From 401f430a2294da6b3ee178d16074080a472a4eac Mon Sep 17 00:00:00 2001 From: bangerth Date: Thu, 19 Oct 2006 22:06:00 +0000 Subject: [PATCH] Move introduction to doxygen file git-svn-id: https://svn.dealii.org/trunk@14029 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-21/doc/intro.dox | 218 ++++++++++++++++++ .../doc/yanli_project_introduction.tex | 218 ------------------ 2 files changed, 218 insertions(+), 218 deletions(-) delete mode 100644 deal.II/examples/step-21/doc/yanli_project_introduction.tex diff --git a/deal.II/examples/step-21/doc/intro.dox b/deal.II/examples/step-21/doc/intro.dox index f693a95d3c..c232b6bcef 100644 --- a/deal.II/examples/step-21/doc/intro.dox +++ b/deal.II/examples/step-21/doc/intro.dox @@ -1 +1,219 @@

Introduction

+ +This program grew out of a student project by Yan Li at Texas A&M +University. Most of the work for this program is by her. + +In this project, we propose a numerical simulation for two phase +flow problem in porous media. The two phase flow system includes one +elliptic equation and one nonlinear transport equation. We apply +mixed finite element method and Discontinuous Galerkin method for +this system. Some numerical results for two dimensional case are +given by $RT_{0}\times DQ_{0}\times DQ_{0}$. + +The numerical computation is based on $dealII$. We use vector shape +functions from step9, DG method from step12, mixed method and Schur +complement from step20 and many many useful tools from the library. + +

Two Phase Flow Problem

+ +The modeling of two phase flow in porous media is important for both +environmental rededication and the management of petroleum +reservoirs. Practical situations involving two phase flow include +the dispersal of a nonaqueous phase liquid in an aquifer or the +displacement of a non-aqueous heterogeneity on the flow and +transport. Simulation models, if they are to provide realistic +predictions, must accurately account for these effects. + +In our project,we consider a kind of periodic permeability,our +numerical result shows that the heterogeneity effects are simulated +accurately. + +Consider two phase flow in a reservoir $\Omega$ under the assumption +that the displacement is dominated by viscous effects; i.e. we +neglect the effects of gravity, compressibility, and capillary +pressure. Porosity will be considered to be constant. The two phase +will be referred to as water and oil, designated by subscripts $w$ +and $o$, respectively. We write Darcy's for each phase as follows: +@f{eqnarray*} +\mathbf{u}_{j} = \frac{k_{rj}(S)}{\mu_{j}} \mathbf{K} \cdot \nabla p +@f} + where, $\mathbf{u}_{j}$ is the phase velocity, $K$ is the +permeability tensor, $k_{rj}$ is the relative permeability to phase +$j$($j=o,w$),$S$ is the water saturation(volume fraction), $P$ is +pressure and $\mu_{j}$ is the viscosity of phase $j$($j=o,w$). + +Combining Darcy's law with a statement of conservation of mass +allows us to express the governing equations in terms of the +so-called pressure and saturation equations: +@f{eqnarray*} +\nabla \cdot (\mathbf{K}(x,y)\lambda(S) \nabla p)= q(x,y) && \forall(x,y)\in\Omega\\ + S_{t} + \mathbf{u} \cdot \nabla F(S) = 0&& \forall(x,y)\in\Omega +@f} + + + where, $\lambda$ is the total mobility, f is the fractional +flow of water, $q$ is a source term and $\mathbf{u}$ is the total +velocity, which are respectively given by: +@f[\mathbf{u} = +\mathbf{u}_{o} + \mathbf{u}_{w} = -\lambda(S) \mathbf{K}\cdot\nabla +p@f] +@f[\lambda(S) = \frac{k_{rw}(S)}{\mu_{w}}+\frac{k_{ro}(S)}{\mu_{o}}@f] +@f[F(S) = \frac{k_{rw}(S)/\mu_{w}}{k_{rw}(S)/\mu_{w} + k_{ro}(S)/\mu_{o}}@f] + + +

Discretization

+ + For simplicity, in our project we will assume no +source $q=0$ and the heterogeneous porous medium is isotropic +$\mathbf{K}(x,y) = +k(x,y) \mathbf{I}$. + +Our two dimensional numerical simulation will be done on unit cell +$\Omega = [0,1]\times [0,1]$ for $t\in [0,T]$. +\f{eqnarray} +\mathbf{u}(x,y)+\mathbf{K}(x,y)\lambda(S) \nabla p= 0 && \forall(x,y)\in\Omega, \forall t\in [0,T]\\ +\nabla \cdot\mathbf{u}(x,y)= 0 && \forall(x,y)\in\Omega, \forall t \in [0,T] \\ +S_{t} + \mathbf{u} \cdot \nabla F(S) = 0&& \forall(x,y)\in\Omega, +\forall t \in [0,T] +\f} + Boundary conditions are: +\f[ +\begin{array}{cr} +p(x,y)=1 & \forall(x,y)\in \Gamma_{1}:=\{(x,y)\in \partial \Omega: x=0\}\\ +p(x,y)=0 & \forall(x,y)\in \Gamma_{2}:=\{(x,y)\in \partial \Omega: x=1\}\\ +\mathbf{u}(x,y)\cdot \mathbf{n}=0 & \forall(x,y)\in +\partial\Omega \setminus(\Gamma_{1}\bigcup \Gamma_{2}) +\end{array} +\f] + +Initial conditions are: +\f[ +\begin{array}{cr} +S(x,y,t=0)= 1& \forall (x,y) \in \Gamma_{1}\\ +S(x,y,t=0) = 0 & \forall(x,y)\in \partial \Omega \setminus +\Gamma_{1} +\end{array} +\f] + +We apply mixed finite method on velocity and pressure. To be +well-posed, we choose Raviart-Thomas spaces $RT_{k}$ for +$\mathbf{u}$ and discontinuous elements of class $DQ_{k}$ for $p$, +then the mixed +system is: + +Find $(\mathbf{u},p)\in RT_{k}\times DQ_{k}$ such that: +@f{eqnarray*} +\sum_{\kappa}\{ \int _{\kappa}(K \lambda)^{-1} \mathbf{u}\cdot +\mathbf{v} dx - \int_{\kappa} p \nabla \cdot \mathbf{v} dx\} + =- \int_{\Gamma _{1}} \mathbf{v}\cdot \mathbf{n}&& \forall\mathbf{v}\in RT_{k}(\Omega)\\ +\sum_{\kappa}\{\int (\nabla \cdot \mathbf{u}) q dx\} = 0 && \forall +q\in DQ_{k}(\Omega) +@f} +For saturation, we also use discontinuous finite element method. +i.e. Find $S^{n+1} \in DQ_{k}$ such that for all $ \phi \in DQ_{k}$, +the following formulation holds: +@f{eqnarray*} +\sum_{\kappa}\{\int_{\kappa}\frac{S^{n+1}-S^{n}}{\triangle t} \phi +dx + \int_{\kappa} (\mathbf{u}^{n+1}\cdot \nabla F(S^{n})) \phi +dx\} =0 +@f} +Integrating by parts: +@f{eqnarray*} +\nonumber + \sum_{\kappa}\{\int_{\kappa}S^{n+1} \phi dx +\triangle t +\int_{\partial \kappa}F(S^{n})( \mathbf{u}^{n+1}\cdot \mathbf{n} ) +\phi dx &-\triangle t\int_{\kappa} F(S^{n})( \mathbf{u^{n+1}}\cdot +\nabla +\phi )dx\}\\ +&= \sum_{\kappa}\int_{\kappa} S^{n} \phi dx +@f} +where $\mathbf{n}$ denotes the unit outward normal to the +boundary $\partial \kappa$. And here we can use $u^{n+1}$ instead of +$u^{n}$ is because that we view $(u^{n+1},p^{n+1},S^{n+1})$ as +a block vector,$u^{n+1}$ could be implement in the coefficient function for saturation. +We believe the saturation is computed more accurately in this way. + +Considering the discontinuity of the discrete function $S_h$ on +interelement faces, the flux $\mathbf{u}^{n+1}\cdot \mathbf{n} $ is +computed as: + @f{eqnarray*} +&&\int_{\partial \kappa}F(S^{n}) (\mathbf{u}^{n+1}\cdot \mathbf{n}) +\phi dx =\\ +\nonumber && \int_{\partial \kappa _{+}} +F(S^{n,+})(\mathbf{u}^{n+1,+}\cdot \mathbf{n})\phi dx ++\int_{\partial \kappa _{-}} F(S^{n,-})(\mathbf{u}^{n+1,-}\cdot +\mathbf{n})\phi dx +@f} + +where, $\partial \kappa _{-}:= \{x\in +\partial\kappa , \mathbf{u}(x) \cdot \mathbf{n}<0\}$ denotes the inflow boundary +and$\partial \kappa _{+}:= \{\partial \kappa \setminus \partial +\kappa_{-}\}$ is the outflow part of the boundary. By the +discontinuity of$ S_{h}$ , $F(S^{n,-})$ takes the value of +neighboring cell,$F(S^{n+})$ takes the value of cell $\kappa$. + +

Implementation

+ +We use +$dealII$ to implement our mixed and DG system. The main idea is same +with step-20 but there are some new problems we have to consider: + + $(1)$ We has the three blocks vector $(u,p,S)$ , in which +all the functions are dependent on time. i.e. At each time step we +need project the $solution$ into $old-solution$, using +$old-solution$ to get a new $solution$. +Keep doing this until the last time step; + +At time $t=t^{n+1}$ , suppose $old-solution=(u^{n},p^{n},S^{n})$ is +known, in $assemble-system()$ part, we assemble system matrix as: +\f[ +\begin{array}{cccccccccccc} +\lceil &M(S^{n}) &B^{T}& 0 &\rceil & \lceil& \mathbf{u}^{n+1}&\rceil& &\lceil& 0 &\rceil\\ +| &B& 0 & 0 & | &| & p^{n+1} &| &=&| & q &|\\ +\lfloor&\triangle t \nabla F(S^n)& 0& I & \rfloor & \lfloor +&S^{n+1} & \rfloor & & \lfloor& S^{n}& \rfloor +\end{array} +\f] + +In $solve()$ part, we solve the first two equations independent of +the third equation, since $M( S^n)$ is already known. As in step-20, +using vector base functions, Schur complement with a +preconditioner and CG method, we get $u^{n+1}$and $p^{n+1}$. + +Then, with the above $u^{n+1}$ and $p^{n+1}$, we could compute +$S^{n+1}$ by : +@f{eqnarray*} +\sum_{\kappa}\int_{\kappa}S^{n+1} \phi dx&&= +\sum_{\kappa}\{\int_{\kappa} S^{n} \phi dx+\Delta t\int_{\kappa} +F(S^{n}) \mathbf{u^{n+1}}\cdot \nabla \phi dx\\ +\nonumber && -\Delta t \int_{\partial \kappa_{-}}F(S^{n,-}) +\mathbf{u}^{n+1,-}\cdot \mathbf{n} \phi dx -\Delta t \int_{\partial +\kappa_{+}}F(S^{n}) \mathbf{u}^{n+1}\cdot \mathbf{n} \phi dx\} +@f} +Now, project solution $(u^{n+1},p^{n+1},S^{n+1})$ into +$old-solution$, do the above process for next time step. + + +$(2)$ The numerical flux term is related with neighbor cells. In our +implementation $solve( )$, we do the following on each cell: + +For each face, compute the flux $\mathbf{u}\cdot F(S)$, the flux is +negative means it is an in-flow face. Then if this in-flow face is +on the boundary $\Gamma_{1}$:$F(S^{-})=F(1)$; If the in-flow +face is not on boundary, $F(S^{-})=F(S|_{neighbor})$. +Flux is positive means it is an out-flow face, we just use $ F(S)$ on +current cell. + +All the other functions are commented in code, please see next part +- the commented program. + +

Test Case

+ +Our two phase flow are chosen as water and oil. The total mobility +is : @f[\lambda (S) = \frac{1.0}{\mu} S^2 +(1-S)^2@f] Permeability is +: +@f[K(x,y)=\mathbf{k}(x,y)I=\frac{1.0}{2+1.99\sin(2\pi\frac{2x-y}{\epsilon})} +I@f] + Fractional flow of water is: @f[F(S)=\frac{S^2}{S^2+\mu (1-S)^2}@f] +Choose $\epsilon=0.05$ , viscosity $\mu=0.2$. +The resulting solution will be shown in result part. diff --git a/deal.II/examples/step-21/doc/yanli_project_introduction.tex b/deal.II/examples/step-21/doc/yanli_project_introduction.tex deleted file mode 100644 index 64779eb700..0000000000 --- a/deal.II/examples/step-21/doc/yanli_project_introduction.tex +++ /dev/null @@ -1,218 +0,0 @@ -\documentclass[12pt]{article} -\topmargin 0pt \oddsidemargin 0pt \evensidemargin 0pt -\textwidth=14truecm \textheight=21.5truecm -\renewcommand{\baselinestretch}{1.5} -\begin{document} -\author{Yan Li} -\title{A Numerical Simulation for Two Phase Flow} -\maketitle -\section{Introduction} -\subsection{Abstract} - -In this project, we propose a numerical simulation for two phase -flow problem in porous media. The two phase flow system includes one -elliptic equation and one nonlinear transport equation. We apply -mixed finite element method and Discontinuous Galerkin method for -this system. Some numerical results for two dimensional case are -given by $RT_{0}\times DQ_{0}\times DQ_{0}$. -\\ -The numerical computation is based on $dealII$. We use vector shape -functions from step9, DG method from step12, mixed method and Schur -complement from step20 and many many useful tools from the library. - -\subsection{Two Phase Flow Problem } -The modeling of two phase flow in porous media is important for both -environmental rededication and the management of petroleum -reservoirs. Practical situations involving two phase flow include -the dispersal of a nonaqueous phase liquid in an aquifer or the -displacement of a non-aqueous heterogeneity on the flow and -transport. Simulation models, if they are to provide realistic -predictions, must accurately account for these effects. -%However, -%because permeability heterogeneity occurs at many different length -%scales, numerical flow models cannot in general resolve all of the -%scales of variation.Therefore, approaches are needed for -%representing the effects of subgrid scale variations on larger scale -%flow results. Typically, upscaled or multiscale models are employed -%for such systems. \\ -In our project,we consider a kind of periodic permeability,our -numerical result shows that the heterogeneity effects are simulated accurately.\\ -Consider two phase flow in a reservoir $\Omega$ under the assumption -that the displacement is dominated by viscous effects; i.e. we -neglect the effects of gravity, compressibility, and capillary -pressure. Porosity will be considered to be constant. The two phase -will be referred to as water and oil, designated by subscripts $w$ -and $o$, respectively. We write Darcy's for each phase as follows: -\begin{eqnarray} -\mathbf{u}_{j} = \frac{k_{rj}(S)}{\mu_{j}} \mathbf{K} \cdot \nabla p -\end{eqnarray} -\indent where, $\mathbf{u}_{j}$ is the phase velocity, $K$ is the -permeability tensor, $k_{rj}$ is the relative permeability to phase -$j$($j=o,w$),$S$ is the water saturation(volume fraction), $P$ is -pressure and $\mu_{j}$ is the viscosity of phase $j$($j=o,w$).\\ -Combining Darcy's law with a statement of conservation of mass -allows us to express the governing equations in terms of the -so-called pressure and saturation equations: -\begin{eqnarray} -\nabla \cdot (\mathbf{K}(x,y)\lambda(S) \nabla p)= q(x,y) && \forall(x,y)\in\Omega\\ - S_{t} + \mathbf{u} \cdot \nabla F(S) = 0&& \forall(x,y)\in\Omega -\end {eqnarray} - - -\indent where, $\lambda$ is the total mobility, f is the fractional -flow of water, $q$ is a source term and $\mathbf{u}$ is the total -velocity, which are respectively given by: -$$\mathbf{u} = -\mathbf{u}_{o} + \mathbf{u}_{w} = -\lambda(S) \mathbf{K}\cdot\nabla -p$$ -$$\lambda(S) = \frac{k_{rw}(S)}{\mu_{w}}+\frac{k_{ro}(S)}{\mu_{o}}$$ -$$F(S) = \frac{k_{rw}(S)/\mu_{w}}{k_{rw}(S)/\mu_{w} + k_{ro}(S)/\mu_{o}}$$ - - -\subsection{Discretization} - - For simplicity, in our project we will assume no -source $q=0$ and the heterogeneous porous medium is isotropic -$\mathbf{K}(x,y) = -k(x,y) \mathbf{I}$. \\ -Our two dimensional numerical simulation will be done on unit cell -$\Omega = [0,1]\times [0,1]$ for $t\in [0,T]$. -\begin {eqnarray} -\mathbf{u}(x,y)+\mathbf{K}(x,y)\lambda(S) \nabla p= 0 && \forall(x,y)\in\Omega, \forall t\in [0,T]\\ -\nabla \cdot\mathbf{u}(x,y)= 0 && \forall(x,y)\in\Omega, \forall t \in [0,T] \\ -S_{t} + \mathbf{u} \cdot \nabla F(S) = 0&& \forall(x,y)\in\Omega, -\forall t \in [0,T] -\end {eqnarray} - Boundary conditions are: -\[ -\begin {array}{cr} -p(x,y)=1 & \forall(x,y)\in \Gamma_{1}:=\{(x,y)\in \partial \Omega: x=0\}\\ -p(x,y)=0 & \forall(x,y)\in \Gamma_{2}:=\{(x,y)\in \partial \Omega: x=1\}\\ -\mathbf{u}(x,y)\cdot \mathbf{n}=0 & \forall(x,y)\in -\partial\Omega \setminus(\Gamma_{1}\bigcup \Gamma_{2}) -\end {array} -\] -\\ -Initial conditions are: -\[ -\begin {array}{cr} -S(x,y,t=0)= 1& \forall (x,y) \in \Gamma_{1}\\ -S(x,y,t=0) = 0 & \forall(x,y)\in \partial \Omega \setminus -\Gamma_{1} -\end {array} -\] -\\ -We apply mixed finite method on velocity and pressure. To be -well-posed, we choose Raviart-Thomas spaces $RT_{k}$ for -$\mathbf{u}$ and discontinuous elements of class $DQ_{k}$ for $p$, -then the mixed -system is:\\ -Find $(\mathbf{u},p)\in RT_{k}\times DQ_{k}$ such that: -\begin {eqnarray} -\sum_{\kappa}\{ \int _{\kappa}(K \lambda)^{-1} \mathbf{u}\cdot -\mathbf{v} dx - \int_{\kappa} p \nabla \cdot \mathbf{v} dx\} - =- \int_{\Gamma _{1}} \mathbf{v}\cdot \mathbf{n}&& \forall\mathbf{v}\in RT_{k}(\Omega)\\ -\sum_{\kappa}\{\int (\nabla \cdot \mathbf{u}) q dx\} = 0 && \forall -q\in DQ_{k}(\Omega) -\end {eqnarray} -For saturation, we also use discontinuous finite element method. -i.e. Find $S^{n+1} \in DQ_{k}$ such that for all $ \phi \in DQ_{k}$, -the following formulation holds: -\begin {eqnarray} -\sum_{\kappa}\{\int_{\kappa}\frac{S^{n+1}-S^{n}}{\triangle t} \phi -dx + \int_{\kappa} (\mathbf{u}^{n+1}\cdot \nabla F(S^{n})) \phi -dx\} =0 -\end {eqnarray} -Integrating by parts: -\begin {eqnarray} -\nonumber - \sum_{\kappa}\{\int_{\kappa}S^{n+1} \phi dx +\triangle t -\int_{\partial \kappa}F(S^{n})( \mathbf{u}^{n+1}\cdot \mathbf{n} ) -\phi dx &-\triangle t\int_{\kappa} F(S^{n})( \mathbf{u^{n+1}}\cdot -\nabla -\phi )dx\}\\ -&= \sum_{\kappa}\int_{\kappa} S^{n} \phi dx -\end {eqnarray} - -\indent where,$\mathbf{n}$ denotes the unit outward normal to the -boundary $\partial \kappa$. And here we can use $u^{n+1}$ instead of -$u^{n}$ is because that we view $(u^{n+1},p^{n+1},S^{n+1})$ as -a block vector,$u^{n+1}$ could be implement in the coefficient function for saturation. -We believe the saturation is computed more accurately in this way.\\ -Considering the discontinuity of the discrete function $S_h$ on -interelement faces, the flux $\mathbf{u}^{n+1}\cdot \mathbf{n} $ is -computed as: - \begin{eqnarray} -&&\int_{\partial \kappa}F(S^{n}) (\mathbf{u}^{n+1}\cdot \mathbf{n}) -\phi dx =\\ -\nonumber && \int_{\partial \kappa _{+}} -F(S^{n,+})(\mathbf{u}^{n+1,+}\cdot \mathbf{n})\phi dx -+\int_{\partial \kappa _{-}} F(S^{n,-})(\mathbf{u}^{n+1,-}\cdot -\mathbf{n})\phi dx -\end{eqnarray} - -where, $\partial \kappa _{-}:= \{x\in -\partial\kappa , \mathbf{u}(x) \cdot \mathbf{n}<0\}$ denotes the inflow boundary -and$\partial \kappa _{+}:= \{\partial \kappa \setminus \partial -\kappa_{-}\}$ is the outflow part of the boundary. By the -discontinuity of$ S_{h}$ , $F(S^{n,-})$ takes the value of -neighboring cell,$F(S^{n+})$ takes the value of cell $\kappa$. - -\subsection{Implementation} -We use -$dealII$ to implement our mixed and DG system. The main idea is same -with step-20 but there are some new problems we have to consider:\\ -\indent $(1)$ We has the three blocks vector $(u,p,S)$ , in which -all the functions are dependent on time. i.e. At each time step we -need project the $solution$ into $old-solution$, using -$old-solution$ to get a new $solution$. -Keep doing this until the last time step;\\ -At time $t=t^{n+1}$ , suppose $old-solution=(u^{n},p^{n},S^{n})$ is -known, in $assemble-system()$ part, we assemble system matrix as: -\[ -\begin {array}{cccccccccccc} -\lceil &M(S^{n}) &B^{T}& 0 &\rceil & \lceil& \mathbf{u}^{n+1}&\rceil& &\lceil& 0 &\rceil\\ -| &B& 0 & 0 & | &| & p^{n+1} &| &=&| & q &|\\ -\lfloor&\triangle t \nabla F(S^n)& 0& I & \rfloor & \lfloor -&S^{n+1} & \rfloor & & \lfloor& S^{n}& \rfloor -\end {array} -\] -\\ -In $solve()$ part, we solve the first two equations independent of -the third equation, since $M( S^n)$ is already known. As in step-20, -using vector base functions, Schur complement with a -preconditioner and CG method, we get $u^{n+1}$and $p^{n+1}$. \\ -Then, with the above $u^{n+1}$ and $p^{n+1}$, we could compute -$S^{n+1}$ by : -\begin {eqnarray} -\sum_{\kappa}\int_{\kappa}S^{n+1} \phi dx&&= -\sum_{\kappa}\{\int_{\kappa} S^{n} \phi dx+\Delta t\int_{\kappa} -F(S^{n}) \mathbf{u^{n+1}}\cdot \nabla \phi dx\\ -\nonumber && -\Delta t \int_{\partial \kappa_{-}}F(S^{n,-}) -\mathbf{u}^{n+1,-}\cdot \mathbf{n} \phi dx -\Delta t \int_{\partial -\kappa_{+}}F(S^{n}) \mathbf{u}^{n+1}\cdot \mathbf{n} \phi dx\} -\end {eqnarray} -Now, project solution $(u^{n+1},p^{n+1},S^{n+1})$ into -$old-solution$, do the above process for next time step.\\ - \indent -$(2)$ The numerical flux term is related with neighbor cells.In our -implementation $solve( )$, we do the following on each cell: \\ -For each face, compute the flux $\mathbf{u}\cdot F(S)$, the flux is -negative means it is an in-flow face. Then if this in-flow face is -on the boundary $\Gamma_{1}$:$F(S^{-})=F(1)$; If the in-flow -face is not on boundary, $F(S^{-})=F(S|_{neighbor})$. -Flux is positive means it is an out-flow face, we just use $ F(S)$ on current cell.\\ -All the other functions are commented in code, please see next part -- the commented program. - -\subsection{Test Case} -Our two phase flow are chosen as water and oil. The total mobility -is : $$\lambda (S) = \frac{1.0}{\mu} S^2 +(1-S)^2$$ Permeability is -: -$$K(x,y)=\mathbf{k}(x,y)I=\frac{1.0}{2+1.99\sin(2\pi\frac{2x-y}{\epsilon})} -I$$ - Fractional flow of water is: $$F(S)=\frac{S^2}{S^2+\mu (1-S)^2}$$ -Choose $\epsilon=0.05$ , viscosity $\mu=0.2$. \\ -The resulting solution will be shown in result part. - -\end{document} -- 2.39.5