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From: Blue Venom on 4 May 2010 09:54
Let f be a continuous real valued function such that tf(t) >= 0 for all
t in R.
Then the CP:

y''+e^(-x)f(y)=0
y(0)=y'(0)=0

has the unique solution y=0.

Any hint?
From: Rob Johnson on 11 May 2010 07:04
In article <4be02709$0$824$4fafbaef(a)reader5.news.tin.it>,
Blue Venom <mandalayray1985(a)gmail.com> wrote:
>Let f be a continuous real valued function such that tf(t) >= 0 for all
>t in R.
>Then the CP:
>
>y''+e^(-x)f(y)=0
>y(0)=y'(0)=0
>
>has the unique solution y=0.

Let F(x) be the integral of f from 0 to x. Since xf(x) >= 0, we
have that F(x) is decreasing for x < 0 and increasing for x > 0.
Thus, F(x) >= F(0) = 0.

Multiply the DE by y' and integrate wrt x from 0 to t:

y'' + exp(-x) f(y) = 0

y'y'' + exp(-x) f(y) y' = 0

2 |\t
1/2(y'(t)) + | exp(-x) dF(y) = 0
\| 0

2 |\t
1/2(y'(t)) + exp(-t) F(y(t)) + | exp(-x) F(y(x)) dx = 0
\| 0

For t >= 0, each term on the LHS is non-negative, yet they sum to 0.
Therefore, each must be 0 for t >= 0. This means that y(t) = 0 for
all t >= 0.

f is continuous; therefore, its integral, F, is also continuous. y
is twice differentiable; therefore, it is continuous. Thus, F(y(x))
is continuous. Thus, for any M > 0, F(y(x)) must attain a maximum on
[-M,0]. If the maximum is attained at 0, then F(y(x)) is identically
0 on [-M,0]. Otherwise, there is some s in [-M,0] so that
F(y(s)) >= F(y(x)) for all x in [-M,0]. Then

|\s
exp(-s) F(y(s)) + | exp(-x) F(y(x)) dx
\| 0

|\0
= exp(-s) F(y(s)) - | exp(-x) F(y(x)) dx
\| s

|\0
>= exp(-s) F(y(s)) - F(y(s)) | exp(-x) dx
\| s

= exp(-s) F(y(s)) + F(y(s)) (1 - exp(-s))

= F(y(s))

Therefore,

2 |\s
0 = 1/2(y'(s)) + exp(-s) F(y(s)) + | exp(-x) F(y(x)) dx
\| 0
2
>= 1/2(y'(s)) + F(y(s))

Since both terms on the RHS are non-negative and their sum is less
than or equal to 0, each must be 0. Thus, the maximum of F(y(t)) on
[-M,0] is 0. This means that y = 0 on [-M,0]. M was arbitrary, so
we get that y(t) = 0 for all t < 0.

Rob Johnson <rob(a)trash.whim.org>
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