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hustesewzw

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[求助] 求助matlab高手指點：大型非線性方程組（牛頓拉佛森法）已有3人參與

最近想完成一個生化模型問題，模型主體方程為微分方程組，用差分法轉(zhuǎn)化為多元非線性方程組。擬采用牛頓拉佛森法求解，方程組比較大：99個未知數(shù)，99個方程，matlab程序已寫好，問題是求解到最后有部分方程的y值始終比較大（我設(shè)置的是1e-4），x的收斂條件也為1e-4（可以滿足）。jacobi迭代次數(shù)設(shè)置為30次。請問我的y值不能滿足條件的原因是什么呢？是我的初始值取得有問題嗎？有沒有高手為我指點一下迷津，在此深表感謝�。�！獻上我的全部金幣。

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★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★
感謝參與，應(yīng)助指數(shù) +1
hustesewzw: 金幣+30, ★★★很有幫助 2014-10-08 12:06:16

分享一個NewtonRaphson的算法給你，

function [x, resnorm, F, exitflag, output, jacob] = newtonraphson(fun, x0, options)
% NEWTONRAPHSON Solve set of non-linear equations using Newton-Raphson method.
%
% [X, RESNORM, F, EXITFLAG, OUTPUT, JACOB] = NEWTONRAPHSON(FUN, X0, OPTIONS)
% FUN is a function handle that returns a vector of residuals equations, F,
% and takes a vector, x, as its only argument. When the equations are
% solved by x, then F(x) == zeros(size(F(

, 1)).
%
% Optionally FUN may return the Jacobian, Jij = dFi/dxj, as an additional
% output. The Jacobian must have the same number of rows as F and the same
% number of columns as x. The columns of the Jacobians correspond to d/dxj and
% the rows correspond to dFi/d.
%
% EG: J23 = dF2/dx3 is the 2nd row ad 3rd column.
%
% If FUN only returns one output, then J is estimated using a center
% difference approximation,
%
% Jij = dFi/dxj = (Fi(xj + dx) - Fi(xj - dx))/2/dx.
%
% NOTE: If the Jacobian is not square the system is either over or under
% constrained.
%
% X0 is a vector of initial guesses.
%
% OPTIONS is a structure of solver options created using OPTIMSET.
% EG: options = optimset('TolX', 0.001).
%
% The following options can be set:
% * OPTIONS.TOLFUN is the maximum tolerance of the norm of the residuals.
% [1e-6]
% * OPTIONS.TOLX is the minimum tolerance of the relative maximum stepsize.
% [1e-6]
% * OPTIONS.MAXITER is the maximum number of iterations before giving up.
% [100]
% * OPTIONS.DISPLAY sets the level of display: {'off', 'iter'}.
% ['iter']
%
% X is the solution that solves the set of equations within the given tolerance.
% RESNORM is norm(F) and F is F(X). EXITFLAG is an integer that corresponds to
% the output conditions, OUTPUT is a structure containing the number of
% iterations, the final stepsize and exitflag message and JACOB is the J(X).
%
% See also OPTIMSET, OPTIMGET, FMINSEARCH, FZERO, FMINBND, FSOLVE, LSQNONLIN
%
%% initialize
% There are no argument checks!
x0 = x0(

; % needs to be a column vector
% set default options
oldopts = optimset( ...
'TolX', 1e-12, 'TolFun', 1e-6, 'MaxIter', 100, 'Display', 'iter');
if nargin<3
options = oldopts; % use defaults
else
options = optimset(oldopts, options); % update default with user options
end
FUN = @(x)funwrapper(fun, x); % wrap FUN so it always returns J
%% get options
TOLX = optimget(options, 'TolX'); % relative max step tolerance
TOLFUN = optimget(options, 'TolFun'); % function tolerance
MAXITER = optimget(options, 'MaxIter'); % max number of iterations
DISPLAY = strcmpi('iter', optimget(options, 'Display')); % display iterations
TYPX = max(abs(x0), 1); % x scaling value, remove zeros
ALPHA = 1e-4; % criteria for decrease
MIN_LAMBDA = 0.1; % min lambda
MAX_LAMBDA = 0.5; % max lambda
%% set scaling values
% TODO: let user set weights
weight = ones(numel(FUN(x0)),1);
J0 = weight*(1./TYPX'); % Jacobian scaling matrix
%% set display
if DISPLAY
fprintf('\n%10s %10s %10s %10s %10s %12s\n', 'Niter', 'resnorm', 'stepnorm', ...
'lambda', 'rcond', 'convergence')
for n = 1:67,fprintf('-'),end,fprintf('\n')
fmtstr = '%10d %10.4g %10.4g %10.4g %10.4g %12.4g\n';
printout = @(n, r, s, l, rc, c)fprintf(fmtstr, n, r, s, l, rc, c);
end
%% check initial guess
x = x0; % initial guess
[F, J] = FUN(x); % evaluate initial guess
Jstar = J./J0; % scale Jacobian
if any(isnan(Jstar(

)) || any(isinf(Jstar(

))
exitflag = -1; % matrix may be singular
else
exitflag = 1; % normal exit
end
if issparse(Jstar)
rc = 1/condest(Jstar);
else
if any(isnan(Jstar(

))
rc = NaN;
elseif any(isinf(Jstar(

))
      rc = Inf;
else
      rc = 1/cond(Jstar); % reciprocal condition
end
end
resnorm = norm(F); % calculate norm of the residuals
dx = zeros(size(x0));convergence = Inf; % dummy values
%% solver
Niter = 0; % start counter
lambda = 1; % backtracking
if DISPLAY,printout(Niter, resnorm, norm(dx), lambda, rc, convergence);end
while (resnorm>TOLFUN || lambda<1) && exitflag>=0 && Niter<=MAXITER
if lambda==1
      %% Newton-Raphson solver
      Niter = Niter+1; % increment counter
      dx_star = -Jstar\F; % calculate Newton step
      % NOTE: use isnan(f) || isinf(f) instead of STPMAX
      dx = dx_star.*TYPX; % rescale x
      g = F'*Jstar; % gradient of resnorm
      slope = g*dx_star; % slope of gradient
      fold = F'*F; % objective function
      xold = x; % initial value
      lambda_min = TOLX/max(abs(dx)./max(abs(xold), 1));
end
if lambda<lambda_min
      exitflag = 2; % x is too close to XOLD
      break
elseif any(isnan(dx)) || any(isinf(dx))
      exitflag = -1; % matrix may be singular
      break
end
x = xold+dx*lambda; % next guess
[F, J] = FUN(x); % evaluate next residuals
Jstar = J./J0; % scale next Jacobian
f = F'*F; % new objective function
%% check for convergence
lambda1 = lambda; % save previous lambda
if f>fold+ALPHA*lambda*slope
      if lambda==1
         lambda = -slope/2/(f-fold-slope); % calculate lambda
      else
         A = 1/(lambda1 - lambda2);
         B = [1/lambda1^2,-1/lambda2^2;-lambda2/lambda1^2,lambda1/lambda2^2];
         C = [f-fold-lambda1*slope;f2-fold-lambda2*slope];
         coeff = num2cell(A*B*C);
         [a,b] = coeff{:};
         if a==0
            lambda = -slope/2/b;
         else
            discriminant = b^2 - 3*a*slope;
            if discriminant<0
                  lambda = MAX_LAMBDA*lambda1;
            elseif b<=0
                  lambda = (-b+sqrt(discriminant))/3/a;
            else
                  lambda = -slope/(b+sqrt(discriminant));
            end
         end
         lambda = min(lambda,MAX_LAMBDA*lambda1); % minimum step length
      end
elseif isnan(f) || isinf(f)
      % limit undefined evaluation or overflow
      lambda = MAX_LAMBDA*lambda1;
else
      lambda = 1; % fraction of Newton step
end
if lambda<1
      lambda2 = lambda1;f2 = f; % save 2nd most previous value
      lambda = max(lambda,MIN_LAMBDA*lambda1); % minimum step length
      continue
end
%% display
resnorm0 = resnorm; % old resnorm
resnorm = norm(F); % calculate new resnorm
convergence = log(resnorm0/resnorm); % calculate convergence rate
stepnorm = norm(dx); % norm of the step
if any(isnan(Jstar(

)) || any(isinf(Jstar(

))
      exitflag = -1; % matrix may be singular
      break
end
if issparse(Jstar)
      rc = 1/condest(Jstar);
else
      rc = 1/cond(Jstar); % reciprocal condition
end
if DISPLAY,printout(Niter, resnorm, stepnorm, lambda1, rc, convergence);end
end
%% output
output.iterations = Niter; % final number of iterations
output.stepsize = dx; % final stepsize
output.lambda = lambda; % final lambda
if Niter>=MAXITER
exitflag = 0;
output.message = 'Number of iterations exceeded OPTIONS.MAXITER.';
elseif exitflag==2
output.message = 'May have converged, but X is too close to XOLD.';
elseif exitflag==-1
output.message = 'Matrix may be singular. Step was NaN or Inf.';
else
output.message = 'Normal exit.';
end
jacob = J;
end

function [F, J] = funwrapper(fun, x)
% if nargout<2 use finite differences to estimate J
try
[F, J] = fun(x);
catch
F = fun(x);
J = jacobian(fun, x); % evaluate center diff if no Jacobian
end
F = F(

; % needs to be a column vector
end

function J = jacobian(fun, x)
% estimate J
dx = eps^(1/3); % finite difference delta
nx = numel(x); % degrees of freedom
nf = numel(fun(x)); % number of functions
J = zeros(nf,nx); % matrix of zeros
for n = 1:nx
% create a vector of deltas, change delta_n by dx
delta = zeros(nx, 1); delta(n) = delta(n)+dx;
dF = fun(x+delta)-fun(x-delta); % delta F
J(:, n) = dF(

/dx/2; % derivatives dF/d_n
end
end

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2樓2014-10-08 10:18:20

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zhchh008

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專業(yè): 資源化工

笑臉修改為

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3樓2014-10-08 10:19:34

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zhchh008

金蟲 (正式寫手)

應(yīng)助: 13 (小學生)
金幣: 1602.2
散金: 871
紅花: 13
帖子: 835
在線: 202.7小時
蟲號: 99456
注冊: 2005-11-11
性別: GG
專業(yè): 資源化工

笑臉修改為冒號加右括號 “

"....建議木蟲修改代碼，讓網(wǎng)頁可以識別程序代碼。

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4樓2014-10-08 10:21:17

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dingd

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【答案】應(yīng)助回帖

★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★
感謝參與，應(yīng)助指數(shù) +1
hustesewzw: 金幣+30, ★★★很有幫助 2014-10-08 12:06:05

牛頓拉佛森法是典型的局部最優(yōu)算法，對初值方程敏感。99個方程99個未知數(shù)，想給出合適的初值對一般人來說幾乎不可能。如果僅僅想得到結(jié)果，建議用1stOpt來求，全局最優(yōu)，不依賴初值，使用也很簡單。

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5樓2014-10-08 11:09:25

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hustesewzw

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引用回帖:

5樓: Originally posted by dingd at 2014-10-08 11:09:25
牛頓拉佛森法是典型的局部最優(yōu)算法，對初值方程敏感。99個方程99個未知數(shù)，想給出合適的初值對一般人來說幾乎不可能。如果僅僅想得到結(jié)果，建議用1stOpt來求，全局最優(yōu)，不依賴初值，使用也很簡單。

我目前能想到的也只有這個可能了，謝謝，另外請問您使用過1stopt這個軟件嗎，我能向您請教一些問題嗎？希望站內(nèi)聯(lián)系！

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6樓2014-10-08 12:05:03

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tutu6287

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專業(yè): 結(jié)構(gòu)工程

【答案】應(yīng)助回帖

★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★
感謝參與，應(yīng)助指數(shù) +1
hustesewzw: 金幣+20, ★★★很有幫助 2014-10-13 16:54:11

jacobi迭代整300次試試，99個未知數(shù)不算大

[ 發(fā)自手機版 http://www.gaoyang168.com/3g ]

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7樓2014-10-09 19:34:20

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qxq_6

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樓主后來解決了嗎，我現(xiàn)在遇到一個類似的問題，希望可以幫忙解答

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[求助] 求助matlab高手指點：大型非線性方程組（牛頓拉佛森法） 已有3人參與

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[求助] 求助matlab高手指點：大型非線性方程組（牛頓拉佛森法）已有3人參與

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