integro-differential equations
Leslaw Bieniasz
Hi,
I am a bit unfamiliar with integro-differential equations, and I want to
make sure if I understand things properly. My question is:
Is it possible to solve integro-differential equations (or systems)
without specifying initial conditions, or initial conditions must always
be supplied? In principle, it seems to me that at least for some unknown
functions (in a system) their initial values might be obtainable from the
integrals in the integro-differential equations. If anybody can point me
to the relevant discussions in the literature, it would be nice.
Leslaw
Michael Schuster
> I am a bit unfamiliar with integro-differential equations, and I want to
> make sure if I understand things properly. My question is:
> Is it possible to solve integro-differential equations (or systems)
> without specifying initial conditions, or initial conditions must always
> be supplied?
Michael
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Leslaw Bieniasz
>> I am a bit unfamiliar with integro-differential equations, and I want to
>> make sure if I understand things properly. My question is:
>> Is it possible to solve integro-differential equations (or systems)
>> without specifying initial conditions, or initial conditions must always
>> be supplied?
> Yes, always. See any book about mathematics in your lib.
>
> Michael
Hi,
I doubt whether ANY book about mathematics gives the answer.
How about the following example of an integro-differential system with
unknowns x(t) and y(t), and given function p(t):
t
I x(t) dt + x(t) - p(t) = 0
0
dy(t)/dt + x(t) * y(t) = 0
or
t
I x(t) dt + y(t) * x(t) - p(t) = 0
0
dy(t)/dt + x(t) * y(t) = 0
Do we really need to supply the initial condition for x(t) ?
Leslaw
Peter Spellucci
In article <Pine.GHP.4.64.09...@kinga.cyf-kr.edu.pl>,
yes of course, how could you otherwise start to compute "next x"?
youre case above can be directly transformed into a system of integral
equations, maybe you mean this only as an example. In principle here anything
goes as usual: you discretize all the equations , use the initial conditions
and compute the solution on a grid: using BDF2 and the trapezoidal rule
you get of course implicit, possibly nonlinear equations to be solved in every
time step, and for more complicated integrals it also might be necessary to
handle the complete history backwards.
hth
peter
Leslaw Bieniasz
Well, I think this is possible, in fact I am doing such calculations
currently with success. In the first example above, x(t) is a solution of
a purely integral equation, hence initial condition is not needed:
t
I x(t) dt + x(t) - p(t) = 0
0
Let us divide the t interval into steps of length h. Then, for the first
step we can write (in the spirit of the trapezium quadrature):
(x(0) + x(h)) * h/2 + x(h) - p(h) = 0
and
(x(0) + x(h/2)) * h/4 + x(h/2) - p(h/2) = 0
where I take x(h/2) = (x(0) + x(h))/2
This gives a system of two independent equations for x(o) and x(1),
allowing a simultaneous determination of both. Hence, initial condition
for x(t) is not needed.
Leslaw
> youre case above can be directly transformed into a system of integral
> equations, maybe you mean this only as an example. In principle here anything
> goes as usual: you discretize all the equations , use the initial conditions
> and compute the solution on a grid: using BDF2 and the trapezoidal rule
> you get of course implicit, possibly nonlinear equations to be solved in every
> time step, and for more complicated integrals it also might be necessary to
> handle the complete history backwards.
>
> hth
> peter
>
>
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Peter Spellucci
In article <Pine.GHP.4.64.09...@kinga.cyf-kr.edu.pl>,
Leslaw Bieniasz <nbbi...@cyf-kr.edu.pl> writes:
>On Mon, 19 Jan 2009, Peter Spellucci wrote:
>
>Well, I think this is possible, in fact I am doing such calculations
>currently with success. In the first example above, x(t) is a solution of
>a purely integral equation, hence initial condition is not needed:
>
> t
> I x(t) dt + x(t) - p(t) = 0
> 0
>
>Let us divide the t interval into steps of length h. Then, for the first
>step we can write (in the spirit of the trapezium quadrature):
>
>(x(0) + x(h)) * h/2 + x(h) - p(h) = 0
>
>and
>
>(x(0) + x(h/2)) * h/4 + x(h/2) - p(h/2) = 0
>
>where I take x(h/2) = (x(0) + x(h))/2
>
>This gives a system of two independent equations for x(o) and x(1),
>allowing a simultaneous determination of both. Hence, initial condition
>for x(t) is not needed.
>
>Leslaw
you are right here, of course you can apply analytic computation to a
numerical scheme, but in linear cases (as this one) the usefulness
is doubtful (you are constructing an approximation to the "general solution"
in terms of the initial value as parameter), the formula for
a general grid point simply solves the recursive scheme into a
much for complicated formula which also might be quite disadvantageous
with respect to rounding errors.
and what in the general nonlinear case? even the simplest nonlinear
systems of ODE's don't allow an analytical solution in terms of
known elemnetary functions.
remark:
a search for "integro differential" and "books" in Zentralblatt Mathematik
gives you 38 hits,
for example
Tarang, Mare
Stability of the spline collocation method for Volterra
integro-differential equations. (English)
Dissertationes Mathematicae Universitatis Tartuensis 34.
Tartu: Tartu University Press. Tartu:
Univ. Tartu, Faculty of Mathematics and Computer Science (Thesis).
(2004)
Badea, Claudia Lidia
Integro-Differentialgleichungen. Eine Einführung. Theorie - Anwendungen -
Programmierung.
(Integro-differential equations. An introduction. Theory - Applications -
Programming). (German)
Schriftenreihe der Österreichischen Computer Gesellschaft.
47. München, Wien: R. Oldenbourg, Österreichische Computer Gesellschaft
. VIII, 440 p. (1990).
hth
peter
Greg Heath
Does your question also apply to boundary conditions?
If the highest derivative of a variable is n, then n auxiliary
conditions (initial or boundary) are needed.
Hope this helps.
Greg
Greg Heath
No.
The existence of y' requires the specification of y0.
Since there are no derivatives of x, the specification
of x0 is not required.
For the 1st set:
Obviously, x0 = p0 and since the 1st equation is
independent of y, it can be solved for x. Then,
given x and y0, the second equation can be solved
for y.
For the 2nd set:
The second equation needs y0 specified.
If y0 = 0, then y(t) == 0, x(t) = p'(t).
In general, y'(0) = -x0*y0 = -p0. Then
x can be eliminated to obtain a second
order equation for y with y0 and y'(0)
specified.
Hope this helps.
Greg
Greg Heath
> On Mon, 19 Jan 2009, Peter Spellucci wrote:
>
Impying that x0 = p0. Then x'(t)+x(t) = p'(t)
yields an obvious solution either analytically
or numerically.
Hope this helps.
Greg
Leslaw Bieniasz
Good! Thanks. At last someone understands what I was asking about,
and agrees with my view of the problem. Now, the main issue
in my question (perhaps reformulated a little bit) is:
is there any general method to determine, given an arbitrary
integro-differential system, which of the unknowns require the initial
condition, and which do not require? Can it be that simple that only those
unknowns the derivatives of which occur in the system, need initial
conditions? No kind of dependencies between the equations has to be
taken into account?
The examples that I wrote were supposed to illustrate the issue
in question; certainly I am not interested in solving them, neither
analytically nor numerically.
Leslaw
> For the 1st set:
>
> Obviously, x0 = p0 and since the 1st equation is
> independent of y, it can be solved for x. Then,
> given x and y0, the second equation can be solved
> for y.
>
> For the 2nd set:
>
> The second equation needs y0 specified.
> If y0 = 0, then y(t) == 0, x(t) = p'(t).
>
> In general, y'(0) = -x0*y0 = -p0. Then
> x can be eliminated to obtain a second
> order equation for y with y0 and y'(0)
> specified.
>
> Hope this helps.
>
> Greg
>
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