Prove that the statements (1) and (2) are true.

Question

Question: Let $u:\mathbb{R}\to \mathbb{R}$ be a twice continuously differentiable function such that $u(0)>0$ and $u'(0)>0$. Suppose $u$ satisfies$$u''(x)=\frac{u(x)}{x^2+1}\quad \forall x\in \mathbb{R}.$$Prove that the following statements are true:

(1) The function $uu'$ is monotonically increasing on $[0,\infty )$.

(2) The function $u$ is monotonically increasing on $[0,\infty )$.

This was one of the MCQ type question in my exam which I did by exemplifying $u(x)=1+\tan ^{-1}x$, luckily it made me discard the other options. I am looking for the proofs of above statements. I tried to solve it being a second order ordinary differential equation, couldn't do it with existing methods that I know.

Any help? Thanks.

Answer 1

For the first one: $$(u(x)u(x)')'=u(x)'^2+u(x)u(x)''=u(x)'^2+\frac{u(x)^2}{x^2+1}>0$$ for every $x\in [0,\infty)$, so 1) is true.

Answer 2

For any non negative $x,y$ such that $x<y$, note by FTC that:

$uu’(y)-uu’(x)=\int_y^x(uu’)’(t)dt=\int_x^yu’’(t)u(t)+(u’(t))^2 dt\ge 0$ as each summand in the integrand is non negative. It follows that$f$ is monotonically increasing.

For part 2), $u^2(y)-u^2(x)=2\int_x^y uu’(t)dt\ge 2uu’(x)(y-x)\ge 2uu’(0)(y-x)\gt 0. $ It follows that $u^2$ is strictly increasing. Let's call this conclusion $(A)$.

Note that $u(x)\gt u(0)$ for every $x\gt 0$. To see this, suppose on the contrary that there exists a $t_0\gt 0$ such that $u(t_0)\le u(0)$. If $u(t_0)=u(0)$, then $u^2(t_0)-u^2(0)=0$ violating $(A)$. So suppose that $u(t_0)<u(0)$.

Since $u''(0)>0$, by continuity of $u''$, there exists a $p\in (0,t_0)$ such that $u''>0$ on the interval $[0,p]$. This implies that $u'$ is positive on $[0,p/2]$. Hence, there is an $x_0\in (0,p/2)$ such that $u(x_0)>u(0)$.

By IVT, there is a $z_0\in (x_0,t_0)$ such that $u(z_0)\in (\max(0,u(t_0)),u(0)) $. It follows that $u^2(z_0)-u^2(x_0)\lt u^2(0)-u^2(x_0)<0$, which contradicts $(A)$.

So $u(x)\gt u(0)$ for every $x\gt 0$, i.e., $u$ is positive on $[0,\infty)$.

It follows that $u(y)-u(x)=\frac{u^2(y)-u^2(x)}{u(y)+u(x)}\gt 0$ for every non negative $x<y$, whence it follows that $u$ is increasing.

Answer 3

In order to solve both question 1) and 2), we rely on the fact that a function $f(x)$ is monotonically increasing on $[0,+\infty)$ if and only if $f'(x)\geq0 \:\forall x\in[0,+\infty)$.

1. Consider the function $g(x):[0,+\infty) \to \mathbb R$ such that $g(x) = u(x)u'(x)$. Its derivative can be computed using the chain rule: $$ g'(x) = (u(x)u'(x))' = u'(x)u'(x)+u(x)u''(x) = u'(x)^2 + u(x)u''(x) = u'(x)^2 + \frac{u(x)^2}{x^2+1} > 0 $$ for every $x\in[0,+\infty)$, since $u'(x)^2>0$ and $\frac{u(x)^2}{x^2+1}>0$. Therefore the function is monotonically increasing on $[0,+\infty)$.

2. Let's consider the function $u(x)$. We want to prove that $u'(x)>0,\forall x\in[0,+\infty)$. Since $u'(x)$ satisfies $u'(0)>0$ then if we manage to prove that $u'(x)$ is also increasing in $[0,+\infty)$, that is $u''(x)\geq0\:\forall x\in[0,+\infty)$ we have proved also the first statement, since in this case $u'(x)>0,\forall x\in[0,+\infty)$. We know that: $$ (u'(x))' = u''(x) = \frac{u(x)}{x^2+1} $$ If $u(x)\geq0$, then $\frac{u(x)}{x^2+1}\geq0$ and $u'(x)$ is increasing at $x$. If $u(x)\lt 0$ then $\frac{u(x)}{x^2+1}\lt0$ and $u'(x)$ is decreasing at $x$. Since $u(0)>0$, by the above reasoning this implies that $u''(0)>0$, and, since $u'(0)>0$, this implies that both $u(x)$ and $u'(x)$ are increasing at $0$, so that $\exists c>0:u(c)>u(0)>0, u'(c)>u'(0)>0$. By this last fact and by reiterating this reasoning, we get that $u(x)>0,\forall x\in[0,+\infty)$. Therefore, $u''(x)>0 \:\forall x\in[0,+\infty)$ and this implies $u'(x)>0 \:\forall x\in[0,+\infty)$. This is what we wanted to prove, that is the function $u(x)$ is increasing.

Answer 4

\begin{align}u''(x)(x^2+1)=u(x)\end{align} This means that sign of $u(x)$ and $u''(x)$ are same for all $x$, as $\forall x\in R,x^2+1>0 $.

Now for $u'(x)\le0$ (decreasing) to be true for some $x$ we should have $u''(x)<0$ for some $x\in [0,\infty)$.

That itself will conclude $u(x)<0$ for some $x$ in this interval (as $u(x)$ and $u''(x)$ have same sign). For this to be true, by Intermediate Value Theorem, $u(x_1)=0$ for some $x_1<x$. This requires $u''(x_2)<0$ for some $x_2<x_1$. However, $\forall x_2<x_1,u(x_2)>0$, so $u''(x_2)>0$ (since they have same signs).

Thus, $\forall x\in[0,\infty),u'(x)>0$ , hence the function is monotonically increasing.