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Message |
| Freddy |
 Posted: Thu Mar 08, 2007 12:47 pm |
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Guest
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Hi All,
I have posted this message in the sci.math forum but I thought I would
try this forum to ask this quick question, hopefully the response is
faster.
I'm trying to solve a complex integro-differential equation which has
the following form:
_ xmax
dy |
--- = | f(x ; a) . y(x,t) dx
dt _|
xmin
where "a" is a matrix of parameters.
so I'm trying to use the levenberg marquardt to get the best estimate
of the parameters.
the function f(x) is actually a very complex one too and it's a set of
several functions multiplying each others:
f(x ; a) = g(x ; a) . h(x) . n(x)
but in order to be able to do so, I will need to find the values of
dy/
da to return them to the routine.
Any thoughts on how can I achieve that?
it might be some elementary question for many..but I would like some
help if possible.
Thank you,
Freddy |
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| Fred Krogh |
| Posted: Thu Mar 08, 2007 6:19 pm |
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Guest
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Freddy wrote:
[quote:1f7d5b3f25]Hi All,
I have posted this message in the sci.math forum but I thought I would
try this forum to ask this quick question, hopefully the response is
faster.
I'm trying to solve a complex integro-differential equation which has
the following form:
_ xmax
dy |
--- = | f(x ; a) . y(x,t) dx
dt _|
xmin
where "a" is a matrix of parameters.
so I'm trying to use the levenberg marquardt to get the best estimate
of the parameters.
the function f(x) is actually a very complex one too and it's a set of
several functions multiplying each others:
f(x ; a) = g(x ; a) . h(x) . n(x)
but in order to be able to do so, I will need to find the values of
dy/
da to return them to the routine.
Any thoughts on how can I achieve that?
it might be some elementary question for many..but I would like some
help if possible.
Thank you,
Freddy
I was going to respond to your earlier message, and almost changed my[/quote:1f7d5b3f25]
mind after all the repeats.
One way of getting what you want (if you can do it) is to also solve the
differential equation
d(dy/da)/dt = \int_{xmin}^{xmax} [ (\partial f(x:a)/\partial a) y(x,t) +
f(x:a) dy/da ] dx
with initial conditions dy/da = 0.
Note I have mainly followed your notation, but y more properly should be
written y(x,t,a). If y is a vector then dy/da is a matrix, else dy/da is
a vector. I'm assuming that "a" can be thought of as a vector rather
than as a matrix as you describe it.
Regards,
Fred
with initial conditions (dy/da) = 0. |
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| Freddy |
| Posted: Thu Mar 08, 2007 10:29 pm |
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Guest
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On Mar 8, 6:19 pm, Fred Krogh <fkr...@mathalacarte.com> wrote:
[quote:c6f4b30b8f]Freddy wrote:
Hi All,
I have posted this message in the sci.math forum but I thought I would
try this forum to ask this quick question, hopefully the response is
faster.
I'm trying to solve a complex integro-differential equation which has
the following form:
_ xmax
dy |
--- = | f(x ; a) . y(x,t) dx
dt _|
xmin
where "a" is a matrix of parameters.
so I'm trying to use the levenberg marquardt to get the best estimate
of the parameters.
the function f(x) is actually a very complex one too and it's a set of
several functions multiplying each others:
f(x ; a) = g(x ; a) . h(x) . n(x)
but in order to be able to do so, I will need to find the values of
dy/
da to return them to the routine.
Any thoughts on how can I achieve that?
it might be some elementary question for many..but I would like some
help if possible.
Thank you,
Freddy
I was going to respond to your earlier message, and almost changed my
mind after all the repeats.
One way of getting what you want (if you can do it) is to also solve the
differential equation
d(dy/da)/dt = \int_{xmin}^{xmax} [ (\partial f(x:a)/\partial a) y(x,t) +
f(x:a) dy/da ] dx
with initial conditions dy/da = 0.
Note I have mainly followed your notation, but y more properly should be
written y(x,t,a). If y is a vector then dy/da is a matrix, else dy/da is
a vector. I'm assuming that "a" can be thought of as a vector rather
than as a matrix as you describe it.
Regards,
Fred
with initial conditions (dy/da) = 0.
[/quote:c6f4b30b8f]
Hi Fred,
Thank you for your reply...I really appreciate your answer. I will
certainly give it a shot and see how it works....even though computing
that at every step is going to be very consuming in my case..but
that's my only resort so far.
thanks again...but i'm not sure what was the reason for your
hesitation...you talked about repeats...if you mean the posts I'm
sorry but sometimes a person can get very frustrated looking for an
answer..if that was offending I apologize.
Fred, |
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| Peter Spellucci |
| Posted: Fri Mar 09, 2007 7:50 am |
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Guest
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In article <1173452229.704156.190360@p10g2000cwp.googlegroups.com>,
"Freddy" <zfreddyzzz@gmail.com> writes:
[quote:2792d60172]Hi Again,
I was just wondering as Freddy wrote that y should be written as
y(x,t,a) actually the only reason that I said "a" is a matrix is
because I have 4 parameters in my function. a = [a1, a2, a3, a4]
so could that be written as y = y(x,t,a1,a2,a3,a4).
and if so how accurate is this representation
[dy(x,t,a1,a2,a3,a4)/da1] = [y(x,t,a1+eps,a2,a3,a4)-y(x,t,a1-
eps,a2,a3,a4)]/(2.eps)
[/quote:2792d60172]
this would be a finite difference replacement, end assuming everything
else is smooth, with an error of O(eps^2)
but I cannot see what this should help. if you are able to
integrate the equation for dy/dt, then you will also be able
to integrate the variational equation for dy/da , provided you are
able to write down the Jacobian of f with respect to a, but again:
the x on the left hand side, where it is gone?
or have you simply a misprint and y does not depend on x?
in that case the integral on the right would turn over in a factor for y(t)
and anything would become pretty easy. but as you wrote it it is
not well defined, even if oe is willing to read this as an integro differential
equation (then again a variable would be missing, this time under the integral
clearly something is wrong
hth
peter
[quote:2792d60172]
is that possible or I just deserve to be shot for writing
that..lol...and broken many rules.
Cheers,
-F-
Freddy
I was going to respond to your earlier message, and almost changed my
mind after all the repeats.
One way of getting what you want (if you can do it) is to also solve the
differential equation
d(dy/da)/dt = \int_{xmin}^{xmax} [ (\partial f(x:a)/\partial a) y(x,t) +
f(x:a) dy/da ] dx
with initial conditions dy/da = 0.
Note I have mainly followed your notation, but y more properly should be
written y(x,t,a). If y is a vector then dy/da is a matrix, else dy/da is
a vector. I'm assuming that "a" can be thought of as a vector rather
than as a matrix as you describe it.
Regards,
Fred
with initial conditions (dy/da) = 0.
[/quote:2792d60172] |
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| Freddy |
| Posted: Fri Mar 09, 2007 10:57 am |
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Guest
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Hi Again,
I was just wondering as Freddy wrote that y should be written as
y(x,t,a) actually the only reason that I said "a" is a matrix is
because I have 4 parameters in my function. a = [a1, a2, a3, a4]
so could that be written as y = y(x,t,a1,a2,a3,a4).
and if so how accurate is this representation
[dy(x,t,a1,a2,a3,a4)/da1] = [y(x,t,a1+eps,a2,a3,a4)-y(x,t,a1-
eps,a2,a3,a4)]/(2.eps)
is that possible or I just deserve to be shot for writing
that..lol...and broken many rules.
Cheers,
-F-
[quote:8c32aec2d8]
Freddy
I was going to respond to your earlier message, and almost changed my
mind after all the repeats.
One way of getting what you want (if you can do it) is to also solve the
differential equation
d(dy/da)/dt = \int_{xmin}^{xmax} [ (\partial f(x:a)/\partial a) y(x,t) +
f(x:a) dy/da ] dx
with initial conditions dy/da = 0.
Note I have mainly followed your notation, but y more properly should be
written y(x,t,a). If y is a vector then dy/da is a matrix, else dy/da is
a vector. I'm assuming that "a" can be thought of as a vector rather
than as a matrix as you describe it.
Regards,
Fred
with initial conditions (dy/da) = 0.[/quote:8c32aec2d8] |
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| Freddy |
| Posted: Fri Mar 09, 2007 10:59 pm |
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Guest
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On Mar 9, 1:50 pm, spellu...@fb04373.mathematik.tu-darmstadt.de (Peter
Spellucci) wrote:
[quote:e07511eae9]In article <1173452229.704156.190...@p10g2000cwp.googlegroups.com>, "Freddy" <zfreddy...@gmail.com> writes:
Hi Again,
I was just wondering as Freddy wrote that y should be written as
y(x,t,a) actually the only reason that I said "a" is a matrix is
because I have 4 parameters in my function. a = [a1, a2, a3, a4]
so could that be written as y = y(x,t,a1,a2,a3,a4).
and if so how accurate is this representation
[dy(x,t,a1,a2,a3,a4)/da1] = [y(x,t,a1+eps,a2,a3,a4)-y(x,t,a1-
eps,a2,a3,a4)]/(2.eps)
this would be a finite difference replacement, end assuming everything
else is smooth, with an error of O(eps^2)
but I cannot see what this should help. if you are able to
integrate the equation for dy/dt, then you will also be able
to integrate the variational equation for dy/da , provided you are
able to write down the Jacobian of f with respect to a, but again:
the x on the left hand side, where it is gone?
or have you simply a misprint and y does not depend on x?
in that case the integral on the right would turn over in a factor for y(t)
and anything would become pretty easy. but as you wrote it it is
not well defined, even if oe is willing to read this as an integro differential
equation (then again a variable would be missing, this time under the integral
clearly something is wrong
hth
peter
is that possible or I just deserve to be shot for writing
that..lol...and broken many rules.
Cheers,
-F-
Freddy
I was going to respond to your earlier message, and almost changed my
mind after all the repeats.
One way of getting what you want (if you can do it) is to also solve the
differential equation
d(dy/da)/dt = \int_{xmin}^{xmax} [ (\partial f(x:a)/\partial a) y(x,t) +
f(x:a) dy/da ] dx
with initial conditions dy/da = 0.
Note I have mainly followed your notation, but y more properly should be
written y(x,t,a). If y is a vector then dy/da is a matrix, else dy/da is
a vector. I'm assuming that "a" can be thought of as a vector rather
than as a matrix as you describe it.
Regards,
Fred
with initial conditions (dy/da) = 0.
[/quote:e07511eae9]
Actually I'm trying to solve the population balance equation for
flocculation.
so I'm trying to solve for y (number density of particles of a certain
diameter "x" at a time "t")
and the RHS does take into account the diameter, the number density,
and various other functions like the flocculation frequency for
example. and it's on the RHS that we have the model constants that I'm
trying to match to experimental data.
I'm really still not sure what might be missing under the integral
though as you mentioned. But hope that gives an idea about the problem
I'm dealing with.
Thanks a lot for the help you've been providing guys.
Freddy |
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