Übungsblatt 15

Übungsblatt 15
Analysis II, FS15
Ausgabe Mittwoch, 3. Juni. Das Blatt dient als Wiederholung von Teilen des Analysis
II Vorlesungsstoffs. Keine Abgabe, Musterlösungen werden online gestellt.
Übung 1.
Seien p, q > 0. Wir definieren eine Funktion f : R2 → R, sodass
(
f (x, y) =
xp y q
,
x2 −xy+y 2
(x, y) 6= (0, 0).
0,
(x, y) = (0, 0).
(i) Zeigen Sie, dass für jede (x, y) ∈ R2 gilt
|xy| ≤ x2 − xy + y 2 .
(ii) Untersuchen Sie die Stetigkeit von f im Punkt (0, 0).
(iii) Ist die Funktion differenzierbar an der Stelle (0, 0)?
Lösung 1.
(i) Since (|x| − |y|)2 = |x|2 − 2|xy| + y 2 ≥ 0.
(ii) Using the inequality in (i), we find that
|f (x, y)| ≤ |x|p−1 |y|q−1 .
If p + q > 2, then |x|p−1 |y|q−1 tends to zero when (x, y) → (0, 0) (which means
|(x, y)| → 0), so f is continuous at (0, 0).
If p + q ≤ 2, we consider f (x, x) = xp+q−2 for x ∈ R. If p + q = 2, then
limx→0 f (x, x) = 1; if p + q < 2, then limx↓0 f (x, x) = +∞. So f (x, y) does not
converges to zero when (x, y) converges to (0, 0) along the line x = y. We conclude
that f is not continuous at (0, 0).
(iii) If p + q ≤ 2, then f is not differentiable since f is not continuous.
Now we consider p + q > 2. We find that the partial derivatives of f exists at (0, 0):
∂f
f (0, y) − f (0, 0)
(0, 0) = lim
=0
y→0
∂y
|y|
1
∂f
f (x, 0) − f (0, 0)
(0, 0) = lim
=0
x→0
∂x
|x|
For every vector h = (hx , hy ) ∈ R2 , we set
∂f
∂f
f (hx , hy ) − f (0, 0) − ∂x (0, 0)hx + ∂y (0, 0)hy |hpx hqy |
=
R(h) := .
|h|
|h||h2x − hx hy + h2y |
If p + q > 3, then using inequality in (i) and inequality |hx |, |hy | ≤ |h|, we find that
R(h) ≤ |h|p+q−3
So we have limh→0 R(h) = 0 and thus the function f is differentiable by Definition
2.2.1.
√
If 2 < p + q ≤ 3, then let hn = ( n1 , n1 ), we find that R(hn ) = 22 n3−(p+q) . Suppose
that f is differentiable, then it follows from by Lemma 2.3.3 and√Definition 2.2.1
that limh→0 R(h) = 0 and a fortiori limn→∞ R(hn ) = limn→∞ 22 n3−(p+q) = 0,
which is absurd as p + q ≤ 3. Therefore f is not differentiable at (0, 0).
Übung 2.
Berechnen Sie im Folgenden
(i) die kritischen Punkte von f (x, y) = y(x2 +y 2 )−2y 2 +1 über x2 +y 2 < 2. Entscheide,
ob sie Minima, Maxima oder Sattelpunkte sind.
(ii) das Integral
Z 1Z 1 p
0
√
x3 + 1 dxdy
y
(iii) das Volumen der Menge zwischen den Oberflächen x2 + y 2 < z < x2 /2 + y 2 /2 + 1
(iv) das Linienintegral
Z
(x − y)
γ
über das Liniensegment von (1, 3) bis (5, −2)
(v) ∇ × F und ∇ · F für F = (x2 , 2z, −y).
Lösung 2.
(i)
∇f =
∂f ∂f
,
∂x ∂y
= (2xy, x2 + 3y 2 − 4y).
The critical points of f correspond to ∇f = 0. They are (0, 0) and (0, 4/3).
We next compute the Hessian matrix
2
The Hessian matrix is
"
Hf (x, y) =
2y
2x
2x 6y − 4
#
Its determinant is D(x, y) = 2y(6y − 4) − 4x2 = 12y 2 − 8y − 4x2 . At the first
critical point D(0, 0) = 0. So, no statement can be made based on Satz 3.5.5. The
function f (0, 0) = 1, and 0 < f (x, y) < 1 on [−0.5, 0.5] × [−0.5, 0.5]. This point is
a local maximum. For the other point, D(0, 4/3) = 32/3 > 0 and the first minor
2y = 2 × 4/3 = 8/3 > 0 is also positive. The matrix is thus positive definite and
726F
May 2O,2O14
Math
Second Midterm
by Silvester’s criteria this point is a local minimum.
(ii) It is easiest to first change the order of integration (Fubini Theorem). To do this,
points)
5.we{Q
draw
the domain.
√
Z 1 p
Z 1(u)ffi-,ru*
Z 1 Z x2 p
Z 1 p
the
integral
2(2
2 − 1)
2
x x3 + 1dx =
x3 + 1 dxdy =
x3 + 1 dydx =
.
√
9
0
0
0
0
y
I,'I**Td'xd'v
t /'
:
[[ffi't
-(l*J,|il
x-S
d*
:lx1 w J,
7bv"o ,= , ,rt/'
+J, l, \ +
4
g= T'
-/o
J
,
Abbildung 1: We: are integrating under the curve.
27
Lt=Xat
: ?(rr/=,) *Eits-
(iii) We find that the two paraboloids intersect over the circle x2 + y 2 = 2. We can then
write
7yz / *
dq=
Z
Z √2 Z √2−x2
√
√
Z
Z
Y= =tr--xQt=l
(u$*>-trN u = r,dzdydx
√
√
− 2 − 2−x2
x2 +y 2
x2 /2+y 2 /2+1
I
andz:****
2−x2
!
x2 y 2
+
+ 1 − x2 − y 2 dxdy
2
2
√
√
− 2 − 2−x2
the
!
Z √2solid
Z 2π bounded
r2
: n2 + y2 ftom
by the paraboloids
"
#√z
2
2
4
r
r
= 1 from above.
1 − (Hint:
rdrdθdraw
= 2π a picture.)
−
=π
(b) Compute the volume of
below
2
2
2
8
{-L-'* { : xaf
o?-Z?
€L
;
5 xH-*:
'( J:J
z-0
0
3
v- I{e' f
t
.D-(x"oJ
0
#I
(iv) The curve can be parametrized as γ = (1 − t)(1, 3) + t(5, −2) = (1 + 4t, 3 − 5t)
with t ∈ [0, 1]. The integral then is
Z
(x − y) =
γ
Z 1
0
√
5√
(9t − 2) 16 + 25dt =
41.
2
(v)
∇·F =
.
∇×F =
∂F
∂F
∂F
+
+
= 2x
∂x
∂y
∂z
∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x
∂y
= (−3, 0, 0),
where F = (F1 , F2 , F3 ) = (x2 , 2z, −y).
Übung 3.
Seien U ⊂ R und f : U × [a, b] → R, c : U → [a, b] C 1 -Abbildungen.
(i) Sei
G : U × [a, b] → R,
(x, c) 7→
Z c
f (x, t) dt.
a
Zeigen Sie, dass G eine C 1 -Abbildung ist.
(ii) Zeigen Sie, dass gilt:
∂
∂x
Z c(x)
f (x, t) dt = f (x, c(x))c0 (x) +
a
Z c(x)
∂f
a
∂x
(x, t) dt.
Lösung 3.
(i) We start by computing the partial derivatives: The derivative in x,
∂G
(x, c) =
∂x
Z c
∂f
a
∂x
(x, t) dt,
exists by the ‘Differentiationssatz’, Satz 4.1.1. The derivative in c,
∂G
(x, c) = f (x, c),
∂c
exists by the fundamental theorem of calculus, Theorem I.10.3.2.
Now, the derivative in c is certainly continuous because f is. However, although the
derivative in x is guaranteed to be continuous in x by Bemerkung 4.2.2, this result
from the notes is not enough to show continuity as a function of two variables.
We therefore do it by hand. This will get a bit messy but it is a fairly standard
method and is very close to how the Differentiationssatz itself is proved. We will
write g(x, c) = ∂f
∂x (x, c) for brevity.
4
Let (x, c) ∈ U × [a, b], and pick > 0. Since U is open and g is continuous, we
may use Lemma 4.1.2 or Lemma 4.1.4 to pick δ1 > 0 such that K̄δ1 (x) ⊂ U and,
if |x − y| < δ1 , then
max |g(x, t) − g(y, t)| <
.
2(b − a)
t∈[a,b]
Again using continuity, specifically Korollar 1.10.3 and Korollar 1.11.1, let
0 ≤ M := max{|g(y, t)| : |y − x| ≤ δ1 , t ∈ [a, b]} < ∞.
Now pick δ2 = /(2M ) if M 6= 0, or just if M = 0. Let δ = min(δ1 , δ2 ) and
(y, d) ∈ U × [a, b] be such that |(y, d) − (x, c)| < δ. Certainly |x − y| < δ1 and
|c − d| < δ2 . We estimate as follows (using the triangle inequality and properties
of integrals)
Z d
Z d
Z c
Z c
Z c
Z c
g(x, t) dt
g(y, t) dt −
g(y, t) dt +
g(y, t) dt −
g(x, t) dt = g(y, t) dt −
a
a
a
a
a
a
Z c
Z d
|g(x, t) − g(y, t)| dt
g(y, t) dt +
≤ a
c
≤ |c − d| max |g(y, t)| + (c − a) max |g(x, t) − g(y, t)|
t∈[a,b]
t∈[a,b]
<
M+
(c − a) ≤ ,
2M
2(b − a)
and we are done.
So, the partial derivatives of G in both x and c exist and are continuous, and so
G is in C 1 .
(ii) Let
F : U → R,
x 7→
Z c(x)
f (x, t) dt.
a
We are going to show that F is differentiable with the derivative given in the
question.
Define
h : U → U × [a, b],
x 7→ (h1 (x), h2 (x)) := (x, c(x)).
The function h is differentiable because its components are (Satz 5.1.3). The
‘Hauptkriterium für Differenzierbarkeit’, Satz 2.6.2, tells us that G is differentiable because each of the partial derivatives is continuous (as a function of two
variables).
We observe that F = G ◦ h. The chain rule, Satz 5.3.1, tells us that F is differentiable, as the composition of two differentiable functions, and even gives us its
derivative:
∂G
∂G
F 0 (x) =
(h(x))h01 (x) +
(h(x))h02 (x)
∂x
∂c
Z c(x)
∂f
=
(x, t) dt + f (x, c(x))c0 (x).
∂x
a
5
This completes the proof. Notice that we got for free from the proof that F is a
C 1 function. Also, the proof would work just fine if we had U ⊂ Rn and replaced
all x-derivatives with a partial derivative with respect to xi , or some directional
derivative.
Übung 4.
Berechnen Sie das Taylorpolynom bis zur zweiten Ordnung zu den im folgenden gelisteten Funktionen, mit x = (x1 , x2 , ..., xn ) und x0 = (x1,0 , x2,0 , ..., xn,0 ), x, x0 ∈ Rn .
Schätzen Sie das Restglied Ra2 f (x) = f (x) − Ta2 f (x) unter Verwendung der “klein-o” or
“groß-O” Notation ab:
i) f1 (x) = x1 − x31 − x42 in x0 = (−1, 0)
ii) f2 (x) = (ex1 + e−2x1 ) sin(2 x2 ) in x0 = (0, π)
iii) f3 (x) = cos(2 x1 + x2 ) + sin(x1 − x2 ) in x0 = (π, π)
Lösung 4.
The Taylor polynomial up to second order reads
1
Tx20 f (x) =f (x0 ) + h∇f (x0 ), (x − x0 )i + (x − x0 )T Hf (x)|x=x0 (x − x0 ) ,
2
see Bemerkung 3.4.1 in the script. H is the Hesse matrix. In component notation and
for x = (x1 , x2 , ..., xn ) and x0 = (x1,0 , x2,0 , ..., xn,0 ), so x, x0 ∈ Rn , it reads
Tx20 f (x) =f (x0 ) +
+
1
2
n
X
∂
i=1
n X
n
X
∂xi
f (x)|x=x0 (xi − xi,0 )
∂2
f (x)|x=x0 (xi − xi,0 )(xj − xj,0 )
∂xi ∂xj
i=1 j=1
(1)
Use Korollar 3.4.2 to estimate its remainder Ra2 f (x) with f ∈ C 3 (Ω) and Ω ⊂ Rn .
i) Here, x = (x1 , x2 ) and x0 = (−1, 0) and x, x0 ∈ R2 . Computing all partial derivatives of Eq. (1), we get
∂
∂
f1 (x)|(−1,0) (x1 + 1) +
f1 (x)|(−1,0) (x2 − 0)
∂x1
∂x2
1 ∂2
1 ∂2
2
+
f
(x)|
(x
+
1)
+
f1 (x)|(−1,0) (x1 + 1)(x2 − 0)
1
(−1,0) 1
2 ∂x21
2 ∂x1 ∂x2
1 ∂2
+
f1 (x)|(−1,0) (x2 − 0)2
2 ∂x22
2
T(−1,0)
f1 (x) =f1 (−1, 0) +
=0 − 2(1 + x1 ) + 3(1 + x1 )2 = 1 + 4 x1 + 3 x21
6
and the remainder is
2
2
R(−1,0)
f1 (x) = f1 (x) − T(−1,0)
f1 (x)
= −(1 + 3 x1 + 3 x21 + x31 + x42 ) ,
so we have
f1 (x) = 1 + 4 x1 + 3 x21 + O(|(x1 + 1)3 + x42 |) .
ii) Computing each term of Eq. (1) and collecting all results, one finds
f2 (x) = −2(x1 − 2)(x2 − π) + O(|(x1 , x2 ) − (2, π)|3 ) .
iii) To get a simpler calculation of the derivatives, the function can be written as
f3 (x) = cos(2 x1 + x2 ) + sin(x1 − x2 ) = cos(x1 )2 cos(x2 ) + cos(x2 ) sin(x1 )
− cos(x2 ) sin(x1 )2 − cos(x1 ) sin(x2 ) − 2 cos(x1 ) sin(x1 ) sin(x2 ) .
Again, we can expand the two functions independently and collect the terms up
to the second order. One finds
1
f3 (x) = −1 + x1 − x2 + (2 x1 + x2 − 3π)2 + O(|(x1 , x2 ) − (π, π)|3 ) .
2
Übung 5.
Sei f : R → R eine C 1 -Funktion, so dass
|f 0 (t)| ≤ k < 1
für t ∈ R.
Betrachte die Funktion
ϕ : R2 → R2 ,
(x, y) 7−→ (x + f (y), y + f (x)).
Zeigen Sie, dass ϕ ein Diffeomorphismus ist.
Lösung 5.
Since f and ϕ are functions of class C 1 , the Jacobian matrix of ϕ is given by
Jϕ(x, y) =
1
f 0 (y)
0
f (x)
1
!
.
We observe that Jϕ(x, y) is invertible since detJϕ(x, y) = 1 − f 0 (x)f 0 (y) ≥ 1 − k 2 > 0;
Therefore, the inverse function theorem (Satz 5.6.1.) implies that ϕ is a diffeomorphism
in the neighborhood of every point (x, y) ∈ R2 . To prove that ϕ is a diffeomorphism from
R2 to R2 , it is enough to show that ϕ is a bijective function. Because in a neighborhood
7
of any point local and global inverse must coincide.
We take (u, v) ∈ R2 . Then, the value of (x, y) ∈ R2 such that ϕ(x, y) = (u, v) is determined by the relationship
y = v − f (x)
and
u = x + f (v − f (x)).
Consider the function g(x) = x − f (v − f (x)) for x ∈ R. Clearly, g is a function of class
C 1 and its derivative is given by (we apply the chain rule)
g 0 (x) = 1 − f 0 (v − f (x))f 0 (x).
Moreover, we have that g 0 (x) ≥ 1 − k 2 > 0 from our assumption over f . Thus, g is a
bijective and strictly increasing function from R to R. Hence there exists a unique x ∈ R
such that g(x) = u, and therefore a unique (x, y) ∈ R2 such that ϕ(x, y) = (u, v), which
prove the bijectivity of ϕ.
8