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Solving Integro-differential equation with limited integral

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ash
ash on 25 Jun 2015
Commented: SOZHAESWARI P on 5 Sep 2021 at 3:09
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

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.

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
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?

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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)));
  4 Comments
SOZHAESWARI P
SOZHAESWARI P on 5 Sep 2021 at 3:09
How to solve the numerical solution of nonlinear parabolic integro differential equation for two grid finite element method example MATLAB codings

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