A posteriori error estimates for wave maps into spheres

Abstract

We provide a posteriori error estimates in the energy norm for temporal semi-discretisations of wave maps into spheres that are based on the angular momentum formulation. Our analysis is based on novel weak–strong stability estimates which we combine with suitable reconstructions of the numerical solution. We present time-adaptive numerical simulations based on the a posteriori error estimators for solutions involving blow-up.

Appendix

This appendix is devoted to the proof of Lemma 2. Here, each quantity that is defined continuously in time is to be understood as its restriction to $(t_n,t_{n+1})$. Let us begin by giving explicit formulae for certain parts of $r_u$ and $r_\omega$. The first such expressions make explicit the difference between some piecewise quadratic, globally continuous function and its piecewise linear interpolations:

$$\begin{aligned} \begin{aligned} u^*(t) - \widehat{u}(t)&= \frac{1}{2} \frac{(t-t_n)(t_{n+1}-t)}{t_{n+1}-t_n} \left( u^{n+1} \times \omega ^{n+1} - u^n \times \omega ^n\right) ,\\ \tilde{w}(t) - \widehat{\omega }(t)&= \frac{1}{2} \frac{(t-t_n)(t_{n+1}-t)}{t_{n+1}-t_n} \left( \Delta u^{n+1} \times u^{n+1} - \Delta u^n \times u^n\right) ,\\ \widehat{u}(t) \times \widehat{w}(t) - I_1(\widehat{u} \times \widehat{w})(t)&= - \frac{(t-t_n)(t_{n+1}-t)}{(t_{n+1}-t_n)^2} (u^{n+1} - u^n) \times (\omega ^{n+1} - \omega ^n). \end{aligned} \end{aligned}$$

(53)

1.1 Controlling $|\widehat{u}|$, $|u^*|$

Let us now study how far away $\widehat{u}, u^*$ are from maps into the sphere: If $u^{n+1}-u^n$ is sufficiently small, a geometric argument implies

$$\begin{aligned} \Big \vert 1-|\widehat{u}\vert \Big \vert \le |u^{n+1}-u^n|^2= (A^u)^2, \ \Big \vert 1-|u^* |\Big \vert \le \Big \vert 1- |\widehat{u}|\Big \vert + \Big \vert \widehat{u} - u^*\Big \vert \le (A^u)^2 + \tau _n B^u \end{aligned}$$

and, therefore,

$$\begin{aligned} \Big |\tilde{u}- u^*\Big \vert = \Big |\tilde{u}\Big (1-\Big \vert u^*\Big |\Big )\Big |\le |u^{n+1}-u^n|^2= (A^u)^2+ \tau _n B^u. \end{aligned}$$

(54)

Thus, the conditions in Lemma 2 imply

$$\begin{aligned} \frac{3}{4} \le |\widehat{u}|\le \frac{5}{4} , \qquad \frac{1}{2} \le |u^* |\le \frac{3}{2}. \end{aligned}$$

(55)

1.2 Estimating $r_{u,1}$

Since $\tilde{u}(t_n)=\widehat{u}(t_n)$ and $\tilde{w}(t_n)=\widehat{\omega }(t_n)$ we may rewrite $r_{u,1}$ as

$$\begin{aligned} r_{u,1} = \tilde{u} \times (\tilde{w}- \widehat{\omega }) + (\tilde{u}- u^*) \times \widehat{\omega }+ ( u^*- \widehat{u}) \times \widehat{\omega }+ \widehat{u} \times \widehat{w} - I_1(\widehat{u} \times \widehat{w}) \end{aligned}$$

(56)

such that

$$\begin{aligned} \begin{aligned}&\Big \vert r_{u,1} \Big \vert \le \tau _n (\Delta \tilde{u}^{n+1} \times \tilde{u}^{n+1} - \Delta \tilde{u}^{n} \times \tilde{u}^{n} ) 2 |u^{n+1} - u^n|\, |\omega ^{n+1} - \omega ^n|\\&\ + \max \{|\omega ^n|,|\omega ^{n+1}|\} \left[ |u^{n+1}-u^n|^2+ |t_{n+1}-t_n|\ |u^{n+1} \times \omega ^{n+1} - u^n \times \omega ^n|\right] \, . \end{aligned} \end{aligned}$$

(57)

This proves (44).

We obtain (45) by applying the product rule to (56) since $|\widehat{u}(t,x)|\le 1$. We also use the fact that $\nabla \tilde{u}$ is the projection of $\nabla u^*$ onto the tangent space of the sphere, so that, provided $|u^{n+1}-u^n|< 1/2$ holds, we have the point-wise estimate

$$\begin{aligned} |\nabla \tilde{u}|\le 2 |\nabla u^*|. \end{aligned}$$

(58)

1.3 Estimating $r_{u,3}$

The key to estimating $r_{u,3}$ is to control $\partial _t u^* \cdot u^*$. We notice that

$$\begin{aligned} \partial _t u^* \cdot u^* = (I_1[\widehat{u} \times \widehat{\omega }] + a_u) \cdot (\widehat{u} + (u^* - \widehat{u})) \end{aligned}$$

(59)

and, thus,

$$\begin{aligned} \vert \partial _t u^* \cdot u^* \vert \le \vert I_1[\widehat{u} \times \widehat{\omega }]\cdot \widehat{u} \vert + A^u A^\omega (1 + \tau _n B^u) + C^\omega \tau _n B^u \end{aligned}$$

(60)

so that it remains to understand $I_1[\widehat{u} \times \widehat{\omega }]\cdot \widehat{u}$. Using orthogonality, we obtain

$$\begin{aligned} \begin{aligned} I_1[\widehat{u} \times \widehat{\omega }]\cdot \widehat{u}&= [ \ell _n^0(t) u^n\times \omega ^n + \ell _n^1(t) u^{n+1}\times \omega ^{n+1} ] \cdot [ \ell _n^0(t) u^n + \ell _n^1(t) u^{n+1} ] \\&= \ell _n^0(t) \ell _n^1(t) [ u^{n+1} \cdot (u^n\times \omega ^n ) + u^{n} \cdot (u^{n+1}\times \omega ^{n+1} )]\\&= \ell _n^0(t) \ell _n^1(t)\det (u^{n+1},u^n,\omega ^{n+1}-\omega ^n)\\&= \ell _n^0(t) \ell _n^1(t)\det (u^{n+1}-u^n,u^n,\omega ^{n+1}-\omega ^n). \end{aligned} \end{aligned}$$

(61)

We insert (61) into (59) and obtain

$$\begin{aligned} \vert \partial _t u^* \cdot u^* \vert \le A^u A^\omega (2 + \tau _n B^u) + C^\omega \tau _n B^u\,. \end{aligned}$$

(62)

Moreover, due to (55) we arrive at

$$\begin{aligned} \left| 1 - \frac{1}{|u^* |} \right| = \left| 1 - \frac{1}{|\widehat{u}\vert } - \frac{ \vert u^* \vert - \vert \widehat{u}\vert }{ \vert u^* \vert \vert \widehat{u}\vert }\right| \le \frac{4}{3} (A^u)^2 +\frac{8}{3} \tau _n B^u \end{aligned}$$

(63)

Using (63) and (62) we obtain:

$$\begin{aligned} \vert r_{u,3}\vert \le (C^\omega + \frac{1}{4} A^u A^\omega ) \left( \frac{4}{3} (A^u)^2 +\frac{8}{3} \tau _n B^u\right) + 4 A^u A^\omega (2 + \tau _n B^u) +4 C^\omega \tau _n B^u. \end{aligned}$$

(64)

Let us note that for

$$\begin{aligned} \begin{aligned} \nabla r_{u,3}&= \nabla \partial _t u^* \left( 1 - \frac{1}{|u^* |}\right) + \nabla (I_1[\widehat{u}\times \widehat{\omega }] \cdot \widehat{u}) + \nabla a_u \cdot u^* \\&\qquad +a_u \cdot \nabla u^* + \nabla (I_1[\widehat{u}\times \widehat{\omega }] \cdot (u^* - \widehat{u})) + \partial _t u^* \nabla \left( 1 - \frac{1}{|u^* |}\right) \end{aligned} \end{aligned}$$

(65)

we have suitable bounds for all terms on the right hand side of (65) except for $\nabla \left( 1 - \frac{1}{|u^* |}\right)$, e.g., $\nabla (I_1[\widehat{u}\times \widehat{\omega }] \cdot \widehat{u})$ can be estimated by applying the product rule to (61). As a first step towards estimating for $\nabla \left( 1 - \frac{1}{|u^* |}\right) = - \frac{u^* \cdot \nabla u^*}{|u^* |^3}$, we compute

$$\begin{aligned} \begin{aligned} |u^* |^3\partial _{x_j}\! \left( 1 - \frac{1}{|u^* |}\right)&= \partial _{x_j} \widehat{u} \cdot \widehat{u} + \partial _{x_j} (u^* - \widehat{u}) \cdot \widehat{u} \\&\qquad \qquad + \partial _{x_j} \widehat{u}\cdot \! (u^* - \widehat{u}) + \partial _{x_j} (u^*- \widehat{u})\cdot (u^* - \widehat{u}). \end{aligned} \end{aligned}$$

(66)

We recall $\vert u^n\vert =1$, which implies $\partial _{x_j} u^n \cdot u^n=0$, for all n, so that

$$\begin{aligned} \begin{aligned} \partial _{x_j} \widehat{u} \cdot \widehat{u}&= \partial _{x_j} (\ell _n^0 (t) u^n + \ell _n^1 (t) u^{n+1}) \cdot (\ell _n^0 (t) u^n + \ell _n^1 (t) u^{n+1})\\&= \ell _n^0 (t)\ell _n^1(t) ( \partial _{x_j} u^n \cdot u^{n+1} + \partial _{x_j} u^{n+1} \cdot u^{n} )\\&= - \ell _n^0 (t)\ell _n^1(t) \partial _{x_j} ( u^{n+1} - u^n )\cdot ( u^{n+1} - u^n ). \end{aligned} \end{aligned}$$

(67)

We insert (67) into (66) and obtain

$$\begin{aligned} \left| \nabla \left( 1 - \frac{1}{|u^* |}\right) \right| \le 8[ A^u_x A^u + \tau _n B^u_x + C^u_x \tau _n B^u + \tau _n^2 B^u_x B^u] \end{aligned}$$

(68)

Thus, we obtain

$$\begin{aligned} \begin{aligned} \vert \nabla r_{u,3}\vert&\le \left( C^u_x C^\omega + C^\omega _x + \frac{1}{4} A^u_x A^\omega +\frac{1}{4} A^u A^\omega _x\right) \left( \frac{4}{3} (A^u)^2 +\frac{8}{3} \tau _n B^u\right) \\&\quad + \frac{3}{2} A^u_x A^\omega +2 A^u C^u_x A^\omega + \frac{3}{2} A^u A^\omega _x + \left( C^u_x C^\omega + C^\omega _x \right) \tau _n B^u + C^\omega \tau _n B^u_x\\&\quad + 8\left( C^\omega + \frac{1}{4} A^u A^\omega \right) \left( A^u_x A^u + \tau _n B^u_x + C^u_x \tau _n B^u + \tau _n^2 B^u_x B^u\right) . \end{aligned} \end{aligned}$$

(69)

This completes providing bounds for the different components of $r_u$ and $\nabla r_u$.

1.4 Estimating $r_\omega$

Obviously, $\vert r_{\omega ,2}\vert = \vert a^\omega \vert \le \frac{1}{4} A^u A^u_{xx}$ and

$$\begin{aligned} \begin{aligned} r_{\omega ,1}&= (\Delta \tilde{u}- \Delta u^*)\times \tilde{u}+ (\Delta u^* - \Delta \widehat{u})\times \tilde{u}+ \Delta \widehat{u} \times (\tilde{u}- u^*) \\&\qquad \qquad + \Delta \widehat{u} \times (u^*-\widehat{u}) + \Delta \widehat{u} \times \widehat{u} - I_1[\Delta \widehat{u} \times \widehat{u}]. \end{aligned} \end{aligned}$$

(70)

We insert (54) and (53) into (70) and obtain

$$\begin{aligned} \vert r_{\omega ,1}\vert \le \vert (\Delta \tilde{u}- \Delta u^*) \vert + \tau _n B^u + C^u_{xx} (A^u)^2 + C^u_{xx}\tau _n B^u + A^u_{xx} A^u\,, \end{aligned}$$

(71)

where we have used that

$$\begin{aligned} \Delta \widehat{u} \times \widehat{u} - I_1[\Delta \widehat{u} \times \widehat{u}] =- \ell ^0_n(t)\ell ^1_n(t) (\Delta u^{n+1}-\Delta u^n) \times (u^{n+1} - u^n). \end{aligned}$$

It remains to provide an estimate for $|\Delta \tilde{u}- \Delta u^*\vert$. We note that $u^*=\tilde{u}|u^* |$ and, thus,

$$\begin{aligned} \Delta (\tilde{u}- u^*) = \Delta u^* \left( 1 - \frac{1}{|u^* |} \right) - \frac{\sum _j \partial _{x_j} \tilde{u}\cdot \partial _{x_j} |u^* |}{ |u^* |} - \frac{ \tilde{u}\Delta |u^* |}{ |u^* |}, \end{aligned}$$

(72)

so that

$$\begin{aligned} \begin{aligned} \vert \Delta (\tilde{u}- u^*) \vert&\le ( C^u_{xx}+ \tau _n B^u_{xx}) \left( \frac{4}{3} (A^u)^2+ \frac{8}{3} \tau _n B^u\right) + |\nabla \tilde{u}\vert \frac{\vert u^* \cdot \nabla u^*\vert }{ \vert u^*|^2 } + \frac{ |\Delta |u^* |\, \vert }{ |u^* |}\\&\le ( C^u_{xx}+ \tau _n B^u_{xx}) \left( \frac{4}{3} (A^u)^2+ \frac{8}{3} \tau _n B^u\right) \\&+(C^u_x + \tau _n B^u_x)2[ A^u_x A^u + \tau _n B^u_x + C^u_x \tau _n B^u + \tau _n^2 B^u_x B^u] + \frac{ |\Delta |u^* |\, \vert }{ |u^* |} \end{aligned} \end{aligned}$$

(73)

where we have used (68). Orthogonality $\partial _{x_j} u^n \perp u^n$ for all j and all n implies

$$\begin{aligned} \Delta |u^* |= - \ell _n^0 (t)\ell _n^1(t)[(u^{n+1} -u^n)\cdot \partial _{x_j}^2 (u^{n+1} -u^n) + |\partial _{x_j} (u^{n+1} -u^n)|^2] . \end{aligned}$$

(74)

Inserting (74) into (73) implies

$$\begin{aligned} \begin{aligned} \vert \Delta (\tilde{u}- u^*) \vert&\le ( C^u_{xx}+ \tau _n B^u_{xx}) \left( \frac{4}{3} (A^u)^2+ \frac{8}{3} \tau _n B^u\right) + A^u_{xx} A^u + (A^u_x)^2\\&\quad +(C^u_x + \tau _n B^u_x)2[ A^u_x A^u + \tau _n B^u_x + C^u_x \tau _n B^u + \tau _n^2 B^u_x B^u] \end{aligned} \end{aligned}$$

(75)

Inserting (75) into (71) completes the bound for $r_\omega$.

1.5 Estimating $r_g$

We note that orthogonality $u^n \perp \omega ^n$ implies

$$\begin{aligned} \begin{aligned} \tilde{u}\cdot \tilde{w}&= \tilde{u}\cdot (\tilde{w}- \widehat{\omega }) - (\tilde{u}- \widehat{u})\cdot \widehat{\omega }+ \widehat{u} \cdot \widehat{\omega }\\&= \tilde{u}\cdot (\tilde{w}- \widehat{\omega }) - (\tilde{u}- \widehat{u})\cdot \widehat{\omega }- \ell _n^0 (t)\ell _n^1(t)(u^{n+1} -u^n)\cdot (\omega ^{n+1} -\omega ^n) \end{aligned} \end{aligned}$$

(76)

so that

$$\begin{aligned} \vert \tilde{u}\cdot \tilde{w}\vert \le \tau _n B^\omega + C^\omega ( (A^u)^2 + \tau _n B^u) + A^u A^\omega . \end{aligned}$$

(77)

Thus, using the definition of $r_g$

$$\begin{aligned} \vert r_g \vert&\le (C^\omega + \tau _n B^\omega )) [\tau _n B^\omega + C^\omega ( (A^u)^2 + \tau _n B^u) + A^u A^\omega ] \\&\qquad \qquad + [\tau _n B^\omega + C^\omega ( (A^u)^2 + \tau _n B^u) + A^u A^\omega ]^2. \end{aligned}$$