# Fall 2017, problem 53

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a two times differentiable function satisfying $f(0) = 1$, $f^\prime(0) = 0$ and for all $x \in [0, \infty)$, it satisfies $$f^{\prime \prime}(x) - 5 f^\prime(x) + 6 f(x) \geq 0.$$ Prove that, for all $x \in [0,\infty)$, $$f(x) \geq 3e^{2x} - 2e^{3x}.$$

Hubert
4 years ago

$g=f'-2f\Rightarrow~g'-3g\ge 0\Rightarrow~h_t=g_te^{-3t}$ is not decreasing (by $h'\ge 0$)

$\Rightarrow~f'_t-2f_t=g_t\ge g_0e^{3t}=-2e^{3t}\Rightarrow~(f_te^{-2t})'\ge -2e^t\Rightarrow~$

$\int_0^t LHS\ge \int_0^t RHS\Rightarrow~f_te^{-2t}-1\ge-2(e^t-1)\Rightarrow~$

$\quad\quad f_x\ge 3e^{2x}-2e^{3x}$, done.

TME102475
4 years ago

This problem can be easily solved via a Laplace Transform with the given initial conditions. Let:

$f''(x) - 5f'(x) + 6f(x) = C; C \ge 0$

whose Laplace Transform computes to:

$F(s) = L[f(x)] = (s^2F(s) - sf(0) - f'(0)) - 5(sF(s) - f(0)) + 6F(s) = \frac{C}{s}$;

or $(s^2 - 5s + 6)F(s) - s + 5 = \frac{C}{s}$;

or $F(s) = \frac{s^2 - 5s + C}{s(s^2 - 5s + 6)} = \frac{\frac{C}{6}}{s} + \frac{3 - \frac{C}{2}}{s - 2} + \frac{\frac{C}{3} - 2}{s - 3}$;

Taking the inverse Laplace Transform now produces:

$f(x) = L^{-1}[F(s)] = 3e^{2x} - 2e^{3x} + C(\frac{1}{6} - \frac{1}{2}e^{2x} + \frac{1}{3}e^{3x}) = \boxed{3e^{2x} - 2e^{3x} + C \cdot g(x)}$;

Now, let's examine the behavior of $g(x)$ over $x \in [0, \infty)$. Taking the first & second derivatives of $g$ yields:

$g'(x) = -e^{2x} + e^{3x}$;

$g''(x) = -2e^{2x} + 3e^{3x}$

where $g'(x) = e^{2x}(e^{x} - 1) = 0 \Rightarrow x = 0$ is the only critical point and $g''(0) = 3 - 2 = 1 > 0$ (hence, a global minimum over $[0, \infty)$). This result shows that the function $C \cdot g(x)$ is zero for $C = 0$ or $x = 0$ and is monotonically strictly increasing for $x, C > 0$. Hence,

$f(x) = 3e^{2x} - 2e^{3x} + C(\frac{1}{6} - \frac{1}{2}e^{2x} + \frac{1}{3}e^{3x}) \ge 3e^{2x} - 2e^{3x}$

holds for all $x, C \ge 0$.

$\mathbb{Q}. \mathbb{E}. \mathbb{D}.$