The calculation for characteristic multiplier of Hill’s equation in case q(t) < 0

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Abstract

We have given the calculation for characteristic multiplier of Hill’s equation x + q(t)x = 0 in case q(t) > 0 in the work of Jinlin Shi et al. [Appl. Math. Comput., in press (AMC8559)]. In this paper, we consider the calculation for characteristic multiplier in case q(t) < 0.

Section snippets

The discriminant of Hill’s equation in case q(t) < 0

Consider Hill’s equation:x+q(t)x=0,where q(t)  c1, q(t + π) = q(t).

If q(t) > 0, we had given the discriminant of (1.1) as follows: (see [2])Δ=2cos0πq(s)ds+n=1124n-10π0t10t2n-1cosΨ(t1t2n)j=12nq(tj)q(tj)dt2ndt1,whereΨ(t1t2n)=0πq(s)ds-2t2t1q(s)ds--2t2nt2n-1q(s)ds.If q(t) < 0, we will prove that (1.2) still holds. In this case, (1.2) can be written asΔ=2cosi0π-q(s)ds+n=1124n-10π0t10t2n-1cosiΨ(t1t2n)j=12nq(tj)q(tj)dt2ndt1,where i is imaginary unit andΨ(t1t2n)=0π-q(s)ds-2t2t1-q(s

The calculation of Δ1 and Δ2

From (1.26),Δ1=1230π0t1q(t1)q(t1)q(t2)q(t2)cosi(Φ(π)-2Φ(t1,t2))dt2dt1=cosiΦ(π)1230π0t1q(t1)q(t1)q(t2)q(t2)cos2i(Φ(t1)-Φ(t2))dt2dt1+siniΦ(π)1230π0t1q(t1)q(t1)q(t2)q(t2)sin2i(Φ(t1)-Φ(t2))dt2dt1=123cosiΦ(π)0π0t1q(t1)q(t1)q(t2)q(t2)cos2iΦ(t1)cos2iΦ(t2)dt2dt1+123cosiΦ(π)0π0t1q(t1)q(t1)q(t2)q(t2)sin2iΦ(t1)sin2iΦ(t2)dt2dt1+123siniΦ(π)0π0t1q(t1)q(t1)q(t2)q(t2)sin2iΦ(t1)cos2iΦ(t2)dt2dt1-123siniΦ(π)0π0t1q(t1)q(t1)q(t2)q(t2)cos2iΦ(t1)sin2iΦ(t2)dt2dt1=defI1+I2+I3+I4.It follows

The estimation of error

Since Δ-Δ˜=n=3Δn, so the extreme error do not exceed n=3Δn. By (1.26),Δn=124n-10π0t10t2n-1cosi(Φ(π)-2Φ(t1,t2)--2Φ(t2n-1,t2n))j=12nq(tj)q(tj)dt2ndt1.First, we estimatecosi(Φ(π)-2Φ(t1,t2)--2Φ(t2n-1,t2n))=12expΨ(t1t2n)+exp-Ψ(t1t2n),whereΨ(t1t2n)=Φ(π)-2Φ(t1,t2)--2Φ(t2n-1,t2n)=0π-q(τ)dτ-2t2t1-q(τ)dτ--2t2nt2n-1-q(τ)dτ.Noting that 0  t2n  t2n−1    t2  t1  π, we can concludeΨ(t1t2n)=0t2n-q(τ)dτ+t2n-1t2n-2-q(τ)dτ++t1π-q(τ)dτ-t2t1-q(τ)dτ++t2nt2n-1-q(τ)dτ=defα1+α2.It follows from -q

An example

We consider the differential equationx-1+12cos2t-12sin2t+116cos22tx=0.It is easily proved that x(t)=expt+18sin2t is one of its solutions whose characteristic exponent is 1, we denote the characteristic multiplier of (4.1) by ρ1, ρ2, then we can obtain ρ1 = eπ.

We denote the discriminant of (4.1) by Δ, then the characteristic multiplier ρ satisfyρ2-Δρ+1=0andρ1ρ2=1,ρ1+ρ2=Δso ρ2 = eπ, Δ = eπ + eπ  23.1838.

According to (2.13), (2.20), Δ˜ may be obtained in the following way. In the interval [0, π] choose n

References (2)

  • J. Shi, M. Lin, J. Chen, The calculation for characteristic multiplier of Hill’s equation, Appl. Math. Comput., in...
  • J. Shi

    A new form of discriminant for Hill’s equation

    Ann. Diff. Eqs.

    (1999)

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