Solution of Wave Equation with Zero Initial Condition
Tidy up
Let’s say that we have the following wave equation. where $\Delta_{\mathbf{x}}$ is Laplacian for the variable $\mathbf{x}$.
$$ \begin{align} \partial_{t}^{2} p(\mathbf{x}, t) &= \Delta_{\mathbf{x}} p(\mathbf{x}, t) &\text{on } \mathbb{R} \times [0, \infty) \\ p(\mathbf{x}, 0) &= f(\mathbf{x}) &\text{on } \mathbb{R} \\ \partial_{t} p(\mathbf{x}, 0) &= 0 &\text{on } \mathbb{R} \end{align} $$
The solution of the above partial differential equation is as follows.
$$ \begin{equation} p(\mathbf{x}, t) = \dfrac{1}{(2\pi)^{n}} \int\limits_{\mathbb{R}^{n}} \hat{f} (\boldsymbol{\xi}) \cos (t \left| \boldsymbol{\xi} \right|) e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} \end{equation} $$
Here, $\hat{f}$ is the Fourier transform of $f$(…/1086). Now let’s consider the wave equation with the initial conditions as follows.
$$ \begin{align*} \partial_{t}^{2} p(\mathbf{x}, t) &= \Delta_{\mathbf{x}} p(\mathbf{x}, t) &\text{on } \mathbb{R} \times [0, \infty) \\ p(\mathbf{x}, 0) &= 0 &\text{on } \mathbb{R} \\ \partial_{t} p(\mathbf{x}, 0) &= g(\mathbf{x}) &\text{on } \mathbb{R} \end{align*} $$
The solution of the above partial differential equation is as follows.
$$ p(\mathbf{x}, t) = \dfrac{1}{(2\pi)^{n}} \int\limits_{\mathbb{R}^{n}} \hat{g} (\boldsymbol{\xi}) \dfrac{\sin (t \left| \boldsymbol{\xi} \right|)}{\left| \boldsymbol{\xi} \right|} e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} $$
Description
Let’s put the definitions of Fourier transform and it’s Inverse transform as below.
$$ \hat{f}(\boldsymbol{\xi}) = \int\limits_{\mathbb{R}^{n}} f(\mathbf{x}) e^{\mathrm{i} \boldsymbol{\xi} \cdot \mathbf{x}} \mathrm{d} \mathbf{x}, \qquad f(\mathbf{x}) = \dfrac{1}{(2\pi)^{n}}\int\limits_{\mathbb{R}^{n}} f(\mathbf{x}) e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} $$
The latter method of proof is the same as the former, so it is omitted.
Proof of proof
Just check that $(4)$ satisfies $(1)$, $(2)$, and $(3)$. Let’s first calculate the second derivative of time,
$$ \partial_{t}^{2} p(\mathbf{x}, t) = -\left| \boldsymbol{\xi} \right|^{2} \dfrac{1}{(2\pi)^{n}} \int\limits_{\mathbb{R}^{n}} \hat{f} (\boldsymbol{\xi}) \cos (t \left| \boldsymbol{\xi} \right|) e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} $$
The Laplacian is calculated as follows.
$$ \begin{align*} \Delta_{\mathbf{x}} p(\mathbf{x}, t) &= \dfrac{1}{(2\pi)^{n}} \int\limits_{\mathbb{R}^{n}} \hat{f} (\boldsymbol{\xi}) \cos (t \left| \boldsymbol{\xi} \right|) (\Delta_{\mathbf{x}} e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}}) \mathrm{d} \boldsymbol{\xi} \\ &= (- \left| \boldsymbol{\xi} \right|^{2}) \dfrac{1}{(2\pi)^{n}} \int\limits_{\mathbb{R}^{n}} \hat{f} (\boldsymbol{\xi}) \cos (t \left| \boldsymbol{\xi} \right|) e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} \\ \end{align*} $$
Thus, $(1)$ is established. When $p(\mathbf{x}, 0)$ is calculated, $(2)$ is established because it is as follows.
$$ \begin{align*} p(\mathbf{x}, 0) &= \dfrac{1}{(2\pi)^{n}} \int\limits_{\mathbb{R}^{n}} \hat{f} (\boldsymbol{\xi}) \cos ( 0 \left| \boldsymbol{\xi} \right|) e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} \\ &= \dfrac{1}{(2\pi)^{n}} \int\limits_{\mathbb{R}^{n}} \hat{f} (\boldsymbol{\xi}) e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} \\ &= f(\mathbf{x}) \end{align*} $$
It is also easy to see that $(3)$ is established.
$$ \begin{align*} \partial_{t}p(\mathbf{x}, 0) &= - \left| \boldsymbol{\xi} \right| \int\limits_{\mathbb{R}^{n}} \hat{f} (\boldsymbol{\xi}) \sin ( 0 \left| \boldsymbol{\xi} \right|) e^{\mathrm{i} \mathbf{x} \cdot \boldsymbol{\xi}} \mathrm{d} \boldsymbol{\xi} \\ &= 0 \end{align*} $$
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