Periodic Waves in the Fractional Modified Korteweg–de Vries Equation

Abstract

Periodic waves in the modified Korteweg–de Vries (mKdV) equation are revisited in the setting of the fractional Laplacian. Two families of solutions in the local case are given by the sign-definite dnoidal and sign-indefinite cnoidal solutions. Both solutions can be characterized in the general fractional case as global minimizers of the quadratic part of the energy functional subject to the fixed $L^4$ norm: the sign-definite (sign-indefinite) solutions are obtained in the subspace of even (odd) functions. Morse index is computed for both solutions and the spectral stability criterion is derived. We show numerically that the family of sign-definite solutions has a generic fold bifurcation for the fractional Laplacian of lower regularity and the family of sign-indefinite solutions has a generic symmetry-breaking bifurcation both in the fractional and local cases.

Appendices

Appendix A. Stokes Expansion of General Small-Amplitude Waves

Here we generalize the Stokes expansion of Sect. 5.1 in order to prove that $\sigma _0 > 0$ for $\alpha > \alpha _0$ and $\sigma _0 < 0$ for $\alpha < \alpha _0$, where $\sigma _0 = \langle {\mathcal {L}}^{-1} 1, 1 \rangle$ is computed on the small-amplitude wave (5.1) and (5.3) in terms of the small amplitude A.

Let $\psi$ satisfy Eq. (1.1) for (c, b) defined in an open neighborhood ${\mathcal {I}} \subset {\mathbb {R}}^2$ of the point $\left( \frac{1}{2},0\right)$. We generalize the decomposition (5.1) by setting

$$\begin{aligned} \psi (x) = \psi _0 + \varphi (x), \end{aligned}$$

(6.4)

where $\psi _0 = \psi (c,b)$ is a root of the cubic equation $b + c \psi _0 = 2 \psi _0^3$ and $\varphi$ is not required to satisfy the zero-mean property. Since three roots exist for $\psi _0$ at $b = 0$, we are picking uniquely the positive root by using the expansion

$$\begin{aligned} \psi _0(c,b) = \frac{1}{2} \sqrt{2c} + \frac{b}{2c} + {\mathcal {O}}(b^2). \end{aligned}$$

(6.5)

Eq. (1.1) is written in the equivalent form:

$$\begin{aligned} D^{\alpha } \varphi + (c - 6 \psi _0^2) \varphi = 2 \varphi ^3 + 6 \psi _0 \varphi ^2, \end{aligned}$$

(6.6)

which generalizes (5.2). By using the Stokes expansion in terms of small amplitude A:

$$\begin{aligned} \left\{ \begin{array}{l} \varphi (x) = A \varphi _1(x) + A^2 \varphi _2(x) + A^3 \varphi _3(x) + {\mathcal {O}}(A^4), \\ c - 6 \psi _0^2 = -1 + A^2 \omega _2 + {\mathcal {O}}(A^4), \end{array} \right. \end{aligned}$$

(6.7)

we obtain recursively: $\varphi _1(x) = \cos (x)$,

$$\begin{aligned}&\varphi _2(x) = -3 \psi _0 + \frac{3 \psi _0}{2^{\alpha }-1} \cos (2x),\\&\varphi _3(x) = \frac{1}{3^{\alpha } - 1} \left[ \frac{1}{2} + \frac{18 \psi _0^2}{2^{\alpha }-1} \right] \cos (3x), \end{aligned}$$

and

$$\begin{aligned} \omega _2 = \frac{3}{2} - 36 \psi _0^2 + \frac{18 \psi _0^2}{2^{\alpha }-1}. \end{aligned}$$

By substituting $\omega _2$ to the expansion for c in (6.7) and using expansion (6.5), we obtain

$$\begin{aligned} \gamma _2 A^2 = 2c - 1 + 6 b + {\mathcal {O}}((2c-1)^2 + b^2), \end{aligned}$$

(6.8)

where $\gamma _2 = -\omega _2 |_{\psi _0 = \frac{1}{2}}$ is the same as in (5.3). By using (6.8), we obtain perturbatively:

$$\begin{aligned} a= & {} \frac{1}{2\pi } \int _{-\pi }^{\pi } \psi (x) dx = \psi _0 \left[ 1 - 3 A^2 + {\mathcal {O}}(A^4) \right] \\= & {} \frac{1}{2} - a_1(2c-1) - a_2 b + {\mathcal {O}}((2c-1)^2 + b^2),\\ \omega= & {} c - 6 a^2 \\= & {} -1 + \frac{1}{2} (1 + 12 a_1) (2c-1) + 6 a_2 b + {\mathcal {O}}((2c-1)^2 + b^2), \end{aligned}$$

and

$$\begin{aligned} \beta= & {} b + c a - 2 a^3 \\= & {} \frac{1}{4} (1 + 4 a_1)(2c-1) + (1+ a_2) b + {\mathcal {O}}((2c-1)^2 + b^2), \end{aligned}$$

where

$$\begin{aligned} a_1 := \frac{3}{8 \gamma _2} \frac{4 - 2^{\alpha }}{2^{\alpha } - 1}, \quad a_2 := \frac{3}{2 \gamma _2} \frac{2 + 2^{\alpha }}{2^{\alpha } - 1}. \end{aligned}$$

If $\gamma _2 \ne 0$ for $\alpha \ne \alpha _0$ given by (5.4), the transformation ${\mathcal {I}} \ni (c,b) \mapsto (\omega ,a) \in {\mathcal {O}}$ is $C^1$ and invertible with the inverse transformation

$$\begin{aligned} c= & {} \frac{1}{2} + (\omega + 1) + 6 (a -\frac{1}{2}) + {\mathcal {O}}((\omega + 1)^2 + (2a-1)^2),\\ b= & {} -\frac{1}{a_2} \left[ 2 a_1 (\omega + 1) + \frac{1}{2} (1 + 12 a_1) (2a - 1) + {\mathcal {O}}((\omega + 1)^2 + (2a-1)^2) \right] , \end{aligned}$$

from which we obtain $\beta = \beta (\omega ,a)$:

$$\begin{aligned} \beta = \frac{a_2 - 4 a_1}{2a_2} (\omega + 1) + \frac{2 a_2 - 1 - 12 a_1}{2a_2} (2a - 1) + {\mathcal {O}}((\omega + 1)^2 + (2a-1)^2) \end{aligned}$$

and

$$\begin{aligned} s_0 = \omega - \partial _a \beta + 12 a \partial _{\omega } \beta = \frac{1}{a_2} + {\mathcal {O}}((\omega + 1)^2 + (2a-1)^2). \end{aligned}$$

Since $\sigma _0 = \frac{2\pi }{s_0}$, we have $\mathrm{sign}(\sigma _0) = \mathrm{sign}(a_2) = \mathrm{sign}(\gamma _2)$, from which it follows that $\sigma _0 > 0$ for $\alpha > \alpha _0$ and $\sigma _0 < 0$ for $\alpha < \alpha _0$.

Furthermore, it follows from (6.8) that

$$\begin{aligned} \gamma _2 A^2 = \frac{2 (a_2 - 6 a_1)}{a_2} (\omega + 1) + \frac{3(2a_2 - 1 - 12 a_1)}{a_2} (2 a - 1) + {\mathcal {O}}((\omega +1)^2 + (2a-1)^2). \end{aligned}$$

Explicit computation shows that $2a_2 - 1 - 12 a_1 = 0$, hence $\Vert \phi \Vert ^2_{L^2} = \pi A^2 + {\mathcal {O}}(A^4)$ as a function of $(\omega ,a)$ satisfies

$$\begin{aligned} \frac{\partial }{\partial \omega } \Vert \phi \Vert _{L^2}^2 = \frac{2\pi (a_2 - 6 a_1)}{\gamma _2 a_2} + {\mathcal {O}}(A^2) = \frac{2\pi }{3} \frac{2^{\alpha } - 1}{2^{\alpha } + 2} + {\mathcal {O}}(A^2) = \frac{d}{d \omega } \Vert \phi \Vert _{L^2}^2 + {\mathcal {O}}(A^2), \end{aligned}$$

in agreement with (5.5). In other words, although a is defined by $\omega$ at the periodic waves satisfying $b = 0$ by

$$\begin{aligned} 2a - 1 = -\frac{4 a_1}{1 + 12 a_1} (\omega + 1) + {\mathcal {O}}((\omega + 1)^2), \end{aligned}$$

this dependence does not result in the discrepancy between partial and ordinary derivatives of $\Vert \phi \Vert _{L^2}^2$ in $\omega$ along the family of even periodic waves in the limit $A \rightarrow 0$.

Appendix B. On the Variational Problem (6.3) with Two Constraints

We show that the periodic solutions to Eq. (1.1) with two parameters (c, b) can be recovered from the ground state of the variational problem (6.3).

Proposition 6.5

Fix $\alpha > \frac{1}{2}$ and $m_0 := (2\pi )^{-\frac{1}{4}}$. For every $m \in [-m_0,m_0]$ and every $c \in (-\,1,\infty )$, there exists the ground state (minimizer) $\chi \in H^{\frac{\alpha }{2}}_{\mathrm{per}}$ of the variational problem (6.3). If $m \in (-\,m_0,m_0)$, the ground state has the single-lobe profile and there exists $C > 0$ such that $\psi (x) = C \chi (x)$ satisfies Eq. (1.1) with some b.

Proof

The bound $|m| \le m_0$ follows by the Hölder’s inequality

$$\begin{aligned} \left| \int _{{\mathbb {T}}} u dx \right| \le \left( \int _{{\mathbb {T}}} 1^{\frac{4}{3}} dx \right) ^{\frac{3}{4}} \left( \int _{{\mathbb {T}}} u^4 dx \right) ^{\frac{1}{4}} = (2\pi )^{\frac{3}{4}}. \end{aligned}$$

The quadratic functional $B_c(u)$ in (2.3) is bounded from below by the Poincaré inequality:

$$\begin{aligned} B_c(u) \ge \frac{1}{2} \Vert u - m \Vert ^2_{L^2} + \frac{1}{2} c \Vert u \Vert ^2_{L^2} = \frac{1}{2} (1 + c) \Vert u \Vert ^2_{L^2} - \pi m^2. \end{aligned}$$

By the same analysis as in the proof of Theorem 2.1, for every $m \in [-m_0,m_0]$ and every $c \in (-\,1,\infty )$, there exists the ground state $\chi \in H^{\frac{\alpha }{2}}_{\mathrm{per}}$ of the variational problem (6.3). Moreover, by the symmetric rearrangements, the ground state is either constant or has the single-lobe profile. The constant solution corresponds to $|m| = m_0$, hence the ground state has the single-lobe profile if $m \in (-\,m_0,m_0)$.

With two Lagrange multipliers $\mu$ and $\nu$, the ground state $\chi \in H^{\frac{\alpha }{2}}_{\mathrm{per}}$ satisfies the stationary equation

$$\begin{aligned} D^{\alpha } \chi + c \chi + \nu = \mu \chi ^3. \end{aligned}$$

(6.9)

Lagrange multipliers satisfy two relations due to the constraints in (6.3):

$$\begin{aligned} \mu = 2 B_c(\chi ) + 2 \pi m \nu , \quad \mu \int _{{\mathbb {T}}} \chi ^3 dx = 2 \pi (c m + \nu ). \end{aligned}$$

(6.10)

Eliminating $\nu$ yields

$$\begin{aligned} \left[ 1 - m \int _{{\mathbb {T}}} \chi ^3 dx \right] \mu = 2 \left[ B_c(\chi ) - \pi c m^2 \right] . \end{aligned}$$

(6.11)

The left-hand side of (6.11) can be rewritten in the equivalent symmetrized form:

$$\begin{aligned} 1 - m \int _{{\mathbb {T}}} \chi ^3 dx= & {} \frac{1}{2\pi } \left[ \left( \int _{{\mathbb {T}}} dx \right) \left( \int _{{\mathbb {T}}} \chi ^4 dx \right) - \left( \int _{{\mathbb {T}}} \chi dx \right) \left( \int _{{\mathbb {T}}} \chi ^3 dx \right) \right] \\= & {} \frac{1}{16 \pi } \int _{{\mathbb {T}}} \int _{{\mathbb {T}}} \left( \left[ \chi (x) - \chi (y) \right] ^4 + 3 \left[ \chi ^2(x) - \chi ^2(y) \right] ^2 \right) dx dy, \end{aligned}$$

from which it follows that it is strictly positive if $\chi (x)$ is not constant. Similarly, the right-hand side of (6.11) is strictly positive if $\chi (x)$ is not constant due to the following inequality

$$\begin{aligned} 2 B_c(\chi ) - 2 \pi c m^2 = \Vert D^{\frac{\alpha }{2}} \chi \Vert ^2_{L^2} + c \Vert u - m \Vert ^2_{L^2} \ge (1 + c) \Vert u - m \Vert ^2_{L^2} > 0. \end{aligned}$$

Since the ground state is non-constant if $m \in (-\,m_0,m_0)$, we obtain the unique $\mu > 0$ from (6.11) such that the transformation $\psi = C \chi$ with $C := \sqrt{\mu }/\sqrt{2}$ reduces (6.9) to Eq. (1.1) with $b = \nu \sqrt{\mu }/\sqrt{2}$, where $\nu$ is uniquely found from (6.10). $\square$