FEM

FEM.tex

@@ -120,12 +120,18 @@ \subsubsection{Example}
 
 \numericsubsection{Elliptic}
 
-We have a set of given local basis functions $v_1(x), \ldots, v_n(x)$ that are continuous and piecewise differentiable.
-We subdivide $\Omega$ into meshes and assign each nodal point a nodal variable $a_k$.
+\makebox[\columnwidth]{\includegraphics[width=0.5\columnwidth]{images/FEM_Geometry}}
+
+Introduce on problem domain nodal points that divide the geometry into cells or meshes.
+Associate to each nodal variable $a_k$ of a nodal point a local function $v_k$ from 
+a set of given local basis functions $v_1(x), \ldots, v_n(x)$ that are continuous and piecewise differentiable.
+Thus, the ansatz $\utild(x) = a_0v_0(x)+a_1v_1(x)+\ldots+a_nv_n(x)$ has shape functions $v_k$ that are one on
+the respective nodal point $k$.
 Then, we use shape functions to represent the PDE on the mesh,
 e.g. using triangular functions $l_1(x)=1-x$ and $l_2(x) = x$ to obtain \emph{local} element matrices
+\begin{snugshade*}
 \begin{align*}
-	E = 
+	E_\text{step} = 
 	\begin{bmatrix}
 		\int_0^1 l_1^\prime(s)\cdot l_1^\prime(s)\ ds & \int_0^1 l_1^\prime(s)\cdot l_2^\prime(s)\ ds \\
 		\int_0^1 l_2^\prime(s)\cdot l_1^\prime(s)\ ds & \int_0^1 l_2^\prime(s)\cdot l_2^\prime(s)\ ds
@@ -135,7 +141,9 @@ \subsubsection{Example}
 		1 & -1 \\
 		-1 & 1
 	\end{bmatrix}
-\end{align*}
+\end{align*}    
+\end{snugshade*}
+
 
 that can then be used to construct the mesh matrix $M=1/h\cdot E$ using mesh size $h$.
 Afterwards, the global Ritz matrix can be computed, e.g. for a one-dimensional mesh with 4 nodal points and 2 unknown inner points,
@@ -162,4 +170,253 @@ \subsubsection{Example}
 
 Finally, we have obtained the Ritz system: $R^4\cdot\vec{a}=\vec{r^4}$.
 That yields the approximation function (as defined by the ansatz): $\tilde{u}(x)=\sum_{i=0}^3 a_i\cdot v_i(x)$
-3with $a_0$ and $a_3$ fulfilling the boundary conditions.
+with $a_0$ and $a_3$ fulfilling the boundary conditions.
+
+This shape function approach can also be applied to the Ritz vector.
+We get $\tilde{f}(x) = f(a_0)\cdot v_0(x)+\ldots+f(a_n)v_n(x)$ and we approximate 
+$r_{k}^{n}=\int_{0}^{1}f(x)\cdot v_{k}(x)\;dx$ by $\tilde{r}_{k}^{n}=\int_{0}^{1}\tilde{f}(x)\cdot v_{k}(x)\;dx$
+which essentially is $\tilde{r}^n = S^n \cdot \vec{f}^n$ where $\vec{f}^n$ is the vector of nodal values
+and $S_{k,j}^n=\int_0^1 v_j(x)\cdot v_k(x)\ dx$
+
+\subsubsection{Example With Inhomogeneous B.C.}
+
+\textbf{Given} A real function $u(x)$ on interval $\Omega=[0,1]$ that satisfies the differential equation $u''(x) + 8 = 0$
+and the boundary conditions $u(0) = 1$ and $u(1) = 2$. We want to find an approximation function $\utild(x)$ using the meshes
+$[0, 0.5], [0.5, 0.75], [0.75, 1]$ and linear shape functions.
+\\[1em]
+\textbf{Discretisation of Geometry}
+\makebox[\columnwidth]{\includegraphics[width=0.8\columnwidth]{images/FEM_Example}}
+
+\textbf{Solution}
+
+We get the following mesh matrices:
+
+Mesh 1 (step size = 1/2):
+\begin{align*}
+    {\color{red}
+    \frac{1}{h}\begin{bmatrix}
+        1 & -1 \\
+        -1 & 1
+    \end{bmatrix}
+    = \sfrac{1}{2}^{-1} \begin{bmatrix}
+        1 & -1 \\
+        -1 & 1
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        2 & -2 \\
+        -2 & 2
+    \end{bmatrix}
+    }
+\end{align*}
+
+Mesh 2 (step size = 1/4):
+\begin{align*}
+    {\color{blue}
+    \frac{1}{h}\begin{bmatrix}
+        1 & -1 \\
+        -1 & 1
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        4 & -4 \\
+        -4 & 4
+    \end{bmatrix}
+    }
+\end{align*}
+
+Mesh 3 (step size = 1/4):
+\begin{align*}
+    {\color{green}
+    4\begin{bmatrix}
+        1 & -1 \\
+        -1 & 1
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        4 & -4 \\
+        -4 & 4
+    \end{bmatrix}
+    }
+\end{align*}
+
+resulting in the global Ritz-Matrix and Ritz-Vector:
+
+\begin{align*}
+    & \overbrace{
+        \begin{bmatrix}
+            {\color{red}2} & {\color{red}-2} & 0 & 0 \\
+            {\color{red}-2} & {\color{red}2} + {\color{blue}4} = 6 & {\color{blue}-4} & 0 \\
+            0 & {\color{blue}-4} & {\color{blue}4} + {\color{green}4} = 8 & {\color{green}-4} \\
+            0 & 0 & {\color{green}-4} & {\color{green}4}
+        \end{bmatrix}
+    }^R
+    \overbrace{
+        \begin{bmatrix}
+            a_0 \\
+            a_1 \\
+            a_2 \\
+            a_3
+        \end{bmatrix}
+    }^{\vec{a}}
+    =
+    \overbrace{
+        \begin{bmatrix}
+            \int_0^1 f(x)v_0(x)\ dx \\
+            \int_0^1 f(x)v_1(x)\ dx \\
+            \int_0^1 f(x)v_2(x)\ dx \\
+            \int_0^1 f(x)v_3(x)\ dx
+        \end{bmatrix}
+    }^{\vec{r}} \\
+    & = 
+    \begin{bmatrix}
+        \frac{\sfrac{1}{2}\cdot 8}{2} = 2\quad\text{(area of {\color{red} red shape function})} \\
+        \frac{\sfrac{3}{4}\cdot 8}{2} = 3\quad\text{(area of {\color{blue} blue shape function})} \\
+        \frac{\sfrac{1}{2}\cdot 8}{2} = 2\quad\text{(area of {\color{violet} violet shape function})} \\
+        \frac{\sfrac{1}{4}\cdot 8}{2} = 1\quad\text{(area of {\color{green} green shape function})} \\
+    \end{bmatrix}
+\end{align*}
+
+to satisfy inhomogeneous boundary conditions, we set $a_0$ and $a_3$ to the boundary condition in the vector $\vec{a}$ and
+eliminate the corresponding equations, leading to the reduced Ritz system:
+
+\begin{align*}
+    \begin{bmatrix}
+        -2 & 6 & -4 & 0 \\
+        0 & -4 & 8 & -4
+    \end{bmatrix}
+    \begin{bmatrix}
+        1\;\text{\color{gray} (b.c.)} \\
+        a_1 \\
+        a_2 \\
+        2\;\text{\color{gray} (b.c.)}
+    \end{bmatrix}
+    = \begin{bmatrix}
+        3 \\ 2
+    \end{bmatrix}
+\end{align*}
+meaning
+\begin{align*}
+    \begin{bmatrix}
+        6 & -4 \\
+        -4 & 8
+    \end{bmatrix}
+    \begin{bmatrix}
+        a_1 \\ a_2
+    \end{bmatrix}
+    +
+    \begin{bmatrix}
+        -2 \\ -8
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        3 \\ 2
+    \end{bmatrix}
+\end{align*}
+
+giving $a_1 = \sfrac{5}{2}$ and $a_2 = \sfrac{5}{2}$.
+
+\subsubsection{Example With Homogeneous Boundary Conditions}
+
+\textbf{Given} the differential equation $u''(x) + 1 = 0$ on $\Omega = [0,1]$ with homogeneous boundary conditions.
+We want to approximate using meshes $[0,0.25], [0.25, 0.75], [0.75, 1]$.
+\\[1em]
+\textbf{Discretisation of Geometry}
+
+Since we have homogeneous boundary conditions, we can ignore $v_0$ and $v_3$:
+\makebox[\columnwidth]{\includegraphics[width=0.4\columnwidth]{images/FEM_Example2}}
+giving $\utild(x) = 0 v_0(x)+a_1v_1(x)+a_2v_2(x)+0v_3(x)$.
+\\[1em]
+\textbf{Solution}
+
+For steps 1, 2 and 3, we get the mesh matrices
+\begin{align*}
+    M_1 = 4\cdot E, M_2=2\cdot E, M_3 = 4\cdot E
+\end{align*}
+using element matrix $E$ of step function and thus the Ritz matrix
+\begin{align*}
+    R=\begin{bmatrix}
+        4 & -4 & 0 & 0 \\
+        -4 & 6 & -2 & 0 \\
+        0 & -2 & 6 & -1 \\
+        0 & 0 & -4 & 4
+    \end{bmatrix}
+\end{align*}
+
+due to homogeneous boundary conditions, we cancel the first and last row and columns, resulting in the reduced Ritz system:
+\begin{align*}
+    \begin{bmatrix}
+        6 & -2 \\
+        -2 & 6
+    \end{bmatrix}
+    \begin{bmatrix}
+        a_1 \\ a_2
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        \sfrac{3}{8} \\ \sfrac{3}{8}
+    \end{bmatrix}
+\end{align*}
+
+\subsubsection{Example With Funky $f(x)$}
+
+Problem: $-u''(x) = f(x)$ in $\Omega = [0,1]$ with
+\begin{align*}
+    f(x) = \left\{
+    \begin{array}{ll}
+        8 & \text{if } x\in [0,\sfrac{1}{4}), \\
+        24 & \text{if } x\in [\sfrac{1}{4},\sfrac{1}{2}), \\
+        40 & \text{if } x\in [\sfrac{1}{2},\sfrac{3}{4}), \\
+        16 & \text{if } x\in [\sfrac{3}{4},1),
+    \end{array}
+    \right.
+\end{align*}
+and homogeneous b.c. and $h = 1/4$.
+
+\makebox[\columnwidth]{\includegraphics[width=0.5\columnwidth]{images/FEM_Example3}}
+
+The Ritz vector then is:
+
+For $v_1$: $8\cdot\frac{1}{4}\frac{1}{2} = 1$; $24\cdot \frac{1}{4}\frac{1}{2} = 3\rightarrow 1+3=4$
+
+For $v_2$: $24\cdot\frac{1}{4}\frac{1}{2} = 3$; $40\cdot \frac{1}{4}\frac{1}{2} = 5\rightarrow 3+5=8$
+
+For $v_3$: $40\cdot\frac{1}{4}\frac{1}{2} = 5$; $16\cdot \frac{1}{4}\frac{1}{2} = 2\rightarrow 5+2=7$
+
+\numericsubsection{p-Strategy}
+
+Using quadratic shape functions $q_1(x) = (1-x)(1-2x), q_2(x) = 4x(1-x), q_3(x) = -x(1-2x)$ as shape functions
+
+\makebox[\columnwidth]{\includegraphics[width=0.4\columnwidth]{images/FEM_pStrategy}}
+
+The v-functions for three nodes then look like
+
+\makebox[\columnwidth]{\includegraphics[width=0.5\columnwidth]{images/FEM_v_functions}}
+
+with an affine transformation of the x values onto the nodal points.
+
+We get a new element matrix:
+\begin{snugshade*}
+    \begin{align*}
+        E = 
+        \begin{bmatrix}
+            7 & -8 & 1 \\
+            -8 & 16 & -8 \\
+            1 & -8 & 7
+        \end{bmatrix}
+    \end{align*}
+\end{snugshade*}
+
+and the associated mesh matrix $M = 1/h\cdot E$.
+The Ritz vector can be computed analogous to the previous methods:
+\begin{align*}
+    \vec{r} = \begin{bmatrix}
+        \int_0^1 f(x)\cdot v_0(x)\ dx \\
+        \int_0^1 f(x)\cdot v_1(x)\ dx \\
+        \vdots \\
+        \int_0^1 f(x)\cdot v_n(x)\ dx
+    \end{bmatrix}
+\end{align*}
+
+with the difference that the integral is not as straight forward as with the triangle function
+({\color{blue}\faInfo}\ except for baby Laplace problem $u''(x) = 0$, since there the Ritz-vector is zero and no integration needed)