B********e
发帖数: 10014 | 1 一个貌似应该很easy的题怎末也搞不定
a(t) bounded continuous, y(t) is a nonzero solution of y''+a(t)y=0 such that
lim_{t->\infty} y=0. show there is a solution on [0,\infty) which is not b
ounded.
证明或者证伪
3x |
G******i
发帖数: 163 | 2 Let z(t) be another solution that is linearly independent from y(t).
y(t)->0 => y''(t)=-a(t)y(t)->0 => y'(t) ->0 as t->infinity.
Suppose z(t) is bounded on [0,infinity).
We have z''(t)=-a(t)z(t) bounded and hence z'(t) bounded.
but (yz' -y'z)'=0 => yz'-y' z=C (nonzero).
Contradiction! |
B********e
发帖数: 10014 | 3 多谢啊
还在吗？
我想过这么做，可是死活搞不明白这一步：
y(t)->0 => y''(t)=-a(t)y(t)->0 => y'(t) ->0 as t->infinity
能再点一下吗？
【在 G******i 的大作中提到】
: Let z(t) be another solution that is linearly independent from y(t).
: y(t)->0 => y''(t)=-a(t)y(t)->0 => y'(t) ->0 as t->infinity.
: Suppose z(t) is bounded on [0,infinity).
: We have z''(t)=-a(t)z(t) bounded and hence z'(t) bounded.
: but (yz' -y'z)'=0 => yz'-y' z=C (nonzero).
: Contradiction!
|
B********e
发帖数: 10014 | 4 还有why z''bounded =>z' bounded ?
all day worked on this problem, felt so stupid!
3x!
【在 G******i 的大作中提到】
: Let z(t) be another solution that is linearly independent from y(t).
: y(t)->0 => y''(t)=-a(t)y(t)->0 => y'(t) ->0 as t->infinity.
: Suppose z(t) is bounded on [0,infinity).
: We have z''(t)=-a(t)z(t) bounded and hence z'(t) bounded.
: but (yz' -y'z)'=0 => yz'-y' z=C (nonzero).
: Contradiction!
|
G******i
发帖数: 163 | 5 I meant z,z'' bounded => z' bounded.
Proof: z(t+1)=z(t)+z'(t)+1/2*z''(s) for some t
【在 B********e 的大作中提到】
: 还有why z''bounded =>z' bounded ?
: all day worked on this problem, felt so stupid!
: 3x!
|
B********e
发帖数: 10014 | 6 sorry, 一个意思
多谢多谢！
【在 B********e 的大作中提到】
: 还有why z''bounded =>z' bounded ?
: all day worked on this problem, felt so stupid!
: 3x!
|
