2.2. LAPLACE’S EQUATION 39 Combining this calculation with estimate (36), we deduce |u(x)−g(x0)| 2ε, provided |x x0| is sufficiently small. c. Green’s function for a ball. To construct Green’s function for the unit ball B(0, 1), we will again employ a kind of reflection, this time through the sphere ∂B(0, 1). DEFINITION. If x Rn {0}, the point ˜ x = x |x|2 is called the point dual to x with respect to ∂B(0, 1). The mapping x ˜ x is inversion through the unit sphere ∂B(0, 1). We now employ inversion through the sphere to compute Green’s func- tion for the unit ball U = B0(0, 1). Fix x B0(0, 1). Remember that we must find a corrector function φx = φx(y) solving (37) Δφx = 0 in B0(0, 1) φx = Φ(y x) on ∂B(0, 1) then Green’s function will be (38) G(x, y) = Φ(y x) φx(y). The idea now is to “invert the singularity” from x B0(0, 1) to ˜ x / B(0, 1). Assume for the moment n 3. Now the mapping y Φ(y ˜) x is harmonic for y = ˜. x Thus y |x|2−nΦ(y ˜) x is harmonic for y = ˜, x and so (39) φx(y) := Φ(|x|(y ˜)) x is harmonic in U. Furthermore, if y ∂B(0, 1) and x = 0, |x|2|y ˜|2 x = |x|2 |y|2 2y · x |x|2 + 1 |x|2 = |x|2 2y · x + 1 = |x y|2. Thus (|x||y ˜|)−(n−2) x = |x y|−(n−2). Consequently (40) φx(y) = Φ(y x) (y ∂B(0, 1)), as required.
Previous Page Next Page