Monotonicity preserving bicubic interpolation: A progress report
In this paper are considered some data \(f_{ij}=f(x_ i,y_ j)\) on a rectangular mesh: \(x_ 1<x_ 2<...<x_ m\); \(y_ 1<y_ 2<...<y_ n\). The authors want to find a function \(p(x,y)\in C^{1,1}(R)\), \(R=[x_ 1,x_ m]\times [y_ 1,y_ n]\), which interpolates the data: \(p(x_ i,y_ j)=f_{ij}\), with \(i=1,...,m\), \(j=1,...,n\), and which verifies: \(sig(p(x,\bar y)-p(x,y))=s_ 1\) if \(y_ 1\leq y<\bar y\leq y_ n\), \(\forall x\in[x_ 1,x_ m]\) and \(sig(p(\bar x,y)-p(x,y))=s_ 2\) if \(x_ 1\leq x<\bar x\leq x_ m\), \(\forall y\in [y_ 1,y_ n]\). It is chosen to use the class if piecewise bicubic Hermite functions on the mesh, which are completely determined by: p(i,j), \(p_ x(i,j)\), \(p_ y(i,j)\), \(p_{xy}(i,j)\), \(i=1,...,m\), \(j=1,...,n\) where \(p(i,j)=p(x_ i,y_ j)\). We shall use the notation \(h_ i=x_{i+1}-x_ i\), \(k_ j=y_{j+1}-y_ j\), \(\Delta_{1,j}(x)=(p(x,y_{j+1})-p(x,y_ j))/k_ j\), \(j=1,...,n-1\), \(\Delta_{2,i}(y)=(p(x_{i+1},y)-p(x_ i,y))/h_ i\), \(i=1,...,m-1\). In this paper are given sufficient conditions for monotonic piecewise bicubic Hermite surfaces: \[ (1)\quad -s_ 1a_{ij}/h_ i\leq s_ 1(p_ x(i,j+1)-p_ x(i,j))\leq s_ 1a_{ij}/h_{i-1}\quad where \] \[ a_{ij}=3k_ j\Delta_{1,j}(x_ i),\quad i=1,...,m,\quad j=1,...,n-1. \] \[ (2)\quad -s_ 2b_{ij}/k_ j\leq s_ 2(p_ y(i+1,j)-p_ y(i,j))\leq s_ 2b_{ij}/k_{j-1}\quad where \] \[ b_{ij}=3h_ i\Delta_{2,i}(y_ j),\quad i=1,...,m- 1,\quad j=1,...,n \] and an algorithm to provide a more reasonable interpolant without expending an excessive amount of time. The procedure employed is a five step process. Step 1: Verify that the data are monotone; compute \(s_ 1\) and \(s_ 2\). Step 2: Computer initial values for \(p_ x(i,j)\) and \(p_ y(i,j)\), \(i=1,...,m\), \(j=1,...,n\). Step 3: Check the sufficient conditions (1) and (2) that \(\Delta_{1,j}(x)\) and \(\Delta_{2,i}(y)\) don't change sign. Step 4: Construct an interval of acceptable values for \(p_{xy}(i,j)\). Step 5: Computer \(p_{xy}(i,j)\), \(i=1,...,m\), \(j=1,...,n\). This algorithm has one drawback: It is not symmetric with respect to the independent variables. This lack of symmetry may occur in Step 4, and only happens when \(p_ x\) and \(p_ y\) are both changed at the same point. The authors conjecture that this undesirable feature might be eliminated by a different approach, based on linear programming technology.
scientific article; zbMATH DE number 3921798