MATH529: L14 Fourier transform PDE solution

Overview

• Fourier integral

• Fourier transform

Fourier transform

• Often used for steady-state problems

• Recall $f(t+T)=f\left(t\right)$

  ${\displaystyle f\left(t\right)=\frac{A_0}{2}+\sum _{k=1}^{\infty }[A_k\cos (2\pi k\frac{t}{T})+B_k\sin (2\pi k\frac{t}{T})]}$

• The Fourier coefficients $A_k,B_k$ are obtained as scalar products

  ${\displaystyle A_k=\frac{2}{T}\underset{0}{\overset{T}{\int }}f\left(t\right)\cos (2\pi k\frac{t}{T})dt,B_k=\frac{2}{T}\underset{0}{\overset{T}{\int }}f\left(t\right)\sin (2\pi k\frac{t}{T})dt}$

• By analogy define a Fourier transform

  ${\displaystyle F\left(\alpha \right)=ℱ\left\{f\left(x\right)\right\}=\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{i\alpha x}dx,f\left(x\right)={ℱ}^{-1}\left(F\left(\alpha \right)\right)=\underset{-\infty }{\overset{\infty }{\int }}F\left(\alpha \right)e^{-i\alpha x}d\alpha }$

Fourier transform of derivatives

• $F\left(\alpha \right)=ℱ\left\{f\left(x\right)\right\}={\int }_{-\infty }^{\infty }f\left(x\right)e^{i\alpha x}dx,f\left(x\right)={ℱ}^{-1}\left(F\left(\alpha \right)\right)={\int }_{-\infty }^{\infty }F\left(\alpha \right)e^{-i\alpha x}d\alpha$

• Evaluate $ℱ\left\{f^{\text{'}}\left(x\right)\right\}$

  ${\displaystyle ℱ\left\{f^{\text{'}}\left(x\right)\right\}=\underset{-\infty }{\overset{\infty }{\int }}{f}^{\text{'}}\left(x\right)e^{i\alpha x}dx={[f\left(x\right)e^{i\alpha x}]}_{x\to -\infty }^{x\to \infty }-i\alpha \underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{i\alpha x}dx}$

• Impose condition of $f$ to allow Fourier transform

  ${\displaystyle L_2:\underset{-\infty }{\overset{\infty }{\int }}{f}^2\left(x\right)dx\mathrm{is}\mathrm{finite}\Rightarrow {[f\left(x\right)e^{i\alpha x}]}_{x\to -\infty }^{x\to \infty }=0}$

• $ℱ\left\{f^{\text{'}}\left(x\right)\right\}=-i\alpha F\left(\alpha \right)$, $ℱ\left\{f^{\text{'}\text{'}}\left(x\right)\right\}=-{\alpha }^{2}F\left(\alpha \right)$, $ℱ\left\{f^{\text{'}\text{'}\text{'}}\left(x\right)\right\}=i{\alpha }^{3}F\left(\alpha \right)$

Heat equation BVP solved by Fourier transform

Apply to a PDE, $u_t=k{u}_{xx}$ for $0<x<\pi$, $t>0$, with initial conditions $u(x,0)=-\sin 3x$, boundary conditions $u(0,t)=u(\pi ,t)=0$.

• Apply Fourier transform, obtain a family of ODEs labeled by $\alpha$

  ${\displaystyle \begin{array}{rcl}\frac{\partial }{\partial t}U(\alpha ,t)& =& k(-{\alpha }^2)U(\alpha ,t)\Rightarrow \end{array}}$

• Solve the ODE $\Rightarrow$$U(\alpha ,t)=c_1{e}^{-{\alpha }^{2}kt}$

• Inverse Fourier transform