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A prolongation operator transfers a coarse-grid correction to a finer mesh. For nested one-dimensional grids of spacings $2h$ and $h$, linear interpolation sets $(Pe{)}_{2j}=e_j$ and $(Pe{)}_{2j+1}=(e_j+e_{j+1})/2$, with zero endpoint values for homogeneous correction boundary conditions. The associated full-weighting restriction is $R=P^T/2$, and the Galerkin coarse-grid operator is $R{A}_{h}P$. The correction equation acts on the residual so that a coarse representation of the error can improve the fine-grid approximation.

A coarse-grid operator represents the equation solved for an error correction on the coarser level of a multigrid method. It may be obtained by rediscretizing the differential equation at the coarse spacing, or through the Galerkin coarse-grid operator formula displayed above. Here $R$ transfers residuals to the coarse grid and $P$ transfers coarse corrections to the fine grid. The right-side and operator scalings must be consistent with the chosen transfer convention.

Historical core-paper title on the 2020 examination cover.

A salt of morphine. Formulation, route and concentration matter to dose interpretation.

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

Historical pharmaceutical preparation; its composition and clinical use must be identified by edition. ^[ref]

Let $U\subseteq \mathbb{C}$ be open, let $L$ be a straight line, and let $f:U\to \mathbb{C}$ be continuous on $U$ and holomorphic on $U\setminus L$. Then $f$ is holomorphic on all of $U$. Near a point of $L$, work in a disk whose closure lies in $U$. Split a small filled triangle along $L$ and triangulate the two resulting convex polygons. Translate each triangle slightly into its corresponding open half-plane, keeping it in $U$. Its contour integral is zero by Cauchy's integral theorem. On each segment the integral is the endpoint difference times the integral of $f$ along the unit-interval parametrization; uniform continuity on a containing compact disk therefore makes the contour integral continuous as the vertices move back. Each untranslated piece consequently has zero integral. Internal oriented edges cancel when the pieces are added, so every small triangle has zero integral. Morera's theorem proves the claim locally along the line. The continuity across $L$ is essential; holomorphicity on the two sides alone is insufficient.

A map $A:X\to X^{\ast }$ on a real Banach space is a monotone operator exactly when the displayed scalar function is nondecreasing on $[0,1]$ for every $u,v\in X$. Applying operator monotonicity to $u+tv$ and $u+sv$, $s<t$, gives $(t-s)(\phi (t)-\phi (s))\ge 0$. Conversely take $u=y$, $v=x-y$ and compare the endpoints to recover $⟨A(x)-A(y),x-y⟩\ge 0$. No continuity or differentiation of $A$ is needed.

A map from a subset of a real Banach space to its continuous dual space is strongly monotone if the displayed inequality holds for every pair in its domain, with one constant $m>0$. The brackets denote the dual pairing. This implies strict monotonicity for distinct arguments. Fixing $v=u_0$ and applying the dual norm inequality gives $⟨A(u),u-u_0⟩\ge m\Vert u-u_0{\Vert }^2-\Vert A(u_0)\Vert \Vert u-u_0\Vert$, so it is a nonlinear coercive operator with respect to $u_0$.