Converting symbolic matrix to numeric matrix with while loop

I am trying to convert several 1x11 symbolic matrix to a 1x11 numeric matrix.

WhenI run the code it gives the error:

Reference to non-existent field 'Ni'.

Error in Carbon_IB_HW (line 50)

solN1(1,i) = double(sol.Ni);

Does anyone know how I can define the elements in the matricies so I don't get this error? Or if there is some other way I can set this up that will avoid this problem entirely? Thanks.

The relevent code is:

clc
clear all
%Defining know values
T = linspace(10^8,10^9,11);
l10T = log10(T);
a1 = [5.91e-15 4.42e-15 3.30e-15 2.46e-15 1.83e-15 1.36e-15 1.01e-15 7.53e-16 5.60e-16 4.17e-16 2.10e-16];
a2 = [4.22e-15 3.02e-15 2.16e-15 1.55e-15 1.12e-15 8.05e-16 5.83e-16 4.24e-16 3.10e-16 2.28e-16 1.69e-16];
a3 = [4.49e-15 3.27e-15 2.39e-15 1.74e-15 1.28e-15 9.36e-16 6.89e-16 5.09e-16 3.77e-16 2.81e-16 2.11e-16];
a4 = [3.14e-14 2.25e-14 1.61e-14 1.15e-14 8.17e-15 5.82e-15 4.15e-15 2.96e-15 2.11e-15 1.50e-15 1.07e-15];
a5 = [6.95e-14 4.98e-14 3.57e-14 2.56e-14 1.84e-14 1.32e-14 9.43e-15 6.76e-15 4.84e-15 3.48e-15 2.50e-15];
a6 = [1.33e-14 1.01e-14 7.56e-15 5.65e-15 4.21e-15 3.12e-15 2.30e-15 1.69e-15 1.24e-15 9.11e-16 6.63e-16];
k = 1.38e-23;
m = 1.9944235e-26;
V = sqrt((2*k.*(T.*11600))./(m));
I1 = 10.6; A11 = 1.829; A12 = -1.975; A13 = 1.149; A14 = -3.583; A15 = 2.451;
I2 = 24.4; A21 = 8.390e-1; A22 = -7.95e-1; A23 = 3.263; A24 = -5.382; A25 = 3.476;
I3 = 41.4; A31 = 4.009e-1; A32 = -3.518e-1; A33 = 2.375; A34 = -3.992; A35 = 2.794;
I4 = 64.5; A41 = 1.350; A42 = -8.748e-1; A43 = -1.444; A44 = 2.330; A45 = -2.730;
I4b = 285; A41b = -2.777; A42b = 5.376; A43b = -8.748; A44b = 1.766e1; A45b = -9.086;
I5 = 392; A51 = 9.205e-1; A52 = -6.297e-1; A53 = 1.316; A54 = -9.156e-2; A55 = 0;
I6 = 490; A61 = 2.489e-1; A62 = 1.847e-1; A63 = 4.475e-2; A64 = -9.432e-2; A65 = 5.122e-1;
E = 6.242e18.*(.5*m.*V.^2);
%Solving for collisional cross section
Si1 = (1e-13./(I1.*E)).*((A11.*log(E./I1))+((A12.*(1-(I1./E).^1))+(A13.*(1-(I1./E).^2))+(A14.*(1-(I1./E).^3))+(A15.*(1-(I1./E).^4))));
Si2 = (1e-13./(I2.*E)).*((A21.*log(E./I2))+((A22.*(1-(I2./E).^1))+(A23.*(1-(I2./E).^2))+(A24.*(1-(I2./E).^3))+(A25.*(1-(I2./E).^4))));
Si3 = (1e-13./(I3.*E)).*((A31.*log(E./I3))+((A32.*(1-(I3./E).^1))+(A33.*(1-(I3./E).^2))+(A34.*(1-(I3./E).^3))+(A35.*(1-(I3./E).^4))));
Si4 = (1e-13./(I4.*E)).*((A41.*log(E./I4))+((A42.*(1-(I4./E).^1))+(A43.*(1-(I4./E).^2))+(A44.*(1-(I4./E).^3))+(A45.*(1-(I4./E).^4))));
Si4b = (1e-13./(I4b.*E)).*((A41b.*log(E./I4))+((A42b.*(1-(I4b./E).^1))+(A43b.*(1-(I4b./E).^2))+(A44b.*(1-(I4b./E).^3))+(A45b.*(1-(I4b./E).^4))));
Si5 = (1e-13./(I5.*E)).*((A51.*log(E./I5))+((A52.*(1-(I5./E).^1))+(A53.*(1-(I5./E).^2))+(A25.*(1-(I5./E).^3))+(A55.*(1-(I5./E).^4))));
Si6 = (1e-13./(I6.*E)).*((A61.*log(E./I6))+((A62.*(1-(I6./E).^1))+(A63.*(1-(I6./E).^2))+(A64.*(1-(I6./E).^3))+(A65.*(1-(I6./E).^4))));
S1 = Si1.*V;
S2 = Si2.*V;
S3 = Si3.*V;
S4 = (Si4.*V)+(Si4b.*V);
S5 = Si5.*V;
S6 = Si6.*V;
%Solving system of equations to find populations at level N'x'
n = 11;
N1 = sym('N1',[1 n]); 
N2 = sym('N2',[1 n]); 
N3 = sym('N3',[1 n]); 
N4 = sym('N4',[1 n]); 
N5 = sym('N5',[1 n]); 
N6 = sym('N6',[1 n]);
eqns = [N1 == N2.*(a2./S1), N2 == N3.*(a3./S2), N3 == N4.*(a4./S3),...
    N4 == N5.*(a5./S4), N5 == N6.*(a6./S5), 1 == N1+N2+N3+N4+N5+N6];
sol = solve(eqns,[N1 N2 N3 N4 N5 N6]);
solN1 = zeros(1,n); 
solN2 = zeros(1,n); 
solN3 = zeros(1,n); 
solN4 = zeros(1,n); 
solN5 = zeros(1,n); 
solN6 = zeros(1,n);
i=1;
while i<=11
    solN1(1,i) = double(sol.N1i);
    solN2(1,i) = double(sol.N2i);
    solN3(1,i) = double(sol.N3i);
    solN4(1,i) = double(sol.N4i);
    solN5(1,i) = double(sol.N5i);
    solN6(1,i) = double(sol.N6i);
    i=i+1;
end

Answer 1

[Torsten]

First solve the equations for all variables involved being scalars:

syms N1 N2 N3 N4 N5 N6 a2sym a3sym a4sym a5sym a6sym S1sym S2sym S3sym S4sym S5sym
eqns = [N1 == N2*(a2sym/S1sym), N2 == N3*(a3sym/S2sym), N3 == N4*(a4sym/S3sym),...
        N4 == N5*(a5sym/S4sym), N5 == N6*(a6sym/S5sym), 1 == N1+N2+N3+N4+N5+N6];
sol = solve(eqns,[N1 N2 N3 N4 N5 N6]);

Then use "subs" to insert the values from the arrays for a2,a3,a4,a5,a6,...

N1num = double(subs(sol.N1,{a2sym,a3sym,a4sym,a5sym,a6sym,S1sym,S2sym,S3sym,S4sym,S5sym},{a2,a3,a4,a5,a6,...
                           S1,S2,S3,S4,S5}));
N2num = double(subs(sol.N2,{a2sym,a3sym,a4sym,a5sym,a6sym,S1sym,S2sym,S3sym,S4sym,S5sym},{a2,a3,a4,a5,a6,...
                           S1,S2,S3,S4,S5}));
N3num = double(subs(sol.N3,{a2sym,a3sym,a4sym,a5sym,a6sym,S1sym,S2sym,S3sym,S4sym,S5sym},{a2,a3,a4,a5,a6,...
                           S1,S2,S3,S4,S5}));
N4num = double(subs(sol.N4,{a2sym,a3sym,a4sym,a5sym,a6sym,S1sym,S2sym,S3sym,S4sym,S5sym},{a2,a3,a4,a5,a6,...
                           S1,S2,S3,S4,S5}));
N5num = double(subs(sol.N5,{a2sym,a3sym,a4sym,a5sym,a6sym,S1sym,S2sym,S3sym,S4sym,S5sym},{a2,a3,a4,a5,a6,...
                           S1,S2,S3,S4,S5}));
N6num = double(subs(sol.N6,{a2sym,a3sym,a4sym,a5sym,a6sym,S1sym,S2sym,S3sym,S4sym,S5sym},{a2,a3,a4,a5,a6,...
                           S1,S2,S3,S4,S5}));    

But why do you set up your problem with symbolic variables ? You have a simple linear system in N1,...,N6 that can be solved by numerical backslash much more efficient than with symbolic calculus.

[Hugh Currah]

Thanks, this worked. I am not sure why I used symbolic variables, it is just what I am used to in Matlab.