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Research article Special Issues

Analysis of relative controllability and finite-time stability in nonlinear switched fractional impulsive systems

  • Received: 21 January 2025 Revised: 14 March 2025 Accepted: 25 March 2025 Published: 07 April 2025
  • MSC : 34A08, 34K10, 34K37, 37C25

  • This article investigated a class of switched impulsive fractional control systems with delays occurring at different time instants in both the state and control input. First, we analyzed the state response behavior and established sufficient conditions ensuring the system's stability over a finite time horizon. Next, we demonstrated the system's relative controllability using the fixed-point approach. Finally, a numerical simulation was presented to validate the theoretical findings.

    Citation: P. K. Lakshmi Priya, K. Kaliraj, Panumart Sawangtong. Analysis of relative controllability and finite-time stability in nonlinear switched fractional impulsive systems[J]. AIMS Mathematics, 2025, 10(4): 8095-8115. doi: 10.3934/math.2025371

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  • This article investigated a class of switched impulsive fractional control systems with delays occurring at different time instants in both the state and control input. First, we analyzed the state response behavior and established sufficient conditions ensuring the system's stability over a finite time horizon. Next, we demonstrated the system's relative controllability using the fixed-point approach. Finally, a numerical simulation was presented to validate the theoretical findings.



    In [1,pp. 42 and 55], we find the Maclaurin power series expansions

    cosx=k=0(1)kx2k(2k)!=1x22+x424x6720+x840320,xR

    and

    lncosx=k=122k1(22k1)k(2k)!|B2k|x2k=x22x412x64517x82520 (1)

    for x2<π24, where B2k denotes the Bernoulli numbers which can be generated by

    zez1=k=0Bkzkk!=1z2+k=1B2kz2k(2k)!,|z|<2π.

    Let

    F(x)={ln2(1cosx)x2,0<|x|<2π;0,x=0 (2)

    and

    R(x)={ln2(1cosx)x2lncosx,0<|x|<π2;16,x=0;0,x=±π2. (3)

    In the recent paper [2], Li and Qi obtained the following two results:

    1. The function F(x) defined by (2) can be expanded into the Maclaurin power series expansion

    F(x)=n=1E2n(2n)!x2n=x212x41440x690720x84838400 (4)

    for |x|<2π, where

    E2n=|A2n1,1B2n1,2n1ωnC1,2n+1|,n1,ωk=(1)k(k+1)(2k+1),k0,A2n1,1=(0ω10ω20ωn10)=(ai,j)1i2n1j=1,n1,ai,1={0,1i=2k12n1;ωk,2i=2k2n2,B2n1,2n1=|(00)ω00000(11)ω000(20)ω10(22)ω000(31)ω100(2n40)ωn20(2n42)ωn300(2n31)ωn200(2n20)ωn10(2n22)ωn1(2n22n2)ω0|=(bi,j)1i,j2n1,n1,bi,j={0,1i<j2n1;(ij)ωk,0ij=2k2n2;0,1ij=2k12n3,C1,2n1=(0(2n11)ωn10(2n13)ωn20(2n12n3)ω10)=(ci,j)i=11j2n1,n1,c1,j={0,1j=2k12n1;(2n12k1)ωnk,2j=2k2n2.

    2. The function R(x) defined by (3) decreasingly maps [0,π2] onto [0,16].

    We now introduce several new even functions as follows:

    1. The first function is

    F0(x)=lncosx,xk=0(±2kππ2,±2kπ+π2). (5)

    2. The second function is

    F1(x)={ln2(1cosx)x2,xR{±2kπ,k=1,2,};0,x=0. (6)

    It is clear that F1(x)=F(x) on (2π,2π).

    3. Generally, the third function we are introducing is

    Fn(x)={ln((1)n(2n)!x2n[cosxn1k=0(1)kx2k(2k)!]),x00,x=0 (7)

    for n2.

    Since the double inequality

    0<(1)n[cosxn1k=0(1)kx2k(2k)!]<x2n(2n)! (8)

    is valid for n2 and xR{0}, see [3,p. 326], the function Fn(x) is significantly defined for n2 and xR.

    As a stronger version of the double inequality (8), the following positive, nonnegative, decreasing, and concave properties of the normalized tail

    CosRn(x)={(1)n(2n)!x2n[cosxn1k=0(1)kx2k(2k)!],x01,x=0 (9)

    for x(0,) and n1 were discovered in the paper [4]:

    (a) the normalized tail CosR1(x) is nonnegative on (0,) and is decreasing on [0,2π];

    (b) the normalized tail CosRn(x) for n2 is decreasing and positive on (0,);

    (c) the normalized tail CosR1(x) is concave on (0,x0), where x0(π2,π) is the first positive zero of the equation

    (x22)sinx+2xcosx=0

    and the normalized remainder CosRn(x) for n2 is concave on (0,π).

    Comparing the definition in (9) with those in (6) and (7) leads to the following conclusions:

    (a) the function F1(x) is decreasing and negative on (0,2π), and is concave on (0,x0);

    (b) the function Fn(x) for n2 is decreasing and negative on (0,), and is concave on (0,π).

    4. The fourth function Rm,n(x) for n>m0 is defined

    (a) when n>m=0, by

    R0,n(x)={Fn(x)F0(x),0<|x|<π2;1(n+1)(2n+1),x=0;0,x=±π2, (10)

    (b) when n>m=1, by

    R1,n(x)={Fn(x)F1(x),xR{±2kπ,k=1,2,};6(n+1)(2n+1),x=0;0,x{±2kπ,k=1,2,},

    (c) when n>m2, by

    Rm,n(x)={Fn(x)Fm(x),x0;(m+1)(2m+1)(n+1)(2n+1),x=0.

    It is easy to see that R0,1(x)=R(x).

    We now propose the following three problems:

    1. Is the function Fn(x) for n0 decreasing and concave?

    2. What is the Maclaurin power series expansion of Fn(x) for n0 around the origin x=0?

    3. Is the function Rm,n(x) for n>m0 decreasing?

    The first problem for the case n=0 is immediate: the even function F0(x)=lncosx is decreasing and, by virtue of the series expansion (1), is concave in x(0,π2). The first problem for the case n1 was solved in the paper [4], as mentioned above. In a word, the first problem has been thoroughly solved.

    The second problem for n=0 is just the Maclaurin power series expansion (1). The second problem for n=1 was solved by the Maclaurin series expansion (4), which was established in [2,Section 3].

    The third problem for (m,n)=(0,1) was solved in [2,Section 4], as mentioned above.

    In this paper, we will give a full answer to the second problem for all cases n2, solve the first problem on the interval (0,π2) once again, and discuss the third problem for the case n2.

    In this section, we solve the second problem: what is the Maclaurin power series expansion of Fn(x) for n0 around the origin x=0?

    Theorem 1. For n0, let

    ei,j(n)={(1)i/21+(1)i21(2n+i2n),1i2m,j=1(1)(ij+1)/21+(1)ij+12(i1j2)(2n+ij+12n),1i2m,2j2m

    and

    D2m(n)=|ei,j(n)|(2m)×(2m).

    Then the even function Fn(x) for n0 can be expanded into

    Fn(x)=m=1D2m(n)(2m)!x2m,|x|<{π2,n=0;2π,n=1;,n2. (11)

    Proof. Let u(x) and v(x)0 be two n-time differentiable functions on an interval I for a given integer n0. Then the nth derivative of the ratio u(x)v(x) is

    dndxn[u(x)v(x)]=(1)n|W(n+1)×(n+1)(x)|vn+1(x),n0, (12)

    where the matrix

    W(n+1)×(n+1)(x)=(U(n+1)×1(x)V(n+1)×n(x))(n+1)×(n+1),

    the matrix U(n+1)×1(x) is an (n+1)×1 matrix whose elements satisfy uk,1(x)=u(k1)(x) for 1kn+1, the matrix V(n+1)×n(x) is an (n+1)×n matrix whose elements are v,j(x)=(1j1)v(j)(x) for 1n+1 and 1jn, and the notation |W(n+1)×(n+1)(x)| denotes the determinant of the (n+1)×(n+1) matrix W(n+1)×(n+1)(x). This is a slight reformulation of [5,p. 40,Exercise 5].

    Let

    un(x)=k=0(1)k+1(2k+2n+22n)x2k+1(2k+1)!andvn(x)=k=0(1)k(2k+2n2n)x2k(2k)!.

    Then, straightforward differentiation yields

    u(2+1)n(0)=limx0k=0(1)k+1(2k+2n+22n)2k+12+1x2k2(2k+1)!=(1)+1(2+2n+22n),v(2)n(0)=limx0k=0(1)k(2k+2n2n)2k2x2k2(2k)!=(1)(2+2n2n),u(2)n(0)=0,v(2+1)n(0)=0

    for 0. Considering Expression (14) and applying the derivative formula (12) for the ratio of two differentiable functions, we acquire

    F(2m)n(0)=limx0[k=0(1)k+1(2k+2n+22n)x2k+1(2k+1)!k=0(1)k(2k+2n2n)x2k(2k)!](2m1)=limx0[un(x)vn(x)](2m1)=(1)2m1v2mn(0)|un(0)vn(0)00un(0)vn(0)(11)vn(0)0un(0)vn(0)(21)vn(0)0u(3)n(0)v(3)n(0)(31)vn(0)0u(4)n(0)v(4)n(0)(41)v(3)n(0)0u(2m1)n(0)v(2m1)n(0)(2m11)v(2m2)n(0)(2m12m2)vn(0)|=|0(00)0001(2n+22n)0(11)000(20)(2n+22n)0(22)01(2n+42n)0(31)(2n+22n)0(33)0(40)(2n+42n)0(42)(2n+22n)0(1)m1(2n+2m22n)0(1)m2(2m31)(2n+2m42n)0(1)m3(2m33)(2n+2m62n)0(1)m1(2m20)(2n+2m22n)0(1)m2(2m22)(2n+2m42n)0(1)m(2n+2m2n)0(1)m1(2m11)(2n+2m22n)0(1)m2(2m13)(2n+2m42n)000000000000000000000000000000(2m32m7)(2n+42n)0(2m32m5)(2n+22n)0(2m32m3)00(2m22m6)(2n+42n)0(2m22m4)(2n+22n)0(2m22m2)(2m12m7)(2n+62n)0(2m12m5)(2n+42n)0(2m12m3)(2n+22n)0|

    for m1. In other words, for m1,

    F(2m)n(0)=D2m(n)=|ei,j(n)|(2m)×(2m).

    Consequently, the even function Fn(x) can be expanded into

    Fn(x)=k=0F(k)nxkk!=m=1F(2m)nx2m(2m)!=m=1D2m(n)(2m)!x2m.

    The proof of Theorem 1 is completed.

    Remark 1. When n=0, a direct computation gives

    D4(0)=|0(00)0010(11)00(20)0(22)10(31)0|=2andD2(0)=|0(00)10|=1.

    Then, the first two terms of the Maclaurin power series expansion of the function lncosx are

    D2(0)2!x2D4(0)4!x4=12!x224!x4=12x2112x4,

    which coincide with the first two terms in the series expansion (1).

    When n=1, straightforward computation shows

    D4(1)=|0(00)001(42)0(11)00(20)(42)0(22)1(62)0(31)(42)0|=160andD2(1)=|0(00)1(42)0|=16.

    Then, the first two terms of the Maclaurin power series expansion of the function F(x) defined by (2) are

    D2(1)2!x2D4(1)4!x4=112x211440x4,

    which coincide with the first two terms in the series expansion (4).

    Remark 2. Comparing the Maclaurin series expansions (1) and (4) with the series expansion (11) reveals

    |B2m|=m22m1(22m1)D2m(0)andE2m=D2m(1) (13)

    for m1. The first formula in (13) is a new determinantal expression for the Bernoulli numbers B2m with m1.

    Additionally, we point out that, in the papers [2,6,7,8,9,10], there have been many related results, but different from and more complicated than the first one in (13), and plenty of closely-related references on closed-form formulas and determinantal expressions for the Bernoulli numbers and polynomials B2m and Bm(x) with mN.

    In this section, we give an alternative and united proof of a modification of the first problem: is the function Fn(x) for n0 decreasing and concave on (0,π2)?

    Theorem 2. For n=0 and n2, the even function Fn(x) defined by (5) and (7) is decreasing and concave on (0,π2). The even function F1(x) defined in (6) is decreasing on (0,2π) and (xk,2(k+1)π) for kN, while it is increasing on (2kπ,xk) for kN, where xk(2kπ,2(k+1)π) for kN is the zero of the equation tanx2=x2 on (0,).

    Proof. In the first section of this paper, it has been immediately verified that the function F0(x)=lncosx is decreasing and concave on (0,π2).

    The derivative of F1(x) can be written as

    F1(x)=1tanx21x2,x±2kπ,kN.

    Therefore, the derivative F1(x) is negative on (0,2π), is positive on (2kπ,xk), and is negative on (xk,2(k+1)π) for kN, where xk(2kπ,2(k+1)π) for kN is the zero of the equation tanx2=x2 on (0,). Accordingly, the function F1(x) is decreasing on (0,2π) and (xk,2(k+1)π), while it is increasing on (2kπ,xk) for kN.

    On the interval (0,π2) and for n2, the function Fn(x) can be written as

    Fn(x)=lnk=n(1)kn(2n)!(2k)!x2k2n=lnk=0(1)k(2k+2n2n)x2k(2k)!.

    Its first derivative is

    Fn(x)=k=0(1)k+1(2k+2n+22n)x2k+1(2k+1)!k=0(1)k(2k+2n2n)x2k(2k)!. (14)

    By virtue of [11,Theorem 7.6], or in view of the results at the site https://math.stackexchange.com/a/477549 (accessed on 18 January 2024), we derive the integral representation

    cosxn1k=0(1)kx2k(2k)!=(1)nx2nk=0(1)k(2n+2k)!x2k=(1)n(2n2)!x0(xt)2n2sintdt (15)

    for n1 and xR. From the integral representation (15), it follows that

    (1)n(2n)!x2n[cosxn1k=0(1)k(2k)!x2k]=k=0(1)k(2k+2n2n)x2k(2k)!=2n(2n1)x2nx0(xt)2n2sintdt=2n(2n1)10(1u)2n2sin(ux)xdu>0,0<x<π (16)

    and

    [k=0(1)k(2k+2n2n)x2k(2k)!]=k=0(1)k+1(2k+2n+22n)x2k+1(2k+1)!=2n(2n1)10(1u)2n2ddx[sin(ux)x]du=2n(2n1)10(1u)2n2cos(ux)(ux)tan(ux)x2du<0,0<x<π2

    for n1, where we used the inequalities cosx>0 and xtanx<0 in x(0,π2). This means that

    Fn(x)=10(1u)2n2cos(ux)(ux)tan(ux)x2du10(1u)2n2sin(ux)xdu<0,0<x<π2

    for n1. In conclusion, the function Fn(x) for n1 is decreasing on (0,π2).

    It is known that

    sinxx=k=0(1)kx2k(2k+1)!,|x|<.

    Straightforward differentiating and simplifying give

    (sinxx)=k=0(1)k+1(2k+2)(2k+1)(2k+3)!x2k=k=0(4k+4)(4k+3)(4k+5)![(4k+1)(4k+2)(4k+5)(4k+3)x2]x4k<0,x(0,103)(0,π2].

    Therefore, a direct differentiation and simplification yield

    Fn(x)=[10(1u)2n2ddx(sin(ux)x)du10(1u)2n2sin(ux)xdu]=1[10(1u)2n2sin(ux)xdu]2[(10(1u)2n2ddx(sin(ux)x)du)210(1u)2n2d2dx2(sin(ux)x)du10(1u)2n2sin(ux)xdu]=1[10(1u)2n2sin(ux)xdu]2[(10u(1u)2n2ddx(sin(ux)ux)du)210u(1u)2n2d2dx2(sin(ux)ux)du10u(1u)2n2sin(ux)uxdu]<0

    on (0,π2) for n1. Accordingly, the function Fn(x) for n1 is concave on (0,π2). The proof of Theorem 2 is thus complete.

    Remark 3. We note that a concave function must be a logarithmically concave function, but the converse is not true. However, a logarithmically convex function must be a convex function, but the converse is not true.

    In [2,Section 4], the function R0,1(x)=R(x) defined by (3) or (10) for n=1 was proved to be decreasing on [0,π2] onto [0,16].

    Theorem 3. The even function R0,2(x) defined by (10) for the case n=2 is decreasing on [0,π2].

    Proof. For n2, direct differentiation gives

    Fn(x)F0(x)=[lnCosRn(x)](lncosx)=CosRn(x)CosRn(x)cosxsinx

    and

    [Fn(x)F0(x)]=[CosRn(x)CosRn(x)cosxsinx]=(CosRn(x)CosRn(x)[CosRn(x)]2)cosxsinxCosRn(x)CosRn(x)[CosRn(x)sinx]2.

    Taking n=2 and simplifying lead to

    [F2(x)F0(x)]=72x10[CosR2(x)sinx]2[4x(x46x2+12)+4(7x216)xcosx+16xcos(2x)4x3cos(3x)+(3x4+4x216)sinx+2(x410x2+12)sin(2x)(x22x4)(x2+2x4)sin(3x)+4sin(4x)]=72x10[CosR2(x)sinx]2k=6(1)kQ(k)x2k+1(2k+1)!=72x10[CosR2(x)sinx]2k=3[Q(2k)Q(2k+1)(4k+3)!(4k+1)!x2]Q(2k+1)(4k+3)!x4k+1,

    where

    Q(k)=42k+24(4k428k3+107k2+61k+324)32k3+(4k44k3+39k2+53k+64)22k+4(12k468k37k217k20),k6.

    From the facts that

    12k468k37k217k20=12(k6)4+220(k6)3+1361(k6)2+2923(k6)+490490,k6,4k44k3+39k2+53k+64=4(k6)4+92(k6)3+822(k6)2+3437(k6)+57825782,k6,

    and, by induction,

    42k+24(4k428k3+107k2+61k+324)32k3=16×32k[(43)2k4k428k3+107k2+61k+324108]>0,k7,

    we conclude, together with Q(6)=3871296, that Q(k)3871296 for k6.

    Let

    Q(k)=Q(2k)Q(2k+1)(4k+3)!(4k+1)!,k3.

    The inequality

    Q(k+1)>Q(k),k3

    is equivalent to

    (2k+3)(4k+7)Q(2k+1)Q(2k+2)>(2k+1)(4k+3)Q(2k)Q(2k+3)

    for k3, that is,

    Q(k)=216[73728(k2)9+1508352(k2)8+13174784(k2)7+63787136(k2)6+185680928(k2)5+329304964(k2)4+345900612(k2)3+210339955(k2)2+94117995(k2)+42704050]+27×24k{(142293×24k+9+93239336)k5+4(228963×24k+7315163)k4+(1730269×24k+577141534)k3+(590587×24k+561162271)k2+31(39633×24k+3705961)k+72(32581×16k50315)+4096(57×24k+3+847)k8+2048(24k+7540k1635)k9+32k6[24k+6(3328k3377)+1151464k+2801469]}+27×212k+7[256k(24k+545k5)+9×24k+10+12352k5+24016k4+33804k3+78984k2+84261k+33915]+16×34k[159784960(k4)9+2908177408(k4)8+31203008512(k4)7+218332899456(k4)6+1039043941024(k4)5+3394763402820(k4)4+7477586709444(k4)3+10528669786283(k4)2+8416641010791(k4)+2806689073644]+8×38k+3(8192k9+9216k8+30720k7+148608k6+179232k5+77220k4+1027780k3+2247363k2+1992375k+751032)+64k[107481088(k4)9+1948553216(k4)8+20892092672(k4)7+146632542496(k4)6+702801488840(k4)5+2321298590084(k4)4+5179870312810(k4)3+7367488384493(k4)2+5849176490661(k4)+1817172122508]+28k+334k[313600(k13)6+20524992(k13)5+540906544(k13)4+7219464756(k13)3+49643220604(k13)2+151463526603(k13)+98734469508]>0,k3.

    By virtue of the above expression for Q(k), we see that the sequence Q(k) is positive for k13. On the other hand, it is straightforward that

    Q(3)=352780960860822574080,Q(4)=249398534923066892200578048,Q(5)=61305423558540152150473900185600,Q(6)=8665426549135034591879376586377676800,Q(7)=896760345938903204692495757195177830886400,Q(8)=77679040214402381392287613848479095562672133120,Q(9)=6092826722565745296376115137707251410869680486780928,Q(10)=453017430243005416828735039714559150312936188925226332160,Q(11)=32749134150707011053456859834163170645192635726524315359872000,Q(12)=2332073744679555396812875864144181190554219273032749874293655526400.

    Consequently, the sequence Q(k) is positive for all k3. As a result, the sequence Q(k) is increasing in k3. It is immediate that Q(3)=423115=3.678. Hence, we acquire

    Q(k)423115=3.678,k3.

    Accordingly, when

    0<xπ2=1.570<423115=3.678,

    the derivative [F2(x)F0(x)] is negative, and then the derivative ratio F2(x)F0(x) is decreasing in x(0,π2].

    In [12,pp. 10-11,Theorem 1.25], a monotonicity rule for the ratio of two functions was established as follows.

    For a,bR with a<b, let p(x) and q(x) be continuous on [a,b], differentiable on (a,b), and q(x)0 on (a,b). If the ratio p(x)q(x) is increasing on (a,b), then both p(x)p(a)q(x)q(a) and p(x)p(b)q(x)q(b) are increasing in x(a,b).

    With the help of this monotonicity rule and in view of the decreasing property of the derivative ratio F2(x)F0(x) in x(0,π2], we derive that the ratio F2(x)F0(x)=R0,2(x) is decreasing in x(0,π2]. The required proof of Theorem 3 is completed.

    Remark 4. How to verify the decreasing property of the function R0,n(x) for n3 on (0,π2], of the function R1,n(x) on (0,2π), and of the function Rm,n(x) for n>m2 on (0,)? The ideas, approaches, techniques, and methods used in the proof of Theorem 3 should not be valid again, so we need to discover new ideas, approaches, techniques, and methods for verifying the decreasing property mentioned above.

    Remark 5. Let

    fα(x)=10(1u)αcos(ux)du,αR.

    Prove that the function fα(x) is positive in x(0,) if and only if α>1, while it is decreasing in x(0,) if and only if α2.

    This paper and the articles [2,13,14] are siblings, because some results in [2] have been generalized in this paper, and the results in [13,14] are about the Maclaurin power series expansions of logarithmic expressions involving normalized tails of the tangent and sine functions.

    In this paper, we presented the following results.

    1. The function Fn(x) for n0 was expanded into the Maclaurin power series expansion (11) in Theorem 1.

    2. The function Fn(x) defined by (7) for n0 was proved in Theorem 2 to be decreasing and concave on (0,π2).

    3. A new determinantal expression (13) of the Bernoulli numbers B2m for m1 was derived.

    4. The ratio R0,2(x) was proved in Theorem 3 to be decreasing in x[0,π2].

    In order to verify the decreasing property of the function R0,n(x) for n3 on (0,π2), of the function R1,n(x) on (0,2π), and of the function Rm,n(x) for n>m2 on (0,), we need new ideas, novel approaches, creative techniques, and innovative methods.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported in part by the National Natural Science Foundation of China (Grant No. 12061033).

    The authors declare there is no conflicts of interest.



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