# Solving Integro-differential equation with limited integral

30 ビュー (過去 30 日間)
ash 2015 年 6 月 25 日

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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### 採用された回答

Torsten 2015 年 6 月 25 日
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.

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### その他の回答 (2 件)

ash 2015 年 6 月 28 日
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 件のコメント

ash 2015 年 6 月 29 日
Any help?
Torsten 2015 年 6 月 30 日
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 2015 年 6 月 30 日
(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.

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Claudio Gelmi 2017 年 1 月 6 日
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

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