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z05MJ^1c>H8X(#;tnMD4Ldkg|g#1JU?geE=_LYzf_>52u}1nV$BktYm9AR7KiuP62g^dmkK!9q?>NP)ZmP!J1+qX{zr;C90H2dGeR z!ZHW|2FR~}@&h^;D2^~K0dwuY-mmpf!N8nMl=^`F1sLB67y@U6M3o1Ih7+#?!u;Rv z{BPd{1CDG6cLzY4SP0OG0jI)$S_TCaWkT1XfVqYcGpPSHWCQa#(E$w#IE^EScPI*m zAVN9{I5s5&*GWf27!fGoSeVeAKtX{K4cRE-5(RW@2%^aYXns&c3IYeZL_0u(7?eb!NX`MQ4Iy8 z!Z27o-3CbJh9X1`g#zY(IV@lXCA0zx2xo+XJQSt?M=Js7zbyU#U13gh`7+IIFDp+k WUr#%G8W=Eaz#%k(f{L0-H2)7kC&S(V diff --git a/doc/doc.tex b/doc/doc.tex index 15122ff..39359e6 100644 --- a/doc/doc.tex +++ b/doc/doc.tex @@ -200,12 +200,12 @@ Damit ist $\phi_{\ell}$ die Galerkinapproximation an $\phi$ % \subsection{Vorkonditionieren} % \begin{align} -% V \phi_{\ell/2} &= b\\ +% V \hat \phi_{\ell} &= b\\ % D &= diag(V)\\ % A &= D \cdot V \cdot D\\ % c & = D\cdot b\\ % A\cdot y &= c\\ -% \phi_{\ell/2} &= D \cdot y\\ +% \hat \phi_{\ell} &= D \cdot y\\ % \end{align} \subsection{Netze} @@ -294,36 +294,36 @@ sollte \end{defi} \begin{defi}[Saturationsannahme] \begin{align} - \norm{\phi -\phi_{\ell/2}} &\leq C_{sat} \cdot \norm{\phi - \phi_{\ell}} & 0 < C_{sat} < 1 + \norm{\phi -\hat \phi_{\ell}} &\leq C_{sat} \cdot \norm{\phi - \phi_{\ell}} & 0 < C_{sat} < 1 \end{align} \end{defi} \begin{defi}[$\ell-\ell/2$ - Schätzer] \begin{align} - \eta_{\ell} &:= \norm{\phi_{\ell/2} - \phi_{\ell}}_{H^{-1/2}} + \eta_{\ell} &:= \norm{\hat \phi_{\ell} - \phi_{\ell}}_{H^{-1/2}} \end{align} Der Fehlerschätzer ist aber nur unter der Saturationsannahme zuverlässig und ist effizient mit $C_{\tt eff} = 1$ \end{defi} Da aber die $\norm{\cdot}_{H^{1/2}}$ schlecht zu berechnen ist, wenden wir die $L_2$-Projektion an. \begin{defi}[ersetzen von $\norm{\cdot}_{H^{1/2}}$] \begin{align} - \mu_{\ell} &:= \norm{ \varrho^{1/2} (\phi_{\ell/2} - {\phi_{\ell}})}_{L_2(\Gamma)} + \mu_{\ell} &:= \norm{ \varrho^{1/2} (\hat \phi_{\ell} - {\phi_{\ell}})}_{L_2(\Gamma)} \end{align} $\mu_{\ell}$ ist da $\mu_{\ell} \approx \eta_{\ell}$ gilt noch immer zuverlässig und effizient. \end{defi} Um sich unnötige Berechnungen zu sparen ist es sinnvoll $\phi_{\ell}$ zu ersetzen. \begin{defi}[ersetzen von $\phi_{\ell}$] \begin{align} - \tilde \mu_{\ell} &:= \norm{\varrho^{1/2}(\phi_{\ell/2} - \Pi_{\ell}\phi_{\ell/2})}_{L_2(\Gamma)} + \tilde \mu_{\ell} &:= \norm{\varrho^{1/2}(\hat \phi_{\ell} - \Pi_{\ell}\hat \phi_{\ell})}_{L_2(\Gamma)} \end{align} wobei $\Pi_{\ell}$ die $L_2$ Projektion auf $P^0(\T_{\ell})$ ist. \end{defi} \begin{sat}[A-posteriori Fehlerschätzer] Seien also: \begin{align} - \eta_{\ell} &= \enorm{\phi_{\ell/2} - \phi_{\ell}}\\ - \tilde\eta_{\ell} &= \enorm{\phi_{\ell/2} - \Pi_{\ell}\phi_{\ell/2}}\\ - \mu_{\ell} &= \norm{\varrho^{1/2}(\phi_{\ell/2} - \phi_{\ell})}_{L^2(\Gamma)}\\ - \tilde\mu_{\ell} &= \norm{\varrho^{1/2}(\phi_{\ell/2} - \Pi_{\ell}\phi_{\ell/2})}_{L^2(\Gamma)} + \eta_{\ell} &= \enorm{\hat \phi_{\ell} - \phi_{\ell}}\\ + \tilde\eta_{\ell} &= \enorm{\hat \phi_{\ell} - \Pi_{\ell}\hat \phi_{\ell}}\\ + \mu_{\ell} &= \norm{\varrho^{1/2}(\hat \phi_{\ell} - \phi_{\ell})}_{L^2(\Gamma)}\\ + \tilde\mu_{\ell} &= \norm{\varrho^{1/2}(\hat \phi_{\ell} - \Pi_{\ell}\hat \phi_{\ell})}_{L^2(\Gamma)} \end{align} Dann gilt auf isotropen Netzen: \begin{itemize} @@ -397,13 +397,13 @@ Liegen die beiden Elemente parallel zueinander lassen sie sich Folgendermaßen d Sei: \Ta = $ \v + [(0,l_1) \times (0,l_2) \times \{0\}]$ und \Tb = $ \tilde \v + [(0,\tilde l_1) \times (0,\tilde l_2) \times \{0\}]$, wobei $\v,\tilde \v \in \R^3$ ist. Weiterhin sei $\boldsymbol{\delta} = \tilde \v - \v \in \R^3$ definiert.\\ Damit können wir zeigen, dass \begin{eqnarray*} -&&-\frac{1}{4\pi} \int_T \int_{\tilde T} \frac{1}{|x-y|} ds_y ds_x\\ -&=&-\frac{1}{4\pi} \int_T \int_{\tilde T} +&&\frac{1}{4\pi} \int_T \int_{\tilde T} \frac{1}{|x-y|} ds_y ds_x\\ +&=&\frac{1}{4\pi} \int_T \int_{\tilde T} \left ( (x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2 \right)^{-1/2} ds_y ds_x\\ -&=&-\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_1}\int_0^{\tilde l_2} +&=&\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_1}\int_0^{\tilde l_2} \left ( ((x_1 + v_1)-(y_1 +\tilde v_1))^2+((x_2 + v_2)-(y_2 +\tilde v_2))^2+( v_3- \tilde v_3)^2 \right)^{-1/2} dy_2 dy_1 dx_2 dx_1\\ -&=&-\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_1}\int_0^{\tilde l_2} +&=&\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_1}\int_0^{\tilde l_2} \left ( (x_1-y_1-\delta_1)^2+(x_2-y_2-\delta_2)^2+\delta_3^2 \right)^{-1/2} dy_2 dy_1 dx_2 dx_1 \end{eqnarray*} @@ -418,13 +418,13 @@ F_{par}(x_1,x_2,y_1,y_2,\delta_1,\delta_2,\delta_3) &:=& \int \int \int \int \le $\v,\tilde \v \in \R^3$\\ $\boldsymbol{\delta} = \tilde \v - \v \in \R^3$\\ \begin{eqnarray*} -&&-\frac{1}{4\pi} \int_T \int_{\tilde T} \frac{1}{|x-y|} ds_y ds_x\\ -&=&-\frac{1}{4\pi} \int_T \int_{\tilde T} +&&\frac{1}{4\pi} \int_T \int_{\tilde T} \frac{1}{|x-y|} ds_y ds_x\\ +&=&\frac{1}{4\pi} \int_T \int_{\tilde T} \left ( (x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2 \right)^{-1/2} ds_y ds_x\\ -&=&-\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_2}\int_0^{\tilde l_3} +&=&\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_2}\int_0^{\tilde l_3} \left ( ((x_1 + v_1)-\tilde v_1)^2+((x_2 + v_2)-(y_2 +\tilde v_2))^2+( v_3- (y_3 +\tilde v_3))^2 \right)^{-1/2} dy_3 dy_2 dx_2 dx_1\\ -&=&-\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_2}\int_0^{\tilde l_3} +&=&\frac{1}{4\pi} \int_0^{l_1}\int_0^{l_2}\int_0^{\tilde l_2}\int_0^{\tilde l_3} \left ( (x_1-\delta_1)^2+(x_2-y_2-\delta_2)^2+(y_3-\delta_3)^2 \right)^{-1/2} dy_3 dy_2 dx_2 dx_1 \end{eqnarray*} @@ -436,13 +436,13 @@ dy_3 dy_2 dx_2 dx_1 \subsection{Bestimmtes Integral} \begin{eqnarray*} -&&-\frac{1}{4\pi} \int_T \int_{\tilde T} \frac{1}{|x-y|} ds_y ds_x\\ -&\approx& -\frac{1}{4\pi} \int_0^{k_1}\int_0^{k_2}\int_0^{\tilde k_1}\int_0^{\tilde k_2} +&&\frac{1}{4\pi} \int_T \int_{\tilde T} \frac{1}{|x-y|} ds_y ds_x\\ +&\approx& \frac{1}{4\pi} \int_0^{k_1}\int_0^{k_2}\int_0^{\tilde k_1}\int_0^{\tilde k_2} \dif{}{y_2} \dif{}{y_1} \dif{}{x_2} \dif{}{x_1} F_{par/ort}(x_1,x_2,y_1,y_2) dy_2 dy_1 dx_2 dx_1\\ % -&=& -\frac{1}{4\pi}\big( \int_0^{k_1}\int_0^{k_2}\int_0^{\tilde k_1} +&=& \frac{1}{4\pi}\big( \int_0^{k_1}\int_0^{k_2}\int_0^{\tilde k_1} \dif{}{y_1} \dif{}{x_2} \dif{}{x_1} F_{par/ort}(x_1,x_2,y_1,\tilde k_2) - @@ -450,7 +450,7 @@ dy_2 dy_1 dx_2 dx_1\\ F_{par/ort}(x_1,x_2,y_1,0) \big) dy_1 dx_2 dx_1\\ % -&=& -\frac{1}{4\pi}\big( \int_0^{k_1}\int_0^{k_2} +&=& \frac{1}{4\pi}\big( \int_0^{k_1}\int_0^{k_2} \dif{}{x_2} \dif{}{x_1} F_{par/ort}(x_1,x_2,\tilde k_1,\tilde k_2) - @@ -464,7 +464,7 @@ dy_2 dy_1 dx_2 dx_1\\ F_{par/ort}(x_1,x_2,0,0)\big) dx_2 dx_1\\ % -&=& -\frac{1}{4\pi}\big( \int_0^{k_1} +&=& \frac{1}{4\pi}\big( \int_0^{k_1} \dif{}{x_1} F_{par/ort}(x_1,k_2,\tilde k_1,\tilde k_2) - @@ -490,7 +490,7 @@ dy_2 dy_1 dx_2 dx_1\\ F_{par/ort}(x_1,0,0,0)\big) dx_1\\ % -&=& -\frac{1}{4\pi}\big( +&=& \frac{1}{4\pi}\big( F_{par/ort}(k_1,k_2,\tilde k_1,\tilde k_2) - F_{par/ort}(k_1,k_2,\tilde k_1,0) @@ -626,7 +626,7 @@ Relevant zum Verfeinern eines Netzes sind also die Koordinaten $COO$, Elemente $ Da wir später einen Fehlerschätzer berechnen wollen, ist es wichtig sich zu jedem Element seine Teilelemente zu merken. Dazu legen wir während der Teilung eine $M \times 4$ Matrix an, in der die maximal vier Elementindizes gespeichert sind. Wenn wir also ein Element in vier gleich große Teile verfeinern, so wird das neue Element links unten das erste sein und alle weiteren folgen gegen den Uhrzeigersinn. Teilen wir ein Element in zwei gleich große Elemente, so werden die doppelt belegten Quadranten auch doppelt eingetragen. Ein gar nicht geteiltes Element wird also vier mal den alten Indizes speichern. Dadurch wird sicher gestellt, dass das arithmetische Mittel über die Elemente immer gültig auszuführen ist. (Siehe Figur:\ref{exmpl13:f2s}) -$[\T_{\ell/2}, F2S ] = refineQuad(\T_{\ell}, marked);$\\ +$[\\hat T_{\ell}, F2S ] = refineQuad(\T_{\ell}, marked);$\\ $[COO_{fine}, ELE_{fine}, NEI_{fine}, F2S ] = refineQuad(COO, ELE, NEI, marked);$ \begin{figure}[ht] @@ -698,7 +698,7 @@ $marked = mark(xF2S, mu, theta, nu);$ %Es seien \Ta, \Tb ~$\subseteq$~ $\R^3$ zwei beschränkte, achsenorientierte rechteckige Seiten in $\R^3$. %Berechnet werden soll: \begin{eqnarray*} -V(j,k) &=& -\frac{1}{4\pi} \int_{T_j} \int_{T_k} \frac{1}{|x-y|} ds_y ds_x \in \R^3,T_j,T_k\in\T +V(j,k) &=& \frac{1}{4\pi} \int_{T_j} \int_{T_k} \frac{1}{|x-y|} ds_y ds_x \in \R^3,T_j,T_k\in\T \end{eqnarray*} Wobei $\zeta$ die Zulässigkeitsbedingung und $type$ die Berechnungsart bestimmt. $V = mex\_build\_AU(\T,zeta,type)$\\ @@ -711,9 +711,9 @@ Mithilfe der oben Definierten Funktionen ist es uns nun möglich den Ablauf der $\theta \in (0,1),i =0$ \begin{enumerate} \renewcommand{\theenumi}{(\roman{enumi})} - \item Verfeinere $T_{\ell}^{(i)}$ um $T_{\ell/2}$ zu erhalten - \item Berechne die Galerkinlösung $\phi_{\ell/2} \in P^0(\T_{\ell/2})$ - \item Berechne Fehlerschätzer $\tilde \mu_{i} := \norm{\varrho^{\ell/2}(\phi_{\ell/2} - \Pi_{\ell} \phi_{\ell/2} )}$ + \item Verfeinere $T_{\ell}^{(i)}$ um $\hat T_{\ell}$ zu erhalten + \item Berechne die Galerkinlösung $\hat \phi_{\ell} \in P^0(\\hat T_{\ell})$ + \item Berechne Fehlerschätzer $\tilde \mu_{i} := \norm{\varrho^{\ell/2}(\hat \phi_{\ell} - \Pi_{\ell} \hat \phi_{\ell} )}$ \item Wähle $M_{\ell} \subseteq T_{\ell}^{(i)}$ mit minimaler Kardinalität, so dass \begin{align} \theta \sum_{T\in \T^{(i)}_{\ell}} \tilde\mu_{i}^2 & \leq \sum_{T\in M_{\ell}} \tilde\mu_{i}^2 @@ -725,14 +725,14 @@ $\theta \in (0,1),i =0$ Zum Plotten (Abb.\ref{exmplAA_2DQuad})werden noch folgende Schritte ausgeführt \begin{itemize} \item Berechne Galerkinlösung $\phi_{l} \in P^0(\T_{\ell}^{(i)})$ -% \item $\enorm{\phi_{\ell/2}}$ +% \item $\enorm{\hat \phi_{\ell}}$ % \item $\enorm{\phi_{\ell}}$ \item $error_{i} = \sqrt{\enorm{\phi}^2 - \enorm{\phi_{l}}^2}$ - \item $\mu_{i} = \norm{\varrho^{1/2}(\phi_{\ell/2} - \phi_{l} )}$ - \item $\eta_{i} = \enorm{\phi_{\ell/2} - \phi_{l}}$ - \item $\kappa_{i} = \enorm{\phi_{\ell/2}^{(i)}}-\enorm{\phi_{\ell/2}^{(i-1)}}$ + \item $\mu_{i} = \norm{\varrho^{1/2}(\hat \phi_{\ell} - \phi_{l} )}$ + \item $\eta_{i} = \enorm{\hat \phi_{\ell} - \phi_{l}}$ + \item $\kappa_{i} = \enorm{\hat \phi_{\ell}^{(i)}}-\enorm{\hat \phi_{\ell}^{(i-1)}}$ \item $\kappa2_{i} = \enorm{\phi_{l}^{(i)}}-\enorm{\phi_{l}^{(i-1)}}$ - \item $\kappa3_{i} = \enorm{\phi_{\ell/2}^{(i)}-\phi_{\ell/2}^{(i-1)}}$ + \item $\kappa3_{i} = \enorm{\hat \phi_{\ell}^{(i)}-\hat \phi_{\ell}^{(i-1)}}$ \end{itemize} -- 2.47.3