how to solve integral in the defining the function command?

clc

%defining constant

ti = 0; %inital time

tf = 10E-5;% final time

tspan=[ti tf];

o = 10E6; % detuning frequency

tc = 70E-9; %photon life time in cavity

tf = 240E-6; %flouroscence lifetime

a = 0.02; %round trip loss

P = 1; %pump strenght

k = 0.2; %critical coupling strength

l= 0.5;

% define function

%y(1) = I

%y(2) = G

%y(3) = phase difference

f = @(t,y) [

((y(2)-a-l.*(abs(cos(y(3)+ pi/4)))).*y(1) + k.*y(1).*cos(y(3)- pi/2)).*(2/tc);

(P - (y(2).*(y(1) + 1))) / tf;

o - (k / tc).*2.* sin(y(3));

];

%initial consitions

[T,Y] = ode45(f,tspan,[1;1;1]*10E-5);

%plotting the graphs

plot(T,Y(:,3));

ylim([0 30])

in this the program I represent y(1) as I(Φ), in the equation 2 intead of y(1) i want to use integral of I(Φ)dΦ , where the we assume I(Φ) is gaussian distribution with a variable mean Φ0 and constant rms width σ = 0.1

is it possible to do this, if yes, how ?

Answer 1

[Torsten]

integral_{0}^(t) I(phi) dphi = normcdf(t,0,0.1) - 0.5

So you can work with

y(1) = normcdf(t,0,0.1) - 0.5

in your equations.

[SAHIL SAHOO]

but how ill you write this in this?

f = @(t,y) [

((y(2)-a-l.*(abs(cos(y(3)+ pi/4)))).*y(1) + k.*y(1).*cos(y(3)- pi/2)).*(2/tc);

(P - (y(2).*(y(1) + 1))) / tf;

o - (k / tc).*2.* sin(y(3));

];

if you integrate outside it will give different result.

[Torsten]

I don't know the background of your equations.

If you want to take y1 as integral of I(Φ)dΦ , where we assume I(Φ) is gaussian distribution with a variable mean Φ0 and constant rms width σ = 0.1, the equations are

f = @(t,y) [  (P - (y(1).*(normcdf(t,0,0.1) - 0.5 + 1))) / tf; o - (k / tc).*2.* sin(y(2)) ];

Since the equation for y(1) is obsolete, y(2) became y(1) and y(3) became y(2) in the function handle.