18.156 Lecture 4 Notes

Lecture 4 Removable Singularity Theorem Theorem 1 Let u be harmonic in Ω \ {x0 }, if � o(|x − x0 |2−n ) u(x) = o(ln |x − x0 |) , , n > 2, n=2 as x → x0 , then u extends to a harmonic function in Ω. Proof: Without loss of generality, we can assume Ω = B(0, 2), then u|∂B(0,1) is contin­ uous. Thus by Poisson Integral formula, ∃v ∈ C(B(0, 1)) ∩ C ∞ (B(0, 1)) to be harmonic function with boundary condition v = u on ∂B(0, 1). Choose � > 0 and δ > 0 small, consider � u(x) − v(x) − �(|x|2−n − 1) , n > 2, ω(x) = u(x) − v(x) + � log |x|) , n = 2, then ω(x) is harmonic on B1 (0) \ BBδ (0), and ω(x) = 0 on ∂B1 (0). On ∂Bδ (0), −�|x|2−n is the dominate term, thus ω ≤ 0 on ∂Bδ (0) for δ small enough. Now by maximum principle, ω ≤ 0 on B1 (0) \ Bδ (0), i.e. u(x) ≤ v(x) + �(|x|2−n − 1), Thus by letting � → 0, we get u(x) ≤ v(x), ∀x ∈ B1 (0) \ Bδ (0). This is true for any δ small, so it is true for ∀x ∈ B1 (0) \ {0}. By reverting u and v, we can get v(x) ≤ u(x), ∀x ∈ B1 (0) \ {0}, thus v(x) = u(x), ∀x ∈ B1 (0) \ {0}. Now we can define u(0) = v(0), and extend u to be a harmonic function on B(0, 1), thus a harmonic function on Ω = B(0, 2). � Example This gives an example of Dirichlet problem that is NOT solvable: Take Ω = B(0, 1) \ {0}, then ∂Ω = ∂B(0, 1) ∪ {0}. Consider the Dirichlet problem ⎧ ⎨ Δu = 0 , in Ω, u=0 , on ∂B(0, 1), ⎩ u=1 , at 0 1 If this is solvable, then the solution u can be extend to a bounded harmonic function on B(0, 1). Now by MVP, u(0) = 0, which is a contradiction. Laplacian in general coordinate systems Theorem 2 Let gij be the metric component of a coordinate system, then � 1 Δu = � ∂k (g kj ∂j u det(grs )). det(grs ) Proof: Take any ϕ ∈ C0∞ , we have � � � ϕΔu det(gij )dy = � = � = � = � = Thus the formula follows. ϕΔudx < �ϕ, �u > dx � ij g ∂i ϕ∂j u det(gij )dy � ij ϕ∂i (g ∂j u det(gij ))dy � � ∂i (g ij ∂j u det(gij )) � dy. ϕ det(gij ) det(gij ) � Laplacian in spherical coordinates (r, ω) Now g = dr2 + r2 gS n−1 , so � � � � 1 0 1 0 ij ⇒ ) = . (gij ) = = (g 0 r2 gS n−1 0 r−2 gSijn−1 so � det(gij ) = � r2(n−1) det(gS n−1 ) = rn−1 thus 2 � det(gS n−1 ), Δu = 1 rn−1 ∂1 (g 1j ∂j u rn−1 � � det(gS n−1 )) det(gS n−1 ) � � 1 � ∂k (g kj ∂j u rn−1 det(gS n−1 )) + rn−1 det(gS n−1 ) k>1 � � 1 1 = n−1 ∂r (∂r u rn−1 ) + � ∂k (r−2 gSijn−1 ∂j u det(gS n−1 )) r det(gS n−1 ) k>1 = = 1 ∂r (∂r u rn−1 1 rn−1 rn−1 ) + r−2 ΔS n−1 u (urr rn−1 + ur (n − 1)rn−2 ) + r−2 ΔS n−1 u = urr + (n − 1) 1 ur + 2 ΔS n−1 u. r r If u(r, θ) = f (r)B(θ) is variables separated, then Δu(r, θ) = (frr + (n − 1) fr f (r) )Bθ + 2 ΔS n−1 B(θ). r r Proposition 1 Let B(θ) be a homogeneous harmonic polynomial of degree k restricted to S n−1 , then ΔS n−1 B(θ) = −k(k + n − 2)B(θ). Remark 1 Let Pk be the set of homogeneous polynomials of degree k on Rn , Hk be the set of harmonic homogeneous polynomials of degree k on Rn , then Pk = Hk ⊕ r2 Pk−2 . It’s not hard to prove dimPk = so dimHk = (k + n − 1)! , k!(n − 1)! (k + n − 3)! (k + n − 1)! (k + n − 3)! − = (2k + n − 2) . k!(n − 1)! (k − 2)!(n − 1)! k!(n − 2)! For such a B(θ) ∈ Hk , we have Δ(f (r)B(θ)) = (frr + fr n−1 fr − k(k + n − 2) 2 )B(θ). r r For the solution of the equation frr + n−1 fr fr − k(k + n − 2) 2 = 0, r r let f = rp , then fr = p rp−1 , frr = p(p − 1)rp−2 , we get 0 = p(p − 1)rp−2 + p(n − 2)rp−2 − k(k + n − 2)rp−2 = (p − k)(p + k + n − 2)rp−2 . 3 Thus p = k or p = −k − n + 2. For p = k, we get u(r, θ) = rk B(θ), where B(θ) ∈ Hk , thus u is just the homogeneous k harmonic polynomial on Rn . For those p = −k − n + 2, if k = 0, then p = 2 − n and B(θ) = constant, thus u = c · r2−n , which is the fundamental solution. if k > 0, then p < 2 − n, note that B(θ) is defined on the compact set S n−1 , thus B is bounded, so u grows faster than the fundamental solution near the origin. From above we get a degree gap of harmonic function: · · · · · · , −n, −(n − 1), −(n − 2), �, 0, 1, 2, · · · · · · Notice that we have to have the gap in view of our removable singularity theorem. Homogeneous expansions Theorem 3 Any harmonic function in B(0, 1) can be expressed as an infinite sum u(x) = ∞ � pk (x), pk ∈ Hk . k=0 � Proof: Take the Taylor expansion of u, u = pk , where pk ∈ Pk , we have � 0 = Δu = Δpk , but Δpk ∈ Pk−2 , thus Δpk = 0 for all k, i.e. pk ∈ Hk . 4 �