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$f\in C^{1}[0,\infty)$, $f(0)=0$ and $$ f(x)+f'(x)-\frac{1}{x+1}\int_{0}^{x}f(t)dt=0 $$ then $f'(x)=$ ?

I'v tried in the following ways. First, let $F(x)=\int_{0}^{x}f(t)dt$, then we are left to solve a second order ODE with initial condition $F(0)=0$, and $F'(0)=0$, but the problem is it seems to me that it's not that easy to solve it.

I hope there are some other ways to handle it that a one year students can understand. (I have tried to write $g(x)=f(x)e^x$ to rewrite the equation, but it doesn't make it easier)

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    $\begingroup$ Write $F(x) = (1+x) g(x)$ and plug in. This reduces the equation to first order ODE. $\endgroup$
    – Marek
    Commented Jan 9, 2013 at 11:13
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    $\begingroup$ $f(x)\equiv 0$ is a solution. $\endgroup$
    – Paul
    Commented Jan 9, 2013 at 12:09

2 Answers 2

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Multiply the both sides by (x+1). Then differentiate. You'll get

$$(x+1) f''(x)+(x+2) f'(x)=0$$

Solve this equation. The solution is

$$f(x)=C_1 \text{Ei}(-x-1)+C_2$$

Now find derivative:

$$f'(x)=\frac{C_1 e^{-x}}{x+1}$$

From your condition that $f(0)=0$, it follows that $f'(0)=0$ and as such, $C_1=0$. So $f(x)$ is a constant zero.

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  • $\begingroup$ are you sure your first equation is correct? I don't think it is. $\endgroup$ Commented Jan 9, 2013 at 15:49
  • $\begingroup$ the second term should be $f'(x)$. But here f is only assumed to be in $C^1$, so I think you can't do this way $\endgroup$
    – Tomas
    Commented Jan 9, 2013 at 15:56
  • $\begingroup$ @nbubis I fixed the typo, thanks. The result does not change, it was only a typo when I copied it here. $\endgroup$
    – Anixx
    Commented Jan 9, 2013 at 16:29
  • $\begingroup$ @sun I fixed the typo, thanks. The result does not change, it was only a typo when I copied it here. $\endgroup$
    – Anixx
    Commented Jan 9, 2013 at 16:30
  • $\begingroup$ @nbubis I also corrected the conclusion: actually there ARE non-trivial solutions, yet the condition does not specify a unique one. $\endgroup$
    – Anixx
    Commented Jan 9, 2013 at 16:36
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You can try expanding $f(x)$ as a series around $x=-1$: $$f(x) = \sum a_n(x+1)^n$$ So that your equation becomes: $$a_n (x+1)^n + n a_n (x+1)^{n-1} - \frac{a_n (x+1)^n}{n+1} + \frac{a_n}{(n+1) (x+1)}=0$$ $$(n+1)a_n(x+1)^{n+1} + n (n+1)a_n (x+1)^{n} - a_n (x+1)^{n+1} + a_n=0$$ Since you know that $f(x)=0$, $$(n+1)a_n + n (n+1)a_n = 0 \ \ \rightarrow \ \ \forall n:a_n = 0$$ Thus the only analytic solution is $f(x)=0$, but of course ther may exist non analytic solutions as well, though I doubt it.

Edit: As @Anixx has shown there are no non analytic solutions as well.

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  • $\begingroup$ why the downvote? $\endgroup$ Commented Jan 9, 2013 at 18:37

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