I was trying to solve the following problem(where "$u\in D'(I)$" means that $u$ is a distribution on the interval $I$):

Find all solutions $u\in D'(\mathbb{R})$ to the following equation:


I tried to solve it in the following way:

$$\begin{align} &tu'+u=t &\iff \\&(tu)'=(t^2/2+C_1)'&\iff \\&(tu-t^2/2-C_1)'=0&\iff \\&tu-t^2/2-C_1=C_2&\iff \\&t(u-t/2)=C&\iff \\&u-t/2=\underline{t}^{-1}C &\iff \\&u=\underline{t}^{-1}C+t/2 \end{align}$$ where $C_1,C_2\in\mathbb{C}$ are arbitrary constants, and $C=C_1+C_2$.

However, according to the solution manual, I should get that

$u=\underline{t}^{-1}C+t/2+D\delta$, where $C$ and $D$ are arbitrary complex constants, and $\delta$ is the Dirac delta function.

Where in my solution attempt did I make a mistake, which made me lose the $D\delta$-term in my answer?


1 Answer 1


When you solve the equation $$xT=0$$ for $T\in D'(\Bbb R)$, the set of solutions is $k\delta_0$ with $k$ being an arbitrary constant (there are several answers on this site explaining this effect). Hence, when you solve $t(u-t/2)=C$, you obtain a family of solutions $u-t/2=Cv.p.\left(\frac 1t\right)+k\delta_0$, and so forth.

  • $\begingroup$ Thank you for a good answer. I realized this myself yesterday right before I went to bed, so I didn't feel like I had the time to write the answer myself then. It's good to see that I was able to understand things properly. $\endgroup$
    – Scounged
    Sep 9, 2015 at 5:12

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