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Let me start with an example. If we base our calculations on the Newton's second law without any further mathematical treatment, then our equation describes equilibrium of forces, i.e. it is of the kind "force = force (e.g. as the product of mass and acceleration)". Similarly it could be momenta, angular momenta, electric charges, voltages, various quantities like "speed change per kilogram and tesla" etc. etc. etc...

How is that with the wave equation? It's crucial for interpretations of source terms. So, could you describe like that inhomogenous string equation:

$$ \frac{\partial^2y}{\partial x^2}-\frac{1}{c_0^2}\frac{\partial^2y}{\partial t^2}=q $$

Wave equation for the acoustic pressure:

$$ \nabla^2 p-\frac{1}{c_0^2}\frac{\partial^2p}{\partial t^2}=q $$

and for the acoustic velocity:

$$ \nabla^2 \vec{v}-\frac{1}{c_0^2}\frac{\partial^2 \vec{v}}{\partial t^2}=q $$

where $q$ are source terms with required dimension and units. What is the physical meaning of those terms?

Note: If that would be considered helpful for the site, let's broaden the list of wave equations.

Edit to clarify: I am not looking for the steps in mathematical solution or the general principle of pressence a pulsating body in a fluid (in the case of acoustics). Question is how exactly represent $q$ that pulsation? What is the intuition and meaning of the term with dimension of $time^{-1} \times distance^{-1}$ (that's for acoustical velocity)? Etc.

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    $\begingroup$ The source term means something external to the wave-supporting medium is pushing on that medium. For example, if a speaker vibrates and pushes on the air, it is represented as a source term in the wave equation. Is that all you're asking? $\endgroup$
    – DanielSank
    Commented Oct 16, 2015 at 15:30
  • $\begingroup$ No, that's "too shallow". I'll make an edit cause it is obviously not clear (and fault can only be mine). $\endgroup$ Commented Oct 16, 2015 at 15:38
  • $\begingroup$ Ok. Are you trying to get an example of how a physical source is represented in the wave equation? $\endgroup$
    – DanielSank
    Commented Oct 16, 2015 at 15:40
  • $\begingroup$ See the edit, please. $\endgroup$ Commented Oct 16, 2015 at 15:42

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To solve a non homogeneous wave equation you need to use Green's function. The Green function of the inhomogeneous wave equation is defined as $$ (\nabla^2-\frac{1}{c^2}\partial_t^2)G(x,x';t,t') =\delta(x-x')\delta(t-t') $$ It means that $G$ is the solution of the wave equation for a Dirac source localized at the position $r'$ and at a time $t'$. Why is this Green function important ? Because once you have it you can have any solution $\Phi(x,t)$ to the equation $$ (\nabla^2-\frac{1}{c^2}\partial_t^2)\Phi(x,t)=\rho(x,t) $$ by computing $$ \Phi=\int G(x,x';t,t')\rho(x',t')dx'dt' \space \space\space\space\space(1) $$

because

$$ (\nabla^2-\frac{1}{c^2}\partial_t^2)\Phi=\int (\nabla^2-\frac{1}{c^2}\partial_t^2)G(x,x';t,t')\rho(x',t')dx'dt' $$ using the definition of the Green function: $$ =\int \delta(x-x')\delta(t-t')\rho(x',t')dx'dt'= \rho(x,t) $$ In equation (1) lies the answer to your question: To obtain your solution you have to sum up the influence of every infinitesimal part of your source term $\rho$. This influence is given by the Green function, which is exactly the solution of the same problem for an infinitesimal pointlike source.
Its a little bit simpler in frequency domain, or even in static (where you have Poisson or Laplace equation instead of the wave equation) but the idea is always the same: obtain a Green function for a point-like (Dirac source) and then obtain the solution by integration.

A good example to physically illustrate this idea is the Huygens principle: every part of the phase front of a wave is a little source of a spherical wave. If you want the solution of a problem where your source ($\rho$) is a circular hole cut in an opaque screen you'll have to sum up little spherical wave (Green function) over all the hole surface. That's how diffraction problem are solved.

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    $\begingroup$ Thanks for your time and effort. This solution is well-known, mathematically very elegant with nice physical meaning. BUT. What are these Dirac sources? Sources of what exactly, if you know what I mean? I've tried to explain that in my own answer. If you can correct me or elaborate your own for that, I would be grateful. $\endgroup$ Commented Oct 16, 2015 at 15:26
  • $\begingroup$ Did you understand the example with the Huygens principle ? They are the sources of a wave, if it's the wave equation, or the sources of a static electric field if it's Poisson's equation. Take a look at the form of the Green's function, amybe it will help : phy.duke.edu/~rgb/Class/phy319/phy319/node75.html $\endgroup$
    – EigenDavid
    Commented Oct 16, 2015 at 15:29
  • $\begingroup$ Yes, I did. The question is the interpretation or the physical intuition on maths of the source itself. What should represent the $q$ (or $\rho$) with the dimension of length times time (example for acoustic velocity source term)? $\endgroup$ Commented Oct 16, 2015 at 15:36
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The source term $q$ just adds to the canonical solutions $A \cos(k(x-ct)+\phi)$ or $A \exp(\pm k(x-ct))$ a polynomial term of degree 2 in x and t (more precisely, $a x^2 +b t^2 +c x t +d x +e t +f$, with $2(a+b)=q$).

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  • $\begingroup$ Thanks, but it's not what I was asking about. Yes, of course, that's mathematically correct. But what is the physical interpretation? $\endgroup$ Commented Oct 15, 2015 at 22:24
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    $\begingroup$ The differential equation =0 means an average balance, with kinetics energy transforming into potential energy and back (especially for the oscillating solution). With =q, you force to add a global inbalance corresponding to a constant acceleration in time or the equivalent in space, i.e., you force additionnal kinetic or potential energy (nb: 'additional' can be positive or negative). $\endgroup$ Commented Oct 16, 2015 at 7:59
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Equation of the string: Let's multiply the equation by $c_0^2$ and rearrange:

$$ \frac{\partial^2 y}{\partial t^2}=c_0^2\frac{\partial^2 y}{\partial x^2} + c_0^2q_y $$

i.e. acceleration of the point $y$ located at $x$ in time $t$ is the acceleration caused by string tension plus the "source acceleration" given by external forces: $q^* \equiv c_0^2q_y$.

Acoustic pressure: Usually the $q$ here (let's denote it $q_p$) could be written:

$$ q_p=\nabla\cdot \vec{F}-\rho_0\frac{\partial Q}{\partial t} $$

where $Q$ is volume source term (from inhomogenous continuity equation) and $\vec{F}$ is the sum of external volume forces (e.g. gravity - from momentum equation). With usual neglecting of the $\vec{F}$, the source term $q_p$ is the "speed of the rate of increase of fluid volume per unit volume of the fluid times the density" and that's something like "local density acceleration" ($kg \cdot m^{-3} \cdot s^{-2}$). The typical example is a small pulsating body in a fluid.

Acoustic velocity: the dimension of $q_v$ is $length \times time$ which corresponds to the order of magnitude of the source pulsation in the fluid: $kg \cdot m^{-2} \cdot m \cdot s^{-1}$ - the last two terms are speed of pulsation and the first two is surface density of the fluid on the control surface surrounding the pulsating body.

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Perhaps clarity is enhanced by viewing this source term as an excitation of a wave. This term represents the process of imparting energy to a medium or system in a manner that generates a disturbance propagating through the medium. Mathematically, it is evident as the quantity added to the homogeneous solution, acting as an amplifier

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