verification: find a function $f$ such that $\frac{df}{d\phi} = \sin(\theta+\beta)\, \delta(\phi-\theta-\beta)$

Question

Comments

I initially posed the problem statement incorrectly. The problem statement has been altered.

Problem Statement:

By $\delta$ I denote the Dirac delta function. By $\frac{df}{d\phi} : \mathbb{R} \rightarrow \mathbb{R}$ I denote a function defined by $$\frac{df}{d\phi} = \sin(\theta+\beta)\, \delta(\phi-\theta-\beta),$$ which is valid for $\phi \in [0, \pi]$, $\theta \in [0, \pi]$ and $\beta \in [0, 2\,\pi)$. Find a function $f$ that satisfies the equation above?

Solution:

Let $f$ be a function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $$f(\phi) = H(\phi-\theta-\beta)\,\sin(\theta+\beta) + K,$$ which is valid for $\phi \in [0,\pi]$, $\phi \in [0,\pi]$, and $\beta \in [0,2\pi]$. I assert that $f$ satisfies the equation $\frac{df}{d\phi} = \sin(\theta+\beta)\, \delta(\phi-\theta-\beta)$.

Demonstration

To prove this, I'm simply going to the derivative of my solution.

\begin{align} \dfrac{d}{d\phi}[f(\phi)] &= \dfrac{d}{d\phi}[H(\phi-\theta-\beta)\,\sin(\theta+\beta) + K] \\ &= \dfrac{d}{d\phi}[H(\phi-\theta-\beta)\,\sin(\theta+\beta)] \\ &= \sin(\theta+\beta)\,\dfrac{d}{d\phi}[H(\phi-\theta-\beta)] \\ &= \sin(\theta+\beta)\,\delta(\phi-\theta-\beta)\dfrac{d}{d\phi}[\phi-\theta-\beta] \\ &= \sin(\theta+\beta)\,\delta(\phi-\theta-\beta) \end{align}

Answer 1

It appears my answer is correct. Post closed

Answer 2

The distribution $\delta(\phi+\beta)\sin(\phi+\beta)$ is defined by multiplication of smooth functions and deltas which would just be $\sin(-\beta+\beta)\delta(\phi+\beta) = 0$ so your solution isn't correct.

Another way to see this is to take the derivative of your solution: $$(H(\phi+\beta)\cos(\phi+\beta))'= \delta(\phi+\beta)\cos(\phi+\beta) - H(\phi+\beta)\sin(\phi+\beta)$$ $$ = \delta(\phi+\beta) - H(\phi+\beta)\sin(\phi+\beta)$$ which is definitely not equal to the original distribution we started off with.