diff --git a/web/content/docs/benchmarks/elliptic/elliptic-robin.pandoc b/web/content/docs/benchmarks/elliptic/elliptic-robin.pandoc
index 2606b938090df621c91db52a68e3c7789e87025f..aed9b30b3280717b2eb91d1e069a324088d15a94 100644
--- a/web/content/docs/benchmarks/elliptic/elliptic-robin.pandoc
+++ b/web/content/docs/benchmarks/elliptic/elliptic-robin.pandoc
@@ -2,7 +2,7 @@
 date = "2017-02-15T11:46:49+01:00"
 title = "Robin boundary condition"
 project = "Elliptic/line_1_GroundWaterFlow/line_1e1_robin_left_picard.prj"
-author = "Dmitri Naumov"
+author = "Thomas Fischer, Dmitri Naumov"
 weight = 104
 
 [menu]
@@ -17,32 +17,142 @@ weight = 104
 
 We start with simple linear homogeneous elliptic problem:
 $$
-\begin{equation}
+\begin{equation*}
 k\; \Delta h = 0 \quad \text{in }\Omega
-\end{equation}$$
+\end{equation*}$$
 w.r.t boundary conditions
 $$
 \eqalign{
-h(x) = g_D(x) &\quad \text{on }\Gamma_D,\cr
-k{\partial h(x) \over \partial n} = g_N(x) &\quad \text{on }\Gamma_N,
+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, the subscripts $D$ and $N$ denote the Dirichlet- and Neumann-type boundary conditions, $n$ is the normal vector pointing outside of $\Omega$, and $\Gamma = \Gamma_D \cup \Gamma_N$ and $\Gamma_D \cap \Gamma_N = \emptyset$.
+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 $\Gamma_D \cap \Gamma_N \cap \Gamma_R = \emptyset$.
 
-## Problem specification and analytical solution
+## First benchmark: Problem specification
 
-We solve the Laplace equation on a line domain $[0\times 1]^2$ with $k = 1$ w.r.t. the specific boundary conditions:
+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`](https://github.com/ufz/ogs/tree/master/Tests/Data/Elliptic/line_1_GroundWaterFlow/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
+$$
+\begin{equation*}
+\left.\frac{\partial h}{\partial n}\right\rvert_{x=0} = -h'(x)|_{x=0}.
+\end{equation*}
+$$
+From the evaluation of the 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
+$$
+\begin{equation*}
+h(x)|_{x=1} = A + B = -\alpha (h_0 - B) + B = -\alpha h_0 + (1+\alpha) B = g_D
+\end{equation*}
+$$
+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.
+
+{{< img src="../line_1e1_robin_left.png" >}}
+
+## 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:
 $$
 \eqalign{
-h(x,y) = 1 &\quad \text{on } (x=0,y) \subset \Gamma_D,\cr
-h(x,y) = 1 &\quad \text{on } (x,y=0) \subset \Gamma_D,\cr
-k {\partial h(x,y) \over \partial n} = 1 &\quad \text{on } (x=1,y) \subset \Gamma_N,\cr
-k {\partial h(x,y) \over \partial n} = 0 &\quad \text{on } (x,y=1) \subset \Gamma_N.
+h(x) = g_D &\quad \text{for } x=0,\cr
+{\partial h \over \partial n} = \alpha (h_0 - h(x)) &\quad \text{for } x=1,
 }$$
+see
+[`line_1e1_robin_right_picard.prj`](https://github.com/ufz/ogs/tree/master/Tests/Data/Elliptic/line_1_GroundWaterFlow/line_1e1_robin_right_picard.prj).
 
-The solution of this problem is
+One particular solution is
+$$
+\begin{equation*}
+h(x) = A x + B.
+\end{equation*}
+$$
+Due to the Dirichlet boundary condition it follows:
+$$
+\begin{equation*}
+h(0) = g_D = B \quad \Rightarrow \quad h(x) = A x + g_D.
+\end{equation*}
 $$
-\begin{equation}
-h(x,y) = 1 + \sum_{k=1}^\infty A_k \sin\bigg(C_k y\bigg) \sinh\bigg(C_k x\bigg),
-\end{equation}
+From the Robin-type boundary condition we get
 $$
+\begin{equation*}
+h'(x)|_{x=1} = A = \alpha \left(h_0 - h(x)|_{x=1} \right)
+    = \alpha \left.\left(h_0 - (Ax+g_D)\right)\right\rvert_{x=1}
+    = \alpha (h_0 - g_D) - \alpha A.
+\end{equation*}
+$$
+$$
+\begin{equation*}
+\Rightarrow A = \frac{\alpha (h_0 - g_D)}{1+\alpha}
+\end{equation*}
+$$
+$$
+\begin{equation*}
+h(x) = \frac{\alpha (h_0 - g_D)}{1+\alpha} x + g_D.
+\end{equation*}
+$$
+The values from the project file are: $\alpha = -2,$ $h_0 = 1.5$, $g_D = 1$ yielding
+$$
+\begin{equation*}
+h(x) = x + 1.
+\end{equation*}
+$$
+
diff --git a/web/content/docs/benchmarks/elliptic/line_1e1_robin_left.png b/web/content/docs/benchmarks/elliptic/line_1e1_robin_left.png
new file mode 100644
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+++ b/web/content/docs/benchmarks/elliptic/line_1e1_robin_left.png
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