New solitary wave solutions in a perturbed generalized BBM equation

Abstract

In this manuscript, based on the geometric singular perturbation theory, several new solitary wave solutions in a perturbed generalized Benjamin–Bona–Mahony (BBM) equation are detected by the explicit calculation of the associated Melnikov integrals. These solitary wave solutions are homoclinic to non-trivial steady states and have not been found before. We also determine the zeroth-order approximations to the speeds of these solitary waves explicitly. In the calculations of the Melnikov integrals, the explicit expressions of the unperturbed homoclinic orbits play an important role.

Appendix

1.1 The explicit expressions of homoclinic orbits for the unperturbed system (11)

Here, we only give the detailed derivation of the explicit expression of homoclinic orbit for the unperturbed system (11) when $C\in (-\frac{4}{27}, 0)$. The other cases are similar and hence are omitted.

The homoclinic orbit in this case is determined by

$$\begin{aligned} \frac{dx}{\sqrt{-(x-x_{1})(x-x_{5})(x-x_{3})^{2}}}=\frac{\sqrt{2}}{2}\mathrm{d}\tau , \end{aligned}$$

where $x=x_{3}$ is solved from $x^3-x^2-C=0$, and $x=x_{1}, x=x_{5}$ are determined by $-\frac{1}{3}x^{3}+\frac{1}{4}x^{4}-Cx=h$ where $-\frac{4}{27}<C<0$, $h=u(x_{3})$.

If $x>x_{3}$, let $x-x_{3}=\frac{1}{u}>0$, we get

$$\begin{aligned} \frac{-du}{\sqrt{(x_{3}u+1-x_{1}u)(x_{5}u-x_{3}u-1)}}=\frac{\sqrt{2}}{2}\mathrm{d}\tau . \end{aligned}$$

Then, setting $x_{3}u+1-x_{1}u=v$ gives

$$\begin{aligned} \frac{-\frac{1}{x_{3}-x_{1}}dv}{\sqrt{v[(\frac{v-1}{x_{3}-x_{1}})(x_{5}-x_{3})-1]}}=\frac{\sqrt{2}}{2}\mathrm{d}\tau . \end{aligned}$$

Now let $\sqrt{v}=w$, one obtains

$$\begin{aligned} \frac{-2}{\sqrt{(x_{5}-x_{3})(x_{3}-x_{1})}}\frac{dw}{\sqrt{w^{2} -\frac{x_{5}-x_{1}}{x_{5}-x_{3}}}}=\frac{\sqrt{2}}{2}\mathrm{d}\tau , \end{aligned}$$

i.e.,

$$\begin{aligned}&\frac{-2}{\sqrt{(x_{5}-x_{3})(x_{3}-x_{1})}}\ln \left| w+\sqrt{w^{2} -\frac{x_{5}-x_{1}}{x_{5}-x_{3}}}\right| \\&\quad =\frac{\sqrt{2}}{2}\tau +\sigma , \end{aligned}$$

where $\sigma$ is a constant of integral.

Without loss of generality, let $X_{5}(x_{5},0)$ be the initial point, that is, $x(0)=x_{5}$, then

$$\begin{aligned} \sigma =\frac{-2}{\sqrt{(x_{5}-x_{3})(x_{3}-x_{1})}}\ln \sqrt{\frac{x_{5}-x_{1}}{x_{5}-x_{3}}}. \end{aligned}$$

Denote

$$\begin{aligned} \rho =\frac{\sqrt{2(x_{5}-x_{3})(x_{3}-x_{1})}}{4}, \end{aligned}$$

we obtain

$$\begin{aligned} x(\tau )=\frac{2(x_{3}-x_{1})(x_{5}-x_{3})}{(x_{5}-x_{1}) \cosh (2\rho \tau )+2x_{3}-x_{1}-x_{5}}+x_{3}. \end{aligned}$$

Similarly, if $x<x_{3}$, we can derive

$$\begin{aligned} x(\tau )=x_{3}-\frac{2(x_{3}-x_{1})(x_{5}-x_{3})}{(x_{5}-x_{1}) \cosh (2\rho \tau )+2x_{3}-x_{5}-x_{1}}. \end{aligned}$$

1.2 The detailed calculations of the Melnikov integrals

Case 1 When $C=0$.

In this case, the Melnikov integral is

$$\begin{aligned} M(c)= & {} \int ^{+\infty }_{-\infty } \left[ (-2x+3x^{2})y^{2}-\frac{1}{c^{2}}y^{2}\right] \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty } \left( -2x+3x^{2}-\frac{1}{c^{2}}\right) \left( \frac{2}{3}x^{3}-\frac{1}{2}x^{4}\right) \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty } \left( -\frac{2}{3c^{2}}x^{3} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) x^{4}\right. \nonumber \\&\left. +3x^{5} -\frac{3}{2}x^{6}\right) \mathrm{d}\tau , \end{aligned}$$

where $x(\tau )$ is given in (21).

Consider

$$\begin{aligned} I_{n}= & {} \int ^{+\infty }_{-\infty }x^{n}\mathrm{d}\tau =\int ^{+\infty }_{-\infty }\frac{1}{(a+b\tau ^{2})^{n}}\mathrm{d}\tau \\= & {} 2n\int ^{+\infty }_{-\infty }\frac{b\tau ^{2}+a-a}{(a+b\tau ^{2})^{n+1}}\mathrm{d}\tau \\= & {} 2nI_{n}-2naI_{n+1}, \end{aligned}$$

we thus have

$$\begin{aligned} I_{3}=\frac{3}{4a}I_{2} \quad I_{4}=\frac{5}{6a}I_{3} \quad I_{5}=\frac{7}{8a}I_{4} \quad I_{6}=\frac{9}{10a}I_{5}. \end{aligned}$$

Here, $x(\tau )=\frac{1}{\frac{3}{4}+\frac{\tau ^{2}}{6}}$, hence we have $a=3/4$ and $b=1/6$, and thus,

$$\begin{aligned} M(c)= & {} \int ^{+\infty }_{-\infty } \left[ -\frac{2}{3c^{2}}\frac{3}{4a} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) \frac{5}{8a^{2}}\right. \nonumber \\&\left. +\frac{105}{64a^{3}}-\frac{189}{256a^{4}}\right] x^{2}\mathrm{d}\tau \\= & {} \left( \frac{2}{27}-\frac{1}{9c^{2}}\right) \int ^{+\infty }_{-\infty }x^{2}\mathrm{d}\tau \\= & {} \left( \frac{2}{27}-\frac{1}{9c^{2}}\right) \frac{4\sqrt{2}\pi }{3}. \end{aligned}$$

Case 2 When $C=-\frac{2}{27}$.

The Melnikov integral is

$$\begin{aligned} M(c)= & {} \int ^{+\infty }_{-\infty }\left[ (-2x+3x^{2})y^{2}-\frac{1}{c^{2}}y^{2}\right] \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty }\left( -2x+3x^{2}-\frac{1}{c^{2}}\right) \\&\left( \frac{5}{162}+\frac{2}{3}x^{3}-\frac{1}{2}x^{4}-\frac{4}{27}x\right) \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty } \left[ -\frac{5}{162c^{2}}+\left( \frac{4}{27c^{2}}-\frac{5}{81}\right) x\right. \\&\left. +\frac{7}{18}x^{2}-\left( \frac{2}{3c^{2}}+\frac{4}{9}\right) x^{3} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) x^{4}\right. \\&\left. +3x^{5}-\frac{3}{2}x^{6}\right] \mathrm{d}\tau , \end{aligned}$$

in which

$$\begin{aligned} x(\tau )=\frac{1}{3}-\frac{\sqrt{6}}{3}\mathrm{sech}\left( \frac{\sqrt{3}}{3}\tau \right) . \end{aligned}$$

Denote $\chi =\frac{\sqrt{6}}{3}\mathrm{sech}(\frac{\sqrt{3}}{3}\tau )$, then

$$\begin{aligned} M(c)= & {} \int ^{+\infty }_{-\infty } \left[ -\frac{5}{162c^{2}} +\left( \frac{4}{27c^{2}}-\frac{5}{81}\right) x +\frac{7}{18}x^{2}\right. \\&\left. - \left( \frac{2}{3c^{2}}+\frac{4}{9}\right) x^{3} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) x^{4}\right. \\&\left. +3x^{5}-\frac{3}{2}x^{6}\right] \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty }\left[ \left( -\frac{1}{3c^{2}}-\frac{1}{9}\right) \chi ^{2}\right. \\&\left. +\left( \frac{1}{2c^{2}}+\frac{7}{6}\right) \chi ^{4}-\frac{3}{2}\chi ^{6}\right] \mathrm{d}\tau . \end{aligned}$$

Let

$$\begin{aligned} I_{k}=\int ^{+\infty }_{-\infty }\chi ^{k}\mathrm{d}\tau =\int ^{+\infty }_{-\infty } \left( \frac{\sqrt{6}}{3}\mathrm{sech}\left( \frac{\sqrt{3}}{3}\tau \right) \right) ^{k}\mathrm{d}\tau , \end{aligned}$$

it follows from $\mathrm{sech} p^{2}=1-\mathrm{tanh}p^{2}$ and $d\mathrm{tanh}p=\mathrm{sech}p^{2}dp$ that

$$\begin{aligned} I_{k}=\eta \int ^{+\infty }_{-\infty }((1-\mathrm{tanh}p^{2}))^{\frac{k-2}{2}}d\mathrm{tanh}p, \end{aligned}$$

where $p=\frac{\sqrt{3}}{3}\tau$, $\eta =\sqrt{3}(\frac{\sqrt{6}}{3})^{k}$, which is equivalent to

$$\begin{aligned} I_{k}=\eta \int ^{1}_{-1}((1-\mu ^{2}))^{\frac{k-2}{2}}\mathrm{d}\mu {=}2\eta \int ^{1}_{0}((1-\mu ^{2}))^{\frac{k-2}{2}}\mathrm{d}\mu , \end{aligned}$$

where $\mu =\mathrm{tanh}p$.

Let

$$\begin{aligned} a_{m}=\int ^{1}_{0}(1-\theta ^{2})^{m}d\theta , \end{aligned}$$

which becomes

$$\begin{aligned} a_{m}=\int ^{\frac{\pi }{2}}_{0}(\cos \gamma )^{2m+1}\mathrm{d}\gamma , \end{aligned}$$

where $\theta =\sin \gamma$. When $m=0, 1, 2, 3, \cdots$, it can be calculated that

$$\begin{aligned} a_{m}= & {} \int _{0}^{\frac{\pi }{2}}\left( cos\gamma \right) ^{2m+1}\mathrm{d}\gamma {=}\frac{2m}{2m+1}\int _{0}^{\frac{\pi }{2}}cos\gamma ^{2m-1}d\gamma \\ {}= & {} \cdots =\frac{2m}{2m+1}\frac{2m-2}{2m-11}\cdots \frac{2}{3}\int _{0}^{\frac{\pi }{2}}cos\gamma d\gamma \\= & {} \frac{(2m)!!}{(2m+1)!!}, \end{aligned}$$

and when $m=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}, \cdots$,

$$\begin{aligned} a_{m}= & {} \int _{0}^{\frac{\pi }{2}}\left( cos\gamma \right) ^{2m+1}\mathrm{d}\gamma {=}\frac{2m}{2m+1}\int _{0}^{\frac{\pi }{2}}cos\gamma ^{2m-1}\mathrm{d}\gamma \\ {}= & {} \cdots =\frac{2m}{2m+1}\frac{2m-2}{2m-11}\cdots \frac{1}{2}\int _{0}^{\frac{\pi }{2}}cos^{0}\gamma d\gamma \\= & {} \frac{(2m)!!}{(2m+1)!!}\frac{\pi }{2}. \end{aligned}$$

Obviously,

$$\begin{aligned} I_{k}=2\eta \times a_{\frac{k-2}{2}}, \end{aligned}$$

thus

$$\begin{aligned} I_{2}= & {} \int ^{+\infty }_{-\infty }\chi ^{2}\mathrm{d}\tau =\int ^{+\infty }_{-\infty } \left( \frac{\sqrt{6}}{3}\mathrm{sech}\left( \frac{\sqrt{3}}{3}\tau \right) \right) ^{2}\mathrm{d}\tau \\= & {} 2\sqrt{3}\times \left( \frac{\sqrt{6}}{3}\right) ^{2}\times 1{,}\\ I_{4}= & {} \int ^{+\infty }_{-\infty }\chi ^{4}\mathrm{d}\tau =\int ^{+\infty }_{-\infty } \left( \frac{\sqrt{6}}{3}\mathrm{sech}\left( \frac{\sqrt{3}}{3}\tau \right) \right) ^{4}\mathrm{d}\tau \\= & {} 2\sqrt{3}\times \left( \frac{\sqrt{6}}{3}\right) ^{4}\times \frac{2}{3} \end{aligned}$$

and

$$\begin{aligned} I_{6}= & {} \int ^{+\infty }_{-\infty }\chi ^{6}\mathrm{d}\tau =\int ^{+\infty }_{-\infty } \left( \frac{\sqrt{6}}{3}\mathrm{sech}\left( \frac{\sqrt{3}}{3}\tau \right) \right) ^{6}\mathrm{d}\tau \\= & {} 2\sqrt{3}\times \left( \frac{\sqrt{6}}{3}\right) ^{6}\times \frac{2}{3}\times \frac{4}{5}, \end{aligned}$$

which finally gives

$$\begin{aligned} M(c)= & {} \int ^{+\infty }_{-\infty } \left[ -\frac{5}{162c^{2}}+\left( \frac{4}{27c^{2}}-\frac{5}{81}\right) x +\frac{7}{18}x^{2}\right. \\&\left. -\left( \frac{2}{3c^{2}}+\frac{4}{9}\right) x^{3} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) x^{4}\right. \\&\left. +3x^{5}-\frac{3}{2}x^{6}\right] \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty } \left[ \left( -\frac{1}{3c^{2}}-\frac{1}{9}\right) \chi ^{2}+\left( \frac{1}{2c^{2}}+\frac{7}{6}\right) \chi ^{4}\right. \\&\left. -\frac{3}{2}\chi ^{6}\right] \mathrm{d}\tau \\= & {} 2\sqrt{3}\times \left( -\frac{2}{27}c^{2}+\frac{14}{405}\right) . \end{aligned}$$

Case 3 When $C=-\frac{4}{27}$.

In this case, $x(\tau )=\frac{2}{3}-\frac{1}{\frac{3}{4}+\frac{\tau ^{2}}{3}}$. Denote $\chi (\tau )=\frac{1}{\frac{3}{4}+\frac{\tau ^{2}}{3}}$; direct calculation gives the Melnikov integral

$$\begin{aligned} M(c)= & {} \int ^{+\infty }_{-\infty }\left[ -\frac{8}{81c^{2}} +\left( \frac{8}{27c^{2}}-\frac{16}{81}\right) x +\frac{8}{9}x^{2}\right. \\&\left. -\left( \frac{2}{3c^{2}}+\frac{8}{9}\right) x^{3} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) x^{4}\right. \\&\left. +3x^{5}-\frac{3}{2}x^{6}\right] \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty } \left[ (-\frac{1}{3c^{2}}\chi ^{3} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) \chi ^{4}\right. \\&\left. +3\chi ^{5}-\frac{3}{2}\chi ^{6}\right] \mathrm{d}\tau . \end{aligned}$$

By further calculation, we get

$$\begin{aligned} M(c)= & {} \int ^{+\infty }_{-\infty } \left[ (-\frac{1}{3c^{2}}\chi ^{3} +\left( \frac{1}{2c^{2}}-\frac{4}{3}\right) \chi ^{4}\right. \\&\left. +3\chi ^{5}-\frac{3}{2}\chi ^{6}\right] \mathrm{d}\tau \\= & {} \int ^{+\infty }_{-\infty }\left[ \frac{2}{27}-\frac{1}{9c^{2}}\right] \chi ^{2}\mathrm{d}\tau \\= & {} \left( \frac{2}{27}-\frac{1}{9c^{2}}\right) \frac{4\pi }{3}. \end{aligned}$$