Fundamental solution How I can find the fundamental solution to $∂u+au=f$ where $a$ is a constant? I know that for $Lu=f$ I want to find $E$ such that $LE= δ$. I am trying to solve $(∂+a)E= δ$. I am stuck and I don't know what I am doing wrong because I can't find an $E$ that works. 
For example. $E=H$ gives $(∂+a)H= \delta + aH \neq \delta$
 A: Here's my intuition: away from $x=0$, the fundamental solution should solve $\partial E + a E = 0$ and thus should be of the form $E(x) = C e^{-ax};$ but it should jump up by $1$ at $x=0$. Thus I guess the piecewise smooth function
$$ E(x) = \begin{cases}e^{-ax} & x < 0 \\ 2 e^{-ax} & x \ge 0 \end{cases}.$$
To check that this is indeed a solution, we need to verify $$\int(E' + aE)\phi = \int_{-\infty}^\infty (a E \phi - E \phi') = \phi(0) \;\;\;\;\;\left(=\int \delta\,\phi\right)$$ for an arbitrary test function $\phi \in C^\infty_c(\mathbb R).$
To do this, we can break the integral up into two pieces and integrate by parts, remembering that $\phi,\phi'$ must vanish at infinity: on the domain $(-\infty,0)$ we get $$ \int_{-\infty}^0(ae^{-ax}\phi(x)-e^{-ax}\phi'(x))\,dx= \int_{-\infty}^0 -\frac d{dx}(e^{-ax} \phi(x))dx=-\phi(0),$$ while on $(0,\infty)$ we get $$2\int_0^\infty (ae^{-ax}\phi(x)-e^{-ax}\phi'(x))\,dx= 2\int_0^\infty -\frac d{dx}(e^{-ax}\phi(x))= 2\phi(0),$$ so adding these we find that $\int(E'+aE)\phi = \phi(0)$ as desired.
