1、Broyden方法求解非线性方程组的Matlab实现 注：matlab代码来自网络，仅供学习参考。1. 把以下代码复制在一个.m文件上function sol, it_hist, ierr = brsola(x,f,tol, parms)% Broydens Method solver, globally convergent% solver for f(x) = 0, Armijo rule, one vector storage% This code comes with no guarantee or warranty of any kind.% function sol, it_his
2、t, ierr = brsola(x,f,tol,parms)% inputs:% initial iterate = x% function = f% tol = atol, rtol relative/absolute% error tolerances for the nonlinear iteration% parms = maxit, maxdim% maxit = maxmium number of nonlinear iterations% default = 40% maxdim = maximum number of Broyden iterations% before re
3、start, so maxdim-1 vectors are % stored% default = 40% output:% sol = solution% it_hist(maxit,3) = scaled l2 norms of nonlinear residuals% for the iteration, number function evaluations,% and number of steplength reductions% ierr = 0 upon successful termination% ierr = 1 if after maxit iterations% t
4、he termination criterion is not satsified.% ierr = 2 failure in the line search. The iteration% is terminated if too many steplength reductions% are taken.% internal parameter:% debug = turns on/off iteration statistics display as% the iteration progresses% alpha = 1.d-4, parameter to measure suffic
5、ient decrease% maxarm = 10, maximum number of steplength reductions before% failure is reported % set the debug parameter, 1 turns display on, otherwise off%debug=1;% initialize it_hist, ierr, and set the iteration parameters%ierr = 0; maxit=40; maxdim=39; it_histx=zeros(maxit,3);maxarm=10;%if nargi
6、n = 4 maxit=parms(1); maxdim=parms(2)-1; endrtol=tol(2); atol=tol(1); n = length(x); fnrm=1; itc=0; nbroy=0;% evaluate f at the initial iterate% compute the stop tolerance%f0=feval(f,x);fc=f0;fnrm=norm(f0)/sqrt(n);it_hist(itc+1)=fnrm;it_histx(itc+1,1)=fnrm; it_histx(itc+1,2)=0; it_histx(itc+1,3)=0;f
7、nrmo=1;stop_tol=atol + rtol*fnrm;outstat(itc+1, :) = itc fnrm 0 0;% terminate on entry?%if fnrm stop_tol sol=x; returnend% initialize the iteration history storage matrices%stp=zeros(n,maxdim);stp_nrm=zeros(maxdim,1);lam_rec=ones(maxdim,1);% Set the initial step to -F, compute the step norm%lambda=1
8、;stp(:,1) = -fc;stp_nrm(1)=stp(:,1)*stp(:,1);% main iteration loop%while(itc = (1 - lambda*alpha)*fnrmo & iarm maxarm% lambda=lambda*lrat; if iarm=0 lambda=lambda*lrat; else lambda=parab3p(lamc, lamm, ff0, ffc, ffm); end lamm=lamc; ffm=ffc; lamc=lambda; x = xold + lambda*stp(:,nbroy); fc=feval(f,x);
9、 fnrm=norm(fc)/sqrt(n); ffc=fnrm*fnrm; iarm=iarm+1; end% set error flag and return on failure of the line search% if iarm = maxarm disp(Line search failure in brsola ) ierr=2; it_hist=it_histx(1:itc+1,:); sol=xold; return; end% How many function evaluations did this iteration require?% it_histx(itc+
10、1,1)=fnrm; it_histx(itc+1,2)=it_histx(itc,2)+iarm+1; if(itc = 1) it_histx(itc+1,2) = it_histx(itc+1,2)+1; end; it_histx(itc+1,3)=iarm;% terminate?% if fnrm stop_tol sol=x; rat=fnrm/fnrmo; outstat(itc+1, :) = itc fnrm iarm rat; it_hist=it_histx(1:itc+1,:);% it_hist(itc+1)=fnrm; if debug=1 disp(outsta
11、t(itc+1,:) end return end% modify the step and step norm if needed to reflect the line % search% lam_rec(nbroy)=lambda; if lambda = 1 stp(:,nbroy)=lambda*stp(:,nbroy); stp_nrm(nbroy)=lambda*lambda*stp_nrm(nbroy); end% it_hist(itc+1)=fnrm; rat=fnrm/fnrmo; outstat(itc+1, :) = itc fnrm iarm rat; if deb
12、ug=1 disp(outstat(itc+1,:) end% if theres room, compute the next search direction and step norm and% add to the iteration history % if nbroy 1 for kbr = 1:nbroy-1 ztmp=stp(:,kbr+1)/lam_rec(kbr+1); ztmp=ztmp+(1 - 1/lam_rec(kbr)*stp(:,kbr); ztmp=ztmp*lam_rec(kbr); z=z+ztmp*(stp(:,kbr)*z)/stp_nrm(kbr);
13、 end end% store the new search direction and its norm% a2=-lam_rec(nbroy)/stp_nrm(nbroy); a1=1 - lam_rec(nbroy); zz=stp(:,nbroy)*z; a3=a1*zz/stp_nrm(nbroy); a4=1+a2*zz; stp(:,nbroy+1)=(z-a3*stp(:,nbroy)/a4; stp_nrm(nbroy+1)=stp(:,nbroy+1)*stp(:,nbroy+1);% else% out of room, time to restart% stp(:,1)
14、=-fc; stp_nrm(1)=stp(:,1)*stp(:,1); nbroy=0;% end% end whileend% Were not supposed to be here, weve taken the maximum% number of iterations and not terminated.%sol=x;it_hist=it_histx(1:itc+1,:);ierr=1;if debug=1 disp( outstat)end function lambdap = parab3p(lambdac, lambdam, ff0, ffc, ffm)% Apply thr
15、ee-point safeguarded parabolic model for a line search.% This code comes with no guarantee or warranty of any kind.% function lambdap = parab3p(lambdac, lambdam, ff0, ffc, ffm)% input:% lambdac = current steplength% lambdam = previous steplength% ff0 = value of | F(x_c) |2% ffc = value of | F(x_c +
16、lambdac d) |2% ffm = value of | F(x_c + lambdam d) |2% output:% lambdap = new value of lambda given parabolic model% internal parameters:% sigma0 = .1, sigma1=.5, safeguarding bounds for the linesearch% % set internal parameters%sigma0=.1; sigma1=.5;% compute coefficients of interpolation polynomial
17、% p(lambda) = ff0 + (c1 lambda + c2 lambda2)/d1% d1 = (lambdac - lambdam)*lambdac*lambdam 0 we have negative curvature and default to% lambdap = sigam1 * lambda%c2 = lambdam*(ffc-ff0)-lambdac*(ffm-ff0);if c2 = 0 lambdap = sigma1*lambdac; returnendc1=lambdac*lambdac*(ffm-ff0)-lambdam*lambdam*(ffc-ff0
18、);lambdap=-c1*.5/c2;if (lambdap sigma1*lambdac) lambdap=sigma1*lambdac; end2. 应用举例把以下代码复制在command 窗口中x=1 2 3;f=(x)3*x(1)-cos(x(2)*x(3)-1/2;x(1)2-81*(x(2)+0.1)2+sin(x(3)+1.06;exp(-x(1)*x(2)+20*x(3)+(10*pi-3)/3;tol=3,-5;sol, it_hist, ierr = brsola(x,f,tol)说明：以上应用举例只是给出了上文中代码的一个应用实例，具体能否得到方程的满意数值解还需要进一步调节初始给的x和tol的值。
