Difference between revisions of "Tutorial/Coupled problems"

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== The considered problem ==
 
== The considered problem ==
  
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Here, the two academics problem of the Laplace equation with [[Academic example: Laplace equation with Neumann boundary condition | Neumann boundary condition]] and with [[Academic example: Laplace equation with Neumann boundary condition | Dirichlet boundary condition]] are coupled, in te sense that the solution of the first problem is data for the second problem. The computation domain is still the unit square $\Omega = [0,1]\times[0,1]$ with boundary $\Gamma$ and unit outwardly directed normal $\mathbf{n}$. The problems read as follows : firstly, find $u$ such that
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Here is explained how to use the solution of a first problem as a data for a second problem. Two kind of problem are studied here, depending if the solution of the first problem directly appears in the weak formulation (for example, as a source or a Neumann boundary), or if it is used a constraint (Dirichlet boundary condition). Let us begin by the first example with the very simple following example. The computation domain is the unit square $\Omega = [0,1]\times[0,1]$ with boundary $\Gamma$ and unit outwardly directed normal $\mathbf{n}$.  
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The first coupled problem read as follows :
 
\begin{equation}
 
\begin{equation}
 
\begin{cases}\label{eq:problemU}
 
\begin{cases}\label{eq:problemU}
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-\Delta u + u = f & \text{in } \Omega,\\
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u = C & \text{in } \Omega,\\
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\displaystyle{\frac{\partial u}{\partial \mathbf{n}} = 0} & \text{on }\Gamma,
 
 
\end{cases}
 
\end{cases}
 
\end{equation}
 
\end{equation}
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where $\displaystyle{\Delta = \frac{\partial^2}{\partial x_1^2} + \frac{\partial^2}{\partial x_2^2} }$ is the Laplace operator and $f$ is given by
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where $C$ is a constant. Then, find the solution $v$ of the second problem
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$$
 
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\forall (x,y)\in [0,1]^2,\qquad f(x,y) = (1+2\pi^2)\cos(\pi x)\cos(\pi y).
 
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$$
 
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Then, find the solution $v$ of the second problem
 
 
\begin{equation}
 
\begin{equation}
 
\begin{cases}\label{eq:problemV}
 
\begin{cases}\label{eq:problemV}
 
-\Delta v + v = 0 & \text{in } \Omega,\\
 
-\Delta v + v = 0 & \text{in } \Omega,\\
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\displaystyle{v = u} & \text{on }\Gamma,
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\displaystyle{\frac{\partial v}{\partial \mathbf{n}} = u} & \text{on }\Gamma,
 
\end{cases}
 
\end{cases}
 
\end{equation}
 
\end{equation}
 
where $u$ is the solution of problem (\ref{eq:problemU}).
 
where $u$ is the solution of problem (\ref{eq:problemU}).

Revision as of 20:21, 26 November 2011

The considered problem

Here is explained how to use the solution of a first problem as a data for a second problem. Two kind of problem are studied here, depending if the solution of the first problem directly appears in the weak formulation (for example, as a source or a Neumann boundary), or if it is used a constraint (Dirichlet boundary condition). Let us begin by the first example with the very simple following example. The computation domain is the unit square $\Omega = [0,1]\times[0,1]$ with boundary $\Gamma$ and unit outwardly directed normal $\mathbf{n}$.

The first coupled problem read as follows : \begin{equation} \begin{cases}\label{eq:problemU} u = C & \text{in } \Omega,\\ \end{cases} \end{equation} where $C$ is a constant. Then, find the solution $v$ of the second problem \begin{equation} \begin{cases}\label{eq:problemV} -\Delta v + v = 0 & \text{in } \Omega,\\ \displaystyle{\frac{\partial v}{\partial \mathbf{n}} = u} & \text{on }\Gamma, \end{cases} \end{equation} where $u$ is the solution of problem (\ref{eq:problemU}).

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