## 6 Phase Field Lattice Boltzmann Method
### Solving the Cahn-Hillard Equation for Interface Capturing with the WENO Scheme
By splitting the flux term, l, into positive and negative terms, we have:  
\[\
abla.(\upomega)=\
abla.(\text{l})=\frac{1}{\Delta x}\left(\widetilde{\text {l}}_{\text{i}+\frac{1}{2}}^{+}-\widetilde{\text{l}}_{\text{i}+\frac{1}{2}} \right)+\frac{1}{\Delta x}\left(\widetilde{\text{l}}_{\text{i}-\frac{1}{2}}^{+ }-\widetilde{\text{l}}_{\text{i}-\frac{1}{2}}^{-}\right) \tag{25}\]where \(\widetilde{l}_{i+\frac{1}{2}}^{+}\) and \(\widetilde{l}_{i-\frac{1}{2}}^{+}\) are approximated by five points as:  
\[\widetilde{l}_{i+\frac{1}{2}}^{+}=\omega_{0}^{+}\Big{(}\frac{2}{6}\widetilde{l }_{i-2}^{1+}-\frac{7}{6}\widetilde{l}_{i-1}^{+}+\frac{11}{6}\widetilde{l}_{i}^ {+}\Big{)}+\omega_{1}^{+}\Big{(}-\frac{1}{6}\widetilde{l}_{i-1}^{+}-\frac{5}{6 }\widetilde{l}_{i}^{+}+\frac{2}{6}\widetilde{l}_{i+1}^{+}\Big{)}+\omega_{2}^{+ }\Big{(}\frac{2}{6}\widetilde{l}_{i}^{+}+\frac{5}{6}\widetilde{l}_{i+1}^{+}- \frac{1}{6}\widetilde{l}_{i+2}^{+}\Big{)} \tag{26}\]  
And:  
\[\widetilde{l}_{i+\frac{1}{2}}^{-}=\omega_{2}^{-}\Big{(}-\frac{1}{6}\widetilde{ l}_{i-1}^{-}+\frac{5}{6}\widetilde{l}_{i}^{-}+\frac{2}{6}\widetilde{l}_{i+1}^ {-}\Big{)}+\omega_{1}^{-}\Big{(}\frac{2}{6}\widetilde{l}_{i}^{-}+\frac{5}{6} \widetilde{l}_{i+1}^{-}-\frac{1}{6}\widetilde{l}_{i+2}^{-}\Big{)}+\omega_{0}^ {-}\Big{(}\frac{11}{6}\widetilde{l}_{i+1}^{-}-\frac{7}{6}\widetilde{l}_{i+2}^ {-}+\frac{2}{6}\widetilde{l}_{i+3}^{-}\Big{)} \tag{27}\]  
The stencil weights can be calculated as:  
\[\omega_{0}^{\pm}=\frac{\alpha_{0}^{\pm}}{\alpha_{0}^{\pm}+\alpha_{1}^{\pm}+ \alpha_{2}^{\pm}}\quad\omega_{1}^{\pm}=\frac{\alpha_{1}^{\pm}}{\alpha_{0}^{ \pm}+\alpha_{1}^{\pm}+\alpha_{2}^{\pm}}\quad\omega_{2}^{\pm}=\frac{\alpha_{2} ^{\pm}}{\alpha_{0}^{\pm}+\alpha_{1}^{\pm}+\alpha_{2}^{\pm}} \tag{28}\]  
where:  
\[\alpha_{0}^{\pm}=\frac{1}{10}\Bigg{(}\frac{1}{\epsilon+\mathrm{IS}_{0}^{\pm}} \Bigg{)}\quad\alpha_{1}^{\pm}=\frac{6}{10}\Bigg{(}\frac{1}{\epsilon+\mathrm{ IS}_{1}^{\pm}}\Bigg{)}\quad\alpha_{2}^{\pm}=\frac{3}{10}\Bigg{(}\frac{1}{ \epsilon+\mathrm{IS}_{2}^{\pm}}\Bigg{)} \tag{29}\]  
The parameter, \(\mathrm{IS}_{k}\), is defined as:  
\[ \begin{array}{l}\mathrm{IS}_{0}^{+}=\frac{13}{12}\big{(}\widetilde{l}_{i-2}^ {+}-2\widetilde{l}_{i-1}^{+}+\widetilde{l}_{i}^{+}\big{)}^{2}+\frac{1}{4}\big{(} \widetilde{l}_{i-2}^{+}-4\widetilde{l}_{i-1}^{+}+3\widetilde{l}_{i}^{+}\big{)} ^{2}\\ \mathrm{IS}_{1}^{+}=\frac{13}{12}\big{(}\widetilde{l}_{i-1}^{+}-2\widetilde{l}_ {i}^{+}+\widetilde{l}_{i+1}^{+}\big{)}^{2}+\frac{1}{4}\big{(}\widetilde{l}_{i- 1}^{+}-\widetilde{l}_{i+1}^{+}\big{)}^{2}\\ \mathrm{IS}_{2}^{+}=\frac{13}{12}\big{(}\widetilde{l}_{i}^{+}-2\widetilde{l}_ {i+1}^{+}+\widetilde{l}_{i+2}^{+}\big{)}^{2}+\frac{1}{4}\big{(}\widetilde{l}_{ i-1}^{+}-4\widetilde{l}_{i+1}^{+}+\widetilde{l}_{i+2}^{+}\big{)}^{2}\end{array}  \tag{30}\]  
And:  
\[ \begin{array}{l}\mathrm{IS}_{0}^{-}=\frac{13}{12}\big{(}\widetilde{l}_{i+1}^ {-}-\frac{7}{6}\widetilde{l}_{i+2}^{-}+\frac{11}{6}\widetilde{l}_{i+3}^{-} \big{)}^{2}+\frac{1}{4}\big{(}\widetilde{l}_{i+1}^{-}-4\widetilde{l}+3\widetilde {l}_{i+3}^{-}\big{)}^{2}\\ \mathrm{IS}_{1}^{-}=\frac{13}{12}\big{(}\widetilde{l}_{i}^{-}-2\widetilde{l}_{i+ 1}^{-}+\widetilde{l}_{i+2}^{-}\big{)}^{2}+\frac{1}{4}\big{(}\widetilde{l}_{i- 1}^{-}+\widetilde{l}_{i+2}^{-}\big{)}^{2}\\ \mathrm{IS}_{2}^{-}=\frac{13}{12}\big{(}\widetilde{l}_{i-1}^{-}-2\widetilde{l}_ {i+1}^{-}-\widetilde{l}_{i+1}^{-}\big{)}^{2}+\frac{1}{4}\big{(}\widetilde{l}_{ i-1}^{-}-4\widetilde{l}_{i}^{-}+3\widetilde{l}_{i+1}^{-}\big{)}^{2}\end{array}  \tag{31}\]  
In this way, \(\widetilde{l}_{i+\frac{1}{2}}^{+}\) in Equation (25) can be calculated. The value of \(\widetilde{l}_{i-\frac{1}{2}}^{-}\) can be calculated in the same way. Next, we focus on the diffusion term in Equation (23).