Index exceeds the number of array elements. Index must not exceed 1.
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I'm trying to apply runge kutta method to my problem but i messed up the code at line34(k1_ivme) and matlab giving this error. What should i do for fix this?
z(1)=-314;
v(1)=0;
h=0.1;
tson=10;
t=0:h:tson;
L=1.5;
vi=0.000001;
n = 100;
for i=1:n-1;
Re(i)=abs(v(i))*L/vi;
Cf(i)=0.075/((log(Re)-2).^2);
k1_hiz(i)=h*v(i);
k1_ivme=h*(5207.3-3.47*Cf(i)*v(i)^2);
k2_hiz(i)=h*(v(i)+0.5*k1_ivme(i));
k2_ivme(i)=h*(5207.3-3.47*(Cf(i)+0.5*k1_ivme)*(v(i)+0.5*k1_ivme))^2;
k3_hiz(i)=h*(v(i)+0.5*k2_ivme(i));
k3_ivme(i)=h*(5207.3-3.47*(Cf(i)+0.5*k2_ivme)*(v(i)+0.5*k2_ivme))^2;
k4_hiz(i)=h*(v(i)+k3_ivme(i));
k4_ivme(i)=h*(5207.3-3.47*(Cf(i)+k3_ivme)*(v(i)+k3_ivme))^2;
z(i+1)=z(i)+(1/6*(k1_hiz(i)+2*k2_hiz(i)+2*k3_hiz(i)+k4_hiz(i)));
v(i+1)=v(i)+(1/6*(k1_ivme(i)+2*k2_ivme(i)+2*k3_ivme(i)+k4_ivme(i)));
end
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Answers (2)
VBBV
on 17 Jan 2022
z(i+1)=z(i)+(1/6*(k1_hiz(i)+2*k2_hiz(i)+2*k3_hiz(i)+k4_hiz(i)));
v(i+1)=v(i)+(1/6*(k1_ivme(i)+2*k2_ivme(i)+2*k3_ivme(i)+k4_ivme(i)));
Look at these two lines. you have declared n as 100 and trying to access 101th element . Could that be possible ?
Torsten
on 17 Jan 2022
z(1)=-314;
v(1)=0;
h=0.1;
tson=10;
t=0:h:tson;
L=1.5;
vi=0.000001;
n = 100;
for i=1:n
Re(i)=abs(v(i))*L/vi;
Cf(i)=0.075/((log(Re(i))-2).^2);
k1_hiz(i)=h*v(i);
k1_ivme(i)=h*(5207.3-3.47*Cf(i)*v(i)^2);
k2_hiz(i)=h*(v(i)+0.5*k1_ivme(i));
k2_ivme(i)=h*(5207.3-3.47*(Cf(i)+0.5*k1_ivme(i))*(v(i)+0.5*k1_ivme(i)))^2;
k3_hiz(i)=h*(v(i)+0.5*k2_ivme(i));
k3_ivme(i)=h*(5207.3-3.47*(Cf(i)+0.5*k2_ivme(i))*(v(i)+0.5*k2_ivme(i)))^2;
k4_hiz(i)=h*(v(i)+k3_ivme(i));
k4_ivme(i)=h*(5207.3-3.47*(Cf(i)+k3_ivme(i))*(v(i)+k3_ivme(i)))^2;
z(i+1)=z(i)+(1/6*(k1_hiz(i)+2*k2_hiz(i)+2*k3_hiz(i)+k4_hiz(i)));
v(i+1)=v(i)+(1/6*(k1_ivme(i)+2*k2_ivme(i)+2*k3_ivme(i)+k4_ivme(i)));
end
Now the code runs, but there must be something wrong with your RK4 integrator.
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