ℝ³中相交正则曲面上1/||x_A - x_B||^k型二重积分收敛的k值求解问询

Hey there! Let's work through this convergence problem for your double integral over intersecting regular surfaces in 3D space. I'll break it down based on how the two surfaces intersect, since the behavior near the intersection is what determines whether the integral converges.

First, let's restate your problem clearly:

• We have two regular surfaces (A) and (B) in (\mathbb{R}^3) that intersect (either at points, along a curve, or in a surface region).
• The total energy integral is:
  $$E = \int_A \int_B \frac{1}{|x_A - x_B|^k} dA dB$$
• We need to find all values of (k) for which this integral converges.

The key here is analyzing the singularity of the integrand (\frac{1}{|x_A - x_B|^k}) near the intersection set (S = A \cap B) — since (|x_A - x_B| \to 0) as (x_A, x_B) approach (S), the integral's convergence depends entirely on how the integrand behaves in these neighborhoods.

Case 1: The surfaces intersect in a 2-dimensional region (they overlap partially or fully) #

If (A) and (B) share a 2D sub-surface, then near this overlap, the integral reduces to integrating over pairs of points on the same surface:
$$\int_S \int_S \frac{1}{|x - y|^k} dS(x) dS(y)$$
Here, (|x - y|) behaves like the planar Euclidean distance between points on the surface. The local integral approximates to a 4-dimensional integral of (\frac{1}{r^k}) (where (r) is planar distance), and this converges only when (1 - k > -1) — in other words, when (k < 2). For (k \geq 2), the singularity becomes too strong for the integral to converge.

Conclusion for this case: Integral converges if and only if (k < 2).

Case 2: The surfaces intersect along a 1-dimensional curve (transverse or tangential intersection along a curve) #

When (A) and (B) cross along a curve (C), near any point on (C), we can use local coordinates where (C) is the (x)-axis, (A) is the (xy)-plane, and (B) is the (xz)-plane (for transverse intersection). The distance (|x_A - x_B| = \sqrt{(u-s)^2 + t^2 + v^2}) behaves like a 3D Euclidean distance near the curve.

The local integral is 4-dimensional, and the singularity lies along a 2-dimensional set ((C \times C) in (A \times B)). For convergence, we need (k < 3): when (k = 3), a logarithmic singularity causes divergence, and for (k > 3) the divergence is even faster.

Conclusion for this case: Integral converges if and only if (k < 3).

Case 3: The surfaces intersect at isolated 0-dimensional points (only finitely many intersection points) #

If (A) and (B) only meet at isolated points, near each such point (p), the surfaces are locally disjoint except at (p). The distance (|x_A - x_B|) behaves like the 3D distance from (p) as both points approach (p).

The local integral approximates to a 4-dimensional integral of (\frac{1}{r^k}) (where (r) is 3D distance to (p)). Similar to the 2D overlap case, this converges only when (k < 2); for (k \geq 2), the singularity at (p) makes the integral blow up.

Conclusion for this case: Integral converges if and only if (k < 2).

Summary #

To wrap it up:

• If the surfaces intersect in a 2D region or at isolated points: Converges when (k < 2)
• If the surfaces intersect along a 1D curve: Converges when (k < 3)

备注：内容来源于stack exchange，提问作者cheng