| Since the 20th century, the development of nonlinear functional analysis has achieved the great breakthrough, Firstly,in recent years, many people have studied the solutions of systems of operator equations on the basis of investigating the solutions of operator equations, Thus, the relations among different operator equations are investigated,which have an extensive application in mathematics fields. Secondly, the study of boundary value problems for nonlinear ordinary differential equations has interested people for a long, time and the research in this field is still very active. Recently, the existence of positive solutions of boundary value problems for singular nonlinear ordinary differential equations attracts close attention.In the first chapter of this thesis, We investigate the existence of solutions of a class of systems of nonlinear non-monotone two binary operator equations, by using the cone theory and monotone iterative technique in nonlinear functional analysis. The results obtained in the first chapters extend and improve the conclusions in many recent work. In the latter three chapters, we consider the existence of multiple positive solutions to three sorts of boundary value problems for nonlinear singular differential equations. The tools used in this thesis are the method of lower and upper solutions and the fixed point theorem of, cone expansion and compression and so on. The paper is divided into four-chapters according to contents.In the first chapter, we shall utilize the cone theory and monotone iterativetechnique to study the systems of nonlinear operator equations in Banach space= Ai{x,y),(1.2.2)^ = A2(x,y),where AUA2 : E x E -> E.Using the following lemmas and conditions:(Hi) uo -T(u2 -A2(uu v2) - A2(u2, vi) > T(v2 - vx).(H3) There is A e (0,1] such that (XI + T)l e L(E), and (XI + T)x > ieP.For any uq < u < v < vq, we have A\(u, v) < A2(u, v).(H6) For any countable bounded set C\,C2C D with a(C\) +a(C2) ^ 0. we have a(Ai(Ci,C2)) < max.{a(Ci),a(C2)},i = 1,2. where a is Kuratowsi measure of noncompactness or Hausdorff measure of noncompactness.(H7) For any countable bounded set Cu C2 C D with 0(Ci) +P(C2) j- 0. we have j3(A{(Ci, C2)) < max{/3(Ci), f3(C2)}, i = 1, 2. where /? is weak measure of noncompactness.(H8) Every completely ordering subset in D is relatively compact.We obtain the following-result:Theorem 1.3.1 Assume that E is a real Banach space, P is a normal cone, Uo, vq € E, uq < vo, D = [uq, vo] = {x € E \ uq < x < Vq} is a order interval in E. Suppose that A\, A2 : D x D —>■E is semi-continuous in each variableand conditions {Ht), (H2), (H3), (#4), (H5), (#6) 1& (H,), (H2), (H3), (#4), (H5) (H7) are satisfied. Then system of equations (1.2.2) has a minimal and maximal solution (u*,v*) ￡ D x D, such thatlim un — u*, lim vn — v*,n-$oo n->-oowhereUn = (XI + TJ-'lAiiK.!,^,) + Tlin-i] ,vn = (XI + Ty^XAtiun-x,un0 + Tvn^],n = 1,2, ? ? ■, satisfyuq < ui < ■■■< un < u* < v* < vn < ■? ■< vi < vq.Remark 1.3.1 It is worth noting that we require that operator A\ only exist a lower solution and operator A2 exist a upper solution in Theorem 1.3.1, Theorem 1.3.1 employs the weaker compact conditions to obtain the solutions of the system of equations (1.2.2).Remark 1.3.2 In the Theorem 2.4 of [4], the autor obtained the existence of solutions of A(x,x)=x under the condition that A(x,x)=x is continuous in x. However, in Theorem 1.3.1, A(x,x)=x must have solutions when A — A\ — A2 holds and continuity isn't assumed. Thus Theorem 1.3.1 generalizes and improves the Theorem 2.4 of [4].Remark 1.3.3 It is obvious that the condition (i/5) in Theorem 1.3.1 is satisfied when we assume that T = MI(M > 0), A = A\ = A2- So Theorem 1 of [10] can get from Theorem 1.3.1. What's more, Theorem 1.3.1 obtained the existence of solutions of A(x,x)=x without the condition of continuity, so Theorem 1.3.1 generalizes and improves the main results of [10].Theorem 1.3.2 Assume that E is a real Banach space, P is a normal cone, Uo.^o € E,u0 < vo,D = [uo,vQ] = {x G E | u0 < x < vQ} is a orderinterval in E. Suppose that A\,A2 satisfy (Hi), (H2), (#3), (H$) of conditions of Theorem 1.3.1. Then system of equations (1.2.2) must have solutions.Remark 1.3.4 Theorem 1.3.2 improves the Theorem 2.2 of [4] and differs with it in the method of proof in essence.Application 1.4 As an application of Theorem 1.3.1, we study the solutions for a class of systems of nonlinear integral equations of Volterra type in Banach space.( ftx(t)=xo{t)+ / H1(t,s,x(s),y(s))ds,j J% (1-4.1)= xo(t) + H2(t,s,x(s),y(s))ds,Ii(i = 1,2) € C[DxExE,E], D = {(t, s) G JxJ : t > s},x0 eC[J,E], Using the following lemmas and conditions: (Li) There exists u0, vQ G C[J,E],u0 < u0, such thatuQ(t) < xo(t) + / Hi(t, s,UQ(s),vQ(s))ds,t G J, Jt0f*ZQ\t) + / H2(t,s,Uo(s),fo(s))(is ^ Vo\t)yt G J.Jtowhere fi = {u G E : uQ(t) -m{t)(y2(t) -Jt0(L^ For any x, y with x < y and (t, s) G D, we have Hi(t,s,x,y) < H2(t,s,x,y).(L5) There are k ￡ C[JxJ, R+], L ￡ C[JxJ, R+], satisfying max Jt' k(t, s) ds < I, max/4o L(t, s)ds < 1, such that for any countable sets Dx C ￡l, D2 C ￡1, and (t, s) ￡ J x J, we havea(Hi(t, s, Di, D2)) < k(t, s) max{a(￡>1), a(￡>2)},a(H2{t, s, DUD2)) < L(t, s) max{a(A), *x(D2)}. wherert rtmax/ [k(t, s)]ds < 1, max/ [L(t, s)]ds < 1.Lemma 1.4.l'12J Let i?(4, s,x,y)be continuous uniformly on J x J x Q x Q, and Bi,B2 C [moi^o] be bounded uniformly and equicontinuous. Then For any fixed t ￡ J, H(t, s, Bi(s), ^(s)) is bounded uniformly and equicontinuous at s on J.Lemma 1.4.2l13' Let B ￡ C[J,E] be bounded uniformly and equicontinuous. Then m(t) = a(B(t)) is continuous on J anda ( [B(s)ds) < f a(B(s))ds.Lemma 1.4.3^ Let B ￡ C[J, E] be bounded uniformly and equicontinuous. Then a(B) = max.a(B(t)).We have the following conclusion: 'Theorem 1.4.1 Let E be a Banach space, P be a normal cone in E, Suppose in addition that conditions (Lx) — (L5) are satisfied. Then system of equations (1.4.1) must have continuous solutions (u*(t), v*(￡)) ￡ Q, x Q, and there exist {un(t)}and {vn(t)} are convergent uniformly to u*(t) and v*(t) onJ, where1 A + m(t)1 A 4- m(t)/where A € (0,1], t € J, n = 1,2, ? ? ?. What's more, for any t € J, we have< ? < u*(t) < v*(t) <■■< vn(t) <■■■< vo(t).In the second chapter, by constructing a special cone and using the fixed point theorems of cone expansion and cone compression, we investigate the existence of positive solutions of a class of boundary value problem of second order singular differential equations.u"{t) + f{t, u(t)) + g{t, u{t)) = 0,0 < t < 1,(2.1.1) w(0) = u(l) = 0,where nonlinear term /(￡, u),g(t, u) are allowed to be singular at t=0, t=l and u=0, and f(t,u) and g(t,u) have different properties.Using the following lemmas and conditions:(Si) f, ge C[(0,1) x (0, +oo)], R+], and for any R > r > 0, we have0 < / s(l - s)fr,R(s)ds < +oo, Jo0 < / s(l — s)griu(s)ds < +oo, ./owhere for any t e (0,1), we have frtR(t) = max{f{t,x),x G [rz{t),R]},griR(t) =msx{g{t,x)tx e [rz{t),R]}.(52) There exist 4> G L[0,1], ip{t), andf0 s(l — s)ip(s)ds < +00 hold.(53) There exists R > 0, such that?1/Jo(54) There exist a ,b e C[I, R+] where a(t) does not vanish identically on any subinterval of I, such that /(￡, x) + g(t, x) > a(t)x — 6(t), V t e I.(55) There exists [c, d] C (0,1), such that for any t ￡ [c, d], (i)lim inf> N;or (ii)liminf - ' > N, uniformly with respect to t G [c, d], wherex—H-oo 2;(56) -y(L) > 1, where (Ia;)(?) = /^ G(t, s)a(s)o;(5)ds, V x e C[L, R}7 -y(L) denotes the spectral radius of L, G(t, s) is Green funcdaon.Lemma 2.1.1^ Let K be a cone of a real Banach space E and B : K —>? K a completely continuous operator. Assume that B is order-preserving and positively homogeneous of degree 1 and that there exist v ￡ K\{9},\ > 0. such that Bv > Xv. Then j(B) > A, where j(B) denotes the spectral radius ofB.Lemma 2.1.2t14'(the fixed point theorem of cone expansion and cone compression,) Let P be a cone of a real Banach space E and Pr>s = {x G P : f < H^ll < s},s > r > Q,A: PTtS -VP is completely continuous mapping, such that one of the following two conditions is satisfied :(l)Ax ￡ x,V x € P,\\x\\ =r;Ax^x,y xeP, \\x\\ = s. or1,2) Ax ^i,VieP, ||a;|| = r;Ax ￡ x,V x e P, \\x\\ ="5. Then,A has a fixed point x E P and that r < \\x\\ < s.Lemma 2.2.1 A(P\{9}) C P.Lemma 2.2.2 For any R2 > Ri > 0,A : TR2\PRl -> P is completely continuous operator, where Pr = {x ￡ P : ||rr|| < r}(r > 0).Remark 2.2.1 Since nonlinear terms f(t,u) and g(t,u) may be unbounded at u = 0 and /(￡, u) $J g(t,u) have different properties, the proof of Lemma 2.2.1 and Lemma 2.2.2 has a larger difference with one of the recentpapers.We have the following main results:Theorem 2.3.1 Let (Sl),(S2),(Sz),(Si),(S6) be satisfied. Then BVP (2.1.1) has at least two positive solutions.Theorem 2.3.2 Let (Si), (S2), (S3), (S5) be satisfied, Then BVP (2.1.1) has at least two positive solutions.Remark 2.3.1 Condition (Si) is feasible since f(t,u) and g(t,u) are allowed to be singular at u = 0.Remark 2.3.2 It is worth noting that nonlinear terms /(￡, u) and g(t, u) of Theorem 2.3.1 and Theorem 2.3.2 have different properties and the main results of the paper include some known results.By using Lemma 2.1.2, we also can obtain the following corollary : Corollary 2.3.1 If one of the following conditions holds:(m)(5x),(53),(55). Then BVP (2.1.1) has at least one positive solution.Example 2.4.1 Consider the following singular BVPu"(t) + fft, u(t)) + g(t, u(t)) = 0,0 < t < 1;(2.4.1)u(0)=u(l) = 0, where1 , , 1Remark 2.4.1 It is worth noting that f(t, u) satisfies the condition (Si), where g(t,u) does not satisfie (Si), that is, f(t,u) and g(t,u) have different properties.Example 2.4.2 Consider the following singular BVP?"(<) + /(*, ?(*)) + 9(t, u(t)) = 0,0 < t < 1;(2.4.2)u(0) = u(l) = 0,whereRemark 2.4.2 It is worth noting that f(t,u) satisfies (Si) and (5s), where g(t,u) does not satisfy (Si) and (S5), that is, f(t,u) and #(￡,u) have different properties.In the third chapter, by constructing lower and upper solutions and with the maximal theorem, we investigate the existence of positive solutions of a class of Fourth singular boundary value problemsu(4)(t) = \a(t)f(t,u(t),-u"(t)),0< t < 1, u(Q) = u(l) = 0, ?"(0). = ?"(1) = 0.where A > 0 is a real parameter, a(t) ￡ C((0,1), [0,+oo)),a(i) is allowed to be singular at t = 0 #1 t = 1, /(t, 0,0) and / € C((0,l) x [0,+oo) x -{0, +oo), [0, H-oo)) is continuous and does not vanishldentically on any subin-terval of (0,1) and (0,1) x [0, +oo) x [0, +oo).Using the following definitions and lemmas :Definition 3.2.1 u(t) ￡ C2[0, l]nC4(0,1) is said to be a positive solution of (3.2.1) if for any t G (0,1), u(t) > 0,u"(t) < Q,(t,u(t),-u"(t)) € D,u^(t) = F{t,u(t), -?"(<)), and u(0) = u(l) = 0 and u"(0) = ?"(1) = 0.Definition 3.2.2 Let a <= C2[0,1] n C4(0,1), satisfies that< F(t,a(t)t-a"(t)),te (0,1),a(0) < 0,a(l) < 0,a"(0) > 0,a"(l) > 0Then a is the lower solution of singular BVP (3.2.1).Definition 3.2.3 Let 0 E C2[0,1] n C4(0,1), satisfies thatpW(t) > F(t,P(t),-P"(t)),te (0,1), 0(0) > 0,0(1) > O,0"(O) < O,0"(l) < 0.Then (3 is the upper solution of singular BVP (3.2.1).Lemma 3.2.1(The maximal theorem) If x e F satisfies x^(t) >0,te (a, 6).Thenx{t) >0,x"{t) < 0,te [a,b].Lemma 3.2.2 Assume that a(t) and )3(t) are the lower and upper solutions of singular (3.2.1) respectively and satisfy that(ai) 0 < a(t) < P(t),O< -a"(t) < -/?"(<),*€ (0,l),and(a2)D?cD. , ■In addition,suppose that there is a function p G C((0,1), [0, +oo)) which satisfies(03) For any (t,u,v) ￡ D^,\F(t,u,v)\ < p(t), and F(t,u,v) is monotone increasing on u and monotone increasing on v, and(04) Jo t(l - t)p(t)dt < +00.Then (3.2.1) has at least a solution u e C2[0,1] n C4(0,1), which for any t 6 [0,1], satisfies thata(t) < u(t) "Using the following assumptions(Ci) f(t, u, v) is monotone increasing on u and monotone increasing on v, and there exists a function q 6 C((0,1), [0, +00)) satisfies(i) 0 < Jo1 t(l - t)q(t)a(t)dt < +00, and(ii) For any rj > 0,8 > 0, there exists a positive contant M^g satisfies f(t,u,v) < Mr,,oq(t) in (0,1) x [0,77] x [0,0], that is, f(t,u,v) may be singular at * = 0 or t = 1.We have the following main results:Theorem 3.3.1 Assume that condition (Ci) are satisfied, Then there exists a positive contant A*, such that BVP(3.1.1) has a positive solution u € C2[0,1] n C4(0,1). for any A with 0 < A < A*.Remark 3.3.1 if we suppose that f(t,u, -u") = f(t,u), theorem 3.3.1 is the Theorem 2.1 of [25], so theorem 3.3.1 generates and includes the results of [25].In the fourth chapter, by using the fixed point theorems of cone expansion and cone compression, we investigate the existence of positive solutions of fourth order singular boundary value problem.) = a(t)f{t, u(t), ?"(*)) + b(t)g(t, u(t), ?"(*)), 0 < t < 1, ti(0) = ?(l) = 0, (4.1.2)where a,b e C((0,1), [0, oo)), a(t) and b(t) are allowed to be singular at t = 0 and t = 1;/ and g : [0,1] x [0,oo) x (—oo, 0] —> [0, oo) are continuous;a > 0, P > 0,7 > 0,5 > 0 and A = 07 + a5 + /3y > 0.Using the following cnditions and lemmas:<sub> (Gi) a and b : (0,1) —> [0, do) are continuous, / Mid g : [0,1] x [0,00) x (—00,0] -> [0,oo) are continuous, and a(t) and b(t) do not vanish identically on any subinterval of (0,1);(G2) There exists tQ e (0,1), such that a(t0) > Q,b(tQ) > 0, 3. 0 < Jo G(s,s)(a(s) + b(s))ds < 00;where G(s,s) is Green function.(G3) 7o mug0 mi;(G^ /Q > muJn < mo,^ > mi,^ < mQ;/0Mr1/Jowhere M = max{Mi, M2},Mi = max{/(￡,x,y) : 0 < t < 1,0 < x < 1,-1 < y <0},M2 = max{# (￡,:z,y) : 0 < i < 1,0 < x < 1, -1 < y < 0}. where in (G2),0 + at)(5 + 7 - 75), 0 < t < s < 1,From the condition (G2), we can get a 9 e (0,1/2) which satisfies t0 € (0,1-In (GO - (Gfl),liminf min ^^ ' //. = limsuP maxax TTt\ o,i] \x\ + \y\,. ? f ? 9(t,x,y) - ,. 9(t,x,y)g = liminf mm -r-1^—r-f , 0,. = lunsup max 7-^—r-, -v- ■)x}+\y\^litel0,i-B] \x\ + \y\ " \s\+\y\^te[o,i] \x\ + \y\m0 = -?i=(of G(\,s){a{s) + b(s))ds^ ,m2 = I 9 G(—rs)a(s)c\ h 2where (j, is 0 or 00, 0 € (0,1/2) is the 9 obtained above.Lemma 4.2.1 Suppose that (Gr), (G2) are satisfied. Then A : C2[0,1] -> C2[0,1] is completely continuous .Lemma 4.2.2f14J Let P be a cone of a real Banach space E and Oi and il2 C E are bounded sets, 8 e f^, fli C ￡22, -4 : Pf)(S72\!T21) -> P a completely continuous mapping, one of the following two conditions is satisfied:i)\\Au\\ < \\u\\,VuePndni;\\Au\\ > \\u\\,V u € PDThen A has a fixed point in P n (Q2\^i)-Lemma 4.2.3 Suppose that conditions (Gx) and (G2), operator A defined by (4.2.4) satisfies A(P) C P and A : P —> P is completely continuous.Remark 4.2.1 Since a(t) and b(t) may be unbounded at t = 0 and ￡ = 1, and / and g include the u" term, the proofs of Lemma 4.2.1 and Lemma 4.2.3 have larger difference with those of the similar-result of recent papers(such as [30,32]).We have the following main results:Theorem 4.3.1 Assume that (Gi),(G2),(G3) or (Gl),(G2),{G4) are satisfied. Then problem (4.1.2) has at least one positive solution.Remark 4.3.1 Result of Theorem 4.3.1 cancels the monotonicity condition of [26] and its method has larger difference with [24].Theorem 4.3.2 Assume that (Gi),(G2),{G5) or (Gi), (G2), (G6) are satisfied. In addition, if (G7) is satisfied. Then problem (4.1.2) has at least two positive solutions.Remark 4.3.2 We improve the conditions of the articles [21,22,24], by only using the conditions of / > m2, g. > m^ or /^ > m?, g > m% instead of defining the the increasing orders of / and g and requiring the monotonicity of / and g.
|
PDF Full Text Request