18.034 Honors Differential Equations

advertisement
MIT OpenCourseWare
http://ocw.mit.edu
18.034 Honors Differential Equations
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
18.034 Solutions to Problemset 5
Spring 2009
1. (a) Since fn → f uniformly on [a, b], for � > 0 given, there exists
N ∈ Z+ such that |fn (t) − f (t)| < �/(b − a) for t ∈ [a, b] whenever
n ≥ N . For n ≥ N ,
�� b
� � b
� b
�
�
�
�≤
f
(t)
dt
−
f
(t)
dt
|fn (t) − f (t)| dt < �
n
�
�
a
a
a
(b) fn (t) is given by
n
0
�
1
fn (t) dt =
�0 1
1
1/2n 1/n
1
for all n,
2
f (t) dt = 0.
0
2. In the course of the local existence theorem,
1
|xk (t) − xk−1 (t)| ≤
M Lk−1
|t − t0 |k
for k = 1, 2, . . .
k!
�
�
� �
�
∞
�
�
|
x(t) − xn (t)|
=
�
(xk (t) − xk−1 (t))�
�
k=n+1
�
∞
�
≤
|xk (t) − xk−1 (t)|
≤
=
=
k=n+1
∞
M � (LT )k
L
k!
k=n+1
∞
M (LT )n+1 � (LT )k
L (n+1)!
k!
k=0
T n+1 LT
n
M L (n+1)! e
3. (a) (⇒) Let F (t) = f (t, φ(t)) and solve x�� = F (t) when x(t0 ) = x0 ,
x� (t0 ) = x1 .
�
�
t
d
t
∂f
(⇐
) Use
f (s, t) ds = f (t, t) +
(s, t) ds.
dt
t0
t0 ∂t
(b) Repeat the proof of the local existence theorem by showing
�� t
�
�
�
�
(1) |xn (t) − x0 | = |x1 | |t − t0 | + � (t − s)f (s, xn−1 (s) ds��
t0�
t
≤ |x1 | |t − t0 | + M
|t − s| ds
t0
2
= |x1 | |t − t0 | + M
2 |t − t0 |
≤ B |
t − t0 |
� t
(2) |xn (t) − xn−1 (t)| ≤
(t − s)|
f (s, xn−1 (s)) − f (s, xn−2 (s))| ds
t0�
t
≤ L (t − s)|xn−1 − xn−2 | ds
t0
|t − t0 |2n
(2n)!
where L is the Lipshitz constant.
�
�
�
∞
√
�
∞
1
−st 1
−x2 s
2x
4. (a) L
√ =
e √ dt =
e
dx, where x
2 = st.
x s
t
�
t
0
0
�
2
∞ −x2
=√
e
dx = π/s
s 0
�
�
∞
√
√
√ e−st ∞ �
∞ e−st 1
−st
√ dt by parts.
(b) L[ t] =
e
t dt = t
−
−s 0
−s
2 t
0
�0 ∞
√
1
1
= 0 +
e−st √ dt = π/2s3/2
2s 0
t
≤ M Ln−1
2
5. (a)
2
L[e
t ]
�
∞
=
et
2 −st
dt is indefinite for every real value of s, no
0
matter how large, since t2 − st > 0 for t > s and so
�
∞
�
∞
�
∞
t2 −st
t2 −st
e
dt >
e
dt >
e
0 dt = ∞
0
s
s
�
�
�
∞
1
1
(b) L
k =
e−st k dt, (s > 0). The trouble here is when t = 0.
t
t
0
� τ
� τ
dt
−st 1
−st
Near t = 0,
e
≈ 1 and therefore
e
dt �
=
k
k
t
0
0 t
⎧
⎪
⎪ 1−k �τ
�
�
⎨
�
t
1
� k
=
� 1 . Therefore, L k exists for k < 1.
�
⎪
1
−
k
t
⎪
0
⎩
log t|τ0
k = 1
6. (a) 5 cos 2t − 3 sin 2t + 2.
√
(b) e−t/3 cos 2t.
3
Download