From: wolf Date: Thu, 11 Sep 2003 19:52:17 +0000 (+0000) Subject: Fix documentation for corrected state equation. X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=cba87f72618dd2ddb5a438c8baa67b76c17f0a8e;p=dealii-svn.git Fix documentation for corrected state equation. git-svn-id: https://svn.dealii.org/trunk@7976 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.html b/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.html index e7dab3e8d5..6074cbf92d 100644 --- a/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.html +++ b/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.html @@ -1,7 +1,6 @@

Introduction

-

In real life, most partial differential equations are really systems of equations. Accordingly, the solutions are usually @@ -17,376 +16,480 @@ describes the displacement in each space direction of a rigid body which is subject to a force. Of course, the force is also vector-valued, meaning that in each point it has a direction and an absolute value. The elastic equations are the following: -

-
- - - +\end{displaymath} + --> +

+
\begin{displaymath}-
+ WIDTH= -
-
-

-where the values cijkl are the stiffness coefficients and +$"> +

+where the values $ c_{ijkl}$ are the stiffness coefficients and will usually depend on the space coordinates. In many cases, one knows that the material under consideration is isotropic, in which case by introduction of the two coefficients $\lambda$ -and $\mu$ -the coefficient tensor reduces to -

-
- - - +\end{displaymath} + --> +

+
\begin{displaymath}c_{ijkl}
+ WIDTH= -
-
-

+$"> +

+

The elastic equations can then be rewritten in much simpler a form: -

-
- - - + \vec f, +\end{displaymath} + --> +

+
\begin{displaymath}-
-\nabla (\lambda+\mu) (\div \vec u)
+ WIDTH= -
-
-

+\vec f, +$"> +

and the respective bilinear form is then -

-
- - - + + + \sum_{i,j} + \left( + \mu \partial_i u_j, \partial_j v_i + \right)_\Omega, +\end{displaymath} + --> +

+
\begin{displaymath}a(\vec u, \vec v) =
+ WIDTH= -
-
-

-or as a sum over components: -

-
- - - +\end{displaymath} + --> +

+
\begin{displaymath}a(\vec u, \vec v) =
-\sum_{k,l}
+ WIDTH= -
-
-

+$"> +

+

How do we now assemble the matrix for such an equation? The first thing we need is some knowledge about how the shape functions work in the case of vector-valued finite elements. Basically, this comes down to the following: -let n be the number of shape functions for the scalar finite element of +let $ n$ be the number of shape functions for the scalar finite element of which we build the vector element (for example, we will use bilinear functions for each component of the vector-valued finite element, so the scalar finite element is the FEQ1 element which we have used in previous examples -already, and n=4 in two space dimensions). Further, let N be the number of -shape functions for the vector element; in two space dimensions, we need nshape functions for each component of the vector, so N=2n. Then, the ith +already, and $ n=4$ in two space dimensions). Further, let $ N$ be the number of +shape functions for the vector element; in two space dimensions, we need $ n$ +shape functions for each component of the vector, so $ N=2n$. Then, the $ i$th shape function of the vector element has the form -

-
- - - -\begin{displaymath}\Phi_i(\vec x) = \varphi_{base(i)}(\vec x)\ \vec e_{comp(i)},
-\end{displaymath} -
-
+\end{displaymath} + -->

-where el is the lth unit vector, comp(i) is the function that tells +
+$\displaystyle \Phi_i(\vec x) = \varphi_{base(i)}(\vec x) \vec e_{comp(i)},
+$ +

+where $ e_l$ is the $ l$th unit vector, $ comp(i)$ is the function that tells us which component of $\Phi_i$ -is the one that is nonzero (for + WIDTH="21" HEIGHT="29" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img18.png" + ALT="$ \Phi_i$"> is the one that is nonzero (for each vector shape function, only one component is nonzero, and all others are -zero). - +zero). +$ \varphi_{base(i)}(x)$ describes the space dependence of the shape +function, which is taken to be the $ base(i)$-th shape function of the scalar +element. Of course, while $ i$ is in the range $\varphi_{base(i)}(x)$ -describes the space dependence of the shape -function, which is taken to be the base(i)-th shape function of the scalar -element. Of course, while i is in the range - + WIDTH="89" HEIGHT="29" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img21.png" + ALT="$ 0,\ldots,N-1$">, the functions $0,\ldots,N-1$, -the functions -comp(i) and base(i) have the ranges 0,1 (in 2D) and - + WIDTH="58" HEIGHT="31" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img17.png" + ALT="$ comp(i)$"> and $ base(i)$ have the ranges $ 0,1$ (in 2D) and $0,\ldots,n-1$, + WIDTH="84" HEIGHT="28" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img23.png" + ALT="$ 0,\ldots,n-1$">, respectively.

For example (but this layout is not mandatory, and you should not rely on it), the following layout could be used by the library:

-
+

+
\begin{multline*}\Phi_0(\vec x) =
+ WIDTH= -
-where here -

+\end{multline*}">
+
+

- - - -\begin{displaymath}comp(0)=0, \quad comp(1)=1, \quad comp(2)=0, \quad comp(3)=1, \quad \ldots
-\end{displaymath}
-
+where here +

-

- - - \begin{displaymath}base(0)=0, \quad base(1)=0, \quad base(2)=1, \quad base(3)=1, \quad \ldots
-\end{displaymath} -
-
+ WIDTH="459" HEIGHT="31" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img25.png" + ALT="$\displaystyle comp(0)=0, \quad comp(1)=1, \quad comp(2)=0, \quad comp(3)=1, \quad \ldots +$"> +

+

+
+$\displaystyle base(0)=0, \quad base(1)=0, \quad base(2)=1, \quad base(3)=1, \quad \ldots
+$ +

+

In all but very rare cases, you will not need to know which shape function - - + $\varphi_{base(i)}$ -of the scalar element belongs to a shape function $\Phi_i$of the vector element. Let us therefore define -

-
- - - -\begin{displaymath}\phi_i = \varphi_{base(i)}
-\end{displaymath} -
-
+\end{displaymath} + -->

-by which we can write the vector shape function as -

- - - \begin{displaymath}\Phi_i(\vec x) = \phi_{i}(\vec x)\ \vec e_{comp(i)}.
-\end{displaymath} -
-
+ WIDTH="90" HEIGHT="29" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img28.png" + ALT="$\displaystyle \phi_i = \varphi_{base(i)} +$"> +

+by which we can write the vector shape function as +

-You can now safely forget about the function base(i), at least for the rest +
+$\displaystyle \Phi_i(\vec x) = \phi_{i}(\vec x) \vec e_{comp(i)}.
+$ +

+You can now safely forget about the function $ base(i)$, at least for the rest of this example program.

Now using this vector shape functions, we can write the discrete finite element solution as -

-
- - - -\begin{displaymath}\vec u_h(\vec x) =
-\sum_i \Phi_i(\vec x)\ u_i
-\end{displaymath} -
-
+ \sum_i \Phi_i(\vec x)\ u_i +\end{displaymath} + -->

-with scalar coefficients ui. If we define an analog function $\vec v_h$ -as -test function, we can write the discrete problem as follows: Find coefficients -ui such that -

- - - +\end{displaymath} + --> +

+
\begin{displaymath}a(\vec u_h, \vec v_h) = (\vec f, \vec v_h)
+ WIDTH= -
-
-

+$"> +

+

If we insert the definition of the bilinear form and the representation of $\vec u_h$ -and $\vec v_h$ -into this formula: + SRC="intro/img34.png" + ALT="$ \vec u_h$"> and $ \vec v_h$ into this formula:

-
+

+
\begin{multline*}\sum_{i,j}
+ WIDTH= -
-We note that here and in the following, the indices k,l run over spatial -directions, i.e. - +\end{multline*}"> +
+

+
+
+We note that here and in the following, the indices $ k,l$ run over spatial +directions, i.e. $0\le k,l \le d-1$, -and that indices i,j run over degrees + WIDTH="84" HEIGHT="29" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img37.png" + ALT="$ 0\le k,l < d$">, and that indices $ i,j$ run over degrees of freedoms.

-The local stiffness matrix on cell K therefore has the following entries: -

-
- - - +\end{displaymath} + --> +

+
\begin{displaymath}A^K_{ij}
+ WIDTH= -
-
-

-where i,j now are local degrees of freedom and therefore - +$"> +

+where $ i,j$ now are local degrees of freedom and therefore $0\le i,j \le N-1$. + WIDTH="89" HEIGHT="29" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img41.png" + ALT="$ 0\le i,j < N$">. In these formulas, we always take some component of the vector shape functions $\Phi_i$, -which are of course given as follows (see their definition): -

-
- - - -\begin{displaymath}(\Phi_i)_l = \phi_i \delta_{l,comp(i)},
-\end{displaymath} -
-
+\end{displaymath} + -->

-with the Kronecker symbol - -$\delta_{nm}$. -Due to this, we can delete some of -the sums over k and l:
-
\begin{align*}A^K_{ij}
-&=
-\sum_{k,l}
-\Bigl\{
-\left(
-(\lambda+\mu) \partial_...
-...abla \phi_i,
-\nabla \phi_j
-\right)_K
-\ \delta_{comp(i),comp(j)}.
-\end{align*} -
+ WIDTH="139" HEIGHT="31" ALIGN="MIDDLE" BORDER="0" + SRC="intro/img42.png" + ALT="$\displaystyle (\Phi_i)_l = \phi_i \delta_{l,comp(i)}, +$"> +

+with the Kronecker symbol +$ \delta_{nm}$. Due to this, we can delete some of +the sums over $ k$ and $ l$: +
+

+
+ + + + + + + + + + + + + + + + +
$\displaystyle A^K_{ij}$$\displaystyle = \sum_{k,l} \Bigl\{ \left( \lambda \partial_l \phi_i \delta_{l,comp(i)}, \partial_k \phi_j \delta_{k,comp(j)} \right)_K$ +   
 $\displaystyle \qquad\qquad + \left( \mu \partial_l \phi_i \delta_{k,comp(i)}, ...
+..._i \delta_{k,comp(i)}, \partial_k \phi_j \delta_{l,comp(j)} \right)_K \Bigr\}$ +   
 $\displaystyle = \left( \lambda \partial_{comp(i)} \phi_i, \partial_{comp(j)} \p...
+...j)} + \left( \mu \partial_{comp(j)} \phi_i, \partial_{comp(i)} \phi_j \right)_K$ +   
 $\displaystyle = \left( \lambda \partial_{comp(i)} \phi_i, \partial_{comp(j)} \p...
+...)} + \left( \mu \partial_{comp(j)} \phi_i, \partial_{comp(i)} \phi_j \right)_K.$ +   
+

+
+
+

-Likewise, the contribution of cell K to the right hand side vector is +Likewise, the contribution of cell $ K$ to the right hand side vector is

-
-\begin{align*}f^K_j
-&=
-\sum_l
-\left(
-f_l,
-(\Phi_j)_l
-\right)_K
-\\
-&=
-\s...
-...mp(j)}
-\right)_K
-\\
-&=
-\left(
-f_{comp(j)},
-\phi_j
-\right)_K.
-\end{align*} -
+

+
+ + + + + + + + + + + + +
$\displaystyle f^K_j$$\displaystyle = \sum_l \left( f_l, (\Phi_j)_l \right)_K$ +   
 $\displaystyle = \sum_l \left( f_l, \phi_j \delta_{l,comp(j)} \right)_K$ +   
 $\displaystyle = \left( f_{comp(j)}, \phi_j \right)_K.$ +   
+

+
+
+

This is the form in which we will implement the local stiffness matrix and right hand side vectors. diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.tex b/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.tex index 3a316e1d9e..e744bfc739 100644 --- a/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.tex +++ b/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro.tex @@ -291,12 +291,12 @@ the sums over $k$ and $l$: \mu \nabla \phi_i, \nabla \phi_j \right)_K + \ \delta_{comp(i),comp(j)} + \left( \mu \partial_{comp(j)} \phi_i, \partial_{comp(i)} \phi_j - \right)_K - \ \delta_{comp(i),comp(j)}. + \right)_K. \end{align*} \end{center} diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro/img1.gif b/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro/img1.gif deleted file mode 100644 index e9ba12a0f2..0000000000 Binary files a/deal.II/doc/tutorial/chapter-2.step-by-step/step-8.data/intro/img1.gif and 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