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Hand in y our solutions un til W ednesda y , 05/03/17, 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 II, 04/26/17

Exercise II.1 (3+2 Points) Assume u(t, x) =v

x2 t

for t >0and x∈R. (a) Prove thatu solves the equation

tu−∂2xu= 0 in(0,∞)×R if and only if forz≥0

4zv00(z) + (2 +z)v0(z) = 0. (1) (b) Show that the general solution of (1) is

v(z) =c1

z

Z

0

e−s/4s−1/2ds+c2.

Exercise II.2 (5 Points)

Prove the following statement (Proposition 1.6): Letf ∈C(Rd)∩L1(Rd). Then for every x0 ∈Rd

Z

Rd

f(x)dx=

Z

0

 Z

∂Br(x0)

fdO

 dr.

Furthermore, for everyr >0 andx0∈Rd

d ds

Z

Bs(x0)

f(x)dx s=r

= Z

∂Br(x0)

fdO.

Exercise II.3 (3+2 Points)

LetQ= (0,1)3andS={(x, y, z)∈R3|x2+y2+z2 = 1}. Compute the following integrals with the help of the divergence theorem:

(a)

Z

Q

div(x2, yz, y)T dxdydz,

(b)

Z

S

3x2z2+y4+ 3y2x2+zy2x2dO.

(2)

Exercise II.4 (5 Points) Let

A={(x, y, z)∈R3|(x, y)∈B1(0)⊂R2,0< z < h(x, y)}

for some positive function h∈C1(B1(0)).

Prove the divergence theorem for this specific choice ofA, i.e., forf ∈C1(R3,R3) prove Z

A

divf = Z

∂A

(f, ν)dO.

2

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