Robin boundary condition

• Elliptic/line_1_SteadyStateDiffusion/line_1e1_robin_left_picard.prj

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

$$ \begin{equation*} h(x) = x + 1. \end{equation*} $$