From 601f25f67bbd4f1e02aad0bfa05374f6481a1e24 Mon Sep 17 00:00:00 2001 From: Ludwig Sebastian Glane Date: Thu, 28 Jan 2016 11:46:42 +0000 Subject: [PATCH] modified step-34 tutorial introduction --- examples/step-34/doc/intro.dox | 26 ++++++++++++++++++++------ 1 file changed, 20 insertions(+), 6 deletions(-) diff --git a/examples/step-34/doc/intro.dox b/examples/step-34/doc/intro.dox index 26075a0aae..d296c1cbff 100644 --- a/examples/step-34/doc/intro.dox +++ b/examples/step-34/doc/intro.dox @@ -138,13 +138,13 @@ the second Green %identity: \f[\label{green} \int_{\omega} - (-\Delta u)v\,dx + \int_{\partial\omega} \frac{\partial u}{\partial \mathbf{n} }v \,ds + (-\Delta u)v\,dx + \int_{\partial\omega} \frac{\partial u}{\partial \tilde{\mathbf{n}} }v \,ds = \int_{\omega} - (-\Delta v)u\,dx + \int_{\partial\omega} u\frac{\partial v}{\partial \mathbf{n}} \,ds, + (-\Delta v)u\,dx + \int_{\partial\omega} u\frac{\partial v}{\partial \tilde{\mathbf{n}}} \,ds, \f] -where $\mathbf{n}$ is the normal to the surface of $\omega$ pointing +where $\tilde{\mathbf{n}}$ is the normal to the surface of $\omega$ pointing outwards from the domain of integration $\omega$. In our case the domain of integration is the domain @@ -158,7 +158,8 @@ $\mathbb{R}^n\backslash\Omega$, whose boundary is $ \Gamma_\infty \cup In our program the normals are defined as outer to the domain $\Omega$, that is, they are in fact inner to the integration domain, and some care is required in defining the various integrals -with the correct signs for the normals. +with the correct signs for the normals, i.e. replacing $\tilde{\mathbf{n}}$ +by $-\mathbf{n}$. If we substitute $u$ and $v$ in the Green %identity with the solution $\phi$ and with the fundamental solution of the Laplace equation @@ -271,8 +272,21 @@ $\Omega$ can be defined using the double layer potential itself: The reason why this is possible can be understood if we consider the fact that the solution of a pure Neumann problem is known up to an arbitrary constant $c$, which means that, if we set the Neumann data -to be zero, then any constant $\phi = \phi_\infty$ will be a solution, -giving us an the explicit expression above for $\alpha(\mathbf{x})$. +to be zero, then any constant $\phi = \phi_\infty$ will be a solution. +Inserting constant solution and the Neumann boundary condition in the +boundary integral equation, we have +@f{align*} +\alpha\left(\mathbf{x}\right)\phi\left(\mathbf{x}\right) +&=\int_{\Omega}\phi\left(\mathbf{y}\right)\delta\left(\mathbf{y}-\mathbf{x}\right)\, dy\\ +\Rightarrow +\alpha\left(\mathbf{x}\right)\phi_\infty +&=\phi_\infty\int_{\Gamma\cup\Gamma_\infty}\frac{ \partial G(\mathbf{y}-\mathbf{x}) }{\partial \mathbf{n}_y} \, ds_y +=\phi_\infty\left[\int_{\Gamma_\infty}\frac{ \partial G(\mathbf{y}-\mathbf{x}) }{\partial \mathbf{n}_y} \, ds_y ++\int_{\Gamma}\frac{ \partial G(\mathbf{y}-\mathbf{x}) }{\partial \mathbf{n}_y} \, ds_y +\right] +@f} +The integral on $\Gamma_\infty$ is unity, see above, division by the constant $\phi_\infty$ gives us the explicit +expression above for $\alpha(\mathbf{x})$. While this example program is really only focused on the solution of the boundary integral equation, in a realistic setup one would still need to solve -- 2.39.5