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Integrating evolution equations using Fredholm determinants

  • We outline the construction of special functions in terms of Fredholm determinants to solve boundary value problems of the string spectral problem. Our motivation is that the string spectral problem is related to the spectral equations in Lax pairs of at least three nonlinear evolution equations from mathematical physics.

    Citation: Feride Tığlay. Integrating evolution equations using Fredholm determinants[J]. Electronic Research Archive, 2021, 29(2): 2141-2147. doi: 10.3934/era.2020109

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  • We outline the construction of special functions in terms of Fredholm determinants to solve boundary value problems of the string spectral problem. Our motivation is that the string spectral problem is related to the spectral equations in Lax pairs of at least three nonlinear evolution equations from mathematical physics.



    Our main goal is to outline how to find special functions in terms of Fredholm determinants for the string spectral problem

    fxx=λmf, (1)

    in order to write solutions of integrable PDE with associated spectral problems.

    The motivation in considering (1) is that the three integrable equations of interest to us, namely Hunter-Saxton, Camassa-Holm and μHS equations, all have spectral problems related to the string spectral problem (1).

    The μHS equation

    utxx2μ(u)ux+2uxuxx+uuxxx=0,

    where u=u(t,x), xT=R/Z,tR+ and μ(u)=Tu dx, is introduced first by Khesin, Lenells and Misiołek in [8]. The equation μHS is a well-defined PDE rather than a mixed integro-differential equation since μ(u) is constant in time and is completely determined by the initial condition.

    Several different names have been used in the literature for this equation such as μHS, μCH, generalized Hunter-Saxton equation etc. All of these names can be justified by different aspects of this equation, some closer to Camassa-Holm equation [2], [5]

    ututxx+3uux2uxuxxuuxxx=0,

    and some closer to Hunter-Saxton equation [6]

    utxx+2uxuxx+uuxxx=0,

    from mathematical physics.

    Like (CH) and (HS), the μHS equation is "integrable" in the following sense: it has a Lax pair representation, a bi-Hamiltonian structure and peaked soliton (peakon) solutions. The Lax pair given in [8] is

    ψxx=λψ(μ(u)uxx), (2)
    ψt=(12λu)ψx+12uxψ (3)

    where λ is a spectral parameter. In other words, assuming that (2)-(3) are satisfied for all nonzero λC, the relation ψtxx=ψxxt holds if and only if u(t,x) satisfies the μHS equation.

    Moreover (μHS) can be written as a Hamiltonian system mt=BiδHiδm in two compatible ways where m=μ(u)uxx is called the momentum. The Hamiltonian functionals are

    H1=12umdxandH2=(μ(u)u2+12uu2x)dx

    and the corresponding Hamiltonian operators are B1=(mx+xm) and B2=3x respectively. These two compatible Hamiltonian structures can be used to generate infinitely many conserved quantities [8]. Another manifestation of integrability is the existence of soliton-like solutions. In this respect the μHS equation resembles Camassa-Holm equation rather than the Hunter-Saxton equation, all of whose periodic solutions have finite lifespan. The μHS equation admits peaked traveling waves that are analogues of solitons [8], [14].

    Like Camassa-Holm and Hunter-Saxton equations, (μHS) is a geodesic equation with respect to a right invariant Riemannian metric. This metric is induced by the μ inner product: u,vμ=S1udxS1vdx+S1uxvxdx. All three equations, namely Camassa-Holm, Hunter-Saxton and (μHS), fit into the same Lie theoretic framework as well. They can be written as Euler-Arnold equations [1], [10] on the regular dual of the Lie algebra TeDiff(S1)=vect(S1) of smooth vector fields on the circle

    mt=adA1mm=umx2uxm,m=Au. (4)

    with the following choices of the inertia operator A:

    A={12xfor CH,μ2xfor μHS,2xfor HS. (5)

    The invariance of coadjoint orbits for these equations [18] leads to the conserved quantity

    Adγm=(mγ)(γ)2=const. (6)

    Note that, a transformation λλ changes the spectral problem (2) to the string spectral problem fxx=λmf studied in [7]. The name "string spectral problem" derives from the fact that the equation can be seen as the equation for the amplitude of free oscillations with frequency λ of a string with a mass distribution M where xM=m.

    In this section we follow closely the discussion in [3] and adapt a similar notation.

    The string equation can be reformulated as a first order system

    Vx=[01λm0]V

    where V=[ffx]. Then the isospectral deformation of the string of Zakharov-Shabat type Vt=AV where the entries of the 2×2 matrix A are functions of x,t and λ, satisfy the zero curvature condition

    (AV)x=([01λm0]V)t.

    Then we can write all entries of A in terms of one of them as

    A=[12bx+βb12bxxλmb12bx+β]

    and we also have an equation for the evolution of m:

    λmt=12bxxx+λmxb+2λmbx.

    Note that different boundary conditions impose different restrictions on b. Dirichlet boundary conditions on the string problem f(0)=f(1)=0 imply

    b(0)=b(1)=0, (7)
    bx(0)=bx(1)=0. (8)

    whereas Neumann boundary conditions fx(0)=fx(1)=0 imply

    12bxx(0)+λm(0)b(0)=0, (9)
    12bxx(1)+λm(1)b(1)=0. (10)

    If we assume that b is a linear function of λ and set b=b0+b1λ this leads to the Harry-Dym equation. For our purposes it is a lot more interesting to assume that b is a linear function of 1λ. We set b=b0+b1λ and this leads to the set of equations

    mt2mb0,xmxb0=0 (11)
    12b0,xxx+2mb1,x+mxb1=0 (12)
    b1,xxx=0. (13)

    In this case Dirichlet boundary conditions on the string problem give (CH) after a Liouville transformation [BSS]. On the other hand if we impose Neumann boundary conditions then we obtain the conditions

    b0(0)=b0(1)=0, (14)
    b1,xx(0)=b1,xx(1)=0, (15)
    12b0,xx(0)+mb1(0)=0, (16)
    12b0,xx(1)+mb1(1)=0. (17)

    which lead to (HS). One open problem is what boundary conditions lead to (μHS). If the boundary value problem for (1) that gives (μHS) has a simple spectrum than the next section outlines how to integrate (μHS).

    When the spectral problem

    fxx=λmf,0<x<1 (18)

    associated with an evolution equation has purely discrete and simple spectrum the flow for this evolution equation is a superposition of commuting individual flows.

    We construct and express the flows in terms of theta functions following McKean's recipe introduced in [15] for (CH). The breakdown of solutions can be determined using these expressions for the solutions and properties of the theta functions as in the case of (CH) [16].

    3.1. Individual flows. We consider any eigenvalue λ0 of the spectral problem (18) and its associated eigenfunction f0 normalized by

    f02=S1f20=λ0S1mf20=1. (19)

    The reciprocal of any eigenvalue H=1/λ0 is a conserved quantity and the normalization condition (19) implies that δH/δm=f20. Hence the flow is regulated by ˙m=Xm=(mD+Dm)f20.

    We apply the vector field X to the spectral equation and solve for Xf0. This gives the formula

    Xf0=λ0f0(x)x0m(y)f20(y)dy+f0(x)2. (20)

    Our immediate goal is solving for f0(ˉx) in terms of m(x) and f0(x) where the new variable ˉx is determined by ˙ˉx=f20. We consider from the first summand on the right hand side of (20) evaluated at ˉx: I(ˉx)=λ0ˉx0m(y)f20(y)dy. We differentiate I(ˉx) in t and observe that it satisfies the transport equation

    ddt(I(ˉx))=I2(ˉx)+I(ˉx).

    We solve it and get

    I(ˉx)=λ0ˉx0m(t,y)f20(t,y)dy=etλ0x0m(0,y)f20(0,y)dy1+(et1)λ0x0m(0,y)f20(0,y)dy. (21)

    Differentiating both sides of the second equality above gives

    f0(t,ˉx)=et/2ˉxf0(0,x)1+(et1)λ0x0m(0,y)f20(0,y)dy. (22)

    Let us now consider the integral of the spectral equation (18) from 0 to ˉx and use the following notation

    J(ˉx)=λ0ˉx0m(t,y)f0(t,y)dy=f0(t,x)+f0(t,0). (23)

    Applying d/dt to the first equality leads to the differential equation ddtJ(ˉx)=(12I(ˉx))J(ˉx) whose solution is

    J(ˉx)=λ0ˉx0m(t,y)f0(t,y)dy=et/2λ0x0m(0,y)f0(0,y)dy1+(et1)λ0x0m(0,y)f20(0,y)dy.

    Differentiating this equality gives, after some algebraic manipulations,

    ˉx=1+(et1)λ0x0m(0,y)f20(0,y)dy1+(et1)(λ0x0m(0,y)f20(0,y)dy+f0(0,x)f0(0,x)).

    We use the following notation for the theta functions that appear repeatedly in the above formulas

    ϑ=1+(et1)λ0x0m(0,y)f20(0,y)dy (24)

    and

    ϑ=1+(et1)(λ0x0m(0,y)f20(0,y)dy+f0(0,x)f0(0,x)). (25)

    With this notation we have f0(t,ˉx)=et/2ˉxf0(0,x)/ϑ and ˉx=ϑ2/ϑ2 hence f0(ˉx)=et/2f0(0,x)/ϑ. The theta functions ϑ and ϑ are related by the identity

    ϑ2ϑ2=(et1)2(f0(0,x)f0(0,x))22f0(0,x)f0(0,x)ϑ.

    Then we have

    ˉx=x+(et1)f20(0,x)ϑ. (26)

    Observe that ϑ=1+(et1)x0(f0(0,y))2dy never vanishes. Therefore both formulas (26) and f0(ˉx)=et/2f0(0,x)/ϑ make sense for all time.

    On the other hand, by (6), we have m(t,ˉx)=m(0,x)/ˉx2=m(0,x)ϑ4/ϑ4.

    Summary:

    ˉx=ϑ2ϑ2,m(t,ˉx)=m(0,x)ϑ4ϑ4,f0(t,ˉx)=et/2f0(0,x)ϑ.

    It is easy to check that f0(ˉx)=λ0m(ˉx)f0(ˉx). It is all set up so that f0(0,0)=0 or set the antiderivatives starting at x0 so that f0(0,x0)=0.

    Note that ϑ=1+(et1)x0(f0(0,y))2dy never vanishes. Hence the first and third formulas above always make sense. Hence it is natural to wonder whether breakdown of solutions occur when ϑ vanishes.

    Furthermore the action of the vector field X on the eigenfunctions fn corresponding to the other eigenvalues λn,n0 is determined as follows. Note that the vector field X is given by m=Xf=(mD+Dm)f20. We apply it to the spectral equation fn=λnmfn and use the identity f0fnfnf0=(λnλ0)mf0fn to obtain

    Xfn=λnf0x0mf0fn (27)

    3.2. Composite flows. We now consider the composite flow etX=etnXn with parameters tn,nZ0. Here tn denote the running time of each flow. For convenience we denote the action of tX=tnXn by a bullet . Note that the individual flows generated by the Hamiltonians Hn=1λn commute [Hi,Hj]=0 for ij.

    We implement the notation ff for the matrix [fifj:i,jZ0] and denote by t and λ the diagonal matrices [tn:nZ0] and [λn:nZ0] respectively.

    Next we construct the special functions to prove our main theorem:

    Theorem 3.1. The solution u(t,x) of μHS with initial data u(0,x) is expressed in the Lagrangian scale ˉx(t,x) as

    u(t,ˉx)=ˉx=AϑAϑϑ2. (28)

    where ϑ,ϑ and A satisfy the identity ϑ2=ϑ2ϑA+ϑA.

    Proof. Applying tX to the spectral equation and integrating twice in x leads to

    fi=tj(λifjx0mfifj+δijfi(x)/2). (29)

    In analogy with individual flows we consider the term I(ˉx)ˉx0m(t,y)f(t,y)f(t,y)dy. The solution of the resulting equation I=ItλI+12(tIIt) gives

    I=ˉx0m(t,y)f(t,y)f(t,y)dy=et/2M(1+(et1)λM)1et/2 (30)

    where M=x0m(0,y)f(0,y)f(0,y)dy is the value of I=I(ˉx) evaluated at the initial time. Note that for the inverse of the matrix 1+(et1)λM to exist we assume that ϑ:=det(1+(et1)λM) does not vanish.

    We proceed as for the individual flows and differentiate (30) to obtain

    f(ˉx)=ˉxet/2(1+MC)1f(0,x) (31)

    where C=(et1)λ. In order to achieve our goal of expressing f(ˉx) in terms of m(0,x) and f(0,x) it would be sufficient to derive a formula for ˉx in terms of these initial values. For this purpose we integrate the spectral equation from 0 to ˉx and introduce the notation J(ˉx)ˉxx0mf such that λJ=f. Applying tX to J(ˉx) leads to the equation J=ItλJ+12tJ. The solution of this equation is J(ˉx)=et/2(1+MC)1xx0m(0,y)f(0,y)dy. Then differentiating both the definition and this last formula for λJ(ˉx) in x leads to

    1ˉx=det(1+CM+(et1)f(0,x)f(0,x))det(1+CM). (32)

    We introduce the notation for the theta functions for the composite flow as

    ϑ=det(1+CM+(et1)f(0,x)f(0,x))=det(1+(et1)x0f(0,x)f(0,x))

    and

    ϑ=det(1+CM). (33)

    Note that if we set

    Q1+(et1)x0f(0,y)f(0,y)dy

    we have the identity ϑ2=ϑ2ϑA+ϑA where

    A=detQ×f(0,x).Q1(et1)f(0,x)

    and the theta functions can be written as ϑ=detQ and ϑ=detQ(1f(0,x).Q1f(0,x)). With this notation we have ϑ2/ϑ2=1(A/ϑ) hence

    ˉx=xdetQ×fQ1(et1)fϑ=xfQ1(et1)f.

    Both theta functions ϑ=detQ and ϑ=detQ(1f(0,x).Q1f(0,x)) converge [4] and ϑ is always positive.

    3.3. Breakdown of solutions. Note that the formula (28) makes sense for all times since ϑ is strictly positive. On the other hand, for the derivative of the solution u we have

    u(t,ˉx)=1ˉxˉx=2ϑϑ2ϑϑ. (34)

    Note that ϑ can not vanish when ϑ does. Therefore u(t,ˉx) blows up when ϑ vanishes.

    Since Cx0mffC is compact with finite absolute trace with respect to f1,f2=f1f2dx the condition ϑ0 means that the spectrum of Cx0mffC lies below 1 for all time t0 and xT. This requires that the associated quadratic form ζ2x0m|nζnCnfn|2 be positive for ζ0.



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