MATLAB Answers

Solving Integro-differential equation with limited integral

31 views (last 30 days)
ash
ash on 25 Jun 2015
Hi,
How can I solve this equation numerically using matlab
w''''=w''*int(w'^2,0,1)
I tried using the standard form of ODE function, the only problem I faced is how to represent that limited integral Thanks

  0 Comments

Sign in to comment.

Accepted Answer

Torsten
Torsten on 25 Jun 2015
Write your integro-differential equation as
w1'=w2
w2'=w3
w3'=w4
w4'=w3*integral_{t=0}^{t=1}w2^2(t') dt'
Then discretize the interval [0:1] in n subintervals 0=t(1)<t(2)<...<t(n)=1.
Compute the derivatives as
wj'(t(i))=(wj(t(i+1))-wj(t(i)))/dt (j=1,2,3,4)
and compute the integral using the trapezoidal rule.
You'll arrive at a polynomial system (order 3) of equations for the unknowns
wj(t(2)),wj(t(3)),...,wj(t(n)) (j=1,2,3,4)
which can be solved by fsolve, e.g.
No chance to use ODE45 in this case.
Another way might be to use ODE45 and iteratively adjust the value of the integral, but I'm not sure whether this method will converge.
Good luck !
Best wishes
Torsten.

  0 Comments

Sign in to comment.

More Answers (2)

Claudio Gelmi
Claudio Gelmi on 6 Jan 2017
Take a look at this solver:
Article "IDSOLVER: A general purpose solver for nth-order integro-differential equations": http://dx.doi.org/10.1016/j.cpc.2013.09.008
Best wishes,
Claudio

  2 Comments

Fernando Fernandes
Fernando Fernandes on 14 Jan 2021
Gelmi help me! How can I use your method to solve this equation?
Fernando Fernandes
Fernando Fernandes on 14 Jan 2021
I've downloaded your paper, but i'm a beginner in Matlab. Do I need the solver in http://cpc.cs.qub.ac.uk/summaries/AEQU_v1_0.html ???
How can I install this?

Sign in to comment.


ash
ash on 28 Jun 2015
Thanks, sorry for the late reply
I tried to apply the technique you suggested (as much as I understood) using Euler 1st order, also my problem is a BVP (3 initial conditions and 1 BC) kindly find my code bellow The problem is that the code is too slow, and I can only solve small number of points
Is that what you advised me to do in our previous comment?, are there any enhancement for that code?
Thanks
syms a b
w=10;
T=30;
L=500;
F=1;
E=160e3;
I=2500;
A=T*w;
Npnts=11;
x=linspace(0,L/2,Npnts);
q1=sym(zeros(1,length(x)));
q2=sym(zeros(1,length(x)));
q3=sym(zeros(1,length(x)));
q4=sym(zeros(1,length(x)));
h=x(2)-x(1);
q1(1)=0;
q2(1)=0;
q3(1)=b;
q4(1)=-F/2/E/I;
for i=1:length(x)
q1(i+1)=q1(i)+h*q2(i);
q2(i+1)=q2(i)+h*q3(i);
q3(i+1)=q3(i)+h*q4(i);
q4(i+1)=q4(i)+h*q3(i)*a*A/2/L/I;
end
integ_a=sum(q2.^2)*h-a/2;
sol_ab=solve(integ_a==0,q2(i+1)==0,a,b);
sol_a=sol_ab.a;
sol_b=sol_ab.b;
sol_index=1;
q1=subs(q1,a,sol_a(sol_index));
q1=double(subs(q1,b,sol_b(sol_index)));

  3 Comments

Torsten
Torsten on 30 Jun 2015
The system to solve is
(w1(t(i+1))-w1(t(i)))/h = w2(t(i))
(w2(t(i+1))-w2(t(i)))/h = w3(t(i))
(w3(t(i+1))-w3(t(i)))/h = w4(t(i))
(w4(t(i+1))-w4(t(i)))/h = [sum_{j=1}^{j=Npnts-1}(w2(t(i+1))+w2(t(i)))*h/2]*w3(t(i))
(i=1,Npnts-1)
These are 4*(Npts-1) equations in which you will have to include the boundary conditions.
You can use fsolve to solve this system of polynomial equations.
Best wishes
Torsten.
Torsten
Torsten on 30 Jun 2015
Sorry, should read
(w4(t(i+1))-w4(t(i)))/h = [sum_{j=1}^{j=Npnts-1}(w2(t(j+1))+w2(t(j)))*h/2]*w3(t(i))
Best wishes
Torsten.

Sign in to comment.

Community Treasure Hunt

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

Start Hunting!

Translated by