# Problem 24

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• May 28th 2007, 12:04 PM
ThePerfectHacker
Problem 24
Let $f$ be continously differenciable on $\mathbb{R}$. Solve the integral equation:

$[f(x)]^2 = \int_0^x [f(t)]^2+[f'(t)]^2dt$
• May 28th 2007, 01:22 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Let $f$ be continously differenciable on $\mathbb{R}$. Solve the integral equation:

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

Nice.

RonL
• May 28th 2007, 01:25 PM
janvdl
So whats the answer CaptainBlack? :D
• May 28th 2007, 01:37 PM
CaptainBlack
Quote:

Originally Posted by janvdl
So whats the answer CaptainBlack? :D

Not telling; that would spoil the fun:D

RonL
• May 28th 2007, 01:40 PM
ecMathGeek
Quote:

Originally Posted by ThePerfectHacker
Let $f$ be continously differenciable on $\mathbb{R}$. Solve the integral equation:

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

I would much rather deal with a differential equation. Hopefully this works.

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

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

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

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

$f(x)-f'(x)=0$
$f(x)=f'(x)$
$f(x) = Ae^x$

Am I right?
• May 28th 2007, 03:41 PM
ecMathGeek
Is there a way to apply color and font to LaTeX? I want to try to white-out my solution, but as far as I can tell that's impossible.
• May 30th 2007, 06:58 AM
topsquark
Quote:

Originally Posted by ecMathGeek
I would much rather deal with a differential equation. Hopefully this works.

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

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

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

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

$f(x)-f'(x)=0$
$f(x)=f'(x)$
$f(x) = Ae^x$

Am I right?

The LHS is
$A^2e^{2x}$

and the RHS is
$(A^2 - 1)e^{2x}$

so I would say "no." :(

-Dan
• May 30th 2007, 11:49 AM
Jhevon
Quote:

Originally Posted by CaptainBlack
No(No)

The right hand side is not what you say! :eek:

RonL

it would be $\left( e^{2x} - 1\right)A^2$ ?

if so, it means there is something wrong with what ecMathGeek did, but i didn't really see anything wrong the first time i read it, i'll read it again
• May 30th 2007, 11:51 AM
CaptainBlack
Quote:

Originally Posted by Jhevon
it would be $\left( e^{2x} - 1\right)A^2$ ?

if so, it means there is something wrong with what ecMathGeek did, but i didn't really see anything wrong the first time i read it, i'll read it again

ecMathGeeks proposed solution satisfies the integral equation (ish)

RonL
• May 30th 2007, 06:29 PM
topsquark
Okay, I oopsied. But still:

$f(x) = Ae^x$

Thus
$f(t) = Ae^t$

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

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

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

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

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

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

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

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

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

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

-Dan
• May 30th 2007, 08:18 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Okay, I oopsied. But still:

$f(x) = Ae^x$

Thus
$f(t) = Ae^t$

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

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

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

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

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

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

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

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

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

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

-Dan

So you have shown that if $f(t) = Ae^t$, then:

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

RonL
• May 30th 2007, 09:04 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Okay, I oopsied. But still:

$f(x) = Ae^x$

Thus
$f(t) = Ae^t$

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

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

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

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

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

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

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

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

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

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

-Dan

If you don't like ecMathGeeks approach try a power series solution.

RonL
• May 30th 2007, 11:19 PM
JakeD
Quote:

Originally Posted by ecMathGeek
$f(x) = Ae^x$

Quote:

Originally Posted by CaptainBlack
ecMathGeeks proposed solution satisfies the integral equation (ish)

Quote:

Originally Posted by CaptainBlack
So you have shown that if $f(t) = Ae^t$, then:

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

Then $A = 0.$ So ecMathGeeks proposed solution does not satisfy the integral equation for any $A$, which the second post seems to imply, only when $f(x) = Ae^x = 0.$ Is that what is going on here?
• May 31st 2007, 12:38 AM
CaptainBlack
Quote:

Originally Posted by JakeD
Then $A = 0.$ So ecMathGeeks proposed solution does not satisfy the integral equation for any $A$, which the second post seems to imply, only when $f(x) = Ae^x = 0.$ Is that what is going on here?

The integral equation when converted to a diffrential equation is in fact
an initial value problem as of necessity $f(0)=0$

ecMathGeek gave a general solution to the DE that you get by differentiating
the integral equation but omitted the initial condition that forces $A=0$.

RonL
• May 31st 2007, 05:11 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
If you don't like ecMathGeeks approach try a power series solution.

RonL

Not every continously differenciable function is analytic.*

*)By analytic I mean it has a power series expansion on $\mathbb{R}$.
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