# Thread: Problem 24

1. ## Problem 24

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

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

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

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

RonL

3. So whats the answer CaptainBlack?

4. Originally Posted by janvdl
So whats the answer CaptainBlack?
Not telling; that would spoil the fun

RonL

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

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

6. 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.

7. Originally Posted by ecMathGeek
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?
The LHS is
$\displaystyle A^2e^{2x}$

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

so I would say "no."

-Dan

8. Originally Posted by CaptainBlack
No

The right hand side is not what you say!

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

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

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

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

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

11. Originally Posted by topsquark
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
So you have shown that if $\displaystyle f(t) = Ae^t$, then:

$\displaystyle A^2e^{2x}$$\displaystyle = A^2(e^{2x} - 1) RonL 12. Originally Posted by topsquark Okay, I oopsied. But still: \displaystyle f(x) = Ae^x Thus \displaystyle f(t) = Ae^t \displaystyle \frac{df}{dt} = Ae^t Thus \displaystyle \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2 \displaystyle = \int_0^x dt \, (A^2e^{2t} + A^2e^{2t} ) \displaystyle = \int_0^x dt \, 2A^2e^{2t} \displaystyle = 2A^2 \int_0^x dt \, e^{2t} \displaystyle = 2A^2 \frac{1}{2}e^{2t}|_0^x \displaystyle = A^2(e^{2x} - 1) And \displaystyle [ f(x) ] ^2 = A^2e^{2x} So \displaystyle [ f(x) ] ^2 \neq \int_0^x dt \, [ f(t) ]^2 + \left [ \frac{df}{dt} \right ] ^2 How can this then be the solution of the integral equation? I'm missing something here... -Dan If you don't like ecMathGeeks approach try a power series solution. RonL 13. Originally Posted by ecMathGeek \displaystyle f(x) = Ae^x Originally Posted by CaptainBlack ecMathGeeks proposed solution satisfies the integral equation (ish) Originally Posted by CaptainBlack So you have shown that if \displaystyle f(t) = Ae^t, then: \displaystyle A^2e^{2x}$$\displaystyle = A^2(e^{2x} - 1)$
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?

14. Originally Posted by JakeD
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?
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

15. 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 $\displaystyle \mathbb{R}$.

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