Übungsblatt 8

Übungsblatt 8
Analysis II, FS15
Ausgabe Donnerstag, 16. April.
Abgabe Donnerstag, 23. April, 17 Uhr. Die Lösungen bitte in den Briefkasten oder in
die Ablage der jeweiligen Übungsgruppe am J- oder K-Geschoss von Bau Y27 legen.
Übung 1.
Sei f : R2 → R durch
f (x, y) =

2
2

 xy(x − y ) ,
(x, y) 6= (0, 0),

0,
(x, y) = (0, 0).
(x2 + y 2 )3
gegeben. Rechnen Sie die (möglicherweise uneigentliche) Integrale
Z 2Z 1
Z 1Z 2
f (x, y) dx dy
0
f (x, y) dy dx
und
0
0
0
aus. Schliessen Sie, dass f im Punkt (0, 0) nicht stetig ist.
Lösung 1.
R
We start with the inner integral in the first expression. We compute g(y) = 02 f (x, y) dx
for y ∈ [0, 1]. If y = 0, then f (x, y) = 0 for all x ∈ [0, 2], so g(0) = 0. So suppose that
y 6= 0. The function f (·, y) is continuous on [0, 2], and is therefore a regulated function.
We can therefore compute
Z 2
f (x, y) dx =
g(y) =
Z 2
xy(x2 − y 2 )
0
0
=
(x2 + y 2 )3
dx
Z 4+y2
y(u − 2y 2 )
y2
2u3
y
y3
= −
+ 2
2u 2u
du
4+y2
=−
y2
2y
,
(4 + y 2 )2
y 6= 0.
where for the third equality, we used the substitution u = x2 + y 2 in conjunction with
Satz 10.4.3 from Analysis I. To finish computing the first integral, we just have to find
1
R1
g(y) dy. The function g is continuous on its whole domain, so it is a regulated function.
We calculate:
0
Z 1
Z 1Z 2
f (x, y) dx dy =
0
g(y) dy = −
Z 1
0
0
0
2y
1
dy =
2
2
(4 + y )
4 + y2
1
=−
0
1
.
20
R1
We now move on to the second iterated integral. Let h(x) = 0 f (x, y) dy for x ∈ [0, 1].
As before, h(0) = 0. So we suppose that x 6= 0, and again f (x, ·) is a regulated function.
Hence:
Z 1
f (x, y) dy =
h(x) =
Z 1
xy(x2 − y 2 )
(x2 + y 2 )3
0
0
=
dy
Z 1+x2
x(2x2 − v)
2v 3
x2
x3
x
− 2
=
2v 2v
dv
1+x2
=
x2
x
,
2(1 + x2 )2
x 6= 0.
using the substitution v = x2 + y 2 in the third equality. Just as before, we observe that
h is continuous, so
h(x) dx =
f (x, y) dy dx =
0
0
Z 2
Z 2
Z 2Z 1
0
0
x
1
dx = −
2
2
2(1 + x )
4(1 + x2 )
2
0
1
= .
5
We observe that the two iterated integrals are not equal. Finally we connect it to the
last part of the question. The function f is certainly continuous at every point apart
from (0, 0). Suppose (for a contradiction) that f is also continuous at (0, 0). Then it is
continuous on its whole domain. Fubini’s theorem (Satz 4.3.1) then implies that the two
iterated integrals are equal, but we know that they are not!
This exercise is taken from http://www.math.jhu.edu/~jmb/note/nofub.pdf by
J. M. Boardman.
Übung 2.
Seien U ⊂ Rn offen und f ∈ C 1 (U, Rk ). Wir definieren
|f (u) − f (v)|
.
|u − v|
u,v∈U,
[f ]U = sup
u6=v
Zeigen Sie dass, falls U konvex ist, gilt
[f ]U = sup df |z O ,
z∈U
wobei k · kO die Operatornorm auf L (Rn , Rk ) (siehe Definition 2.1.6) ist.
2
(1)
Lösung 2.
We prove ‘≤’ and ‘≥’. The first inequality is straightforward: by the Schrankensatz (Satz
5.4.2, or Korollar 5.4.5), we obtain, for arbitrary u, v ∈ U ,
|f (u) − f (v)| ≤ sup df |z O |u − v| ≤ supdf |z O |u − v|,
z∈U
z∈[u,v]
and after dividing through by |u − v|, which is non-zero so long as u 6= v, we obtain the
‘≤’ direction.
The other direction needs a bit more work. For h ∈ Rn , we have, using the definition
of the differential and the continuity of | · |,
df |z (h) = lim |f (z + th) − f (z)|
|t|
|f (z + th) − f (z)| |th|
= lim
t→0
|z + th − z|
|t|
|th|
≤ [f ]U lim
= |h|[f ]U .
t→0 |t|
t→0
We then compute, using the definition of the operator norm and the above calculation:
df |z O
= sup df |z (h) ≤ [f ]U .
|h|≤1
This immediately leads to supz∈U df |z O ≤ [f ]U , which completes the proof.
Note that there is no reason for either side of (1) to be finite, but every part of the
solution above also works if one of the sides is infinite.
Übung 3.
Berechnen Sie mithilfe der Kettenregel (Satz 5.3.1 oder Korollar 5.3.5) das Taylorpolynom erster Ordnung Ta1 (f ◦ g) für folgende f , g und a:
(i) g(x, y) = (x2 + 2y, xy + arctan x + 1) und f (u, v) = uv + ln(1 + v 2 ) mit a = (0, 0).
(ii) g(x, y, z) = (−z cos x + y sin z + 1/2, e2x+y+3z ) mit f (u, v) = u2 + ln v mit
a = (2π, 1, 0).
Lösung 3.
(i) We start with the partial derivatives of f :
∂f
(u, v) = v,
∂u
∂f
2v
(u, v) = u +
,
∂v
1 + v2
and the Jacobian of g:
Jg(x, y) =
∂g1
∂x (x, y)
∂g2
∂x (x, y)
∂g1
∂y (x, y)
∂g2
∂y (x, y)
3
!
=
2x
1
y + 1+x
2
!
2
.
x
With this information we can compute the partial derivatives of f ◦ g using the
chain rule (or its ‘concrete’ formulation, Korollar 5.3.5):
∂(f ◦ g)
∂f
∂g1
∂f
∂g2
(x, y) =
(g(x, y))
(x, y) +
(g(x, y))
(x, y).
∂x
∂u
∂x
∂v
∂x
Since we are only interested in the point (x, y) = (0, 0) we can simplify the calculation for ourselves:
∂(f ◦ g)
∂f
∂g1
∂f
∂g2
(0, 0) =
(g(0, 0))
(0, 0) +
(g(0, 0))
(0, 0)
∂x
∂u
∂x
∂v
∂x
∂f
∂g1
∂f
∂g2
=
(0, 1)
(0, 0) +
(0, 1)
(0, 0) = 1.
∂u
∂x
∂v
∂x
Similarly we find
∂f
∂g1
∂f
∂g2
∂(f ◦ g)
(0, 0) =
(0, 1)
(0, 0) +
(0, 1)
(0, 0) = 2.
∂y
∂u
∂y
∂v
∂y
Using the definition of the Taylor polynomial we therefore find
1
T(0,0)
(f ◦ g)(x, y) = ln 2 + (x − 0) + 2(y − 0) = ln 2 + x + 2y.
{There is an alternative way to write out this question, if you are happy with
combining linear maps using their matrix representation. Namely, Satz 5.3.1 states
that d(f ◦ g)|a = df |g(a) ◦ dg|a , and on the level of the matrix representation (the
Jacobian matrices) this says that Jf ◦g (a) = Jf (g(a))Jg (a); see also Bemerkung
5.3.3. We can therefore easily calculate
Jf ◦g (0, 0) = Jf (0, 1)Jg (0, 0) = 1 1
0 2
!
1 0
= 1 2 ,
and this gives us the partial derivatives used for the Taylor polynomial.}
(ii) We calculate the same quantities:
∂f
(u, v) = 2u,
∂u
∂f
1
(u, v) = ,
∂v
v
!
Jg(x, y, z) =
z sin x
sin z
y cos z − cos x
.
2e2x+y+3z e2x+y+3z
3e2x+y+3z
We then use the chain rule again:
∂(f ◦ g)
∂f 1 4π+1 ∂g1
∂f 1 4π+1 ∂g2
(2π, 1, 0) =
(2, e
)
(2π, 1, 0) +
( ,e
)
(2π, 1, 0) = 2
∂x
∂u
∂x
∂v 2
∂x
and (omitting details) we obtain also
∂(f ◦ g)
(2π, 1, 0) = 1,
∂y
4
∂(f ◦ g)
(2π, 1, 0) = 3.
∂z
Thus we obtain
1
T(2π,1,0)
(f ◦ g)(x, y, z) =
5
1
+ 4π + 2(x − 2π) + (y − 1) + 3(z − 0) = + 2x + y + 3z.
4
4
{Using the Jacobian matrix representation we obtain the same result:
Jf ◦g (2π, 1, 0) = Jf ( 21 , e4π+1 )Jg (2π, 1, 0)
= 1 e−4π−1
0
0
0
2e4π+1
e4π+1
3e4π+1
!
= 2 1 3 ,
and this can be used in the Taylor polynomial calculation.}
Übung 4.
(i) Seien D, E ⊂ R Intervalle. Sei f : D → E eine differenzierbare, bijektive Funktion
deren Ableitung nie Null ist. Sei F : D → R, so dass F 0 = f . Wir definieren
G : E → R mit
G(y) = yf −1 (y) − F (f −1 (y)),
y ∈ E.
Zeigen Sie, dass G differenzierbar ist mit G0 = f −1 .
[Hinweis: Analysis I, Satz 9.2.6 und Analysis II, Übungsblatt 1, Übung 1.]
(ii) Sei f wie oben, und sei auch 0 ∈ D. Sei ϕ : D × R → R2 , ϕ(x, y) = (f (x) + y 2 , y 3 ).
Zeigen Sie, dass ϕ injektiv ist, und finden Sie die Jacobi-Matrix von ϕ im Punkt
(0, 0). Finden Sie einen Ausdruck für ϕ−1 : ϕ(D × R) → R2 . Berechnen Sie auch
die Jacobi-Matrix von ϕ−1 in jedem Punkt, an dem die Funktion differenzierbar
ist. Zeigen Sie, dass ϕ−1 nicht differenzierbar ist im Punkt (f (0), 0).
Lösung 4.
(i) Before beginning, let’s think about why F exists at all. This is a consequence of
the fundamental theorem of calculus (Hauptsatz der Differential- und Integralrechnung, Analysis I, Theorem 10.3.2);
since f is differentiable on D, it is certainly
R
continuous, so we may take F (x) = ax f (t) dt for any a ∈ D.
Now we need to show that G is differentiable. The first step is to show that f −1
is differentiable. This is a consequence of the inverse function theorem (Analysis
I, Satz 9.2.6) and the result given in Übungsblatt 1, Übung 1 of this course. Since
f 0 (x) 6= 0 for every x ∈ D, it follows that f −1 is differentiable at every point y ∈ E,
and the derivative (using the formula from the inverse function theorem) is given
by
1
(f −1 )0 (y) = 0 −1
.
f (f (y))
So G is a composition of differentiable functions, and hence is also differentiable,
and in fact we can easily compute its derivative using the product and chain rules:
G0 (y) = f −1 (y) +
y
f 0 (f −1 (y))
5
−
f (f −1 (y))
= f −1 (y),
f 0 (f −1 (y))
which was to be shown.
(ii) We start with injectivity. If ϕ(x, y) = ϕ(x0 , y 0 ), then y 3 = (y 0 )3 , and since the map
R 3 y 7→ y 3 is injective, it follows that y = y 0 . We then observe that f (x) + y 2 =
f (x0 ) + (y 0 )2 = f (x0 ) + y 2 , so f (x) = f (x0 ). But f is injective, so x = x0 , and we
are done.
The Jacobian of ϕ at any point is given by
!
Jϕ(x, y) =
so at zero we have
f 0 (x) 2y
,
0
3y 2
!
Jϕ(0, 0) =
f 0 (0) 0
.
0
0
The inverse map is given by
!
−1
ϕ
(u, v) =
f −1 (u − v 2/3 )
,
v 1/3
(u, v) ∈ ϕ(D × R).
We observe that (as a composition of differentiable functions) each component
(ϕ−1 )i of ϕ−1 is differentiable at every point (u, v) where v 6= 0. By Satz 5.1.3(1),
it follows that ϕ−1 is also differentiable at those points. Its Jacobian is given by

1
 0 −1
2/3

Jϕ−1 (u, v) =  f (f (u − v ))

0

2v −1/3
− 0 −1

3f (f (u − v 2/3 )) 
.

1 −2/3
v
3
Finally we make the following observation. Assume that ϕ−1 is differentiable at
−1
(f (0), 0). Then its partial derivative ∂(ϕ∂v )2 (f (0), 0) must exist; but v 7→ (ϕ−1 )2 (f (0), v) =
v 1/3 is not differentiable at v = 0, so we have a contradiction.
The point of part (ii) is that ϕ is a function whose differential (Jacobian) at the
point (0, 0) is non-invertible but non-zero, but whose inverse is nonetheless nondifferentiable at ϕ(0, 0). This is in contrast to the one-dimensional case of part (i),
where ‘non-zero’ and ‘invertible’ mean the same thing. We will see the full story
for higher dimensions in sections 5.5 and 5.6 of this course.
6