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A numerical method to obtain positive solution for classes of sublinear semipositone problems
Applied Mathematics and Computation 184 (2007) 445–450 www.elsevier.com/locate/amc
A numerical method to obtain positive solution for classes of sublinear semipositone problems G.A. Afrouzi *, Z. Naghizadeh, S. Mahdavi Department of Mathematics, Faculty of Basic Sciences, Mazandaran University, Babolsar, Iran
Abstract Using a numerical method based on sub- super-solution, we will show the existence of positive solution for the problem Du = kf(u(x)) for x 2 X, with Dirichlet boundary condition. 2006 Elsevier Inc. All rights reserved. Keywords: Positive solutions; Sub- and super-solutions
1. Introduction In this paper, we consider the semipositone problem
DuðxÞ ¼ kf ðuðxÞÞ;
x 2 X;
uðxÞ ¼ 0;
x 2 oX;
ð1Þ
where k > 0 is a constant, X is a bounded region in Rn with smooth boundary, and f is a smooth function such that f 0 (u) is bounded below, f ð0Þ < 0;
ð1:1Þ
and lim
u!1
f ðuÞ ¼ 0: u
ð1:2Þ
We prove under some additional conditions the existence of a positive solution for k 2 I, where I is an interval close to the smallest eigenvalue of D with Dirichlet boundary condition. It was shown that if k is small then (1)–(1.2) has no positive solution. In fact, if k is small, and if uk is a solution, then uk 6 0 in X. Next we *
Corresponding author. E-mail address: [email protected] (G.A. Afrouzi).
0096-3003/$ - see front matter 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.06.052
446
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
consider an anti-maximum principle by Clement and Peletier [3], from which we obtain the existence of a d = d(X) > 0 and a solution zk, positive in X, of Dz kz ¼ 1; x 2 X; ð2Þ z ¼ 0; x 2 oX for k 2 (k1, k1 + d), where k1 is the smallest eigenvalue of the problem D/ðxÞ ¼ k/ðxÞ; x 2 X; /ðxÞ ¼ 0; x 2 oX:
ð3Þ
Let I = [a, c], where a > k1 and c < k1 + d, and let r :¼ max kzk k;
ð1:3Þ
k2I
where kÆk denotes the supremum norm. Now assuming that there exists a m > 0 such that f ðxÞ P x m;
8x 2 ½0; mcr
ð1:4Þ
(note here that (cr > 1)) we prove that, (1)–(1.2) has a non-negative solution for k 2 I. Here the solution zk of (2) will be used to create a sub-solution, that is, a function w(x) such that DwðxÞ 6 kf ðwðxÞÞ; x 2 X; ð4Þ wðxÞ 6 0; x 2 oX: Theorem 1. There exists k0 > 0 such that if 0 < k < k0 and u is a solution of (1)–(1.2) then u 6 0 in X. The above theorem (proved in [1]) shows that if k is small then (1)–(1.2) has no positive solution. In fact, if k is small, and if uk is a solution, then uk 6 0 in X. Theorem 2. Let k 2 I and assume that f satisfies (1.1), (1.2) and (1.4). Then (1) has a non-negative solution. Theorem 2 also proved in [1] that we give a proof for the sake of completeness. Let w(x):¼(cm)z(x), where c, m and z are as before. Then DwðxÞ ¼ ðcmÞkz 1;
x 2 X;
w ¼ 0;
x 2 oX:
But kf ðwðxÞÞ ¼ kf ðcmzðxÞÞ P kcmzðxÞ m P cmkz 1 ¼ DwðxÞ: Thus w(x) is a sub-solution of (1) for k 2 I. Now let v(x) to be the unique positive solution of DvðxÞ ¼ 1; x 2 X; x 2 oX;
vðxÞ 6 0;
ð5Þ
and consider /(x) = J v(x); J > 0 (to be chosen). Then / satisfies D/ðxÞ ¼ J ;
x 2 X;
/ðxÞ ¼ 0;
x 2 oX:
But by (1.2), there exists a J0 > 0 such that for J > J0, D/ðxÞ ¼ J P kf ðJvðxÞÞ;
x 2 X;
ð1:5Þ
and thus /(x) will be a super-solution for (1). Now choose J > J0 large enough that /ðxÞ ¼ JvðxÞ P ðcmÞzðxÞ ¼ wðxÞ:
ð1:6Þ
This is clearly possible by Hopfs maximum principle, and hence by the well-known method of sub- and super-solutions, (1) has a non-negative solution u(x) such that cm z(x) 6 u(x) 6 J v(x), and Theorem 2 is proven. Here we give a simple example that satisfies (1.1), (1.2), and (1.4). Consider 1
f ðsÞ ¼ mðs þ 1Þ2
3m ; 2
m > 0:
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
447
Then f ð0Þ ¼ m2 < 0 and lims!1 f ðsÞ ¼ 0. Thus (1.1), (1.2) are satisfied. Now let p > 0 be such that f(p) = s p m. That is, 3m ¼ p m; 2 n mo2 m2 ðp þ 1Þ ¼ p þ ; 2 3m2 ¼ 0; p2 þ ðm m2 Þp 4 1
mðp þ 1Þ2
and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 4 3 2 2 ðm mÞ þ m mðm 1Þ þ 3 ðm mÞ þ m 2m þ m þ 3m ¼ p¼ : 2 2 2
Hence in order that (1.4) is satisfied, we must have qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m2 m þ m ðm 1Þ2 þ 3 P mðcaÞ; 2 that is, ðm 1Þ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðm 1Þ þ 3 P 2ðcaÞ
ð1:7Þ
(Note here that c and a are quantities that depend on only X). Clearly for a given c and a, there exists a m0 such that if m > m0 then (1.7) is satisfied and equivalently (1.4) will be satisfied. We investigate numerically positive solutions. Our numerical method is based on monotone iteration. We say that u is a positive solution of (1) if u is a classical solution with u(x) 2 I for all x 2 X and u(x) > 0 for all x 2 X. 2. Numerical results It is well known that there must always exists a solution for problems such as (2) between a sub-solution v and a super-solution u such that v 6 u for all x 2 X (see [2]). Consider the boundary value problem DuðxÞ þ f ðx; uðxÞÞ ¼ 0 on X; ð6Þ uðxÞ ¼ 0 on oX: u P v as well as Let u, v 2 C 2 ðXÞ satisfy D uðxÞ þ f ðx; uðxÞÞ 6 0 on X; DvðxÞ þ f ðx; vðxÞÞ P 0
on X;
u P 0; 6 0: u
ðx;uÞ > 0 8ðx; uÞ 2 X ½v; u and such that the operator (D c) with Choose a number c > 0 such that c þ of ou Dirichlet boundary condition has its spectrum strictly contained in the open left-half complex plane. Then the mapping
T : / ! w;
w ¼ T /;
/ 2 C 2 ðXÞ;
/ðxÞ 2 ½v; u;
where w(x) is the unique solution of the BVP DwðxÞ cwðxÞ ¼ ½c/ðxÞ þ f ðx; /ðxÞÞ wðxÞ ¼ 0
on X; on oX
8x 2 X;
ð3:1Þ
ð7Þ
is monotone, i.e. for any /1, /2 satisfying (3.1) and /1 6 /2, we have T/1, T/2 satisfies (3.1), and T /1 6 T/2 on X.
448
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
Consequently, by letting fc(x, u) = cu + f(x, u), the iterations 8 u0 ðxÞ ¼ uðxÞ > > > < ðD cÞu ðxÞ ¼ f ðx; u ðxÞÞ on X; nþ1 c n > n ¼ 0; 1; 2; . . . ; > > : on oX; unþ1 ðxÞ ¼ 0 and
8 v0 ðxÞ ¼ vðxÞ > > > < ðD cÞv ðxÞ ¼ f ðx; v ðxÞÞ nþ1 c n > > > : vnþ1 ðxÞ ¼ 0
ð8Þ
on X; n ¼ 0; 1; 2; . . . ; on oX;
ð9Þ
yield iteration un and vn satisfying v ¼ v0 6 v1 6 6 vn 6 6 un 6 6 u1 6 u0 ¼ u; so that the limits u1 ðxÞ ¼ lim un ðxÞ; n!1
v1 ðxÞ ¼ lim vn ðxÞ; n!1
2
exists in C ðXÞ. We have (i) v1(x) 6 u1(x) on X; (ii) u1 and v1 are, respectively, stable from above and below; (iii) if u1 6 v1 and both u1 and v1 are asymptotically stable, then there exists an unstable solution such that v1 6 / 6 u1, / 2 C 2 ðXÞ we use following algorithm Sub- and super-solution algorithm 1. Find a subsolution v0 and a super-solution u0. Choose a number c > 0; 2. Solve the boundary value problem Dwnþ1 ðxÞ cwnþ1 ðxÞ ¼ fc ðx; wn ðxÞÞ on X; uðxÞ ¼ 0 on oX
ð10Þ
for wn = vn and wn = un, respectively; 3. If kwn+1 wnk < , output and stop. Else go to step 2. For doing step 1, we solve the problem Dz kz ¼ 1; x 2 X; z ¼ 0; x 2 oX
ð11Þ
to obtain subsolution. We know from Section 1 that problem (11) has a positive solution for k 2 (k1, k1 + d). The obtained results shows there is an array of positive solution for k 2 (17, 35) so k1 is around 17. For brevity we express just some of those numerical results. Approximation of zk for k = 15 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
0.268 0.447 0.513 0.505
0.423 0.701 0.753 0.528
0.431 0.718 0.778 0.497
0.283 0.493 0.636 0.345
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
449
Approximation of zk for k = 17 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
1.895 3.266 3.789 3.727
3.067 5.199 5.626 3.912
3.130 5.341 5.818 3.676
2.022 3.625 4.712 2.514
Approximation of zk for k = 30 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
0.002 0.028 0.050 0.054
0.017 0.062 0.087 0.060
0.021 0.068 0.093 0.056
0.008 0.041 0.070 0.033
Approximation of zk for k = 36 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
0.001 0.024 0.048 0.053
0.012 0.056 0.084 0.060
0.016 0.063 0.091 0.056
0.005 0.038 0.068 0.033
Let w(x) = cmzk(x) as a subsolution in our algorithm where c and m obtained from Section 1. And to obtain super-solution of (1) for k 2 I (I = [a, c] where a > k1 and c < k1 + d) we solve DvðxÞ ¼ 1; x 2 X; ð12Þ vðxÞ ¼ 0; x 2 oX by finite difference (see [4,5]). We choose J such that (1.5), (1.6) are satisfied. We execute algorithm for k 2 I = [17.1, 34.9] for brevity we express just some of those numerical results Approximation of u for k = 17.1 x/y 0.2 0.4 0.6 0.8
0.2
0.4 5
1.68 · 10 2.51 · 105 2.51 · 105 1.68 · 105
0.6 5
0.8 5
2.51 · 10 3.78 · 105 3.78 · 105 2.51 · 105
2.51 · 10 3.78 · 105 3.78 · 105 2.51 · 105
1.68 · 105 2.51 · 105 2.51 · 105 1.68 · 105
0.4
0.6
0.8
Approximation of u for k = 25 x/y 0.2 0.4 0.6 0.8
0.2 5
3.59 · 10 5.37 · 105 5.37 · 105 3.59 · 105
5
5.37 · 10 8.09 · 105 8.09 · 105 5.37 · 105
5
5.37 · 10 8.09 · 105 8.09 · 105 5.37 · 105
3.59 · 105 5.37 · 105 5.37 · 105 3.59 · 105
450
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
Approximation of u for k = 30 x/y 0.2 0.4 0.6 0.8
0.2
0.4 6
0.51 · 10 0.77 · 106 0.77 · 106 0.51 · 106
0.6 6
0.8 6
0.77 · 10 1.66 · 106 1.66 · 106 0.77 · 106
0.77 · 10 1.66 · 106 1.66 · 106 0.77 · 106
0.51 · 106 0.77 · 106 0.77 · 106 0.51 · 106
0.4
0.6
0.8
Approximation of u for k = 34.9 x/y 0.2 0.4 0.6 0.8
0.2 6
0.70 · 10 1.04 · 106 1.04 · 106 0.70 · 106
6
1.04 · 10 1.57 · 106 1.57 · 106 1.04 · 106
6
1.04 · 10 1.57 · 106 1.57 · 106 1.04 · 106
0.70 · 106 1.04 · 106 1.04 · 106 0.70 · 106
References [1] A. Castro, J.B. Garner, R. Shivaji, Existence results for class of sublinear semipositone problems, Results in Mathematics 23 (1993) 214–220. [2] G. Chen, J. Zhou, Wei-Ming Ni, Algorithms and visualization for solutions of nonlinear equations, International Journal of Bifurcation and Chaos 10 (2000) 1565–1612. [3] P. Clement, L.A. Peletier, An anti-maximum principle for second order elliptic operators, Differential Equations 34 (1979) 218–229. [4] M. Dehghan, Numerical procedures for a boundary value problem with a non-linear boundary condition, Applied Mathematics and Computation 147 (2004) 291–306. [5] I.G. Petrovsky, Lectures on Partial Differential Equations, Dover Publications, Inc., 1991.
A numerical method to obtain positive solution for classes of sublinear semipositone problems G.A. Afrouzi *, Z. Naghizadeh, S. Mahdavi Department of Mathematics, Faculty of Basic Sciences, Mazandaran University, Babolsar, Iran
Abstract Using a numerical method based on sub- super-solution, we will show the existence of positive solution for the problem Du = kf(u(x)) for x 2 X, with Dirichlet boundary condition. 2006 Elsevier Inc. All rights reserved. Keywords: Positive solutions; Sub- and super-solutions
1. Introduction In this paper, we consider the semipositone problem
DuðxÞ ¼ kf ðuðxÞÞ;
x 2 X;
uðxÞ ¼ 0;
x 2 oX;
ð1Þ
where k > 0 is a constant, X is a bounded region in Rn with smooth boundary, and f is a smooth function such that f 0 (u) is bounded below, f ð0Þ < 0;
ð1:1Þ
and lim
u!1
f ðuÞ ¼ 0: u
ð1:2Þ
We prove under some additional conditions the existence of a positive solution for k 2 I, where I is an interval close to the smallest eigenvalue of D with Dirichlet boundary condition. It was shown that if k is small then (1)–(1.2) has no positive solution. In fact, if k is small, and if uk is a solution, then uk 6 0 in X. Next we *
Corresponding author. E-mail address: [email protected] (G.A. Afrouzi).
0096-3003/$ - see front matter 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.06.052
446
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
consider an anti-maximum principle by Clement and Peletier [3], from which we obtain the existence of a d = d(X) > 0 and a solution zk, positive in X, of Dz kz ¼ 1; x 2 X; ð2Þ z ¼ 0; x 2 oX for k 2 (k1, k1 + d), where k1 is the smallest eigenvalue of the problem D/ðxÞ ¼ k/ðxÞ; x 2 X; /ðxÞ ¼ 0; x 2 oX:
ð3Þ
Let I = [a, c], where a > k1 and c < k1 + d, and let r :¼ max kzk k;
ð1:3Þ
k2I
where kÆk denotes the supremum norm. Now assuming that there exists a m > 0 such that f ðxÞ P x m;
8x 2 ½0; mcr
ð1:4Þ
(note here that (cr > 1)) we prove that, (1)–(1.2) has a non-negative solution for k 2 I. Here the solution zk of (2) will be used to create a sub-solution, that is, a function w(x) such that DwðxÞ 6 kf ðwðxÞÞ; x 2 X; ð4Þ wðxÞ 6 0; x 2 oX: Theorem 1. There exists k0 > 0 such that if 0 < k < k0 and u is a solution of (1)–(1.2) then u 6 0 in X. The above theorem (proved in [1]) shows that if k is small then (1)–(1.2) has no positive solution. In fact, if k is small, and if uk is a solution, then uk 6 0 in X. Theorem 2. Let k 2 I and assume that f satisfies (1.1), (1.2) and (1.4). Then (1) has a non-negative solution. Theorem 2 also proved in [1] that we give a proof for the sake of completeness. Let w(x):¼(cm)z(x), where c, m and z are as before. Then DwðxÞ ¼ ðcmÞkz 1;
x 2 X;
w ¼ 0;
x 2 oX:
But kf ðwðxÞÞ ¼ kf ðcmzðxÞÞ P kcmzðxÞ m P cmkz 1 ¼ DwðxÞ: Thus w(x) is a sub-solution of (1) for k 2 I. Now let v(x) to be the unique positive solution of DvðxÞ ¼ 1; x 2 X; x 2 oX;
vðxÞ 6 0;
ð5Þ
and consider /(x) = J v(x); J > 0 (to be chosen). Then / satisfies D/ðxÞ ¼ J ;
x 2 X;
/ðxÞ ¼ 0;
x 2 oX:
But by (1.2), there exists a J0 > 0 such that for J > J0, D/ðxÞ ¼ J P kf ðJvðxÞÞ;
x 2 X;
ð1:5Þ
and thus /(x) will be a super-solution for (1). Now choose J > J0 large enough that /ðxÞ ¼ JvðxÞ P ðcmÞzðxÞ ¼ wðxÞ:
ð1:6Þ
This is clearly possible by Hopfs maximum principle, and hence by the well-known method of sub- and super-solutions, (1) has a non-negative solution u(x) such that cm z(x) 6 u(x) 6 J v(x), and Theorem 2 is proven. Here we give a simple example that satisfies (1.1), (1.2), and (1.4). Consider 1
f ðsÞ ¼ mðs þ 1Þ2
3m ; 2
m > 0:
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
447
Then f ð0Þ ¼ m2 < 0 and lims!1 f ðsÞ ¼ 0. Thus (1.1), (1.2) are satisfied. Now let p > 0 be such that f(p) = s p m. That is, 3m ¼ p m; 2 n mo2 m2 ðp þ 1Þ ¼ p þ ; 2 3m2 ¼ 0; p2 þ ðm m2 Þp 4 1
mðp þ 1Þ2
and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 4 3 2 2 ðm mÞ þ m mðm 1Þ þ 3 ðm mÞ þ m 2m þ m þ 3m ¼ p¼ : 2 2 2
Hence in order that (1.4) is satisfied, we must have qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m2 m þ m ðm 1Þ2 þ 3 P mðcaÞ; 2 that is, ðm 1Þ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðm 1Þ þ 3 P 2ðcaÞ
ð1:7Þ
(Note here that c and a are quantities that depend on only X). Clearly for a given c and a, there exists a m0 such that if m > m0 then (1.7) is satisfied and equivalently (1.4) will be satisfied. We investigate numerically positive solutions. Our numerical method is based on monotone iteration. We say that u is a positive solution of (1) if u is a classical solution with u(x) 2 I for all x 2 X and u(x) > 0 for all x 2 X. 2. Numerical results It is well known that there must always exists a solution for problems such as (2) between a sub-solution v and a super-solution u such that v 6 u for all x 2 X (see [2]). Consider the boundary value problem DuðxÞ þ f ðx; uðxÞÞ ¼ 0 on X; ð6Þ uðxÞ ¼ 0 on oX: u P v as well as Let u, v 2 C 2 ðXÞ satisfy D uðxÞ þ f ðx; uðxÞÞ 6 0 on X; DvðxÞ þ f ðx; vðxÞÞ P 0
on X;
u P 0; 6 0: u
ðx;uÞ > 0 8ðx; uÞ 2 X ½v; u and such that the operator (D c) with Choose a number c > 0 such that c þ of ou Dirichlet boundary condition has its spectrum strictly contained in the open left-half complex plane. Then the mapping
T : / ! w;
w ¼ T /;
/ 2 C 2 ðXÞ;
/ðxÞ 2 ½v; u;
where w(x) is the unique solution of the BVP DwðxÞ cwðxÞ ¼ ½c/ðxÞ þ f ðx; /ðxÞÞ wðxÞ ¼ 0
on X; on oX
8x 2 X;
ð3:1Þ
ð7Þ
is monotone, i.e. for any /1, /2 satisfying (3.1) and /1 6 /2, we have T/1, T/2 satisfies (3.1), and T /1 6 T/2 on X.
448
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
Consequently, by letting fc(x, u) = cu + f(x, u), the iterations 8 u0 ðxÞ ¼ uðxÞ > > > < ðD cÞu ðxÞ ¼ f ðx; u ðxÞÞ on X; nþ1 c n > n ¼ 0; 1; 2; . . . ; > > : on oX; unþ1 ðxÞ ¼ 0 and
8 v0 ðxÞ ¼ vðxÞ > > > < ðD cÞv ðxÞ ¼ f ðx; v ðxÞÞ nþ1 c n > > > : vnþ1 ðxÞ ¼ 0
ð8Þ
on X; n ¼ 0; 1; 2; . . . ; on oX;
ð9Þ
yield iteration un and vn satisfying v ¼ v0 6 v1 6 6 vn 6 6 un 6 6 u1 6 u0 ¼ u; so that the limits u1 ðxÞ ¼ lim un ðxÞ; n!1
v1 ðxÞ ¼ lim vn ðxÞ; n!1
2
exists in C ðXÞ. We have (i) v1(x) 6 u1(x) on X; (ii) u1 and v1 are, respectively, stable from above and below; (iii) if u1 6 v1 and both u1 and v1 are asymptotically stable, then there exists an unstable solution such that v1 6 / 6 u1, / 2 C 2 ðXÞ we use following algorithm Sub- and super-solution algorithm 1. Find a subsolution v0 and a super-solution u0. Choose a number c > 0; 2. Solve the boundary value problem Dwnþ1 ðxÞ cwnþ1 ðxÞ ¼ fc ðx; wn ðxÞÞ on X; uðxÞ ¼ 0 on oX
ð10Þ
for wn = vn and wn = un, respectively; 3. If kwn+1 wnk < , output and stop. Else go to step 2. For doing step 1, we solve the problem Dz kz ¼ 1; x 2 X; z ¼ 0; x 2 oX
ð11Þ
to obtain subsolution. We know from Section 1 that problem (11) has a positive solution for k 2 (k1, k1 + d). The obtained results shows there is an array of positive solution for k 2 (17, 35) so k1 is around 17. For brevity we express just some of those numerical results. Approximation of zk for k = 15 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
0.268 0.447 0.513 0.505
0.423 0.701 0.753 0.528
0.431 0.718 0.778 0.497
0.283 0.493 0.636 0.345
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
449
Approximation of zk for k = 17 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
1.895 3.266 3.789 3.727
3.067 5.199 5.626 3.912
3.130 5.341 5.818 3.676
2.022 3.625 4.712 2.514
Approximation of zk for k = 30 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
0.002 0.028 0.050 0.054
0.017 0.062 0.087 0.060
0.021 0.068 0.093 0.056
0.008 0.041 0.070 0.033
Approximation of zk for k = 36 x/y
0.2
0.4
0.6
0.8
0.2 0.4 0.6 0.8
0.001 0.024 0.048 0.053
0.012 0.056 0.084 0.060
0.016 0.063 0.091 0.056
0.005 0.038 0.068 0.033
Let w(x) = cmzk(x) as a subsolution in our algorithm where c and m obtained from Section 1. And to obtain super-solution of (1) for k 2 I (I = [a, c] where a > k1 and c < k1 + d) we solve DvðxÞ ¼ 1; x 2 X; ð12Þ vðxÞ ¼ 0; x 2 oX by finite difference (see [4,5]). We choose J such that (1.5), (1.6) are satisfied. We execute algorithm for k 2 I = [17.1, 34.9] for brevity we express just some of those numerical results Approximation of u for k = 17.1 x/y 0.2 0.4 0.6 0.8
0.2
0.4 5
1.68 · 10 2.51 · 105 2.51 · 105 1.68 · 105
0.6 5
0.8 5
2.51 · 10 3.78 · 105 3.78 · 105 2.51 · 105
2.51 · 10 3.78 · 105 3.78 · 105 2.51 · 105
1.68 · 105 2.51 · 105 2.51 · 105 1.68 · 105
0.4
0.6
0.8
Approximation of u for k = 25 x/y 0.2 0.4 0.6 0.8
0.2 5
3.59 · 10 5.37 · 105 5.37 · 105 3.59 · 105
5
5.37 · 10 8.09 · 105 8.09 · 105 5.37 · 105
5
5.37 · 10 8.09 · 105 8.09 · 105 5.37 · 105
3.59 · 105 5.37 · 105 5.37 · 105 3.59 · 105
450
G.A. Afrouzi et al. / Applied Mathematics and Computation 184 (2007) 445–450
Approximation of u for k = 30 x/y 0.2 0.4 0.6 0.8
0.2
0.4 6
0.51 · 10 0.77 · 106 0.77 · 106 0.51 · 106
0.6 6
0.8 6
0.77 · 10 1.66 · 106 1.66 · 106 0.77 · 106
0.77 · 10 1.66 · 106 1.66 · 106 0.77 · 106
0.51 · 106 0.77 · 106 0.77 · 106 0.51 · 106
0.4
0.6
0.8
Approximation of u for k = 34.9 x/y 0.2 0.4 0.6 0.8
0.2 6
0.70 · 10 1.04 · 106 1.04 · 106 0.70 · 106
6
1.04 · 10 1.57 · 106 1.57 · 106 1.04 · 106
6
1.04 · 10 1.57 · 106 1.57 · 106 1.04 · 106
0.70 · 106 1.04 · 106 1.04 · 106 0.70 · 106
References [1] A. Castro, J.B. Garner, R. Shivaji, Existence results for class of sublinear semipositone problems, Results in Mathematics 23 (1993) 214–220. [2] G. Chen, J. Zhou, Wei-Ming Ni, Algorithms and visualization for solutions of nonlinear equations, International Journal of Bifurcation and Chaos 10 (2000) 1565–1612. [3] P. Clement, L.A. Peletier, An anti-maximum principle for second order elliptic operators, Differential Equations 34 (1979) 218–229. [4] M. Dehghan, Numerical procedures for a boundary value problem with a non-linear boundary condition, Applied Mathematics and Computation 147 (2004) 291–306. [5] I.G. Petrovsky, Lectures on Partial Differential Equations, Dover Publications, Inc., 1991.