1 2In this section, we deal with only with the FirstOrderType2R case. 3 4 5 \begin{equation} 6 \begin{array}{l} 7 \label{eq:full-toto1-ter} 8 M x_{k+1} = M x_{k} +h \theta f(x_{k+1},t_{k+1}) +h(1-\theta)f(x_{k},t_{k}) + h r_{k+\gamma} \\[2mm] 9 y_{k+\gamma} = h(t_{k+\gamma},x_{k+\gamma},\lambda _{k+\gamma}) \\[2mm] 10 r_{k+\gamma} = g(t_{k+\gamma},\lambda_{k+\gamma})\\[2mm] 11 \end{array} 12\end{equation} 13 14 \paragraph{Newton's linearization of the first line of~(\ref{eq:full-toto1-ter})} The first line of the problem~(\ref{eq:full-toto1-ter}) can be written under the form of a residue $\mathcal R$ depending only on $x_{k+1}$ and $r_{k+\gamma}$ such that 15\begin{equation} 16 \label{eq:full-NL3} 17 \mathcal R (x_{k+1},r _{k+\gamma}) =0 18\end{equation} 19with $$\mathcal R(x,r) = M(x - x_{k}) -h\theta f( x , t_{k+1}) - h(1-\theta)f(x_k,t_k) - h r. $$ 20The solution of this system of nonlinear equations is sought as a limit of the sequence $\{ x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma} \}_{\alpha \in \NN}$ such that 21 \begin{equation} 22 \label{eq:full-NL7} 23 \begin{cases} 24 x^{0}_{k+1} = x_k \\ \\ 25 r^{0}_{k+\gamma} = (1-\gamma ) r_{k} + \gamma r^0_{k+1} = r_k \\ \\ 26 \mathcal R_L( x^{\alpha+1}_{k+1},r^{\alpha+1}_{k+\gamma}) = \mathcal 27 R(x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma}) + \left[ \nabla_{x} \mathcal 28 R(x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma})\right] (x^{\alpha+1}_{k+1}-x^{\alpha}_{k+1} ) + \\[2mm] 29 \qquad\qquad\qquad\qquad\qquad\qquad\left[ \nabla_{r} \mathcal R(x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma})\right] (r^{\alpha+1}_{k+\gamma} - r^{\alpha}_{k+\gamma} ) =0 30 \end{cases} 31\end{equation} 32\begin{ndrva} 33 What about $r^0_{k+\gamma}$ ? 34\end{ndrva} 35 36The residu free is also defined (useful for implementation only): 37\[\mathcal R _{\free}(x) \stackrel{\Delta}{=} M(x - x_{k}) -h\theta f( x , t_{k+1}) - h(1-\theta)f(x_k,t_k).\] 38We get 39\begin{equation} 40 \mathcal R (x^{\alpha}_{k+1},r^{\alpha}_{k+\gamma}) = \fbox{$\mathcal R^{\alpha}_{k+1} \stackrel{\Delta}{=} \mathcal R_{\free}(x^{\alpha}_{k+1} ) - h r^{\alpha}_{k+\gamma}$}\label{eq:full-rfree-1} 41\end{equation} 42 43\[ \mathcal R 44_{\free}(x^{\alpha}_{k+1} )=\fbox{$ \mathcal R _{\free, k+1} ^{\alpha} \stackrel{\Delta}{=} M(x^{\alpha}_{k+1} - x_{k}) -h\theta f( x^{\alpha}_{k+1} , t_{k+1}) - h(1-\theta)f(x_k,t_k)$}\] 45 46The computation of the Jacobian of $\mathcal R$ with respect to $x$, denoted by $ W^{\alpha}_{k+1}$ leads to 47\begin{equation} 48 \label{eq:full-NL9} 49 \begin{array}{l} 50 W^{\alpha}_{k+1} \stackrel{\Delta}{=} \nabla_{x} \mathcal R (x^{\alpha}_{k+1})= M - h \theta \nabla_{x} f( x^{\alpha}_{k+1}, t_{k+1} ).\\ 51 \end{array} 52\end{equation} 53At each time--step, we have to solve the following linearized problem, 54\begin{equation} 55 \label{eq:full-NL10} 56 \mathcal R^{\alpha}_{k+1} + W^{\alpha}_{k+1} (x^{\alpha+1}_{k+1} - 57 x^{\alpha}_{k+1}) - h (r^{\alpha+1}_{k+\gamma} - r^{\alpha}_{k+\gamma} ) =0 , 58\end{equation} 59By using (\ref{eq:full-rfree-1}), we get 60\begin{equation} 61 \label{eq:full-rfree-2} 62 \mathcal R _{\free}(x^{\alpha}_{k+1}) - h r^{\alpha+1}_{k+\gamma} + W^{\alpha}_{k+1} (x^{\alpha+1}_{k+1} - 63 x^{\alpha}_{k+1}) =0 64\end{equation} 65 66%\fbox 67{ 68 \begin{equation} 69 \boxed{ x^{\alpha+1}_{k+1} = h(W^{\alpha}_{k+1})^{-1}r^{\alpha+1}_{\gamma+1} +x^\alpha_{\free}} 70 \end{equation} 71} 72with : 73\begin{equation} 74 \boxed{x^\alpha_{\free}\stackrel{\Delta}{=}x^{\alpha}_{k+1}-(W^{\alpha}_{k+1})^{-1}\mathcal R_{\free,k+1}^{\alpha} \label{eq:full-rfree-12}} 75\end{equation} 76 77The matrix $W$ is clearly non singular for small $h$. 78 79Note that the linearization is equivalent to the case (\ref{eq:rfree-2}) and (\ref{eq:rfree-12}) with $\gamma=1$ and replacing $r_{k+1}$ by $r_{k+\gamma}$. 80 81 \paragraph{Newton's linearization of the second line of~(\ref{eq:full-toto1-ter})} 82The same operation is performed with the second equation of (\ref{eq:full-toto1-ter}) 83\begin{equation} 84 \begin{array}{l} 85 \mathcal R_y(x,y,\lambda)=y-h(t_{k+\gamma},\gamma x + (1-\gamma) x_k ,\lambda) =0\\ \\ 86 \end{array} 87\end{equation} 88which is linearized as 89\begin{equation} 90 \label{eq:full-NL9} 91 \begin{array}{l} 92 \mathcal R_{Ly}(x^{\alpha+1}_{k+1},y^{\alpha+1}_{k+\gamma},\lambda^{\alpha+1}_{k+\gamma}) = \mathcal 93 R_{y}(x^{\alpha}_{k+1},y^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma}) + 94 (y^{\alpha+1}_{k+\gamma}-y^{\alpha}_{k+\gamma})- \\[2mm] \qquad \qquad \qquad \qquad \qquad \qquad 95 \gamma C^{\alpha}_{k+1}(x^{\alpha+1}_{k+1}-x^{\alpha}_{k+1}) - D^{\alpha}_{k+\gamma}(\lambda^{\alpha+1}_{k+\gamma}-\lambda^{\alpha}_{k+\gamma})=0 96 \end{array} 97\end{equation} 98 99This leads to the following linear equation 100\begin{equation} 101 \boxed{y^{\alpha+1}_{k+\gamma} = y^{\alpha}_{k+\gamma} 102 -\mathcal R^{\alpha}_{y,k+1}+ \\ 103 \gamma C^{\alpha}_{k+1}(x^{\alpha+1}_{k+1}-x^{\alpha}_{k+1}) + 104 D^{\alpha}_{k+\gamma}(\lambda^{\alpha+1}_{k+\gamma}-\lambda^{\alpha}_{k+\gamma})}. \label{eq:full-NL11y} 105\end{equation} 106with, 107\begin{equation} 108 \begin{array}{l} 109 C^{\alpha}_{k+\gamma} = \nabla_xh(t_{k+1}, x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma} ) \\ \\ 110 D^{\alpha}_{k+\gamma} = \nabla_{\lambda}h(t_{k+1}, x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma}) 111 \end{array} 112\end{equation} 113and 114\begin{equation}\fbox{$ 115\mathcal R^{\alpha}_{yk+1} \stackrel{\Delta}{=} y^{\alpha}_{k+\gamma} - h(x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma})$} 116 \end{equation} 117 118Note that the linearization is equivalent to the case (\ref{eq:NL11y}) by replacing $\lambda_{k+1}$ by $\lambda_{k+\gamma}$ and $x_{k+1}$ by $x_{k+\gamma}$. 119 120 \paragraph{Newton's linearization of the third line of~(\ref{eq:full-toto1-ter})} 121The same operation is performed with the third equation of (\ref{eq:full-toto1-ter}) 122\begin{equation} 123 \begin{array}{l} 124 \mathcal R_r(r,\lambda)=r-g(\lambda,t_{k+1}) =0\\ \\ \end{array} 125\end{equation} 126which is linearized as 127\begin{equation} 128 \label{eq:full-NL9} 129 \begin{array}{l} 130 \mathcal R_{L\lambda}(r^{\alpha+1}_{k+\gamma},\lambda^{\alpha+1}_{k+\gamma}) = \mathcal 131 R_{r,k+\gamma}^{\alpha} + (r^{\alpha+1}_{k+\gamma} - r^{\alpha}_{k+\gamma}) - B^{\alpha}_{k+\gamma}(\lambda^{\alpha+1}_{k+\gamma} - 132 \lambda^{\alpha}_{k+\gamma})=0 133 \end{array} 134 \end{equation} 135\begin{equation} 136 \label{eq:full-rrL} 137 \begin{array}{l} 138 \boxed{r^{\alpha+1}_{k+\gamma} = g(\lambda ^{\alpha}_{k+\gamma},t_{k+\gamma}) -B^{\alpha}_{k+\gamma} 139 \lambda^{\alpha}_{k+\gamma} + B^{\alpha}_{k+\gamma} \lambda^{\alpha+1}_{k+\gamma}} 140 \end{array} 141\end{equation} 142with, 143\begin{equation} 144 \begin{array}{l} 145 B^{\alpha}_{k+\gamma} = \nabla_{\lambda}g(\lambda ^{\alpha}_{k+\gamma},t_{k+\gamma}) 146 \end{array} 147\end{equation} 148and the residue for $r$: 149\begin{equation} 150\boxed{\mathcal 151 R_{rk+\gamma}^{\alpha} = r^{\alpha}_{k+\gamma} - g(\lambda ^{\alpha}_{k+\gamma},t_{k+\gamma})} 152 \end{equation} 153Note that the linearization is equivalent to the case (\ref{eq:rrL}) by replacing $\lambda_{k+1}$ by $\lambda_{k+\gamma}$ and $x_{k+1}$ by $x_{k+\gamma}$. 154 155\paragraph{Reduction to a linear relation between $x^{\alpha+1}_{k+1}$ and 156$\lambda^{\alpha+1}_{k+\gamma}$} 157 158Inserting (\ref{eq:full-rrL}) into~(\ref{eq:full-rfree-12}), we get the following linear relation between $x^{\alpha+1}_{k+1}$ and 159$\lambda^{\alpha+1}_{k+1}$, 160 161\begin{equation} 162 \begin{array}{l} 163 x^{\alpha+1}_{k+1} = h(W^{\alpha}_{k+1} )^{-1}\left[g(\lambda^{\alpha}_{k+\gamma},t_{k+\gamma}) + 164 B^{\alpha}_{k+\gamma} (\lambda^{\alpha+1}_{k+\gamma} - \lambda^{\alpha}_{k+\gamma}) \right ] +x^\alpha_{free} 165\end{array} 166\end{equation} 167that is 168\begin{equation} 169 \begin{array}{l} 170\boxed{x^{\alpha+1}_{k+1}=x_p + h (W^{\alpha}_{k+1})^{-1} B^{\alpha}_{k+\gamma} \lambda^{\alpha+1}_{k+\gamma}} 171 \end{array} 172 \label{eq:full-rfree-13} 173\end{equation} 174with 175\begin{equation} 176 \boxed{x_p \stackrel{\Delta}{=} h(W^{\alpha}_{k+1} )^{-1}\left[g(\lambda^{\alpha}_{k+\gamma},t_{k+\gamma}) -B^{\alpha}_{k+\gamma} (\lambda^{\alpha}_{k+\gamma}) \right ] +x^\alpha_{free}} 177\end{equation} 178 179 180\paragraph{Reduction to a linear relation between $y^{\alpha+1}_{k+\gamma}$ and 181$\lambda^{\alpha+1}_{k+\gamma}$} 182 183Inserting (\ref{eq:full-rfree-13}) into (\ref{eq:full-NL11y}), we get the following linear relation between $y^{\alpha+1}_{k+1}$ and $\lambda^{\alpha+1}_{k+1}$, 184\begin{equation} 185 \begin{array}{l} 186 y^{\alpha+1}_{k+1} = y_p + \left[ h \gamma C^{\alpha}_{k+\gamma} ( W^{\alpha}_{k+1})^{-1} B^{\alpha}_{k+1} + D^{\alpha}_{k+1} \right]\lambda^{\alpha+1}_{k+1} 187 \end{array} 188\end{equation} 189with 190\begin{equation} 191y_p = y^{\alpha}_{k+1} -\mathcal R^{\alpha}_{yk+1} + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1} 192\end{equation} 193that is 194\begin{equation}\boxed{ 195y_p = h(x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma}) + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1} } 196\end{equation} 197\textcolor{red}{ 198 \begin{equation} 199 \boxed{ x_q=(x_p -x^{\alpha}_{k+1})\label{eq:full-xqq}} 200 \end{equation} 201} 202 203 204\paragraph{The linear case} 205\begin{equation} 206 \begin{array}{lcl} 207 y_p &=& h(x^{\alpha}_{k+\gamma},\lambda^{\alpha}_{k+\gamma}) + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1}\\ 208 &=& C^{\alpha}_{k+1} x^{\alpha}_{k+\gamma} + D^{\alpha}_{k+1}\lambda^{\alpha}_{k+\gamma} + \gamma C^{\alpha}_{k+1}(x_q) - D^{\alpha}_{k+1} \lambda^{\alpha}_{k+1} \\ 209 &=& C^{\alpha}_{k+1} (x^{\alpha}_{k+\gamma} + \gamma x_p - \gamma x^{\alpha}_{k+1} ) \\ 210 &=& C^{\alpha}_{k+1} ((1-\gamma) x_{k} + \gamma x_{free} ) \text {since } x_p =x_{free} 211\end{array} 212\end{equation} 213 214 215 216 217\paragraph{Implementation details} 218 219For the moment (Feb. 2011), we set $x_q=(1-\gamma) x_{k} + \gamma x_{free} $ in the linear case. 220The nonlinear case is not yet implemented since we need to 221change the management of \texttt{ H_alpha} Relation to be able to compute the mid--point values. 222% things that remain to do 223% 224% \begin{itemize} 225% \item implement the function \texttt{BlockVector computeg(t,lambda)} and \texttt{SimpleVector computeh(t,x,lambda)} which takes into account the values of the argument and return and vector 226% \item remove temporary computation in Relation of {\verb Xq, \verb g_alpha and \verb H_alpha }. This should be stored somewhere else. (in the node of the graph) 227% \end{itemize} 228 229 230 231 232 233 234 235 236\clearpage 237 238 239%%% Local Variables: 240%%% mode: latex 241%%% TeX-master: "DevNotes" 242%%% End: 243