我需要解决七个差异系统。通过fmin研究相对于实验值的最小差异，找出最佳系数集（ k 的8个值和 y0 的7个值）。
function dY = KIN241(t,y,k)
%
% version 2: direct formation of intermediates of the type {L}2M from MR3 and L (ligand)
% MR3 + 2L --> {L}2M
% simultaneous formation of 4 intermediates A, B, C and D in four parallel reactions,
% followed by 4 reactions to afford only one product P
%
% dL/dt = -2*k_tot*MR3*(L)^2, where k_tot = k_LA + k_LB + k_LC + k_LD
% dMR3/dt = -k_tot*MR3*(L)^2
% with :
% Variables: x(1) = L, x(2) = MR3
% Parameters: B(1) = k_LA, B(2) = k_LB, B(3) = k_LC, B(4) = k_LD
%
% dP/dt = k_AP*A + k_BP*B + k_CP*C + k_DP*D
% dA/dt = 2*k_LA*MR3*(L)^2 - k_AP*A
% dB/dt = 2*k_LB*MR3*(L)^2 - k_BP*B
% dC/dt = 2*k_LC*MR3*(L)^2 - k_CP*C
% dD/dt = 2*k_LA*MR3*(L)^2 - k_DP*D
%
% with :
% Variables: y(3) = P, y(4) = A, y(5) = B, y(6) = C, y(7) = D
% Parameters: B(5) = k_AP, B(6) = k_BP, B(7) = k_CP, B(8) = k_DP
% dL = dY(1);
% dMR3 = dY(2);
% dP = dY(3);
% dA = dY(4);
% dB = dY(5);
% dC = dY(6);
% dD = dY(7);
%y = indata;
%B = rand(1,8);
%disp(B);
k_LA = k(1);
k_LB = k(2);
k_LC = k(3);
k_LD = k(4);
k_AP = k(5);
k_BP = k(6);
k_CP = k(7);
k_DP = k(8);
%
dY(1) = -2.*(k_LA + k_LB + k_LC + k_LD).*y(2).*(y(1))^2;
dY(2) = -(k_LA + k_LB + k_LC + k_LD).*y(2).*(y(1))^2;
dY(3) = k_AP.*y(4) + k_BP.*y(5) + k_CP.*y(6) + k_DP.*y(7);
dY(4) = 2.*k_LA.*y(2).*(y(1))^2 - k_AP.*y(4);
dY(5) = 2.*k_LB.*y(2).*(y(1))^2 - k_BP.*y(5);
dY(6) = 2.*k_LC.*y(2).*(y(1))^2 - k_CP.*y(6);
dY(7) = 2.*k_LD.*y(2).*(y(1))^2 - k_DP.*y(7);
dY = [dY(1); dY(2); dY(3); dY(4); dY(5); dY(6); dY(7)];
调用ode的函数如下：
function SSE = SSE_minimization(tspan_inp,val_exp,par)
k_inp = par{1};
y0_inp = par{2};
f = @(Tt,Ty) KIN241(Tt,Ty,k_inp); %function to call ode45
%size_limit = length(k_inp);
%options = odeset('NonNegative',1:size_limit,'RelTol',1e-8,'AbsTol',[1e-8 1e-8 1e-8 1e-8 1e-8 1e-8 1e-8]);
[ts,val_theo] = ode45(f, tspan_inp, y0_inp); %Cexp is the state variable predicted by the model
err = val_exp - val_theo;
SSE = sum(sum(err.^2)); %sum squared-error
实验数据在那里：
Time_M =
0
60
381
702
1023
1344
1665
1986
2307
2628
2949
3270
3591
3912
4233
4554
4875
5196
5517
5838
6159
6480
6801
7122
7443
7764
8085
和
浓度_M =
0.0385 0.0224 0.0385 0.0385 0.0385 0.0385 0.0064
0.0384 0.0343 0.0381 0.0351 0.0359 0.0262 0.0301
0.0374 0.0377 0.0251 0.0207 0.0377 0.0348 0.0369
0.0364 0.0379 0.0184 0.0264 0.0376 0.0366 0.0373
0.0358 0.0380 0.0163 0.0291 0.0375 0.0371 0.0375
0.0353 0.0380 0.0162 0.0300 0.0374 0.0373 0.0376
0.0348 0.0381 0.0166 0.0305 0.0374 0.0374 0.0376
0.0344 0.0381 0.0170 0.0312 0.0373 0.0374 0.0377
0.0339 0.0381 0.0177 0.0312 0.0372 0.0376 0.0378
0.0336 0.0382 0.0184 0.0317 0.0372 0.0375 0.0378
0.0331 0.0382 0.0190 0.0317 0.0372 0.0376 0.0379
0.0327 0.0382 0.0194 0.0319 0.0372 0.0376 0.0379
0.0324 0.0382 0.0201 0.0322 0.0372 0.0376 0.0379
0.0320 0.0382 0.0206 0.0324 0.0371 0.0377 0.0379
0.0317 0.0382 0.0212 0.0327 0.0371 0.0376 0.0380
0.0313 0.0382 0.0217 0.0328 0.0371 0.0377 0.0380
0.0309 0.0382 0.0222 0.0329 0.0371 0.0377 0.0380
0.0307 0.0382 0.0228 0.0333 0.0371 0.0377 0.0380
0.0304 0.0383 0.0233 0.0335 0.0371 0.0377 0.0381
0.0301 0.0383 0.0237 0.0337 0.0371 0.0377 0.0381
0.0299 0.0383 0.0242 0.0339 0.0371 0.0378 0.0381
0.0296 0.0383 0.0248 0.0341 0.0371 0.0378 0.0381
0.0291 0.0383 0.0250 0.0336 0.0370 0.0378 0.0381
0.0288 0.0383 0.0255 0.0339 0.0370 0.0382 0.0381
0.0285 0.0383 0.0258 0.0340 0.0370 0.0378 0.0382
0.0283 0.0383 0.0262 0.0341 0.0370 0.0378 0.0382
0.0283 0.0383 0.0268 0.0346 0.0370 0.0379 0.0381
当我从主代码打电话时：
y0 = (rand(1,7))';
k0 = (rand(1,8))';
par0{1} = k0; par0{2} = y0;
[solution, fval, exitflag, output] = fminsearch(@(par) SSE_minimization(tspan,Concentration_M,par0),y0);
k_opt = solution(1);
y0 = solution(2);
一个周期后，我收到以下错误：
Attempted to access k(2); index out of bounds because numel(k)=1.
Error in KIN241 (line 34)
k_LB = k(2);
实际上，代码只找到k_opt的一个值而不是八个，而y0只找到一个而不是七个......
请您帮我设置/更正代码，以优化此系统方程的k和y0参数。
提前多多谢谢！