非线性方程组的边值问题

szwstar

求助方程组的边值问题
function [T,Z]=rks41(F,a,b,Za,M)

%Input  -F is the system input as a string 'F'
%       -a and b are the end points of the interval
%       -Za=[x(a) y(a)] are the initial conditions
%       -M is the number of steps
%Output -T is the vector of steps
%       -Z=[x1(t) ...xn(t)]; where xk(t) is the approximation
%        to the kth dependent variable

h=(b-a)/M;
T=zeros(1,M+1);
Z=zeros(M+1,length(Za));
T=a:h:b;
Z(1,:)=Za;

for j=1:M
    k1=h*feval(F,T(j),Z(j,:));
    k2=h*feval(F,T(j)+h/2,Z(j,:)+k1/2);
    k3=h*feval(F,T(j)+h/2,Z(j,:)+k2/2);
    k4=h*feval(F,T(j)+h,Z(j,:)+k3);
    Z(j+1,:)=Z(j,:)+(k1+2*k2+2*k3+k4)/6;
   
end

function Z=F1(x,Z)
y1=Z(1);
y2=Z(2);
y3=Z(3);
y4=Z(4);
Z=[y2,-2*y1*y3-2*3*cos(4*x)*y1+2*1.2*y1,y4,0.5*(y3-y1^2)];
%Z=[y2;-2*y1+3*y2;y4;-2*y3+3*y4];

function Z=F2(x,Z)
y1=Z(1);
y2=Z(2);
y3=Z(3);
y4=Z(4);
Z=[y2,-2*y1*y3-2*3*cos(4*x)*y1+2*1.2*y1,y4,0.5*(y3)];
%Z=[y2;x-2*y1+3*y2;y4;x-2*y3+3*y4];

function L=linsht1(F1,F2,a,b,alpha1,alpha2,beta1,beta2,M)

%线性打靶法 求解非线性方程组的边值问题
%Input     -F1 and F2 are the systems of first-order equations
%           representing the I.V.P.'s (1) and (2), respectively;
%           设w(t),n(t)为边值问题
%           w''(t)=-2*w(t)*n(t)-2*3*cos(4*t)*w(t)+2*1.2*w(t), w(a)=0, w(b)=0 (1)
%           n''(t)=0.5*(n(t)-w(t)^2),      n(a)=0, n(b)=0     (2)
%           
%          input as strings 'F1','F2'
%          - a and b are the end points of the interval
%          -alpha=x(a) and beta=x(b); boundary conditions
%          -M is the number of steps
% Output   -L=[T' X]; where T' is the (M+1)x1 vector of ordinates

%Solve the system F1

Za=[alpha1,0,alpha2,0];
[T,Z]=rks41(F1,a,b,Za,M);
U1=Z(:,1);
U2=Z(:,3);

%Solve the sysstem F2

Za=[0,1,0,1];
[T,Z]=rks41(F2,a,b,Za,M);
V1=Z(:,1);
V2=Z(:,3);

%Calculate the solution to the boundary value problem
X1=U1+(beta1-U1(M+1))*V1/V1(M+1);
X2=U2+(beta2-U2(M+1))*V2/V2(M+1);

%L1=[T' U1 U2];
L=[T' X1 X2];

程序运行无错误，可是结果不正确，大家帮忙看看有什么问题，如有做过类似程序，能否提供参考（szwstar@163.com），谢谢