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.
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