Approximations for period-1 rotation of vertically and horizontally excited parametric pendulum

Abstract

We obtain analytical approximations for period-1 rotations of both vertically and horizontally excited pendulum using Galerkin projections with elliptic functions (GP), elliptic averaging (EA), method of multiple scales (MMS) and harmonic balance (HB). The application of GP and EA has been extended to parametrically excited pendulum for the first time in this paper, while the results from MMS and HB have been adapted from the existing literature. We compare these approximations with the numerical solution to ascertain their accuracy and identify the correct approximate solution to be used for determining other properties like stability of the solution. This comparison has been made for two different forcing frequencies: one closer to the natural frequency of the pendulum while the other at a higher frequency. We find that the appearance of the period-1 rotation at smaller amplitudes of forcing as a saddle-node bifurcation is best captured by GP and EA, but the approximation for the initial angular displacement and velocity as well as the root- mean-square error over the entire time period using GP and EA deteriorates at larger forcing amplitudes. In the large amplitude regime, HB with one-term approximation gives the best result for lower forcing frequency, while MMS results in the best approximation for the higher frequency. Inclusion of more terms in the harmonic balance approximation leads to an increased accuracy in the entire frequency and amplitude range. This observation holds for both vertical and horizontal excitation. We have also used these approximations to ascertain the stability of period-1 rotation and find that the HB approximation results in the most consistent prediction of stable parameter regimes. Hence, a multi-term HB analysis, which is the simplest and straight forward technique among the methods studied in this paper, leads to the best approximation for period-1 rotation of parametric pendula.

Appendices

Appendix 1: Evaluation of integrals arising in Galerkin projection using elliptic functions

We have outlined here the procedure for carrying out the Galerkin projection using elliptic functions and have also performed the evaluation of a representative integral using contour integration in detail.

1.1 Vertical excitation

To get Eq. (4.8), we multiply both sides of Eq. (4.7) with $\text {sn}\, \tau \, \text {cn}\, \tau$ and integrate w.r.t. $\tau$ from $\tau =0$ to 2K. This yields

$$\begin{aligned}&\left[ 1-\left( \frac{\omega }{\omega _0}\right) ^2\right] \underbrace{\int _0^{2K}(\text {sn}\, \tau \, \text {cn}\, \tau )^2 \mathrm{d}\tau }_{\text {First Integral}}\nonumber \\&\qquad + \left( \frac{\beta \, \omega }{k \omega _0}\right) \underbrace{\int _0^{2K} \text {sn}\, \tau \, \text {cn}\, \tau \, \text {dn}\, \tau \mathrm{d}\tau }_{\text {Second Integral}}\nonumber \\&\quad = p\cos \phi \, \underbrace{\int _0^{2K} (\text {sn}\, \tau \, \text {cn}\, \tau )^2 \cos \dfrac{\pi \tau }{K} \mathrm{d}\tau }_{\text {Third Integral}} \nonumber \\&\qquad +\, p \sin \phi \, \underbrace{\int _0^{2K} (\text {sn}\, \tau \, \text {cn}\, \tau )^2 \sin \dfrac{\pi \tau }{K} \mathrm{d}\tau }_{\text {Fourth Integral}}. \end{aligned}$$

(8.1)

There are four integrals in Eq. (8.1). The first integral can be directly taken from the handbook [43] and is

$$\begin{aligned}&\int _0^{2K}(\text {sn}\, \tau \, \text {cn}\, \tau )^2 \mathrm{d}\tau \nonumber \\&\quad = \dfrac{2}{3k^4}\left[ (2-k^2)E-2(1-k^2)K\right] . \end{aligned}$$

(8.2)

Using the integration identity $\displaystyle \int _0^{2a_1}h(x) \mathrm{d}x=\int _0^{a_1} h(x) \mathrm{d}x + \int _0^{a_1} h(2a_1-x) \mathrm{d}x$ for a real function h(x) with $a_1$ as a real number, and the properties of the Jacobi elliptic functions [43, 44], one can easily show that

$$\begin{aligned}&\int _0^{2K} \text {sn}\, \tau \, \text {cn}\, \tau \, \text {dn}\, \tau \mathrm{d}\tau = 0, \text {and} \end{aligned}$$

(8.3)

$$\begin{aligned}&\int _0^{2K} (\text {sn}\, \tau \, \text {cn}\, \tau )^2 \sin \dfrac{\pi \tau }{K} \mathrm{d}\tau = 0. \end{aligned}$$

(8.4)

We are left with evaluating the third integral in Eq. (8.1), i.e., $\displaystyle I=\int _0^{2K} (\text {sn}\, \tau \, \text {cn}\, \tau )^2 \cos \dfrac{\pi \tau }{K} \mathrm{d}\tau$ which is not straightforward. We will use the method of contour integral for the same. Toward this end, we introduce a complex variable z in place of the real variable $\tau$. With the complex variable z, our integrand has some poles in the complex plane which can be exploited to evaluate the desired integral on the real axis. The contour we use for integration is shown in Fig. 22.

Our integrand $(\text {sn}\, z \, \text {cn}\, z)^2 \cos \dfrac{\pi z}{K}$ has two simple poles at $z=iK^{\prime }$ (which lies on $C_1$) and $z=2K+i2K^{\prime }$ (which lies on $C_3$), where $i=\sqrt{-1}$. These two simple poles will contribute half of their respective residues to the evaluation of the integral over the contour. The residues⁴ at the two poles are the same and are given by

$$\begin{aligned} \dfrac{i\pi (4K^2-2k^2K^2-\pi ^2)}{6k^4K^3} \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) . \end{aligned}$$

Using the residue theorem [35], we have

$$\begin{aligned}&\text {P.V. of} \left( \int _{C_1}+\int _{C_2}+\int _{C_3}+\int _{C_4}\right) (\text {sn}\, z \, \text {cn}\, z)^2 \cos \dfrac{\pi z}{K} \mathrm{d}z \nonumber \\&\quad = \dfrac{\pi ^2(2k^2K^2-4K^2+\pi ^2)}{3k^4K^3} \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) , \end{aligned}$$

(8.5)

where P.V. stands for “Principal Value.” Using the periodicity of Jacobi elliptic functions, one can easily show that

$$\begin{aligned} \text {P.V. of}\left( \int _{C_1}+\int _{C_3}\right) (\text {sn}\, z \, \text {cn}\, z)^2 \cos \dfrac{\pi z}{K} \mathrm{d}z =0. \end{aligned}$$

By definition, we have $\displaystyle \int _{C_2}(\text {sn}\, z \, \text {cn}\, z)^2 \cos \dfrac{\pi z}{K} \mathrm{d}z = I$. To evaluate the integral along $C_4$, we make a transformation $z=u+2iK^{\prime }$ with u varying from 2K to 0. With this transformation and again using the periodicity of the Jacobi elliptic functions, we can evaluate

Substituting the above integrals into Eq. (8.5) gives us

$$\begin{aligned}&I\left( 1- \text {cosh}\dfrac{2 \pi K^{\prime }}{K}\right) \nonumber \\&\quad =\dfrac{\pi ^2(2k^2K^2-4K^2+\pi ^2)}{3k^4K^3} \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) , \nonumber \\&\implies \quad I = \int _0^{2K} (\text {sn}\, \tau \, \text {cn}\, \tau )^2 \cos \dfrac{\pi \tau }{K} \mathrm{d}\tau \nonumber \\&\quad = \dfrac{\pi ^2(4K^2-2k^2K^2-\pi ^2)}{6k^4K^3 \, \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) }. \end{aligned}$$

(8.6)

Equation (4.8) is obtained by substituting Eqs. (8.2)–(8.4) and (8.6) in Eq. (8.1).

Fig. 22

The chosen contour for evaluating the desired integral

Similarly for obtaining Eq. (4.9), we multiply both sides of Eq. (4.7) with $\text {dn}\, \tau$ and integrate w.r.t. $\tau$ from 0 to 2K. This results in

$$\begin{aligned}&\left[ 1-\left( \frac{\omega }{\omega _0}\right) ^2\right] \underbrace{\int _0^{2K}\text {sn}\, \tau \, \text {cn}\, \tau \, \text {dn} \, \tau \mathrm{d}\tau }_{\text {First Integral}}\nonumber \\&\qquad + \left( \frac{\beta \, \omega }{k \omega _0}\right) \underbrace{\int _0^{2K} \text {dn}^2\, \tau \mathrm{d}\tau }_{\text {Second Integral}}\nonumber \\&\quad = p\cos \phi \, \underbrace{\int _0^{2K} \text {sn}\, \tau \, \text {cn} \, \tau \, \text {dn} \tau \, \cos \dfrac{\pi \tau }{K} \mathrm{d}\tau }_{\text {Third Integral}} \nonumber \\&\qquad +\, p \sin \phi \, \underbrace{\int _0^{2K} \text {sn}\, \tau \, \text {cn} \, \tau \, \text {dn} \tau \, \sin \dfrac{\pi \tau }{K} \mathrm{d}\tau }_{\text {Fourth Integral}}. \end{aligned}$$

(8.7)

The first integral of Eq. (8.7) has already been shown to be zero in Eq. (8.3). The second integral of Eq. (8.7) can be found in the handbook [43].

$$\begin{aligned} \int _0^{2K} \text {dn}^2\, \tau \mathrm{d}\tau = 2E. \end{aligned}$$

(8.8)

Using the same integral identity as before and the properties of the Jacobi elliptic functions [43, 44], one can easily show that

$$\begin{aligned} \int _0^{2K}\text {sn}\, \tau \, \text {cn} \, \tau \, \text {dn} \tau \, \cos \dfrac{\pi \tau }{K} \mathrm{d}\tau = 0, \end{aligned}$$

(8.9)

i.e., the third integral vanishes. For the evaluation of the fourth integral, we again use contour integration with the same contour as shown in Fig. 22 to get

$$\begin{aligned}&\int _0^{2K}\text {sn}\, \tau \, \text {cn} \, \tau \, \text {dn} \tau \, \sin \dfrac{\pi \tau }{K} \mathrm{d}\tau \nonumber \\&\quad = \dfrac{\pi ^3}{2k^2 K^2 \, \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) }. \end{aligned}$$

(8.10)

1.2 Horizontal excitation

The relevant equation for getting Eq. (5.5) is

$$\begin{aligned}&2\left[ 1-\left( \frac{\omega }{\omega _0}\right) ^2\right] \underbrace{\int _0^{2K}(\text {sn}\, \tau \, \text {cn}\, \tau )^2 \mathrm{d}\tau }_{\text {First Integral}} \nonumber \\&\qquad +\, 2\left( \frac{\beta \, \omega }{k \omega _0}\right) \underbrace{\int _0^{2K}\text {sn}\, \tau \, \text {cn}\, \tau \, \text {dn}\, \tau \, \mathrm{d}\tau }_{\text {Second Integral}}\nonumber \\&\quad = -p \underbrace{\int _0^{2K}\text {sn}\, \tau \, \text {cn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \mathrm{d}\tau }_{\text {Third Integral}}\nonumber \\&\qquad +\,2p \underbrace{\int _0^{2K}\text {sn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \, \text {cn}^3\, \tau \, \mathrm{d}\tau }_{\text {Fourth Integral}}. \end{aligned}$$

(8.11)

The first and second integrals of Eq. (8.11) have already been evaluated in Eqs. (8.2) and (8.3), respectively. The third and fourth integrals can be evaluated using the method of contour integral by taking the same contour as shown in Fig. 22. The respective results are

$$\begin{aligned}&\int _0^{2K}\text {sn}\, \tau \, \text {cn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \mathrm{d}\tau \nonumber \\&\quad = \dfrac{\pi ^2 \sin \phi }{k^2 K \, \text {cosh}\left( \dfrac{\pi K^{\prime }}{K}\right) } \text {and} \end{aligned}$$

(8.12)

$$\begin{aligned}&\int _0^{2K}\text {sn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \, \text {cn}^3\, \tau \, \mathrm{d}\tau \nonumber \\&\quad = \dfrac{\pi ^2(5K^2k^2-4K^2+\pi ^2) \sin \phi }{6k^4 K^3 \, \text {cosh}\left( \dfrac{\pi K^{\prime }}{K}\right) }. \end{aligned}$$

(8.13)

Similarly, to get Eq. (5.6), we need to consider the equation

$$\begin{aligned}&2\left[ 1-\left( \frac{\omega }{\omega _0}\right) ^2\right] \underbrace{\int _0^{2K}\text {sn}\, \tau \, \text {cn}\, \tau \text {dn}\, \tau \, \mathrm{d}\tau }_{\text {First Integral}} \nonumber \\&\quad +\,2\left( \frac{\beta \, \omega }{k \omega _0}\right) \underbrace{\int _0^{2K} \text {dn}^2\, \tau \, \mathrm{d}\tau }_{\text {Second Integral}}\nonumber \\&\quad = -p \underbrace{\int _0^{2K} \text {dn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \mathrm{d}\tau }_{\text {Third Integral}}\nonumber \\&\qquad +\,2p \underbrace{\int _0^{2K}\text {dn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \, \text {cn}^2\, \tau \, \mathrm{d}\tau }_{\text {Fourth Integral}}. \end{aligned}$$

(8.14)

Again, the first and second integrals of Eq. (8.14) are given in Eqs. (8.3) and (8.8), respectively. The third and fourth integrals evaluated using contour integral are

$$\begin{aligned}&\int _0^{2K} \text {dn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \mathrm{d}\tau = \dfrac{\pi \cos \phi }{\text {cosh}\left( \dfrac{\pi K^{\prime }}{K}\right) } \text {and}\nonumber \\ \end{aligned}$$

(8.15)

$$\begin{aligned}&\int _0^{2K}\text {dn}\, \tau \, \cos (k\omega _0 \tau - \phi ) \, \text {cn}^2\, \tau \, \mathrm{d}\tau \nonumber \\&\quad = \dfrac{\pi (\pi ^2+K^2 k^2) \cos \phi }{2k^2 K^2 \, \text {cosh}\left( \dfrac{\pi K^{\prime }}{K}\right) }. \end{aligned}$$

(8.16)

Appendix 2: Evaluation of integrals arising in elliptic averaging

We have outlined here the procedure for evaluating the integrals for carrying out the elliptic averaging. The subscript “av” has been omitted to simplify the presentation.

1.1 Vertical excitation

Integrating the r.h.s. of Eq. (4.12) w.r.t. t from 0 to $T=\dfrac{2\pi }{\omega }$ and dividing it by T leads to

$$\begin{aligned} {\dot{k}}= & {} \dfrac{\epsilon \beta _1 k}{T} \int _0^\mathrm{T} \text {dn}^2\, v \, \mathrm{d}t \nonumber \\&- \dfrac{\epsilon k^2 p_1}{T} \int _0^\mathrm{T} \text {sn}\, v \, \text {cn}\, v \, \text {dn} \, v \cos \omega t \, \mathrm{d}t, \nonumber \\ \implies \quad {\dot{k}}= & {} \dfrac{\epsilon \beta _1 k}{2K} \int _{\Delta }^{\Delta +2K} \text {dn}^2\, v \, \mathrm{d}v \nonumber \\&- \dfrac{\epsilon k^2 p_1}{2K} \int _{\Delta }^{\Delta +2K} \text {sn}\, v \, \text {cn}\, v \, \text {dn} \, v \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v, \nonumber \\ \implies \quad {\dot{k}}= & {} \dfrac{\epsilon \beta _1 k}{2K} \underbrace{\int _0^{2K} \text {dn}^2\, v \, \mathrm{d}v}_{\text {First Integral}} \nonumber \\&- \dfrac{\epsilon k^2 p_1}{2K} \underbrace{\int _0^{2K} \text {sn}\, v \, \text {cn}\, v \, \text {dn} \, v \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Second Integral}},\nonumber \\ \end{aligned}$$

(9.1)

where $\Delta =\dfrac{K \phi }{\pi }$. Since the integrands are periodic with period $=2K$, changing the limits of integration as done in Eq. (9.1) does not change the value of the integrals.

The first integral of Eq. (9.1) has already been evaluated in Eq. (8.8). The second integral can be evaluated using contour integral with the same contour as shown in Fig. 22. The result is

$$\begin{aligned}&\int _0^{2K}\text {sn}\, v \, \text {cn} \, v \, \text {dn} v \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \mathrm{d}v \nonumber \\&\quad = \dfrac{\pi ^3 \sin \phi }{2k^2 K^2 \, \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) }. \end{aligned}$$

(9.2)

Similarly, integrating the r.h.s. of Eq. (4.13) w.r.t. t from 0 to T and dividing by T yields

$$\begin{aligned} {\dot{\phi }}&= \dfrac{\pi }{kK} -\omega + \dfrac{\epsilon \pi k}{{k^{\prime }}^2 K T} \Bigg [ -\beta _1 k \int _0^\mathrm{T} \text {sn}\, v \, \text {cn} \, v \, \text {dn} \, v \, \mathrm{d}t \nonumber \\&\quad + \dfrac{\beta _1}{k}\int _0^\mathrm{T} Z(v) \, \text {dn}^2 v \, \mathrm{d}t \nonumber \\&\quad + k^2 p_1 \int _0^\mathrm{T} (\text {sn}\, v \, \text {cn} \, v)^2 \cos \omega t \, \mathrm{d}t \nonumber \\&\quad - p_1 \int _0^\mathrm{T} \text {sn}\, v \, \text {cn} \, v \, \text {dn}\, v \, Z(v) \cos \omega t \, \mathrm{d}t \Bigg ], \nonumber \\ \implies {\dot{\phi }}&= \dfrac{\pi }{kK} -\omega + \dfrac{\epsilon \pi k}{{2k^{\prime }}^2 K^2} \Bigg [ -\beta _1 k \int _{\Delta }^{\Delta +2K} \text {sn}\, v \, \text {cn} \, v \, \text {dn} \, v \, \mathrm{d}v \nonumber \\&\quad + \dfrac{\beta _1}{k}\int _{\Delta }^{\Delta +2K} Z(v) \, \text {dn}^2 v \, \mathrm{d}v \nonumber \\&\quad + k^2 p_1 \int _{\Delta }^{\Delta +2K} (\text {sn}\, v \, \text {cn} \, v)^2 \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v \nonumber \\&\quad - p_1 \int _{\Delta }^{\Delta +2K} \text {sn}\, v \, \text {cn} \, v \, \text {dn}\, v \, Z(v) \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v \Bigg ], \nonumber \\ \implies {\dot{\phi }}&={} \dfrac{\pi }{kK} -\omega + \dfrac{\epsilon \pi k}{{2k^{\prime }}^2 K^2} \Bigg [ -\beta _1 k \underbrace{\int _0^{2K} \text {sn}\, v \, \text {cn} \, v \, \text {dn} \, v \, \mathrm{d}v}_{\text {First Integral}} \nonumber \\&\quad +\dfrac{\beta _1}{k}\underbrace{\int _{0}^{2K} Z(v) \, \text {dn}^2 v \, \mathrm{d}v}_{\text {Second Integral}} \nonumber \\&\quad + k^2 p_1 \underbrace{\int _{0}^{2K} (\text {sn}\, v \, \text {cn} \, v)^2 \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Third Integral}}\nonumber \\&\quad - p_1 \underbrace{\int _{0}^{2K} \text {sn}\, v \, \text {cn} \, v \, \text {dn}\, v \, Z(v) \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Fourth Integral}} \Bigg ], \end{aligned}$$

(9.3)

where $\Delta =\dfrac{K \phi }{\pi }$. The first integral is given in Eq. (8.3). For the second integral, we use the integral identity $\displaystyle \int _0^{2a_1}h(x) \mathrm{d}x=\int _0^{a_1} h(x) \mathrm{d}x + \int _0^{a_1} h(2a_1-x) \mathrm{d}x$ and the properties of the Jacobi elliptic and zeta functions [43, 44] to show that

$$\begin{aligned} \int _{0}^{2K} Z(v) \, \text {dn}^2 v \, \mathrm{d}v = 0. \end{aligned}$$

(9.4)

The third and fourth integrals can be evaluated using the method of contour integration as

$$\begin{aligned}&\int _{0}^{2K} (\text {sn}\, v \, \text {cn} \, v)^2 \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v \nonumber \\&\quad =\dfrac{\pi ^2(4K^2-2k^2K^2-\pi ^2) \cos \phi }{6k^4K^3 \, \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) } \text {and} \end{aligned}$$

(9.5)

$$\begin{aligned}&\int _{0}^{2K} \text {sn}\, v \, \text {cn} \, v \, \text {dn}\, v \, Z(v) \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v \nonumber \\&\quad = \dfrac{\cos \phi }{12K^3 k^2}\left[ \dfrac{3 \pi ^4 \text {cosh}\left( \dfrac{\pi K^{\prime }}{K}\right) }{\text {sinh}^2\left( \dfrac{\pi K^{\prime }}{K}\right) } \right. \nonumber \\&\qquad \left. - \dfrac{(2\pi ^4+4\pi ^2K^2k^2-8\pi ^2K^2+12\pi ^2KE)}{\text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) }\right] . \end{aligned}$$

(9.6)

1.2 Horizontal excitation

The relevant equations for the case of horizontal excitation are

$$\begin{aligned} {\dot{k}}= & {} \dfrac{\epsilon }{4K}\left[ 2\beta _1 k \underbrace{\int _0^{2K} \text {dn}^2\, v \, \mathrm{d}v}_{\text {First Integral}} - 2 p_1 k^2 \right. \nonumber \\&\quad \underbrace{\int _0^{2K} \text {cn}^2 \, v \, \text {dn}\, v \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Second Integral}} \nonumber \\&\left. +\,p_1 k^2 \underbrace{\int _0^{2K} \text {dn}\, v \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Third Integral}}\right] . \end{aligned}$$

(9.7)

$$\begin{aligned} {\dot{\phi }}= & {} \dfrac{\pi }{kK} -\omega + \dfrac{\epsilon \pi k}{{4k^{\prime }}^2 K^2}\left[ -p_1 k^2 \right. \nonumber \\&\quad \left. \underbrace{\int _0^{2K} \text {sn} \, v \, \text {cn}\, v \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {First Integral}} \right. \nonumber \\&+ 2 p_1 k^2 \underbrace{\int _0^{2K}\text {sn} \, v \, \text {cn}^3\, v \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Second Integral}}\nonumber \\&+ p_1\underbrace{\int _0^{2K} \text {dn} \, v \, Z(v) \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Third Integral}} \nonumber \\&\left. -\,2p_1\underbrace{\int _0^{2K} \text {dn} \, v \, \text {cn}^2\, v \, Z(v) \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v}_{\text {Fourth Integral}} \right] .\nonumber \\ \end{aligned}$$

(9.8)

The values of the first, second and third integrals of Eq. (9.7) are available in Eqs. (8.8), (8.16) and (8.15), respectively. There were more integrals involved in Eq. (9.8) which vanish as per Eqs. (8.3) and (9.4). The first and second integrals of Eq. (9.8) are evaluated in Eqs. (8.12) and (8.13), respectively. Finally, the third and the fourth integrals in Eq. (9.8) can be evaluated using the method of contour integration as

$$\begin{aligned}&\int _0^{2K} \text {dn} \, v \, Z(v) \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v \nonumber \\&\quad = \sin \phi \left[ \dfrac{\pi ^2}{K \, \text {cosh}\left( \dfrac{\pi K^{\prime }}{K}\right) }-\dfrac{\pi ^2 \, \text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) }{2K\, \text {cosh}^2 \left( \dfrac{\pi K^{\prime }}{K}\right) } \right] , \text {and}\nonumber \\ \end{aligned}$$

(9.9)

$$\begin{aligned}&\int _0^{2K} \text {dn} \, v \, \text {cn}^2\, v \, Z(v) \, \cos \left( \dfrac{\pi v}{K}-\phi \right) \, \mathrm{d}v \nonumber \\&\quad = \dfrac{\sin \phi }{12 K^3 k^2}\left[ \dfrac{\pi ^2(10K^2k^2-8K^2+2\pi ^2+12KE)}{\text {cosh}\left( \dfrac{\pi K^{\prime }}{K}\right) } \right. \nonumber \\&\qquad \left. -\dfrac{3\pi ^2(\pi ^2+K^2k^2)\text {sinh}\left( \dfrac{\pi K^{\prime }}{K}\right) }{\text {cosh}^2 \left( \dfrac{\pi K^{\prime }}{K}\right) }\right] . \end{aligned}$$

(9.10)