Problem in solving a differential equations system

Hi, I'm trying to solve non linear propagation equations for second harmonic generation (lasers)

Here is my code

function ypoint  = EquaDiff( y,z )

c=3e8;
eps0=8.85e-12;
lw=1.030;
w=2*pi()*c/(lw*1e-6);
deff=0.828e-12; 
n= 1.605; 
q=w*deff/(n*c);
deltak=0;

ypoint=zeros(2,1);

ypoint(1)=j*q*conj(y(1))*y(2);

ypoint(2)=j*q*(y(1)^2);

ypoint=ypoint';

clear all;

[y,z]=ode45 (@EquaDiff, [0:0.001:0.03],[1 0]);

y1=y(:,1);

y2=y(:,2);

plot (z,abs(y1).^2,z,abs(y2).^2);

I get the following error

?? Attempted to access y(2); index out of bounds because numel(y)=1.

Error in ==> EquaDiff at 16
ypoint(1)=j*q*conj(y(1))*y(2);

Error in ==> funfun\private\odearguments at 110
f0 = feval(ode,t0,y0,args{:});   % ODE15I sets args{1} to yp0.

Error in ==> ode45 at 173
[neq, tspan, ntspan, next, t0, tfinal, tdir, y0, f0, odeArgs, odeFcn, ...

Error in ==> f at 4
[y,z]=ode45 (@EquaDiff, [0:0.001:0.03],[1 0]);

Does somebody has an idea what i did wrong?

Answer 1

[Walter Roberson]

The supplied function (e.g., EquaDiff) is expected to have pass a scalar t as its first argument, and a column vector y as its second argument. Therefore accessing element #2 of the first argument is an error.

Your function EquaDiff only uses one of its two arguments. Is your system independent of time ?

Answer 2

[DEYRA]

It is indeed independant of time (further equations will take z in account)

I found some typos and right now it seems to be working (apparently the ypoint=ypoint' is not needed)

Thanks anyway for your answer

Answer 3

[Ros Rego]

Hello, I'm with a similar error: This is my code:

function J = f(Q, h)
    A = 1.54e-2; %area of both tanks
    c12 = 6e-4; %Flow constant of the interconnectiong pipe
    c2 = 13e-4;
    y1 = (1/A)*(Q(1) - c12*sqrt(h(1) - h(2)));
    y2 = (1/A)*(Q(2) + c12*sqrt(h(1) - h(2)) - c2*sqrt(h(2)));
    J = [y1; y2];
end

Nt = 101;                 %Step Size of time
ti = 0;                   %Initial time (sec)
tf = 10;                  %Final time (sec)
t = linspace(ti,tf,Nt);   %Time vector (sec)

h = [1.7 0.3];                %Initial liquid levels
Q= [1 1];                   %Initial pump flow
% Q1=1;
% Q2=1;
%Initial value for ode45 routine
initial1 = [Q h];   

[t1,y1] = ode45(@f,t,[Q, h]);

And I get this error: Attempted to access Q(2); index out of bounds because numel(Q)=1.

[Torsten]

I guess you mean

Nt = 101;                 %Step Size of time
ti = 0;                   %Initial time (sec)
tf = 10;                  %Final time (sec)
tspan = linspace(ti,tf,Nt);   %Time vector (sec)

h = [1.7 0.3];                %Initial liquid levels
Q= [1 1];                   %Initial pump flow

[t1,y1] = ode45(@(t,y)f(t,y,h),tspan,Q);

function J = f(t, Q, h)
    A = 1.54e-2; %area of both tanks
    c12 = 6e-4; %Flow constant of the interconnectiong pipe
    c2 = 13e-4;
    y1 = (1/A)*(Q(1) - c12*sqrt(h(1) - h(2)));
    y2 = (1/A)*(Q(2) + c12*sqrt(h(1) - h(2)) - c2*sqrt(h(2)));
    J = [y1; y2];
end