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Small Amplitude Traveling Waves in the Full-Dispersion Whitham Equation

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Abstract

In this article, we provide an alternative way to construct small amplitude traveling waves for general Whitham type equations, in both periodic and whole line contexts. More specifically, Fourier analysis techniques allow us to reformulate the problem to the study of waves that are small and regular perturbations of well-understood ODE’s. In addition, rigorous spectral stability of these waves is established.

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Notes

  1. In principle, one could compute explicitly the next terms, up to any degree of accuracy.

  2. In this case, the only symmetry is the translation in the x variable.

  3. Note that the precise form for \(\nu \) is primarily for convenience. We could let \(\nu = m(0) + \kappa \epsilon ^2\) for any \(\kappa >0\) and all that would really change are some nonessential coefficients.

  4. And in fact, for the Whitham example, where \(m(k)=\sqrt{\frac{\tanh (k)}{k}}\) it is not smoothing.

  5. The most common example one should think is \(X=H^s\), a Sobolev space, with dual \(X^*=H^{-s}\).

  6. A much more precise result is contained in Theorem 2.3, [14], but we state this corollary, as it is enough for our purposes.

  7. And hence \(k_0^{\le 0}({{\mathscr {L}}}_\epsilon )=1\), since the left hand side of (24) is non-negative.

  8. And in fact it has a single negative eigenvalue.

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Author information

Authors and Affiliations

  1. Department of Mathematics, University of Kansas, 1460 Jayhawk Boulevard, Lawrence, KS, 66045–7523, USA

    Atanas Stefanov

  2. Department of Mathematics, Drexel University, 3141 Chestnut St, Philadelphia, PA, 19104, USA

    J. Douglas Wright

Authors
  1. Atanas Stefanov
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  2. J. Douglas Wright
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Corresponding author

Correspondence to J. Douglas Wright.

Additional information

Stefanov is partially supported by NSF-DMS under Grant # 1614734. Wright is partially supported by NSF-DMS under Grant # 1511488.

Appendix: Assorted Proofs

Appendix: Assorted Proofs

Proof

(Lemma 3) Take \(K>k_*/\epsilon \). Then we clearly have

$$\begin{aligned} \left| {1 \over {1 \over 2} m''(0)(1+ K^2)}\right| \le {2 \epsilon ^2 \over |m''(0)|k_*} \le C\epsilon ^2. \end{aligned}$$

Since \(K>k_* /\epsilon \), we know that \(m(\epsilon K) \le m_1\) by (5). Thus we have

$$\begin{aligned} m(0) - {1 \over 2} m''(0) \epsilon ^2 - m(\epsilon K)> m(0) - m_1 > 0. \end{aligned}$$

Here we have used that fact that \(m''(0) < 0\), which is implied by (4). This in turn implies:

$$\begin{aligned} \left| {\epsilon ^2 \over m(0)-{1 \over 2} m''(0) \epsilon ^2 - m(\epsilon K)} \right| \le {\epsilon ^2 \over m(0) - m_1} \le C \epsilon ^2. \end{aligned}$$

The triangle inequality gives:

$$\begin{aligned} \sup _{|K| \ge k_*/\epsilon } \left| {\epsilon ^2 \over m(0)-{1 \over 2} m''(0) \epsilon ^2 - m(\epsilon K)} + {1 \over {1 \over 2}m''(0)(1+ K^2)}\right| \le C\epsilon ^2. \end{aligned}$$
(32)

Now suppose that \(|K| \le k_*/\epsilon \). We have

$$\begin{aligned}&{\epsilon ^2 \over m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(\epsilon K)} + {1 \over {1 \over 2}m''(0)(1+ K^2)} \nonumber \\&\qquad = { m(0) + {1 \over 2}m''(0) \epsilon ^2 K^2- m(\epsilon K) \over [m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(\epsilon K)][ {1 \over 2}m''(0)(1+ K^2)]} \end{aligned}$$
(33)

The fact that m is even and \(C^{3,1}\) implies, by way of Taylor’s theorem, that there exists \(C>0\) such that

$$\begin{aligned} \left| m(0) + {1 \over 2}m''(0) k^2- m(k)\right| \le C{k^4} \end{aligned}$$

when \(|k| \le k_*\). This implies that

$$\begin{aligned} \left| m(0) + {1 \over 2}m''(0) \epsilon ^2 K^2- m(\epsilon K)\right| \le C \epsilon ^4 K^4 \end{aligned}$$
(34)

when \(|K| \le k_*/\epsilon \).

The fundamental theorem of calculus implies that

$$\begin{aligned} m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(k) = -{1 \over 2} m''(0) \epsilon ^2 - \int _0^k \int _0^s m''(\sigma ) d\sigma ds. \end{aligned}$$

Here we used the fact that m(k) is even. Then we use (4) to see that

$$\begin{aligned} m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(k) \ge -{1 \over 2} m''(0) \epsilon ^2 - {1 \over 2} m_2 k^2 \end{aligned}$$

so long as \(|k| \le k_*\). Thus, for \(|K| \le k_*/\epsilon \) we have

$$\begin{aligned} m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(\epsilon K) \ge \epsilon ^2\left( -{1 \over 2} m''(0) - {1 \over 2} m_2 K^2\right) \end{aligned}$$

Since \(m''(0)\) and \(m_2\) are both negative this implies:

$$\begin{aligned} \left| m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(\epsilon K)\right| \ge C\epsilon ^2\left( 1+ K^2\right) \end{aligned}$$
(35)

when \(|K| \le k_*/\epsilon \).

Thus we can control the left hand side of  (33) using (34) and (35) as:

$$\begin{aligned} \left| {\epsilon ^2 \over m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(\epsilon K)} + {1 \over {1 \over 2}m''(0)(1+ K^2)} \right| \le {C \epsilon ^2 K^4 \over (1+K^2)^2} \end{aligned}$$

when \(K\le k_*/\epsilon \). Since \(K^4/(1+K^2)^2 \le 1\) we have

$$\begin{aligned} \sup _{|K| \le k_*/\epsilon } \left| {\epsilon ^2 \over m(0)- {1 \over 2}m''(0) \epsilon ^2 - m(\epsilon K)} + {1 \over {1 \over 2}m''(0)(1+ K^2)} \right| \le C\epsilon ^2 \end{aligned}$$

\(\square \)

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Stefanov, A., Wright, J.D. Small Amplitude Traveling Waves in the Full-Dispersion Whitham Equation. J Dyn Diff Equat 32, 85–99 (2020). https://doi.org/10.1007/s10884-018-9713-8

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