A convolution integral involving Bessel functions

INFIDEL

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Feb 19, 2006, 2:06:59 PM2/19/06

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I've been trying to evaluate the following integral analytically, and
am now well and truly stuck. Does anybody have any ideas?

The domain of integration is the intersection of two unit circles with
centres at (x,y)= (+-w,0). The integrand is a product of two functions,
which are each Bessel functions of the first kind (J_0) squared. Let a
denote the first zero of J_0(z).

For w a given real number, 0<=w<=2, the double integral is

f(w/2) = Integrate[Integrate[ J_0(a sqrt((x - w/2)^2 + y^2))^2 J_0(
a sqrt((x + w/2)^2 + y^2))^2 , {x, 0, sqrt(1-y^2) - w/2}, {y, 0,
sqrt(1 - (w/2)^2)}]].

Thanks for any suggestions!

INFIDEL

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Feb 19, 2006, 7:45:00 PM2/19/06

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Actually, the centres of the circles are at (x,y)= (+-w/2,0).

junoexpress

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Feb 20, 2006, 2:15:02 PM2/20/06

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I'm a little confused.

Are you saying that the domain of integration is over the INTERSECTION
of two circles where:
one circle has its center at (-w/2,0),
the other has its center at (+w/2,0)
and both circles have radius w?

Or do you mean to say that the domain of integration is over the UNION
of two circles where:
one circle has its center at (-w/2,0),
the other has its center at (+w/2,0)
and both circles have radius w/2?

Juno

Igor Khavkine

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Feb 20, 2006, 9:45:01 PM2/20/06

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I don't know if this will help. But there is a nice change of
coordinates that may make the problem more tractable. We can use the
so-called bipolar coordinates, where the variable r1 and r2 represent
the distances from two select points in the plane. Usually, these are
taken to be at (0,+1) and (0,-1). The parameters of your integral can
be rescaled to place the centers of the two circles at the latter
coordinates and vary their radii instead.

The relation between bipolar and cartesian coordinates are:

r1^2 = (x-1)^2 + y^2,
r2^2 = (x+1)^2 + y^2;

x = (r2^2 - r1^2)/4, where s = (r1+r2+2)/2.
y = sqrt(s(s-r1)(s-r2)(s-2)),

The coordinates (r1,r2) take values in a semi-infinite strip extending
diagonally from around the origin in the first quadrant. It is bounded
by the lines r1+r2=2, r1-r2=2, and r2-r1=2. The integration region now
acquires a simple form. It is the intersection of the range of the
coordinates and the half planes r1 <= r and r2 <= r, with r being the
radius of the two circles.

To write down the new integral we need the Jacobian of the cartesian
coordinates with respect to the bipolar ones. It turns out to have the
nice form

1 dr1 dr2
dx dy = - --------, where y is as defined above.
2 y(r1,r2)

Beside the Jacobian, the integrand becomes J_0(a r1/r)^2 J_0(a r2/r)^2.
Unfortunately, the denominator of the Jacobian does not separate
cleanly. But this form may be a starting point for some sort of series
expansion.

An alternate direction is another change of variables:

r1 = p1 + p2, so that dr1 dr2 = 2 dp1 dp2,
r2 = p1 - p2, and y = sqrt((p1^2-1)(1-p2^2)).

The advantage is that the Jacobian of the previous transformation is
now factored in the (p1,p2) coordinates. The disadvantage is that the
arguments of the Bessel functions become slightly more complicated.
There are some addition formulas for Bessle functions [1], but
unfortunately they give an infinite number of terms.

Hope some of this helps.

Igor

[1]
http://functions.wolfram.com/BesselAiryStruveFunctions/BesselJ/16/02/

INFIDEL

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Feb 21, 2006, 11:45:00 AM2/21/06

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It's the intersection of the two circles of unit radius.

If w=0, the centres coincide at (0,0) and the intersection is the
entire unit disc.

If w=2, the centres are at (-1,0) and (1,0) and the unit radius
discs just touch. (The integral is zero.)