New type of ode45 question: coupled equations but with large summations on dependent variables
1 次查看(过去 30 天)
显示 更早的评论
I am trying to replicate the following set of equations. i am getting wrong values. Can anyone see the code and please tell me if the method of ode45 implementation is correct? If correct, then it would mean there is something wrong with the input values that is causing wrong results for me. I have commented everything for better understanding.
% Here, I define the zspan and p_initial and call the following function
function dpdz= lorenz(z,p)
%%
ws=[1520:10:1560]*1e-9; F=3e8./ws;
f_p1= 3e8/1427e-9; f_p2=3e8/1495e-9;
gr=.4/1e3; %Raman gain spectrum
alphaa=[0.1832 0.1932 0.1142 0.0746 0.1781 0.1432 0.1969]./4.343/1e3 ;
%attenuation coefficient encountered by each frequency channel
channels=numel(F);
m=length(p);
for i=1: channels %For each Pi. The first equation
A=0; B=0; f_i=F(i);
for k= i+1:channels %I separately wrote the components of each eqn. and then added them accordingly
A= A-((f_i/F(k))*p(k)*p(i))*gr;
end
for k=1:(i-1)
B= B+(p(k)*p(i)*gr);
end
alpha_i=alphaa(i);
C= p(i)*alpha_i;
D= p(m-1)*p(i)*gr;
E=p(m)*p(i)*gr;
dpdz(i)= A+B-C+D+E; %this gives one equation. As 5 channels, dpdz(1) to dpdz(5), five eqns.
end %end of the every channel P loop
%=======================================================%
%for the pump 1 eqn.
A=0; B=0; C=0;
for k=1:channels
A= A+(((f_p1/F(k))*p(k)*p(m-1))*gr) ;
end
B= -p(m)*p(m-1)*gr;
C=p(m-1)*alpha_i;
dpdz(m-1)= A+B+C;
%=======================================================%
%for the pump 2 eqn.
A=0; B=0; C=0;
for k=1:channels
A= A+ (((f_p2/F(k))*p(k)*p(m))*gr) ;
end
B= (f_p2/f_p1)*p(m)*p(m-1)*gr; %need to add the gr term later
C=p(m)*alpha_i; %later you will have to change the alpha notation
dpdz(m)= A+B+C;
dpdz=dpdz';
%=======================================================%
end
0 个评论
回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Ordinary Differential Equations 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!