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I have some doubts regarding the following problem from Apostol's Calculus I book (problem 31 from section 8.28). I posted my solution attempt below. Can anyone please verify the solution and give me some hints concerning the doubts?

Problem

Given a function $f$ which satisfies the differential equation

$$xf''(x) + 3x[f'(x)]^2 = 1 - e^{-x}$$

for all real $x$.

(a) If $f$ has an extremum at a point $c \neq 0$, show that this extremum is a minimum.

(b) If $f$ has an extremum at $0$, is it a maximum or a minimum?

(c) If $f(0) = f'(0) = 0$, find the smallest constant $A$ such that $f(x) \leq Ax^2$ for all $x \geq 0$.

Solution

(a) If $f$ has an extremum at a point $c \neq 0$, then $f'(c) = 0$. Plugging this into the differential equation, we get:

$$cf''(c) = 1 - e^{-c} \implies f''(c) = \dfrac{1 - e^{-c}}{c}$$

By cases:

  • If $c > 0$, then $1 - e^{-c} > 0$, so $f''(c)$ is positive.
  • Otherwise, if $c < 0$, then $1 - e^{-c} < 0$, and so $f''(c)$ is positive.

Either way, the second derivative, $f''(c)$, is positive. Therefore, $f$ has a minimum at $c$.

(b) If $x \neq 0$, we can divide both sides of the differential equation by $x$ to obtain:

$$f''(x) + 3[f'(x)]^2 = \dfrac{1}{x} - \dfrac{1}{xe^x}$$

Consider the limit of the above expression as $x \to 0$. If $f$ has an extremum at $0$, then we have $f'(0) = 0$, so the limit of the left-hand side expression is $\lim_{x \to 0} [f''(x) + 3[f'(x)]^2] = \lim_{x \to 0} f''(0)$. Furthermore, the limit of the right-hand side expression is:

$$\begin{aligned} \lim_{x \to 0} \left( \dfrac{1}{x} - \dfrac{1}{xe^x} \right) &= \lim_{x \to 0} \left( \dfrac{e^x - 1}{xe^x} \right) \\ &= \lim_{x \to 0} \left( \dfrac{e^x}{e^x + xe^x} \right) & \text{(L'Hôpital)} \\ &= \lim_{x \to 0} \left( \dfrac{1}{1 + x} \right) = 1 \end{aligned}$$

Therefore, we have that $f''(0) \to 1$ as $x \to 0$. Since this is a positive value, we can conclude that $f$ has a minimum at $0$.

DOUBT: Above, I am concluding that $f''(0)$ has a positive value based on the fact that $f''(0) \to 1$. However, I think that this depends on assuming that $f''$ is continuous at $0$. How can I justify this assumption?

(c) DOUBT: I couldn't figure this one out. Any hint?

EDIT: For item (c), here is the beginning of an initial attempt.

The Taylor expansion of $f(x)$ near $x = 0$ is:

$$f(x) = f(0) + f'(0)x + f''(0)\dfrac{x^2}{2} + o(x^2)$$

We are given that $f(0) = f'(0) = 0$. Also, from item (b), we concluded that $f''(x) \to 1$ as $x \to 0$. If $f''(0)$ is continuous at zero (again: how can I justify this claim?), then we have $f''(0) = 1$. So:

$$f(x) = \dfrac{x^2}{2} + o(x^2)$$

If we can show that the remainder of this expansion is $\leq 0$ for all $x > 0$, we will have $f(x) \leq x^2/2$ for all $x > 0$, and so $A = \frac{1}{2}$. Is this correct, and is there any way to proceed from here?

favq
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2 Answers2

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For $(c)$:

We are given: $$xf''(x) + 3x[f'(x)]^2 = 1 - e^{-x}, \;\;f(0)=f'(0)=0$$

Find $A: f(x) \leq Ax^2\;\;\forall x\geq 0$

You've already showed that an extremum at $x=0$ will be a minimum, with $f''(0) = 1$. We are told $f'(0)=f(0)=0$, so we have $f$ taking on a local minimum of $0$ at $x=0$.

Therefore, we have it acting locally like $x^2$ in the neighborhood around $x=0$ (second order approximation).

We need to find a multiplier $A$ of $x^2$ so $f(x) \leq Ax^2$ for $x\geq 0$.

Let $g(x)=Ax^2$ then $g'(x)=2Ax,\;g''(x)=2A$

I want see if $f'(x)$ is bounded from above by $g'(x)$ for $x>0$, in which case:

$$\int_0^x f'(z)dz \leq \int_0^x g'(z)dz \implies $$ $$f(x)-f(0) = f(x)-0 = f(x) < g(x)-g(0) = g(x)-0 = g(x)$$

Going back to the equation, if $x>0$ we get

$$f''(x) + 3[f'(x)]^2 = \dfrac{1}{x} - \dfrac{1}{xe^x} \implies [f'(x)]^2 = \frac{e^x-xe^xf''(x)-1}{3xe^x}$$

From the non-negativity of the LHS, we can see that:

$$\frac{e^x-xe^xf''(x)-1}{3xe^x} \geq 0 \implies e^x-xe^xf''(x)-1 > 0$$ $$\implies f''(x) < \frac{e^x-1}{xe^x}$$

Since $f''(0)=1>0$ and $f'(0)=0$ we know there cannot be any critical points on $x>0$ because it will have to be a minimum.

So how big can $f'(x)$ be? Well we know a bound on $f''(x)$ on $x>0$

$$f''(x) < \frac{e^x-1}{xe^x} < 1 \;\;\forall x>0 \implies f'(x)=\int_0^x f''(z)dz \leq \int_0^x 1 dz = x$$

So we see that $g'(x)\geq x$ is sufficient for $f(x)\leq g(x)$ on $x>0$:

$$g'(x)\geq x\;\forall x>0 \implies g(x) = \int_0^x g'(z)dz \geq \int_0^x zdz =\frac{x^2}{2} \geq \int_0^x f'(z)dz = f(x)$$

So it looks like $A=\frac{1}{2}$ will be as small as we can go.

  • I don't immediately see how $f''$ has to be continuous at $0$ since isolating it demands a division by $x$. It seems to me that it could have a $x^p,-1<p<0$ type singularity in principle (for example). – Ian Aug 03 '22 at 12:49
  • @Ian see my comment at the top of answer — it is because we are doing arithmetic operations on $f’,f’’$ and this equals a continuous function over the real line. –  Aug 03 '22 at 12:53
  • @Ian https://math.hws.edu/~mitchell/Math130F12/tufte-latex/Notes09.pdf –  Aug 03 '22 at 12:54
  • I'm not convinced, to be honest. Just because $x f$ is continuous at zero doesn't imply that $f$ is, since $f$ could have a singularity tame enough to be dissipated by the multiplication by $x$. Instead of doing that, you can just say that $xf''$ is continuous and changes sign through $0$ so $f''$ doesn't change sign through $0$. – Ian Aug 03 '22 at 12:57
  • @Ian ah — I see your point — I was taking the converse of the continuity theorems — $xf$ Is continuous if $x,f$ are continuous but not necessarily the converse –  Aug 03 '22 at 13:03
  • @Ian -- actually realized I didn't even need the whole continuity bit so just deleted -- the first derivative arguments were sufficient. –  Aug 03 '22 at 19:59
  • Just a question. Why is the part where we conclude that $f'(x) \geq 0$ for all $x > 0$ necessary? Could I just conclude directly, by integrating $f''(x) < (e^x - 1)/(xe^x) < 1$ twice from $0$ to $x$ (and using the fact that $f(0) = 0$ and $f'(0) = 0$), that $f''(x) < 1 \implies f'(x) < x \implies f(x) < x^2/2$ for all $x > 0$? – favq Aug 03 '22 at 20:45
  • I also edited the original question to add an unfinished attempt at item (c) that I had neglected to add before. – favq Aug 03 '22 at 21:04
  • @favq correct -- that part is not necessary (I can remove it) and we can cut to the chase that the derivative is bounded above by $x$. –  Aug 03 '22 at 21:25
  • @favq -- as for your attempt -- I thought along the same lines but didn't have any info on the higher order derivatives (I guess you could get those by further differentiating the equation and seeing if you can get some facts about the sign of the remainder). –  Aug 03 '22 at 21:27
  • @Bey How do we know such A is the smallest one? – S11n Sep 16 '22 at 10:33
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    @S11n its the lowest bound I could get to -- if $A< \frac12$ then in some neighborhood around $0$ we'll have $f$ go above $g(x)$ since $f'(x) \approx x$ near $0$, so anything less than $\frac12$ will result in $g$ growing too slowly compared to $f$. –  Sep 16 '22 at 20:44
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For (b), $f''(x)$ exists for all $x$, or so I assume given the statement of the exercise. You have established that $\lim\limits_{x\to0}f''(x)=1$ (assuming $f'(0)=0$), but worry that $f''(0)$ may exist but not equal $1$. That sort of thing is not possible.

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