APM346-2018S > Quiz-B

Quiz-B P2

(1/3) > >>

Victor Ivrii:
Simplify (write as the sum of $\delta (x)$, $\delta '(x)$, $\delta''' (x)$, ... with the numerical coefficients)
\begin{equation}
\sin(x)\delta'''(x).
\tag{1}
\end{equation}

Jingxuan Zhang:
Observe for nice $\varphi$
$$(\sin(x) \varphi(x))'''=-\cos(x) \varphi(x)-3\sin(x)\varphi'(x)+3\cos(x)\varphi''(x)+\sin(x)\varphi'''(x).$$
Therefore
$$\langle\sin(x)\delta'''(x),\varphi(x)\rangle=\langle\delta'''(x),\sin(x) \varphi(x)\rangle=-\langle\delta(x),(\sin(x) \varphi(x))'''\rangle=\varphi(0)-3\varphi''(0).$$
So $$\sin(x)\delta'''(x)=\delta(x)-3\delta''(x).$$

Victor Ivrii:
Typical error: calculated $\int \sin(x)\delta''' (x)\,dx$ thus applying $\sin(x)\delta''' (x)$ to the single test function $1$ instead of an arbitrary $\varphi$.

Another solution (some thought about it): $\displaystyle{\sin(x)=x-\frac{x^{3}}{3!}+\ldots}$ and therefore $\displaystyle{\sin(x)\delta''=x\delta'''-\frac{x^{3}}{3!}\delta'''}$ since $x^{k}\delta^{(n)}(x)=0$ for $k>n$. But one need to know that
$$
x^k\delta^{(n)}(x)= k! (-1)^k \delta^{(n-k)}(x)\qquad\text{for }\ n\ge k.
\tag{*}$$


Bonus: Prove (*)

Andrew Hardy:
We have the definition that $$  (\partial f)(\phi) = -(f)(\partial\phi) $$

so $$ ( x^k \delta^{(1)}(x) ) = -k  x^{(k-1) }\delta^{(1-1)} (x) $$
assume this holds true for  n-1
 $$ x^{(k+1)} δ^{(n-1)}(x)=(k-1)! (−1)^{(k-1)}δ^{(n-1−k)}(x) $$
via induction  and by the definition $ x^{k+1} δ^{(n-1)}(x) = - x^k δ^{(n)}(x) $exactly, how? and what you make induction with respect to? V.I.
 $$ x^k δ^{(n)}(x)=  k!(−1)^kδ^{(n−k)}(x) $$
 I have to add the condition $$ n\geq k $$ because if n was less than, I'd have a negative derivative which is undefined.

Victor Ivrii:
Actually I imposed $n\ge k$ but the negative derivative are defined as primitives: $\delta^{(-1)}(x)=\theta (x)$ and
$\delta^{(-k)}= \frac{x^{k-1}}{(k-1)!} \theta(x)$ for $k\ge 1$. We can even deal with $k\in \mathbb{C}$...

Navigation

[0] Message Index

[#] Next page