Second order differential inequality and comparison theorem. If x:[0,1] \rightarrow \mathbb R

Question

Second order differential inequality and comparison theorem. If x:[0,1] \rightarrow \mathbb R

yert21trey123z7 2022-05-01 Answered

Second order differential inequality and comparison theorem.
If

x : [0, 1] \Rightarrow R, x \in C^{\infty}

that satisfies

{\begin{cases} \ddot{x} \leq - 2 x \\ x (0) = x (1) = 0 \end{cases}

is it true that

x \geq 0 on [0, 1]

?
I observed that if

x \geq 0

, then

\ddot{x} \leq 0

, so x is concave. However I could not find out any other properties. I think if there exists a solution of the following differential equation

{\begin{cases} \ddot{x} = - 2 x \\ x (0) = x (1) = 0 \end{cases}

then some kind of comparative theorem can be used. However, there is no non-trivial solution for this.
If the above proposition is incorrect, could you give me a counterexample?

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Kendal Kelley

Answered 2022-05-02 Author has 16 answers

Answer: Concavity is enough:

x (t) = x ((1 - t) \cdot 0 + t \cdot 1) \geq (1 - t) \cdot x (0) + t \cdot x (1) = 0

.
Alternatively, by Rolle's theorem,

x^{'} (t_{0}) = 0

for some

t_{0} \in (0, 1)

, ans as

x^{″} \leq 0 if t < t_{0} and x^{'} (t) \leq 0 if t > t_{0}

, so x not decreases on

[0, t_{0}]

-thus is non-negative there, and x not increases on

[t_{0}, 1]

- thus again is non-negative there.

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