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

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.

• 1) show (uu')'>0 then what can you say about positivity of $u$ and $u'$ ?
– zwim
Jun 18, 2022 at 10:15
• So you are saying that these two options are stated as correct in the answers? Jun 18, 2022 at 11:57
• Hey! I found you here. Actually I want to ask you that which problem book should I solve first for real analysis? Because I am finding real analysis too tough :( please tell me. Jan 24 at 10:49

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.

For any non negative $$x,y$$ such that $$x, 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).

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 $$x, whence it follows that $$u$$ is increasing.

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.

• part 2 seems complicated, notice you have $u^2\nearrow$...
– zwim
Jun 18, 2022 at 14:17

\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 $$x_1. This requires $$u''(x_2)<0$$ for some $$x_2. However, $$\forall x_20$$, 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.