Blow-Up of Finite Difference Solutions to Parabolic Equations With Semilinear Dynamical Boundary...
Transcript of Blow-Up of Finite Difference Solutions to Parabolic Equations With Semilinear Dynamical Boundary...
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Blow-Up of Finite Difference Solutions to Parabolic Equationswith Semilinear Dynamical Boundary Conditions ?
Miglena N. Koleva, Lubin G. Vulkov
University of Rousse, Department of Mathematics, 7017 Rousse, Bulgaria, mkoleva{lvalkov}@ru.acad.bg
Abstract. In the present work we analyze semidiscrete and fulldiscrete (finite difference) solutions ofthe semilinear parabolic problem with dynamical boundary conditions that can be posed on the wholedomain boundary as well as only on a part of the boundary. Necessary and sufficient conditions for blow-up of the numerical solutions are found. It is proved that the numerical solutions and the discrete blow-uptimes converge to the corresponding real ones, when mesh size goes to zero. Numerical experiments arealso discussed.
1 Introduction
Let be a bounded domain in RN (N 1) with piecewise smooth boundary = S1 S2, S1 S2 = . Weconsider the following problem:
u
t4u = f(u) in QT = (0, T ), (1)
u
t+ k
u
n= g1(u) on S1 (0, T ), (2)
au
n+ bu = g2(u), a 0, b 0, a+ b = 1, on S2 (0, T ), (3)
u(x, 0) = u0(x) on S1. (4)
Here k > 0, a, b are constants, 4 is the Laplace operator with respect to the space variables and /n is theoutward derivative to S1 (or S2).
Problems of type (1)-(4) model processes of filtration in the hydrology, in heat transfer in a solid in contactwith a fluid, chemistry, semiconductors etc., see [1], [3], [6], [7] and [9] for more information.
We shall consider nonlinearities f , g1 and g2 for which blow-up occurs in finite time. A point x isa blow-up point if there exists a sequence [(xn, tn)] such that tn Tmax < +, xn x and u(xn, tn) as n.
The important questions are:1. When the solutions blow-up?2. Where (BUS) the solutions blow-up?Blow-up results for differential problems with dynamical boundary conditions start with the paper [9],
where the special case is considered:
S1 , i.e. S2 , f(u) = g(u) = u1+, > 0 (5)
Theorem 1. (M. Kirane, 1992) Let u0 W 2p (), 1 p 0, to problem (1)-(4), (5) blows up in afinite time.
Later these results were extended in many papers, see [4], [6] and [8].
? This research was supported by the Bulgarian National Fund of Science under Project VU-MI-106/2005.
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240 M. Koleva and L. Vulkov
In [12] was studied the blow-up of finite difference solution to parabolic problem (1)-(4) in 1D case. In [13]was considered blow-up of continuous and semidiscrete solutions of the problem:
4u = 0 in (0, T ),u
t+ k
u
n= g(u) on S1 (0, T ), (6)
au
n+ bu = 0, a 0, b 0, a+ b = 1 on S2 (0, T ),
u(x, 0) = u0(x) on S1.
In the present work we analyze: Semidiscrete and fulldiscrete (finite difference) solutions of the problem(1)-(4). Necessary and sufficient conditions for blow-up of the numerical solutions are found. It is proved thatthe semidiscrete solutions and the numerical blow-up times converge to the corresponding real ones, whenmesh size goes to zero. Numerical experiments are also discussed.
2 The Semidiscrete Problem
We introduce the following uniform mesh on the domain = [0, 1] [0, 1]
= {(x1,i, x2,j)| x1,i = (i 1)h1, i = 1, . . . , N, (N 1)h1 = 1;x2,j = (j 1)h2, j = 1, . . . ,M, (M 1)h2 = 1}.
and denote y = y(t) = yij(t) = y(x1,i, x2,j , t), i = 1, . . . , N , j = 1, . . . ,M the values of the numericalapproximation at the nodes x = (x1,i, x2,j) in the time t. Define the following finite differences for any meshfunction vi = v(xi) given on by vx,i = (vi+1 vi)/h, vx,i = vx,i1.
We consider the case S2 0, which is more complicated in view of deriving a second order differencescheme. In fact, equation (3) can be obtain from (2), removing the time derivative and associating the sourceterm g1(u) with g2(u) bu.
For semidiscretization of the problem (1)-(4) we use the finite difference method and obtain the followingsystem of ordinary differential equations :
dy
dt=
2k(h2yx1 + h1yx2) + 2(h1 + h2)g1(y) + kh1h2f(y)kh1h2 + 2(h1 + h2)
, i = 1, j = 1, (7)
dy
dt=
1kh2 + 2
(kh2yx1x1 + 2kyx2 + 2g1(y) + kh2f(y)),i = 2, . . . , N 1,j = 1, (8)
dy
dt=
2k(h1yx2 h2yx1) + 2(h1 + h2)g1(y) + kh1h2f(y)kh1h2 + 2(h1 + h2)
, i = N, j = 1, (9)
dy
dt=
1kh1 + 2
(kh1yx2x2 + 2kyx1 + 2g1(y) + kh1f(y)),i = 1,j = 2, . . . ,M 1, (10)
dy
dt= yx1x1 + yx2x2 + f(y), i = 2, . . . , N 1, j = 2, . . . ,M 1, (11)
dy
dt=
1kh1 + 2
(kh1yx2x2 2kyx1 + 2g1(y) + kh1f(y)), i = N,j = 2, ...,M 1, (12)dy
dt=
2k(h2yx1 h1yx2) + 2(h1 + h2)g1(y) + kh1h2f(y)kh1h2 + 2(h1 + h2)
,i = 1,j =M, (13)
dy
dt=
1kh2 + 2
(kh2yx1x1 2kyx2 + 2g1(y) + kh2f(y)), i = 2, . . . , N 1,j =M, (14)dy
dt=2k(h2yx1 + h1yx2) + 2(h1 + h2)g1(y) + kh1h2f(y)
kh1h2 + 2(h1 + h2),i = N,j =M, (15)
y(0) = u0(x), i = 1, . . . , N, j = 1, . . . ,M. (16)
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Blow-Up of Finite Difference Solutions to Parabolic Eqns 241
Let S2 = {x = (x1, x2)| x1 = 1, 0 x2 1}. Then, instead of (7), (10) and (13) we have the followingequations, respectively:
dy
dt=
kh2kh2 + 2
(2h1yx1 +
2h2yx2 +
2ah2
g1(y) +2ah1
(g2(y) by) + f(y)), (17)
dy
dt= yx2x2 +
2h1yx1 +
2bah1
y 2ah1
g2(y) + f(y), (18)
dy
dt=
kh2kh2 + 2
(2h1yx1
2h2yx2
2bah1
y +2ah1
g2(y) +2kh2
g1(y) + f(y)). (19)
The local truncation error is (h) = O(h21 + h22) if u C4,1x,t , x (x1, x2), [16].
2.1 Blow-up of semidiscrete solution
We can rewrite the equations (7)-(16) in the following vector-matrix form:
MY (t) = AY (t) + G(Y (t)) (20)
where
Y (t) = [y11, y21, . . . , yN1 y1(t)
, . . . , y1j , y2j , . . . , yNj yj(t)
, . . . , y1M , y2M , . . . , yNM yM (t)
]T , (21)
Y (0) = [y1(0), y2(0), . . . , yM (0)]T . (22)
In the Theorem 2, we will need the following maximum principle:
Lemma 1. Let the assumptions of Theorem 1 hold. Then Y (t) C1[0, Tmax) and Y (t) 0 for t [0, Tmax).
Theorem 2. Let the assumptions of Theorem 1 hold. Then there exists a time Thb
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242 M. Koleva and L. Vulkov
3 Time Discretization and Methods for Computation
We introduce a nonuniform mesh grid in time
t0 = 0, tn = tn1 +4tn1 =n1k=0
4tn (4tk > 0, k = 0, 1, ...).
The time increment 4tn is chosen to be decreasing according to the growth of the numerical solution, [2, 14].For example, in the case f(u) = g1(u) = g2(u) = u1+, > 0, the time step [14], 1 < p
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Blow-Up of Finite Difference Solutions to Parabolic Eqns 243
0
0.5
1
0
0.5
1
1
1.5
2
2.5
3
x2
u0=sin(pi x1)+sin(pi x2)+1
x1
u 0
0
0.5
1
0
0.5
1
0
10
20
30
40
50
60
x2
T=1.20450e1
x1
y
0
0.5
1
0
0.5
1
0
200
400
600
800
1000
1200
x2
T=1.20661e1
x1
y
Fig. 1. = 2, = 0.01, N =M = 21
0
0.5
1
0
0.5
1
5
10
15
x2
u0=5[sin(pi x1)+sin(pi x2)+1]
x1
u 0
0
0.5
1
0
0.5
1
0
10
20
30
40
50
60
x2
T=2.10021e3
x1
y
0
0.5
1
0
0.5
1
0
200
400
600
800
1000
1200
x2
T=2.29968e3
x1
y
Fig. 2. = 2, = 0.01, N =M = 41
0
0.5
1
0
0.5
1
5
10
15
x2
u0=5[sin(pi x1)+sin(pi x2)+1]
x1
u 0
0
0.5
1
0
0.5
1
0
10
20
30
40
50
60
x2
T=2.68083e7
x1
y
0
0.5
1
0
0.5
1
0
200
400
600
800
1000
1200
x2
T=2.68697e7
x1
y
Fig. 3. = 5, = 0.01, N =M = 41
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244 M. Koleva and L. Vulkov
Numerical experiments for our test examples, show that the BUS consist of points (point blow-up), except thecase of constant initial function for some value of , see Figure 4. BUS depends on the maximum value ofthe initial function and (i.e nonlinear source terms in general case). The solution blows up as fast as themaximum value of the initial function and the value of are bigger, see Figure 1,2,3. In the case of differentsource terms or in more general case, when S2 6 0, BUS depends also on the interaction between nonlinearities.Thus, the blow-up point may not coincides with the point, where the initial function has maximum value.
0
0.5
1
0
0.5
1
0
0.5
1
1.5
2
u0 1
x2x1
u 0
0
0.5
1
0
0.5
1
0
200
400
600
800
1000
1200
x2
=2
x1y
0
0.5
1
0
0.5
1
0
200
400
600
800
1000
1200
x2
=5
x1
y
T=5.02512e1
T=0
T=2.04004e1
T= 0
Fig. 4. Different values of at different time levels T , = 0.01, N =M = 21
4.2 Convergence Results
Now, we check the convergence rate of the numerical solution of the problem (1)-(4), f(u) = u2+F (x1, x2, t),g1(u) = u2 + G1(x1, x2, t), g2(u) = u2 + G2(x1, x2, t) and u0(x1, x2) = sin(pix1) sin(pix2). The functions F ,G1 and G2 are chosen, such that the exact solution of the problem is u(t, x1, x2) = et sin(pix1) sin(pix2). Themesh step sizes in space are h1 = h2 = h (N =M) and the ratio h2 = 1 is fixed.
In Table 1 we give the max errors (EN ) of the numerical solutions computed, using grid with N2 grid nodesin space and convergence rates (CR), calculated with the following formula: CR = log2[EN/E2N ]. Decreasing
Table 1. Max error and convergence rate at time T
T = 0.05 T = 0.1
N Max error CR Max error CR
11 4.36960871e-3 6.12952948e-3
21 1.27949323e-3 1.77193 1.59991442e-3 1.93778
41 3.31186965e-4 1.94985 4.03557181e-4 1.98715
81 8.36224738e-5 1.98568 1.01108231e-4 1.99687
161 2.09256494e-5 1.99862 2.52919547e-5 1.99915
321 5.23390773e-6 1.99931 6.32377763e-6 1.99982
the space mesh step size twice yields to decreasing the error approximately four times, which means that theconvergence rate is O( + h2).
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