Problem 24

**ThePerfectHacker** · May 28th 2007, 12:04 PM

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

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

**CaptainBlack** · May 28th 2007, 01:22 PM

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

**janvdl** · May 28th 2007, 01:25 PM

So whats the answer CaptainBlack?

**CaptainBlack** · May 28th 2007, 01:37 PM

Originally Posted by janvdl

So whats the answer CaptainBlack?

Not telling; that would spoil the fun

RonL

**ecMathGeek** · May 28th 2007, 01:40 PM

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?

**ecMathGeek** · May 28th 2007, 03:41 PM

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.

**topsquark** · May 30th 2007, 06:58 AM

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

**Jhevon** · May 30th 2007, 11:49 AM

Originally Posted by CaptainBlack

No

The right hand side is not what you say!

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

**CaptainBlack** · May 30th 2007, 11:51 AM

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

**topsquark** · May 30th 2007, 06:29 PM

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

**CaptainBlack** · May 30th 2007, 08:18 PM

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

**CaptainBlack** · May 30th 2007, 09:04 PM

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

**JakeD** · May 30th 2007, 11:19 PM

Originally Posted by ecMathGeek

$f(x) = Ae^x$

Originally Posted by CaptainBlack

ecMathGeeks proposed solution satisfies the integral equation (ish)

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?

**CaptainBlack** · May 31st 2007, 12:38 AM

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

**ThePerfectHacker** · May 31st 2007, 05:11 AM

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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