 ## Equations

We start with simple linear homogeneous elliptic problem: $$\begin{equation*} k\;\Delta h = 0 \quad \text{in }\Omega \end{equation*}$$ w.r.t boundary conditions \eqalign{ h = g_D &\quad \text{on }\Gamma_D,\cr k{\partial h \over \partial n} = g_N &\quad \text{on }\Gamma_N,\cr {\partial h \over \partial n} = \alpha (h_0 - h(x)) &\quad \text{on }\Gamma_R, } where $h$ could be hydraulic head, pressure, or temperature and $k$ is the diffusion tensor (hydraulic conductivity, permeability divided by dynamic viscosity, or heat conductivity). The subscripts $D,$ $N,$ and $R$ denote the Dirichlet-, Neumann-, and Robin-type boundary conditions, $n$ is the normal vector pointing outside of $\Omega$, and $\Gamma = \Gamma_D \cup \Gamma_N \cup \Gamma_R$ and $\Gamma_D \cap \Gamma_N \cap \Gamma_R = \emptyset$.

## First benchmark: Problem specification

We solve the Laplace equation on a line domain $[0, 1]$ with $k = 1$ w.r.t. the specific boundary conditions: \eqalign{ {\partial h \over \partial n} = \alpha (h_0 - h(x)) &\quad \text{for } x=0,\cr h(x) = g_D &\quad \text{for } x=1, } see line_1e1_robin_left_picard.prj.

### Analytical solution

One particular solution is $$\begin{equation*} h(x) = A x + B. \end{equation*}$$

The normal direction is facing out of the bulk domain. The Robin-type boundary condition in this example is set on the left side of the line domain. Consequently, in this case the directional derivative is the negative derivative $$\begin{equation*} \left.\frac{\partial h}{\partial n}\right\rvert_{x=0} = -h’(x)\rvert_{x=0}. \end{equation*}$$ From the evaluation of the Robin-type boundary condition it follows $$\begin{equation*} \left.\frac{\partial h}{\partial n}\right\rvert_{x=0} = -A = \alpha (h_0 - h(0)) = \alpha (h_0 - B). \end{equation*}$$ Using the expression for $A$ in the Dirichlet-type boundary condition $$\begin{equation*} h(x)\rvert_{x=1} = A + B = -\alpha (h_0 - B) + B = -\alpha h_0 + (1+\alpha) B = g_D \end{equation*}$$ yields for $\alpha \not= -1$: \begin{align*} B &= \frac{g_D + \alpha h_0}{1 + \alpha} \quad \textrm{and}\ A &= -\alpha \left( h_0 - \frac{g_D + \alpha h_0}{1 + \alpha} \right) = -\alpha \left( \frac{h_0 + \alpha h_0 - g_D - \alpha h_0}{1 + \alpha} \right) = -\alpha \left( \frac{h_0 - g_D}{1 + \alpha} \right). \end{align*} The particular solution is $$\begin{equation*} h(x) = \frac{\alpha (g_D - h_0)}{1 + \alpha} x + \frac{g_D + \alpha h_0}{1 + \alpha}. \end{equation*}$$ Using the values from the project file $\alpha = -2,$ $h_0 = 1.5$, $g_D = 2$ results in $$\begin{equation*} h(x) = \frac{-2 (2 - 1.5)}{1 + (-2)} x + \frac{2+(-2) \times 1.5}{1 + (-2)} = x + 1. \end{equation*}$$

### Results and evaluation

The left figure shows the pressure along the line, in the right figure the difference between the analytical solution and the numerical calculated solution is plotted.

## Second benchmark: Problem specification and analytical solution

We solve the Laplace equation on a line domain $[0, 1]$ with $k = 1$ w.r.t. the specific boundary conditions: \eqalign{ h(x) = g_D &\quad \text{for } x=0,\cr {\partial h \over \partial n} = \alpha (h_0 - h(x)) &\quad \text{for } x=1, } see line_1e1_robin_right_picard.prj.

One particular solution is $$\begin{equation*} h(x) = A x + B. \end{equation*}$$ Due to the Dirichlet boundary condition it follows: $$\begin{equation*} h(0) = g_D = B \quad \Rightarrow \quad h(x) = A x + g_D. \end{equation*}$$ From the Robin-type boundary condition we get $$\begin{equation*} h’(x)\rvert_{x=1} = A = \alpha \left(h_0 - h(x)\rvert_{x=1} \right) = \alpha \left.\left(h_0 - (Ax+g_D)\right)\right\rvert_{x=1} = \alpha (h_0 - g_D) - \alpha A. \end{equation*}$$ $$\begin{equation*} \Rightarrow A = \frac{\alpha (h_0 - g_D)}{1+\alpha} \end{equation*}$$ $$\begin{equation*} h(x) = \frac{\alpha (h_0 - g_D)}{1+\alpha} x + g_D. \end{equation*}$$ The values from the project file are: $\alpha = -2,$ $h_0 = 1.5$, $g_D = 1$ yielding $$\begin{equation*} h(x) = x + 1. \end{equation*}$$

This article was written by Thomas Fischer, Dmitri Naumov. If you are missing something or you find an error please let us know.
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