Übungsblatt 15

Übungsblatt 15
Analysis 1, HS14
Ausgabe 6. Januar 2015.
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Übung 1.
Geben Sie alle Lösungen der gewöhnlichen Differentialgleichung
ty 0 (t) + y(t) = te−2t ,
t ∈ [1, ∞),
an. Wenn wir die Anfangswertbedingung y(1) = 0 einführen, was ist die eindeutige
Lösung vom Anfangswertproblem?
Lösung 1.
We first rewrite the equation as
1
y 0 (t) = − y(t) + e−2t ,
t
which is a linear ODE (ordinary differential equation) of the form given in the script in
equation (11.3), with
1
g(t) = − ,
h(t) = e−2t .
t
Since g and h are continuous on the domain given, we may apply Satz 11.1.5, which
tells us that y is a solution to the ODE if and only if
y(t) = e
Rt
1
g(s) ds
Z t
C+
−
e
Rτ
1
g(s) ds
h(τ ) dτ ,
1
for some C ∈ R.
We evaluate (using the fundamental theorem of calculus, since g is continuous on the
domain of the integral)
Z τ
1
g(s) ds = −
Z τ
1
s−1 ds = − ln(s)|τ1 = − ln(τ ).
1
We then evaluate (using integration by parts, since everything in sight is continuous):
Z t
−
e
Rτ
1
g(s) ds
Z t
h(τ ) dτ =
τ e−2τ dτ
1
1
t
1
1 t −2τ
e
dτ
= − τ e−2τ +
2
2 1
1
1 −2τ 1 −2τ t
= − τe
− e 2
4
1
1 −2t 1 −2t 3 −2
= − te
− e
+ e .
2
4
4
Z
Now we are done. Writing D = C + 43 e−2 to simplify presentation, we have that y is
a solution if and only if
D 1 −2t
1
y(t) =
− e
− e−2t .
t
2
4t
Finally we find the solution to the initial value problem. This amounts to finding the
correct value of D in the formula above. Substituting, we get
1
1
0 = y(1) = D − e−2 − e−2 ,
2
4
from which we obtain that
y(t) =
3 −2 1 −2t
1
e − e
− e−2t
4t
2
4t
solves the initial value problem. This solution is unique, because no other value of D will
satisfy the initial condition.
Übung 2.
Finden Sie eine Lösung von jeder der folgenden gewöhnlichen Differentialgleichungen.
Bestimmen Sie den Definitionsbereich von Ihrer Lösung.
(i) et y 0 (t) − cos2 (y(t)) = 0, mit Bedingung y(0) = 0. In dieser Unteraufgabe soll die
Lösung explizit gegeben werden.
y(t)
(ii) ty 0 (t) = 1+y(t)
, mit Bedingung y(1) = 12 . In dieser Unteraufgabe darf die Lösung in
impliziter Form gegeben werden (d.h., sie darf die Umkehrfunktion einer bestimmten, von Ihnen angegebenen, Funktion enthalten.)
Lösung 2.
This question concerns separable ODEs, so we will put everything in terms of the general
equation given by (11.6).
(i) Rearranging, we obtain
y 0 (t) = e−t cos2 (y(t)),
2
which is (11.6) with g(t) = e−t and F (y) = cos2 (y). The additional condition gives
y0 = 0 and t0 = 0.
We observe that F and g are continuous functions with F (0) = 1. This means that
Satz 11.3.1 can be applied, so we will calculate all the quantities involved.
1
1
=
,
F (y)
cos2 (y)
y ∈ (−π/2, π/2),
and using the fundamental theorem of calculus1
Z y
H(y) =
0
1
dz =
F (z)
Z y
0
1
dz = tan(z)|y0 = tan(y),
cos2 (z)
y ∈ (−π/2, π/2).
Again using the fundamental theorem of calculus
Z t
Z t
g(s) ds =
G(t) =
0
0
t
e−s ds = −e−s 0 = 1 − e−t ,
t ∈ R.
We now require an inverse of the function H in some neighbourhood of the point
y0 = 0. Fortunately, H = tan is injective on (−π/2, π/2) with image R, so we may
define the inverse function H −1 : R → (−π/2, π/2) by H −1 (z) = arctan(z).
This leads us to consider the solution given by (11.8), namely
y(t) = H −1 (G(t) − G(t0 ) + H(y0 )).
(Notice that we have carefully arranged our antiderivatives G, H to ensure that
G(t0 ) = H(y0 ) = 0!) We also notice that G(t) = 1 − e−t always lies in the domain
of H −1 , no matter the value of t ∈ R. We thus obtain
y(t) = arctan(1 − e−t ),
t ∈ R.
If you wish, you can differentiate the expression for y to check that it satisfies the
differential equation.
Remark. The condition y(0) = 0 plays an important role in selecting the correct
solution; in particular, it affects which inverse of H we choose. In the calculation
above, we note that the function tan is not injective, and hence has many partial
inverses, depending on how we restrict its domain. For instance, if we had been given
the condition y(0) = π, corresponding to (t0 , y0 ) = (0, π), we would have needed to
use the inverse of tan|(π/2,3π/2) instead and would have obtained a different solution.
(ii) We first rearrange the ODE to give the form in (11.6), namely
y 0 (t) =
1
1 y(t)
,
t 1 + y(t)
Note: in this sheet we use the convention that, if a < b and f : [a, b] → R is a regulated function, then
Ra
Rb
f (x) dx = − a f (x) dx.
b
3
which corresponds to g(t) = t−1 and F (y) =
y0 = 12 .
y
1+y ,
while the condition gives t0 = 1,
We have that g is a continuous function on any domain not containing 0, and F is
continuous on any domain not containing −1, and F (y0 ) = F ( 21 ) = 13 . As before,
we proceed to calculate the quantities in Satz 11.3.1.
We use the fundamental theorem of calculus, assuming that y > 0:
Z y
H(y) =
1/2
1
dz =
F (z)
Z y
1+z
1/2
z
1
dz = ln(z) + z|y1/2 = ln(y) + y + ln(2) − .
2
The function H : (0, ∞) → R is bijective, and furthermore, H 0 (y) = 1/F (y) 6= 0
for y > 0. Therefore, using Satz 9.2.6 there exists a differentiable inverse function
H −1 : R → (0, ∞).
We now deal with G, again using the fundamental theorem of calculus, to obtain,
for t > 0,
Z t
Z t
1
g(s) ds =
G(t) =
ds = ln(s)|t1 = ln(t).
1 s
1
Finally, we plug everything into (11.8), again observing that G(t0 ) = H(y0 ) = 0
and that G(t) always lies in the domain of H −1 , to obtain the solution
y(t) = H −1 (ln(t)),
t > 0.
If desired, one may differentiate y (using the expression for the derivative of H −1
given in Satz 9.2.6) to check that it satisfies the ODE.
Remark. In this part of the question we were not so lucky with the function H,
and we could not find an explicit inverse function. In general, finding inverses is not
an easy problem, and one often has to be satisfied with the ‘implicit’ form which
we gave above. You may be interested to know that H −1 is related to a special
function called the Lambert W -function,
and in terms of this function the solution
1 may be given by y(t) = W 12 e 2 t for t > 0.
Second remark. Again in this part the condition y(1) = 1/2 was important. If
we had been given something else, say y(1) = −1/2, we would have again obtained
a different solution. Can you find an implicit or explicit form for it?
Übung 3.
(i) Sei C > 0 und sei Y : [0, T ] → R eine differenzierbare Funktion, sodass für alle
t ∈ [0, T ] gilt Y 0 (t) ≤ CY (t). Zeigen Sie, dass
Y (t) ≤ Y (0)eCt ,
[Hinweis: Beweis vom Lemma von Gronwall.]
4
t ∈ [0, T ].
(ii) Seien y : [0, T ] → R and z : [0, T ] → R zwei Lösungen der gewöhnlichen Differentialgleichung
y 0 (t) = F (y(t)),
t ≥ 0,
wobei F : R → R eine differenzierbare Funktion mit stetiger Ableitung ist. Zeigen
Sie, dass ein L > 0 existiert, so dass
y(t) − z(t) ≤ eLt y(0) − z(0) ,
t ∈ [0, T ].
[Hinweis: Benutzen Sie Unteraufgabe (i) und den Mittelwertsatz.]
Lösung 3.
(i) We just reproduce the proof of Lemma 11.4.5, making the necessary changes. Note
that in our version, we make two main changes. First, we do not require that the
codomain of Y be [0, ∞) and we do not require that Y (0) = 0; these are important
weakenings of the conditions. Second, we require that the endpoint of the domain
is included; this will be important when we apply the result in part (ii).
Here we go. Let z(t) = e−Ct Y (t). Then z is certainly differentiable, and furthermore
we obtain
z 0 (t) = e−Ct Y 0 (t) − Ce−Ct Y (t) = e−Ct (Y 0 (t) − CY (t)) ≤ 0.
In particular, this implies that z is a monotone decreasing function (Korollar 9.3.4),
so z(t) ≤ z(0) = Y (0) for all t ∈ [0, T ]. If we substitute the definition of z and
rearrange, we obtain Y (t) ≤ eCt Y (0), and we are done.
(ii) First, we observe that y and z are both continuous functions (because they are
differentiable) on a compact (i.e., closed and bounded) interval. This implies, by
Weierstrass’ theorem, that they are bounded. From this we deduce that there exists
a closed bounded interval J ⊂ R such that, for all t ∈ [0, T ], we have y(t) ∈ J and
z(t) ∈ J.
Now, we assumed that F was differentiable with continuous derivative. Since J is
closed and bounded, we may apply Weierstrass’ theorem again, but this time to
the function F 0 |J . This implies that there exists some number L > 0 such that
F 0 (x) ≤ L
for all
x ∈ J.
(S1)
equation
Now we want to apply the result from part (i). Let Y (t) = y(t) − z(t). Using the
fact that y and z are both solutions, we calculate, for 0 ≤ t ≤ T ,
Y 0 (t) = y 0 (t) − z 0 (t) = F (y(t)) − F (z(t)) =
5
F (y(t)) − F (z(t))
(y(t) − z(t)).
y(t) − z(t)
Now, using the mean value theorem for the function F , we get that for each t there
exists some ξ(t) ∈ J such that
F (y(t)) − F (z(t))
= F 0 (ξ(t)) ≤ L,
y(t) − z(t)
where in the final equality we used (S1). This yields for Y :
Y 0 (t) ≤ L(y(t) − z(t)) = LY (t),
t ∈ [0, T ],
which is precisely the condition required to apply part (i). We deduce that
Y (t) ≤ eLt Y (0) = eLt (y(0) − z(0)),
and we are done.
Übung 4.
Betrachten Sie folgende gewöhnliche Differentialgleichung, wobei c, ω ≥ 0:
y 00 (t) + 2cωy 0 (t) + ω 2 y(t) = 0.
(1)
Wir suchen reellwertige Lösungen y : R → R. Sei V = {y : R → R | y löst (1)}.
(a) Finden Sie das charakteristische Polynom von (1) und bestimmen Sie deren Nullstellen. Es gibt vier Fälle zu unterscheiden: c = 0, 0 < c < 1, c = 1 und c > 1.
(b) Sei 0 < c < 1.
(i) Finden Sie eine Basis für V . Geben Sie eine explizite Formel für eine allgemeine
Lösung von (1).
(ii) Finden Sie die Lösung vom Anfangswertproblem
y 00 (t) + 2cωy 0 (t) + ω 2 y(t) = 0,
y(0) = 0,
y 0 (0) = 1.
(2)
Eklären Sie, warum diese eindeutig ist.
(iii) Skizzen Sie die Lösung.
(c) Sei jetzt c = 1; das heisst, wir betrachten die gewöhnliche Differentialgleichung
y 00 (t) + 2ωy 0 (t) + ω 2 y(t) = 0.
Beantworten Sie Unteraufgaben (i–iii) oben für diese Differentialgleichung.
6
(3)
Lösung 4.
(a) The characteristic polynomial is
P (x) = x2 + 2cωx + ω 2 .
Notice that the coefficients are real, so we will be looking for real zeroes and conjugate
pairs of zeroes. We now distinguish the four cases. The question concerns the ODE
for a damped harmonic oscillator, hence the descriptions below.
(I) c = 0. This case is undamped motion. The two zeroes are ±ωi, each with
multiplicity 1.
(II) 0 < c < 1. This case is underdamped. In this case the discriminant of the
quadratic equation
√ is negative, so we have a pair of distinct conjugate zeroes,
namely ω(−c ± i 1 − c2 ), each with multiplicity 1.
(III) c = 1. This case is critically damped. There is only one zero, namely −ω, with
multiplicity 2.
(IV) c > 1. This case √
is overdamped. Here both the zeroes are real, and they are
given by ω(−c ± c2 − 1). Each has multiplicity 1.
(b) We now find ourselves in the case (II). We will fit √
everything to the notation of
Satz 11.4.9 and
11.4.10.
Let
us
write
z
=
ω(−c
+
i
1 − c2 ) for the first zero and
1
√
z̄1 = ω(−c√− i 1 − c2 ). The real and imaginary parts of z1 are denoted by α1 = −c
and β1 = 1 − c2 , respectively. Since each has multiplicity 1, we have m1 = 1.
(i) From Satz 11.4.10, we have that a basis of V is given by
p
p
{t → e−ωct sin(tω 1 − c2 ), t → e−ωct cos(tω 1 − c2 )},
where each function is regarded as being from R to R.
It follows (from the definition of a basis) that y : R → R is a solution to the
ODE if and only if y has the form
p
p
y(t) = Ae−ωct sin(tω 1 − c2 ) + Be−ωct cos(tω 1 − c2 ),
t ∈ R,
for some constants A, B ∈ R.
(ii) To solve the initial value problem it remains to compute the values of A, B; we
1
obtain A = ω√1−c
and B = 0, giving
2
y(t) =
p
1
√
e−ωct sin(tω 1 − c2 ),
ω 1 − c2
t ∈ R.
Uniqueness follows from Lemma 11.4.4.
(iii) See Abbildung 1, left. The solution oscillates; it exhibits exponential decay as
t → +∞ and exponential growth as t → −∞.
7
Abbildung 1: Left: c = 1/4, ω = 1. Right: c = 1, ω = 1.
(c) We are now in case (III). In the notation of Sätze 11.4.9–10, we have µ1 = −ω and
k1 = 2. Applying the latter theorem, we obtain the basis
{t 7→ e−ωt , t 7→ te−ωt }.
The general solution is given by
y(t) = e−ωt (A + Bt),
t ∈ R.
Substituting the initial condition gives A = 0, B = 1, that is,
y(t) = te−ωt ,
t ∈ R.
The uniqueness again follows from Lemma 11.4.4. It is plotted in Abbildung 1, right.
It comes up rapidly from below, crosses zero at t = 0, and subsequently dies away,
approaching 0 from above as t → ∞.
Übung 5.
Betrachten Sie folgende gewöhnliche Differentialgleichung, wobei ω ≥ 0:
y 000 (t) = −iω 2 y 0 (t).
(4)
Jetzt suchen wir kompexwertige Lösungen y : R → C. Sei V = {y : R → C | y löst (4)}.
(i) Finden Sie das charakteristische Polynom von (4) und bestimmen Sie deren Nullstellen.
(ii) Finden Sie eine Basis für V . Geben Sie eine explizite Formel für eine allgemeine
Lösung von (4).
(iii) Finden Sie die Lösung vom Anfangswertproblem
y 000 (t) = −iω 2 y 0 (t),
y(0) = 0,
Erklären Sie, warum diese eindeutig ist.
8
y 0 (0) = 0,
y 00 (0) = 1.
(5)
Lösung 5.
(i) The characteristic polynomial is
P (x) = x3 + iω 2 x = x(x2 + iω 2 ).
Let q = eiπ/4 = √12 (1 + i), and note that q 2 = i. The zeroes of P are 0 (with
multiplicity 1) and ±ωq (each with multiplicity 1). We list the zeroes as λ1 = 0,
λ2 = ωq, λ3 = −ωq, and their multiplicities k1 = k2 = k3 = 1.
(ii) According to Satz 11.4.10, a basis for V is given by
{t 7→ 1, t 7→ eωqt , t 7→ e−ωqt },
with each function being regarded as from R to C. Thus a generic solution to the
ODE is given by
y(t) = A + Beωqt + Ce−ωqt ,
t ∈ R,
for some choices of A, B, C ∈ C.
(iii) We now determine A, B and C using the initial conditions. They give three simultaneous equations, which we solve to obtain A = ωi2 , B = C = − 2ωi 2 . That is, the
solution to the initial value problem is
y(t) =
i
i
− 2 (eωqt + e−ωqt ),
2
ω
2ω
t ∈ R.
Once again uniqueness follows from Lemma 11.4.4.
If we use the definition of cosh from Sheet 10, we obtain an alternative representation of y, namely
y(t) =
i
(1 − cosh(ωqt)),
ω2
9
t ∈ R.