Solving a differential equation of distributions

Question

Can anyone give me a hint on how to solve the following equation? Suppose $u \in \mathcal{D}'(\mathbb{R})$ and $xu' +u = 0$ then $u = A (1/x) + B\delta$ where $A$ and $B$ are constants and $(1/x)$ is the principal value distribution . All I have done is solve the classical equation $xu' +u = 0$ which gives me that $u = \frac{c_1}{x}$ but I don't know how to conclude from this that the solution uses the pricipal value distribution or the delta function. Any help is highly appreciated!

Answer 1

Note that $(xu)' = xu' + u = 0$, hence $xu = \text{const.}$ But any solution of $xu = A$ is of the form $A \cdot\text{p.v.}(\tfrac{1}{x}) + B\cdot \delta(x)$.

If you haven't seen the latter part: Assume that $xu =1$. We know that $v = \text{p.v.}(\tfrac{1}{x})$ is a solution. Assume that $w$ is another solution. Then $v-w$ is a solution to $xu = 0$. But $$xu=0 \implies \text{supp}(u) \subseteq \{0\}$$

which on the other hand implies that $u = \sum_{k=0}^m \lambda_k \delta^{(k)}$ for some $m\in \mathbb N$ and coefficients $\lambda_k\in\mathbb C$. It follows that $u = B\delta$, hence any solution to $xu=1$ is of the form $\text{p.v.}(\tfrac{1}{x}) + B\delta(x)$