3
$\begingroup$

Let $f$ be a real-valued continuous function on $[0,1]$ which is twice continu-ously differentiable on $(0,1)$. Suppose that $f(0) = f(1) = 0$ and $f$ satisfies the following equation:

$$x^2f''(x) + x^4f'(x) - f(x) = 0$$

(a) If $f$ attains its maximum $M$ at some point $x_0$ in the open interval $(0,1)$, then prove that $M= 0$.

(b)Prove that $f$ is identically zero on$[0,1]$.

My idea: At $x_0$, $f'(x_0) = 0$ and $f''(x_0)$ is negative as it is given that the function is twice differentiable and attains maximum at $x_0$. The given equation at $x_0$ reduces to $$f''(x_0) = \frac{f(x_0)}{x_0^2}$$

$f''(x_0) < 0$ implies $f(x_0) < 0$. That is the maximum value attained by the function $f$ is negative on $[0,1]$. Which is a contradiction as $f$ attains $0$ at the point $x=0$. (It is given $f$ attains maximum at $x_0$)

Hence the only way this can be satisfied is if $f(x_0) = M = 0$. Similarly we can prove that the minimum attained by the function $m = 0 $.

since this is a continous value function on $[0,1]$, $m \le f \le M$.

Hence $ f $ is identically equal to $0$.

$\endgroup$
  • 1
    $\begingroup$ It sounds like you've got it to me. $\endgroup$ – John Martin Apr 26 '16 at 21:27
1
$\begingroup$

Nice insight! Anyway, I prefer to exhibit the contradiction in the following way:

If $f(x_0) \neq 0$, from the differential equation $x_0^2f''(x_0)-f(x_0)=0$ and from the condition that $x_0$ is a maximum we simply get that $f(x_0)<0$.

Finally, from continuity arguments, one can say that from Weierstrass theorem that the inequality $f(x_0)\geq f(x)$ is satisfied over all points of the interval $[0,1]$ so that $f(x_0)\geq 0$, since $f(0)=f(1)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.