Home » Blog » Deduce properties of the solutions of a given differential equation

Deduce properties of the solutions of a given differential equation

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.

3 comments

  1. 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 :)

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):