Could anyone help me with this warning please?

1 visualización (últimos 30 días)
Avan Al-Saffar
Avan Al-Saffar el 5 de Dic. de 2014
Comentada: Torsten el 5 de En. de 2015
This is my code :
function RunLogOscilNumeric3
k =10;
p0 =0.1;
t =(0:0.01:10000);
omega = 1;
N0 = 1;
[t,p]=ode23(@logOscilnumeric3,t,p0,[],omega,k,N0);
Pmax = max(p)
Pmean = mean(p)
figure(1)
plot(t,p)
title('The plot of the system with time')
xlabel ('Time')
ylabel ('The system' )
1;
% function dpdt = logOscilnumeric3(t,p,omega,k,N0)
% dpdt = N0*p - (N0*sin(omega*t)*p.^2/k);
% end
Notes: 1- Warning: Failure at t=5.060889e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.
2- I tried to change the ode solver ,,but I still got this warning.
3- I want to solve this system for the specific values of the parameters and time'' I do not want to change those at all'' ,, because I am trying to solve different systems for the same parameters and time vector.
What should I do please?
Thanks in advance.

Respuesta aceptada

Torsten
Torsten el 5 de Dic. de 2014
The analytical solution for your ODE is given by
p(t)=20*exp(t)/(exp(t)*(cos(t)-sin(t))-201)
This function has a singularity between t=5 and t=5.5.
Best wishes
Torsten.
  7 comentarios
Avan Al-Saffar
Avan Al-Saffar el 28 de Dic. de 2014
Dear Torsten If I have the following system : dx/dt = N0*sin(omega*t)*x - (N0*x.^2 / k)
I tried to solve it analytically but I am getting this formula which I can not continue:
( exp( (-N0/omega)*(cos(omega*t)) )/x)= ( integral( (N0/omega) * (exp( (-N0/omega) * (cos(omega*t)) )) )dt )
can you help me please?
Regards
Torsten
Torsten el 5 de En. de 2015
Try MATLAB's dsolve.
If an explicit solution can not be found, you will have to solve the equation numerically for given values of N0, omega and k.
Best wishes
Torsten.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Ordinary Differential Equations en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by