Help to approximate this integral...

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Peter Clifford
Peter Clifford 2019-3-3
编辑： David Goodmanson 2019-3-15
Hello,
I am wondering if it is possible to have an approximation of this integral
I have only an approximation
of f which is a solution of an ODE.
Many thanks,
Maher.
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Peter Clifford
Peter Clifford 2019-3-6
No f is asolution of an ODE equation, I have an approximation of it by ode45 method.
darova
darova 2019-3-6
编辑：darova 2019-3-6
but what f depends on? If we calculate integral x = [0, 3], what will be f (for example y = 1.56)
have you relation ?

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回答（1 个）

David Goodmanson
David Goodmanson 2019-3-6
编辑：David Goodmanson 2019-3-6
Hi M^2,
I am assuming that y(t) is strictly monotonic increasing or decreasing so that the inverse exists. It's good if the y and t arrays are not too sparse. Depending on how time comsuming the ode solver is, in the case of ode45, for example, you can output a lot of points by specifying a grid of points in the tspan input variable rather than just an upper and lower limit.
Assuming that t and y are reasonably dense an that the y array starts at 0, then
V(x) = cumtrapz(exp(-t),y)
There are better ways to do an indefinite integral, but this gives a reasonable result if exp(-t) is fairly smooth,
  2 个评论
Peter Clifford
Peter Clifford 2019-3-6
Thanks for the help.
You took ?
what are the relations between and t in
V(x) = cumtrapz(exp(-t),y)
I have this
[t, f] = ode45(@odefunc, tspan, [0; 1], options);
How to deduce an approximation of V defined by :
where is the inverse function of f which is strictly increasing and
Maher.
David Goodmanson
David Goodmanson 2019-3-7
编辑：David Goodmanson 2019-3-15
HI Maher,
I should have been more explicit. Yes, I took y = f(t). Then
t = f^-1(y)
This is consistent with just using the line
[t, y] = ode45(@odefunc, tspan, [0; 1], options);
There are two side-by-side output arrays, t and y = f(t). These match up point-by-point. But if both of these arrays are monotonic there is no reason you can't consider y to be the independent variable, t the dependent variable and redefine the arrays as y and t(y), the inverse function of y. Since t and y match up point-by point, t is the same as t(y) and
Int {0,x} e-t(y) dy ---> Int {0,x} e-t dy.
With y as the independent variable you can do the definite integral from 0 to x by trapz (assuming the y array starts at 0) and the indefinite integral by cumtrapz to give the function V(x).
Matlab has plenty of integral evaluation capability when given a function f(x) that can be evaluated at arbitrary points in x.
I hope I'm wrong, but if what you have is an independent and a dependent array at fixed points, I don't know that Matlab has anything easily available that's better than the lamentably primitive trapz and cumtrapz.

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