Ngô Quốc Anh

February 11, 2009

Solving initial value problem for wave equation via Fourier transform

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:43

In this topic, we will show you how can we use Fourier transform to solve initial value problem for wave equation in \mathbb R. Following is the problem

\displaystyle \left\{ \begin{gathered} {u_{tt}} = {u_{xx}}, \qquad x \in \mathbb{R}, \hfill \\ u\left( {x,0} \right) = \varphi \left( x \right), \hfill \\ {u_t}\left( {x,0} \right) = \psi \left( x \right). \hfill \\ \end{gathered} \right.

From the equation by taking Fourier transform to the both sides, we obtain

\displaystyle \widehat{{u_{tt}}\left( {\eta ,t} \right)} - {\left( {i\eta } \right)^2}\widehat{u\left( {\eta ,t} \right)} = 0

This is an ODE, the solution is given by

\displaystyle \widehat{u\left( {\eta ,t} \right)} = A\sin \left( {\eta t} \right) + B\cos \left( {\eta t} \right)

From the initial date we get

\widehat{u\left( {\eta ,0} \right)} = \widehat{\varphi \left( \eta \right)}

which implies that

\widehat{\varphi \left( \eta \right)} = B.

From the initial date we see that

\widehat{{u_t}\left( {\eta ,0} \right)} = \widehat{\psi \left( \eta \right)} = A\eta.

Thus, we obtain

\displaystyle \widehat{u\left( {\eta ,t} \right)} = \frac{{\widehat{\psi \left( \eta \right)}}} {\eta}\sin \left( {\eta t} \right) + \widehat{\varphi \left( \eta \right)}\cos \left( {\eta t} \right).

Note that

\displaystyle \cos \left( {\eta t} \right) = \frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}, \quad \frac{{\sin \left( {\eta t} \right)}} {\eta } = \frac{{{e^{i\eta t}} - {e^{ - i\eta t}}}} {{2i\eta }} = \frac{1} {2}\int\limits_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta }.

Moreover,

\displaystyle \widehat{\delta ( {x - \alpha t} )} = \int\limits_{ - \infty }^\infty {{e^{ - i\eta x}}\delta ( {x - \alpha t} )dx} = {e^{ - i\alpha \eta t}}\int\limits_{ - \infty }^\infty {{e^{ - i\eta ( {x - \alpha t} )}}\delta ( {x - \alpha t} )dx} = {e^{ - i\alpha \eta t}}.

Then

\displaystyle \widehat{u\left( {\eta ,t} \right)} = \left( {\frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}} \right)\widehat{\varphi \left( \eta \right)} + \left( {\frac{1} {2}\int\limits_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta } } \right)\widehat{\psi \left( \eta \right)}.

Since

\displaystyle \left( {\frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}} \right)\widehat{\varphi \left( \eta \right)} = \frac{{\widehat{\delta \left( {x + t} \right)*\varphi \left( x \right)} + \widehat{\delta \left( {x - t} \right)*\varphi \left( x \right)}}} {2}

and

\displaystyle \left( {\frac{1} {2}\int_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta } } \right)\widehat{\psi \left( \eta \right)} = \frac{1} {2}\int\limits_{ - t}^t {\widehat{\delta \left( {x + \theta } \right)*\psi \left( x \right)}d\theta }

then

\displaystyle \widehat{u\left( {\eta ,t} \right)} = \frac{{\widehat{\delta \left( {x + t} \right)*\varphi \left( x \right)} + \widehat{\delta \left( {x - t} \right)*\varphi \left( x \right)}}} {2} + \frac{1} {2}\int\limits_{ - t}^t {\widehat{\delta \left( {x + \theta } \right)*\psi \left( x \right)}d\theta }.

Thus,

\displaystyle u\left( {x,t} \right) = \frac{{\delta \left( {x + t} \right)*\varphi \left( x \right) + \delta \left( {x - t} \right)*\varphi \left( x \right)}} {2} + \frac{1} {2}\int\limits_{ - t}^t {\delta \left( {x + \theta } \right)*\psi \left( x \right)d\theta }

or equivalently,

\displaystyle u\left( {x,t} \right) = \frac{{\varphi \left( {x + t} \right) + \varphi \left( {x - t} \right)}} {2} + \frac{1} {2}\int\limits_{x - t}^{x + t} {\psi \left( y \right)dy}.

This is the so-call D’ Alembert formula.

7 Comments »

  1. Bài này nếu giải bằng hàm Green thì thế nào NQA nhỉ ?

    Comment by viettran — March 24, 2009 @ 20:27

    • Nếu thế cần phải tìm hàm Green đã, việc này chắc ko dễ😦

      Comment by Ngô Quốc Anh — December 14, 2009 @ 17:25

  2. hi , my name nguyen , i come from USA , nice to meet you ,

    Comment by vo khanh nguyen — December 31, 2009 @ 16:54

    • Thanks for coming to my blog🙂

      Comment by Ngô Quốc Anh — January 2, 2010 @ 15:20

  3. In the odd-dimensional space (n \geq 3)

    http://arxiv.org/abs/0904.3252

    and in the even case, implies by using Hadamard’s method of descent.

    Comment by Tuan Minh — January 14, 2010 @ 23:40

    • Aha, that’s a good and new work, thanks Minh🙂

      Comment by Ngô Quốc Anh — January 14, 2010 @ 23:44

  4. In fact, this idea is mentioned in the well-known book of Stein in the case n=3 (by using FT/Bessel function), Stein also gave an exercise for the general case. The Torchinsky’s inverse formula for \dfrac{\sin(R\xi)}{|\xi|} is nice!

    Comment by Tuan Minh — January 15, 2010 @ 1:53


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