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Deduce properties of the solutions of a given differential equation

by RoRi

Let f(x) be a solution for all x \in \mathbb{R} of the second order differential equation

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

Without attempting to solve the equation answer the following questions.

  1. If c \neq 0 and f has an extremum at c, prove that this extremum must be a minimum.
  2. If f has an extremum at 0 decide whether this extremum is a maximum or minimum and prove your assertion.
  3. If the solution f(x) satisfies the conditions

        \[ f(0) = f'(0) = 0 \]

    find the minimum value for the constant A such that

        \[ f(x) \leq Ax^2 \qquad \text{for all } x \geq 0. \]

Incomplete.

7 comments

  1. Anonymous says:

    One way to solve part c is to assume f(x) = Ax^2. By substituting this into the differential equation, we get a quadratic with A as the variable. Use the quadratic equation and we get A as a function of x.

    This equation tells us what A must be for f(x) to be equal to Ax^2 at each x, since the solutions to the differential equation must be equal if the functions are equal.

    Now that we have a function for A, all we must do to find a minimum value is to find the maximum value of A for x>=0, so that Ax^2 >= f(x) for all x>= 0.

    This step is a little tricky, and I cheated and used Desmos and an online limit calculator. Basically, we find that limA as x->inf = 0, and A(0) is the max value of A. We calculate the limit of A as x->0, and we get A=1/2.

  2. Evangelos says:

    a.) We know from our given information that c cannot be zero. And we know that at the point where x=c, we have a local extremum where f'(c) = 0. So our differential equation becomes

        \begin{align*} cf''(c) &= 1 - e^{-c} \end{align*}

    And since c is not zero, we can divide both sides, giving us

        \begin{align*} f''(c) &= \frac{1}{c}(1 - \frac{1}{e^{c}}) \end{align*}

    We now have two cases

    1.) c is greater than zero

    If c is greater than zero, then

        \begin{align*} \frac{1}{c} &> 0\\ \frac{1}{e^{c}} & 0 \end{align*}

    Thus,

        \begin{align*} \frac{1}{c}(1 - \frac{1}{e^{c}}) &> 0 \\ f''(c) &> 0 \end{align*}

    And by Theorem 4.9, since f”(c) is greater than 0, f'(c) must be a local minimum.

    2.) c is less than zero

    If c is less than zero, then

        \begin{align*} \frac{1}{c} & 1 \\ 1 - \frac{1}{e^{c}} & 0 \\ f''(c) &> 0 \end{align*}

    And by Theorem 4.9, since f”(c) is greater than 0, f(c) must be a local minimum.

    As such, for all c not equal to zero, the point f(c) must be a local minimum.

    b.) From our given information, we know that if f has an extremum at x=0, then f'(0) = 0. To determine if this is a local maximum, minimum, or neither, we must compute the value of f”(0). If we were to divide both sides by x and take the limit as x approaches zero

        \begin{align*} \lim_{x \to 0} f''(x) &= \frac{1 - e^{-x}}{x} \end{align*}

    We would get the indeterminate form of 0/0

    But if this is the case, then we can use L’Hopital’s Rule and take the derivative with respect to x of the numerator and denominator, and then take the limit as x approaches zero.

        \begin{align*} \lim_{x \to 0} f''(x) &= \lim_{x \to 0} \frac{e^{-x}}{1} \\ &= 1 \end{align*}

    And since this value is greater than zero, the point f(0) is a local minimum.

    c.) If we can assume that f(0) = f'(0) = 0, we can use the result from part (b.) to show the following:

        \begin{align*} f(x) &\leq Ax^{2} \\ f'(x) &\leq 2Ax \\ f''(x) &\leq 2A \end{align*}

    But with our initial conditions and the result of part (b.), we know that

        \begin{align*} f''(0) &= 1 \end{align*}

    So the smallest value of A that fulfills the condition

        \begin{align*} f(x) &\leq Ax^{2} \\ \end{align*}

    And thus, the conditions

        \begin{align*} f'(x) &\leq 2Ax \\ f''(x) &\leq 2A \end{align*}

    for all x greater than or equal to zero, must be

        \begin{align*} A &= \frac{1}{2} \end{align*}

    Looks like that’s all for now, thanks to RoRi for opening up the comments sections, and thanks to the readers for… reading :)

    • Evangelos says:

      So, the latex didn’t render properly for my cases, it should be as follows

      Case 1.) c > 0

          \begin{align*} \frac{1}{c} > 0 \end{align*}

          \begin{align*} \frac{1}{e^{c}} 0 \end{align*}

      Case 2.) c < 0

          \begin{align*} \frac{1}{c} 1 \end{align*}

          \begin{align*} \frac{1}{c}(1 - \frac{1}{e^{c}}) > 0 \end{align*}

    • favq says:

      Hi. I have one question about item (c): My doubt is: why is it possible, in this case, to conclude that f'(x) <= 2Ax from f(x) <= Ax^2?

      • S says:

        I agree it’s most probably not possible to solve c) that way, because the implication most probably doesn’t hold.

      • Anonymous says:

        Hey, I found that you could put the upper bound 1 on f” over all x>=0, given that f'(0)=0. Express for x>0
        f”(x)=(1-e^(-x))/x-3[f'(x)]^2, and we see that
        f”(x)<=(1-e^(-x))/x0.
        But we showed in (b) that f”(x)–>1 as x–>0, so this is the smallest possible upper bound on f”. Integrating then shows that (1/2)x^2 is larger than f.

        I’m not sure if this is entirely related to what you’re asking. But I think the idea is to reverse the direction; you conclude that f(x)<=Ax^2 from the fact that f''(x)<=2A.

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