can you help me please, to find $u \in \mathcal{D}'(\mathbb{R})$,solutions of the equation $xu=H+c $ where $c$ is an constant, $x$ the function identity, and $H$ is Heaviside function. I purpose this solution.
Let $\varphi \in \mathcal{D}(\mathbb{R}).$ since $x \in C^\infty$,then $\psi=x \varphi \in \mathcal{D}(\mathbb{R})$. We have for all $x \neq 0$ \begin{align*} <u,\psi> & = \displaystyle\int_0^\infty \varphi(x) dx + c \displaystyle\int_0^\infty \varphi(x) dx\\ &= \dfrac{1}{x} \displaystyle\int_0^\infty \psi(x) dx+ c \dfrac{1}{x} \displaystyle\int_0^\infty \psi(x) dx \end{align*} Then, $u= \dfrac{1}{x}(1+c)$ sur $]0,+\infty[$.
But what about $x \in ]-\infty,0]$ in the case where my purpose is true?
Thank you in advance for the help.
In fact, you need to solve three different equations.
First we need to learn how to solve the equation $$xu=0.$$This is a standard result, the solution writes $u=b\delta_0$ with $b$ an arbitrary constant.
Second, the equation $$xu=1$$ has a particular solution $PV(1/x)$ - the principal value.
Finally, the most interesting one is the part $$xu=H.$$ The naive approach would be to say that $$s=\begin{cases}1/x,&x>0\\0,&x\le0.\end{cases}$$ This, however, is not a distribution on $\Bbb R$, hence we need to invent something else.
We can notice that the antiderivative of $s$, namely, $H(x)\ln x$, is a distribution (it is locally integrable) on $\Bbb R$, so we would like to say that $u=-(H(x)\ln x)'$. A quick integration by part shows that indeed in the sens of distributions one has $$-x(H(x)\ln x)'=H(x).$$
We can now combine all these results into the final answer
$$u = -(H(x)\ln x)'+cPV(1/x)+b\delta_0,\quad b\in \Bbb C.$$