Stability of Elliptic Solutions to the sinh-Gordon Equation

Abstract

Using the integrability of the sinh-Gordon equation, we demonstrate the spectral stability of its elliptic solutions. With the first three conserved quantities of the sinh-Gordon equation, we construct a Lyapunov functional. By using such Lyapunov functional, we show that these elliptic solutions are orbitally stable with respect to subharmonic perturbations of arbitrary period.

Appendix

Lemma

For $c>1$, $\frac{P(\zeta _{1})}{\zeta ^2_{1}}>0$ and $\frac{P(\zeta _{2})}{\zeta ^2_{2}}<0$, while for $c<-1$, $\frac{P(\zeta _{1})}{\zeta ^2_{1}}<0$ and $\frac{P(\zeta _{2})}{\zeta ^2_{2}}>0$.

Proof

• For $c>1$,

  $$\begin{aligned}&\frac{P(\zeta _{1})}{\zeta ^2_{1}}=8 c \sqrt{{\mathcal {E}}^2-1} K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) +8 ({\mathcal {E}}+1) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) \nonumber \\&\quad -8 ({\mathcal {E}}+1) E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) . \end{aligned}$$

  (83)

  Since $E(k)<K(k)$, $c>1$ and ${\mathcal {E}}>1$, we have $\frac{P(\zeta _{1})}{\zeta ^2_{1}}>0$.

  $$\begin{aligned}&\frac{P(\zeta _{2})}{\zeta ^2_{2}}= 8 c \left( -\sqrt{{\mathcal {E}}^2-1}\right) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) +8 ({\mathcal {E}}+1) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) \nonumber \\&\quad -8 ({\mathcal {E}}+1) E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) . \end{aligned}$$

  (84)

  Let $\frac{P(\zeta _{2})}{\zeta ^2_{2}}=F(c)$. We note that $F'(c)=8\left( -\sqrt{{\mathcal {E}}^2-1}\right) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) <0$. We have

  $$\begin{aligned}&F(c)<F(1)=8 \left( -\sqrt{{\mathcal {E}}^2-1}\right) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) +8 ({\mathcal {E}}+1) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) \nonumber \\&\quad -8 ({\mathcal {E}}+1) E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) . \end{aligned}$$

  (85)

  Using $\frac{E(k)}{K(k)}>k'=\sqrt{1-k^2}$ , see [1, 19.9.8], we have

  $$\begin{aligned}&8 \left( -\sqrt{{\mathcal {E}}^2-1}\right) +8 ({\mathcal {E}}+1)-8 ({\mathcal {E}}+1) \frac{E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) }{K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) }<8 \left( -\sqrt{{\mathcal {E}}^2-1}\right) \nonumber \\&\quad +8 ({\mathcal {E}}+1)-8\sqrt{2}\sqrt{{\mathcal {E}}+1}. \end{aligned}$$

  (86)

  Let $Q({\mathcal {E}})=8 \left( -\sqrt{{\mathcal {E}}^2-1}\right) +8 ({\mathcal {E}}+1)-8\sqrt{2}\sqrt{{\mathcal {E}}+1}$. We note $Q'({\mathcal {E}})=-\frac{8 {\mathcal {E}}}{\sqrt{{\mathcal {E}}^2-1}}+8-\frac{4 \sqrt{2}}{\sqrt{{\mathcal {E}}+1}}<-\frac{4 \sqrt{2}}{\sqrt{{\mathcal {E}}+1}}<0$. So we have $Q({\mathcal {E}})<Q(1)=0$ for ${\mathcal {E}}>1$. Therefore, we have $\frac{P(\zeta _{2})}{\zeta ^2_{2}}=F(c)<F(1)<K \left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) Q({\mathcal {E}})<0$.

• For $c<-1$,

  $$\begin{aligned} \frac{P(\zeta _{1})}{\zeta ^2_{1}}=8 c \sqrt{{\mathcal {E}}^2-1} K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) +8 ({\mathcal {E}}+1) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) -8 ({\mathcal {E}}+1) E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) .\nonumber \\ \end{aligned}$$

  (87)

  Let $\frac{P(\zeta _{1})}{\zeta ^2_{1}}=G(c)$. We note that $G'(c)=8\left( \sqrt{{\mathcal {E}}^2-1}\right) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) >0$. We have

  $$\begin{aligned}&G(c)<G(-1)=8 \left( -\sqrt{{\mathcal {E}}^2-1}\right) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) +8 ({\mathcal {E}}+1) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) \nonumber \\&\quad -8 ({\mathcal {E}}+1) E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) . \end{aligned}$$

  (88)

  Again, using $\frac{E(k)}{K(k)}>k'=\sqrt{1-k^2}$, we have

  $$\begin{aligned}&8 \left( -\sqrt{{\mathcal {E}}^2-1}\right) +8 ({\mathcal {E}}+1)-8 ({\mathcal {E}}+1) \frac{E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) }{K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) }<8 \left( -\sqrt{{\mathcal {E}}^2-1}\right) \nonumber \\&\quad +8 ({\mathcal {E}}+1)-8\sqrt{2}\sqrt{{\mathcal {E}}+1}. \end{aligned}$$

  (89)

  We know $Q({\mathcal {E}})<Q(1)=0$ for ${\mathcal {E}}>1$. Therefore, $\frac{P(\zeta _{1})}{\zeta ^2_{1}}=G(c)<G(-1)<K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) Q({\mathcal {E}})<0$.

  $$\begin{aligned}&\frac{P(\zeta _{2})}{\zeta ^2_{2}}= 8 c \left( -\sqrt{{\mathcal {E}}^2-1}\right) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) +8 ({\mathcal {E}}+1) K\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) \nonumber \\&\quad -8 ({\mathcal {E}}+1) E\left( \sqrt{\frac{{\mathcal {E}}-1}{{\mathcal {E}}+1}}\right) . \end{aligned}$$

  (90)

  Since $E(k)<K(k)$, $c<-1$ and ${\mathcal {E}}>1$, we have $\frac{P(\zeta _{2})}{\zeta ^2_{2}}>0$. This finishes the proof of the lemma.

$\square$