This immediately leads to the statement
@f[
P(t,\mathbf r) - \frac{1}{c_0^2} \frac{\partial p}{\partial t}
-\; dt
=
\lambda a(\mathbf r) \delta(t),
@f]
\int_{-\epsilon}^{\epsilon} \lambda a(\mathbf r) \delta(t) \; dt.
@f]
If we use the property of the delta function that $\int_{-\epsilon}^{\epsilon}
-\delta(t)\; dt = 1$, and assume that $P$ is a smooth function in time, we find
+\delta(t)\; dt = 1$, and assume that $P$ is a continuous function in time, we find
as we let $\epsilon$ go to zero that
@f[
-- \frac{1}{c_0^2} \left[ p(\epsilon,\mathbf r) - p(-\epsilon,\mathbf r) \right]
+- \lim_{\epsilon\rightarrow 0}\frac{1}{c_0^2} \left[ p(\epsilon,\mathbf r) - p(-\epsilon,\mathbf r) \right]
=
\lambda a(\mathbf r).
@f]
0.
@f]
-Now, let $\epsilon\rightarrow 0$. Assuming that $P$ is a smooth function in
+Now, let $\epsilon\rightarrow 0$. Assuming that $P$ is a continuous function in
time, we see that
@f[
P(\epsilon)-P(-\epsilon) \rightarrow 0,