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Solving an Integro-differential equation numerically

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Freyja
Freyja 2015 年 3 月 28 日
回答済み: Roger Stafford 2017 年 1 月 9 日
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

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回答 (3 件)

Claudio Gelmi
Claudio Gelmi 2017 年 1 月 6 日
編集済み: Claudio Gelmi 2017 年 1 月 9 日
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

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Roger Stafford
Roger Stafford 2017 年 1 月 9 日
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.

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Torsten
Torsten 2015 年 3 月 30 日
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.

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Freyja
Freyja 2015 年 4 月 4 日
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 2015 年 4 月 4 日
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.

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