How is Dirac delta $\delta (x^3 )$ different from $\delta (x)$? It is my understanding that $$ \delta [\psi]:=\langle\delta,\psi\rangle=\int_{\Omega}\delta(x)\psi(x)\,dx= \begin{cases} \psi(0) & \text{if $0\in \Omega$}\\ 0 & \text{otherwise} \end{cases} $$ But $x^3=0$ iff $x=0$ so does this not imply, $$\delta(x^3) = \delta(x) = 0$$ Apologies if I am completely misunderstanding the topic, I am very new to distributions.
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1Good question. The delta distribution with nonlinear arguments is a tricky concept. This answer of robjohn has been extremely useful to me in order to understand that. – Giuseppe Negro Jan 22 '21 at 19:55
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Compare $\int_{\Bbb R}f(x)\delta(x^3),dx=\int_{\Bbb R}u^{-2/3}f(u^{1/3})\delta(u)/3,du$ with $\int_{\Bbb R}f(u)\delta(u),du$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jan 22 '21 at 20:35
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1It appears to me that you are using conflicting notation. The "$\delta(x)$" appearing inside the integral is not the same as the "$\delta$" that is applied to $\psi$ on the left hand side. – pregunton Jan 22 '21 at 20:37
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Maybe too naive but øan't you substitute $y = x^3$ in $\int \delta(x^3) f(x) dx$ and see what happens? – Jeroen van der Meer Jan 22 '21 at 20:54
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I found this https://math.stackexchange.com/questions/2481114/dirac-delta-function-deltafx-of-function-f-with-a-higher-order-zero – Jeroen van der Meer Jan 22 '21 at 21:05
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1@pregunton I've updated the question to fit exactly what is written in my lecture handout – retsek680 Jan 22 '21 at 21:39
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See also this result. – Tob Ernack Jan 23 '21 at 02:44
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Let $\psi$ be a test function, i.e. $\psi \in \mathcal{C}^{\infty} _{c}(\mathbb{R})$, then: $$ \int \delta(x^3)\psi (x) dx = \int \delta(y) \psi (y^{\frac{1}{3}}) \frac{1}{3y^{\frac{2}{3}}} dy $$ Note that our new test function $\psi (y^{\frac{1}{3}}) \frac{1}{3y^{2/3}}$ is not smooth, hence we cannot define $\delta(x^3)$ as a distribution. In general, if $u$ is a distribution, then $u \circ g$ is a distribution as well if both $g$ and its inverse are smooth. In such a case: $\langle u \circ g, \psi \rangle := \langle u, \psi \circ g^{-1} |\text{det } dg^{-1}| \rangle$. Please note that the integration above is an abuse of notation.
Saleh
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