Consider the non-homogeneous wave equation in three dimensions with homogeneous initial conditions: $$ \begin{align} & \square f(\underline{x}, t) = g(\underline{x},t), \hspace{3mm} \underline{x} \in \Bbb R^3, t>0\\ & f(\underline{x},0) = 0, \partial_t f(\underline{x},0), \hspace{3mm} \underline{x} \in \Bbb R^3\\ &||f||_{\infty}^{\Bbb R^3 \times [0,+\infty[} < +\infty \end{align} $$
where the D'Alembertian operator is defined with the $(-,+,+,+)$ convention:
$$ \square := \Delta_{\underline{x}} - \frac{1}{v^2}\partial_t^2 $$
By using the Fourier Transform we can find the associated Green Function explicitly and from its expression, solving the time integral, we find:
$$ f(\underline{x},t) = -\frac{1}{4\pi}\int_{\Bbb R^3}d\underline{x}' \left[\frac{g(\underline{x}', t-|\underline{x}-\underline{x}'|/v) + g(\underline{x}', t+|\underline{x}-\underline{x}'|/v)}{|\underline{x}-\underline{x}'|} \right] $$
In physics $f$ represents an electromagnetic potential and $g$ equals a source of the eletromagnetic field (with an irrelevant constant). In order to preserve causality I was told by my teacher that we need to disregard the term $g(\underline{x}', t+|\underline{x}-\underline{x}'|/v)$ becuase it corresponds to the contribution coming from a source in the future with respect to the instant in which we evaluate the potential. Now, my problem is what guarantees that doing so still yields a solution to the initial wave equation? Do I have to check it manually or does it follow from something in particular?
