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Here is an attempt

Let $y(x)=F^{a}(x)$ then

$$y''y-2(y')^{2}=a(a-1)F^{a-2}(F')^{2}F^{a}+aF^{a-1}F''F^{a}-2(F^{a-1}F')^{2}$$

$$=(a(a-1)-2)F^{2(a-1)}(F')^{2}+aF^{2a-1}F''$$

so for $a(a-1)=2\Rightarrow a=-1,2$ we obtain the differential inequality $F^{2a-1}F''\geq 0$ and so the inequality is true for positive convex functions like $e^{2x}$.

Update

I apologize the previous attempt was wrong,as rightly pointed out by professor Israel.

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4 Answers 4

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That's wrong. With $y(x) = e^{c g(x)}$ you should get

$$ y'' y - 2 (y')^2 = e^{2cg(x)} c (g''- c (g')^2) $$

BTW, with $g' = v$

$$ v' - c v^2 = q $$

is a Riccati D.E. I don't think there is a known solution for general $q$.

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"Solving" the inequality $$y''y-2y'^2\geq0\qquad\tag{1}$$ means finding a "geometric condition" that characterizes the functions satisfying $(1)$. This "geometric condition" will be a convexity condition. The latter is an inequality connecting the function values in three arbitrary points, and does not involve derivatives.

Assume that $x\mapsto y(x)>0$ satisfies $(1)$. Then $$\left({1\over y}\right)''=-{1\over y^3}\bigl(y''y-2y'^2\bigr)\leq0\ .$$This implies that the reciprocal $x\mapsto {\displaystyle{1\over y(x)}}$ is a concave function. Conversely, if $x\mapsto y(x)>0$ has a concave reciprocal then $(1)$ holds.

Similarly, a function $x\mapsto y(x)<0$ satisfies $(1)$ iff its reciprocal is convex.

Now a concave function $x\mapsto {\displaystyle{1\over y(x)}}$ cannot reach $\infty$ at some finite $x$. Therefore we don't have to deal with zeros of the corresponding $x\mapsto y(x)$. On the other hand we have to add the function $y(x):\equiv0$ to the solution set.

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Let $y(x)=F^{a}(x)$ then

$$y''y-2(y')^{2}=a(a-1)F^{a-2}(F')^{2}F^{a}+aF^{a-1}F''F^{a}-2(F^{a-1}F')^{2}$$

$$=(a(a-1)-2)F^{2(a-1)}(F')^{2}+aF^{2a-1}F''$$

so for $a(a-1)=2\Rightarrow a=-1,2$ we obtain the differential inequality $F^{2a-1}F''\geq 0$ and so the inequality is true for positive convex functions like $e^{2x}$.

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$\frac {d}{dt} yy' = y'^2 + yy''\\ \frac {d}{dt} y^{-2}y' = -2y^{-3}y'^2 + y^{-2}y'' = y^{-3} (yy''-2y'^2)$

$yy''-2y'^2 \ge 0\\ y^{-3} (yy''-2y'^2) \ge 0\\ y^{-2} y' \ge C\\ -y^{-1} \ge Ct + D\\ y \ge \frac 1{Ct + D}$

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