Hand in y our solutions un til W ednesda y , 05/24/17, 14:15 (PO b o x of y our T A in V3-128)
total points: 20
Prof. Dr. Moritz Kaßmann Fakultät für Mathematik
Sommersemester 17 Universität Bielefeld
Partial Differential Equations Exercise sheet V, 05/17/17
Exercise V.1 (5 Points)
Let Ω ⊂ Rd be a bounded domain, f ∈ C(Ω) andg ∈C(∂Ω). Let u ∈ C2(Ω)∩C(Ω) satisfy
(−∆u=f inΩ,
u=g on∂Ω. (1)
Prove the existence of a constant µ=µ(d,Ω)only depending on the dimensiondand the domain Ωsuch that
max
Ω
|u| ≤µ
max∂Ω |g|+ max
Ω
|f|
.
Hint: Consider the function eu∈C2(Ω)∩C(Ω),u(x) =e u(x) +λ|x|2 for an appropriate choice of the constantλ.
Exercise V.2 (2+2 Points)
a) Show that the Laplace operator in polar coordinates is given by
∆ =∂r2+1 r∂r+ 1
r2∂ϕ2.
Hint: Let u∈C2(R2) and
T : (0,∞)×[0,2π)→R2\ {0}, T(r, ϕ) = (rcosϕ, rsinϕ)
be the mapping that transforms polar coordinates into cartesian coordinates. Consider v(r, ϕ) = (u◦T)(r, ϕ). Calculate ∂rv and∂ϕv.
b) With help of a) prove that the function u : R2 \ {0} → R, u(x, y) = x2+yy 2 is harmonic.
Exercise V.3 (3+3 Points)
Let g ∈C2(R), h∈ C1(R) and c >0. Find a solution u : [0,∞)×[0,2π]→ R of the problem
∂t2u=c2∂x2u in(0,∞)×(0,2π), u(0,·) =g in[0,2π],
∂tu(0,·) =h in(0,2π), u(·,0) =u(·,2π) = 0 in[0,∞).
for the following functions g andh:
a) g: [0,2π]→R, g(x) = sin(nx), n∈N, and h: [0,2π]→R, h(x) = 0.
b) g: [0,2π]→R, g(x) = 0, n∈N, and h: [0,2π]→R, h(x) = sin(nx), n∈N. Exercise V.4 (3+2 Points)
Letd≥2, g∈C2(Rd)and h∈C1(Rd). Letu∈C2([0,∞)×Rd)be a solution of
∂t2u−∆u= 0 in(0,∞)×Rd, u(0,·) =g inRd,
∂tu(0,·) =h inRd. Forx∈Rd defineUx: [0,∞)×(0,∞)→R,
Ux(t, r) =
∂Br(x)
u(t, y)dO(y).
Show the following for everyx∈Rd: a) Ux∈C2([0,∞)×[0,∞))
b) ∂2tUx−∂r2Ux−d−1r ∂rUx= 0 in(0,∞)×(0,∞)
Hint: Use the identity
r1−d∂r(rd−1∂rUx) = d−1r ∂rUx+∂r2Ux.
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