Consider the following first-order ODE:
$$y'(x) = f'(x) y(x);$$
this has solution $y(x) = C e^{f(x)}$. Now, consider taking
$$f_\lambda(x) = \frac{\alpha}{2}\left(1+\tanh(\lambda x)\right);$$
in the limit $\lambda \to \infty$, $f_\lambda(x) \to \alpha\Theta(x)$ (where $\Theta(x)$ is the Heaviside step function), and the solution becomes $y(x) = C e^{\alpha \Theta(x)}$, which is discontinuous at $x = 0$. Presumably, there is a sense in which one can think of this as some kind of solution to the ODE with delta-function coefficient
$$y'(x) = \alpha \delta(x) y(x).$$
Now, the discontinuity in $y$ at $x=0$ is
$$\Delta y \equiv y(0^+) – y(0^-) = C(e^\alpha – 1).$$
My question is the following: is there a way to extract $\Delta y$ from the singular ODE directly without first finding a solution with smooth $f(x)$ and then taking an appropriate limit? For instance, in analogy with what one does with second-order ODEs with delta function coefficients, one could try to integrate it:
$$\Delta y = \lim_{\epsilon \to 0} \int^\epsilon_{-\epsilon} y' dx = \lim_{\epsilon \to 0} \int^\epsilon_{-\epsilon} \alpha \delta(x) y(x) dx = \alpha y(0),$$
but this has the problem that $y(0)$ is not defined due to the discontinuity in $y(x)$.
Best Answer
A friend of mine figured out a solution, at least to this toy problem. As commenters pointed out, the problem with naively integrating the ODE to get an expression for $\Delta y$ is that it is ill-defined to integrate a delta function against a discontinuous function. The solution is to first divide through by $y$ to get
$$\frac{y'(x)}{y(x)} = \alpha\delta(x).$$
This equation can now be integrated across the delta function to yield
$$\Delta \ln y \equiv \ln y(0^+) - \ln y(0^-) = \alpha,$$
thus
$$\frac{y(0^+)}{y(0^-)} = e^\alpha,$$
which indeed agrees with the $f(x) \to \alpha \Theta(x)$ limit of the solution for general $f(x)$.