From: Wolfgang Bangerth Date: Thu, 3 Sep 2009 19:22:44 +0000 (+0000) Subject: Fix two typos found by Till Heinemann. X-Git-Tag: v8.0.0~7191 X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=556cd702ed3d5c08530b13860fee9740f4f0b088;p=dealii.git Fix two typos found by Till Heinemann. git-svn-id: https://svn.dealii.org/trunk@19381 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/examples/step-34/doc/intro.dox b/deal.II/examples/step-34/doc/intro.dox index c350785413..60ce42b991 100644 --- a/deal.II/examples/step-34/doc/intro.dox +++ b/deal.II/examples/step-34/doc/intro.dox @@ -1,6 +1,6 @@
-This program was contributed by Luca Heltai. +This program was contributed by Luca Heltai. @@ -10,7 +10,7 @@ The three dimensional solution generated by this example program is wrong. There is a factor of $1/2$ that appears, and we haven't figured out where it comes from yet. For the moment this issue is "solved" by -multiplying the wind function in the parameter file by two. +multiplying the wind function in the parameter file by two. If you think you have spotted the mistake, please let us know. @@ -24,16 +24,16 @@ usually modeled by the Euler equations of fluid dynamics: \f{align*} \frac{\partial }{\partial t}\mathbf{v} + (\mathbf{v}\cdot\nabla)\mathbf{v} &= - -\frac{1}{\rho}\nabla p + \mathbf{g} + -\frac{1}{\rho}\nabla p + \mathbf{g} \qquad &\text{in } \mathbb{R}^n \backslash \Omega \\ - \nabla \cdot \mathbf{v}&=0 + \nabla \cdot \mathbf{v}&=0 &\text{in } \mathbb{R}^n\backslash\Omega \f} where the fluid density $\rho$ and the acceleration $\mathbf{g}$ due to external forces are given and the velocity $\mathbf{v}$ and the pressure $p$ are the unknowns. Here $\Omega$ is a closed bounded -region representing the body around which the fluid moves. +region representing the body around which the fluid moves. The above equations can be derived from Navier-Stokes equations assuming that the effects due to viscosity are negligible compared to @@ -55,7 +55,7 @@ flow without external forces: -\frac{1}{\rho}\nabla p \qquad &\text{in } \mathbb{R}^n \backslash \Omega \\ - \nabla \cdot \mathbf{v}&=0 + \nabla \cdot \mathbf{v}&=0 &\text{in } \mathbb{R}^n\backslash\Omega \f} @@ -76,14 +76,14 @@ $\mathbf{v}_\infty$ at infinity. An alternative viewpoint is that our coordinate system moves along with the body whereas the background fluid is at rest at infinity. Notice that we define the normal $\mathbf{n}$ as the outer normal to the domain $\Omega$, which -is the opposite of the outer normal to the integration domain. +is the opposite of the outer normal to the integration domain. For both stationary and non stationary flow, the solution process starts by solving for the velocity in the second equation and substituting in the first equation in order to find the pressure. The solution of the stationary Euler equations is typically performed in order to understand the behavior of the given (possibly complex) -geometry when a prescribed motion is enforced on the system. +geometry when a prescribed motion is enforced on the system. The first step in this process is to change the frame of reference from a coordinate system moving along with the body to one in which the body moves @@ -112,7 +112,7 @@ as the homogenous Laplace equation for the unknown $\phi$: \label{laplace} \Delta\phi &= 0 \qquad &&\text{in}\ \mathbb{R}^n\backslash\Omega, \\ - \mathbf{n}\cdot\nabla\phi &= -\mathbf{n}\cdot\mathbf{v}_\infty + \mathbf{n}\cdot\nabla\phi &= -\mathbf{n}\cdot\mathbf{v}_\infty && \text{on}\ \partial\Omega \f} while the momentum equation reduces to Bernoulli's equation that expresses the @@ -123,21 +123,21 @@ pressure $p$ as a function of the potential $\phi$: So we can solve the problem by solving the Laplace equation for the potential. We recall that the following functions, called fundamental -solutions of the Laplace equation, +solutions of the Laplace equation, -\f[ \begin{aligned} +\f[ \begin{aligned} \label{eq:3} G(\mathbf{y}-\mathbf{x}) = & -\frac{1}{2\pi}\ln|\mathbf{y}-\mathbf{x}| \qquad && \text{for } n=2 \\ G(\mathbf{y}-\mathbf{x}) = & \frac{1}{4\pi}\frac{1}{|\mathbf{y}-\mathbf{x}|}&& \text{for } n=3, -\end{aligned} -\f] +\end{aligned} +\f] -satisfy in a distributional sense the equation: +satisfy in a distributional sense the equation: -\f[ --\Delta_y G(\mathbf{y}-\mathbf{x}) = \delta(\mathbf{y}-\mathbf{x}), -\f] +\f[ +-\Delta_y G(\mathbf{y}-\mathbf{x}) = \delta(\mathbf{y}-\mathbf{x}), +\f] where the derivative is done in the variable $\mathbf{y}$. By using the usual Green identities, our problem can be written on the boundary @@ -147,9 +147,9 @@ the second Green identity: \f[\label{green} \int_{\omega} (-\Delta u)v\,dx + \int_{\partial\omega} \frac{\partial u}{\partial \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 \mathbf{n}} \,ds, \f] where $\mathbf{n}$ is the normal to the surface of $\omega$ pointing @@ -173,8 +173,8 @@ $\phi$ and with the fundamental solution of the Laplace equation respectively, as long as $\mathbf{x}$ is chosen in the region $\mathbb{R}^n\backslash\Omega$, we obtain: \f[ - \phi(\mathbf{x}) - - \int_{\Gamma\cup\Gamma_\infty}\frac{\partial G(\mathbf{y}-\mathbf{x})}{\partial \mathbf{n}_y}\phi(\mathbf{y})\,ds_y + \phi(\mathbf{x}) - + \int_{\Gamma\cup\Gamma_\infty}\frac{\partial G(\mathbf{y}-\mathbf{x})}{\partial \mathbf{n}_y}\phi(\mathbf{y})\,ds_y = -\int_{\Gamma\cup\Gamma_\infty}G(\mathbf{y}-\mathbf{x})\frac{\partial \phi}{\partial \mathbf{n}_y}(\mathbf{y})\,ds_y \qquad \forall\mathbf{x}\in \mathbb{R}^n\backslash\Omega @@ -196,8 +196,8 @@ $\phi_\infty$. It is an easy exercise to prove that \f[ -\int_{\Gamma_\infty} \frac{\partial G(\mathbf{y}-\mathbf{x})} -{\partial \mathbf{n}_y}\phi_\infty \,ds_y = -\lim_{r\to\infty} \int_{\partial B_r(0)} \frac{\mathbf{r}}{r} \cdot \nabla G(\mathbf{y}-\mathbf{x}) +{\partial \mathbf{n}_y}\phi_\infty \,ds_y = +\lim_{r\to\infty} \int_{\partial B_r(0)} \frac{\mathbf{r}}{r} \cdot \nabla G(\mathbf{y}-\mathbf{x}) \phi_\infty \,ds_y = -\phi_\infty. \f] @@ -220,7 +220,7 @@ In our case, we know the Neumann values of $\phi$ on the boundary: $\mathbf{n}\cdot\nabla\phi = -\mathbf{n}\cdot\mathbf{v}_\infty$. Consequently, \f[ - \phi(\mathbf{x}) - (D\phi)(\mathbf{x}) = \phi_\infty + + \phi(\mathbf{x}) - (D\phi)(\mathbf{x}) = \phi_\infty + \left(S[\mathbf{n}\cdot\mathbf{v}_\infty]\right)(\mathbf{x}) \qquad \forall\mathbf{x} \in \mathbb{R}^n\backslash\Omega. \f] @@ -246,21 +246,21 @@ $\alpha(\mathbf{x})=\frac 12$, but the value may be smaller or larger at points where the boundary has a corner or an edge. Substituting the single and double layer operators we get: -\f[ - \alpha(\mathbf{x}) \phi(\mathbf{x}) +\f[ + \alpha(\mathbf{x}) \phi(\mathbf{x}) + \frac{1}{2\pi}\int_{\partial \Omega} \frac{ (\mathbf{y}-\mathbf{x})\cdot\mathbf{n}_y }{ |\mathbf{y}-\mathbf{x}|^2 } - \phi(\mathbf{x}) \,ds_y + \phi(\mathbf{y}) \,ds_y = \phi_\infty -\frac{1}{2\pi}\int_{\partial \Omega} \ln|\mathbf{y}-\mathbf{x}| \, \mathbf{n}\cdot\mathbf{v_\infty}\,ds_y -\f] +\f] for two dimensional flows and -\f[ - \alpha(\mathbf{x}) \phi(\mathbf{x}) +\f[ + \alpha(\mathbf{x}) \phi(\mathbf{x}) + \frac{1}{4\pi}\int_{\partial \Omega} \frac{ (\mathbf{y}-\mathbf{x})\cdot\mathbf{n}_y }{ |\mathbf{y}-\mathbf{x}|^3 }\phi(\mathbf{y})\,ds_y = \phi_\infty + \frac{1}{4\pi}\int_{\partial \Omega} \frac{1}{|\mathbf{y}-\mathbf{x}|} \, \mathbf{n}\cdot\mathbf{v_\infty}\,ds_y -\f] +\f] for three dimensional flows, where the normal derivatives of the fundamental solutions have been written in a form that makes computation easier. In either case, $\phi$ is the solution of an integral equation posed entirely on the @@ -289,7 +289,7 @@ $\phi(\mathbf{x})$ for all $\mathbf{x}\in\partial\Omega$. In the next step, we can compute (analytically, if we want) the solution $\phi(\mathbf{x})$ in all of $\mathbb{R}^n\backslash\Omega$. To this end, recall that we had \f[ - \phi(\mathbf{x}) + \phi(\mathbf{x}) = \phi_\infty + (D\phi)(\mathbf{x}) @@ -300,7 +300,7 @@ of $\mathbb{R}^n\backslash\Omega$. To this end, recall that we had where now we have everything that is on the right hand side ($S$ and $D$ are integrals we can evaluate, the normal velocity on the boundary is given, and $\phi$ on the boundary we have just computed). Finally, we can then recover -the velocity as $\mathbf{\tilde v}=\nabla \phi$. +the velocity as $\mathbf{\tilde v}=\nabla \phi$. Notice that the evaluation of the above formula for $\mathbf{x} \in \Omega$ should yield zero as a result, since the integration of the @@ -336,7 +336,7 @@ convenient if we consider the components of the equation individually &= \partial_j [(\partial_j\phi+v_{\infty,j}) \partial_i\phi] - - \partial_j [(\partial_j\phi+v_{\infty,j})] \partial_i\phi + \partial_j [(\partial_j\phi+v_{\infty,j})] \partial_i\phi \\ &= \partial_j [(\partial_j\phi+v_{\infty,j}) \partial_i\phi] @@ -390,7 +390,7 @@ can merge the first and third term into one: @f} We now only need to massage that last term a bit more. Using the product rule, -we get +we get @f{align*} \partial_j [\partial_i (\partial_j\phi) \phi] &= @@ -410,15 +410,15 @@ form. As a consequence, we can now finally state that \partial_j [\partial_i (\partial_j\phi) \phi] \\ &= - \partial_i + \partial_i \left[ (\partial_j\phi)(\partial_j \phi) + v_{\infty,j} \partial_j \phi - - + - \frac 12 (\partial_j\phi)(\partial_j\phi) \right], \\ &= - \partial_i + \partial_i \left[ \frac 12 (\partial_j\phi)(\partial_j \phi) + v_{\infty,j} \partial_j \phi \right], @@ -427,7 +427,7 @@ or in vector form: @f[ \mathbf{v}\cdot\nabla\mathbf{v} = - \nabla + \nabla \left[ \frac 12 \mathbf{\tilde v}^2 + \mathbf{v}_{\infty} \cdot \mathbf{\tilde v} @@ -437,13 +437,13 @@ or in other words: @f[ p = - -\rho + -\rho \left[ \frac 12 \mathbf{\tilde v}^2 + \mathbf{v}_{\infty} \cdot \mathbf{\tilde v} \right] = - -\rho + -\rho \left[ \frac 12 \mathbf{v}^2 - @@ -479,30 +479,30 @@ manifold $\Gamma = \partial \Omega$ into $M$ line segments if $n=2$, or $M$ quadrilaterals if $n=3$. We will call each individual segment or quadrilateral an element or cell, independently of the dimension $n$ of the surrounding space $\mathbb{R}^n$. -We define the finite dimensional space $V_h$ as +We define the finite dimensional space $V_h$ as \f[ \label{eq:definition-Vh} - V_h := \{ v \in C^0(\Gamma) \text{ s.t. } v|_{K_i} \in \mathcal{Q}^1(K_i), + V_h := \{ v \in C^0(\Gamma) \text{ s.t. } v|_{K_i} \in \mathcal{Q}^1(K_i), \forall i\}, \f] with basis functions $\psi_i(\mathbf{x})$ for which we will use the usual FE_Q finite element, with the catch that this time it is defined on a manifold of codimension one (which we do by using the second template argument that is usually defaulted to equal the first; here, we will create objects -FE_Q@ to indicate that we have dim-1 +FE_Q@ to indicate that we have dim-1 dimensional cells in a dim dimensional space). -An element $\phi_h$ of $V_h$ is uniquely +An element $\phi_h$ of $V_h$ is uniquely identified by the vector $\boldsymbol{\phi}$ of its coefficients $\phi_i$, that is: \f[ \label{eq:definition-of-element} - \phi_h(\mathbf{x}) := \phi_i \psi_i(\mathbf{x}), \qquad + \phi_h(\mathbf{x}) := \phi_i \psi_i(\mathbf{x}), \qquad \boldsymbol{\phi} := \{ \phi_i \}, \f] where summation is implied over repeated indexes. Note that we could use discontinuous elements here — in fact, there is no real reason to use continuous ones since the integral formulation does not -imply any derivatives on our trial functions so continuity is unnecessary, +imply any derivatives on our trial functions so continuity is unnecessary, and often in the literature only piecewise constant elements are used.

Collocation boundary element method

@@ -521,11 +521,11 @@ Given the datum $\mathbf{v}_\infty$, find a function $\phi_h$ in $V_h$ such that the following $n\_dofs$ equations are satisfied: \f{align*} - \alpha(\mathbf{x}_i) \phi_h(\mathbf{x}_i) + \alpha(\mathbf{x}_i) \phi_h(\mathbf{x}_i) - \int_{\Gamma_y} \frac{ \partial G(\mathbf{y}-\mathbf{x}_i)}{\partial\mathbf{n}_y } - \phi_h(\mathbf{y}) \,ds_y = - \int_{\Gamma_y} G(\mathbf{y}-\mathbf{x}_i) \, - \mathbf{n}_y\cdot\mathbf{v_\infty} \,ds_y + \phi_h(\mathbf{y}) \,ds_y = + \int_{\Gamma_y} G(\mathbf{y}-\mathbf{x}_i) \, + \mathbf{n}_y\cdot\mathbf{v_\infty} \,ds_y , \f} @@ -544,17 +544,17 @@ where \f[ \begin{aligned} -\mathbf{A}_{ij}&= +\mathbf{A}_{ij}&= \alpha(\mathbf{x}_i) \psi_j(\mathbf{x}_i) -= 1+\int_\Gamma -\frac{\partial G(\mathbf{y}-\mathbf{x}_i)}{\partial \mathbf{n}_y}\,ds_y += 1+\int_\Gamma +\frac{\partial G(\mathbf{y}-\mathbf{x}_i)}{\partial \mathbf{n}_y}\,ds_y \psi_j(\mathbf{x}_i) \\ \mathbf{N}_{ij}&= - \int_\Gamma \frac{\partial G(\mathbf{y}-\mathbf{x}_i)}{\partial \mathbf{n}_y} - \psi_j(\mathbf{y}) \,ds_y + \psi_j(\mathbf{y}) \,ds_y \\ -\mathbf{b}_i&= \int_\Gamma +\mathbf{b}_i&= \int_\Gamma G(\mathbf{y}-\mathbf{x}_i) \, \mathbf{n}_y\cdot\mathbf{v_\infty} ds_y. \end{aligned} @@ -569,14 +569,14 @@ case, $\psi_j(\mathbf{x}_i)=\delta_{ij}$, and as a consequence the matrix $\mathbf{A}$ is diagonal with entries \f[ \mathbf{A}_{ii} - = - 1+\int_\Gamma - \frac{\partial G(\mathbf{y}-\mathbf{x}_i)}{\partial \mathbf{n}_y}\,ds_y + = + 1+\int_\Gamma + \frac{\partial G(\mathbf{y}-\mathbf{x}_i)}{\partial \mathbf{n}_y}\,ds_y = 1-\sum_j N_{ij}, \f] where we have used that $\sum_j \psi_j(\mathbf{y})=1$ for the usual Lagrange -elements. +elements. With this choice of collocation points, the computation of the entries of the matrices $\mathbf{A}$, $\mathbf{N}$ and of the right hand side $\mathbf{b}$ requires the evaluation of singular integrals on the @@ -587,7 +587,7 @@ $\mathcal{T}_h$ can be expressed as a linear (in two dimensions) or bi-linear (in three dimensions) transformation of the reference boundary element $\hat K := [0,1]^{n-1}$, and we perform the integrations after a change of variables from the real element $K_i$ to the reference -element $\hat K$. +element $\hat K$.

Treating the singular integrals.

@@ -596,7 +596,7 @@ $\mathbf{N}_{ii}$ of the system matrix, since, even if the denominator goes to zero when $\mathbf{x}=\mathbf{y}$, the numerator is always zero because $\mathbf{n}_y$ and $(\mathbf{y}-\mathbf{x})$ are orthogonal (on our polygonal approximation of the boundary of $\Omega$), and -the only singular integral arises in the computation +the only singular integral arises in the computation of $\mathbf{b}_i$ on the i-th element of $\mathcal{T}_h$: \f[ \frac{1}{\pi} @@ -625,7 +625,7 @@ determine which elements are important and which are not, leading to a significantly sparser representation of these matrices that also facilitates rapid evaluations of the scalar product between vectors and matrices. This not being the goal of this program, we leave this for more sophisticated -implementations. +implementations.

Implementation

@@ -643,7 +643,7 @@ equivalent way of writing Triangulation@<2,2@>. However, this doesn't have to be so: in the current example, we will for example want to solve on the surface of a sphere, which is a two-dimensional manifold embedded in a three-dimensional space. Consequently, the right class -will be Triangulation@<2,3@>, and correspondingly we will use +will be Triangulation@<2,3@>, and correspondingly we will use DoFHandler@<2,3@> as the DoF handler class and FE_Q@<2,3@> for finite elements.