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Exact solutions to the fractional nonlinear phenomena in fluid dynamics via the Riccati-Bernoulli sub-ODE method

  • The Riccati-Bernoulli sub-ODE method has been used in recent research to efficiently investigate the analytical solutions of a non-linear equation widely used in fluid dynamics research. By utilizing this method, exact solutions are obtained for the space-time fractional symmetric regularized long-wave equation. These results comprehensively understand the long wave equation widely used in numerous fluid dynamics and wave propagation scenarios. The approach to studying these phenomena and using conceptual representation to understand their essential characteristics opens the door to valuable insights that may help improve both the theoretical and applied aspects of fluid dynamics and similar fields. Thus, as these complex equations demonstrate, the suggested approach is a valuable tool for conducting further research into non-linear phenomena across several disciplines.

    Citation: Waleed Hamali, Abdulah A. Alghamdi. Exact solutions to the fractional nonlinear phenomena in fluid dynamics via the Riccati-Bernoulli sub-ODE method[J]. AIMS Mathematics, 2024, 9(11): 31142-31162. doi: 10.3934/math.20241501

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  • The Riccati-Bernoulli sub-ODE method has been used in recent research to efficiently investigate the analytical solutions of a non-linear equation widely used in fluid dynamics research. By utilizing this method, exact solutions are obtained for the space-time fractional symmetric regularized long-wave equation. These results comprehensively understand the long wave equation widely used in numerous fluid dynamics and wave propagation scenarios. The approach to studying these phenomena and using conceptual representation to understand their essential characteristics opens the door to valuable insights that may help improve both the theoretical and applied aspects of fluid dynamics and similar fields. Thus, as these complex equations demonstrate, the suggested approach is a valuable tool for conducting further research into non-linear phenomena across several disciplines.



    Let H and K be complex infinite-dimensional separable Hilbert spaces, and let B(H,K) (resp., C(H,K), C+(H,K)) be the set of all bounded (resp., closed, closable) operators from H to K. If K=H, we use B(H), C(H), and C+(H) as usual. The domain of TC(H) is denoted by D(T). A closed operator T is said to be left (resp., right) Fredholm if its range R(T) is closed and α(T)< (resp., β(T)<), where α(T) and β(T) denote the dimension of the null space N(T) and the quotient space H/R(T), respectively. T is said to be Fredholm if it is both left and right Fredholm. If T is a left or right Fredholm operator, then we define the index of T by ind(T)=α(T)β(T). T is called Weyl if it is a Fredholm operator of index zero. The left essential spectrum, right essential spectrum, essential spectrum, and Weyl spectrum of T are, respectively, defined by

    σle(T)={λC:TλI is not left Fredholm},σre(T)={λC:TλI is not right Fredholm},σe(T)={λC:TλI is not Fredholm},σw(T)={λC:TλI is not Weyl}.

    Iterates T2,T3, of TC(H) are defined by Tnx=T(Tn1x) for xD(Tn) (n=2,3,), where

    D(Tn)={x:x,Tx,,Tn1xD(T)}.

    It is easy to see that

    N(Tn)N(Tn+1),R(Tn+1)R(Tn).

    If n0, we follow the convention that T0=I (the identity operator on H, with D(I)=H). Then N(T0)={0} and R(T0)=H. It is also well known that if N(Tk)=N(Tk+1), then N(Tn)=N(Tk) for nk. In this case, the smallest nonnegative integer k such that N(Tk)=N(Tk+1) is called the ascent of T, and is denoted by asc(T). If no such k exists, we define asc(T)=. Similarly, if R(Tk)=R(Tk+1), then R(Tn)=R(Tk) for nk. Thus, we can analogously define the descent of T denoted by des(T), and define des(T)= if R(Tn+1) is always a proper subset of R(Tn). We call T left (resp., right) Browder if it is left (resp., right) Fredholm and asc(T)< (resp., des(T)<). The left Browder spectrum, right Browder spectrum, and Browder spectrum of T are, respectively, defined by

    σlb(T)={λC:TλI is not left Browder},σrb(T)={λC:TλI is not right Browder},σb(T)={λC:TλI is not Browder}.

    We write σ(T),σap(T), and σδ(T) for the spectrum, approximate point spectrum, and defect spectrum of T, respectively.

    For the spectrum and Browder spectrum, the perturbations of 2×2 bounded upper triangular operator matrix MC=[AC0B], acting on a Hilbert or Banach space, have been studied by numerous authors (see, e.g., [5,8,9,12,13,18]). However, the characterizations of the unbounded case are still unknown. In this paper, the unbounded operator matrix

    TB=[AB0D]:D(A)D(D)HHHH

    is considered, where AC(H), DC(K), and BC+D(K,H)={BC+(K,H):D(B)D(D)}. It is obvious that TB is a closed operator matrix. Our main goal is to present some sufficient and necessary conditions for TB to be Browder (resp., invertible) for some closable operator BC+D(K,H) applying a space decomposition technique. Further, the sufficient and necessary condition, which is completely described by the diagonal operators A and D, for

    σ(TB)=σ(A)σ(D) (1.1)

    is characterized, where σ is the Browder spectrum (resp., spectrum). This is an extension of the results from [3,7,10,12]. In addition, this result is applied to a Hamiltonian operator matrix from elasticity theory.

    Definition 1.1. [4] Let T:D(T)HK be a densely defined closed operator. If there is an operator T:D(T)KH such that D(T)=R(T)R(T),N(T)=R(T) and

    TTx=P¯R(T)x,xD(T),TTy=P¯R(T)y,yD(T)

    then T is called the maximal Tseng inverse of T.

    For the proof of the main results in the next sections, we need the following lemmas:

    Lemma 1.1. [2] Let TB=[AB0D]:D(A)D(D)HHHH be a closed operator matrix such that A,D are closed operators with dense domains and B is a closable operator. Then,

    (1) σle(A)σre(D)σe(TB)σe(A)σe(D).

    (2) σw(TB)σw(A)σw(D).

    (3) σb(TB)σb(A)σb(D).

    Lemma 1.2. [17] Let TB=[AB0D]:D(A)D(D)HHHH be a closed operator matrix such that A,D are closed operators with dense domains and B is a closable operator. Then,

    (1) asc(A)asc(TB)asc(A)+asc(D),

    (2) des(D)des(TB)des(A)+des(D).

    Lemma 1.3. [16] Let TC(H). Suppose that either α(T) or β(T) is finite, and that asc(T) is finite. Then, α(T)β(T).

    Lemma 1.4. [6] Let TC(H). Then, the following inequalities hold for any non negative integer k:

    (1) α(Tk)asc(TB)α(T);

    (2) β(Tk)des(TB)β(T).

    Lemma 1.5. [11] Suppose that the closed operator T is a Fredholm operator and B is T-compact. Then,

    (1) T+B is a Fredholm operator,

    (2) ind(T+B)=ind(T).

    Lemma 1.6. [16] Let TC(H). Suppose that asc(T) is finite, and that α(T)=β(T)<. Then, des(T)=asc(T).

    In this section, some sufficient and necessary conditions for TB to be Browder for some closable operator B with D(B)D(D) are given and the set BC+D(K,H)σb(TB) is estimated.

    Theorem 2.1. Let AC(H) and DC(K) be given operators with dense domains. If β(A)=, then TB is left Browder for some BC+D(K,H) if and only if A is left Browder.

    Proof. The necessity is obvious by Lemmas 1.1–1.3.

    Now we verify the sufficiency. Let p=asc(A)<, then α(Ap)pα(A)< from Lemma 1.4. Thus, we get dim(R(A)+N(Ap))=, since β(A)=. It follows that there are two infinite dimensional subspaces Δ1 and Δ2 of (R(A)+N(Ap)) such that (R(A)+N(Ap))=Δ1Δ2. Define an operator B:KH by

    B=TB=[0U0]:K[R(A)+N(Ap)Δ1Δ2]

    where U:KΔ1 is a unitary operator. Then, we can claim that TB is left Browder.

    First we prove that N(TB)=N(A){0}, which implies α(TB)<. It is enough to verify that N(TB)N(A){0}. Let (xy)N(TB), then Ax+By=0 and Dy=0. Thus, Ax=ByR(A)Δ1={0}, and hence xN(A) and By=0. From the definition of B, we get y=0, which means N(TB)N(A){0}.

    Next, we check that R(TB) is closed. We only need to check that the range of [U0D] is closed since R(A) is closed. Suppose that

    [U0D]yn(y10y3)(n)

    where ynD(D). Then, Uyny1(n) and Dyny3(n). Thus, we have that ynU1y1(n) from U is a unitary operator. We also obtain U1y1D(D) and DU1y1=y3 since D is a closed operator. Hence,

    (y10y3)=[U0D]U1y1

    which implies R(TB) is closed.

    Last, we verify that N(TpB)=N(Tp+1B0), which induces asc(TB)<. Let (xy)N(Tp+1B), then

    {Ap+1x+ApBy+Ap1BDy++ABDp1y+BDpy=0,Dp+1y=0.

    Thus, DpyN(D)R(Dp) and BDpy(R(A)+N(Ap))R(A). This implies that Ap+1x+ApBy+Ap1BDy++ABDp1y=BDpyR(A)(R(A)+R(Ap))R(A)R(A)={0}, and then

    {Ap+1x+ApBy+Ap1BDy++ABDp1y=0,BDpy=0.

    We can obtain Dpy=0 by the definition of B. According to Ap+1x+ApBy++ABDp1y=A(Apx+Ap1By++BDp1y)=0, we have

    Apx+Ap1By++BDp1yN(A).

    Let x1:=Apx+Ap1By++BDp1y, then

    Apx+Ap1By++ABDp2yx1+BDp1y=0.

    It follows that Apx+Ap1By++ABDp2yx1= BDp1y(R(A)+N(A))(R(A)+N(Ap))(R(A)+N(Ap))(R(A)+N(Ap))={0}. Then, Apx+Ap1By++ABDp2yx1=BDp1y=0, i.e.,

    {Apx+Ap1By+Ap1BDy++ABDp2y=x1,BDp1y=0.

    Let x2:=Ap1x+Ap2By+Ap1BDy++BDp2y, then x2N(A2), as Ax2=x1 and x1N(A). This implies Ap1x+Ap2By+Ap3BDy++BDp3yx2=BDp2y(R(A)+N(A2))(R(A)+N(Ap))(R(A)+N(Ap))(R(A)+N(Ap))={0}, and then

    {Ap1x+Ap2By+Ap3BDy++BDp3y=x2,BDp2y=0.

    Continuing this process, there is xpN(Ap) such that

    {Ax+By=xp,By=0.

    Thus, xN(Ap+1)=N(Ap). This induces that (xy)N(TpB), which means that N(TpB)=N(Tp+1B).

    Theorem 2.2. Let AC(H) and DC(K) be given operators with dense domains. If α(D)=, then TB is right Browder for some BC+D(K,H) if and only if D is right Browder.

    Proof. If TB is right Browder for some BC+D(K,H), then D is right Fredholm and des(D)< by Lemmas 1.1 and 1.2. According to the assumption α(D)=, we can obtain ind(D)0, which means D is right Browder.

    Now, we verify the reverse implication. Denote q=des(D). By Lemma 1.4, we have β(Dq)qβ(D)<, and then R(Dq) is closed. Thus, K=R(Dq)R(Dq). This includes N(D)=[N(D)R(Dq)][N(D)R(Dq)]. According to the assumption dim(N(D))= and dim(N(D)R(Dq))<, we know that dim(N(D)R(Dq))=. Then, there exist two infinite dimensional subspaces Ω1 and Ω2 such that N(D)R(Dq)=Ω1Ω2. Define the operator B0 by

    B=[U000]:[Ω1Ω2N(D)R(Dq)N(D)]H

    where U:Ω1H is a unitary operator. Then, we obtain that R(TB)=[HR(D)] is closed, β(TB)=β(D)<, and α(TB)=α(A)+dimΩ2+dim(N(D)R(Dq))=. Thus, we also get des(TB)< by des(D)=q<. This means TB is right Browder.

    Remark 2.1. Let AC(H) and DC(K) be given operators with dense domains. If TB is Browder for some BC+D(K,H), then α(D)< if and only if β(A)<. In fact, if TB is Browder for some BC+D(K,H), then A is left Fredholm with finite ascent and D is right Fredholm with finite descent, by Lemmas 1.1 and 1.2. It follows that TB admits the following representation:

    TB=[A1B11B120B21B2200D1000]:[D(A)N(D)N(D)D(A)][R(A)R(A)R(D)R(D)]

    where D1:D(D)R(D) is invertible and A1:D(A)R(A1)(=R(A)) is an operator with closed range. Let A1:R(A)H be the maximal Tseng inverse of A1, then A1A1=IR(A). Set P=[I0B12D1100IB22D11000I0000I], Q=[IA1B1100I000I], then

    PTBQ=[A1000B21000D1000]:=TB21.

    Thus, ind(TB)=ind(TB21) since P and Q are both injective. Assume that α(D)< (resp., β(A)<), then B21 is a compact operator. By Lemma 1.5, we also obtain ind(TB)=ind(A)+ind(D)=0. Therefore, α(D)< if and only if β(A)<.

    The next result is given under the hypothesis that β(A) and α(D) are both finite.

    Theorem 2.3. Let AC(H) and DC(K) be given operators with dense domains. Suppose that β(A)< and α(D)<. Then, TB is Browder for some BC+D(K,H) if and only if A is left Browder, D is right Browder, and α(A)+α(D)=β(A)+β(D).

    Proof. Necessity. If TB is Browder for some BC+D(K,H), then A is left Fredholm, D is right Fredholm, asc(A)<, and des(D)< by Lemmas 1.1 and 1.2. This means A is left Browder and D is right Browder. We can also get α(A)+α(D)=β(A)+β(D) by Remark 2.1.

    Sufficiency. There are two cases to consider.

    Case Ⅰ: Assume that α(D)β(A), then we can define a unitary operator J1:N(D)M1, where M1R(A) and dim(M1)=α(D). Set

    B=[00J10]:[N(D)N(D)][R(A)R(A)].

    We have α(TB)=α(A) and β(TB)=β(D)+β(A)α(D). Then, asc(TB)=asc(A)< and α(TB)=β(TB)< by the assumption α(A)+α(D)=β(A)+β(D). Thus, des(TB)< from Lemma 1.6, which means TB is Browder.

    Case Ⅱ: Suppose that α(D)>β(A). Define a unitary operator J2:M2R(A), where M2N(D) and dim(M2)=β(A). Let

    B=[00J20]:[M2M2][R(A)R(A)].

    We get α(TB)=α(A)+α(D)β(A) and β(TB)=β(D). Then, des(TB)=des(D)< and α(TB)=β(TB)< by the assumption α(A)+α(D)=β(A)+β(D). Hence, α(TB)=β(TB)< and asc(TB)=des(TB)< by [1, Proposition 3.1]. Therefore, we get asc(TB)=asc(TB)=des(TB)< by Lemma 1.6. This means TB is Browder.

    According to Theorems 2.1–2.3 and Remark 2.1, we obtain the theorem below, which is the main result of this section.

    Theorem 2.4. Let AC(H) and DC(K) be given operators with dense domains. Then, TB is Browder for some BC+D(K,H) if and only if A is left Browder, D is right Browder, and α(A)+α(D)=β(A)+β(D).

    Immediately, we get the following corollary, in which the set BC+D(K,H)σb(TB) is estimated:

    Corollary 2.1. Let AC(H) and DC(K) be given operators with dense domains. Then,

    BC+D(K,H)σb(TB)=σlb(A)σrb(D){λC:α(AλI)+α(DλI)
    β(AλI)+β(DλI)}.

    Remark 2.2. Theorem 2.4 and Corollary 2.1 extend the results of bounded case in [5,18].

    Theorem 2.5. Let AC(H) and DC(K) be given operators with dense domains. Then,

    σb(TB)=σb(A)σb(D) (2.1)

    for every BC+D(K,H), if and only if λσasc(D)(σlb(A)σrb(D)) implies α(AλI)+α(DλI)β(AλI)+β(DλI), where σasc(D)={λC:asc(DλI)=}.

    Proof. Equation (2.1) holds for every BC+D(K,H) if and only if

    σasc(D)(σlb(A)σrb(D))σb(TB)

    by [2, Theorem 3.3]. That is,

    σasc(D)(σlb(A)σrb(D))BC+D(K,H)σb(TB).

    This induces

    σasc(D)(σlb(A)σrb(D)){λC:α(AλI)+α(DλI)β(AλI)+β(DλI)}.

    Theorem 3.1. Let AC(H) and DC(K) with dense domains. Then, TB is right invertible for some BC+D(K,H) if and only if D is right invertible and α(D)β(A).

    Proof. Necessity. If TB is right invertible for some BC+D(K,H), then D is right invertible and TB has the following representation:

    TB=[A1B11B120B21B2200D1]:[D(A)N(D)N(D)D(A)][R(A)R(A)K].

    Obviously, D1:N(D)D(A)K is a bijection. Set P=[I0B12D110IB22D1100I], then

    PTB=[A1B1100B21000D1]:=ˆTB.

    Thus, R(ˆTB)=HK since P is bijective, and hence α(D)β(A).

    Sufficiency. There are two cases to consider.

    Case Ⅰ: Let α(D)=, then we can define a Unitary operator U:N(D)H. Set

    B=[U0]:[N(D)N(D)]H.

    Clearly, TB is right invertible.

    Cases Ⅱ: Assume that α(D)<, then we can define the right invertible operator R:N(D)R(A). Set

    B=[00R0]:[N(D)N(D)][R(A)R(A)].

    Clearly, TB is right invertible.

    From the above theorem and Theorems 5.2.1, 5.2.3 of [15], we can obtain the next results immediately.

    Theorem 3.2. Let AC(H) and DC(K) with dense domains. Then, TB is invertible for some BC+D(K,H) if and only if A is left invertible, D is right invertible, and α(D)β(A).

    Corollary 3.1. Let AC(H) and DC(K) with dense domains. Then,

    BC+D(K,H)σ(TB)=σap(A)σδ(D){λC:α(DλI)β(AλI)}.

    Remark 3.1. Theorem 3.2 and Corollary 3.1 are also valid for bounded operator matrix BB(K,H). These conclusions extend the results in [8,9,12,13].

    From Corollary 3.1, we have the following theorem, which extends the results of [3,7,10,12].

    Theorem 3.3. Let AC(H) and DC(K) be given operators with dense domains. Then,

    σ(TB)=σ(A)σ(D) (3.1)

    for every BC+D(K,H) if and only if

    λσp,1(D)σr,1(A)α(DλI)β(AλI)

    where σp,1(D)={λC:N(DλI){0},N(DλI)=K} and σr,1(A)={λC:N(AλI)={0}, R(AλI)=¯R(AλI)H}.

    Proof. Equation (3.1) holds for every BC+D(K,H) if and only if

    σp,1(D)σr,1(A)σ(TB)

    by [7, Corollary 2]. That is,

    σp,1(D)σr,1(A)BC+D(K,H)σ(TB).

    This induces

    σp,1(D)σr,1(A){λC:α(DλI)β(AλI)}

    by Corollary 3.1, σr,1(A)σap(A)=, and σp,1(D)σδ(D)=.

    Applying Theorems 2.5 and 3.3 for the Hamiltonian operator matrix, we obtain the next result.

    Theorem 3.4. Let H=[AB0A]:D(A)D(A)HKHK be a Hamiltonian operator matrix, where A is a densely defined closed operator and B is a self adjoint operator. Then, the following assertions hold:

    (1) For every CC+A(K,H),

    σb(H)=σb(A)σb(A) (3.2)

    if and only if

    λσasc(A)(σlb(A)σrb(A))α(AλI)+α(A+¯λI)β(AλI)+β(A+¯λI).

    (2) For every CC+A(K,H),

    σ(H)=σ(A)σ(A) (3.3)

    if and only if

    λ(σr,1(A)σr,1(A)){λC:Reλ=0}β(A+¯λI)β(AλI).

    In particular, if σr,1(A) does not include symmetric points about the imaginary axis, then (3.3) holds.

    Example 3.1. Consider the plate bending equation in the domain {(x,y):0<x<1,0<y<1}

    D(2x2+2y2)2ω=0

    with boundary conditions

    ω(x,0)=ω(x,1)=0,
    2ωx2+2ωy2=0,y=0,1.

    Set

    θ=ωx,q=D(3ωx3+3ωy3),m=D(2ωx2+2ωy2)

    then the equation can be written as the Hamiltonian system (see [19])

    x(ωθqm)=[01002y2001D0002y20010](ωθqm)

    and the corresponding Hamiltonian operator matrix is given by

    H=[0100d2dy2001D000d2dy20010]=[AB0A]

    with domain D(A)D(A)HH, where H=L2(0,1)L2(0,1), A=AC[0,1], and

    A=[01d2dy20],B=[0001D],
    D(A)={(ωθ)H:ω(0)=ω(1)=0,ωA,ωH}.

    By a simple calculation, we get σr,1(A)= and σ(A)={kπ:k=±1,±2,...}. By Theorem 3.4, we get

    σ(H)=σ(A)σ(A)={kπ:k=±1,±2,...}.

    On the other hand, we can easily calculate that

    σ(H)={kπ:k=±1,±2,...}=σ(A)σ(A).

    Certain spectral and Browder spectral properties of closed operator matrix TB=[AB0D] in a Hilbert space are considered in this paper. Specifically, some sufficient and necessary conditions are described under which TB is Browder (resp., invertible) for some closable operator B with D(B)D(D). Additionally, a sufficient and necessary condition is obtained under which the Browder spectrum (resp., spectrum) of TB coincides with the union of the Browder spectrum (resp., spectrum) of its diagonal entries. These results also hold for bounded operator matrix. As an application, the corresponding properties of the Hamiltonian operator matrix from elasticity theory are given.

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

    We are grateful to the referees for their valuable comments on this paper.

    This work is supported by the Basic Scientific Research Funds of Subordinate Universities of Inner Mongolia (Grant No. ZSLJ202213).

    No potential conflict of interest was reported by the authors.



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