# Thread: Derivatives, Integrals, Absolute Values Oh My!

1. ## Derivatives, Integrals, and Absolute Values, Oh My!

This is making me feel quite stupid right now, but I have the following integral that is messing with my head
$\frac{1}{2\gamma} \int_{-\infty}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma |x - y|}\right] q(y)\, dy$
Where $q:\mathbb{R}\rightarrow\mathbb{R}$ is continuous with compact support.
My first bit of instinct is to split the integral, because I need to know how to take that derivative. So there are two cases $x < y$ and $x>y$, but how do I translate this into the limits of the integral? It seems that I would split it like so
$\frac{1}{2\gamma} \left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma |x - y|}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma |x - y|}\right] q(y)\, dy\right)$.
$=\frac{1}{2\gamma} \left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma (x - y)}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma (y - x)}\right] q(y)\, dy\right)$
$=\frac{1}{2\gamma} \left(\int_{-\infty}^{x}\! \gamma^2e^{-\gamma (x - y)} q(y)\, dy + \int_{x}^{\infty}\! \gamma^2 e^{-\gamma (y - x)} q(y)\, dy\right)$
$=\frac{\gamma}{2} \left(\int_{-\infty}^{x}\! e^{-\gamma (x - y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma (y - x)} q(y)\, dy\right)$

Does this seem correct?

2. Looks good to me so far.

3. I was hoping it didn't. I guess I will post the whole problem.

Show that
$u(x) = \frac{1}{2\gamma}\int_{-\infty}^{\infty} e^{-\gamma |x - y|}q(y)\, dy$
satisfies
$-u''(x) + \gamma^2 u(x) = q(x)$.
What I end up getting in the end is $0 = q(x)$ instead of $q(x) = q(x)$.

Now my question is miscategorized. lol

4. Well, show what you get when you plug in the derivative and the original function u(x) into your DE.

5. $-u'' + \gamma^2 u = -\frac{d^2}{dx^2}\left(\frac{1}{2\gamma}\int_{-\infty}^{\infty}\! e^{-\gamma|x-y|} q(y)\, dy\right) + \gamma^2\left(\frac{1}{2\gamma}\int_{-\infty}^{\infty}\! e^{-\gamma|x-y|} q(y)\, dy\right)$
$= -\frac{1}{2\gamma}\int_{-\infty}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma|x-y|}\right] q(y)\, dy+ \frac{\gamma}{2}\int_{-\infty}^{\infty}\! e^{-\gamma|x-y|} q(y)\, dy$
$= -\frac{1}{2\gamma}\left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma(x-y)}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma(y-x)}\right] q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right)$
$= -\frac{1}{2\gamma}\left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma(x-y)}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma(y-x)}\right] q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right)$
$= -\frac{1}{2\gamma}\left(\int_{-\infty}^{x}\! \gamma^2 e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! \gamma^2 e^{-\gamma(y-x)} q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right)$
$= -\frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right) = 0$

6. Where did the minus sign out front come from? It's not in post # 3.

Incidentally, I'm thinking that a Dirac delta function is going to show up any minute. I can feel it in my bones.

7. It is now. That was a typo. It is suppose to be $-u''(x) + \gamma^2 u(x) = q(x)$.

EDIT: Maybe the test had a typo? Hmmm.

8. Find the Green's function. That is, solve

$-G''+\gamma^{2}G=\delta(x).$

I'll bet you anything that the Green's function for this DE is $\frac{1}{2\gamma}\,e^{-\gamma|x-y|}.$

At that point, I think you're pretty much done, since you can write the solution as

$u=\int_{-\infty}^{\infty}G(x,y)g(y)\,dy.$

I'll bet that works for you.

9. That might be. It does look like a Green's Function.

10. Whoops. I think the Green's function satisfies

$-G''+\gamma^{2}G=\delta(x-y).$

Try that.

11. I have to come back to this. I have to get through one more test, before the end of the day and I have spent too much time on this problem. Thank you.

12. Ha, I started the next test and it has the same thing attached to a problem with Green's Functions. So I guess that is what I need to do.