I have to solve the first order differential equation:
$$ f'(x) - \text{i}\,k\,f(x)=\text{i}\,a\, f'(x)\,\delta(x)$$
where $k$ and $a$ are reals and "i" is the complex number: $\text{i}^2=-1$. $\delta(x)$ is the delta function (Dirac function).
We have the solution for $x\neq 0$:
$$f(x)=A\,e^{\text{i}kx}\quad\text{for}\, x>0$$
and
$$f(x)=B\,e^{\text{i}kx}\quad \text{for}\, x<0$$
The solution can be written as
$$f(x)=\text{e}^{\text{i}kx}\, \Big(A\,\Theta(x)+ B\,\Theta(-x)\Big)$$
I need to find a relation between A and B. We can integrate the differential equation around $x=0$ but $f(x)$ and $f'(x)$ are not necessary continious at $x=0$. I saw an average equation:
$$\int_{-\epsilon}^\epsilon f'(x)\,\delta(x)=\dfrac{f'(0^+)+f'(0^-)}{2}$$
Is it always true or we have to take some precautions?
Approximating $δ(x)$ by the rectangular function $\frac1ε \chi_{[s,s+ε]}(x)$ gives an ODE \begin{cases} \qquad f'(x)-ikf(x)=0&\text{ for }x<s\text{ or }x>s+ε\\ (1-ia/ε)f'(x)-ikf(x)=0&\text{ for }s\le x\le s+ε \end{cases} that has a continuous solution $$ f(x)=\begin{cases} c_0e^{ikx}&x<s\\ c_1e^{-kx\frac{ε}{a+iε}}& s\le x\le s+ε\\ c_2e^{ikx}&s+ε<x \end{cases} $$ if for continuity at $s$ and $s+ε$ the constants are connected as $$ c_0=c_1e^{-ks\frac{ia}{a+iε}} \text{ and }c_2=c_1e^{-k(s+ε)\frac{ia}{a+iε}} $$ which in the limit $(s,ε)\to(0,0)$ leads to the function $f(x)=c_0e^{ikx}$ which obviously is not a solution at $x=0$. Thus the equation is not solvable.