZi0 (t0 )
− δ 2 Zi0 (t0 ) + f (Ui0 (t0 ), t0 ) − f (Vi0 (t0 ), t0 ) ≤ 0.
dt
But this inequality contradicts (2.11).
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
http://jipam.vu.edu.au/
4
NABONGO D IABATE AND T HÉODORE K. B ONI
3. A SYMPTOTIC B EHAVIOR
In this section, we show that the solution Uh of (2.1) – (2.3) goes to zero as t → +∞ and
give its asymptotic behavior. Firstly, we prove that the solution tends to zero as t → +∞ by the
following:
Theorem 3.1. The solution Uh (t) of (2.1) – (2.3) goes to zero as t → ∞ and we have the
following estimate
1
0 ≤ kUh (t)k∞ ≤
for t ∈ [0, +∞).
1
1−p
(kUh (0)k∞ + a(p − 1)t) p−1
Proof. We introduce the function α(t) which is defined as
1
α(t) =
1
(kUh (0)k1−p
∞ + a(p − 1)t) p−1
and let Wh be the vector such that Wi (t) = α(t). It is not hard to see that
dWi (t)
− δ 2 Wi (t) + a|Wi (t)|p−1 Wi (t) = 0, 0 ≤ i ≤ I − 1, t ∈ (0, T ),
dt
dWI (t)
2b
− δ 2 WI (t) + a|WI (t)|p−1 WI (t) + |WI (t)|q−1 WI (t) ≥ 0, t ∈ (0, T ),
dt
h
Wi (0) ≥ Ui (0), 0 ≤ i ≤ I,
where (0, T ) is the maximal time interval on which kUh (t)k∞ < ∞. Setting Zh (t) = Wh (t) −
Uh (t) and using the mean value theorem, we see that
dZi (t)
− δ 2 Zi (t) + ap|θi (t)|p−1 Zi (t) = 0, 0 ≤ i ≤ I − 1, t ∈ (0, T )
dt
dZI (t)
2b
2
p−1
q−1
− δ ZI (t) + ap|θI (t)|
+ |θI (t)|
ZI (t) ≥ 0, t ∈ (0, T ),
dt
h
Zi (0) ≥ 0, 0 ≤ i ≤ I,
where θi is an intermediate value between Ui (t) and Wi (t). From Lemma 2.1, we have 0 ≤
Ui (t) ≤ Wi (t) for t ∈ (0, T ). If T < ∞, we have
1
< ∞,
kUh (T )k∞ ≤
1
p−1
(kUh (0)k1−p
+
a(p
−
1)T
)
∞
which leads to a contradiction. Hence T = ∞ and we have the desired result.
Remark 1. The estimate of Theorem 3.1 is a semidiscrete version of the result established in
(1.4) for the continuous problem.
Let us give the statement of the main theorem of this section.
Theorem 3.2. Let Uh be the solution of (2.1) – (2.2). Then we have
1
lim t p−1 kUh (t)k∞ = C0 ,
t→∞
where C0 =
1
a(p−1)
1
p−1
.
The proof of Theorem 3.2 is based on the following lemmas. We introduce the function
µ(x) = −λ(C0 + x) + (C0 + x)p ,
1
p−1
1
.
where C0 = a(p−1)
Firstly, we establish an upper bound of the solution for the semidiscrete problem.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
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D ISCRETIZATIONS OF A S EMILINEAR PARABOLIC E QUATION
5
Lemma 3.3. Let Uh be the solution of (2.1) – (2.3). For any ε > 0, there exist positive times T
and τ such that
Ui (t + τ ) ≤ (C0 + ε)(t + T )−λ + (t + T )−λ−1 ,
0 ≤ i ≤ I.
Proof. Define the vector Wh such that
Wi (t) = (C0 + ε)t−λ + t−λ−1 .
A straightforward computation reveals that
dWi
− δ 2 Wi + a|Wi |p−1 Wi
dt
= −λ(C0 + ε)t−λ−1 − (λ + 1)t−λ−2 + a((C0 + ε)t−λ + t−λ−1 )p
= t−λ−1 (−λ(C0 + ε) − (λ + 1)t−1 + a(C0 + ε + t−1 )p ),
because λp = λ + 1. Using the mean value theorem, we get
(C0 + ε + t−1 )p = (C0 + ε)p + ξi t−1 ,
where ξi (t) is a bounded function. We deduce that
dWi
− δ 2 Wi + a|Wi |p−1 Wi = t−λ−1 (µ(ε) − (λ + 1)t−1 + ξi t−1 ),
dt
2b
dWI
− δ 2 WI + a|WI |p−1 WI + |WI |q−1 WI
dt
h
−λ−1
=t
−1
µ(ε) − (λ + 1)t
−1
+ ξi t
2b −qλ+λ+1
−1 q
+ t
(C0 + ε + t ) .
h
p−q
Obviously −qλ + λ + 1 = p−1
< 0. We also observe that µ(0) = 0 and µ0 (0) = 1, which
implies that µ(ε) > 0. Therefore there exists a positive time T such that
dWi
− δ 2 Wi + a|Wi |p−1 Wi > 0, 0 ≤ i ≤ I − 1, t ∈ [T, +∞),
dt
dWI
2b
− δ 2 WI + a|WI |p−1 WI + |WI (t)|q−1 WI (t) > 0, t ∈ [T, +∞),
dt
h
T −λ C0
Wi (T ) >
.
2
−λ
Since from Theorem 3.1 limt→∞ Ui (t) = 0, there exists τ > T such that Ui (τ ) < T 2 C0 <
Wi (T ). We introduce the vector Zh (t) such that Zi (t) = Ui (t + τ − T ), 0 ≤ i ≤ I. We obtain
dZi
− δ 2 Zi + a|Zi |p−1 Zi > 0, 0 ≤ i ≤ I − 1, t ≥ T,
dt
dZI
2b
− δ 2 ZI + a|ZI |p−1 ZI + |ZI (t)|q−1 ZI (t) > 0, t ≥ T,
dt
h
Zi (T ) = Ui (τ ) < Wi (T ).
We deduce from Lemma 2.2 that Zi (t) ≤ Wi (t), that is to say
(3.1)
Ui (t + τ − T ) ≤ Wi (t) for
t ≥ T,
which leads us to the result.
The lemma below gives a lower bound of the solution for the semidiscrete problem.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
http://jipam.vu.edu.au/
6
NABONGO D IABATE AND T HÉODORE K. B ONI
Lemma 3.4. Let Uh be the solution of (2.1) – (2.3). For any ε > 0, there exists a positive time
τ such that
Ui (t + 1) ≥ (C0 − ε)(t + τ )−λ + (t + τ )−λ−1 , 0 ≤ i ≤ I.
Proof. Introduce the vector Vh such that
Vi (t) = (C0 − ε)t−λ + t−λ−1 .
A direct calculation yields
dVi
− δ 2 Vi + a|Vi |p−1 Vi = −λ(C0 − ε)t−λ−1 − (λ + 1)t−λ−2 + a((C0 − ε)t−λ + t−λ−1 )p
dt
= t−λ−1 (−λ(C0 − ε) − (λ + 1)t−1 + a(C0 − ε + t−1 )p )
because λp = λ + 1. From the mean value theorem, we have
(C0 − ε + t−1 )p = (C0 − ε)p + χi (t)t−1 ,
where χi (t) is a bounded function. We deduce that
dVi
− δ 2 Vi + a|Vi |p−1 Vi = t−λ−1 (µ(−ε) − (λ + 1)t−1 + χi t−1 ),
dt
dVI
2b
− δ 2 VI + a|VI |p−1 VI + |VI |q−1 VI
dt
h
2b −qλ+λ+1
−λ−1
−1
−1
−1 q
=t
µ(ε) − (λ + 1)t + χi t + t
(C0 − ε + t ) .
h
Obviously −qλ + λ + 1 < 0. Also, since µ(0) = 0 and µ0 (0) = 1, it is easy to see that
µ(−ε) < 0. Hence there exists T > 0 such that
dVi
− δ 2 Vi + a|Vi |p−1 Vi < 0, 0 ≤ i ≤ I − 1, t ∈ [T, +∞),
dt
dVI
2b
− δ 2 VI + a|VI |p−1 VI + |VI |q−1 VI < 0, t ∈ [T, +∞).
dt
h
Since Vi (t) goes to zero as t → +∞, there exists τ > max(T, 1) such that Vi (τ ) < Ui (1).
Setting Xi (t) = Vi (t + τ − 1), we observe that
dXi
− δ 2 Xi + a|Xi |p−1 Xi < 0, 0 ≤ i ≤ I − 1, t ≥ 1,
dt
dXI
2b
− δ 2 XI + a|XI |p−1 XI + |XI |q−1 XI < 0, t ≥ 1,
dt
h
Xi (1) = Vi (τ ) < Ui (1).
We deduce from Lemma 2.2 that
(3.2)
Ui (t) ≥ Vi (t + τ − 1) for
t ≥ 1,
which leads us to the result.
Now, we are in a position to give the proof of the main result of this section.
Proof of Theorem 3.2. From Lemma 3.3 and Lemma 3.4, we deduce
Ui (t)
Ui (t)
(C0 − ε) ≤ lim inf
≤ lim sup
≤ (C0 + ε),
t→∞
t→∞
tλ
tλ
and we have the desired result.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
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D ISCRETIZATIONS OF A S EMILINEAR PARABOLIC E QUATION
7
4. C ONVERGENCE
In this section, we will show that for each fixed time interval [0, T ], where u is defined, the
solution Uh (t) of (2.1) – (2.3) approximates u, when the mesh parameter h goes to zero.
Theorem 4.1. Assume that (1.1) – (1.3) has a solution u ∈ C 4,1 ([0, 1] × [0, T ]) and the initial
condition at (2.3) satisfies
kUh0 − uh (0)k∞ = o(1)
(4.1)
as
h → 0,
where uh (t) = (u(x0 , t), . . . , u(xI , t))T . Then, for h sufficiently small, the problem (2.1) – (2.3)
has a unique solution Uh ∈ C 1 ([0, T ], RI+1 ) such that
(4.2)
max kUh (t) − uh (t)k∞ = O(kUh0 − uh (0)k∞ + h2 ) as
0≤t≤T
Proof. Let K > 0 and L be such that
2kuxxx k∞
K
kuxxxx k∞
K
≤ ,
≤ ,
3
2
12
2
kuk∞ ≤ K,
h → 0.
ap(K + 1)p−1 ≤ L,
2q(K + 1)q−1 ≤ L.
(4.3)
The problem (2.1) – (2.3) has for each h, a unique solution Uh ∈ C 1 ([0, Tqh ), RI+1 ). Let t(h)
the greatest value of t > 0 such that
kUh (t) − uh (t)k∞ < 1f ort ∈ (0, t(h)).
(4.4)
The relation (4.1) implies that t(h) > 0 for h sufficiently small. Let t∗ (h) = min{t(h), T }. By
the triangular inequality, we obtain
kUh (t)k∞ ≤ ku(x, t)k∞ + kUh (t) − uh (t)k∞
f or t ∈ (0, t∗ (h)),
which implies that
(4.5)
kUh (t)k∞ ≤ 1 + K,
f or t ∈ (0, t∗ (h)).
Let eh (t) = Uh (t) − uh (x, t) be the error of discretization. Using Taylor’s expansion, we have
for t ∈ (0, t∗ (h)),
d
h2
ei (t) − δ 2 ei (t) = uxxxx (e
xi , t) − apξip−1 ei (t),
dt
12
d
2
2h2
h2
eI (t) − δ 2 eI (t) = qθIq−1 eI +
uxxx (e
xI , t) + uxxxx (e
xI , t) − apξIp−1 eI (t),
dt
h
3
12
where θI ∈ (UI (t), u(xI , t) and ξi ∈ (Ui (t), u(xi , t). Using (4.3) and (4.5), we arrive at
d
(4.6)
ei (t) − δ 2 ei (t) ≤ L|ei (t)| + Kh2 , 0 ≤ i ≤ I − 1,
dt
deI (t) (2eI−1 (t) − 2eI (t))
L|eI (t)|
−
≤
+ L|eI (t)| + Kh2 .
2
dt
h
h
Consider the function
(4.7)
2
z(x, t) = e((M +1)t+Cx ) (kUh0 − uh (0)k∞ + Qh2 )
where M , C, Q are constants which will be determined later. We get
zt (x, t) − zxx (x, t) = (M + 1 − 2C − 4C 2 x2 )z(x, t),
zx (0, t) = 0, zx (1, t) = 2Cz(1, t),
2 z(x, 0) = eCx (Uh0 − uh (0) + Qh).
∞
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
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8
NABONGO D IABATE AND T HÉODORE K. B ONI
By a semidiscretization of the above problem, we may choose M, C, Q large enough that
(4.8)
d
z(xi , t) > δ 2 z(xi , t) + L|z(xi , t)| + Kh2 , 0 ≤ i ≤ I − 1,
dt
(4.9)
L
d
z(xI , t) > δ 2 z(xI , t) + |z(xI , t)| + L|z(xI , t)| + Kh2 ,
dt
h
(4.10)
z(xi , 0) > ei (0), 0 ≤ i ≤ I.
It follows from Lemma 3.4 that
z(xi , t) > ei (t) f or t ∈ (0, t∗ (h)),
0 ≤ i ≤ I.
By the same way, we also prove that
z(xi , t) > −ei (t) f or t ∈ (0, t∗ (h)),
0 ≤ i ≤ I,
which implies that
kUh (t) − uh (t)k∞ ≤ e(M t+C) (Uh0 − uh (0)∞ + Qh2 ), t ∈ (0, t∗ (h)).
Let us show that t∗ (h) = T . Suppose that T > t(h). From (4.4), we obtain
(4.11)
1 = kUh (t(h)) − uh (t(h))k∞ ≤ e(M T +C) (Uh0 − uh (0)∞ + Qh2 ).
Since the term in the right hand side of the inequality goes to zero as h goes to zero, we deduce
from (4.11) that 1 ≤ 0, which is impossible. Consequently t∗ (h) = T , and we obtain the desired
result.
5. F ULL D ISCRETIZATIONS
In this section, we study the asymptotic behavior, using full discrete schemes (explicit and
implicit) of (1.1) – (1.3). Firstly, we approximate the solution u(x, t) of (1.1) – (1.3) by the
(n)
solution Uh = (U0n , U1n , . . . , UIn )T of the following explicit scheme
(n+1)
(n)
Ui
− Ui
(n) p−1 (n+1)
2 (n)
(5.1)
= δ Ui − a Ui Ui
, 0 ≤ i ≤ I − 1,
∆t
(n+1)
(5.2)
UI
(n)
− UI
∆t
=δ
2
(n)
UI
(0)
Ui
(5.3)
(n) p−1 (n+1) 2b (n) q−1 (n+1)
− a UI UI
− UI UI
,
h
= φi > 0,
0 ≤ i ≤ I,
h2
where n ≥ 0, ∆t ≤ 2 . We need the following lemma which is a discrete form of the maximum
principle for ordinary differential equations.
Lemma 5.1. Let f ∈ C 1 (R) and let an and bn be two bounded sequences such that
(5.4)
an+1 − an
bn+1 − bn
+ f (an ) ≥
+ f (bn ),
∆t
∆t
(5.5)
n ≥ 0,
a0 ≥ b 0 .
Then we have an ≥ bn , n ≥ 0 for h small enough.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
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D ISCRETIZATIONS OF A S EMILINEAR PARABOLIC E QUATION
9
Proof. Let Zn = an − bn . We get
Zn+1 − Zn
+ f 0 (ξn )Zn ≥ 0,
∆t
(5.6)
where ξn is an intermediate value between an and bn . Obviously
Zn+1 ≥ Zn (1 − ∆tf 0 (ξn )).
(5.7)
Since an and bn are bounded and f ∈ C 1 (R), there exists a positive M such that |f 0 (ξn )| ≤ M .
Let j be the first integer such that Zj < 0. From (5.5), j ≥ 0. We have Zj ≥ Zj−1 (1 − ∆tM ).
Since ∆tM goes to zero as h → 0 and Zj−1 ≥ 0, we deduce that Zj ≥ 0 as h → 0 which is a
contradiction. Therefore, Zn ≥ 0 for any n and we have proved the lemma.
Now, we may state the following.
(n)
Theorem 5.2. Let Uh be the solution of (5.1) – (5.3). We have Uh ≥ 0 and
1
(n) Uh ≤ 1 ,
p−1
1−p
∞
(0) + A(p − 1)n∆t
Uh ∞
where A =
a p−1 .
(0) 1+a∆tUh ∞
Proof. A straightforward calculation yields
(5.8)
(n+1)
Ui
=
∆t (n)
U
h2 i+1
+ 1−
(n+1)
UI
(5.10)
(n)
(n)
Ui + Ui−1
,
(n) p−1
1 + a∆t Ui (n+1)
U0
(5.9)
2∆t
h2
=
1 ≤ i ≤ I − 1,
2∆t (n)
U1
h2
(n)
+ 1 − 2∆t
U0
h2
,
p−1
(n) 1 + a∆t U0 2∆t (n)
UI−1
h2
(n)
+ 1 − 2∆t
UI
h2
=
p−1
.
(n) (n) q−1
b
1 + a∆t UI + 2 h ∆t UI (n)
Since 1 − 2 ∆t
is nonnegative,
h2
using a recursive argument, it is easy to see that Uh ≥ 0. Let i0
(n)
(n) be such that Ui0 = Uh . From (5.8), we get
∞
(n+1) Uh
∞
≤
∆t (n)
U
h2 i0 +1
(n) (n)
Uh + Ui0 −1
∞
(n) p−1
1 + a∆t Uh + 1−
2∆t
h2
if 1 ≤ i0 ≤ I − 1.
∞
Applying the triangle inequality and the fact that 1 − 2∆t
is nonnegative, we arrive at
h2
(n) Uh (n+1) (5.11)
Uh
≤
∞ p−1 .
(n) ∞
1 + a∆t Uh ∞
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
http://jipam.vu.edu.au/
10
NABONGO D IABATE AND T HÉODORE K. B ONI
We
obtain
the same estimation if i0 = 0 or i0 = I. The
inequality
(5.11) implies that
(n+1) (n) (n) (0) Uh
≤ Uh and by iterating, we obtain Uh ≤ Uh . From (5.11), we
∞
∞
∞
∞
also observe that
p
(n+1) (n) a Uh Uh
− kU (n) k∞
∞
≤−
∞p−1 .
(n) ∆t
1 + a∆t Uh ∞
(n) (0) Using the fact that Uh ≤ Uh , we have
∞
∞
(n+1) (n) Uh
− Uh (n) p
∞
∞
≤ −A Uh .
∆t
∞
We introduce the function α(t) which is defined as follows
1
α(t) = 1 .
p−1
1−p
(0) + A(p − 1)t
Uh ∞
We remark that α(t) obeys the following differential equation
(0) α0 (t) = −Aαp (t), α(0) = Uh .
∞
Using a Taylor’s expansion, we have
(∆t)2 00 ˜
α (tn ),
2
where t˜n is an intermediate value between tn and tn+1 . It is not hard to see that α(t) is a convex
function. Therefore, we obtain
α(tn+1 ) − α(tn )
≥ −Aαp (tn ).
∆t
(n) From Lemma 5.1, we get Uh ≤ α(tn ), which ensures that
∞
1
(n) Uh ≤ 1 ,
p−1
∞
(0) 1−p
+ A(p − 1)n∆t
Uh α(tn+1 ) = α(tn ) + ∆tα0 (tn ) +
∞
and we have the desired result.
Remark 2. The estimate of Theorem 5.2 is the discrete form of the one given in (1.4) for the
continuous problem.
(n)
Now, we approximate the solution u(x, t) of problem (1.1) – (1.3) by the solution Uh of the
following implicit scheme
(n+1)
(n)
Ui
− Ui
(n) p−1 (n+1)
(n+1)
(5.12)
= δ 2 Ui
− Ui Ui
, 0 ≤ i ≤ I − 1,
∆t
(n+1)
(5.13)
(5.14)
UI
(n)
− UI
∆t
(n+1)
= δ 2 UI
(0)
Ui
(n) p−1 (n+1) 2b (n) p−1 (n+1)
− a UI UI
− UI UI
,
h
= φi > 0,
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
0 ≤ i ≤ I,
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D ISCRETIZATIONS OF A S EMILINEAR PARABOLIC E QUATION
11
where n ≥ 0. Let us note that in the above construction, we do not need a restriction on the step
time.
The above equations may be rewritten in the following form:
∆t
2∆t (n+1)
(n) p−1
(n)
(n+1)
U0 = − 2 U1
+ 1 + 2 2 + a∆t U0 U0
,
h
h
∆t (n+1)
∆t
∆t (n+1)
(n) p−1
(n)
(n+1)
Ui = − 2 Ui−1 + 1 + 2 2 + a∆t Ui Ui
− 2 Ui+1 , 1 ≤ i ≤ I − 1,
h
h
h
2∆t (n+1)
∆t
(n) p−1 2b (n) q−1
(n)
(n+1)
UI = − 2 UI−1 + 1 + 2 2 + a∆t UI + ∆t UI UI
,
h
h
h
which gives the following linear system
(n+1)
(n)
A(n) Uh
= Uh
where A(n) is the tridiagonal matrix defined as follows

0
0
···
0
d0 −2∆t
h2
−∆t
 h2
d1 −∆t
0
·
·
·
0
h2

−∆t
−∆t
 0
d
0
···
2
2
h
h2

.
.
..
(n)
.
.
.

..
..
..
..
A =  ..
.
 0
−∆t
−∆t
0
·
·
·
d

I−2
h2
h2
 0
0
0
· · · −∆t
d
I−1
2
h
−2∆t
0
0
0
···
0
h2
0
0
0
..
.





,

0 

−∆t 
h2
dI
with
di = 1 + 2
∆t
(n)
+ a∆t|Ui |p−1
2
h
for 0 ≤ i ≤ I − 1
and
∆t
(n) p−1 2b (n) q−1
+ ∆t UI .
dI = 1 + 2 2 + a∆t UI h
h
(n)
Let us remark that the tridiagonal matrix A satisfies the following properties
(n)
(n)
Aij < 0 i 6= j,
Aii > 0 and
(n) X (n) Aii >
Aij .
i6=j
(n)
These properties imply that Uhn exists for any n and Uh ≥ 0 (see for instance [2]). As we
know that the solution of the discrete implicit scheme exists, we may state the following.
(n)
(n)
Theorem 5.3. Let Uh be the solution of (5.12) – (5.14). We have Uh ≥ 0 and
1
(n) Uh ≤ 1 ,
p−1
∞
(0) 1−p
+ A(p − 1)n∆t
Uh ∞
where A =
a p−1 .
(0) 1+a∆tUh ∞
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
http://jipam.vu.edu.au/
12
NABONGO D IABATE AND T HÉODORE K. B ONI
(n)
Proof. We know that Uh ≥ 0 as we have seenabove.
Now, let us obtain the above estimate to
(n) (n)
complete the proof. Let i0 be such that Ui0 = Uh . Using the equality (5.12), we have
∞
∆t (n)
∆t (n)
∆t
(n) (n+1) (n) 1 + 2 2 + a∆t Uh U
≤
U
h
h + 2 Ui0 −1 + 2 Ui0 +1
h
h
h
∞
∞
∞
if 1 ≤ i0 ≤ I − 1.
Applying the triangle inequality, we derive the following estimate
(n) U
h (n+1) Uh
≤
∞ p−1 .
(n) ∞
1 + a∆t Uh ∞
We obtain the same estimation if we take i0 = 0 or i0 = I. Reasoning as in the proof of
Theorem 5.3, we obtain the desired result.
6. N UMERICAL R ESULTS
In this section, we consider the explicit scheme in (5.1) – (5.3) and the implicit scheme in
(5.12) – (5.14). We suppose that p = 2, q = 3, a = 1, b = 1, Ui0 = 0.8 + 0.8 ∗ cos(πhi) and
2
∆t = h2 . In the following tables, in the rows, we give the first n when
(n)
n∆tUh − 1 < ε,
∞
the corresponding time T n = n∆t, the CPU time and the order(s) of method computed from
s=
log((T4h − T2h )/(T2h − Th ))
.
log(2)
Table 1: (ε = 10−2 ): Numerical times, numbers of iterations, CPU times (seconds), and
orders of the approximations obtained with the implicit Euler method
I
16
32
64
128
256
Tn
674.0820
674.2632
674.3085
674.3278
674.4807
n
345129
1.380890.
5.523.934
22095735
87383041
CPU time
103
660
6020
58290
574823
s
2.01
1.24
2.99
Table 2: (ε = 10−2 ): Numerical times, numbers of iterations, CPU times (seconds) and
orders of the approximations obtained with the explicit Euler method
I
16
32
64
128
256
Tn
674.3281
674.3452
674.3290
674.3187
674.3098
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
n
345.255
1.381.058
5.524.102
22845950
88237375
CPU time
90
720
10820
323528
19457811
s
0.08
0.65
0.21
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D ISCRETIZATIONS OF A S EMILINEAR PARABOLIC E QUATION
13
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[1] L. ABIA, J.C. LÓPEZ-MARCOS AND J. MARTINEZ, On the blow-up time convergence of semidiscretizations of reaction-diffusion equations, Appl. Numer. Math., 26 (1998), 399–414.
[2] T.K. BONI, On the asymptotic behavior of solutions for some semilinear parabolic and elliptic equation of second order with nonlinear boundary conditions, Nonl. Anal. TMA, 45 (2001), 895–908.
[3] T.K. BONI, Extinction for discretizations of some semilinear parabolic equations, C.R.A.S, Serie I,
333 (2001), 795–800.
[4] V.A. KONDRATIEV AND L. VÉRON, Asymptotic behaviour of solutions of some nonlinear parabolic or elliptic equation, Asymptotic Analysis, 14 (1997), 117–156.
[5] R.E. MICKENS, Relation between the time and space step-sizes in nonstandard finite difference
schemes for the fisher equation, Num. Methods. Part. Diff. Equat., 13 (1997), 51–55.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 33, 13 pp.
http://jipam.vu.edu.au/