I am interested in solving the following form of ODE, of arbitrary order $n$ (Equation 1):
$$a_n \frac{d^n y}{dx^n} + a_{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \cdots + a_1 \frac{dy}{dx} + a_0 y = \delta(x)$$
where $\delta(x)$ is the Dirac delta. The $x$ domain extends either finitely or infinitely either side of $x=0$.
What seems to be a fairly common approach is to split the domain about $x=0$ into two subdomains, neither of which no longer contain the Dirac delta term anymore. No problem. However, now that you have effectively two order-$n$ ODEs to solve, an extra $n$ boundary conditions are required, and these BCs come from relating $y$ and its derivatives just left of $x=0$ to those just right of $x=0$. So, to get those BCs, you seem to need to take an integral over an infinitesimal length over the domain that contains $x=0$, yielding the following Equation 2:
$$a_n \left(\frac{d^{n-1}y}{dx^{n-1}}(0^+) - \frac{d^{n-1}y}{dx^{n-1}}(0^-)\right)+\cdots+a_1 \left(y(0^+)-y(0^-)\right) + a_0 \int^{0^+}_{0^-} y \, dx = 1$$
This is where things become less clear for me. It seems as though to obtain the “subdomain-connection” BCs, I need to assume that $\frac{d^{n-1} y}{dx^{n-1}}$ is discontinuous about $x=0$, and that all lower order derivatives are continuous. This then yields the right number of boundary conditions:
$$\frac{d^{n-1} y}{dx^{n-1}}(0^+) = \frac{d^{n-1} y}{dx^{n-1}}(0^-) + \frac{1}{a_n}$$
$$\frac{d^{i} y}{dx^{i}}(0^+) = \frac{d^{i} y}{dx^{i}}(0^-)\quad \text{for } i=0,1\dots (n-2)$$
Now, my question comprises of two interrelated parts:
- Is this assumption about the discontinuity of $\frac{d^{n-1} y}{dx^{n-1}}$ and continuity of lower derivatives correct (in general)?
- If so, why? Is there a rigorous explanation to this approach? (For example, why not assume $\frac{d^{n-2} y}{dx^{n-2}}$ to be discontinuous also, and having an equation that relates the (n-1) and (n-2) derivatives to each other?)
Many thanks in advance!