求壹份用fortran語言編寫的牛頓叠代求解非線性方程組的源代碼。謝謝各位。 郵箱：370147046@qq.com

t=0:tao:T;

N=length(x);

M=length(t);

f=inline(f);

g1=inline(g1);

g2=inline(g2);

u=inline(u);

r=w*tao/(h*h)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%a=0;b=1;h=0.1;tao=0.1;T=1;u='exp(-pi^2*t)*sin(pi*x)';f='sin(pi*x)';g1='0';g2='0';;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%a=0;b=1;h=0.1;tao=0.1;T=1;u='exp(x+t)';f='exp(x)';g1='exp(0+t)';

%g2='exp(1+t)';

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%for i=1:N-2

if i==1

S(i,i)=0;

S(i,i+1)=1;

else if i==N-2

S(i,i)=0;

S(i,i-1)=1;

else

S(i,i)=0;

S(i,i-1)=1;

S(i,i+1)=1;

end

end

end

%%%%%%%%%%%%%%%%%%%%%%

U=zeros(N-2,1);

for i=1:N-2

U(i,1)=f(x(i+1));

end

%%%%%%%%%%%%%%%%%%%%%%%I=eye(N-2);

tic

for i=1:M-1

b(1,1)=r*g1(t(i));

b(N-2,1)=r*g2(t(i));

b3(1,1)=r*g1(t(i+1)); b3(N-2,1)=r*g2(t(i+1));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%隱式差分格式

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

U=inv((1+2*r)*I-r*S)*U+inv((1+2*r)*I-r*S)*[b3];

% A0=((1+2*r)*I-r*S);

% b0=U+[b3]

% a0=zhuiganfa(A0,b0);

% U=a0';

end

toc

UU(1,1)=g1(t(M));

UU(N,1)=g2(t(M));

UU(2:N-1,1)=U(1:N-2,1);

jisuan=UU

for i=1:M

for j=1:N

uu(j,i)=u(t(i),x(j));

end

end

junque=uu(:,M)

E=(abs(UU-uu(:,M)));

EE=max(abs(UU-uu(:,M)))

L2=sqrt(sum(abs(UU-uu(:,M)).^2)*h)

aver=(sum(abs(jisuan-junque))/N)

plot(x,junque)

hold on

plot(x,jisuan,'-r*')