# Inhomogeneous wave equation by fourier in analysis

$$\nabla^2\psi_\omega+\frac{\omega^2}{c^2}\psi_\omega=-g\omega,\tag{14-16}$$ which is similar to Poisson's equation.

We may synthesize the solution of Eq. (14-16) by the superposition of unit point solutions corresponding to a source at the point $x_a'$ given by $g_\omega\left(x_a\right)=\delta\left(x_a-x_a'\right)$ where $\delta\left(x_a-x_a'\right)$ is the Dirac $\delta$-function. Each unit source potential satisfies the equation $$\nabla^2G\left(x_a,\,x_a'\right)+\frac{\omega^2}{c^2}G\left(x_a,\,x_a'\right)=-\delta\left(x_a-x_a'\right),\tag{14-17}$$

What are the unit point solutions? What is this whole paragraph meant to be talking about? Can somebody explain this to me? Thanks.

It is about Green's function. I am guessing originally you had the wave equation $$\nabla^2_x \, \psi(x,t) - \frac{1}{c^2}\,\frac{\partial^2}{\partial t^2} \, \psi(x,t) = - \, g(x,t)$$ where $\nabla_x^2 = \frac{\partial^2}{\partial (x^1)^2} + \frac{\partial^2}{\partial (x^2)^2} + ... + \frac{\partial^2}{\partial (x^n)^2}$ is the Laplacian with respect to the space variables $x = (x^1, x^2,... , x^n) \, \in \, \mathbb{R}^n$ and $g(x,t)$ is a known function in the right hand side of the equation, and then you have applied the Fourier transform or the Fourier series, depending on your initial/boundary value problem, with respect to the time variable $t$. For example something like $$\psi_{\omega}(x) = \int_{\mathbb{R}} \psi(x,t)\, e^{-\, i \, \omega \, t} \, dt \,\,\,\, \text{ and } \,\,\,\, g_{\omega}(x) = \int_{\mathbb{R}} g(x,t)\, e^{-\, i \, \omega \, t} \, dt$$ which means that $$-\omega^2 \, \psi_{\omega}(x) = \int_{\mathbb{R}} \frac{\partial^2 \psi}{\partial t^2}(x,t) \, e^{-\, i \, \omega \, t} \, dt$$ Thus, roughly speaking (I mean heuristically), when you apply the Fourier transform to both sides of the wave equation above, you obtain the resulting elliptic equation $$\nabla^2_x \, \psi_{\omega}(x) + \frac{\omega^2}{c^2} \, \psi_{\omega}(x) = - \, g_{\omega}(x)$$ for the unknown function $\psi_{\omega}(x)$, where both the unknown function $\psi_{\omega}(x)$ and the known one $g_{\omega}(x)$ depend on the frequency parameter $\omega$. The convolution identity $$g_{\omega}(x) = \int_{\mathbb{R}^n}\, \delta(x - y) \, g_{\omega}(y) \, dy$$ prompts us that if for any arbitrary $y = (y^1, y^2,... , y^n) \, \in \, \mathbb{R}^n$ (or whatever domain you are working in) we find the solution $G_{\omega}(x, y)$ to the equation (called Green's function) $$\nabla^2_x \, G_{\omega}(x, y) + \frac{\omega^2}{c^2} \, G_{\omega}(x, y) = -\, \delta(x - y)$$ (observe that the laplacian $\nabla^2_x$ is only with respect the variables $x$ and not $y$) then if we multiply both sides of the equation by $g_{\omega}(y)$ $$\nabla^2_x \, G_{\omega}(x, y)\, g_{\omega}(y) + \frac{\omega^2}{c^2} \, G_{\omega}(x, y) \, g_{\omega}(y) = -\, \delta(x - y) \, g_{\omega}(y)$$ and then integrate with respect to $y$ (and not $x$), we obtain $$\nabla^2_x \, \int_{\mathbb{R}^n}\, G_{\omega}(x, y)\, g_{\omega}(y)\, dy + \frac{\omega^2}{c^2} \, \int_{\mathbb{R}^n}\, G_{\omega}(x, y) \, g_{\omega}(y) \, dy = -\, \int_{\mathbb{R}^n}\, \delta(x - y) \, g_{\omega}(y) \, dy = - \, g_{\omega}(x)$$ which means that the function $$\psi_{\omega}(x) = \int_{\mathbb{R}^n}\, G_{\omega}(x, y)\, g_{\omega}(y)\, dy$$ is the solution we are looking for. Finally, the solution to the original wave equation can be obtained by the inverse Fourier transform
$$\psi(x, t) = \int_{\mathbb{R}} \int_{\mathbb{R}^n}\, G_{\omega}(x,y) \, g_{\omega}(y) \, e^{i\, \omega \, t} \, dy \, d\omega$$