By H.; Beckert, H. Schumann

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**Extra resources for 100 Jahre Mathematisches Seminar der Karl-Marx-Universitaet Leipzig**

**Example text**

32) By the chain rule we obtain wxx = urr rx2 + 2urθ rx θx + uθθ θx2 + ur rxx + uθ θxx , wyy = urr ry2 + 2urθ ry θy + uθθ θy2 + ur ryy + uθ θyy . 32): rx = √ rxx = x x2 +y 2 , y2 (x2 +y 2 )3/2 ry = √ , ryy = y x2 +y 2 , x2 (x2 +y 2 )3/2 θx = −y x2 +y 2 , θxx = , 2xy (x2 +y 2 )2 θy = , ryy = x x2 +y 2 , −2xy (x2 +y 2 )2 . 41 Therefore, wxx + wyy = urr (rx2 + ry2 ) + 2urθ (rx θx + ry θy ) + uθθ (θx2 + θy2 ) 1 1 +ur (rxx + ryy ) + uθ (θxx + θyy ) = urr + ur + 2 uθθ . r r (b) In polar coordinates x = r cos θ, y = r sin θ, 0 < r < √ 6, −π ≤ θ < π we obtain the problem 1 1 urr + ur + 2 uθθ = 0 r √ √r u( 6, θ) = 6 sin θ + 6 sin2 θ 0

Assume first that (ξ, η) = (0, 0). It is easy to check that GR (x, y; 0, 0) |x2 +y2 =R2 = 0. On the other hand, GR (x, y; 0, 0) = Γ(x, y) + constant, therefore, ∆ −1 r ln 2π R = −δ(x, y). Suppose now that (ξ, η) = (0, 0). Then GR (x, y; ξ, η) = Γ(x − ξ, y − η) − Γ(R−1 ˜ y − η˜)). ξ 2 + η 2 (x − ξ, ˜ η˜) ∈ BR , it follows that Γ(R−1 ξ 2 + η 2 (x − ξ, ˜ y − η˜)) is harmonic in BR . On Since (ξ, ˜ y−˜ η )) = Γ(x−ξ, y−η). the other hand, for (x, y) ∈ ∂BR we have Γ(R−1 ξ 2 + η 2 (x−ξ, Therefore, GR (x, y; ξ, η) is the Green function in BR .

Then vx (x0 , y0 ) = vy (x0 , y0 ) = 0, vxx (x0 , y0 ) ≤ 0 , vyy (x0 , y0 ) ≤ 0. Therefore, at this point vxx + vyy + xvx + yvy ≤ 0, which is a contradiction. (b) Let ε > 0 . The function vε satisfies (vε )xx + (vε )yy + x(vε )x + y(vε )y > 0 , and thus according to part (a) the maximum of vε is obtained on ∂D. Let M be the maximum of u on ∂D. For all (x1 , y1 ) ∈ D u(x1 , y1 ) ≤ vε (x1 , y1 ) ≤ max{vε (x, y) | (x, y) ∈ ∂D} ≤ M + επ 2 . Letting ε → 0, we obtain u(x1 , y1 ) ≤ M . (c) Write w(x, y) := u1 (x, y) − u2 (x, y), where u1 (x, y), u2 (x, y) are two solutions of the problem.