三维表面的双三次插值

Question

我正在用matlab编写一个算法，用于表面Psi(x,y)的双三次插值。我在代码中有一个错误，似乎无法跟踪它。我正在尝试使用Psi=X^2-0.25的测试用例，以便更容易找到错误。好像我的插值有一个偏移量。我的评论包含在我的代码中。任何帮助将不胜感激。

蓝色的Psi=X^2图和红色的插值图

绘制Psi的计数行，红点是我计算插值的点。粗红线是插值，它与红点相差很多。

function main()

    epsilon=0.000001;
    xMin=-1+epsilon;
    xMax= 1+epsilon;
    yMin=-1+epsilon;
    yMax= 1+epsilon;

    dx=0.1;  Nx=ceil((xMax-xMin)/dx)+1;
    dy=0.1;  Ny=ceil((yMax-yMin)/dy)+1;

    x=xMin:dx:xMax;  x=x(1:Nx);
    y=yMin:dy:yMax;  y=y(1:Ny);

    [XPolInX,XPolInY]=GetGhostMatricies(Nx,Ny); %Linear extrapolation matrix
    [D0x,D0y]=GetDiffMatricies(Nx,Ny,dx,dy); %derivative matricies: D0x is central differencing in x

    [X,Y]=meshgrid(x,y);
    Psi=X.^2-0.25;    %Note that my algorithm is being written for a Psi that may not have a analytic representation. This Psi is only a test case. 

    psi=zeros(Nx+2,Ny+2);   %linearly extrapolate psi (for solving differential equation not shown here)
    psi(2:(Nx+1),2:(Ny+1))=Psi';
    psi=(XPolInY*(XPolInX*psi)')';

    %compute derivatives of psi
    psi_x =D0x*psi;             psi_x =(XPolInY*(XPolInX*psi_x)')';
    psi_y =(D0y*psi')';         psi_y =(XPolInY*(XPolInX*psi_y)')';
    psi_xy=D0x*psi_y;           psi_xy=(XPolInY*(XPolInX*psi_xy)')'; 
    % i have verified that my derivatives are computed correctly    

    biCubInv=GetBiCubicInverse(dx,dy);  

    i=5;  %lets compute the bicubic interpolation at this x(i), y(j)
    j=1;

    psiVoxel=[psi(   i,j),psi(   i+1,j),psi(   i,j+1),psi(   i+1,j+1),...
              psi_x( i,j),psi_x( i+1,j),psi_x( i,j+1),psi_x( i+1,j+1),...
              psi_y( i,j),psi_y( i+1,j),psi_y( i,j+1),psi_y( i+1,j+1),...
              psi_xy(i,j),psi_xy(i+1,j),psi_xy(i,j+1),psi_xy(i+1,j+1)]';

    a=biCubInv*psiVoxel; %a=[a00 a01 ... a33]; polynomial coefficients; 1st index is power of (x-xi), 2nd index is power of (y-yj)

    xi=x(5); yj=y(1);
    clear x y
    x=(xi-.2):.01:(xi+.2);   %this is a local region about the point we are interpolating
    y=(yj-.2):.01:(yj+.2);
    [dX,dY]=meshgrid(x,y);
    Psi=dX.^2-0.25;   

    figure(2)            %just plotting the 0 level contour of Psi here 
    plot(xi,yj,'.r','MarkerSize',20)
    hold on
    contour(x,y,Psi,[0 0],'r','LineWidth',2)
    set(gca,'FontSize',14)
    axis([x(1) x(end) y(1) y(end)])
    grid on
    set(gca,'xtick',(xi-.2):.1:(xi+.2));
    set(gca,'ytick',(yj-.2):.1:(yj+.2));
    xlabel('x')
    ylabel('y')

    [dX dY]=meshgrid(x-xi,y-yj);
    %P is my interpolating polynomial
    P =   a(1)       + a(5)       *dY + a(9)        *dY.^2 + a(13)       *dY.^3 ...
        + a(2)*dX    + a(6)*dX   .*dY + a(10)*dX   .*dY.^2 + a(14)*dX   .*dY.^3 ...
        + a(3)*dX.^2 + a(7)*dX.^2.*dY + a(11)*dX.^2.*dY.^2 + a(15)*dX.^2.*dY.^3 ...
        + a(4)*dX.^3 + a(8)*dX.^3.*dY + a(12)*dX.^3.*dY.^2 + a(16)*dX.^3.*dY.^3 ;
    [c h]=contour(x,y,P)  
    clabel(c,h)

    figure(3)
    plot(x,x.^2-.25)  %this is the exact function
    hold on
    plot(x,P(1,:),'-r*')

    %See there is some offset here 


end
%-------------------------------------------------------------------------
function [XPolInX,XPolInY]=GetGhostMatricies(Nx,Ny)

    XPolInX=diag(ones(1,Nx+2),0);
    XPolInY=diag(ones(1,Ny+2),0);

    XPolInX(1,1)      =0; XPolInX(1,2)      =2; XPolInX(1,3)    =-1;
    XPolInY(1,1)      =0; XPolInY(1,2)      =2; XPolInY(1,3)    =-1;
    XPolInX(Nx+2,Nx+2)=0; XPolInX(Nx+2,Nx+1)=2; XPolInX(Nx+2,Nx)=-1;
    XPolInY(Ny+2,Ny+2)=0; XPolInY(Ny+2,Ny+1)=2; XPolInY(Ny+2,Ny)=-1;

    fprintf('Done GetGhostMatricies\n')
end
%-------------------------------------------------------------------------
function [D0x,D0y]=GetDiffMatricies(Nx,Ny,dx,dy)

    D0x=diag(ones(1,Nx-1),1)-diag(ones(1,Nx-1),-1); 
    D0y=diag(ones(1,Ny-1),1)-diag(ones(1,Ny-1),-1);
    D0x(1,1)=-3; D0x(1,2)=4; D0x(1,3)=-1;
    D0y(1,1)=-3; D0y(1,2)=4; D0y(1,3)=-1;
    D0x(Nx,Nx)=3; D0x(Nx,Nx-1)=-4; D0x(Nx,Nx-2)=1;
    D0y(Ny,Ny)=3; D0y(Ny,Ny-1)=-4; D0y(Ny,Ny-2)=1;

    %pad with ghost cells which are simply zeros
    tmp=D0x;     D0x=zeros(Nx+2,Nx+2);   D0x(2:(Nx+1),2:(Nx+1))=tmp; tmp=0;
    tmp=D0y;     D0y=zeros(Ny+2,Ny+2);   D0y(2:(Ny+1),2:(Ny+1))=tmp; tmp=0;

    %scale appropriatley by dx & dy
    D0x=D0x/(2*dx);
    D0y=D0y/(2*dy);

end
%-------------------------------------------------------------------------
function biCubInv=GetBiCubicInverse(dx,dy)
    %p(x,y)=a00+a01(x-xi)+a02(x-xi)^2+...+a33(x-xi)^3(y-yj)^3
    %biCubic*a=[psi(i,j) psi(i+1,j) psi(i,j+1) psi(i+1,j+1) psi_x(i,j) ... psi_y(i,j) ... psi_xy(i,j) ... psi_xy(i+1,j+1)] 
    %here, psi_x is the x derivative of psi
    %I verified that this matrix is correct by setting dx=dy=1 and comparing to the inverse here http://en.wikipedia.org/wiki/Bicubic_interpolation

    biCubic=[ 
       %00      10      20      30      01      11      21      31        02      12      22        32          03        13        23          33
        1       0       0       0       0       0       0       0         0       0       0         0           0         0         0           0;
        1       dx      dx^2    dx^3    0       0       0       0         0       0       0         0           0         0         0           0;
        1       0       0       0       dy      0       0       0         dy^2    0       0         0           dy^3      0         0           0;
        1       dx      dx^2    dx^3    dy      dx*dy   dx^2*dy dx^3*dy   dy^2    dx*dy^2 dx^2*dy^2 dx^3*dy^2   dy^3      dx*dy^3   dx^2*dy^3   dx^3*dy^3;

        0       1       0       0       0       0       0       0         0       0       0         0           0         0         0           0;
        0       1       2*dx    3*dx^2  0       0       0       0         0       0       0         0           0         0         0           0;
        0       1       0       0       0       dy      0       0         0       dy^2    0         0           0         dy^3      0           0;
        0       1       2*dx    3*dx^2  0       dy      2*dx*dy 3*dx^2*dy 0       dy^2    2*dx*dy^2 3*dx^2*dy^2 0         dy^3      2*dx*dy^3   3*dx^2*dy^3;

        0       0       0       0       1       0       0       0         0       0       0         0           0         0         0           0;
        0       0       0       0       1       dx      dx^2    dx^3      0       0       0         0           0         0         0           0;
        0       0       0       0       1       0       0       0         2*dy    0       0         0           3*dy^2    0         0           0;
        0       0       0       0       1       dx      dx^2    dx^3      2*dy    2*dx*dy 2*dx^2*dy 2*dx^3*dy   3*dy^2    3*dx*dy^2 3*dx^2*dy^2 3*dx^3*dy^2;

        0       0       0       0       0       1       0       0         0       0       0         0           0         0         0           0;
        0       0       0       0       0       1       2*dx    3*dx^2    0       0       0         0           0         0         0           0;
        0       0       0       0       0       1       0       0         0       2*dy    0         0           0         3*dy^2    0           0;
        0       0       0       0       0       1       2*dx    3*dx^2    0       2*dy    4*dx*dy   6*dx^2*dy   0         3*dy^2    6*dx*dy^2   9*dx^2*dy^2];   

    biCubInv=inv(biCubic);

end

%-------------------------------------------------------------------------

Answer 1

我发现了我的错误。我用鬼细胞填充我的基质，但是我忘了现在i Psi中没有鬼影的i+1在psi P中有幽灵细胞。因此，我应该在xi=x(6); yj=y(2)评估我的插值多项​​式xi=x(5); yj=y(1)，而不是{{1}}。