Asymptotic Behavior for Discretizations of a Semilinear Parabolic

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Asymptotic Behavior for Discretizations of a Semilinear Parabolic Powered By Docstoc
					ASYMPTOTIC BEHAVIOR FOR DISCRETIZATIONS
OF A SEMILINEAR PARABOLIC EQUATION WITH
    A NONLINEAR BOUNDARY CONDITION                                                                                      Discretizations of a
                                                                                                                   Semilinear Parabolic Equation
                                                                                                                       Nabongo Diabate and
                                                                                                                          Théodore K. Boni
 NABONGO DIABATE                                  THÉODORE K. BONI                                                   vol. 9, iss. 2, art. 33, 2008
 Université d’Abobo-Adjamé                        Inst. Nat. Polytech. Houphouët-Boigny de Yamoussoukro
 UFR-SFA, Dépt. de Math. et Informatiques         BP 1093 Yamoussoukro,
 16 BP 372 Abidjan 16, (Côte d’Ivoire).           (Côte d’Ivoire).                                                          Title Page
 EMail: nabongo_diabate@yahoo.fr
                                                                                                                             Contents
Received:                16 October, 2007
Accepted:                17 March, 2008
Communicated by:         C. Bandle
2000 AMS Sub. Class.:    35B40, 35B50, 35K60, 65M06.
                                                                                                                          Page 1 of 26
Key words:               Semidiscretizations, Semilinear parabolic equation, Asymptotic behavior, Conver-
                         gence.                                                                                              Go Back
Abstract:                This paper concerns the study of the numerical approximation for the following initial-
                         boundary value problem:                                                                           Full Screen
                                                      p−1
                                 
                                  ut = uxx − a|u|         u, 0 < x < 1, t > 0,                                                Close
                         (P)         ux (0, t) = 0 ux (1, t) + b|u(1, t)|q−1 u(1, t) = 0, t > 0,
                                     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.
Contents
1   Introduction                 3

2   Semidiscretizations Scheme   5

3   Asymptotic Behavior          8         Discretizations of a
                                      Semilinear Parabolic Equation
                                          Nabongo Diabate and
4   Convergence                  14
                                             Théodore K. Boni
                                        vol. 9, iss. 2, art. 33, 2008
5   Full Discretizations         17

6   Numerical Results            24            Title Page

                                                Contents




                                             Page 2 of 26

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1.      Introduction
Consider the following initial-boundary value problem:
(1.1)                ut = uxx − a|u|p−1 u,                0 < x < 1, t > 0,

                                                                                                     Discretizations of a
                                                           q−1                                  Semilinear Parabolic Equation
(1.2)        ux (0, t) = 0 ux (1, t) + b|u(1, t)|                u(1, t) = 0,    t > 0,             Nabongo Diabate and
                                                                                                       Théodore K. Boni
                                                                                                  vol. 9, iss. 2, art. 33, 2008

(1.3)                     u(x, 0) = u0 (x) > 0,              0 ≤ x ≤ 1,
where a > 0, b > 0, q > p > 1, u0 ∈ C 1 ([0, 1]), u0 (0) = 0 and u0 (1) +                                Title Page

b|u0 (1)|q−1 u0 (1) = 0.                                                                                  Contents
    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 :                                                                                            Page 3 of 26

                                                      1                                                   Go Back
(1.4)      0 ≤ u(x, t)    ∞    ≤                                      1    for   t ∈ [0, +∞),
                                   ( u0 (x)      ∞   + a(p − 1)t) p−1                                   Full Screen

                                                                                                            Close
                                            1
(1.5)                              lim t   p−1   u(x, t)   ∞   = C0 ,
                                   t→∞
                          1
                1        p−1
where C0 = a(p−1)       . The same results have been obtained in [2] in the case
where b > 0 and q > p > 1.
    In this paper we are interested in the numerical study of (1.1) – (1.3). At first, us-
ing 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,                Discretizations of a
                                                                                            Semilinear Parabolic Equation
we take a semidiscrete form of (1.1) – (1.3), and show that the semidiscrete solution           Nabongo Diabate and
goes to zero as t tends to infinity and give its asymptotic behavior. In the fourth                 Théodore K. Boni

section, we show that the semidiscrete scheme of the third section converges. In              vol. 9, iss. 2, art. 33, 2008

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.
                                                                                                     Title Page

                                                                                                      Contents




                                                                                                   Page 4 of 26

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2.      Semidiscretizations Scheme
In this section, we give some lemmas which will be used later. Let I be a pos-
itive 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
                                                                                                         Discretizations of a
            d
(2.1)          Ui (t) = δ 2 Ui (t) − a|Ui (t)|p−1 Ui (t),    0 ≤ i ≤ I − 1, t > 0,                  Semilinear Parabolic Equation
            dt                                                                                          Nabongo Diabate and
                                                                                                           Théodore K. Boni
                                                                                                      vol. 9, iss. 2, art. 33, 2008
           d                                             2b
(2.2)         UI (t) = δ 2 UI (t) − a|UI (t)|p−1 UI (t) − |UI (t)|q−1 UI (t),    t > 0,
           dt                                            h
                                                                                                             Title Page
(2.3)                         Ui (0) =   Ui0   > 0,    0 ≤ i ≤ I,
                                                                                                              Contents
where
                         Ui+1 (t) − 2Ui (t) + Ui−1 (t)
               δ 2 Ui (t) =                            , 1 ≤ i ≤ I − 1,
                                      h2
                       2U1 (t) − 2U0 (t)                2UI−1 (t) − 2UI (t)
          δ 2 U0 (t) =          2
                                         , δ 2 UI (t) =                     .                              Page 5 of 26
                               h                                h2
The following lemma is a semidiscrete form of the maximum principle.                                          Go Back
                                 0             I+1                      1            I+1
Lemma 2.1. Let ah (t) ∈ C ([0, T ], R ) and let Vh (t) ∈ C ([0, T ], R                     ) such           Full Screen
that
                                                                                                                Close
         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,        0 ≤ i ≤ I.
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
                           dVi0 (t0 )       Vi (t0 ) − Vi0 (t0 − k)
(2.6)                                 = lim 0                       ≤ 0,
                             dt         k→0            k
                                                                                                    Discretizations of a
                                                                                               Semilinear Parabolic Equation
                                        V1 (t0 ) − V0 (t0 )                                        Nabongo Diabate and
(2.7)                 δ 2 Vi0 (t0 ) =                       ≥ 0 if    i0 = 0,                         Théodore K. Boni
                                                 h2
                                                                                                 vol. 9, iss. 2, art. 33, 2008



                           Vi0 +1 (t0 ) − 2Vi0 (t0 ) + Vi0 −1 (t0 )
(2.8)    δ 2 Vi0 (t0 ) =                                            ≥ 0 if   1 ≤ i0 ≤ I − 1,            Title Page
                                            h2
                                                                                                         Contents

                                       VI−1 (t0 ) − VI (t0 )
(2.9)                δ 2 Vi0 (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.               Page 6 of 26
A straightforward computation reveals:
                                                                                                         Go Back
                    dZi0 (t0 )
(2.10)                         − δ 2 Zi0 (t0 ) + (ai0 (t0 ) − λ)Zi0 (t0 ) ≥ 0.                         Full Screen
                      dt
                                                                                                           Close
                                             dZi0 (t0 )           2
We observe from (2.6) – (2.9) that           ≤ 0 and δ Zi0 (t0 ) ≥ 0. Using (2.10),
                                                dt
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, ∞),
         dVi (t) 2                        dUi (t) 2
(2.11)          −δ Vi (t)+f (Vi (t), t) <        −δ Ui (t)+f (Ui (t), t),                  0 ≤ i ≤ I,
           dt                               dt

                                                                                                             Discretizations of a
(2.12)                           Vi (0) < Ui (0),      0 ≤ i ≤ I.                                       Semilinear Parabolic Equation
                                                                                                            Nabongo Diabate and
                                                                                                               Théodore K. Boni
Then we have Vi (t) < Ui (t), 0 ≤ i ≤ I, t ∈ (0, ∞).
                                                                                                          vol. 9, iss. 2, art. 33, 2008
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}.
                                                                                                                 Title Page
We observe that
                         dZi0 (t0 )           Zi (t0 ) − Zi0 (t0 − k)                                             Contents
                                    = lim 0                             ≤ 0.
                           dt           k→0              k
                          Z (t )−2Z (t )+Z (t )
                                                   i0 −1 0
                          i0 +1 0
                         
                                          i0 0
                                           h2
                                                           ≥ 0 if       1 ≤ i0 ≤ I − 1,
                         
                         
                                                                                                               Page 7 of 26
                         
         δ 2 Zi0 (t0 ) =   2Z1 (t0 )−2Z0 (t0 )
                                    h2
                                               ≥0                 if    i0 = 0,
                         
                         
                                                                                                                 Go Back
                          2Z (t0 )−2Z (t0 )
                                                 ≥0
                          I−1             I
                                     h2
                                                                  if    i0 = I,                                 Full Screen
which implies:                                                                                                      Close
               dZi0 (t0 )
                          − δ 2 Zi0 (t0 ) + f (Ui0 (t0 ), t0 ) − f (Vi0 (t0 ), t0 ) ≤ 0.
                 dt
But this inequality contradicts (2.11).
3.   Asymptotic Behavior
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
                                                                                                      Discretizations of a
have the following estimate                                                                      Semilinear Parabolic Equation
                                                                                                     Nabongo Diabate and
                                               1                                                        Théodore K. Boni
       0 ≤ Uh (t)     ∞   ≤              1−p                   1    for   t ∈ [0, +∞).             vol. 9, iss. 2, art. 33, 2008
                              ( Uh (0)   ∞     + a(p − 1)t)   p−1



Proof. We introduce the function α(t) which is defined as
                                                                                                          Title Page
                                                      1
                          α(t) =                1−p                 1
                                                                                                           Contents
                                   ( Uh (0)     ∞     + 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 ),               Page 8 of 26
       dt
                                                                                                           Go Back
 dWI (t)                                     2b
         − δ 2 WI (t) + a|WI (t)|p−1 WI (t) + |WI (t)|q−1 WI (t) ≥ 0,             t ∈ (0, T ),           Full Screen
   dt                                        h
                           Wi (0) ≥ Ui (0), 0 ≤ i ≤ I,                                                       Close

where (0, T ) is the maximal time interval on which Uh (t) ∞ < ∞. 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
           − δ 2 ZI (t) + ap|θI (t)|p−1 + |θI (t)|q−1 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
                                                                                                     Discretizations of a
                                                       1                                        Semilinear Parabolic Equation
                 Uh (T )     ∞   ≤              1−p                  1    < ∞,                      Nabongo Diabate and
                                     ( Uh (0)   ∞     + a(p − 1)T ) p−1                                Théodore K. Boni
                                                                                                  vol. 9, iss. 2, art. 33, 2008
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 estab-
                                                                                                         Title Page
lished in (1.4) for the continuous problem.
   Let us give the statement of the main theorem of this section.                                         Contents

Theorem 3.2. Let Uh be the solution of (2.1) – (2.2). Then we have
                                           1
                                 lim t p−1 Uh (t)       ∞   = C0 ,
                                 t→∞
                                                                                                       Page 9 of 26
                        1
                 1     p−1                                                                                Go Back
where C0 =    a(p−1)
                             .
                                                                                                        Full Screen
   The proof of Theorem 3.2 is based on the following lemmas. We introduce the
function                                                                                                    Close
                      µ(x) = −λ(C0 + x) + (C0 + x)p ,
                        1
                 1     p−1
where C0 = a(p−1)        .
  Firstly, we establish an upper bound of the solution for the semidiscrete problem.
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 .                                  Discretizations of a
                                                                                     Semilinear Parabolic Equation
                                                                                         Nabongo Diabate and
A straightforward computation reveals that                                                  Théodore K. Boni
                                                                                       vol. 9, iss. 2, art. 33, 2008
    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                     Title Page

              = t−λ−1 (−λ(C0 + ε) − (λ + 1)t−1 + a(C0 + ε + t−1 )p ),                          Contents

because λp = λ + 1. Using the mean value theorem, we get

                       (C0 + ε + t−1 )p = (C0 + ε)p + ξi t−1 ,
                                                                                           Page 10 of 26
where ξi (t) is a bounded function. We deduce that
                                                                                               Go Back
         dWi
             − δ 2 Wi + a|Wi |p−1 Wi = t−λ−1 (µ(ε) − (λ + 1)t−1 + ξi t−1 ),                  Full Screen
          dt
                                                                                                 Close
  dWI                          2b
      − δ 2 WI + a|WI |p−1 WI + |WI |q−1 WI
   dt                          h
                                                     2b −qλ+λ+1
           = t−λ−1 µ(ε) − (λ + 1)t−1 + ξi t−1 +        t        (C0 + ε + t−1 )q .
                                                     h
                           p−q
Obviously −qλ + λ + 1 = p−1 < 0. We also observe that µ(0) = 0 and µ (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, +∞),                    Discretizations of a
        dt                            h                                                 Semilinear Parabolic Equation
                                                                                            Nabongo Diabate and
                                           T −λ C0                                             Théodore K. Boni
                                 Wi (T ) >         .
                                              2                                           vol. 9, iss. 2, art. 33, 2008
Since from Theorem 3.1 limt→∞ Ui (t) = 0, there exists τ > T such that Ui (τ ) <
T −λ C0
   2
        < Wi (T ). We introduce the vector Zh (t) such that Zi (t) = Ui (t + τ − T ),
                                                                                                 Title Page
0 ≤ i ≤ I. We obtain
                                                                                                  Contents
               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                                                   Page 11 of 26
                             Zi (T ) = Ui (τ ) < Wi (T ).
                                                                                                  Go Back
We deduce from Lemma 2.2 that Zi (t) ≤ Wi (t), that is to say
                                                                                                Full Screen
(3.1)                  Ui (t + τ − T ) ≤ Wi (t) for       t ≥ T,
                                                                                                    Close
which leads us to the result.
   The lemma below gives a lower bound of the solution for the semidiscrete prob-
lem.
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 .
                                                                                               Discretizations of a
                                                                                          Semilinear Parabolic Equation
A direct calculation yields                                                                   Nabongo Diabate and
                                                                                                 Théodore K. Boni
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     vol. 9, iss. 2, art. 33, 2008
dt
                            = t−λ−1 (−λ(C0 − ε) − (λ + 1)t−1 + a(C0 − ε + t−1 )p )
                                                                                                   Title Page
because λp = λ + 1. From the mean value theorem, we have
                                                                                                    Contents
                     (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                                                                                     Page 12 of 26

  dVI                          2b                                                                   Go Back
      − δ 2 VI + a|VI |p−1 VI + |VI |q−1 VI
   dt                          h                                                                  Full Screen
                                                 2b
          = t−λ−1    µ(ε) − (λ + 1)t−1 + χi t−1 + t−qλ+λ+1 (C0 − ε + t−1 )q .                         Close
                                                 h
Obviously −qλ + λ + 1 < 0. Also, since µ(0) = 0 and µ (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                                                                          Discretizations of a
                                                                                      Semilinear Parabolic Equation
             dXI                            2b                                            Nabongo Diabate and
                 − δ 2 XI + a|XI |p−1 XI + |XI |q−1 XI < 0,        t ≥ 1,                    Théodore K. Boni
              dt                             h
                                                                                        vol. 9, iss. 2, art. 33, 2008
                             Xi (1) = Vi (τ ) < Ui (1).
We deduce from Lemma 2.2 that
                                                                                               Title Page
(3.2)                   Ui (t) ≥ Vi (t + τ − 1) for t ≥ 1,                                      Contents

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                               Page 13 of 26

                                  Ui (t)               Ui (t)                                   Go Back
          (C0 − ε) ≤ lim inf               ≤ lim sup            ≤ (C0 + ε),
                       t→∞         tλ        t→∞        tλ                                    Full Screen

and we have the desired result.                                                                   Close
4.      Convergence
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
                                                                                                      Discretizations of a
the initial condition at (2.3) satisfies                                                          Semilinear Parabolic Equation
                                                                                                     Nabongo Diabate and
                               0
(4.1)                         Uh − uh (0)      ∞   = o(1)     as   h → 0,                               Théodore K. Boni
                                                                                                   vol. 9, iss. 2, art. 33, 2008
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
                                                    0                                                     Title Page
(4.2)       max Uh (t) − uh (t)           ∞   = O( Uh − uh (0)         ∞   + h2 ) as    h → 0.
           0≤t≤T
                                                                                                           Contents
Proof. Let K > 0 and L be such that
     2 uxxx   ∞         K        uxxxx    ∞       K
                    ≤     ,                   ≤     ,     u   ∞   ≤ K,      ap(K + 1)p−1 ≤ L,
         3              2          12             2
                                                                                                       Page 14 of 26

(4.3)                                    2q(K + 1)q−1 ≤ L.                                                 Go Back

The problem (2.1) – (2.3) has for each h, a unique solution Uh ∈ C 1 ([0, Tqh ), RI+1 ).                 Full Screen
Let t(h) the greatest value of t > 0 such that
                                                                                                             Close
(4.4)                         Uh (t) − uh (t)     ∞   < 1f ort ∈ (0, t(h)).
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
           Uh (t)   ∞   ≤ u(x, t)     ∞   + Uh (t) − uh (t)        ∞     f or t ∈ (0, t∗ (h)),
which implies that
(4.5)                   Uh (t)   ∞   ≤ 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                                                       Discretizations of a
                     ei (t) − δ 2 ei (t) = uxxxx (xi , t) − apξip−1 ei (t),                   Semilinear Parabolic Equation
                  dt                      12                                                      Nabongo Diabate and
                                                                                                     Théodore K. Boni
  d                      2 q−1       2h2                   h2                    p−1
    eI (t) − δ 2 eI (t) = qθI eI +       uxxx (xI , t) + uxxxx (xI , t) − apξI eI (t),          vol. 9, iss. 2, art. 33, 2008
 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                                                                                                     Title Page

                 d                                                                                      Contents
(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)|
(4.7)              −           2
                                          ≤           + L|eI (t)| + Kh2 .                           Page 15 of 26
              dt             h                 h
Consider the function                                                                                   Go Back
                                            2  0
                   z(x, t) = e((M +1)t+Cx ) ( Uh − uh (0)     ∞   + Qh2 )                             Full Screen

                                                                                                          Close
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),
                                        20
                        z(x, 0) = eCx ( Uh − uh (0)      ∞
                                                             + Qh).
By a semidiscretization of the above problem, we may choose M, C, Q large enough
that
             d
(4.8)           z(xi , t) > δ 2 z(xi , t) + L|z(xi , t)| + Kh2 , 0 ≤ i ≤ I − 1,
             dt

             d                             L                                                         Discretizations of a
(4.9)           z(xI , t) > δ 2 z(xI , t) + |z(xI , t)| + L|z(xI , t)| + Kh2 ,                  Semilinear Parabolic Equation
             dt                            h                                                        Nabongo Diabate and
                                                                                                       Théodore K. Boni
                                                                                                  vol. 9, iss. 2, art. 33, 2008
(4.10)                           z(xi , 0) > ei (0), 0 ≤ i ≤ I.
It follows from Lemma 3.4 that                                                                           Title Page

                  z(xi , t) > ei (t) f or t ∈ (0, t∗ (h)),          0 ≤ i ≤ I.                            Contents

By the same way, we also prove that
                 z(xi , t) > −ei (t) f or t ∈ (0, t∗ (h)),           0 ≤ i ≤ I,
                                                                                                      Page 16 of 26
which implies that
                                                                                                          Go Back
         Uh (t) − uh (t)   ∞
                                             0
                               ≤ e(M t+C) ( Uh − uh (0)   ∞
                                                                  + Qh2 ), t ∈ (0, t∗ (h)).
                                                                                                        Full Screen
Let us show that t∗ (h) = T . Suppose that T > t(h). From (4.4), we obtain
                                                                                                            Close
                                                   (M T +C)        0                      2
(4.11)      1 = Uh (t(h)) − uh (t(h))       ∞   ≤e            (   Uh   − uh (0)   ∞
                                                                                      + Qh ).
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.      Full Discretizations
In this section, we study the asymptotic behavior, using full discrete schemes (ex-
plicit and implicit) of (1.1) – (1.3). Firstly, we approximate the solution u(x, t) of
                                  (n)       n    n      n
(1.1) – (1.3) by the solution Uh = (U0 , U1 , . . . , UI )T of the following explicit
scheme
                                                                                                              Discretizations of a
             (n+1)    (n)
           Ui     − Ui                (n)         (n)
                                                        p−1
                                                               (n+1)                                     Semilinear Parabolic Equation
(5.1)                        =   δ 2 Ui     −a   Ui           Ui     ,      0 ≤ i ≤ I − 1,                   Nabongo Diabate and
                 ∆t                                                                                             Théodore K. Boni
                                                                                                           vol. 9, iss. 2, art. 33, 2008

             (n+1)    (n)
           UI    − UI                 (n)         (n)
                                                        p−1
                                                                (n+1)       2b (n)    q−1
                                                                                             (n+1)
(5.2)                        = δ 2 UI       − a UI            UI        −     U             UI       ,
                ∆t                                                          h I                                   Title Page

                                                                                                                   Contents
                                   (0)
(5.3)                            Ui      = φi > 0,      0 ≤ i ≤ I,
                         2
where n ≥ 0, ∆t ≤ h . We need the following lemma which is a discrete form of
                    2
the maximum principle for ordinary differential equations.                                                     Page 17 of 26

Lemma 5.1. Let f ∈ C 1 (R) and let an and bn be two bounded sequences such that                                    Go Back

                 an+1 − an             bn+1 − bn                                                                 Full Screen
(5.4)                      + f (an ) ≥           + f (bn ),                  n ≥ 0,
                    ∆t                    ∆t                                                                         Close



(5.5)                                         a0 ≥ b 0 .

Then we have an ≥ bn , n ≥ 0 for h small enough.
Proof. Let Zn = an − bn . We get
                                     Zn+1 − Zn
(5.6)                                          + f (ξn )Zn ≥ 0,
                                        ∆t
where ξn is an intermediate value between an and bn . Obviously

(5.7)                                Zn+1 ≥ Zn (1 − ∆tf (ξn )).                                  Discretizations of a
                                                                                            Semilinear Parabolic Equation
                                                                                                Nabongo Diabate and
Since an and bn are bounded and f ∈ C 1 (R), there exists a positive M such that
                                                                                                   Théodore K. Boni
|f (ξn )| ≤ M . Let j be the first integer such that Zj < 0. From (5.5), j ≥ 0. We
                                                                                              vol. 9, iss. 2, art. 33, 2008
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.                                                                        Title Page

   Now, we may state the following.                                                                   Contents

                                                                                    (n)
Theorem 5.2. Let Uh be the solution of (5.1) – (5.3). We have Uh ≥ 0 and

                        (n)                                  1
                       Uh            ≤                                         1    ,
                                ∞                1−p                          p−1                 Page 18 of 26
                                           (0)
                                          Uh           + A(p − 1)n∆t
                                                 ∞                                                    Go Back
                   a
where A =              (0) p−1
                                 .                                                                  Full Screen
            1+a∆t Uh
                            ∞
                                                                                                        Close
Proof. A straightforward calculation yields
                         ∆t (n)              2∆t       (n)       (n)
           (n+1)           U
                         h2 i+1
                                      + 1−    h2
                                                     Ui      + Ui−1
(5.8)     Ui       =                                   p−1             ,   1 ≤ i ≤ I − 1,
                                                 (n)
                                     1 + a∆t Ui
                                                2∆t (n)                2∆t    (n)
                                (n+1)            h2
                                                    U1    + 1−          h2
                                                                            U0
(5.9)                          U0         =                               p−1     ,
                                                                     (n)
                                                   1 + a∆t          U0


                                              2∆t (n)                      2∆t         (n)                               Discretizations of a
                            (n+1)              h2
                                                  UI−1        + 1−          h2
                                                                                  UI                                Semilinear Parabolic Equation
(5.10)                     UI       =                         p−1                             q−1 .                     Nabongo Diabate and
                                                        (n)             b               (n)
                                        1 + a∆t UI                  + 2 h ∆t UI                                            Théodore K. Boni
                                                                                                                      vol. 9, iss. 2, art. 33, 2008

Since 1 −       2 ∆t
                  h2
                       is nonnegative, using a recursive argument, it is easy to see that
 (n)                                      (n)           (n)
Uh      ≥ 0. Let i0 be such that Ui0 = Uh                           . From (5.8), we get                                     Title Page
                                                               ∞
                                                                                                                              Contents
                           ∆t (n)                2∆t          (n)                (n)
                             U
                           h2 i0 +1
                                      + 1−        h2
                                                         Uh            + Ui0 −1
         (n+1)                                                    ∞
       Uh              ≤                                       p−1                           if   1 ≤ i0 ≤ I − 1.
                 ∞                                      (n)
                                        1 + a∆t        Uh
                                                               ∞

Applying the triangle inequality and the fact that 1 −                           2∆t
                                                                                        is nonnegative, we arrive         Page 19 of 26
                                                                                  h2
at                                                                                                                            Go Back
                                                                     (n)
                                                                Uh                                                          Full Screen
                                     (n+1)                                  ∞
(5.11)                              Uh            ≤                               p−1 .
                                              ∞                            (n)                                                  Close
                                                       1 + a∆t Uh
                                                                                  ∞

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)
                   Uh             − U (n)     ∞                a Uh
                              ∞                                             ∞
                                                    ≤−                              p−1 .
                              ∆t                                        (n)
                                                            1 + a∆t    Uh
                                                                                    ∞

                        (n)             (0)
Using the fact that Uh            ≤ Uh            , we have                                      Discretizations of a
                              ∞               ∞                                             Semilinear Parabolic Equation
                                                                                                Nabongo Diabate and
                          (n+1)               (n)
                        Uh             − Uh                                 p
                                                                                                   Théodore K. Boni
                                   ∞                ∞              (n)
                                            ≤ −A                  Uh                .         vol. 9, iss. 2, art. 33, 2008
                               ∆t                                           ∞

We introduce the function α(t) which is defined as follows
                                                                                                     Title Page
                                                        1
                        α(t) =                                           1      .                     Contents
                                              1−p                       p−1
                                        (0)
                                       Uh           + A(p − 1)t
                                              ∞

We remark that α(t) obeys the following differential equation
                                                                  (0)                             Page 20 of 26
                        α (t) = −Aαp (t),           α(0) = Uh                   .
                                                                            ∞                         Go Back
Using a Taylor’s expansion, we have
                                                                                                    Full Screen
                                                   (∆t)2
                   α(tn+1 ) = α(tn ) + ∆tα (tn ) +       α (t˜ ),
                                                             n                                          Close
                                                     2
where t˜ is an intermediate value between tn and tn+1 . It is not hard to see that α(t)
        n
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
                                            ∞

                         (n)                                        1
                        Uh           ≤                                                       1    ,
                                ∞                       1−p                                 p−1
                                                  (0)
                                             Uh               + A(p − 1)n∆t
                                                        ∞
                                                                                                                                    Discretizations of a
and we have the desired result.                                                                                                Semilinear Parabolic Equation
                                                                                                                                   Nabongo Diabate and
Remark 2. The estimate of Theorem 5.2 is the discrete form of the one given in (1.4)                                                  Théodore K. Boni
for the continuous problem.                                                                                                      vol. 9, iss. 2, art. 33, 2008

   Now, we approximate the solution u(x, t) of problem (1.1) – (1.3) by the solution
  (n)
Uh of the following implicit scheme                                                                                                     Title Page
                (n+1)          (n)
              Ui      − Ui                   (n+1)             (n)
                                                                        p−1
                                                                               (n+1)                                                     Contents
(5.12)                               = δ 2 Ui           − Ui                  Ui       ,     0 ≤ i ≤ I − 1,
                     ∆t

           (n+1)         (n)
         UI         − UI                  (n+1)               (n)
                                                                    p−1
                                                                              (n+1)        2b (n)       p−1
                                                                                                                   (n+1)
(5.13)                         = δ 2 UI           − a UI                  UI          −      U                    UI       ,         Page 21 of 26
                   ∆t                                                                      h I
                                                                                                                                         Go Back

                                      (0)                                                                                              Full Screen
(5.14)                               Ui     = φi > 0,           0 ≤ i ≤ I,
                                                                                                                                           Close
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:

              (n)        2∆t (n+1)        ∆t        (n)
                                                                                           p−1
                                                                                                      (n+1)
           U0       =−     2
                             U1    + 1 + 2 2 + a∆t U0                                             U0          ,
                          h               h
 (n)         ∆t (n+1)       ∆t        (n)
                                                      p−1
                                                                  (n+1)       ∆t (n+1)
Ui     =−       Ui−1 + 1 + 2 2 + a∆t Ui                         Ui        −     U      ,    1 ≤ i ≤ I−1,
             h2             h                                                 h2 i+1
      (n)        2∆t (n+1)       ∆t        (n)
                                                            p−1          2b     (n)
                                                                                      q−1
                                                                                             (n+1)
 UI         =−       UI−1 + 1 + 2 2 + a∆t UI                         +      ∆t UI           UI       ,
                  h2             h                                       h
which gives the following linear system
                                                                                                                Discretizations of a
                                            (n+1)         (n)
                                      A(n) Uh       = Uh                                                   Semilinear Parabolic Equation
                                                                                                               Nabongo Diabate and
                                                                                                                  Théodore K. Boni
where A(n) is the tridiagonal matrix defined as follows
                                                                                                             vol. 9, iss. 2, art. 33, 2008

                         d0 −2∆t 0           0    ···   0                        0
                                                                                    
                                h2
                         −∆t          −∆t
                      h2      d1      h2
                                             0    ···   0                        0                                 Title Page
                               −∆t
                      0              d2 −∆t       0   ···                       0
                                                                                  
                                h2          h2
                                                                                   
               (n)
                      .         .    ..    ..    ..     .                       .                                  Contents
            A = .    .         .
                                 .       .     .     .   .
                                                         .                       .
                                                                                 . ,
                                                                                   
                                           −∆t         −∆t
                      0
                               0     ···   h2
                                                 dI−2 h2                        0 
                      0        0      0   · · · −∆t dI−1
                                                   h2
                                                                               −∆t 
                                                                                h2
                                                       −2∆t
                          0     0      0   ···     0    h2
                                                                                dI
                                                                                                                 Page 22 of 26
with
                                ∆t        (n)                                                                        Go Back
                   di = 1 + 2      + a∆t|Ui |p−1          for     0≤i≤I −1
                                h2                                                                                 Full Screen
and
                             ∆t           (n)
                                              p−1   2b       (n)
                                                                 q−1                                                   Close
                    dI = 1 + 2  + a∆t UI          + ∆t UI            .
                             h2                      h
Let us remark that the tridiagonal matrix A(n) satisfies the following properties
                                (n)                 (n)
                           Aii > 0 and Aij < 0 i = j,
                                           (n)                 (n)
                                          Aii       >         Aij .
                                                        i=j
                                      n                                    (n)
These properties imply that     exists for any n and Uh ≥ 0 (see for instance [2]).
                                     Uh
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                              Discretizations of a
                                                                                                   Semilinear Parabolic Equation
                   (n)                      1                                                          Nabongo Diabate and
                 Uh     ≤                                   1 ,                                           Théodore K. Boni
                               ∞                    1−p                             p−1
                                            (0)
                                           Uh             + A(p − 1)n∆t                              vol. 9, iss. 2, art. 33, 2008
                                                    ∞
                  a
where A =             (0) p−1
                                .
            1+a∆t Uh                                                                                        Title Page
                         ∞

                             (n)
Proof. We know that Uh              ≥ 0 as we have seen above. Now, let us obtain the above                  Contents
                                                                          (n)          (n)
estimate to complete the proof. Let i0 be such that     =                Ui0
                                                                    . Using the       Uh
                                                                  ∞
equality (5.12), we have
           ∆t            (n)      (n+1)          (n)    ∆t (n)      ∆t (n)
     1 + 2 2 + a∆t Uh           Uh         ≤ Uh        + 2 Ui0 −1 + 2 Ui0 +1                             Page 23 of 26
           h                 ∞           ∞           ∞  h            h
                                     if 1 ≤ i0 ≤ I − 1.                                                      Go Back

Applying the triangle inequality, we derive the following estimate                                         Full Screen
                                                               (n)
                                                              Uh                                               Close
                                 (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.   Numerical Results
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 =
                                 2
0.8 + 0.8 ∗ cos(πhi) and ∆t = h . In the following tables, in the rows, we give the
                                2
first n when
                                     (n)
                               n∆tUh − 1       < ε,                                        Discretizations of a
                                             ∞                                        Semilinear Parabolic Equation
                                                                                          Nabongo Diabate and
the corresponding time T n = n∆t, the CPU time and the order(s) of method com-               Théodore K. Boni
puted from                                                                              vol. 9, iss. 2, art. 33, 2008
                            log((T4h − T2h )/(T2h − Th ))
                       s=                                 .
                                       log(2)
                                                                                               Title Page
Table 1: (ε = 10−2 ) Numerical times, numbers of iterations, CPU times (seconds),
                                                                                                Contents
    and orders of the approximations obtained with the implicit Euler method

                    I     Tn          n           CPU time     s
                    16    674.0820    345129      103          -
                    32    674.2632    1.380890.   660          -                            Page 24 of 26
                    64    674.3085    5.523.934   6020         2.01
                                                                                                Go Back
                    128   674.3278    22095735    58290        1.24
                    256   674.4807    87383041    574823       2.99                           Full Screen

                                                                                                  Close
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     Tn          n           CPU time   s
                   16    674.3281    345.255     90         -
                   32    674.3452    1.381.058   720        -
                   64    674.3290    5.524.102   10820      0.08                        Discretizations of a
                                                                                   Semilinear Parabolic Equation
                   128   674.3187    22845950    323528     0.65                       Nabongo Diabate and
                   256   674.3098    88237375    19457811   0.21                          Théodore K. Boni
                                                                                     vol. 9, iss. 2, art. 33, 2008




                                                                                            Title Page

                                                                                             Contents




                                                                                         Page 25 of 26

                                                                                             Go Back

                                                                                           Full Screen

                                                                                               Close
References
[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
                                                                                            Discretizations of a
    parabolic and elliptic equation of second order with nonlinear boundary con-       Semilinear Parabolic Equation
                                                                                           Nabongo Diabate and
    ditions, Nonl. Anal. TMA, 45 (2001), 895–908.
                                                                                              Théodore K. Boni

[3] T.K. BONI, Extinction for discretizations of some semilinear parabolic equa-         vol. 9, iss. 2, art. 33, 2008

    tions, C.R.A.S, Serie I, 333 (2001), 795–800.
[4] V.A. KONDRATIEV AND L. VÉRON, Asymptotic behaviour of solutions of                          Title Page
    some nonlinear parabolic or elliptic equation, Asymptotic Analysis, 14 (1997),
                                                                                                 Contents
    117–156.
[5] R.E. MICKENS, Relation between the time and space step-sizes in nonstan-
    dard finite difference schemes for the fisher equation, Num. Methods. Part. Diff.
    Equat., 13 (1997), 51–55.                                                                Page 26 of 26

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