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2. Reduction from ∇ · r(x)∇u(x) = 0 (1) ∆v(x) − q(x)v(x) = 0 (2) to together with Λr to Λq . We consider the problems in a domain Ω ⊂ Rn with a sufficiently regular boundary ∂Ω. Also we assume that r(x) > 0 for x ∈ Ω. Let u(x) be a solution to (1) and consider v = r1/2 u. Then 0 = ∇ r∇ r−1/2 v   = ∆ r1/2 v − ∇ r−1/2 v∇r   = v∆r1/2 + 2 ∇r1/2 (∇v) + r1/2 ∆v − 2∇ v∇r1/2   = v∆r1/2 + 2 ∇r1/2 (∇v) + r1/2 ∆v − 2 ∇r1/2 (∇v) − 2v∆r1/2    = r1/2 ∆v − v∆r1/2 = r1/2 ∆v − r−1/2 ∆r1/2 v ,  which implies (2) for q = r−1/2 ∆r1/2 . (3) Consider the problem ∆v − qv = 0 in Ω, v|∂Ω = f (4) with a general potential q ∈ L∞ (Ω). If the problem (4) with f ≡ 0 has no solution then the following Dirichlet-to-Neumann map Λq : H 1/2 (∂Ω) → H −1/2 (∂Ω) associated with (4) by Λq f = ∂v ∂ν ∂Ω is well defined. The normal derivative on the boundary is understood in the weak sense,   Z ∂v ,h = ∇v · ∇w + qvw, ∂ν ∂Ω Ω where w ∈ H 1 (Ω) and h = w|∂Ω ∈ H 1/2 (∂Ω). Let u ∈ H 1 (Ω) be the unique solution to (1) with u|∂Ω = r−1/2 f ∈ H 1/2 (∂Ω). Then r1/2 u = v 1 solves (4), and for w ∈ H 1 (Ω) with w|∂Ω = h ∈ H 1/2 (∂Ω) we get Z Z Z hLq f, hi = ∇v · ∇w + qvw = ∇v · ∇w + w∆v = ∇ (w∇v) Ω Ω Ω Z Z Z Z    ∂r1/2 1/2 1/2 uw + ∇ r w∇u = = ∇ wu∇r + ∇ r−1/2 w · r∇u ∂ν Ω ∂Ω Ω Ω Z Z Z 1/2  ∂r + ∇ r−1/2 w · (r∇u) + r−1/2 w∇ (r∇u) = uw ∂ν Ω Ω ∂Ω Z Z 1/2  ∂r = r−1/2 f h + r∇u · ∇ r−1/2 w ∂ν ∂Ω Ω   1/2  −1/2 −1/2 −1/2 ∂r ,h . = Lr r f ,r h + r f ∂ν Hence Λq f = r −1/2    1 ∂r Λr + r−1/2 f . 2 ∂ν 2 (5)

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