Solving Integro-differential equation with limited integral

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ash
ash 2015-6-25
评论： Hewa selman 2021-12-22
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

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采纳的回答

Torsten
Torsten 2015-6-25
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.
  1 个评论
Hewa selman
Hewa selman 2021-12-22
Hello
Now are you sure that we can adjust the value of integral, then put it in system and solve it by ode45 or ode15s.

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更多回答（2 个）

Claudio Gelmi
Claudio Gelmi 2017-1-6
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
MATLAB solver here: http://cpc.cs.qub.ac.uk/summaries/AEQU_v1_0.html
Best wishes,
Claudio
ash
ash 2015-6-28
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)));
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Torsten
Torsten 2015-6-30
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.
SOZHAESWARI P
SOZHAESWARI P 2021-9-5
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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