Let $\displaystyle f $ be continously differenciable on $\displaystyle \mathbb{R}$. Solve the integral equation:

$\displaystyle [f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt$

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- May 28th 2007, 12:04 PM #1

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- May 28th 2007, 01:22 PM #2

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- May 28th 2007, 01:25 PM #3

- May 28th 2007, 01:37 PM #4

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- May 28th 2007, 01:40 PM #5
I would much rather deal with a differential equation. Hopefully this works.

$\displaystyle \frac{d}{dx}[f(x)]^2=\frac{d}{dx}\int_0^x [f(t)]^2+[f'(t)]^2dt$

$\displaystyle 2[f(x)]f'(x) = [f(x)]^2+[f'(x)]^2$

$\displaystyle [f(x)]^2-2 [f(x) \cdot f'(x)] + [f'(x)]^2 = 0$

$\displaystyle [f(x) - f'(x)]^2=0$

$\displaystyle f(x)-f'(x)=0$

$\displaystyle f(x)=f'(x)$

$\displaystyle f(x) = Ae^x$

Am I right?

- May 28th 2007, 03:41 PM #6

- May 30th 2007, 06:58 AM #7

- May 30th 2007, 11:49 AM #8

- May 30th 2007, 11:51 AM #9

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- May 30th 2007, 06:29 PM #10
Okay, I oopsied. But still:

$\displaystyle f(x) = Ae^x$

Thus

$\displaystyle f(t) = Ae^t$

$\displaystyle \frac{df}{dt} = Ae^t$

Thus

$\displaystyle \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2$

$\displaystyle = \int_0^x dt \, (A^2e^{2t} + A^2e^{2t} )$

$\displaystyle = \int_0^x dt \, 2A^2e^{2t}$

$\displaystyle = 2A^2 \int_0^x dt \, e^{2t}$

$\displaystyle = 2A^2 \frac{1}{2}e^{2t}|_0^x$

$\displaystyle = A^2(e^{2x} - 1)$

And

$\displaystyle [ f(x) ] ^2 = A^2e^{2x}$

So

$\displaystyle [ f(x) ] ^2 \neq \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2$

How can this then be the solution of the integral equation? I'm missing something here...

-Dan

- May 30th 2007, 08:18 PM #11

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- May 30th 2007, 09:04 PM #12

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- May 30th 2007, 11:19 PM #13
Then $\displaystyle A = 0.$ So ecMathGeeks proposed solution does not satisfy the integral equation for any $\displaystyle A$, which the second post seems to imply, only when $\displaystyle f(x) = Ae^x = 0.$ Is that what is going on here?

- May 31st 2007, 12:38 AM #14

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The integral equation when converted to a diffrential equation is in fact

an initial value problem as of necessity $\displaystyle f(0)=0$

ecMathGeek gave a general solution to the DE that you get by differentiating

the integral equation but omitted the initial condition that forces $\displaystyle A=0$.

RonL

- May 31st 2007, 05:11 AM #15

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