Boundary integral methods for dispersive equations, Airy flow and the modified Korteweg de Vries equation

Abstract

In this paper, we implement interface tracking methods for the evolution of 2-D curves that follow Airy flow, a curvature-dependent dispersive geometric evolution law. The curvature of the curve satisfies the modified Korteweg de Vries equation, a dispersive non-linear soliton equation. We present a fully discrete space-time analysis of the equations (proof of convergence) and numerical evidence that confirms the accuracy, convergence, efficiency, and stability of the methods.

Appendix A

1.1 A.1 Dynamics of curvature k, arc-length variation s _α and angle between the tangent vector and x-axis

By continuity of second derivatives

$$ \begin{array}{ll} X_{\alpha t} &\displaystyle=X_{t \alpha}=\frac{\partial X_{t}}{\partial \alpha}=s_{\alpha}\frac{\partial [V\textbf{n}+T\textbf{s}]}{\partial s}=s_{\alpha}[V_{s}\textbf{n}+V\textbf{n}_{s}+T_{s}\textbf{s}+T\textbf{s}_{s}],\\ s_{\alpha t}&\displaystyle=\frac{x_{\alpha}x_{\alpha t}+y_{\alpha}y_{\alpha t}}{s_{\alpha}}=\frac{X_{\alpha}\cdot X_{\alpha t}}{s_{\alpha}}. \end{array} $$

(A.1)

Now, since X_α = s_αs and using the Frenet Formulas we obtain n_s = ks, s_s = −kn where k represents the curvature. Therefore

$$s_{\alpha t}=s_{\alpha} \textbf{s} \cdot [V_{s}\textbf{n}+Vk\textbf{s}+T_{s}\textbf{s}+T(-k\textbf{n})]=s_{\alpha}(T_{s}+Vk). $$

Also, we can derive the rate of change for $\theta =\arctan (\frac {y_{\alpha }}{x_{\alpha }})$ (angle between the tangent vector to the curve and the x-axis) as follows

$$ \begin{array}{ll} \theta_{t} &\displaystyle=\frac{1}{1+(\frac{y_{\alpha}}{x_{\alpha}})^{2}}\frac{x_{\alpha}y_{\alpha t}-y_{\alpha}x_{\alpha t}}{(x_{\alpha}^{2})}=\frac{x_{\alpha}y_{\alpha t}-y_{\alpha}x_{\alpha t}}{s_{\alpha}^{2}}\\ &=-\textbf{n}\cdot [V_{s}\textbf{n}+Vk\textbf{s}+T_{s}\textbf{s}+T(-k\textbf{n})]=-V_{s}+kT, \end{array} $$

(A.2)

therefore

$$ \begin{array}{l} \theta_{\alpha t}=\theta_{t \alpha}=\frac{\partial [-V_{s}+kT] }{\partial \alpha}=s_{\alpha}\frac{\partial [-V_{s}+k T]}{\partial s}=s_{\alpha}[-V_{ss}+k_{s}T+kT_{s}]. \end{array} $$

(A.3)

Since $k=\frac {x_{\alpha }y_{\alpha \alpha }-x_{\alpha \alpha } y_{\alpha }}{s_{\alpha }^{3}}$ and $k=\frac {\theta _{\alpha }}{s_{\alpha }}$, we get

$$ \begin{array}{ll} k_{t}&\displaystyle=\frac{s_{\alpha} \theta_{\alpha t}-\theta_{\alpha} s_{\alpha t}}{s_{\alpha}^{2}}=\frac{s_{\alpha}^{2}[-V_{ss}+k_{s}T+kT_{s}]-s_{\alpha}^{2}k[T_{s}+kV]}{s_{\alpha}^{2}}\\ & \displaystyle=-V_{ss}+k_{s}T-k^{2}V=-(\frac{\partial^{2}}{\partial s \partial s}+k^{2})V+Tk_{s}. \end{array} $$

(A.4)

1.2 A.2 Linear analysis

Let X(α, t) be given in equation (62). Then,

$$ \begin{array}{ll} X_{\alpha}&=r_{\alpha}(cos\alpha,\sin\alpha)+r(-\sin\alpha,\cos\alpha)\\ X_{\alpha \alpha}&=r_{\alpha \alpha}(\cos\alpha, \sin\alpha)+r(-\cos \alpha,-\sin\alpha)+r_{\alpha}(-2\sin\alpha,2\cos\alpha). \end{array} $$

(A.5)

This implies that

$$ s_{\alpha}=\sqrt{(r_{\alpha}\cos \alpha+r(-\sin\alpha))^{2}+(r_{\alpha}\sin\alpha+r\cos\alpha)^{2}}=\sqrt{r^{2}+r_{\alpha}^{2}}, $$

(A.6)

and

$$ k=\frac{r^{2}+ 2r_{\alpha}^{2}-rr_{\alpha \alpha}}{(\sqrt {r^{2}+r_{\alpha}^{2}})^{3}}. $$

(A.7)

The normal vector and differential of the arc-length are

$$ \begin{array}{ll} \textbf{n}&\displaystyle=\frac{(r_{\alpha}\sin \alpha+r\cos\alpha,r\sin\alpha-r_{\alpha}\cos\alpha)}{\sqrt{r^{2}+r_{\alpha}^{2}}},\\ s_{\alpha}&\displaystyle=\sqrt{R^{2}+ 2R(\delta_{R}\cos(m\alpha)-\delta_{I}\sin(m\alpha))}+O(\delta^{2})\\ &\displaystyle=R[1+\frac{\delta_{R}\cos(m\alpha)-\delta_{I}\sin(m\alpha)}{R}+O(\delta^{2})]. \end{array} $$

(A.8)

Therefore

$$ \frac{1}{s_{\alpha}}=\frac{1}{R}[1-\frac{\delta_{R}\cos(m\alpha)-\delta_{I}\sin(m\alpha)}{R}+O(\delta^{2})]. $$

(A.9)

Using the expression for the curvature (A.7), it follows that the normal velocity is

$$ k_{s}=\frac{1}{s_{\alpha}}\frac{\partial \left[ \frac{r^{2}+ 2r_{\alpha}^{2}-rr_{\alpha \alpha}}{(\sqrt {r^{2}+r_{\alpha}^{2}})^{3}}\right]}{\partial \alpha}=\frac{1}{s_{\alpha}^{4}}(2rr_{\alpha}+ 3r_{\alpha}r_{\alpha\alpha}-rr_{\alpha\alpha\alpha})-\frac{3}{s_{\alpha}^{5}}s_{\alpha\alpha}(r^{2}+ 2r_{\alpha}^{2}-rr_{\alpha\alpha}). $$

(A.10)

A simple computation shows

$$ \begin{array}{ll} &r_{\alpha}=m(-\delta_{R}\sin(m\alpha)-\delta_{I}\cos(m\alpha))\\ & r_{\alpha \alpha}=m^{2}(-\delta_{R}\cos(m\alpha)+\delta_{I}\sin(m\alpha))\\ & r_{\alpha\alpha\alpha}=m^{3}(\delta_{R}\sin(m\alpha)+\delta_{I}\cos(m\alpha))\\ &\displaystyle s_{\alpha\alpha}=\frac{rr_{\alpha}+r_{\alpha}r_{\alpha \alpha}}{s_{\alpha}}. \end{array} $$

(A.11)

Thus, after substitution in (A.10) we can write

$$ \begin{array}{ll} &\displaystyle k_{s}=\frac{1}{s_{\alpha}^{4}}\left[2Rm[-\delta_{R}\sin(m\alpha)-\delta_{I}\cos(m\alpha)]-Rm^{3}(\delta_{R}\sin(m\alpha)+\delta_{I}\cos(m\alpha))\right]\\ &\displaystyle-\frac{3}{s_{\alpha}^{6}}[rr_{\alpha}+r_{\alpha}r_{\alpha \alpha}][R^{2}+ 2R(\delta_{R}\cos(m\alpha)-\delta_{I}\sin(m\alpha))-Rm^{2}(-\delta_{R}\cos(m\alpha)\\ &+\delta_{I}\sin(m\alpha))]\\ &\displaystyle=\frac{1}{R^{3}}\{[-m(\delta_{R}\sin(m\alpha)+\delta_{I}\cos(m\alpha))(2+m^{2})]+ 3m(\delta_{R}\sin(m\alpha))\\ &+\delta_{I}\cos(m\alpha)\}+O(\delta^{2})\\ &\displaystyle=\frac{1}{R^{3}}(\delta_{R}\sin(m\alpha)+\delta_{I}\cos(m\alpha))(m-m^{3})+O(\delta^{2}). \end{array} $$

(A.12)

The tangential velocity can be computed similarly:

$$ \begin{array}{ll} &\displaystyle T=\frac{\left( \frac{r^{2}+ 2r_{\alpha}^{2}-rr_{\alpha \alpha}}{(\sqrt {r^{2}+r_{\alpha}^{2}})^{3}} \right)^{2}}{2}=\frac{1}{2}\left( (\frac{1}{R^{3}}-3\frac{\gamma}{R^{4}})(R^{2}+ 2R\gamma-R(-m^{2})\gamma) \right)^{2}\\ &\displaystyle=\frac{1}{2}\left( \frac{1}{R}-\frac{1}{R^{2}}3\gamma+\frac{\gamma}{R^{2}}(2+m^{2})) \right)^{2}+O(\delta^{2})=\frac{1}{2} \left( \frac{1}{R^{2}}+\gamma(m^{2}-1) \right) \end{array} $$

(A.13)

where γ = δ_R cos(mα) − δ_I sin(mα). Therefore, the velocity is

$$ \begin{array}{ll} &\displaystyle W:=-k_{s}\textbf{n}+T\textbf{s}=\\ &\displaystyle(\frac{1}{R^{3}}(\delta_{R}\sin(m\alpha)+\delta_{I}\cos(m\alpha))(m^{3}-m))\textbf{n}+\frac{1}{2} \left( \frac{1}{R^{2}}+\gamma(m^{2}-1) \right) \textbf{s}+O(\delta^{2}), \end{array} $$

(A.14)

whose projection is

$$ W \cdot (\cos \alpha,\sin\alpha)=\delta_{I} \left( \frac{m^{3}-1.5m}{R^{3}} \right)\cos(m\alpha)+ \delta_{R} \left( \frac{m^{3}-1.5m}{R^{3}} \right)\sin(m\alpha)+O(\delta^{2}). $$

(A.15)

1.3 A.3 Direct calculation 1

We can write the equations based on CN scheme (19) as follows

$$ \widehat{\dot{\theta}_{m}^{j + 1}}={\zeta_{m}^{1}}\widehat{\dot{\theta}_{m}^{j-1}}+ 2{\Delta} t {\zeta_{m}^{2}} \widehat{\dot{ NL}_{m}^{j}}+A_{0}({\Delta} t^{3}), $$

(A.16)

$$ \widehat{\dot{\theta}_{m}^{j-1}}={\zeta_{m}^{1}}\widehat{\dot{\theta}_{m}^{j-3}}+ 2{\Delta} t {\zeta_{m}^{2}} \widehat{\dot{ NL}_{m}^{j-2}}+A_{0}({\Delta} t^{3}), $$

(A.17)

to obtain

$$ \begin{array}{ll} &\langle \widehat{\dot{\theta}^{j + 1}}-\widehat{\dot{\theta}^{j-1}}, \widehat{\dot{\theta}^{j + 1}}+\widehat{\dot{\theta}^{j-1}} \rangle=\\ &{\langle\zeta_{m}^{1}}\widehat{\dot{\theta}_{m}^{j-1}}+ 2{\Delta} t {\zeta_{m}^{2}} \widehat{\dot{ NL}_{m}^{j}}-\left( {\zeta_{m}^{1}}\widehat{\dot{\theta}_{m}^{j-3}}+ 2{\Delta} t {\zeta_{m}^{2}} \widehat{\dot{ NL}_{m}^{j-2}}+A_{0}({\Delta} t^{3}) \right),\\ &{\zeta_{m}^{1}}\widehat{\dot{\theta}_{m}^{j-1}}+ 2{\Delta} t {\zeta_{m}^{2}} \widehat{\dot{ NL}_{m}^{j}}+{\zeta_{m}^{1}}\widehat{\dot{\theta}_{m}^{j-3}}+ 2{\Delta} t {\zeta_{m}^{2}} \widehat{\dot{ NL}_{m}^{j-2}}+A_{0}({\Delta} t^{3})\rangle\\ &=\langle {\zeta_{m}^{1}}\left( \widehat{\dot{\theta}_{m}^{j-1}} -\widehat{\dot{\theta}_{m}^{j-3}}\right)+ 2{\Delta} t {\zeta_{m}^{2}}\left( \widehat{\dot{ NL}_{m}^{j}}- \widehat{\dot{ NL}_{m}^{j-2}}\right)+A_{0}({\Delta} t^{3}),\\ & {\zeta_{m}^{1}}\left( \widehat{\dot{\theta}_{m}^{j-1}} +\widehat{\dot{\theta}_{m}^{j-3}}\right)+ 2{\Delta} t {\zeta_{m}^{2}}\left( \widehat{\dot{ NL}_{m}^{j}}+ \widehat{\dot{ NL}_{m}^{j-2}}\right)+A_{0}({\Delta} t^{3})\rangle. \end{array} $$

(A.18)

1.4 A.4 Nonlinear error estimates

In this proof, the hypothesis 2 ≤ r is used, which is satisfied in both schemes (4 ≤ r for ADB and 6 ≤ r for CN).

Proof Lemma 1

We start computing upper bounds for $|S_{h}\dot {f^{j}}|_{\infty },|\dot {f^{j}}|_{\infty }$. Notice that

$$ h|\dot{f}^{j}|^{2}\leq \sum\limits_{i=-N/2 + 1}^{N/2}|\dot{{f_{i}^{j}}}|^{2}h=||\dot{f^{j}}||_{l^{2}}^{2}, $$

(A.19)

which implies

$$ |\dot{{f_{m}^{j}}}|_{\infty}\leq h^{-1/2}||\dot{f^{j}}||_{l^{2}} \text{ and, } |S_{h}\dot{f^{j}}|_{\infty},|S_{h}\dot{f^{j}}|_{\infty}\leq h^{-3/2}||\dot{f^{j}}||_{l^{2}}. $$

(A.20)

Then, condition 2 ≤ r and the definition of T^∗ shows that

$$ |S_{h}\dot{\theta^{j}}|_{\infty},|\dot{\theta^{j}}|_{\infty}\leq C \text{ are bounded.} $$

(A.21)

Now, we define

$$ NL(\alpha_{m},t_{j}):=\frac{1}{2}(\frac{2\pi}{L})^{3}(\theta_{\alpha}(\alpha_{m},t_{j}))^{3} \text{ and }\: \widetilde{N{L_{m}^{j}}}:=\frac{1}{2}(\frac{2\pi}{\widetilde{L}})^{3}(S_{h} \widetilde{{\theta_{m}^{j}}})^{3}, $$

(A.22)

to write the nonlinear terms as follows:

$$ \dot{NL}_{m}^{j}:=\widetilde{N{L_{m}^{j}}}-NL(\alpha_{m},t_{j})=\dot{\xi}_{m}^{j}{T_{m}^{j}}+\dot{\xi}_{m}^{j}\dot{T}_{m}^{j}+{\xi_{m}^{j}}\dot{T}_{m}^{j}, $$

(A.23)

where

$$ {\xi_{m}^{j}}:=\frac{2\pi}{L}S_{h}{\theta_{m}^{j}},\: \xi(\alpha_{m},t_{j}):=\frac{2\pi}{L} \theta_{\alpha}(\alpha_{m},t_{j}),\: \widetilde{{\xi_{m}^{j}}}:=\frac{2\pi}{\widetilde{L}}S_{h} \widetilde{{\theta_{m}^{j}}}, $$

(A.24)

$$ \dot{\xi}_{m}^{j}:=(\widetilde{{\xi_{m}^{j}}}- {\xi_{m}^{j}})+({\xi_{m}^{j}}-\xi(\alpha_{m},t_{j})), $$

(A.25)

and

$$ {T_{m}^{j}}:=\frac{1}{2}(\frac{2\pi}{L})^{2}(S_{h}{\theta_{m}^{j}})^{2},\: T(\alpha_{m},t_{j}):=\frac{1}{2}(\frac{2\pi}{L})^{2}(\theta_{\alpha}(\alpha_{m},t_{j}))^{2},\: \widetilde{{T_{m}^{j}}}:=\frac{1}{2}(\frac{2\pi}{\widetilde{L}})^{2}(S_{h} \widetilde{{\theta_{m}^{j}}})^{2}. $$

(A.26)

□

1.4.1 A.4.1 Error in tangential velocity

We calculate the error for the tangential velocity

$$ \dot{T}_{m}^{j}:=\widetilde{{T_{m}^{j}}}-T(\alpha_{m},t_{j})=\frac{1}{2}\left[ \widetilde{{\xi_{m}^{j}}}^{2}-({\xi_{m}^{j}})^{2}\right]=\frac{1}{2}\left[ 2\dot{\xi}_{m}^{j} {\xi_{m}^{j}}+(\dot{\xi}_{m}^{j})^{2}\right]. $$

(A.27)

Since the truncation error $({\xi _{m}^{j}}-\xi (\alpha _{m},t_{j}))=O(h^{r + 2})$, it follows that

$$\begin{array}{@{}rcl@{}} \dot{\xi}_{m}^{j}&=&\frac{2\pi}{\widetilde{L}}S_{h}\widetilde{{\theta_{m}^{j}}}-\frac{2\pi}{L}S_{h}{\theta_{m}^{j}}+O(h^{r + 2})= 2\pi S_{h}\dot{\theta}_{m}^{j}\dot{(L^{-1})}+\frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}\\ &&+ 2\pi S_{h}{\theta_{m}^{j}}\dot{(L^{-1})}+O(h^{r + 2}), \end{array} $$

where

$$\dot{L^{-1}}=\frac{1}{\widetilde{L}}-\frac{1}{L}=\frac{-\dot{L}}{L^{2}}+\frac{(\dot{L})^{2}}{L^{2}[L+\dot{L}]}, $$

which implies $|\dot {L^{-1}}|\leq C|\dot {L}|$, and therefore

$$ A_{0}\dot{(L^{-1})}=A_{0}(\dot{L}). $$

(A.28)

Combining (A.21), (A.28) and

$$ S_{h}{\theta_{m}^{j}}=\theta_{\alpha}(\alpha_{m},t_{j})+O(h^{r + 2}). $$

(A.29)

in the expression for $\dot {\xi }_{m}^{j}$ we see that

$$ \dot{\xi}_{m}^{j}=\frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j} +A_{0}(\dot{L})+O(h^{r + 2}). $$

(A.30)

By hypothesis over time (35), (A.45) and computation (A.20) we find that

$$ |A_{0}(\dot{L})|_{\infty}\leq Ch^{-1/2}||\dot{L}||_{l^{2}}=h^{-1/2}O(h^{r + 3})=O(h^{r + 5/2}). $$

(A.31)

Hence, we rewrite

$$ \dot{\xi}_{m}^{j}=\frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j} +O(h^{r + 2})=h^{-1}A_{0}(\dot{\theta}^{j})+O(h^{r + 2}). $$

(A.32)

Also $||O(h^{r + 2})||_{l^{2}}=O(h^{r + 2})$ and the conditions $\frac {{\Delta } t}{h}\leq C$, 2 ≤ r imply that

$$ \begin{array}{ll} &||\dot{\xi}^{j}||_{l^{2}}=O(h^{-1}(h^{r}+{\Delta} t^{2}))+O(h^{r + 2}) =O(h) \text{ and }\\ &|\dot{\xi}|_{\infty}=h^{-1/2}||\dot{\xi}^{j}||_{l^{2}}\leq C\text{ are bounded.} \end{array} $$

(A.33)

With this information (A.33) back to (A.27) is possible to rewrite

$$\dot{T}_{m}^{j}=\frac{1}{2}\left[ 2\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right)\frac{2\pi}{L}S_{h}{\theta_{m}^{j}}+ \left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right)^{2} \right]. $$

Again using (A.29), the fact that θ_α is bounded and the upper bound (A.33), the first term on the right hand side becomes

$$ \begin{array}{ll} &\displaystyle\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right)\frac{2\pi}{L}S_{h}{\theta_{m}^{j}}=\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right) \left( \frac{2\pi}{L}\theta_{\alpha}(\alpha_{m},t_{j})+O(h^{r + 2})\right)\\ &\displaystyle=\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right) \frac{2\pi}{L}\theta_{\alpha}(\alpha_{m},t_{j})+ \dot{\xi}_{m}^{j} O(h^{r + 2})=\theta_{\alpha}(\alpha_{m},t_{j})(\frac{2\pi}{L})^{2}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2}). \end{array} $$

(A.34)

As shown previously $S_{h}\dot {\theta }_{m}^{j},\dot {\xi }_{m}^{j}$, are bounded. Thus, the second term can be computed as

$$ \begin{array}{ll} &\displaystyle\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right)^{2}=\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right)\dot{\xi}_{m}^{j}=\frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j} \dot{\xi}_{m}^{j}+O(h^{r + 2})\\ &\displaystyle=\frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right)+O(h^{r + 2})=\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}\right)^{2}+O(h^{r + 2}) \end{array} $$

(A.35)

Using 2 ≤ r we verify the inequality

$$ ||(S_{h}\dot{\theta}_{m}^{j})^{2}||_{l^{2}}\leq h^{-2}||\dot{\theta}_{m}^{j}||^{2}_{l^{2}}=O(h^{-2})O(h^{r}+{\Delta} t^{2})||\dot{\theta}^{j}||_{l_{2}}=A_{0}(\dot{\theta}^{j}). $$

(A.36)

Therefore

$$ \dot{T}_{m}^{j}=(\frac{2\pi}{L})^{2}\theta_{\alpha}(\alpha_{m},t_{j})S_{h}\dot{\theta}_{m}^{j}+A_{0}(\dot{\theta}^{j})+O(h^{r + 2})=h^{-1}A_{0}(\dot{\theta}^{j})+A_{0}(\dot{\theta}^{j})+O(h^{r + 2}). $$

(A.37)

By hypothesis for time T^∗ and 2 ≤ r, we find that

$$ ||\dot{T^{j}}||_{l^{2}}=O(h^{-1}(h^{r}+{\Delta} t^{2}))+O(h^{r}+{\Delta} t^{2})\leq C,\: |\dot{T^{j}}|_{\infty}\leq h^{-1/2}||\dot{T^{j}}||_{l^{2}}\leq C $$

(A.38)

are bounded quantities.

1.4.2 A.4.2 Error for nonlinear term

Combining equation (A.32), (A.33), (A.37), (A.38), we approximate (A.23) to obtain

$$ \begin{array}{ll} &\dot{NL}_{m}^{j}=\displaystyle\left\{ \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j} +O(h^{r + 2})\right\}{T_{m}^{j}}+\dot{\xi}_{m}^{j}\dot{T}_{m}^{j}+{\xi_{m}^{j}}\left\{ \frac{(2\pi)^{2}}{L^{2}}\theta_{\alpha}(\alpha_{m},t_{j}) S_{h}\dot{\theta}_{m}^{j}\right.\\ &\displaystyle\left.\vphantom{\frac{(2\pi)^{2}}{L^{2}}}+A_{0}(\dot{\theta}^{j})+O(h^{r + 2}) \right\}\\ &\displaystyle=\frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}{T_{m}^{j}}+\dot{\xi}_{m}^{j}\dot{T}_{m}^{j}+{\xi_{m}^{j}}(\frac{2\pi}{L})^{2}{\theta_{\alpha}(\alpha_{m},t_{j})}^{2} S_{h}\dot{\theta}_{m}^{j}+A_{0}(\dot{\theta}^{j})+O(h^{r + 2}). \end{array} $$

(A.39)

The second term on the previous expression can be analyzed using equations (A.21) as follows

$$ \begin{array}{ll} &\displaystyle\dot{\xi}_{m}^{j}\dot{T}_{m}^{j}=\left( \frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}+O(h^{r + 2})\right)\left( (\frac{2\pi}{L})^{2}\theta_{\alpha}(\alpha_{m},t_{j})S_{h}\dot{\theta}_{m}^{j}+A_{0}(\dot{\theta}_{m}^{j})+O(h^{r + 2}) \right)\\ &\displaystyle= \frac{2\pi}{L}\left( S_{h}\dot{\theta}_{m}^{j}\right)^{2} (\frac{2\pi}{L})^{2}\theta_{\alpha}(\alpha_{m},t_{j})+A_{0}(\dot{\theta^{j}})+O(h^{r + 2}), \end{array} $$

(A.40)

which by the estimate (A.36) simplifies to $\dot {\xi }_{m}^{j}\dot {T}_{m}^{j}=A_{0}(\dot {\theta })+O(h^{r + 2})$.

As a consequence of the truncation error for the tangent velocity ${T_{m}^{j}}-T(\alpha _{m},t_{j})=O(h^{r + 2})$ and ${\xi _{m}^{j}}-\xi (\alpha _{m}t_{j})=O(h^{r + 2})$, we obtain

$$ \begin{array}{ll} &\displaystyle\dot{NL}_{m}^{j}=\frac{2\pi}{L}S_{h}\dot{\theta}_{m}^{j}\left( \frac{1}{2}(\frac{2\pi}{L})^{2}\theta_{\alpha}^{2}(\alpha_{m},t_{j})+O(h^{r + 2})\right)\\ &\displaystyle+\left( \frac{2\pi}{L}\theta_{\alpha}(\alpha_{m},t_{j})+O(h^{r + 2}) \right)(\frac{2\pi}{L})^{2}{\theta_{\alpha}(\alpha_{m},t_{j})}^{2} S_{h}\dot{\theta}_{m}^{j}+A_{0}(\dot{\theta}^{j})+O(h^{r + 2}). \end{array} $$

(A.41)

Finally, using equation (A.21) we attain an expression for the nonlinear error

$$ \dot{NL}_{m}^{j}=\frac{3}{2}(\frac{2\pi}{L})^{3}\theta_{\alpha}^{2}(\alpha_{m},t_{j})S_{h}\dot{\theta}_{m}^{j} +A_{0}(\dot{\theta}^{j})+O(h^{r + 2})=h^{-1}A_{0}(\dot{\theta}^{j})+A_{0}(\dot{\theta}^{j})+O(h^{r + 2}), $$

(A.42)

and the upper bounds

$$ \begin{array}{ll} ||\dot{NL}_{m}^{j}||_{l_{2}}&\leq \frac{3}{2}\frac{2\pi}{L}|k^{2}|_{\infty}h^{-1}||\dot{\theta}_{m}^{j}||_{l^{2}}+O(h^{r}+{\Delta} t^{2})+O(h^{r + 3/2})\\ &=O(h^{r-1}+h^{-1}{\Delta} t^{2}),\\ |\dot{NL}^{j}|_{\infty}&=h^{-1/2}||\dot{NL}^{j}||_{l^{2}}\leq C,\\ {\Delta} t ||\dot{NL}_{m}^{j}||_{l_{2}}&=C\frac{{\Delta} t}{h}||\dot{\theta}_{m}^{j}||_{l^{2}}+O(h^{r}+{\Delta} t^{2})=O(h^{r}+{\Delta} t^{2}), \end{array} $$

(A.43)

for j = 1, ... , n, provided that $\frac {{\Delta } t}{h}$ is bounded.

1.5 A.5 Proof of convergence for Adams Bashforth (ADB) discretization

Proof of Theorem 1

The error between the numerical and the exact solution (at a given time t_j) is given by

$$ \dot{\theta}_{m}^{j}:=\widetilde{{\theta_{m}^{j}}}-\theta(\alpha_{m},t_{j}). $$

(A.44)

Defining the auxiliary time,

$$ T^{*}=Sup\{t|t\leq T,|\dot{L}|<h^{r + 3},||\dot{\theta}^{j}||_{l^{2}}=O(h^{r}+{\Delta} t^{2})\},\\ $$

(A.45)

for j = 0, 1, ... , n (we have an overall accuracy of h²) we will show that the error at the step n + 1 also satisfies the estimate $||\dot {\theta }^{n + 1}||_{l^{2}}=O(h^{r}+{\Delta } t^{2})$. Hence T^∗ = T by induction. □

Taylor approximations:

for the first step of the induction argument, we calculate upper bounds for the first step, based on a combination of Euler and integrating factor method (IFM) using the Taylor expansion:

$$ \frac{\partial {\Psi}}{\partial t}=(rNL),\:\: {r_{m}^{t}}(t)=e^{i(2\pi m)^{3}tL^{-3}},\:\: {\Psi}(m,j)={r_{m}^{t}}\widehat{{\theta_{m}^{j}}}. $$

(A.46)

Expanding Ψ around time t₀ and defining $\zeta _{m}=e^{-i(2\pi m)^{3}L^{-3}{\Delta } t}$, we obtain

$$ \widehat{ {\theta_{m}^{1}}}=\zeta_{m} (\widehat{ {\theta_{m}^{0}}}+{\Delta} t \widehat{N{L_{m}^{0}}} )+{\Delta} t^{2}\frac{\zeta_{m}}{2}({\Psi}_{tt})_{m}^{0}+O({\Delta} t^{3}). $$

(A.47)

The numerical solution satisfies at the first step (Euler discretization)

$$ \widehat{\widetilde{{\theta_{m}^{1}}}}=\zeta_{m} (\widehat{ \widetilde{ {\theta_{m}^{0}}}}+{\Delta} t \widehat{\widetilde{N{L_{m}^{0}}}} ) $$

(A.48)

thus, we can write an expression for the error at the first step

$$ \widehat{\dot{\theta}_{m}^{1}}=\zeta_{m} \widehat{\dot{\theta}_{m}^{0}} +{\Delta} t \zeta_{m} \widehat{\dot{NL}_{m}^{0}}+{\Delta} t^{2}\frac{\zeta_{m}}{2}({\Psi}_{t t})_{m}^{0}+O({\Delta} t^{3}). $$

(A.49)

Observe that $||\frac {\zeta }{2}\widehat {({\Psi }_{t t})^{0}}||=\frac {1}{2}||\widehat {({\Psi }_{t t})^{0}}||=\frac {1}{2\sqrt {2\pi }}||({\Psi }_{t t})^{0}||_{l^{2}}$. We will see that the coefficients of Δt² term are bounded (independent of discretization) in l² norm. Since

$$ {{\Psi}_{t}}_{m}=e^{i t(\frac{2\pi m}{L})^{3}}\widehat{NL_{m}}, $$

(A.50)

then

$$ {{\Psi}_{t t}}_{m}=e^{i t(\frac{2\pi m}{L})^{3}}\left( (\widehat{N{L^{0}_{m}}})_{t}+\widehat{N{L^{0}_{m}}}i(\frac{2\pi m}{L})^{3}\right). $$

(A.51)

Because θ is 2 times differentiable with respect to time (so we can commute derivatives), we can write $NL_{t}=-\frac {1}{2s_{\alpha }^{3}}\frac {3}{2}\theta _{\alpha }^{2}\theta _{\alpha t}=-\frac {1}{2s_{\alpha }^{3}}\frac {3}{2}\theta _{\alpha }^{2}\frac {1}{s_{\alpha }^{3}}[\theta _{\alpha \alpha \alpha \alpha }+(\frac {\theta _{\alpha }^{3}}{2})_{\alpha }]$ which involves spatial derivatives of order 4 for theta. Hence by the assumption 4 ≤ r, these derivatives are L² integrable. Also $(\widehat {NL_{m}})_{t}=(\widehat {{{NL_{t}}_{m}}})$ shows that $(\widehat {NL^{0}})_{t}$ is l² integrable.

To control the second term of (A.51) observe that

$$ \widehat{(NL_{sss})^{0}_{m}}=-\widehat{N{L^{0}_{m}}}i(\frac{2\pi m}{L})^{3}, $$

(A.52)

and $NL_{sss}=\frac {1}{2s_{\alpha }^{6}}(\theta _{\alpha }^{3})_{\alpha \alpha \alpha }$ involves L² integrable derivatives of order 4 for theta. This shows that $\widehat {NL^{0}}i(\frac {2\pi m}{L})^{3}$ is bounded in the l² norm and consequently $||({\Psi }_{t t})_{m}^{0}||_{l^{2}}$ is also bounded. In other words

$$ \widehat{ {\theta_{m}^{1}}}=\zeta_{m} (\widehat{ {\theta_{m}^{0}}}+{\Delta} t \widehat{N{L_{m}^{0}}} )+A_{0}({\Delta} t^{2}), $$

(A.53)

and

$$ ||\widehat{\dot{\theta}^{1}}||^{2}=\langle \zeta_{m} \widehat{\dot{\theta}_{m}^{0}} +{\Delta} t \zeta_{m} \widehat{\dot{NL}_{m}^{0}}+A_{0}({\Delta} t^{2}), \zeta_{m} \widehat{\dot{\theta}_{m}^{0}} +{\Delta} t \zeta_{m} \widehat{\dot{NL}_{m}^{0}}+A_{0}({\Delta} t^{2})\rangle. $$

(A.54)

Now, we analyze the error after the second step (1 ≤ j). Using Taylor’s approximation we obtain

$$ \widehat{\theta_{m}^{j + 1}}=\widehat{{\theta_{m}^{j}}}\zeta_{m}+\frac{{\Delta} t}{2}(3\zeta_{m} \widehat{N{L_{m}^{j}}}-(\zeta_{m})^{2}\widehat{NL_{m}^{j-1}})+\frac{5{\Delta} t^{3}}{12}({\Psi}^{(3)})_{m}^{j}+O({\Delta} t^{4}). $$

(A.55)

On the other hand, the numerical solution (17) satisfies:

$$ \widehat{\widetilde{\theta_{m}^{j + 1}}}=\widehat{\widetilde{{\theta_{m}^{j}}}} \zeta_{m}+\frac{{\Delta} t}{2}(3 \zeta_{m}\widehat{\widetilde{N{L_{m}^{j}}}}- (\zeta_{m})^{2}\widehat{\widetilde{NL_{m}^{j-1}}}). $$

(A.56)

Subtracting (A.55) from (A.56) we obtain the following equation for the error in θ:

$$ \widehat{\dot{\theta}_{m}^{j + 1}}=\zeta_{m} \widehat{\dot{\theta}_{m}^{j}}+\frac{{\Delta} t}{2}\widehat{\dot{ \mu}_{m}^{j}}+\frac{5{\Delta} t^{3}}{12}({\Psi}^{(3)})_{m}^{j}+O({\Delta} t^{4}), $$

(A.57)

where

$$ \widehat{\dot{ \mu}_{m}^{j}}= 3 \zeta_{m}\widehat{\dot{NL}_{m}^{j}}- (\zeta_{m})^{2}\widehat{\dot{NL}_{m}^{n-1}}. $$

(A.58)

Similarly way to the first step and for future estimates we show that the coefficient for the Δt³ term in (A.57) is integrable in the l² norm.

From (A.51) we obtain

$$ \begin{array}{ll} &{{\Psi}_{t t t}}_{m}=\\ &\displaystyle e^{i t(\frac{2\pi m}{L})^{3}}\left( (\widehat{NL_{m}})_{t t}\,+\,(\widehat{NL_{m}})_{t}i(\frac{2\pi m}{L})^{3} + i(\frac{2\pi m}{L})^{3}\left( (\widehat{NL_{m}})_{t}\,+\,\widehat{NL_{m}}i(\frac{2\pi m}{L})^{3}\right)\right)\\ &=e^{i t(\frac{2\pi m}{L})^{3}}\left( (\widehat{NL_{m}})_{t t}+ 2\underbrace{(\widehat{NL_{m}})_{t}i(\frac{2\pi m}{L})^{3}}_{T1} +\underbrace{(i(\frac{2\pi m}{L})^{3})^{2}\widehat{NL_{m}}}_{T2}\right). \end{array} $$

(A.59)

Now

$$\begin{array}{ll} &\displaystyle NL_{t t}=((\frac{{\theta_{s}^{3}}}{2})_{t})_{t}=(\frac{1}{s_{\alpha}^{3}}\frac{3}{2}\theta_{\alpha}^{2}\theta_{\alpha t})_{t}=\frac{3}{2s_{\alpha}^{3}}\left( \theta_{\alpha}^{2}(\theta_{sss}+\frac{{\theta_{s}^{3}}}{2})_{\alpha}\right)_{t}\\ &\displaystyle=\frac{3}{2s_{\alpha}^{3}}\left( 2\theta_{\alpha}\theta_{\alpha t}\theta_{t \alpha}+ \theta_{\alpha}^{2}(\theta_{sss}+(\frac{{\theta_{s}^{3}}}{2}))_{\alpha t}\right)\\ &\displaystyle=\frac{3}{2s_{\alpha}^{3}}\left( 2\theta_{\alpha}\theta_{\alpha t}\theta_{t \alpha}+ \theta_{\alpha}^{2}(\theta_{sss}+(\frac{{\theta_{s}^{3}}}{2}))_{\alpha t}\right), \end{array} $$

(A.60)

involves spatial derivatives of order 7 for θ, since 4 ≤ r (by hypothesis) we know these are l² integrable. In addition temporal derivatives, and Fourier transform commute, we conclude that each term in (A.60) is l² integrable. Moreover, (NL_t)_sss, (NL)_ssssss are also l² integrable provided θ is at least 7 times differentiable. Consequently, terms T1 and T2 of (A.59) are l² integrable too. This implies that Ψ_ttt is also l² integrable and we rewrite (A.57) as

$$ \widehat{\dot{\theta}_{m}^{j + 1}}=\zeta_{m} \widehat{\dot{\theta}_{m}^{j}}+\frac{{\Delta} t}{2}\widehat{\dot{ \mu}_{m}^{j}}+A_{0}({\Delta} t^{3}). $$

(A.61)

To estimate the error consider the inner product

$$ \langle \widehat{\dot{\theta}^{j + 1}}-\zeta \widehat{\dot{\theta}^{j-1}}, \widehat{\dot{\theta}^{j + 1}}+\zeta \widehat{\dot{\theta}^{j-1}}\rangle=||\widehat{\dot{\theta}^{j + 1}}||^{2}-|| \widehat{\dot{\theta}^{j-1}}||^{2}+ 2iIm(\langle \widehat{\dot{\theta}^{j + 1}},\zeta \widehat{\dot{\theta}^{j-1}} \rangle), $$

(A.62)

where we have used that |ζ_m| = 1 for each m and ζ = (ζ_{−N/2 + 1},.., ζ_N/2).

Using (A.61) and

$$ \widehat{\dot{\theta}_{m}^{j}}=\zeta_{m} \widehat{\dot{\theta}_{m}^{j-1}}+\frac{{\Delta} t}{2}\widehat{\dot{ \mu}_{m}^{j-1}}+A_{0}({\Delta} t^{3}), $$

(A.63)

into the main equation (A.62) we obtain the right hand side

$$ \begin{array}{ll} &\langle \widehat{\dot{\theta}^{j + 1}}-\zeta \widehat{\dot{\theta}^{j-1}}, \widehat{\dot{\theta}^{j + 1}}+\zeta \widehat{\dot{\theta}^{j-1}}\rangle\\ &\displaystyle=\langle (\zeta-1) \widehat{\dot{\theta}^{j}}+ \frac{{\Delta} t}{2} (\widehat{\dot{\mu}^{j}}+\widehat{\dot{\mu}^{j-1}})+A_{0}({\Delta} t^{3}), (\zeta+ 1) \widehat{\dot{\theta}^{j}}\\ &+ \frac{{\Delta} t}{2}(\widehat{\dot{\mu}^{j}}-\widehat{\dot{\mu}^{j-1}})+A_{0}({\Delta} t^{3}) \rangle. \end{array} $$

(A.64)

By definition (A.58) of $\dot {\mu }_{m}^{j}$, using the estimate for the nonlinear error ${\Delta } t ||\dot {NL}^{j}||_{l^{2}}=O(h^{r}+{\Delta } t^{2})$ (57) and Plancherel theorem we have that

$$ ||{\Delta} t\dot{ \mu}^{j}||_{l^{2}}={\Delta} t|| 3 \zeta\widehat{\dot{NL}^{j}}- (\zeta)^{2}\widehat{\dot{NL}^{n-1}}||\leq C{\Delta} t(||\dot{NL}^{j}||_{l^{2}}+||\dot{NL}^{j-1}||_{l^{2}})=O(h^{r}+{\Delta} t^{2}), $$

(A.65)

for j = 1, ... , n.

Then we rewrite (A.64) as follows

$$ \begin{array}{ll} &\underbrace{\langle (\zeta-1) \widehat{\dot{\theta}^{j}}, (\zeta+ 1) \widehat{\dot{\theta}^{j}}\rangle}_{{J_{1}^{j}}} +\\ &\underbrace{\langle (\zeta-1) \widehat{\dot{\theta}^{j}}, \frac{{\Delta} t}{2}(\widehat{\dot{\mu}^{j}}-\widehat{\dot{\mu}^{j-1}})\rangle+ \langle \frac{{\Delta} t}{2}(\widehat{\dot{\mu}^{j}}+\widehat{\dot{\mu}^{j-1}}),(\zeta+ 1)\widehat{\dot{\theta}^{j}} \rangle}_{{J_{2}^{j}}}\\ &+\underbrace{(\frac{ {\Delta} t}{2})^{2}\langle \widehat{\dot{\mu}^{j}}+\widehat{\dot{\mu}^{j-1}}, \widehat{\dot{\mu}^{j}}-\widehat{\dot{\mu}^{j-1}}) \rangle}_{{J_{3}^{j}}}+\\ & \underbrace{\langle A_{0}({\Delta} t^{3}), (\zeta+ 1) \widehat{\dot{\theta}^{j}}+ \frac{{\Delta} t}{2}(\widehat{\dot{\mu}^{j}}-\widehat{\dot{\mu}^{j-1}})+A_{0}({\Delta} t^{3}) \rangle}_{{J_{4}^{j}}}\\ &+\underbrace{\langle (\zeta-1) \widehat{\dot{\theta}^{j}}+ \frac{{\Delta} t}{2} (\widehat{\dot{\mu}^{j}}+\widehat{\dot{\mu}^{j-1}})+A_{0}({\Delta} t^{3}),A_{0}({\Delta} t^{3}) \rangle}_{{J_{5}^{j}}}. \end{array} $$

(A.66)

Adding those terms (A.62) over time, we obtain a telescopic sum

$$ \sum\limits_{j = 2}^{n}\langle \widehat{\dot{\theta}^{j + 1}}-\zeta \widehat{\dot{\theta}^{j-1}}, \widehat{\dot{\theta}^{j + 1}}+\zeta \widehat{\dot{\theta}^{j-1}}\rangle=||\widehat{\dot{\theta}^{n + 1}}||^{2}+|| \widehat{\dot{\theta}^{n}}||^{2}-\left( ||\widehat{\dot{\theta}^{2}}||^{2}+|| \widehat{\dot{\theta}^{1}}||^{2}\right)+I_{1}, $$

(A.67)

where I₁ is a purely imaginary term.

Now we analyze the sum over time of the right-hand side terms

J ₁ contribution:

a direct calculation shows that

$$ {J_{1}^{j}}=||\widehat{\dot{\theta}^{j}}||^{2}-||\widehat{\dot{\theta}^{j}}||^{2}+ 2iIm(\langle \zeta \widehat{\dot{\theta}^{j}},\widehat{\dot{\theta}^{j}} \rangle). $$

(A.68)

Thus, the sum over time is telescopic

$$ \sum\limits_{j = 2}^{n}{J_{1}^{j}}=|| \widehat{\dot{\theta}^{n-1}}||^{2}+|| \widehat{\dot{\theta}^{n-2}}||^{2}-|| \widehat{\dot{\theta}^{1}}||^{2}-|| \widehat{\dot{\theta}^{0}}||^{2}+I_{2}, $$

(A.69)

where I₂ is a purely imaginary term.

J ₂ contribution:

similarly

$${J_{2}^{j}}= 2Re\underbrace{\left( \langle \zeta \widehat{\dot{\theta}^{j}},\frac{{\Delta} t}{2}\widehat{\dot{\mu}^{j}}\rangle+\langle \widehat{\dot{\theta}^{j}},\frac{{\Delta} t}{2}\widehat{\dot{\mu}^{j-1}} \rangle\right)}_{J_{*}}+ 2iIm\left( \langle \frac{{\Delta} t}{2} \widehat{\dot{\mu}^{j-1}},\zeta \widehat{\dot{\theta}^{j}} \rangle+\langle\frac{{\Delta} t}{2}\widehat{\dot{\mu}^{j}},\widehat{\dot{\theta}^{j}} \rangle\right), $$

where

$$ \begin{array}{ll} J_{*}&= 2Re\left( \langle \zeta \widehat{\dot{\theta}^{j}},\widehat{\dot{\theta}^{j + 1}}-\zeta \widehat{\dot{\theta}^{j}}+A_{0}({\Delta} t^{3}) \rangle+\langle \widehat{\dot{\theta}^{j}},\widehat{\dot{\theta}^{j}}-\zeta\widehat{\dot{\theta}^{j-1}}+A_{0}({\Delta} t^{3})\rangle \right)\\ &= 2Re\left( \langle \zeta \widehat{\dot{\theta}^{j}},\widehat{\dot{\theta}^{j + 1}}\rangle-||\widehat{\dot{\theta}^{j}}||^{2}+||\widehat{\dot{\theta}^{j}} ||^{2}-\langle \widehat{\dot{\theta}^{j}},\zeta\widehat{\dot{\theta}^{j-1}}\rangle \right)+A_{0}({\Delta} t^{3})A_{0}(\dot{\theta^{j}})\\ &= 2Re\left( \langle \widehat{\dot{\theta}^{j + 1}},\zeta \widehat{\dot{\theta}^{j}} \rangle-\langle \widehat{\dot{\theta}^{j}},\zeta\widehat{\dot{\theta}^{j-1}}\rangle \right)+{\Delta} tA_{0}({\Delta} t^{2})A_{0}(\dot{\theta^{j}}). \end{array} $$

(A.70)

The sum over time is

$$ \begin{array}{ll} \displaystyle\sum\limits_{j = 2}^{n}{J_{2}^{j}}&= 2Re\left( \langle \widehat{\dot{\theta}^{n + 1}},\zeta \widehat{\dot{\theta}^{n}} \rangle-\langle \widehat{\dot{\theta}^{2}},\zeta\widehat{\dot{\theta}^{1}}\rangle \right)+A_{0}({\Delta} t^{2})A_{0}(\dot{\theta^{j}})+I_{3}\\ &\displaystyle= 2Re\left( \langle \zeta_{m} \widehat{\dot{\theta}_{m}^{n}}+\frac{{\Delta} t}{2}\widehat{\dot{ \mu}_{m}^{n}}+A_{0}({\Delta} t^{3}) ,\zeta \widehat{\dot{\theta}^{n}} \rangle-\langle \widehat{\dot{\theta}^{2}},\zeta\widehat{\dot{\theta}^{1}}\rangle \right)\\ &\quad+A_{0}({\Delta} t^{2})A_{0}(\dot{\theta^{j}})+I_{3} \end{array} $$

(A.71)

where I₃ is a purely imaginary term.

For the first step, $\dot {\theta }_{m}^{0}$ is zero. In addition (A.54), (57) and Plancherel theorem shows that

$$ ||\dot{\theta}^{1}||_{l^{2}}^{2}= ||\widehat{\dot{\theta}^{1}}||^{2}=||A_{0}({\Delta} t^{2})||^{2}=O({\Delta} t^{4}). $$

(A.72)

For the second step, consider (A.61) and approximation (A.65) to get

$$||\dot{\theta}^{2}||_{l^{2}}^{2}\leq O((h^{r}+{\Delta} t^{2})^{2}), $$

which by induction implies that $||\dot {\theta }^{n}||=O(h^{r}+{\Delta } t^{2})$.

Considering only real terms in (A.71) and approximation (A.65) for the nonlinear error we obtain

$$ \begin{array}{ll} \displaystyle|Re(\sum\limits_{j = 2}^{n}{J_{2}^{j}})|&\displaystyle\leq 2|| \zeta_{m} \widehat{\dot{\theta}_{m}^{n}}+\frac{{\Delta} t}{2}\widehat{\dot{ \mu}_{m}^{n}}+A_{0}({\Delta} t^{3})|| \cdot||\zeta \widehat{\dot{\theta}^{n}}|| +C(h^{r}{\Delta} t^{2}+{\Delta} t^{4})\\ &\displaystyle\leq O(h^{r}+{\Delta} t^{2})O(h^{r}+{\Delta} t^{2})+O((h^{r}+{\Delta} t^{2})^{2})=O((h^{r}+{\Delta} t^{2})^{2}). \end{array} $$

(A.73)

J ₃ contribution:

a direct calculation shows

$$ {J_{3}^{j}}=(\frac{{\Delta} t}{2})^{2}(||\widehat{\dot{\mu}^{j}}||^{2}-||\widehat{ \dot{\mu}^{j-1}}||^{2})+(\frac{{\Delta} t}{2})^{2}2iIm(\langle\widehat{\dot{ \mu}^{j-1}},\widehat{\dot{\mu}^{j}}\rangle). $$

(A.74)

Then, the sum over time is also telescopic

$$ \sum\limits_{j = 2}^{n}{J_{3}^{j}}=(\frac{{\Delta} t}{2})^{2}(||\widehat{\dot{\mu}^{j}}||^{2}-|| \widehat{\dot{\mu}^{j-1}}||^{2})+I_{4}, $$

(A.75)

where I₄ is a purely imaginary term.

J ₄, J ₅ contribution:

by induction and approximation (A.65) we find that $(\zeta + 1) \widehat {\dot {\theta }^{j}}+ \frac {{\Delta } t}{2}(\widehat {\dot {\mu }^{j}}-\widehat {\dot {\mu }^{j-1}})+A_{0}({\Delta } t^{3})=A_{0}(\dot {\theta }^{j}+{\Delta } t^{3})$, then using Cauchy-Schwarz, triangle inequalities and Plancherel theorem we get

$$ |J_{4}|\leq {\Delta} t^{3}||A_{0}(\dot{\theta}^{j}+{\Delta} t^{3})||_{l^{2}}\leq {\Delta} t^{3}O(h^{r}+{\Delta} t^{3}). $$

(A.76)

Similarly,

$$ |J_{5}|={\Delta} t^{3}O(h^{r}+{\Delta} t^{3}). $$

(A.77)

Thus, the sum over time is

$$ |\sum\limits_{j = 2}^{n}{J_{4}^{j}}+{J_{5}^{j}}|=O(h^{r}{\Delta} t^{2}+{\Delta} t^{4}). $$

(A.78)

With this (A.69), (A.73), (A.75), (A.78) information and considering the real part of (A.67) we write

$$ \begin{array}{ll} ||\widehat{\dot{\theta}^{n + 1}}||^{2}+|| \widehat{\dot{\theta}^{n}}||^{2}-\left( ||\widehat{\dot{\theta}^{2}}||^{2}+|| \widehat{\dot{\theta}^{1}}||^{2}\right)& \leq\\ & || \widehat{\dot{\theta}^{n-1}}||^{2}+|| \widehat{\dot{\theta}^{n-2}}||^{2}-|| \widehat{\dot{\theta}^{1}}||^{2}-|| \widehat{\dot{\theta}^{0}}||^{2}\\ &+O((h^{r}+{\Delta} t^{2})^{2})+O(h^{r}{\Delta} t^{2}+{\Delta} t^{4}). \end{array} $$

(A.79)

Therefore,

$$||\dot{\theta}^{n + 1}||_{l^{2}}^{2}=O((h^{r}+{\Delta} t^{2})^{2})\Rightarrow ||\dot{\theta}^{n + 1}||_{l^{2}}\leq C(h^{r}+{\Delta} t^{2}). $$

As a consequence, the upper bound holds for a longer time (j = n + 1) than T^∗ (31), and T^∗ = T as desired.

1.6 A.6 Here we show the conservation of the quantities (72) under Airy Flow over time

M1:

k = θ_s/s_α is a perfect derivative of a periodic function. The result follows by the Fundamental Theorem of Calculus.

M2:

Observe that

$$ I:=\frac{\partial M2}{\partial t}=\int 2kk_{t} ds. $$

(A.80)

We know that for airy flow

$$ k_{t}=k_{sss}+ 3\frac{k^{2}k_{s}}{2}=(k_{ss}+\frac{k^{3}}{2})_{s}, $$

(A.81)

then

$$ I=\int 2kk_{sss}+ 3k^{3}k_{s} ds= 2\int k k_{sss}ds+\int \frac{\partial (\frac{3}{4}k^{4})}{\partial s}ds= 2\int k k_{sss} ds. $$

(A.82)

Again, since (kk_ss)_s = kk_sss + k_ssk_s we have

$$ \int k k_{sss} ds=-\int k_{s}k_{ss} ds=-\frac{1}{2}\int \frac{\partial {k_{s}^{2}}}{\partial s}ds= 0\Rightarrow I = 0, $$

(A.83)

and M2 is conserved over time.

M3:

Similarly, using integration by parts and periodicity of the functions we obtain

$$ \begin{array}{ll} &\displaystyle J:=\frac{\partial M3}{\partial t}=\int (k_{s} k_{st}-\frac{1}{2}k^{3}k_{t} )ds =\int [-k_{ss}-\frac{1}{2}k^{3}]k_{t}ds\\ &\displaystyle=-\int (k_{ss}+\frac{1}{2}k^{3})(k_{ss}+\frac{k^{3}}{2})_{s}ds=\frac{-1}{2}\int\frac{\partial (k_{ss}+\frac{k^{3}}{2})^{2}}{\partial s} ds= 0. \end{array} $$

(A.84)