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Mathematics版 - ode题求助
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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!

1 (共1页)
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话题: bounded话题: infinity话题: solution话题: nonzero话题: infty