yert21trey123z7

2022-05-01

Second order differential inequality and comparison theorem.

If$x:[0,1]\Rightarrow \mathbb{R},x\in {C}^{\mathrm{\infty}}$ that satisfies

$$\{\begin{array}{l}\ddot{x}\le -2x\\ x(0)=x(1)=0\end{array}$$

is it true that$x\ge 0\text{}\text{on}\text{}[0,1]$ ?

I observed that if$x\ge 0$ , then $\ddot{x}\le 0$ , 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

$$\{\begin{array}{l}\ddot{x}=-2x\\ x(0)=x(1)=0\end{array}$$

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?

If

is it true that

I observed that if

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?

Kendal Kelley

Beginner2022-05-02Added 16 answers

Answer: Concavity is enough:

$x\left(t\right)=x((1-t)\cdot 0+t\cdot 1)\ge (1-t)\cdot x\left(0\right)+t\cdot x\left(1\right)=0$ .

Alternatively, by Rolle's theorem,${x}^{\prime}\left({t}_{0}\right)=0$ for some ${t}_{0}\in (0,1)$ , ans as $x{}^{\u2033}\le 0\text{}\text{if}\text{}t{t}_{0}\text{}\text{and}\text{}{x}^{\prime}\left(t\right)\le 0\text{}\text{if}\text{}t{t}_{0}$ , 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.

Alternatively, by Rolle's theorem,

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