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In studying a reflection-transmission problem involving exotic materials, I have come across the following linear first-order differential equation: \begin{equation}\tag{1}\label{eq:1} A\frac{\partial}{\partial t}g(t) + Bg(t) = f(t), \end{equation} where $A$ and $B$ are constants, $g(t)$ is associated with the reflected wave, and $f(t)$ is a (finite) driving function associated with the incident wave. Both $A$ and $B$ may be positive or negative. I am interested in the behavior of the solution in the limit that $A\rightarrow0$.

I know there is an exact solution to Eq. \eqref{eq:1}, which is \begin{equation} g(t) = C e^{-Bt/A} + \frac{1}{A}\int_{-\infty}^t e^{-B(t-t')/A}f(t')dt', \end{equation} where $C=0$ because $g(t)=0$ if $f(t)=0$. However, I do not understand how this exact solution reduces to the case where $A=0$, which is $g(t)=B^{-1}f(t)$. Any insight would be greatly appreciated.

I've seen a lot of documents discussing asymptotic analyses of linear differential equations (for example, see these lecture notes), but they all start with second-order equations. Is this because there is inherently problematic with first-order?

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I presume $A > 0$ and $B > 0$. A change of variables $s = (t-t')/A$ in the integral gives you

$$ g(t) = C e^{-Bt/A} + \int_0^\infty e^{-Bs} f(t - A s) \; ds $$

Now as $A \to 0+$, $f(t-As) \to f(t)$ if $f$ is continuous. Assuming $f$ is bounded, we can use the Dominated Convergence Theorem and this integral goes to

$$f(t) \int_0^\infty e^{-Bs}\; ds = B^{-1} f(t)$$

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    Thanks! This helps a lot. What if A<0, or B<0? One of my concerns is that physically there is nothing dictating the signs of these quantities. – 2017-01-13
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    A change of variables $t \to -t$ in the differential equation changes $A$ to $-A$. Or change $f$ to $-f$ and you switch the signs of both $A$ and $B$. – 2017-01-13
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    Wouldn't a change of variables $t\rightarrow-t$ also change the limits of integration from $(\infty,t)$ to $(-\infty,-t)$, which would make the integral diverge in the limit $A\rightarrow0$? – 2017-01-13
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    It changes the limits of integration from $(-\infty, t)$ to $(-t, +\infty)$, which makes the integral converge when $B/A < 0$. – 2017-01-13
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    Maybe I am missing something. When I assume $A<0$ and $B>0$, and make the substitution $\tau=-t$ in the differential equation, I get the integral $(-A)^{-1}\int_\tau^\infty e^{-B(\tau-\tau')/(-A)}f(-\tau')d\tau'$, which I don't think converges. Making the substitution $s=(\tau-\tau')/(-A)$ leads to the integral $\int_{-\infty}^0 e^{-Bs}f(t + (-A)s)ds$, which doesn't converge. What am I missing? – 2017-01-17
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    You're integrating over the wrong half-line. If $A < 0$, let $\tilde{A}= -A$, $\tau = -t$, $g(t) = \tilde{g}(\tau)$, $f(t) = \tilde{f}(\tau)$, and the differential equation becomes $\tilde{A} \tilde{g}'(\tau) + B \tilde{g}(\tau) = \tilde{f}(\tau)$, which has the same form as your original, and is solved by $$ \tilde{g}(\tau) = C e^{-B \tau/\tilde{A}} + \int_0^\infty e^{-Bs} \tilde{f}(\tau - \tilde{A}s)\; ds$$ i.e. $$ g(t) = C e^{- B t/A} + \int_0^\infty e^{-Bs} f(t-As)\; ds$$ – 2017-01-18