Wave equation for sound waves and moving source Is there a way to take a moving source already into account when one derives the wave equation for sound waves and derive from that using only math the Doppler effect for moving sources?
 A: If I understand you correctly, you want to desribe sound waves from the viewpoint of a moving observer. To do this, you just have to take the standard wave equation and perform a variable substitution $\mathbf{x}\mapsto \mathbf{y}:=\mathbf{x}-\mathbf{v}\cdot t$.
You start with
$$
  \partial_t^2 p(\mathbf{x},t) = c^2\nabla^2 p(\mathbf{x},t)
  = c^2\sum_i\partial_{\!x_i}^2 p(x_1,\ldots,t)
$$
The right side transforms as
$$
  c^2\sum_i\partial_{\!x_i}^2 p(x_1,\ldots,t)
  = c^2\sum_i\partial_{\!x_i}^2 p(y_1+v_1t,\ldots,t)
$$
where we yet have to replace the derivative WRT $x_i$ against the derivative WRT $y_i$, but they are the same:
$$
  \frac{\partial}{\partial x_i} = \frac{\partial y_i}{\partial x_i}\frac{\partial}{\partial y_i} = (1+0)\cdot\frac{\partial}{\partial y_i};
$$
so
$$
  c^2\nabla_{\!\!\mathbf{x}}^2p(\mathbf{x},t) = c^2\nabla_{\!\!\mathbf{y}}^2p(\mathbf{y}+\mathbf{v}t,\ t).
$$
The left side transforms as
$$
  \partial_t^2 p(\mathbf{x},t) = \partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot t,\ t).
$$
Here, we would like to drop the $+\mathbf{v}t$, but before we can do this we need to take into account the derivative of this term:
$$
  \partial_t p(\mathbf{y}+\mathbf{v}\cdot t,\ t)
  = \left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
              + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t)
$$
now the second derivative:
$$
  \partial_t\left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
  = \left.\bigl(\partial_t\partial_\tau p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
  + \left.\bigl(\partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
$$
where
$$
  \left.\bigl(\partial_t\partial_\tau p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
   = \left.\bigl(\partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
$$
and
$$
  \partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t)
  = \mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}
    \left(\left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
              + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t)\right)
$$
$$
  = \left.\bigl(\partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}
  + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}^2\,\mathbf{v}\,p(\mathbf{y}+\mathbf{v}\cdot t,\ t)
$$
All together, we get
$$
  \partial_t^2p(\mathbf{x},t)
  = \left.\left(
    \partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)
   + 2 \partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)
   + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}^2\,\mathbf{v}\,p(\mathbf{y}+\mathbf{v}\cdot t,\ t)
  \right)\right|_{\tau=t}.
$$
Now we may drop the $+\mathbf{v}t$ on both sides of the equation, and get
$$
  \partial_t^2 p(\mathbf{y},t)
   + 2 \partial_t\,\mathbf{v}\!\cdot\!\nabla p(\mathbf{y},t)
   + \mathbf{v}\cdot\nabla^2\,\mathbf{v}\,p(\mathbf{y},t)
  = c^2\nabla^2p(\mathbf{y},t)
$$
