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Solution of Wave Equation with Zero Initial Condition 📂Partial Differential Equations

Solution of Wave Equation with Zero Initial Condition

Tidy up

Let’s say that we have the following wave equation. where Δx\Delta_{\mathbf{x}} is Laplacian for the variable x\mathbf{x}.

∂t2p(x,t)=Δxp(x,t)on R×[0,∞)p(x,0)=f(x)on R∂tp(x,0)=0on R \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.

p(x,t)=1(2π)n∫Rnf^(ξ)cos⁡(t∣ξ∣)eix⋅ξdξ \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, f^\hat{f} is the Fourier transform of ff(…/1086). Now let’s consider the wave equation with the initial conditions as follows.

∂t2p(x,t)=Δxp(x,t)on R×[0,∞)p(x,0)=0on R∂tp(x,0)=g(x)on R \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(x,t)=1(2π)n∫Rng^(ξ)sin⁡(t∣ξ∣)∣ξ∣eix⋅ξdξ 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.

f^(ξ)=∫Rnf(x)eiξ⋅xdx,f(x)=1(2π)n∫Rnf(x)eix⋅ξdξ \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)(4) satisfies (1)(1), (2)(2), and (3)(3). Let’s first calculate the second derivative of time,

∂t2p(x,t)=−∣ξ∣21(2π)n∫Rnf^(ξ)cos⁡(t∣ξ∣)eix⋅ξdξ \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.

Δxp(x,t)=1(2π)n∫Rnf^(ξ)cos⁡(t∣ξ∣)(Δxeix⋅ξ)dξ=(−∣ξ∣2)1(2π)n∫Rnf^(ξ)cos⁡(t∣ξ∣)eix⋅ξdξ \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)(1) is established. When p(x,0)p(\mathbf{x}, 0) is calculated, (2)(2) is established because it is as follows.

p(x,0)=1(2π)n∫Rnf^(ξ)cos⁡(0∣ξ∣)eix⋅ξdξ=1(2π)n∫Rnf^(ξ)eix⋅ξdξ=f(x) \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)(3) is established.

∂tp(x,0)=−∣ξ∣∫Rnf^(ξ)sin⁡(0∣ξ∣)eix⋅ξdξ=0 \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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