diff --git a/web/content/docs/benchmarks/heatconduction/heatconduction-neumann.pandoc b/web/content/docs/benchmarks/heatconduction/heatconduction-neumann.pandoc
index 8b7dd7aa2f862dd5bc0dad513275800ea7d02f9e..570d50b144cd291b214ef0aad962cd13352db822 100644
--- a/web/content/docs/benchmarks/heatconduction/heatconduction-neumann.pandoc
+++ b/web/content/docs/benchmarks/heatconduction/heatconduction-neumann.pandoc
@@ -25,32 +25,28 @@ $$
 \eqalign{
 T(x, t=0) = T_0,\cr
 T(x) = g_D(x) &\quad \text{on }\Gamma_D,\cr
-k{\partial T(x) \over \partial n} = g_N(x) &\quad \text{on }\Gamma_N,
+\lambda {\partial T(x) \over \partial n} = g_N(x) &\quad \text{on }\Gamma_N,
 }
 $$
 where $T$ could be temperature, 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 $\Gamma_D \cap \Gamma_N = \emptyset$.
 
 ## Problem specification and analytical solution
 
-We solve the Parabolic equation on a line domain $[60\times 1]$ with $k = 3.2$ and $\rho C_\textrm{p} = 2.5e10^6$ w.r.t. the specific boundary conditions:
+We solve the parabolic equation on a line domain $[60\times 1]$ with $\lambda  = 3.2$ and $\rho C_\textrm{p} = 2.5 \times 10^6$ w.r.t. the specific boundary conditions:
 
 $$
 \eqalign{
-k {\partial T(x) \over \partial n} = 2 &\quad \text{on } (x=0) \subset \Gamma_N.\cr
-k {\partial T(x) \over \partial n} = 0 &\quad \text{on } (x=60) \subset \Gamma_N.
+\lambda  {\partial T(x) \over \partial n} = 2 &\quad \text{on } (x=0) \subset \Gamma_N.\cr
+\lambda  {\partial T(x) \over \partial n} = 0 &\quad \text{on } (x=60) \subset \Gamma_N.
 }
 $$
 
 The solution of this problem is
-
-TODO: is not rendered correct
-
 $$
 \begin{equation}
-T(x,t) = \frac{2q}{\lambda}\left((\frac{\alpha t}{\pi})^{\frac{1}{2}} e^{-x^2/4\alpha t} - \frac{x}{2}\textrm{erfc}( \frac{x}{2\sqrt{\alpha  t}})\right),
+T(x,t) = \frac{2q}{\lambda}\left(\left(\frac{\alpha t}{\pi}\right)^{\frac{1}{2}} e^{-x^2/4\alpha t} - \frac{x}{2}\textrm{erfc}\left( \frac{x}{2\sqrt{\alpha  t}}\right)\right),
 \end{equation}
 $$
-
 where $T_\textrm{b}$ is the boundary temperature, $\textrm{erfc}$ is the complementary error function and $\alpha = \lambda/(C_p \rho)$ is the thermal diffusivity.
 
 ## Input files