A family of linearly weighted-$\theta$ compact ADI schemes for sine-Gordon equations in high dimensions

Abstract

This paper proposes a family of weighted-$\theta$ high-order ADI schemes for sine-Gordon equations in high dimensions in combination with the discretization of a three-level linearized scheme in temporal direction and a classical compact difference scheme in spatial direction. The unique solvability, convergence and stability of the difference scheme in two dimensions are strictly proved with the convergence order four in space and order two in time under the $L^\infty$-norm. More remarkably, the present numerical scheme can be extended to cope with the three-dimensional case feasibly. Several benchmark problems in two and three dimensions verify theoretical results and the capacity in simulating the dynamic behaviour of ring solitons. Specially, several novel numerical isosurfaces in three dimensions are presented for the first time.

Appendix. Detailed proofs of lemmas and remarks

In this appendix, we will give the detailed proofs for the lemmas and remarks presented in Sects. 2 and 4.2.

• (I). Proof of Lemma 2.5.

Proof

According to the inverse estimate in Lemmas 2.2–2.4, it is straightforward to show that

$$\begin{aligned}&\;\Vert (\alpha \mathcal {A}_y\delta _x^2+\beta \mathcal {A}_x\delta _y^2)u\Vert ^2\\ =&\;\alpha ^2\Vert \mathcal {A}_y\delta _x^2u\Vert ^2+\beta ^2\Vert \mathcal {A}_x\delta _y^2u\Vert ^2 +2\alpha \beta (\mathcal {A}_y\delta _x^2u,\mathcal {A}_x\delta _y^2u)\\ =&\;\alpha ^2\Vert \delta _x^2u\Vert ^2+\beta ^2\Vert \delta _y^2u\Vert ^2+2\alpha \beta \Big [(\delta _x^2u,\delta _y^2u) \\&\;+\frac{h_2^2}{12}(\delta _y^2u,\delta _x^2\delta _y^2u) +\frac{h_1^2}{12}(\delta _x^2u,\delta _x^2\delta _y^2u)+\frac{h_1^2h_2^2}{12^2}\Vert \delta _x^2\delta _y^2u\Vert ^2\Big ]\\ \leqslant&\;\alpha ^2\Vert \delta _x^2u\Vert ^2+\beta ^2\Vert \delta _y^2u\Vert ^2 +2\alpha \beta \Vert \delta _x\delta _yu\Vert ^2\\ \leqslant&\; ( M_{\alpha \beta })^2\cdot \Vert \varDelta _hu\Vert ^2, \end{aligned}$$

which completes the proof.

• (II). Proof of Lemma 2.6.

Proof

On one hand, using Lemmas 2.2–2.3, we can get from Lemma 3.4 in [24]

$$\begin{aligned}&\;(\mathcal {A}_x\mathcal {A}_yu,-\varDelta _h u)\\ =&\;|u|_1^2-\frac{h_1^2}{12}(\Vert \delta _x^2u\Vert ^2+\Vert \delta _y\delta _xu\Vert ^2) -\frac{h_2^2}{12}(\Vert \delta _y^2u\Vert ^2+\Vert \delta _y\delta _xu\Vert ^2)\\&\;\quad +\frac{h_1^2h_2^2}{12^2}(\Vert \delta _y\delta _x^2u\Vert ^2+\Vert \delta ^2_y\delta _xu\Vert ^2)\\ \leqslant&\; |u|_1^2-\frac{h_1^2}{18}(\Vert \delta _x^2u\Vert ^2+\Vert \delta _y\delta _xu\Vert ^2) -\frac{h_2^2}{12}(\Vert \delta _y^2u\Vert ^2+\Vert \delta _y\delta _xu\Vert ^2) \leqslant |u|_1^2, \end{aligned}$$

which completes the proof of the right-hand side of the first inequality. On the other hand, it follows from Lemma 3.5 in [24] that

$$\begin{aligned} \Vert u\Vert _{\alpha \beta }^2&\;=((\alpha \delta _x^2+\beta \delta _y^2)u,(\delta _x^2+\delta _y^2)u)+ \frac{\alpha h_2^2+\beta h_1^2}{12}(\delta _x^2\delta _y^2u,(\delta _x^2+\delta _y^2)u)\\&\;=\alpha \Vert \delta _x^2u\Vert ^2+\beta \Vert \delta _y^2u\Vert ^2+(\alpha +\beta )\Vert \delta _x\delta _yu\Vert ^2 -\frac{\alpha h_2^2+\beta h_1^2}{12}(\Vert \delta _y\delta _x^2u\Vert ^2+\Vert \delta _x\delta _y^2u\Vert ^2)\\&\;\geqslant \frac{2}{3}(\alpha \Vert \delta _x^2u\Vert ^2+\beta \Vert \delta _y^2u\Vert ^2+(\alpha +\beta )\Vert \delta _x\delta _yu\Vert ^2) \geqslant \frac{2}{3} m_{\alpha \beta }\cdot \Vert \varDelta _hu\Vert ^2. \end{aligned}$$

According to the above inequality, it is easy to prove the right-hand side of the second inequality, and in a word, the proof of the second inequality is completed.

• (III). Proof of Lemma 2.11.

Proof

We first of all consider by Lemma 2.3

$$\begin{aligned}&\;(\mathcal {A}_x\mathcal {A}_y\delta _t^2u^k,-\delta _x^2 \varDelta _tu^k)\\ =&\;\frac{1}{2\tau } \Big (\mathcal {A}_x\mathcal {A}_y(\delta _t u^{k+\frac{1}{2}}-\delta _t u^{k-\frac{1}{2}}), -\delta _x^2(\delta _t u^{k+\frac{1}{2}}+\delta _t u^{k-\frac{1}{2}})\Big )\\ =&\;\frac{1}{2\tau }\Big [(\mathcal {A}_x\mathcal {A}_y\delta _t u^{k+\frac{1}{2}}, -\delta _x^2\delta _t u^{k+\frac{1}{2}})-(\mathcal {A}_x\mathcal {A}_y\delta _t u^{k-\frac{1}{2}}, -\delta _x^2\delta _t u^{k-\frac{1}{2}})\Big ]\\&\;+\frac{1}{2\tau }\Big [ \Big (\big (\mathcal {I}+\frac{h_1^2}{12}\delta _x^2+\frac{h_2^2}{12}\delta _y^2 +\frac{h_1^2h_2^2}{12^2}\delta _x^2\delta _y^2\big )\delta _t u^{k+\frac{1}{2}},-\delta _x^2\delta _t u^{k-\frac{1}{2}}\Big )\\&\;-\Big (\big (\mathcal {I}+\frac{h_1^2}{12}\delta _x^2+\frac{h_2^2}{12}\delta _y^2 +\frac{h_1^2h_2^2}{12^2}\delta _x^2\delta _y^2\big )\delta _t u^{k-\frac{1}{2}},-\delta _x^2\delta _t u^{k+\frac{1}{2}}\Big )\Big ]\\ =&\;\frac{1}{2\tau }\Big [ (\mathcal {A}_x\mathcal {A}_y\delta _t u^{k+\frac{1}{2}}, -\delta _x^2\delta _t u^{k+\frac{1}{2}})-(\mathcal {A}_x\mathcal {A}_y\delta _t u^{k-\frac{1}{2}}, -\delta _x^2\delta _t u^{k-\frac{1}{2}})\Big ],\quad k\in \mathcal {T}_{N}. \end{aligned}$$

By analogy with the definition of $\varDelta _h$, the proof is completed.

• (IV). Proof of Lemma 2.12.

Proof

Making use of Lemmas 2.1, 2.3 and 2.7, we see for $k\in \mathcal {T}_{N}$ that

$$\begin{aligned}&\;-\big (\varTheta (\alpha \mathcal {A}_y\delta _x^2+\beta \mathcal {A}_x\delta _y^2)u^{k},-\varDelta _h\varDelta _tu^{k}\big )\\ =&\;\theta \big ((\alpha \mathcal {A}_y\delta _x^2+\beta \mathcal {A}_x\delta _y^2)(u^{k+1}+u^{k-1}),-\varDelta _h\varDelta _tu^{k}\big ) \\&\;+(1-2\theta ) \big ((\alpha \mathcal {A}_y\delta _x^2+\beta \mathcal {A}_x\delta _y^2)u^{k},-\varDelta _h\varDelta _tu^{k}\big )\\\ =&\;-\frac{\theta }{2\tau }\Big [\Vert u^{k+1}\Vert _{\alpha \beta }^2-\Vert u^{k-1}\Vert _{\alpha \beta }^2+(u^{k+1},u^{k-1})_{\alpha \beta }-(u^{k-1},u^{k+1})_{\alpha \beta }\Big ] \\&\;-\frac{1-2\theta }{2\tau }\Big [(u^{k},u^{k+1})_{\alpha \beta }+(u^{k},u^{k-1})_{\alpha \beta }\Big ]\\ =&-\frac{\theta }{2\tau }\Big (\Vert u^{k+1}\Vert _{\alpha \beta }^2-\Vert u^{k-1}\Vert _{\alpha \beta }^2\Big ) -\frac{1-2\theta }{2\tau }\Big [(\Vert u^{k+1}\Vert _{\alpha \beta }^2-\Vert u^{k+1}-u^k\Vert _{\alpha \beta }^2) \\&-(\Vert u^{k-1}\Vert _{\alpha \beta }^2-\Vert u^{k-1}-u^k\Vert _{\alpha \beta }^2)\Big ]\\ =&\;\frac{1}{2\tau }\Big [\frac{1}{2}(\Vert u^{k+1}\Vert _{\alpha \beta }^2+\Vert u^{k}\Vert _{\alpha \beta }^2) -\frac{1}{2}(\Vert u^{k}\Vert _{\alpha \beta }^2+\Vert u^{k-1}\Vert _{\alpha \beta }^2) \\&\;+\frac{2\theta -1}{2}(\Vert u^{k+1}-u^{k}\Vert _{\alpha \beta }^2-\Vert u^{k-1}-u^{k}\Vert _{\alpha \beta }^2) \Big ]. \end{aligned}$$

The lemma is proved.

• (V). Proof of Lemma 2.13.

Proof

Making use of Lemma 2.3, we see that

$$\begin{aligned}&\;(\alpha \beta \theta ^2\tau ^4\delta _x^2\delta _y^2\delta _t^2u^{k},-\varDelta _h\varDelta _tu^{k})\\ =&\;\frac{1}{2\tau } \alpha \beta \theta ^2\tau ^4\big (\delta _x^2\delta _y^2(\delta _tu^{k+\frac{1}{2}}-\delta _tu^{k-\frac{1}{2}}), -\varDelta _h(\delta _tu^{k+\frac{1}{2}}+\delta _tu^{k-\frac{1}{2}})\big )\\ =&\;\frac{1}{2\tau } \alpha \beta \theta ^2\tau ^4\Big [(\Vert \delta _y\delta ^2_x\delta _tu^{k+\frac{1}{2}}\Vert ^2+\Vert \delta ^2_y\delta _x\delta _tu^{k+\frac{1}{2}}\Vert ^2) -( \Vert \delta _y\delta ^2_x\delta _tu^{k-\frac{1}{2}}\Vert ^2+\Vert \delta ^2_y\delta _x\delta _tu^{k-\frac{1}{2}}\Vert ^2)\\&\;-(\delta _x^2\delta ^2_y\delta _tu^{k+\frac{1}{2}},\delta _x^2\delta _tu^{k-\frac{1}{2}}) -(\delta _x^2\delta ^2_y\delta _tu^{k+\frac{1}{2}},\delta _y^2\delta _tu^{k-\frac{1}{2}})\\&\;+(\delta _x^2\delta ^2_y\delta _tu^{k-\frac{1}{2}},\delta _x^2\delta _tu^{k+\frac{1}{2}}) +(\delta _x^2\delta ^2_y\delta _tu^{k-\frac{1}{2}},\delta _y^2\delta _tu^{k+\frac{1}{2}})\Big ]\\ =&\;\frac{1}{2\tau }\alpha \beta \theta ^2\tau ^4 (|\delta _y\delta _x\delta _tu^{k+\frac{1}{2}}|_1^2 - |\delta _y\delta _x\delta _tu^{k-\frac{1}{2}}|_1^2),\quad k\in \mathcal {T}_{N}, \end{aligned}$$

which is exactly Lemma 2.13.

• (VI). Proof of Remark 4.1.

Proof

When $\theta \in \big [0,\frac{1}{4}\big ]$, the conclusion is same to (2.3). When $\theta \in \big (\frac{1}{4},1\big ]$, applying Lemma 2.6 we have

$$\begin{aligned} \mathcal {V}^{k+1}\geqslant&\; \frac{1}{2}(\Vert v^{k+1}\Vert _{\alpha \beta }^2+\Vert v^{k}\Vert _{\alpha \beta }^2)+ ({2\theta -1})(\Vert v^{k+1}\Vert _{\alpha \beta }^2+\Vert v^{k}\Vert _{\alpha \beta }^2)+\frac{1}{6}|\delta _tv^{k+\frac{1}{2}}|_1^2\\ \geqslant&\; \frac{4\theta -1}{2}(\Vert v^{k+1}\Vert _{\alpha \beta }^2+\Vert v^{k}\Vert _{\alpha \beta }^2)\\ \geqslant&\; \frac{4\theta -1}{3} m_{\alpha \beta }(\Vert \varDelta _hv^{k+1}\Vert ^2+\Vert \varDelta _hv^{k}\Vert ^2). \end{aligned}$$

It follows from the above inequality that

$$\begin{aligned} \Vert \varDelta _hv^{k+1}\Vert ^2+\Vert \varDelta _hv^{k}\Vert ^2\leqslant \frac{3}{(4\theta -1) m_{\alpha \beta }} \mathcal {V}^{k+1},\quad k\in \mathcal {T}_{N}\cup \{0\}. \end{aligned}$$

(7.1)