WATANABE, TOSHIKAZU

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RIGHT:URL:CITATION:AUTHOR(S):ISSUE DATE:TITLE:NOTE ON SOLUTIONS OFBOUNDARY VALUE PROBLEMS FORFRACTIONAL ORDER BEAMEQUATIONS (Study on NonlinearAnalysis and Convex Analysis)WATANABE, TOSHIKAZUWATANABE, TOSHIKAZU. NOTE ON SOLUTIONS OF BOUNDARY VALUE PROBLEMS FOR FRACTIONAL ORDER BEAMEQUATIONS (Study on Nonlinear Analysis and Convex Analysis). 数理解析研究所講究録 2021, 2190: 188-1962021-07http://hdl.handle.net/2433/265672188NOTE ON SOLUTIONS OF BOUNDARY VALUE PROBLEMS FOR FRACTIONAL ORDER BEAM EQUATIONS TOSHIKAZU WATANABE SCHOOL OF INTERDISCIPLINARY MATHEMATICAL SCIENCES, MEIJI UNIVERSITY 1. INTRODUCTION Throughout this paper, we denote by R the set of all real numbers. In [19], we consider the boundary value problem for fractional order differential equation (1.1) { Dg+Do+u(t) - f(t, u(t), D0+u(t)) = 0, t E [O, 1] u(0) = A, u(l) = B, D0+u(0) = C, D0+u(l) = D, where Do+ is the Riemann-Liouville derivative of order o: with respect to t, 1 < a:, /3::; 2, A, B, C, Dare constants, and f is a continuous function of [0, 1] x Rx R into R In this paper we propose the following differential equation (1.2) of order o:, 3 < o: ::; 4 with the two point boundary condition involving the form (1.1). For simplicity, we consider the cases of A = B = C = D = 0. (1.2) { D0+u(t) = f(t, u(t), Dg_;3u(t), Do+Dg_;3u(t), Do+Do+Dg_;3u(t)), 0 < t < 1, u(0) = u(l) = 0, Do+Dg_;3u(0) = Do+Dg_;3u(l) = 0, where Do+ is the Riemann-Liouville fractional derivative and f is a function of [0, 1] x Rx Rx Rx R into R Leto:> 0. The Riemann-Liouville fractional integral of order o: of u, denoted Itf+u, is defined by Ig+u(t) = rto:) fo\t - s)°'-1u(s)ds, provided the right-hand side exists. The Riemann-Liouville fractional derivative of order o: of a function u of (0, oo) into R is given by a ( ) 1 dn 1t( )n-a-1 ( ) Do+U t = r( )-d t - s u s ds, n - O'. tn O where n = [o:] + 1 ([o:] denotes the integer part of o:) and r(o:) denotes the gamma function; see [11, 18]. Note that for o: > /3 > 0, we have D°' tf3 = r(/3 + l) tf3-a 0+ r (/3 - o: + 1) · 2010 Mathematics Subject Classification. Primary 34A08, 34B15, 47H10. Key words and phmses. Fixed point theorem, fractional order, differential equations, boundary value problem. 189A function u E C[0, 1] is called a solution of problem (1.2) if Dg+u E C[0, 1], Da_;3u E £ 1[0, 1], Da_;2u E £ 1[0, 1], D°'-1u E £ 1[0, 1], u satisfies the boundary conditions and equality in (1.2) a.e. on [0, 1]. Many researchers have considered the differential equation (1.2) with o: = 4; see [1, 2, 9, 10, 14, 15, 21, 22, 23, 25, 26]. Equation (1.2) with o: = 4 can be used to model the deformations of an elastic beam; see [21, 22] and the references therein. The boundary conditions in (1.2) with respect to normal derivative ensures that both endpoints are simply supported. Meanwhile, fractional differential equations have been of interest recently; see [3, 5, 6, 7, 8, 11, 12, 17, 18, 19, 24]. In particular, for higher order boundary problems, see [17, 19, 20, 24]. However, to the best of our knowledge, there are no results for the boundary value problem represented by (1.2) for 3 < o: :S 4, which we consider in the present paper. We use the several methods to prove the existence and uniqueness of solutions. Moreover we consider the properties of Green function given by (1.2). 2. LEMMAS For a continuous mapping h of [0, 1] into JR, we consider the following fractional differential boundary problems defined by (2.1) {D0+u(t) = h(t), 0 < t < 1, u(0) = u(l) = 0, Do+D0_;3u(0) = Do+D0_;3u(l) = 0, where 3 < o: :S 4. In this section, we show the unique solution to the boundary value problem represented by (2.1). A mapping u of [0, 1] into lR is a solution of that boundary value problem if u is continuous on [0, 1] and u satisfies (2.1). The following lemma can be found in [6]; see [11] also. We denoted by C(0, 1) the set of all continuous mappings of (0, 1) into lR and by L(0, 1) the set of all Lebesgue integrable mappings of [0, 1] into R Lemma 2.1. Leto: > 0. If u(t) E C(0, 1) n L(0, 1) satisfying D0+u(t) E C(0, 1) n L(0, 1), then there exist constants C1, C2, ... , Cn E lR such that Ia+Do+u(t) = u(t) + C1t°'-1 + C2t°'-2 + · · · + Cnt°'-n, where n = [o:] + 1 and I0+u is the Riemann-Liouville fractional integral of order o: of a function u defined by Ia+u(t) = rto:) fo\t - s)°'-1u(s)ds. Using Lemma 2.1, we obtain the following. Lemma 2.2. Leth be a continuous mapping of [0, 1] into JR. Let 3 < o::::; 4. Then the unique solution of the boundary value problem represented by (2.1) is u(t) = fo 1 G(t, s)h(s)ds, 190where (2.2) G(t, s) = Remark 2.3. If a = 4, then G(t s) = {i (s(l - t)(2t- t2 - s2)) (0:::; s:::; t:::; 1), ' ½(t(l-s)(2s-s2-t2))(o:::;t:::;s:::;l). Lemma 2.4. Let 3 < a :::; 4. The function G(t, s) in Lemma 2.2 satisfies the following conditions: (i) m:::; G(t, s) :::; M, where m = {s(t(l - t)(l - t) - (t - s)(t - s)) if (s:::; t), 0 if (t :::; s ), M = {(t - s)2 + (1- s)(2t - t2 - s2) if (s:::; t), (1 - t)(2s - t2 - s2) if (t:::; s). (ii) Ifs and t satisfy 0 :::; t :::; 3- 2\/'5, or 1 2: t 2: 3- 2\/'5 and t - 0(1 - t) :::; s :::; t, then G(t, s) 2: 0. We consider the following. lot G(t, s)h(s)ds = [-G(t, s)v(s)]t + lot G1(t, s)v(s)ds, and 11 G(t, s)h(s)ds = [-G(t, s)v(s)]i + 11 G1(t, s)v(s)ds, where v(s) = J/ h(s)ds and G1(t,s) = i;(t,s) (2.3) r{a) (-(a - l)(t - s)°'-2 + ta-l + (a - 1)(1 - s)°'-2ta-3 - t°'-3) if0:::;s:::;t:::;1 da)ta-1 + (a -1)(1- s)°'-2ta-3 -ta-3 if0:::;t:::;s:::;l. Moreover 191We also have fo 1 ng_;3G(t, s)h(s)ds = lot G2(t, s)v(s)ds, fo 1 Do+Dg_;3G(t, s)h(s)ds = fo 1 G3(t, s)v(s)ds, where (2.4) G t 8 = { r(2) ((t - s) - (1 - s)t°'-3) (0 ::S: s ::S: t ::S: 1), 2( ' ) --1-(1 - s)t°'-3 (0 < t < s < 1) 1'(2) - - - ' and Remark 2.5. The function f01 G(·, s)ds is continuous on [0, 1]. In fact, we have f 1 G(t, s)ds = _l_ (t°' - ~ta-l + (1 - ~) t°'-3) for all 0 ::S: t ::S: 1. lo ar(a) 2 2 3. MAIN RESULT Next we use the method of order reduction to transform (1.2) to a nonlinear integral equation. To do this, let T1v(t) = 1g_;3T2v(t) = fo 1 G1(t, s)v(s)ds, T2v(t) = fo 1 G2(t, s)v(s)ds, T3v(t) = fo 1 G3(t, s)v(s)ds, where G1(t, s), G2(t, s) and G3(t, s) are given by (2.3), (2.4) and (2.5). From the above formulas, it follows that Do+Do+Dg_;3T1v(t) = Do+Do+T2v(t) = Do+T3v(t) = -v(t). Note that since T1v(t) = fo 1 G1(t,s)v(s)ds = fo 1 G(t,s)f(s)ds, we have T1v(0) = T1v(l) = 0. Moreover by definition, T2v(0) = fo 1 G2(0, s)ds = 0, T2v(l) = fo 1 G2(l, s)ds = 0. Boundary value problem (1.1) can be converted into a ternminal value problem Do+v(t) = -f(t, T1v(t), T2v(t), T3v(t), -v(t)), fo 1 v(s)ds = 0. 192From the above formulas, it follows that Do+v(t) = f(t, T1v(t), T2v(t), T3v(t), -v(t)) where Do+T3v(t) = -v(t), Do+T2v(t) = T3v(t), D0_;3T1v(t) = T2v(t). Then we have the following lemma. Lemma 3.1. Let 3 <a:::; 4. The boundary value problem (1.2) is equivalent to the following integral equations forms; {v(t) = J/ f(s, T1v(s), T2v(s), T3v(s), -v(s))ds, T1v(t) = f01 G1(t, s)v(s)ds, T2v(t) = f01 G2(t, s)v(s)ds, T3v(t) = f01 G3(t, s)v(s)ds, where G1(t, s), G2(t, s) and G3(t, s) are given by (2.3), (2.4) and (2.5). Lemma 3.2. G1(t, s), G2(t, s) and G3(t, s) satisfy the followings. (i) -8:::; 3s+3s3-3t-4t2-1:::; G1(t, s):::; 3t2-2s2+3-4s:::; 6, if0:::; s:::; t:::; 1 -5 :::; 3t3 - 4s2 + 2t - 1 :::; G 1 ( t, s) :::; t2 + 3 - 4t :::; 3, if 0 :::; t :::; s :::; 1. (ii) -1:::; G2(t, s):::; 0, -1:::; G3(t, s):::; 1. Next we define an operator A from C[0, 1] into C[0, 1] by Av(t) = 11 f(s, T1v(s), T2v(s), T3v(s), -v(s))ds where v E C[0, 1]. Then the solution of boundary value problem (1.2 ) is a fixed point of mapping A. Also let (T1v)(t) = fo 1 G1(t, s)v(s)ds, (T2v)(t) = fo 1 G2(t, s)v(s)ds, (T3v)(t) = fo 1 v(s)ds. Then the existence of solution of the boundary value problem (1.2) is equivalent to the existence of fixed point of A on C[0, 1]. Take uo(t) = 1 - t. 11 (T1uo)(t)dt:::; r( 1 ) uo(t), 11 (T2uo)(t)dt:::; ( ( 1 ) - !) uo(t). t a+2 t 3a-2 8 11 (T3uo)(t)dt:::; 0.0276515uo.(a = 3.5, t = 1/6(8 - 8/(-109 + 27\/'17)1/3 + (-109 + 27\/'17)1/3) rv 0.5474636625659386) Moreover if a = 4, 11 (T3uo)(t)dt:::; }2 uo = 0.09375uo, (t = 1/4). If a= 3, f/(T3uo)(t)dt:::; iuo(t) = 0.166667uo(t) (t = 0). 1 1 1 (3.1) C1 = r(a + 2), C2 = 3(a _ 2) - 8, C3 = 0.0276515, (a= 3.5). 193If o: = 4, then we have 1 1 Ci = 120' C2 = 24. However if o: = 4, calculate directly, then we can take 291 1 1 1 C1 = 30720 > 120 , C2 = 9v'6 = 0.04536090 > 24 = 0.0416667, 3 C3 = 32 = 0.09375 > 0.0276515. Now we have the following theorem, which is the version of [27, Theorem 1]. Theorem 3.3. Suppose that there exist four nonnegative constants Mi (i 1, 2, 3, 4) with M1 C1 + M2C2 + M3C3 + M4 < 1 such that 4 lf(t, x1, x2, x3, x4) - f(t, YI, Y2, Y3, Y4)1 ::; L Milxi - Yil, Xi, Yi ER i=l Then the boundary value problem (1.2) has a unique solution. Next we consider the Banach contraction principal. Then, there exist constants D1 = - 4- ( o: - l ) D2 = ! + l D3 = ! - 3 - o: r(o:) o:(o:-2) ' 6 2(o:-2)' 2 (o:-l)(o:-2)" Ho:= 4, then 1 5 2 D1 = 4, D2 = 12 , D3 = 3. Note that for o: = 4, if we calculate directly, result is same. In this case we also have the theorem, which is the version of [27, Theorem 3]. Theorem 3.4. Suppose that there exist four nonnegative constants Mi (i = 1, 2, 3, 4) with M1D1 + M2D2 + M3D3 + M4 < 1 such that 4 lf(t, x1, x2, x3, x4) - f(t, Yl, Y2, Y3, Y4)1 ::; L Milxi - Yil, Xi, Yi ER i=l Then the boundary value problem (1.2) has a unique solution. We also have the theorem, which is the version of [27, Theorem 3]. In this section, we consider the existence and uniqueness of solutions of the boundary value problem represented by (1.2). It seems that there are few uniqueness results if the norm of related linear operator is greater than 1. In fact, Theorems 3.3, conclude that r(T) is less than 1, where r(T) is the spectral radius of linear operator T. 1 Note that r(T) = limn➔oo IITnll:;;. By Theorem 3.3, since for any v E C[0.1], 1 Tnv(t)::; NMnuo(t), we have IITnll:;; ::; M < 1, thus we have r(T) < l. Theorem 3.5. Suppose that there exist four nonnegative constants Mi (i = 1, 2, 3, 4) such that 4 lf(t, x1, x2, x3, x4) - f(t, Yl, Y2, Y3, Y4)1 ::; L Milxi - Yil, Xi, Yi E lR, i=l 194and r(T) < l. Then the boundary value problem (1.2) has a unique solution. Finally we consider the method in [28]. In order to do this, we give several lemmas. First put H1(t, s) = ng.;3G(t, s). Thus we have (3.2) { ½(t - s)2 - ½(1 - s)t2 + (a-1)\a-2) (-(1 - s)°'-1 + (1 - s))) if O:::; s:::; t:::; 1, H1(t,s) = 1 2 1 a-1 . - 2(1 - s)t + (a-l)(a-2) (-(1 - s) + (1 - s)) 1f O:::; t:::; s:::; 1. Also put H2(t, s) = Do+Dg.;3G(t, s). (3.3) H2(t s) = {st - s = s(t - 1) if O:::; s:::; t:::; 1, ' s-lifO:::;t:::;s:::;l. (3.4) Hts- - - -{s if O < s < t < l, 3 ( ' ) - 0 if O :::; t :::; s :::; 1. Lemma 3.6. H1(t, s) satisfies the following. (i) -is(l - s) :::; H1(t, s) :::; ½ ((t - s) 2 + (1- s)(s(l - s) + t(l - t))) if O :::; s:::; t:::; 1, ½(1- s)t(l - 3t):::; H1(t, s):::; ½(1- s)(s(l - s) + t(l - t)) if O:::; t:::; s:::; 1. (ii) IfO:::; t:::; 1- J1=s:::; s, then H1(t,s) 2:: 0. For u E C°'[O, 1], put cp(t) = f(t, u(t), ng.;3u(t), ng.;2u(t), ng.;1u(t)). Then the boundary value problem (1.2) becomes where u(t) = fo 1 G(t, s)cp(s)ds, ng.;3u(t) = fo 1 H1(t, s)cp(s)ds, ng.;2u(t) = fo 1 H2(t, s)cp(s)ds, ng.;1u(t) = fo 1 H3 (t, s)cp(s)ds. For 'P, we have the equation Acp = 'P, where A is a non-linear operator defined by with Thus we have the following lemma. 195Lemma 3.7. Let 3 < o: ::; 4. The boundary problem {1.2} is equivalent to the following integral equations forms; { cp(t) = f(t, u'P(t), v'P(t), w'P(t), x'P(t)), v'P(t) = f01 H1(t, s)cp(s)ds, w'P(t) = f01 H2(t, s)cp(s)ds, x'P(t) = f01 H3(t, s)cp(s)ds, where H1(t, s), H2(t, s) and H3(t, s) are given by {3.2}, {3.3} and {3.,4}. By (3.2) and Lemma 3.6, there exists E1, E2, E3 and E4 such that E1 = sup f 1 IG(t, s)lds, E2 = sup r1 IH1(t, s)lds, tE[O,1] Jo tE[O,1] Jo E3 = sup f1 IH2(t, s)lds, E4 = sup f1 IH3(t, s)lds. tE[O,l] Jo tE[O,1] Jo In this case following theorem holds. It is a version of [28]. Theorem 3.8. Suppose that there exist four nonnegative constants Mi (i 1, 2, 3, 4) with M1E1 + M2E2 + M3E3 + M4E4 < 1 such that 4 lf(t, x1, x2, x3, x4) - f(t, Yl, Y2, Y3, Y4)1 ::; L Milxi - Yil, Xi, Yi ER i=l Then the boundary value problem {1.2} has a unique solution. 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NOTE ON SOLUTIONS OF BOUNDARY VALUE PROBLEMS FOR FRACTIONAL ORDER BEAM EQUATIONS (Study on Nonlinear Analysis and Convex Analysis)

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NOTE ON SOLUTIONS OF BOUNDARY VALUE PROBLEMS FOR FRACTIONAL ORDER BEAM EQUATIONS (Study on Nonlinear Analysis and Convex Analysis)

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