Solving an Integro-differential equation numerically

63 Ansichten (letzte 30 Tage)
Freyja
Freyja am 28 Mär. 2015
Kommentiert: Hewa selman am 23 Dez. 2021
Hi, I am interested in writing a code which gives a numerical solution to an integro-differential equation. First off I am very new to integro-differential equations and do not quite understand them so I decided to start simple and would like some help with the first steps. My proposed equation is in the attached picture and the formulas I wish to use are also there though I'm open to suggestions. Even if someone can help me with the first step (just the maths part) where i = 0 I would be very grateful. My goal is to end up with a system of linear algebraic equations which I can then solve with Matlab. Thanks in advance to anyone who takes the time to look at this tricky problem.
Best regards, Freyja

Antworten (3)

Claudio Gelmi
Claudio Gelmi am 6 Jan. 2017
Bearbeitet: Claudio Gelmi am 9 Jan. 2017
Take a look at this solver:
"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

Roger Stafford
Roger Stafford am 9 Jan. 2017
I hate to see numerical approximation methods used when there exists a very simple and precise method done by hand. First we designate by K the integral of t*y(t) from 0 to 1, which is unknown as yet. This gives
y(x) = 1 + (K-1/3)*x
Integrating this w.r. to x gives
y(x) = x + (K-1/3)*x^2/2 + C
where C is the unknown constant of integration. However, since y(0) = 0, this implies that C = 0. Now we have
t*y(t) = t^2 + (K-1/3)*t^3/2
Integrating t*y(t) from 0 to 1 gives t^3/3 + (K-1/3)*t^4/8 evaluated at t = 1 minus its value at t = 0, so that gives
K = 1/3 + (K-1/3)*1/8
which has the unique solution K = 1/3. This in turn gives us our final answer:
y(x) = x.
No need for matlab or numerical approximations.

Torsten
Torsten am 30 Mär. 2015
i=0:
(y(1/2)-y(0))/(1/2)=1-1/3*0+0*integral_0^1(t*y(t))dt
-> 2*y(1/2)=2*y(0)+1-1/3*0+0*integral_0^1(t*y(t))dt
-> 2*y(1/2)=1
-> y(1/2)=1/2
Now do the same for i=1, and you are done.
Best wishes
Torsten.
  2 Kommentare
Freyja
Freyja am 4 Apr. 2015
Hi Torsten, Thanks for replying, unfortunately it's not quite was I was looking for, but either way I've solved it now so it's all good :)
Thanks anyway :D
Roger Stafford
Roger Stafford am 4 Apr. 2015
If you use the trapezoidal approximation to the integral, your exact solution will not quite satisfy your equation. Only if you use an exact integral using 'int' or calculus methods will the equation hold true.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Systems Of Linear Equations finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by