Dirac Delta, higher derivatives with test function

[mishima]

Hi, I am curious about:

$$x^m \delta^{(n)}(x) = (-1)^m \frac {n!} {(n-m)!} \delta^{(n-m)}(x) , m \leq n $$

I understand the case where m=n and m>n but not this. Just testing the left hand side with m=3 and n=4 and integrating by parts multiple times, I get -6. With the same values, the right hand side evaluates to -24.

 

[mishima]

To elaborate, what I'm doing on the left hand side is integration by parts multiple times:

$$\int x^3\delta^{(4)} (x) dx=x^3\delta^{(3)}(x)-\int 3x^2\delta^{(3)}(x)dx$$

Bounds are -infinity to +infinity (Latex ignorance). The non integral term evaluates to zero because of the way test functions are defined nonzero only inside a finite interval. So,

$$\int x^3\delta^{(4)} (x) dx=-\int 3x^2\delta^{(3)}(x)dx$$

Then the idea is to repeat the integration by parts until the dirac delta derivative order is 1.

$$\int 3x^2\delta^{(3)}(x)dx=-\int6x\delta^{(2)}(x)dx$$
$$\int6x\delta^{(2)}(x)dx=-\int6\delta(x)dx=-6$$

Being mindful of the way the negative sign flip flops each iteration. I know there is a formula for this but I'm just showing my understanding.

 

[jasonRF]

You forgot to include a test function in your integral. Let $\phi(x)$ be a test function and try starting with
$$ \int \, \phi(x) \, x^3 \, \delta^{(4)}(x) \, dx$$
and I think you will get the correct result.

Jason

 

[stevendaryl]

Hi, I am curious about:

$$x^m \delta^{(n)}(x) = (-1)^m \frac {n!} {(n-m)!} \delta^{(n-m)}(x) , m \leq n $$

I understand the case where m=n and m>n but not this. Just testing the left hand side with m=3 and n=4 and integrating by parts multiple times, I get -6. With the same values, the right hand side evaluates to -24.

For one thing, the equals sign there isn't correct. It's not that the two sides are equal, but that they integrate to the same thing.

For another thing, I don't understand why it isn't just:

$$\int x^m \frac{d^n}{dx^n} \delta(x) dx = m! \int \frac{d^{n-m}}{dx^{n-m}} \delta(x) dx
$$

 

[jasonRF]

For one thing, the equals sign there isn't correct. It's not that the two sides are equal, but that they integrate to the same thing.

The equality sign is fine since the two sides are equal as distributions (generalized functions). The presence of the delta functions makes that pretty clear, I think. But distributions are not functions, which map numbers to numbers. Instead, they are functionals, which map functions to numbers. Using our standard non-rigorous physics/ engineering approach, this means that when you multiply them by an arbitrary test function and "integrate", they produce a number that in general depends on the particular test function. For the expression in the original post, both sides produce the same number when given the same test function.

All of this is also why the test function ($\phi(x)$ in my example) had to be included in the integral. If you don't includ it, the explicit assumption is that $\phi(x)=1$, and since all of the derivatives of $\phi$ are zero, the original expresion simply says $0=0$. EDIT (again!): both sides are zero only for the $m<n$ case.

jason

 

[stevendaryl]

The equality sign is fine since the two sides are equal as distributions (generalized functions).

Okay. I was thinking that they weren't equal as distributions, because doing integration by parts will produce a derivative of the test function. However, that derivative is multiplied by a power of $x$, so it will evaluate to 0 at $x=0$.

$\int (x \frac{d}{dx} \delta(x)) \phi(x) dx = \int \frac{d}{dx} (x \delta(x) \phi(x)) dx - \int \delta(x) \frac{d}{dx} (x \phi(x)) dx = \int \frac{d}{dx} (x \delta(x) \phi(x)) dx - \int \delta(x) (\phi(x) + x \frac{d}{dx} \phi(x)) dx$
$= 0$ (surface integral) $- \phi(0) - (x \frac{d}{dx} \phi(x))|_{x=0} = - \phi(0)$

 

[jasonRF]

I edited my post at the same time you are posting, so maybe it is more clear...

 

[mishima]

Thanks...I was thinking that the $x^3$ was the test function $\phi(x)$, but I understand now why that's not the case. It has to be the same test function on both sides. I will re-evaluate now.

 

[mishima]

Just confirming I was able to get $24 \phi^{'}(0)$ for both sides with m=3 and n=4. Likewise m=2 and n=4 yields $12 \phi^{''}(0)$ for both. Pretty interesting behavior. Thanks again.

 

[jasonRF]

I've received a PM asking for some details on how these calculations work, so here is a sketch. We need to show
$$
\int \phi(x)\, x^m \, \delta^{(n)}(x)\, dx = (-1)^m \frac {n!} {(n-m)!} \int \phi(x) \, \delta^{(n-m)}(x) \, dx
$$
when $m \leq n$ for every 'nice' test function $\phi$. In this case, 'nice' means a function where the first $n-m$ derivatives are continuous.

We start with the left-hand-side. Using the definition $$\int \psi(x) \delta^{(n)}(x) dx = (-1)^n \psi^{(n)}(0)$$, we have
$$
\begin{eqnarray*}
\int \, \phi(x) \, x^m \, \delta^{(n)}(x) \, dx & = & \left. (-1)^n \frac{d^n }{dx^n} x^m \phi(x) \right|_{x=0}
\end{eqnarray*}
$$
Then we use Liebnitz's rule
$$
\frac{d^n}{dx^n} f(x) g(x) = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} f^{(k)}(x) g^{(n-k)}(x)
$$
with $f=x^m$ and $g=\phi$. Since we will be evaluating this at $x=0$, it is clear that $k=m$ is the only term that is nonzero. So I get
$$
\left. (-1)^n \frac{d^n }{dx^n} x^m \phi(x) \right|_{x=0} = (-1)^n \frac{n!}{(n-m)!} \phi^{(n-m)}(0)
$$
However,
$$
\phi^{(n-m)}(0) = (-1)^{n-m} \int \phi(x) \, \delta^{(n-m)}(x) \, dx
$$

Puting all of this together we get the desired identity.

jason