# Ngô Quốc Anh

## March 26, 2009

### Solving initial value problem for heat equation via Fourier transform

Filed under: Các Bài Tập Nhỏ, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 21:05

Followed by Solving initial value problem for wave equation via Fourier transform, in this topic I will show you how to solve initial value problem for heat equation via Fourier transform.

Following is the problem

$\displaystyle\left\{\begin{gathered} {u_t} = {c^2}{u_{xx}}, \quad x \in \mathbb{R}, \hfill \\ u(x,0) = \varphi (x). \hfill \\ \end{gathered}\right.$

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

$\displaystyle\widehat{{u_t}\left( {\eta ,t} \right)} - c^2 {\left( {i\eta } \right)^2}\widehat{u\left( {\eta ,t} \right)} = 0$.

This is an ODE, the solution is given by

$\displaystyle\widehat{u\left( {\eta ,t} \right)} = C(\eta)e^{-c^2 \eta^2 t}$.

From the initial date we get

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

which implies that

$\displaystyle\widehat{\varphi \left( \eta \right)} = C(\eta)$.

Thus, we obtain

$\displaystyle\widehat{u\left( {\eta ,t} \right)} = \widehat{\varphi \left( \eta \right)}e^{-c^2 \eta^2 t}$.

Finally, we obtain by taking the inverse Fourier transform

$\displaystyle u\left( {x,t} \right) = \frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x}}\widehat u\left( {\eta ,t} \right)d\eta } = \frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x - {c^2}{\eta ^2}t}}\hat \varphi \left( \eta \right)d\eta }$.

## 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.