Problem 24

ThePerfectHacker · #1 $Old$ May 28th, 2007, 02:04 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 · #2 $Old$ May 28th, 2007, 03:22 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

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 · #3 $Old$ May 28th, 2007, 03:25 PM

So whats the answer CaptainBlack?

CaptainBlack · #4 $Old$ May 28th, 2007, 03:37 PM

Quote:

Originally Posted by janvdl $View Post$

So whats the answer CaptainBlack?

Not telling; that would spoil the fun

RonL

ecMathGeek · #5 $Old$ May 28th, 2007, 03:40 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

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 · #6 $Old$ May 28th, 2007, 05:41 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 · #7 $Old$ May 30th, 2007, 08:58 AM

Quote:

Originally Posted by ecMathGeek $View Post$

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 · #8 $Old$ May 30th, 2007, 01:49 PM

Quote:

Originally Posted by CaptainBlack $View Post$

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 · #9 $Old$ May 30th, 2007, 01:51 PM

Quote:

Originally Posted by Jhevon $View Post$

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 · #10 $Old$ May 30th, 2007, 08:29 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 · #11 $Old$ May 30th, 2007, 10:18 PM

Quote:

Originally Posted by topsquark $View Post$

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 · #12 $Old$ May 30th, 2007, 11:04 PM

Quote:

Originally Posted by topsquark $View Post$

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 · #13 $Old$ May 31st, 2007, 01:19 AM

Quote:

Originally Posted by ecMathGeek $View Post$

$f(x) = Ae^x$

Quote:

Originally Posted by CaptainBlack $View Post$

ecMathGeeks proposed solution satisfies the integral equation (ish)

Quote:

Originally Posted by CaptainBlack $View Post$

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 · #14 $Old$ May 31st, 2007, 02:38 AM

Quote:

Originally Posted by JakeD $View Post$

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 · #15 $Old$ May 31st, 2007, 07:11 AM

Quote:

Originally Posted by CaptainBlank $View Post$

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 $\mathbb{R}$ .