A Concurrent Global–Local Numerical Method for Multiscale PDEs

Abstract

We present a new hybrid numerical method for multiscale partial differential equations, which simultaneously captures the global macroscopic information and resolves the local microscopic events over regions of relatively small size. The method couples concurrently the microscopic coefficients in the region of interest with the homogenized coefficients elsewhere. The cost of the method is comparable to the heterogeneous multiscale method, while being able to recover microscopic information of the solution. The convergence of the method is proved for problems with bounded and measurable coefficients, while the rate of convergence is established for problems with rapidly oscillating periodic or almost-periodic coefficients. Numerical results are reported to show the efficiency and accuracy of the proposed method.

Appendix A: Example

To better appreciate the estimates (3.4) and (3.5), which are crucial in our analysis, let us consider a one-dimensional problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\left( a^{\,\varepsilon }(x)u'(x)\right) ^{\prime }=0,&{}\quad x\in (0,1),\\ u(0)=0,&{}\quad a^{\,\varepsilon }(1)u'(1)=1, \end{array}\right. \end{aligned}$$

where $a^{\,\varepsilon }(x)=2+\sin (x/\varepsilon )$. A direct calculation gives that the effective coefficient $\mathcal {A}=\sqrt{3}$ and the solution of the homogenized problem is $u_0(x)=x/\mathcal {A}$.

We consider a uniform mesh given by

$$\begin{aligned} x_0=0<x_1=h<\cdots<x_i=ih<\cdots <x_{2N}=1, \end{aligned}$$

where $h=1/(2N)$. The finite element space $X_h$ is simply the piecewise linear element associated with the above mesh with zero boundary condition at $x=0$.

Case $h \gg \varepsilon$. We firstly consider the case that $h\gg \varepsilon$, while the precise relation between h and $\varepsilon$ will be made clear below. Denote $v_h(x_j)=v_j$ and the interval $I_j=(x_{j-1},x_j)$, the mean of the coefficients $b^{\,\varepsilon }$ over each $I_j$ is denoted by $b_j={\int \negthickspace \negthickspace \negthickspace \negthinspace -}_{I_j}b^{\,\varepsilon }(x)\,\mathrm {d}x$.

We define the transition function $\rho$ as a piecewise linear function that is supported in $(-2L,2L)$, where L is a fixed number with $0<L<1/4$. Without loss of generality, we assume that $L=Mh$ with M an integer. In particular,

$$\begin{aligned} \rho (x)=\left\{ \begin{array}{ll} 0&{}\quad 0\le x\le x_{N-2M},\\ \dfrac{x-x_{N-2M}}{L}&{}\quad x_{N-2K}\le x\le x_{N-M},\\ 1&{}\quad x_{N-M}\le x\le x_{N+M},\\ \dfrac{x_{N+2M}-x}{L}&{}\quad x_{N+M}\le x\le x_{N+2M},\\ 0&{}\quad x_{N+2M}\le x\le x_{2N}=1. \end{array}\right. \end{aligned}$$

By construction, we get the size of the support of $\rho$ is $\left|K\right|=4L$.

We easily obtain the linear system for $\{v_j\}_{j=1}^{2N}$ as

$$\begin{aligned} \left\{ \begin{array}{ll} -\,b_jv_{j-1}+(b_j+b_{j+1})v_j-b_{j+1}v_{j+1}=0,&{}\quad j=1,\cdots ,2N-1,\\ -\,b_{2N}v_{2N-1}+b_{2N}v_{2N}=h.&{} \end{array}\right. \end{aligned}$$

Define $c_j{:}=(v_j-v_{j-1})b_j/h$, we rewrite the above equation as

$$\begin{aligned} c_j-c_{j-1}=0,\quad j=1,\cdots ,2N-1,\qquad c_{2N}=1. \end{aligned}$$

Hence $c_j=1$ for $j=1,\cdots ,2N$, and the above linear system reduces to

$$\begin{aligned} (v_j-v_{j-1})b_j=h. \end{aligned}$$

Using $v_0=0$, we obtain

$$\begin{aligned} v_j=h\sum _{i=1}^j\dfrac{1}{b_i}. \end{aligned}$$

(A.1)

Observing that $v_h(x)=u_0(x)$ for $x\in [0,x_{N-2M}]$ because they are linear functions that coincide at all the nodal points $x_i$ for $i=0,\cdots , N-2M$.

For $x\in I_{N-2M+j+1}$, we obtain

$$\begin{aligned} u_0(x)-v_h(x)=h\sum _{i=1}^j\left( \dfrac{1}{\mathcal {A}}-\dfrac{1}{b_{N-2M+i}}\right) +(x-x_{N-2M+j})\left( \dfrac{1}{\mathcal {A}}-\dfrac{1}{b_{N-2M+j+1}}\right) . \end{aligned}$$

Define $S_j{:}=h\sum _{i=1}^j\left( \dfrac{1}{\mathcal {A}}-\dfrac{1}{b_{N-2M+i}}\right)$, we rewrite the above equation as

$$\begin{aligned} u_0(x)-v_h(x)=\dfrac{x_{N-2M+j+1}-x}{h}S_j+\dfrac{x-x_{N-2M+j+1}}{h}S_{j+1}, \end{aligned}$$

(A.2)

which immediately yields

$$\begin{aligned} \begin{aligned} \int _{x_{N-2M}}^{x_{N-M}}\left|u_0'(x)-v_h'(x)\right|^2\,\mathrm {d}x&=h\sum _{j=1}^M\left|\dfrac{1}{\mathcal {A}}-\dfrac{1}{b_{N-2M+j}}\right|^2\\&\ge \dfrac{h}{27}\sum _{j=1}^M\left|\mathcal {A}-b_{N-2M+j}\right|^2. \end{aligned} \end{aligned}$$

(A.3)

This is the starting point of later derivation. A direct calculation gives

$$\begin{aligned} b_{N-2M+j}-\mathcal {A}&={\int \negthickspace \negthickspace \negthickspace \negthinspace -}_{I_{N-2M+j}}\rho (x)(a^\varepsilon (x)-\mathcal {A})\,\mathrm {d}x\\&=\dfrac{2-\mathcal {A}}{2}\bigl (\rho (x_{N-2M+j-1})+\rho (x_{N-2M+j})\bigr )+{\int \negthickspace \negthickspace \negthickspace \negthinspace -}_{I_{N-2M+j}}\sin \dfrac{x}{\varepsilon }\,\mathrm {d}x\\&=\dfrac{(2-\mathcal {A})h}{2L}(2j-1)+{\int \negthickspace \negthickspace \negthickspace \negthinspace -}_{I_{N-2M+j}}\sin \dfrac{x}{\varepsilon }\,\mathrm {d}x, \end{aligned}$$

and an integration by parts yields

$$\begin{aligned} {\int \negthickspace \negthickspace \negthickspace \negthinspace -}_{I_{N-2M+j}}\sin \dfrac{x}{\varepsilon }\,\mathrm {d}x&=\dfrac{2j\varepsilon }{L}\sin \dfrac{h}{2\varepsilon }\sin \dfrac{x_{N-2M+j-1/2}}{\varepsilon }\\&\quad -\dfrac{\varepsilon }{L}\cos \dfrac{x_{N-2M+j-1}}{\varepsilon }+\dfrac{\varepsilon ^2}{Lh}\left( \cos \dfrac{x_{N-2M+j-1}}{\varepsilon } -\cos \dfrac{x_{N-2M+j}}{\varepsilon }\right) . \end{aligned}$$

Combining the above two equations, we obtain

$$\begin{aligned} b_{N-2M+j}-\mathcal {A}=\dfrac{(2-\mathcal {A})h}{2L}(2j-1)+\dfrac{2j\varepsilon }{L}\sin \dfrac{h}{2\varepsilon }\sin \dfrac{x_{N-2M+j-1/2}}{\varepsilon }+\text {REM}, \end{aligned}$$

(A.4)

where the remainder term

$$\begin{aligned} \text {REM}{:}=-\dfrac{\varepsilon }{L}\cos \dfrac{x_{N-2M+j-1}}{\varepsilon }+\dfrac{\varepsilon ^2}{Lh}\left( \cos \dfrac{x_{N-2M+j-1}}{\varepsilon } -\cos \dfrac{x_{N-2M+j}}{\varepsilon }\right) , \end{aligned}$$

which can be bounded as

$$\begin{aligned} \left|\text {REM}\right|&\le \dfrac{\varepsilon }{L}+\dfrac{2\varepsilon ^2}{Lh}\left|\sin \dfrac{h}{2\varepsilon }\right|\left|\cos \dfrac{x_{N-2M+j-1/2}}{\varepsilon }\right|\\&\le \dfrac{\varepsilon }{L}+\dfrac{2\varepsilon ^2}{Lh}\dfrac{h}{2\varepsilon }=\dfrac{2\varepsilon }{L}. \end{aligned}$$

Note that $\sum _{j=1}^M(2j-1)^2=M(4M^2-1)/3,$ and

$$\begin{aligned} \sum _{j=1}^Mj^2\sin ^2\dfrac{x_{N-2M+j-1/2}}{\varepsilon } \le \sum _{j=1}^Mj^2=\dfrac{1}{6}M(M+1)(2M+1). \end{aligned}$$

Summing up all the above estimates and using the elementary inequality

$$\begin{aligned} (a+b+c)^2+b^2+c^2\ge \dfrac{a^2}{3}\quad \text {for any}\quad a,b,c\in \mathbb {R}, \end{aligned}$$

we have, for $M\ge 3$,

$$\begin{aligned} \sum _{j=1}^M\left|\mathcal {A}-b_{N-2M+j}\right|^2&\ge \dfrac{1}{3}\dfrac{(2-\mathcal {A})^2h^2}{4L^2}\sum _{j=1}^M(2j-1)^2- \dfrac{4\varepsilon ^2}{L^2}\sin ^2\dfrac{h}{2\varepsilon }\sum _{j=1}^Mj^2\\&\quad \sin ^2\dfrac{x_{N-2M+j-1/2}}{\varepsilon }-\dfrac{4Mh\varepsilon ^2}{L^2}\\&\ge \dfrac{(2-\mathcal {A})^2h^2}{36L^2}M(4M^2-1)-\dfrac{2\varepsilon ^2}{3L^2}M(M+1)(2M+1)-\dfrac{4M\varepsilon ^2}{L^2}\\&\ge \dfrac{(2-\mathcal {A})^2h^2}{36L^2}M(4M^2-1)-\dfrac{2\varepsilon ^2}{3L^2}M(4M^2-1)\\&\ge \dfrac{(2-\mathcal {A})^2h^2}{72L^2}M(4M^2-1) \end{aligned}$$

provided that $\varepsilon /h\le (2-\mathcal {A})/(4\sqrt{3})$. Substituting the above estimate into (A.3), we obtain

$$\begin{aligned} \int _{x_{N-2M}}^{x_{N-M}}\left|u_0'(x)-v_h'(x)\right|^2\,\mathrm {d}x&\ge \dfrac{(2-\mathcal {A})^2h^3}{1944L^2}M(4M^2-1)\\&\ge \dfrac{(2-\mathcal {A})^2h^3}{648L^2}M^3=\dfrac{(2-\mathcal {A})^2}{648}L. \end{aligned}$$

This implies

$$\begin{aligned} \Vert u_0'-v_h'\Vert _{L^2(1/2-2L,1/2-L)}\ge \dfrac{2-\mathcal {A}}{18\sqrt{2}}L^{1/2}=\dfrac{2-\mathcal {A}}{36\sqrt{2}}\left|K\right|^{1/2}. \end{aligned}$$

This shows that the factor $\left|K\right|^{1/2}$ in (3.4) is sharp. The same argument shows the size-dependence of $\left|K\right|$ in the estimate (3.5).

Case $h \ll \varepsilon$. We next consider the case when $h\ll \varepsilon$. In fact, we may employ coarser mesh with mesh size H outside the defect region with $H\gg h$, while a finer mesh with mesh size h inside the defect region. The above derivation remains true and we still have $v_h(x)=u_0(x)$ for $x\in [0,1/2-2L]$. We start from the inequality (A.3). Notice that the dominant term in the expression of $b_{N-2M+j}-\mathcal {A}$ is the oscillatory one in (A.4). Denote $\phi =2h/\varepsilon$. A direct calculation gives

$$\begin{aligned}&\sum _{j=1}^Mj^2\sin ^2\dfrac{x_{N-2M+j-1/2}}{\varepsilon }=\dfrac{1}{2}\sum _{j=1}^Mj^2-\dfrac{1}{2}\sum _{j=1}^M\cos \dfrac{x_{2N-4M+2j-1}}{\varepsilon }\\&\quad =\dfrac{1}{12}M(M+1)(2M+1)\\&\qquad -\biggl \{\dfrac{M(M+1)}{4\sin (\phi /2)}\sin [(N-M)\phi ]+\dfrac{M+1}{4\sin ^2\phi /2}\cos [(N-M)\phi ]\cos \dfrac{\phi }{2}\\&\qquad -\dfrac{\cos [(N-3M/2-1)\phi ]\cos \dfrac{\phi }{2}\sin \dfrac{M+1}{2}\phi }{4\sin ^3(\phi /2)}\biggr \}. \end{aligned}$$

We assume that

$$\begin{aligned} \sin \dfrac{\phi }{2}\ge \dfrac{5}{M}. \end{aligned}$$

(A.5)

Denote the terms in the curled bracket by I. Given (A.5), using the elementary inequalities $2x/\pi \le \sin x\le x$ for $x\in [0,\pi /2]$, we bound I as

$$\begin{aligned} \left|I\right|&\le \dfrac{(M+1)M}{4\sin (\phi /2)}+\dfrac{M+1}{4\sin ^2(\phi /2)}+\dfrac{(M+1)\phi /2}{4\sin ^3(\phi /2)}\\&\le \dfrac{(M+1)M}{4\sin (\phi /2)}+\dfrac{M+1}{4\sin ^2(\phi /2)}+\dfrac{(M+1)\pi }{8\sin ^2(\phi /2)}\\&\le \dfrac{M^2(M+1)}{12}, \end{aligned}$$

which immediately yields

$$\begin{aligned} \sum _{j=1}^Mj^2\sin ^2\dfrac{x_{N-2M+j-1/2}}{\varepsilon }\ge \dfrac{M^3}{12}. \end{aligned}$$

This implies

$$\begin{aligned} \dfrac{4\varepsilon ^2}{L^2}\sin ^2\dfrac{h}{2\varepsilon }\sum _{j=1}^Mj^2\sin ^2\dfrac{x_{N-2M+j-1/2}}{\varepsilon }\ge \dfrac{4\varepsilon ^2}{L^2}\left( \dfrac{2}{\pi }\dfrac{h}{2\varepsilon }\right) ^2\dfrac{M^3}{12}=\dfrac{M}{3\pi ^2}. \end{aligned}$$

Note also

$$\begin{aligned} \dfrac{(2-\mathcal {A})^2h^2}{4L^2}\sum _{j=1}^M(2j-1)^2\le \dfrac{(2-\mathcal {A})^2}{3}M. \end{aligned}$$

Combining the above two estimates, we obtain

$$\begin{aligned} \sum _{j=1}^M\left|\mathcal {A}-b_{N-2M+j}\right|^2&\ge \dfrac{1}{2}\sum _{j=1}^M \left( \dfrac{2\varepsilon }{L}\sin \dfrac{h}{2\varepsilon }j\sin \dfrac{x_{N-2M+j-1/2}}{\varepsilon }+\dfrac{(2-\mathcal {A})h}{2L}(2j-1)\right) ^2\\&\quad -\dfrac{4M\varepsilon ^2}{L^2}\\&\ge \left( \dfrac{1}{6}\left( 1/\pi +\mathcal {A}-2\right) ^2-\dfrac{4\varepsilon ^2}{L^2}\right) M>0 \end{aligned}$$

provided that

$$\begin{aligned} h>\dfrac{\sqrt{6}\varepsilon }{(1/\pi +\mathcal {A}-2)M}. \end{aligned}$$

This condition suffices for the validity of (A.5), which is satisfied under a weaker condition $h>{5\pi \varepsilon }/(2M).$

Substituting the above estimate into (A.3), we may find that there exists C depending only on $\mathcal {A}$ such that

$$\begin{aligned} \Vert u_0'-v_h'\Vert _{L^2(1/2-2L,1/2-L)}\ge CL^{1/2}=\dfrac{C}{2}\left|K\right|^{1/2}. \end{aligned}$$

This proves that the factor $\left|K\right|^{1/2}$ is sharp for (3.4). The same argument shows the size-dependence of $\left|K\right|$ in the estimate (3.5).