Existence and uniqueness of an ODE solution across simple discontinuities

Question

I am studying differential equations using MIT's publicly available materials. One of the exercises runs as follows:

Let $I = (a,b)$ be an open interval containing $0$, and consider the ODE \begin{align} y^{(n)} + a_{n-1}y^{n-1} + \cdots + a_ny = f(t) \end{align} where has a finite number of simple discontinuities on $I$ and $|f(t)| \leq Ae^{Bt}$ for all $t \in I$ where it is defined for some constants $A,B \in \mathbb{R}^+$. Moreover, let $f$ be continuous at $0$. Establish the existence and uniqueness theorem for the IVP $y(0) = y_0, y'(0) = y_1, \ldots, y_{(n-1)}(0) = y_{n-1}$ in the class of so-called generalized solutions of the ODE; i.e. solutions $\phi$ where $\phi, \phi', \ldots, \phi^{(n-1)}$ are all continuous on $I$ and $\phi(t)$ solves the ODE whereever $f$ is continuous.

Here's what I have so far: We may assume without loss of generality that $f$ is discontinuous at only one point $c$, and that $c > 0$. Let $I_0 = (a,c)$, $I_1,(c, b)$. Since $Ae^{Bx}$ is Lipschitz on $(a,c)$, $f$ must be as well, and so there exists a unique solution $\phi$ to our IVP satisfying our continuity conditions on $I_0$. The difficulty, for me, consists in 'extending' $\phi$ to $I_1$; it seems to me that we don't have enough information about the behavior of $y$ on $I_1$ --say, initial conditions-- to know that $\phi$ is going to be continuous (and have the requisite continuous deriviatives) at $c$.

Any help would be appreciated.

Answer 1

Let $I = (a,b)$. We assume that $0 \in I$, and that $f$ is continuous at $0$. Let our simple discontinuities occur at $c_1, \ldots, c_{m-1}$, where $a =c_0 < c_1 < \cdots < c_{m-1} < c_m = b$. Choose $k$ such that $c_k > 0$ and there exists no $c_i$ such that $0 < c_i < c_k$, let $I_k = [0,c_k)$, and let $I_k^* = [0,c_k]$. Let $f_k$ be the restriction of $f$ to $I_k$, $f_k^*$ its extension to $I_k^*$ with $f_k^*(c_k)$ given by $\lim_{t \to c_k^-}f_k(t)$. Since $|f_k(t)| \leq Ae^{Bx}$, it is Lipschitz on $I_k$. Moreover, $f_k^*$ is continuous by hypothesis, so we can find a solution $\phi_k$ to the ODE (with $f_k^*$ standing in in for $f$) on $I_k^*$ satisfying our initial value and continuity conditions. Then for $0 \leq i \leq n-1$, the $\lim_{t \to c_k^-}\phi^{(i)}$ exists. If $k \neq m$, we can use these limits as values for a new initial value problem and repeat the process above, thus generating $\phi_{k+1}$ which once again satisfies our continuity conditions on $I_{k+1}$. Continue in this way until $k = m$. We now do the same thing, except taking limits from the right instead of the left, to generate $\phi_{k-1}, \ldots, \phi_1$. Define $\phi(t) = \phi_j(t)$ for $t \in I_j$. By construction, the $\phi_j$ are unique. Furthermore, the $\phi^{(i)}_{j}(c_j) = \phi^{(i)}_{j+1}(c_j)$ for all $i$ such that $0 \leq i \leq n-1$ and $j$ such that $1 \leq j \leq m-1$, so $\phi^{(i)}$ is continuous for $0 \leq i \leq n-1$. Thus $\phi$ exists, is unique, and satisfies the continuity conditions of the exercise's statement.