Hello everyone!
I have to solve the following equation by using the Runge-Kutta method:
all parameters are known, furthermore, Pin (function of t) and the timespan are vectors of dimensions 1x4096, so N(t) should be of the same dimension.
My code is the following:
n_dz=10;
dz=linspace(0,L,n_dz);
for i=2:n_dz
Dz=dz(i)-dz(i-1)
Fun=@(N) (I/(q*V))-A*N-B*N.^2-C*N.^3-Gamma*(S*(N-N_tran)...
-k*(lambda-lambda_peak)^2).*(Pin/Dz)...
*(exp(((Gamma*(S*(N-N_tran)-k*(lambda-lambda_peak)^2))-alfa_s)*dz(i))...
-exp(((Gamma*(S*(N-N_tran)-k*(lambda-lambda_peak)^2))-alfa_s)*dz(i-1)))...
/((Gamma*(S*(N-N_tran)-k*(lambda-lambda_peak)^2))-alfa_s)...
*Dz/(V*h*f);
y=Runge_Kutta_4(time,Dt,N_start,Fun);
...
...
end
with Runge_Kutta_4 implemented as:
function y=Runge_Kutta_4(timespan,h,N_start,fun)
t=timespan;
n=length(t);
y=zeros(n,1);
y(1)=N_start;
for j=1:n-1
k1 = h*feval(fun,t(j),y(j));
k2 = h*feval(fun,t(j)+h/2,y(j)+k1/2);
k3 = h*feval(fun,t(j)+h/2,y(j)+k2/2);
k4 = h*feval(fun,t(j)+h,y(j)+k3);
y(j+1) = y(j)+1/6*(k1+2*k2+2*k3+k4);
end
end
Notice that in my code, the analysis over the total length "L" is divided into span of length "dz", therefore the last part of "Fun" is different from the formula shown above because it is due to an integral operation.
The following errors are shown:
??? Error using ==>
@(N)(I/(q*V))-A*N-B*N.^2-C*N.^3-Gamma*(S*(N-N_tran)-k*(lambda-lambda_peak)^2).*(Pin/Dz)*(exp(((Gamma*(S*(N-N_tran)-k*(lambda-lambda_peak)^2))-alfa_s)*dz(i))-exp(((Gamma*(S*(N-N_tran)-k*(lambda-lambda_peak)^2))-alfa_s)*dz(i-1)))/((Gamma*(S*(N-N_tran)-k*(lambda-lambda_peak)^2))-alfa_s)*Dz/(V*h*f)
Too many input arguments.
Error in ==> Runge_Kutta_4 at 7
k1 = h*feval(fun,t(j),y(j));
Error in ==> SOA_v1 at 47
y=Runge_Kutta_4(time,Dt,N_start,Fun);
Error in ==> RZ_source at 29
How can i solve this problem?
Thank you for your attention
Best regards
