21
$\begingroup$

Nice Question:

let $f(x)$ have two derivative on $[0,1]$,and such $$f(0)=2,f'(0)=-2,f(1)=1$$ show that:

there exist $c\in(0,1)$,such $$f(c)f'(c)+f''(c)=0$$

my try: since $$f(0)=2,f'(0)=-2,f(1)=1$$ so we easy $$f(x)=x^2-2x+2$$ such this condition,But we can't use this function prove this problem.

and other idea: we can find this ODE $$yy'+y''=0?$$

Thank you for you help!

$\endgroup$
  • 2
    $\begingroup$ Are you sure that there is no condition for the $f'(1)$...? $\endgroup$ – passenger Nov 30 '13 at 17:01
  • 1
    $\begingroup$ $y(x)= c_1 \tanh \left(c_1 x+c_2\right)$ $\endgroup$ – Norbert Nov 30 '13 at 17:22
  • 4
    $\begingroup$ $ff'+f''=D(f^2/2+f')$, but I don't immediately see a way of taking advantage (possibly passenger was hoping to apply Rolle's theorem to $f^2/2+f'$ :-) $\endgroup$ – Jyrki Lahtonen Nov 30 '13 at 17:26
  • 2
    $\begingroup$ @passenger Consider $f(x)=2-2x+x^2(1+(x-1) g(x))$ where $g$ is an arbitrary analytic function in $[0,1]$. The derivative $f'(1)$ turns out to be $g(1)$ which is arbitrary as well. And as far as I can tell the statement holds no matter what $g$ is taken. $\endgroup$ – user1337 Dec 1 '13 at 9:41
  • 1
    $\begingroup$ This problem is not easy,+1,and I hope someone can slove $\endgroup$ – math110 Dec 1 '13 at 12:07
15
$\begingroup$

Following Jyrki Lahtonen's observation, let

$$g(x) = \frac{f(x)^2}{2} + f'(x). $$

Since $g'(x) = f(x)f'(x) + f''(x)$, it is sufficient to prove that $g$ has a stationary point. If this is false, then $g$ is either increasing or decreasing on $(0,1)$.

Suppose first of all that $g(x)$ is increasing. Let $d = 1$ if $f(x) > 0$ for all $x \in [0,1]$, and otherwise let $d$ be smallest such that $f(d) = 0$. For $x \in [0,d)$ we have that $-f'(x) \le f(x)^2/2$. Dividing through by the strictly positive number $f(x)^2$ we get $(\star$):

$$ \frac{-f'(x)}{f(x)^2} = \Bigl( \frac{1}{f(x)} \Bigr)' \le \frac{1}{2} \quad\text{for $x\in [0,d)$.} $$

Since $f(0) = 2$, integrating gives $1/f(x) \le 1/2 + x/2$, or equivalently,

$$ f(x) \ge \frac{2}{1+x} \quad\text{for $x \in [0,d)$.} $$

It follows that $d=1$. Moreover, using the continuity of $f'$, we see that

$$f(1) > \frac{2}{1+1}$$

unless equality holds in ($\star$) for all $x \in (0,1)$. Hence $f(x) = \frac{2}{1+x}$ and $f(x)f'(x) + f''(x) = 0$ for all $x \in (0,1)$.

Now suppose that $g(x)$ is decreasing. Observe that if $f(d) = 0$ for some $d \in (0,1)$ then, since $f(1) = 1$, there exists $e \in (d,1)$ such that $f'(e) > 0$. Hence $g(e) > 0$. Since $g(0) = 0$ this is impossible. Therefore the same argument as before shows that

$$ f(x) \le \frac{2}{1+x} $$

for $x \in [0,1]$ and as before we get $f(x) = \frac{2}{1+x}$ for all $x \in [0,1]$.

$\endgroup$
2
$\begingroup$

I don't know, probably I do not understand the problem but it seems that one can do the following:

Consider the differential equation $$ f''(x) + f(x) f'(x) =0 $$ with boundary conditions $f(0) = 2, f'(0) = -2, f(1) = 1$.

Let's integrate it first time, thus we get: $$ f'(x) + \frac{1}{2} f^2(x) = C_1 $$ $C_1$ we will find by applying the conditions for $x=0$: $$ f'(0)+\frac{1}{2} f^2(0) = -2 + 2 = C_1 $$ Therefore, the new equation is: $$ f'(x) + \frac{1}{2} f^2(x) = 0 $$ And it can easily be integrated: $$ \frac{2}{f(x)} = x + C_2 $$ Applying the last condition which is $f(1)=1$, one can obtain: $$ f(x) = \frac{2}{x+1} $$ Seems like this function satisfies the equation $f'' + f f' = 0$ in any point of $(0,1)$, which means that one can always find a point $c$ such that $f(c)f'(c)+f''(c)=0$.

$\endgroup$
  • $\begingroup$ the proof would be over when we show that the differential equation with the given boundary conditions is well posed $\endgroup$ – vidyarthi Nov 18 '16 at 4:34

Your Answer

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