A numerical method to obtain positive solution for classes of sublinear semipositone systems

Applied Mathematics and Computation 186 (2007) 1113–1119 www.elsevier.com/locate/amc

A numerical method to obtain positive solution for classes of sublinear semipositone systems G.A. Afrouzi *, S. Mahdavi, Z. Naghizadeh Department of Mathematics, Faculty of Basic Sciences, Mazandaran University, Babolsar, Iran

Abstract Using a numerical method based on sub–super solution, we will obtain positive solution to the coupled-system of boundary value problems of the form  DuðxÞ ¼ kf ðx; u; vÞ;

x 2 X;

 DvðxÞ ¼ kgðx; u; vÞ;

x 2 X;

uðxÞ ¼ 0 ¼ vðxÞ;

x 2 oX;

where f, g are C1 functions with at least one of f(x0, 0, 0) or g(x0, 0, 0) being negative for some x0 2 X (semipositone).  2006 Elsevier Inc. All rights reserved. Keywords: Positive solutions; Sub- and super-solutions

1. Introduction Consider positive solutions 8 DuðxÞ ¼ kf ðx; u; vÞ; > > < DvðxÞ ¼ kgðx; u; vÞ; > > : uðxÞ ¼ 0 ¼ vðxÞ;

to the coupled-system of boundary value problems x 2 X; x 2 X;

ð1Þ

x 2 oX;

where k > 0, D is the Laplacian operator, X is a bounded region in Rn, (N P 1) with smooth boundary oX, and f, g are C1 functions with at least one of f(x0, 0, 0) or g(x0, 0, 0) being negative for some x0 2 X (semipositone). In this paper, we want to investigate numerically positive solution of (1) by using the method of sub–super solutions. A super-solution to (1) is defined as an ordered pair of smooth functions ðu; vÞ on X satisfying *

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.2006.08.036

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8 u P kf ðx;  u; vÞ; x 2 X; > < D Dv P kgðx;  u; vÞ; x 2 X; > :   u P 0; v P 0; x 2 oX:

ð2Þ

1   ½q2 ; q 2  where q1 = Sub-solutions are similarly defined with inequalities reversed. Let D ¼ ½q1 ; q 1 ¼ supf 2 ¼ supfvðxÞ : x 2 oXg. inf{u(x) : x 2 oX}, q uðxÞ : x 2 oXg, q2 = inf{v(x) : x 2 oX}, q Theorem 1. Let ð u; vÞ, (u, v) be ordered pairs of super- and sub-solutions of (1), respectively. Suppose that of og ; P 0 on X  D ðcooperative systemÞ: ov ou u and v 6 v on X, then there is solutions (u, v) of (1) such that u 6 u 6 u and v 6 v 6 v on X. If u 6  In [1] for the first time in the literature, the authors consider a class of semipositone systems. In particular they extend many of the results discussed for the positive solutions of single equation in [2] to semipositone systems. It was shown positive solutions to (1) for either k near the first eigenvalue k1 of the operator D subject to Dirichlet boundary conditions, or for k large exists. We consider following assumptions: f, g are C1 functions satisfying either f ðx0 ; 0; 0Þ < 0 or

gðx0 ; 0; 0Þ < 0 for some x0 2 X;

ð3Þ

lim

f ðx; u; vÞ ¼ 0 uniformly in x; v u

ð4Þ

lim

gðx; u; vÞ ¼0 v

ð5Þ

u!1

and u!1

uniformly in x; u:

To introduce additional hypotheses to prove existence results near k1, first we recall the anti-maximum principal by Clement Pletier (see [4]), namely, if Zk is the unique solution of  Dz  kz ¼ 1; x 2 X; ð6Þ 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:

ð7Þ

Let I = [a, c] where a > k1 and c < k1 + d, and let r :¼ max kzk k; k2I

where kÆk denotes the supremum norm. Now assuming that there exists a m1 > 0 such that f ðx; u; vÞ P u  m1

8x 2 X; u 2 ½0; m1 cr; v P 0

ð8Þ

and exists a m2 > 0 such that gðx; u; vÞ P v  m2

8x 2 X; v 2 ½0; m2 cr; u P 0:

ð9Þ

Finally to prove existence results for k large, in addition to (3)–(5), we assume 9f1 ðuÞ 6 f ðx; u; vÞ 8x 2 X; u P 0; v P 0 such that f 1 ðr1 Þ ¼ 0; f10 ðr1 Þ < 0; Z r1 f1 ðsÞds > 0 for some r1 > 0

ð10Þ

0

and 9g2 ðvÞ 6 gðx; u; vÞ 8x 2 X; u P 0; v P 0 Z r2 g2 ðsÞds > 0 for some r2 > 0: 0

such that g2 ðr2 Þ ¼ 0; g02 ðr2 Þ < 0; ð11Þ

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2. Existence results Theorem 2. Let k 2 I and assume (3)–(5), and (8), (9) hold, Then (1) has a positive solution. It was shown in [1] (u, v) is a sub-solution of (1) where u(x) = cm1Zk and v(x) = cm2Zk. Now let w(x) to be the unique positive solution of  DwðxÞ ¼ 1; x 2 X; ð12Þ wðxÞ 6 0; x 2 oX: ð u; vÞ is a super-solution that  u ¼ JwðxÞ and v ¼ e J wðxÞ where J ; e J > 0 are sufficiently large such that 1 f ðx; J kwk; vÞ gðx; u; e J kwkÞ P ; e kkwk J kwk J kwk

ð13Þ

and ð130 Þ

and vðxÞ P vðxÞ on X:

 uðxÞ P uðxÞ on X

Theorem 3. Assume (3)–(5) and (10), (11) hold. Then there exists a k* > 0 such that for every k > k*, (1) has a positive solutions. Here we give a simple example that satisfies the hypotheses of Theorems 2 and 3. Consider 1

hðx; u; vÞ ¼ mðu þ 1Þ2 

3m v þ evþ1 2

8u P 0; v P 0;

ð14Þ

where m > 0 is a constant. Let f ðx; u; vÞ ¼ hðx; u; vÞ; gðx; u; vÞ ¼ hðx; v; uÞ: ¼ 0 uniformly in v. Also, Here f ðx; 0; 0Þ ¼ 1  m2 < 0 for m > 2, f is increasing in u, v, and limu!1 f ðx;u;vÞ u gðx; 0; 0Þ ¼ 1  m2 < 0 for m > 2, g is increasing in u, v, and limv!1 gðx;u;vÞ ¼ 0 uniformly in u. Now, to show that v 1

(8) and (9) are satisfied, it suffices to show that h1 ðuÞ ¼ mðu þ 1Þ2  3m satisfies (8) since h(x, u, v) P h1(u) 2 "u P 0, v P 0. Let p > 0 be such that h1(p) = 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 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 4 3 2 ðm  mÞ þ m mðm  1Þ þ 3 ðm  mÞ þ m  2m þ 4m ¼ p¼ : 2 2 Hence in order that (8) be satisfied, we must have qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 m2  m þ m ðm  1Þ þ 3 P mðraÞ; 2 that is, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðm  1Þ þ ðm  1Þ2 þ 3 P 2ðraÞ: 2

ð15Þ

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Since r and a are quantities that depend only on X, clearly for a given r and a, there exists and m0 sufficiently large such that if m > m0, then (15) is satisfied and equivalently (8) will be satisfied. Thus, (9) is also satisfied for m P m0. Note that this example satisfies the hypotheses of Theorem 3 also since h(x, u, v) P h1(u) "u P 0, v P 0 and one can construct a function f1(u) 6 h1(u) satisfying (10). 3. Numerical results We see in Section 2 that there must always exists a solution for problems such as (1) between a sub-solution (u,v) and a super-solution ð u; vÞ when of P 0, og P 0. ou ov Consider the coupled-system boundary value problems 8 DuðxÞ ¼ kf ðx; u; vÞ; x 2 X; > > < DvðxÞ ¼ kgðx; u; vÞ; x 2 X; > > : uðxÞ ¼ 0 ¼ vðxÞ; x 2 oX:

ð16Þ

Since f, g are C1 functions, there exists positive constants K1, K2 such that X  D. Thus we can study the equivalent system 8 DuðxÞ þ kK 1 uðxÞ ¼ kf ðx; u; vÞ þ kK 1 uðxÞ ¼ kf^ ðx; u; vÞ; x 2 X; > > < gðx; u; vÞ; x 2 X; DvðxÞ þ kK 2 vðxÞ ¼ kgðx; u; vÞ þ kK 2 vðxÞ ¼ k^ > > : uðxÞ ¼ 0 ¼ vðxÞ; x 2 oX:

of ou

P K 1 , and

og ov

P K 2 on

ð17Þ

The mapping T : (u1, v1) ! (u2, v2), (u2, v2) = T(u1, v1) ðu1 ; v1 Þ 2 ½u;  u  ½v; v

8x 2 X;

where (u2, v2) is the unique solution of the coupled-system 8 Du2 ðxÞ þ kK 1 u2 ðxÞ ¼ kf ðx; u1 ; v1 Þ þ kK 1 u1 ðxÞ; x 2 X; > > < Dv2 ðxÞ þ kK 1 v2 ðxÞ ¼ kgðx; u1 ; v1 Þ þ kK 1 v1 ðxÞ; x 2 X; > > : u2 ðxÞ ¼ 0 ¼ v2 ðxÞ; x 2 oX;

ð18Þ

satisfied the hypotheses of Schauder fixed point theorem, and then we can conclude that 9ðu; vÞ 2 D

T ðu; vÞ ¼ ðu; vÞ;

so (u, v) is a solution of (1) (see [3]). By letting f^ ðx; u; vÞ ¼ kf ðx; u; vÞ þ kK 1 u and g^ðx; u; vÞ ¼ kgðx; u; vÞ þ kK 2 v, we use the following iteration to obtain solution: 8 u0 ðxÞ ¼ u; v0 ðxÞ ¼ v; > > > > > < ðD  kK 1 Þunþ1 ¼ f^ ðx; un ; vn Þ;

x 2 X; n ¼ 0; 1; 2; . . . ;

> > ðD  kK 2 Þvnþ1 ¼ ^ gðx; unþ1 ; vn Þ; x 2 X; > > > : unþ1 ¼ 0 ¼ vnþ1 ; x 2 oX: We can also use u0 ðxÞ ¼  u, v0(x) = v as initial guesses. We use following algorithm:

ð19Þ

G.A. Afrouzi et al. / Applied Mathematics and Computation 186 (2007) 1113–1119

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Sub- and super-solution algorithm 1. Find u0 = u and v0 ¼ v. Choose numbers k1, k2 > 0. 2. Solve the boundary value system (19). 3. If kun+1  unk <  and kvn+1  vnk < , output and stop. Else go to step 2. Now we want to apply the algorithm 8 pffiffiffiffiffiffiffiffiffiffiffi v DuðxÞ ¼ km u þ 1  3m þ evþ1 ; > 2 > < pffiffiffiffiffiffiffiffiffiffiffi u þ euþ1 ; DvðxÞ ¼ km v þ 1  3m 2 > > : uðxÞ ¼ 0 ¼ vðxÞ;

for x 2 X; ð20Þ

x 2 X; x 2 oX:

For doing step 1, we solve the problem ( Dz  kz ¼ 1; x 2 X; z ¼ 0;

ð21Þ

x 2 oX;

to obtain u. We know from Section 2 that problem (20) 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. x

y 0.2

0.4

0.6

0.8

Approximation of zk for k = 15 0.2 0.268 0.4 0.447 0.6 0.513 0.8 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

Approximation of zk for k = 17 0.2 1.895 0.4 3.266 0.6 3.789 0.8 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 0.2 0.002 0.4 0.028 0.6 0.050 0.8 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 0.2 0.001 0.4 0.024 0.6 0.048 0.8 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 u = cm1zk(x) where c and m obtained from Sections 1 and 2 and to obtain v for k 2 I (I = [a, c] where a > k1 and c < k1 + d) we solve  DvðxÞ ¼ 1; x 2 X; ð22Þ vðxÞ ¼ 0; x 2 oX; by finite difference (see [5,6]). We choose J such that (13) and (13 0 ) are satisfied. We execute algorithm for k 2 I = [17.1, 34.9]. It is easy to see that u = v for problem (20). For brevity we express just some of those numerical results.

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x

G.A. Afrouzi et al. / Applied Mathematics and Computation 186 (2007) 1113–1119

y 0.4

0.6

0.8

Approximation of u for k = 17.1 0.2 1.009 · 104 0.4 1.511 · 104 0.6 1.511 · 104 0.8 1.009 · 104

0.2

1.511 · 104 2.277 · 104 2.277 · 104 1.511 · 104

1.511 · 104 2.277 · 104 2.277 · 104 1.511 · 104

1.009 · 104 1.511 · 104 1.511 · 104 1.009 · 104

Approximation of u for k = 25 0.2 2.174 · 104 0.4 3.253 · 104 0.6 3.253 · 104 0.8 2.174 · 104

3.253 · 104 4.903 · 104 4.903 · 104 3.253 · 104

3.253 · 104 4.903 · 104 4.903 · 104 3.253 · 104

2.174 · 104 3.253 · 104 3.253 · 104 2.174 · 104

Approximation of u for k = 30 0.2 3.139 · 104 0.4 4.697 · 104 0.6 4.697 · 104 0.8 3.139 · 104

4.697 · 104 7.079 · 104 7.079 · 104 4.697 · 104

4.697 · 104 7.079 · 104 7.079 · 104 4.697 · 104

3.139 · 104 4.697 · 104 4.697 · 104 3.139 · 104

Approximation of u for k = 34.9 0.2 4.257 · 104 0.4 6.369 · 104 0.6 6.369 · 104 0.8 4.257 · 104

6.369 · 104 9.597 · 104 9.597 · 104 6.369 · 104

6.369 · 104 9.597 · 104 9.597 · 104 6.369 · 104

4.257 · 104 6.369 · 104 6.369 · 104 4.257 · 104

Our numerical results (in following tables) show that there exist k* > 0 such that for every k > k*, (20) has a positive solution. In this case k* = 166.696 with decimal accuracy. x

y 0.2

0.4

0.6

0.8

Approximation of u for k = 170 0.2 1.019 · 106 0.4 1.524 · 106 0.6 1.524 · 106 0.8 1.019 · 106

1.524 · 106 2.297 · 106 2.297 · 106 1.524 · 106

1.524 · 106 2.297 · 106 2.297 · 106 1.524 · 106

1.019 · 106 1.524 · 106 1.524 · 106 1.019 · 106

Approximation of u for k = 500 0.2 0.883 · 107 0.4 1.321 · 107 0.6 1.321 · 107 0.8 0.883 · 107

1.321 · 107 1.990 · 107 1.990 · 107 1.321 · 107

1.321 · 107 1.990 · 107 1.990 · 107 1.321 · 107

0.883 · 107 1.321 · 107 1.321 · 107 0.883 · 107

Approximation of u for k = 1000 0.2 3.543 · 107 0.4 5.286 · 107 0.6 5.286 · 107 0.8 3.543 · 107

5.286 · 107 7.963 · 107 7.963 · 107 5.286 · 107

5.286 · 107 7.963 · 107 7.963 · 107 5.286 · 107

3.543 · 107 5.286 · 107 5.286 · 107 3.543 · 107

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[4] P. Clement, L.A. Peletier, An anti-maximum principle for second order elliptic operators, Differential Equations 34 (1979) 218–229. [5] M. Dehghan, Numerical procedures for a boundary value problem with a non-linear boundary condition, Applied mathematics and computation 147 (2004) 291–306. [6] I.G. Petrovsky, Lectures on partial differential equations, Dover publications, Inc., 1991.