Tailored Finite Point Method for Diffusion Equations with Interfaces on Distorted Meshes

Abstract

Diffusion processes is usually coupled with other physical processes such as the fluid equation. The meshes are determined by the fluid that can be distorted as time goes on. Classical finite difference schemes and finite element method are sensitive of mesh deformation. We propose a new tailored finite point method (TFPM) for 2D diffusion equation with tensor diffusion coefficient on highly distorted meshes. Second order convergence is demonstrated numerically with and without interfaces. TFPM is a finite difference method that makes full use of the analytical properties of local solutions. The main advantages of TFPM is that no modifications have to be made for problems with strongly discontinuous coefficients, where most other methods require special treatment at the interfaces. This advantage is important for distorted meshes, since the designing of numerical discretizations near interfaces is more delicate for distorted meshes.

Appendix: Convergence Analysis for 1D Case

We give the convergence order analysis for the 1D case to present the main ideas of the convergence order analysis.

1.1 1D TFPM

Let the computational domain be [0, L] and $K(x)\in C^1[0,L]$ be a scalar function that satisfies $0<\zeta _1<K_1(x)<\zeta _2$, we consider the following 1D diffusion equation:

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\partial _x(K(x)\partial _x u)=f,\quad x\in (0,L) \\ u(0)=u_0,\quad u(L)=u_L. \end{array} \right. \end{aligned}$$

(31)

In this subsection, without any confusion, we use the same notations for the diffusion coefficient and solution as for 2D TFPM.

Let the stencil be as in Fig. 13. The interval [0, L] is divided into I sub-intervals that are denoted by $L_i=[x_{i-1},x_i]$ with $i=1,2,\cdots , I$ and $x_0=0$, $x_I=L$. Let $x_{i-1/2}=\frac{x_{i}+x_{i-1}}{2}$ be the center of $L_i$ and $h_i=x_i-x_{i-1}$ be the length of ith interval. On $L_i$, (31) can be approximated by

$$\begin{aligned} -\partial _x({\bar{K}}_{i}(x)\partial _x {\bar{u}})=\bar{f}_{i}, \end{aligned}$$

(32)

where

$$\begin{aligned} {\bar{K}}_{i}(x)=k_{xi}(x-x_{i-1/2})+k_i, \quad \text{ with } \,\quad k_i=\frac{K(x_{i-1})+K(x_{i})}{2}\, \hbox { and }\, k_{xi}=\frac{K(x_i)-K(x_{i-1})}{h_i},\nonumber \\ \end{aligned}$$

(33)

and

$$\begin{aligned} \bar{f}_{i}=f_i+f_{xi}(x-x_{i-1/2}) \quad \text{ with } \, f_i=\frac{f(x_{i-1})+f(x_{i})}{2} \, \hbox {and }\, f_{xi}=\frac{f(x_i)-f(x_{i-1})}{h_i}.\nonumber \\ \end{aligned}$$

(34)

Piecing ${\bar{K}}_i(x)$ together gives ${\bar{K}}\in C[0,L]$. If we piece ${\bar{u}}|_{L_i}$ together by the continuity of ${\bar{u}}$ and ${\bar{K}}(x)\partial _x{\bar{u}}$ at the grid points, we find an approximation to the solution of (31).

Fig. 13

The stencil for 1D TFPM

Let

$$\begin{aligned} {\tilde{v}}_i(x)=\left\{ \begin{array}{ll}-\frac{x-x_{i-1/2}}{k_{xi}}f_i,\quad &{}k_{xi}\ne 0,\nonumber \\ -\frac{(x-x_{i-1/2})^2}{2k_{i}}f_i, \quad &{} k_{xi}=0, \end{array}\right. \quad \begin{array}{ll}W_{1,i}(x)=1,\quad W_{2,i}(x)=e^{-\frac{k_{xi}(x-x_{i-1/2})}{k_i}},\quad &{}k_{xi}\ne 0,\\ W_{1,i}(x)=1,\quad W_{2,i}(x)=x,\quad &{}k_{xi}=0.\end{array} \nonumber \\ \end{aligned}$$

(35)

Then for any constants $c_{1i}$, $c_{2i}$, $u_{hi}(x)=\sum _{k=1}^2 c_{ki}W_{k,i}(x)+{{\tilde{v}}}_i(x)$ exactly satisfies

$$\begin{aligned} -(k_i\partial _{xx} u_{hi}+k_{xi}\partial _x u_{hi})={f}_{i}. \end{aligned}$$

(36)

Here $c_{1i}$, $c_{2i}$ can be determined by the interface conditions:

$$\begin{aligned} \begin{array}{l} u_{hi}(x_i)= u_{hi+1}(x_i),\qquad ({\bar{K}}_{i} \partial _x {\bar{u}}_{hi})\big |_{x_i}=({\bar{K}}_{i+1} \partial _x u_{hi+1})\big |_{x_i}. \end{array} \end{aligned}$$

(37)

That is

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \sum _{k=1}^{2}c_{k,i}W_{k,i}(x)+{{\tilde{v}}}_{i}(x_i) =\sum _{k=1}^{2}c_{k,i+1}W_{k,i+1}(x_i)+{{\tilde{v}}}_{i+1}(x_i), \\ \displaystyle {\bar{K}}_{i}(x_i)\big (\sum _{k=1}^{2}c_{k,i}\partial _x W_{k,i}+\partial _x {\tilde{v}}_{i}\big )|_{x_i} ={\bar{K}}_{i+1}(x_{i+1})\big (\sum _{k=1}^{2}c_{k,i+1}\partial _x W_{k,i+1}+\partial _x {\tilde{v}}_{i+1}\big )|_{x_i}. \end{array} \right. \end{aligned}$$

(38)

Combing the boundary conditions $u_{h1}(0)=u_0$, $u_{hI}(L)=u_L$ and interface conditions (38), we get a linear system of 2I equations and the coefficients $c_{1i}$, $c_{2i}$ can be determined. Piecing together all $u_{hi}$, we find an approximation to the solution of (31).

1.2 Error Estimate for 1D TFPM

We give an $L^2$ error estimate for the above proposed 1D TFPM. We emphasis that the proof does not depend on the ration between $\max _i \{h_i\}$ and $\min _i \{h_i\}$. Assume that $h=\max _i \{h_i\}$. u satisfies (31), $u_h$ is the solution to the 1D TFPM, then

$$\begin{aligned} \parallel u-u_h\parallel _2=(\int _{L}|u-u_h|^2 \, \mathrm{d} x)^{1/2}\le \Vert u-{\bar{u}}\Vert _{2}+\Vert {\bar{u}}-{\bar{u}}_h\Vert _2+\Vert {\bar{u}}_h- u_h\Vert _2, \end{aligned}$$

where

• ${\bar{u}}\in C^1[0,L]$ and satisfies (32) in each interval $L_i$. The boundary conditions are ${\bar{u}}(0)=u(0)$, ${\bar{u}}(L)=u(L)$.

• For all $i=1,2 \cdots I$, ${\bar{u}}_h|_{L_i}={\bar{u}}_{hi}=\sum _{k=1}^{2}{\bar{c}}_{k,i} W_{k,i}+{\tilde{v}}_i$ with $W_{k,i}$, ${{\tilde{v}}}_i$ being as in (35). ${\bar{c}}_{k,i}$ are constants that are determined by the following boundary conditions: $\partial _x{\bar{u}}_{hi}|_{x_{i}}=\partial _x{\bar{u}}_i|_{x_{i}}$, and ${\bar{u}}_{hi}({x_{i}})={\bar{u}}_i({x_{i}})$.

• $u_h$ is the solution to the 1D TFPM.

We prove that each of the three terms $\Vert u-{\bar{u}}\Vert _{2}$, $\Vert {\bar{u}}-{\bar{u}}_h\Vert _2$, $\Vert {\bar{u}}_h- u_h\Vert _2$ can be controlled by $Ch^2$ in the subsequent part.

1.2.1 Bound for $\Vert u-{\bar{u}}\Vert _2$

From the definitions of $f_i$ and $f_{xi}$ in (34), when $f(x)|_{L_i}\in C^2(L_i)$ for all $i\in \{1,\cdots , I\}$, $f(x)|_{L_i}={\bar{f}}_i+O(h_i^2)$. Then

$$\begin{aligned} \begin{aligned} \Vert f-{\bar{f}}\Vert _2^2=\sum _{i=1}^{I}\int _{L_i}(f-{\bar{f}}_i)^2 \, \mathrm{d} x\le C \sum _{i=1}^{I} h_i^5\le CL h^4. \end{aligned} \end{aligned}$$

(39)

That is $\Vert f-{\bar{f}}\Vert _2\le C h^2$. Similarly, $\Vert K-{\bar{K}}\Vert _2<Ch^2$. We have the following lemma:

Lemma 1

Let $w=u-{\bar{u}}$, then

$$\begin{aligned} \Vert \partial _x w\Vert _2<Ch^2,\quad \quad \Vert w\Vert _2<Ch^2, \end{aligned}$$

(40)

where C is a constant independent of h.

Proof

Multiplying both sides of Eq. (31) by w and integrating over [0, L], from $w\in C^1[0,L]$ and $w(0)=w(L)=0$, one gets

$$\begin{aligned} \int _L(\partial _x w)\cdot K\partial _x u \, \mathrm{d} x=\int _L wf \, \mathrm{d} x. \end{aligned}$$

(41)

Similarly, from Eq. (32),

$$\begin{aligned} \int _L (\partial _x w)\cdot {\bar{K}}\partial _x {\bar{u}} \, \mathrm{d} x=\int _L w{\bar{f}} \, \mathrm{d} x. \end{aligned}$$

(42)

Subtracting (42) from (41) yields

$$\begin{aligned} \begin{aligned} \zeta _1\Vert \partial _x w\Vert _2^2\le \int _\Omega (\partial _x w)\cdot K\partial _x w \, \mathrm{d} x&=- \int _L(\partial _x w)\cdot (K-{\bar{K}})\partial _x {\bar{u}} \, \mathrm{d} x+\int _L w(f-{\bar{f}}) \, \mathrm{d} x\\&\le \Vert \partial _x w\Vert _2\Vert (K-{\bar{K}})\partial _x{\bar{u}}\Vert _2+\Vert w\Vert _2 \Vert f-{\bar{f}}\Vert _2. \end{aligned} \end{aligned}$$

(43)

where the last inequality is from Holder’s inequality. Since $w\in C^1[0,L]$, $w(0)=w(L)=0$, from Friedrichs inequality we have $\Vert w\Vert _2\le C\Vert \partial _x w\Vert _2$. Therefore there exists C independent of h that satisfies

$$\begin{aligned} \Vert \partial _x w\Vert _2\le Ch^2. \end{aligned}$$

Then Friedrichs inequality gives

$$\begin{aligned} \Vert w\Vert _2\le C\Vert \partial _x w\Vert _2\le Ch^2. \end{aligned}$$

$\square$

1.2.2 Bound for $\Vert {\bar{u}}-{\bar{u}}_h\Vert _2$

Let $u_h|_{L_i}=u_{hi}$, then ${\bar{u}}_{hi}$ satisfies

$$\begin{aligned} (k_i\partial _{xx}{\bar{u}}_{hi}+k_{xi}\partial _x {\bar{u}}_{hi})+{f}_{i}=0, \end{aligned}$$

(44)

Then

$$\begin{aligned} \partial _x \cdot \big ({\bar{K}}_i\partial _x {\bar{u}}_{hi}\big )+{\bar{f}}_i=R.H.S.\end{aligned}$$

(45)

with

$$\begin{aligned} \begin{aligned} R.H.S.=&k_{xi}\partial _x^2{\bar{u}}_{hi}(x)(x-x_{i-1/2})+f_{xi}(x-x_{i-1/2}). \end{aligned} \end{aligned}$$

(46)

Since $\frac{1}{h_i}\int _{L_i}x \, \mathrm{d} x=x_{i-1/2}$ and ${\bar{u}}_{hi}\in C^\infty (L_i)$, we have

$$\begin{aligned} \begin{aligned}&\int _{L_i}R.H.S. \, \mathrm{d} x=\int _{L_i}k_{xi}(x-x_{i-1/2})\big (\partial ^2_x{\bar{u}}_h(x_{i-1/2})+(x-x_{i-1/2})\partial ^3_x{\bar{u}}_h(\xi (x))\big ) \, \mathrm{d} x=O(h_i^3). \end{aligned} \end{aligned}$$

(47)

The difference between ${\bar{u}}$ and ${\bar{u}}_h$ is given by the following lemma:

Lemma 2

Let ${\bar{w}}={\bar{u}}-{\bar{u}}_h$, then ${\bar{w}}_i={\bar{w}}|_{L_i}\in C^2(L_i)$ satisfies

$$\begin{aligned} \Vert {\bar{w}}_i\Vert _{\infty }\le Ch_i^3,\quad \quad |({\bar{K}}_i\partial {\bar{w}}_i)|_{x_{i-1}}|\le C h_i^3. \end{aligned}$$

where C is a constant independent of $h_i$.

Proof

Since $\partial _x{\bar{u}}_{hi}|_{x_{i}}=\partial _x{\bar{u}}_{i}|_{x_{i}}$, ${\bar{u}}_{hi}({x_{i}})={\bar{u}}_{i}({x_{i}})$, we have $\partial _x{\bar{w}}_i|_{x_{i}}=0$, ${\bar{w}}_i({x_{i}})=0$. It is easy to check that ${\bar{w}}_i\in C^2(L_i)$, then $\partial _x{\bar{w}}_i(x)=\partial _x{\bar{w}}_i(x_i)+(x-x_i)\partial ^2_x{\bar{w}}(\xi )=(x-x_i)\partial ^2_x{\bar{w}}(\xi )$ for some $\xi \in L_i$. (32) and (45) indicates that $\Vert k_i\partial ^2 _x{\bar{w}}_i+k_{xi}\partial _x{\bar{w}}_i\Vert _{L^\infty }<Ch_i$, thus $\Vert \partial ^2 _x{\bar{w}}_i\Vert _{L^\infty }<Ch_i$, and $\Vert \partial _x{\bar{w}}_i\Vert _{L^\infty }<Ch_i^2$. From ${\bar{w}}_i|_{x_{i}}=0$, $\partial _x{\bar{w}}_i(x_i)=0$, we have ${\bar{w}}_i={\bar{w}}_i(x_i)+h_i\partial _x{\bar{w}}_i(x_i)+h_i^2\partial _x^2{\bar{w}}_i(\xi )=h_i^2\partial _x^2{\bar{w}}(\xi )$ and thus $\Vert {\bar{w}}_i\Vert _{L^\infty }<Ch_i^3$. On the other hand, (32) and (45) gives

$$\begin{aligned} \begin{aligned} \partial _x({\bar{K}}_i \partial {\bar{w}}_i(x))=-R.H.S. \end{aligned} \end{aligned}$$

(48)

Integrating over $L_i$ we find

$$\begin{aligned} \begin{aligned} {\bar{K}}_i(x_i) \partial {\bar{w}}_i|_{x_i}-{\bar{K}}_i(x_{i-1}) \partial {\bar{w}}_i|_{x_{i-1}}=O(h_i^3). \end{aligned} \end{aligned}$$

(49)

Together with $\partial _x{\bar{w}}_i|_{x_{i}}=0$, we obtain

$$\begin{aligned} |({\bar{K}}_i\partial {\bar{w}}_i)|_{x_{i-1}}|\le C h_i^3. \end{aligned}$$

$\square$

1.2.3 Bound for $\Vert {\bar{u}}_{h}-u_h\Vert _2$

The function values ${\bar{u}}_{hi}(x_{i})$ and ${\bar{u}}_{hi+1}(x_{i})$ are not necessarily the same. Since ${\bar{u}}_{i+1}(x_{i})= {\bar{u}}_{i}(x_{i})$ and $\big ({\bar{K}}_{i+1}\partial _x {\bar{u}}_{i+1}\big )|_{x_{i}}= \big ({\bar{K}}_{i}\partial _x{\bar{u}}_{i}\big )|_{x_{i}}$, we have

$$\begin{aligned} \begin{aligned}&{\bar{u}}_{hi+1}(x_{i})- {\bar{u}}_{hi}(x_{i})\\ =&{\bar{u}}_{i+1}(x_{i})-{\bar{w}}_{i+1}(x_{i})- {\bar{u}}_{i}(x_{i})+{\bar{w}}_{i}(x_{i}) =-{\bar{w}}_{i+1}(x_{i})=O(h_{i+1}^3).\\&\big ({\bar{K}}_{i+1}\partial _x {\bar{u}}_{hi+1}\big )|_{x_{i}}- \big ({\bar{K}}_{i}\partial _x{\bar{u}}_{hi}\big )|_{x_{i}}\\ =&\big ({\bar{K}}_{i+1}\partial _x {\bar{u}}_{i+1}\big )|_{x_{i}}-\big ({\bar{K}}_{i+1}\partial _x {\bar{w}}_{i+1}\big )|_{x_{i}}- \big ({\bar{K}}_{i}\partial _x{\bar{u}}_{i}\big )|_{x_{i}}+\big ({\bar{K}}_{i}\partial _x{\bar{w}}_{i}\big )|_{x_{i}} =O(h_{i+1}^3). \end{aligned} \end{aligned}$$

(50)

Then ${\bar{u}}_h|_{L_i}={\bar{u}}_{hi}=\sum _{k=1}^{2}{\bar{c}}_{k,i} W_{k,i}+{\tilde{v}}_i$ and satisfies the interface condition as in (50). At the boundary, ${\bar{u}}_{hI}(L)=u_L$, ${\bar{u}}_{h1}(0)={\bar{u}}(0)-{\bar{w}}(0)=u_0+O(h_1^3)$. The following lemma gives the distance between ${\bar{u}}_h$ and $u_h$.

Lemma 3

Let $w_h={\bar{u}}_h-u_h$, then $w_{hi}(x)=w_h|_{L_i}\in C^\infty (L_i)$ and satisfies

$$\begin{aligned} \Vert \partial _x w_h\Vert _{\infty }<Ch^2,\qquad \Vert w_h\Vert _\infty \le Ch^2, \end{aligned}$$

where C is a constant independent of h.

Proof

For $i=1,\cdots , I$, $k_i\partial _x^2w_{hi}+k_{xi}\partial _xw_{hi}=0$. From (37) and (50), for $i=1,\cdots , I-1$,

$$\begin{aligned} w_{hi}(x_i)-w_{hi+1}(x_i)=O(h_{i+1}^3),\qquad ({\bar{K}}_{i}\partial _xw_{hi})|_{x_i}-({\bar{K}}_{i+1}\partial _x w_{hi+1})|_{x_i}=O(h_{i+1}^3). \end{aligned}$$

(51)

At the boundary $w_{h}(L)=0$, $w_h(0)=O(h_1^3)$. First of all, we show that there exists $\xi \in [0,L]$ such that $|\partial _xw_h(\xi )|\le Ch^2$. There are three situations:

• For some $j\in \{1,2,\cdots , I\}$, there exists $\xi \in (x_{j-1},x_j)$ such that $\partial _xw_{hj}(\xi )=0$.

• For any given $j\in \{1,\cdots , I\}$, $\forall x\in L_j$, $\partial _xw_{hj}(x)$ has the same sign, and there exists $j\in \{1,\cdots , I-1\}$ such that

  $$\begin{aligned} \partial _xw_{hj}(x_j)\partial _xw_{hj+1}(x_j)\le 0. \end{aligned}$$

  From (51) and the fact that ${\bar{K}}_i(x_i)={\bar{K}}_{i+1}(x_i)>0$, we find

  $$\begin{aligned} |\partial _x w_{hj}(x_j)|\le Ch_{i+1}^3 \end{aligned}$$

• For any given $j\in \{1,\cdots , I\}$, $\forall x\in L_j$, $\partial _xw_{hj}(x)$ has the same sign and for all $j\in \{1,\cdots , I-1\}$

  $$\begin{aligned} \partial _xw_{hj}(x_j)\partial _xw_{hj+1}(x_j)\ge 0. \end{aligned}$$

  Assume that for $\forall i\in \{1,\cdots , I-1\}$, $\partial _xw_{hi}(x_i)\ge 0$, $\partial _xw_{hi+1}(x_i)\ge 0$, then $w_{hi}(x_i)\ge w_{hi}(x_{i-1})$ for $\forall i\in \{1,\cdots , I\}$ and

  $$\begin{aligned} \Big |\sum _{i=1}^I\big ( w_{hi}(x_i)-w_{hi}(x_{i-1})\big )\Big |\ge \min _{i=1,\cdots ,I}\big |\partial _xw_{hi}(\xi _{i})\big |\sum _{i=1}^Ih_{i}, \end{aligned}$$

  where $\xi _i\in L_i$ and satisfies $w_{hi}(x_i)-w_{hi}(x_{i-1})=h_i\partial _xw_{hi}(\xi _i)$. On the other hand,

  $$\begin{aligned} \begin{aligned} \Big |\sum _{i=1}^{I}\int _{L_i}\partial _x w_h \, \mathrm{d} x\Big |&=\Big |\sum _{i=1}^I\big ( w_{hi}(x_i)-w_{hi}(x_{i-1})\big )\Big |\\&=\Big |\sum _{i=1}^{I-1}\big (w_{hi}(x_i)-w_{hi+1}(x_i)\big )-w_{h1}(0)+w_{hI}(L)\Big |\\&\le C\sum _{i=1}^Ih_i^3. \end{aligned} \end{aligned}$$

  Similar results hold when $\forall i\in \{1,\cdots , I-1\}$, $\partial _xw_{hi}(x_i)\le 0$, $\partial _xw_{hi+1}(x_i)\le 0.$ Thus

  $$\begin{aligned} \min _{i=1,\cdots ,I}|\partial _xw_{hi}(\xi _i)|<Ch.^2 \end{aligned}$$

We have shown that there exists $\xi \in L_s$, such that $\big |\partial _xw_h(\xi )\big |\le Ch^2$, we then prove by induction that $\forall x\in [0,L]$, $\big |\partial _xw_h(x)\big |<Ch^2$. We first consider the interval $[\xi ,x_{s}]$. Since $|\partial _xw_h(\xi )|\le Ch^2$ and $k_s\partial _x^2w_h(\xi )+k_{xs}\partial _xw_h(\xi )=0$, we have $|\partial _x^2w_h(\xi )|\le Ch^2$. Let

$$\begin{aligned} \alpha =\frac{|k_{xs}|}{|k_s|}, \end{aligned}$$

then for $\forall x\in [\xi ,x_s]$,

$$\begin{aligned} \begin{aligned}|\partial _x^2w_{hs}(x)|&= \alpha \big |\partial _xw_h(x)\big |\le \alpha \Big |\partial _xw_{hs}(\xi )+(x-\xi )\partial _x^2w_{hs}(\xi ')\Big | \\&\le \alpha \Big |\partial _xw_{hs}(\xi )-(x-\xi ) \alpha \partial _xw_{hs}(\xi ')\Big | \\&\le Ch^2 \end{aligned} \end{aligned}$$

where $\xi '\in [\xi ,x]$. Then for $\forall x\in [\xi ,x_s]$,

$$\begin{aligned} \begin{aligned} \big |\partial _xw_{hs}(x)\big |&=\big |\partial _xw_{hs}(\xi )+(x-\xi )\partial _{x}^2w_{hs}(\xi )+\frac{1}{2}(x-\xi )^2\partial _{x}^3w_{hs}(\eta _s)\big |\\&=\big |\partial _xw_{hs}(\xi )+(x-\xi )\partial _{x}^2w_{hs}(\xi )-\frac{1}{2}(x-\xi )^2 \alpha \partial _{x}^2w_{hs}(\eta _s)\big |\\&<Ch^2\big (1+h_s \alpha +h_s^2 \alpha \big ), \end{aligned} \end{aligned}$$

where we have used $k_s\partial _x^3w_h(\eta _s)+k_{xs}\partial ^2_xw_h(\eta _s)=0$ in the second equality. Since ${\bar{K}}_s(x_s)={\bar{K}}_{s+1}(x_s)$, we have $\partial _xw_{hs}(x_s)-\partial _xw_{hs+1}(x_s)=O(h_{s+1}^3)$ from (51), then

$$\begin{aligned} \big |\partial _xw_{hs+1}(x_s)\big |<Ch^2\big (1+h_s \alpha +h^2_s \alpha \big )+Ch_{s+1}^3. \end{aligned}$$

Similar discussions for the interval $L_{s+1}$, we get

$$\begin{aligned} \begin{aligned} \big |\partial _xw_{hs+1}(x_{s+1})\big |=&\big |\partial _xw_{hs+1}(x_s)+(x-x_s)\partial _{x}^2w_{hs+1}(x_s)+\frac{1}{2}(x-x_s)^2\partial _{x}^3w_{hs+1}(\eta _{s+1})\big |\\ \le&\Big (Ch^2\big (1+ \alpha h_s+ h_s^2\alpha \big )+Ch_{s+1}^3\Big )\big (1+h_{s+1} \alpha +h_{s+1}^2 \alpha \big ) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \big |\partial _xw_{hs+2}(x_{s+1})\big |\le \Big (Ch^2\big (1+ \alpha h_s+ h_s^2\alpha \big )+Ch_{s+1}^3\Big )\big (1+h_{s+1} \alpha +h_{s+1}^2 \alpha \big )+Ch_{s+2}^3 \end{aligned}$$

By induction, we find for $\forall s'\in \{s+1,\cdots ,I\}$ and $\forall x\in L_{s'}$,

$$\begin{aligned} \begin{aligned} \big |\partial _xw_{hs'}(x)\big |\le&Ch^2\Pi _{i=s}^{s'}\big (1+h_{i} \alpha +h_{i}^2 \alpha \big )+Ch_{s+1}^3\Pi _{i=s+1}^{s'}\big (1+h_{i} \alpha +h_{i}^2 \alpha \big )\\&+Ch_{s'}^3\big (1+h_{s'} \alpha +h_{s'}^2 \alpha \big ) \end{aligned} \end{aligned}$$

(52)

Since the geometrical mean is less than the arithmetic mean, we have for arbitrary $s_1,s_2\in \{s,s+1,\cdots ,I\}$ and $s_1<s_2$,

$$\begin{aligned} \Pi _{i=s_1}^{s_2}\big (1+h_{i}\alpha +h_{i}^2\alpha \big )\le \Big (\frac{\sum _{i=s_1}^{s_2}\big (1+h_{i}\alpha +h_{i}^2\alpha \big )}{s_2-s_1}\Big )^{s_2-s_1}\le \big (1+\frac{CL}{s_2-s_1}\big )^{s_2-s_1}\le C \end{aligned}$$

where C is a constants independent of I and $h_i$. Then (52) gives that for $\forall x\in [\xi ,L]$, $\big |\partial _xw_{h}(x)\big |<Ch^2$. Similar result holds for the interval $[0,\xi ]$ and we find $\Vert \partial _xw_h\Vert _{\infty }\le Ch^2$. Then since $w_h(L)=0$, it is easy to check $\Vert w_h\Vert _\infty \le Ch^2$ by induction. $\square$