As I am trying to gain some fluency with Green's functions, and there use in non-relativistic PDE's, I thought it'd be a good exercise to work out some expressions -- or "Magic Rules" -- for the Green's function of the wave equation:
\begin{eqnarray}
\left(\frac{\partial^2}{\partial t^2}-c_s^2 \frac{\partial^2}{\partial x_i \partial x_i}\right)\mathbf{\psi}(\vec{x},t)&=& 0
\end{eqnarray}
Let's assume that $c_s$ is our "speed of sound" in the medium, and that $\psi$ some scalar function of space and time. If we assume that $\psi \to 0$ as we go to $\vert x \vert \to \infty$, or, the boundaries of our space of interest, we are well equipped to perform a fourier transform on this PDE:
\begin{eqnarray}
\left(\frac{\partial^2}{\partial t^2}+c_s^2 k^2\right)\tilde{\psi}(\vec{k},t)&=& 0
\end{eqnarray}
We can now laplace transform in the time variable, swapping $t$ for $\omega$, and write
\begin{eqnarray}
\left(\omega^2+c_s^2 k^2\right)\tilde{\psi}(\vec{k},\omega)&=& \omega \tilde{\psi}(\vec{k},0)+ \tilde{\psi}(\vec{k},0)^\prime
\end{eqnarray}
We can see now that our Differential equation has been rendered algebraic in this "doubly" transformed space. We invert the PDE and write
\begin{eqnarray}
\tilde{\psi}(\vec{k},\omega)&=& \frac{ \omega \tilde{\psi_0}(k) + \tilde{\psi_0^{\prime}}(k)}{\left(\omega^2+c_s^2 k^2\right)}
\end{eqnarray}
We can immediately see, once we invert the laplace transform, that our initial conditions are built into the homogeneous solution:
\begin{eqnarray}
\tilde{\psi}(\vec{k},t)&=& \oint_{\gamma-i \infty}^{\gamma+i \infty}\frac{ \omega \tilde{\psi_0}(k)e^{\omega t}}{\left(\omega^2+c_s^2 k^2\right)}\frac{d\omega}{2\pi i}+\oint_{\gamma-i \infty}^{\gamma+i \infty}\frac{ \tilde{\psi_0^{\prime}}(k)e^{\omega t}}{\left(\omega^2+c_s^2 k^2\right)}\frac{d\omega}{2\pi i}
\end{eqnarray}
Both contour integrals have identical poles: $\omega = \pm i c_s \vert \vec{k} \vert $, and so we use the residue theorem to write down the inverted PDE in k-space:
\begin{eqnarray}
\tilde{\psi}(\vec{k},t) &=& \tilde{\psi_0}(k) \cos \left(c_s k t \right)+ \frac{\tilde{\psi_0^{\prime}}(k)\sin \left(c_s k t \right)}{c_s k}
\end{eqnarray}
Which is incredibly simple but difficult to transform back into x-space. First of all, notice that we are multiplying our initial conditions -- or spectra -- with two separate kernels, sine-over-k and cosine. In the x-space, this means we will be "convolving" our initial conditions with these two separate kernels -- or the x-space inverse fourier transforms -- in order to create our homogeneous solution.
Furthermore, remember that our time-transform variable $\omega$ had poles that were only dependent upon the modulus of $\vec{k}$, meaning that all of the k's you see above should be replaced by $abs(\vec{k})$. These kernels exhibit isotropy, which makes sense since the wave equation is a spatially isotropic and homogenous differential equation to begin with -- the laplacian operator has no preference in direction or orientation.
But, such radial functions of $k$ will prove difficult to inverse fourier transform in higher and higher dimensions of $x$.
If we start out with 1-D, we know that the cosine will turn into an evenly spaced pair of dirac delta distributions, and the sin over k term will turn into a top hat function. And so we are left with
\begin{eqnarray}
\psi(x,t)^{1D} &=& \frac{\psi_0(x-c_s t)+\psi_0(x+c_s t)}{2}+\frac{\Pi(c_s t)}{c_s}\ast \psi^{\prime}_0(x) \\
\psi(x,t)^{1D} &=& \frac{\psi_0(x-c_s t)+\psi_0(x+c_s t)}{2}+\int_{x-c_s t}^{x+c_s t} \frac{\psi^{\prime}_0(x^\prime)}{2 c_s} dx^\prime
\end{eqnarray}
This is a wave front uniformly expand with speed $c_s$, independent of whether the problem is sit up with Neumman or Dirichlet boundary conditions. And, it seems I have stumbled upon -- by the laplace transform -- d'Alembert's formula for the Cauchy problem.
----------------------------------------------------------------------
Now if we move into 2-D, we find that the we need to regularize the inverse fourier transforms, which I will leave for a later post...
\begin{eqnarray}
\left(\frac{\partial^2}{\partial t^2}-c_s^2 \frac{\partial^2}{\partial x_i \partial x_i}\right)\mathbf{\psi}(\vec{x},t)&=& 0
\end{eqnarray}
Let's assume that $c_s$ is our "speed of sound" in the medium, and that $\psi$ some scalar function of space and time. If we assume that $\psi \to 0$ as we go to $\vert x \vert \to \infty$, or, the boundaries of our space of interest, we are well equipped to perform a fourier transform on this PDE:
\begin{eqnarray}
\left(\frac{\partial^2}{\partial t^2}+c_s^2 k^2\right)\tilde{\psi}(\vec{k},t)&=& 0
\end{eqnarray}
We can now laplace transform in the time variable, swapping $t$ for $\omega$, and write
\begin{eqnarray}
\left(\omega^2+c_s^2 k^2\right)\tilde{\psi}(\vec{k},\omega)&=& \omega \tilde{\psi}(\vec{k},0)+ \tilde{\psi}(\vec{k},0)^\prime
\end{eqnarray}
We can see now that our Differential equation has been rendered algebraic in this "doubly" transformed space. We invert the PDE and write
\begin{eqnarray}
\tilde{\psi}(\vec{k},\omega)&=& \frac{ \omega \tilde{\psi_0}(k) + \tilde{\psi_0^{\prime}}(k)}{\left(\omega^2+c_s^2 k^2\right)}
\end{eqnarray}
We can immediately see, once we invert the laplace transform, that our initial conditions are built into the homogeneous solution:
\begin{eqnarray}
\tilde{\psi}(\vec{k},t)&=& \oint_{\gamma-i \infty}^{\gamma+i \infty}\frac{ \omega \tilde{\psi_0}(k)e^{\omega t}}{\left(\omega^2+c_s^2 k^2\right)}\frac{d\omega}{2\pi i}+\oint_{\gamma-i \infty}^{\gamma+i \infty}\frac{ \tilde{\psi_0^{\prime}}(k)e^{\omega t}}{\left(\omega^2+c_s^2 k^2\right)}\frac{d\omega}{2\pi i}
\end{eqnarray}
Both contour integrals have identical poles: $\omega = \pm i c_s \vert \vec{k} \vert $, and so we use the residue theorem to write down the inverted PDE in k-space:
\begin{eqnarray}
\tilde{\psi}(\vec{k},t) &=& \tilde{\psi_0}(k) \cos \left(c_s k t \right)+ \frac{\tilde{\psi_0^{\prime}}(k)\sin \left(c_s k t \right)}{c_s k}
\end{eqnarray}
Which is incredibly simple but difficult to transform back into x-space. First of all, notice that we are multiplying our initial conditions -- or spectra -- with two separate kernels, sine-over-k and cosine. In the x-space, this means we will be "convolving" our initial conditions with these two separate kernels -- or the x-space inverse fourier transforms -- in order to create our homogeneous solution.
Furthermore, remember that our time-transform variable $\omega$ had poles that were only dependent upon the modulus of $\vec{k}$, meaning that all of the k's you see above should be replaced by $abs(\vec{k})$. These kernels exhibit isotropy, which makes sense since the wave equation is a spatially isotropic and homogenous differential equation to begin with -- the laplacian operator has no preference in direction or orientation.
But, such radial functions of $k$ will prove difficult to inverse fourier transform in higher and higher dimensions of $x$.
If we start out with 1-D, we know that the cosine will turn into an evenly spaced pair of dirac delta distributions, and the sin over k term will turn into a top hat function. And so we are left with
\begin{eqnarray}
\psi(x,t)^{1D} &=& \frac{\psi_0(x-c_s t)+\psi_0(x+c_s t)}{2}+\frac{\Pi(c_s t)}{c_s}\ast \psi^{\prime}_0(x) \\
\psi(x,t)^{1D} &=& \frac{\psi_0(x-c_s t)+\psi_0(x+c_s t)}{2}+\int_{x-c_s t}^{x+c_s t} \frac{\psi^{\prime}_0(x^\prime)}{2 c_s} dx^\prime
\end{eqnarray}
This is a wave front uniformly expand with speed $c_s$, independent of whether the problem is sit up with Neumman or Dirichlet boundary conditions. And, it seems I have stumbled upon -- by the laplace transform -- d'Alembert's formula for the Cauchy problem.
----------------------------------------------------------------------
Now if we move into 2-D, we find that the we need to regularize the inverse fourier transforms, which I will leave for a later post...
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