matlab - 在没有内置 MATLAB 函数的情况下构造具有自然边界条件的三次样条

Question

作为项目的一部分，我必须在不使用任何内置 MATLAB 函数（如 spline 或 csape）的情况下构建具有自然边界条件的三次样条。我尝试编写以下功能。

虽然我很确定它在计算系数 q 时是正确的，但我无法弄清楚我最终将如何得到实际的三次多项式。我现在在调用函数时得到的输出是 S 的 9 个不同值。

任何帮助或提示将不胜感激。

function S=cubic_s(x,y)

n=length(x);
%construction of the tri-diagonal matrix 
for j=1:n
    V(j,1)=1;
    V(j,2)=4;
    V(j,3)=1;
end
%the first row should be (1,0,...,0) and the last (0,0,...,0,1)
V(1,2)=1; V(n,2)=1; V(2,3)=0; V(n-1,1)=0;  
d=[-1 0 1];
A=spdiags(V,d,n,n);

%construction of the vector b
b=zeros(n,1);
%the first and last elements of b must equal 0
b(1)=0; b(n)=0;
%distance between two consecutive points 
h=(x(n)-x(1))/(n-1);
for j=2:n-1
    b(j,1)=(6/h^2)*(y(j+1)-2*y(j)+y(j-1));
end

%solving for the coefficients q
q=A\b;

%finding the polynomials with the formula for the cubic spline
for j=1:n-1
    for z=x(j):0.01:x(j+1)
        S(j)=(q(j)/(6*h))*(x(j+1)-z)^3+(q(j+1)/(6*h))*(z-x(j))^3+(z-x(j))* (y(j+1)/h-(q(j+1)*h)/6)+(x(j+1)-z)*(y(j)/h-(q(j)*h)/6);
    end
end

标签： matlabfunctionnumerical-methodssplinecubic-spline

Answer 1

您应该在每个 z 时间保存 S，请参见下面的图片和代码

function plot_spline
x = (0:10);
y = [1 4 3 7 1 5 2 1 6 2 3];
xx = x(1):0.01:x(2);


[XX,YY]=cubic_s(x,y);
plot(x,y,'*r', XX,YY,'-k')


function [XX,YY]=cubic_s(x,y)

n=length(x);
%construction of the tri-diagonal matrix 
for j=1:n
    V(j,1)=1;
    V(j,2)=4;
    V(j,3)=1;
end
%the first row should be (1,0,...,0) and the last (0,0,...,0,1)
V(1,2)=1; V(n,2)=1; V(2,3)=0; V(n-1,1)=0;  
d=[-1 0 1];
A=spdiags(V,d,n,n);

%construction of the vector b
b=zeros(n,1);
%the first and last elements of b must equal 0
b(1)=0; b(n)=0;
%distance between two consecutive points 
h=(x(n)-x(1))/(n-1);
for j=2:n-1
    b(j,1)=(6/h^2)*(y(j+1)-2*y(j)+y(j-1));
end

%solving for the coefficients q
q=A\b;

%finding the polynomials with the formula for the cubic spline
enum = 1;
for j=1:n-1
    for z=x(j):0.01:x(j+1)
        YY(enum)=(q(j)/(6*h))*(x(j+1)-z)^3+(q(j+1)/(6*h))*(z-x(j))^3+(z-x(j))* (y(j+1)/h-(q(j+1)*h)/6)+(x(j+1)-z)*(y(j)/h-(q(j)*h)/6);
        XX(enum)=z;
        enum = enum+1;
    end
end