MATH529: L15 Fourier transform PDE solution

Overview

• Fourier integral

• Fourier transforms:

  • General Fourier transform

  • Sine and cosine transforms

  • Finite Fourier transform

  • Fast Fourier transform

• Fourier analysis of PDE solutions

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

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

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

Cosine, Sine transforms

• Cosine transform, inverse cosine transform

  ${\displaystyle F_c\left(\alpha \right)={ℱ}_c\left\{f\right\}=\underset{0}{\overset{\infty }{\int }}f\left(x\right)\cos \alpha xdx,}$ ${\displaystyle f\left(x\right)={ℱ}_c^{-1}\left\{F_c\right\}=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}{F}_c\left(\alpha \right)\cos \alpha xd\alpha }$

• Sine transform, inverse sine transform

  ${\displaystyle F_s\left(\alpha \right)={ℱ}_c\left\{f\right\}=\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sin \alpha xdx,}$ ${\displaystyle f\left(x\right)={ℱ}_s^{-1}\left\{F_s\right\}=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}{F}_s\left(\alpha \right)\sin \alpha xd\alpha }$

Cosine transform of derivatives

• $F_c\left(\alpha \right)={ℱ}_c\left\{f\left(x\right)\right\}={\int }_0^{\infty }f\left(x\right)\cos \alpha xdx$

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

  ${\displaystyle {ℱ}_c\left\{f^{\text{'}}\left(x\right)\right\}=\underset{0}{\overset{\infty }{\int }}{f}^{\text{'}}\left(x\right)\cos \alpha xdx={[f\left(x\right)\sin \alpha x]}_{x=0}^{x\to \infty }+\alpha \underset{0}{\overset{\infty }{\int }}f\left(x\right)\sin \alpha xdx}$ ${\displaystyle {ℱ}_c\left\{f^{\text{'}}\left(x\right)\right\}=\alpha {ℱ}_s\left\{f\left(x\right)\right\}}$

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

  ${\displaystyle {ℱ}_c\left\{f^{\text{'}\text{'}}\left(x\right)\right\}=\underset{0}{\overset{\infty }{\int }}{f}^{\text{'}\text{'}}\left(x\right)\cos \alpha xdx={[f^{\text{'}}\left(x\right)\sin \alpha x]}_{x=0}^{x\to \infty }+\alpha \underset{0}{\overset{\infty }{\int }}{f}^{\text{'}}\left(x\right)\sin \alpha xdx}$ ${\displaystyle {ℱ}_c\left\{f^{\text{'}\text{'}}\left(x\right)\right\}={[f^{\text{'}}\left(x\right)\cos \alpha x]}_{x=0}^{x\to \infty }-{\alpha }^2\underset{0}{\overset{\infty }{\int }}f\left(x\right)\cos \alpha xdx=-f^{\text{'}}\left(0\right)-{\alpha }^2{F}_c\left(\alpha \right)}$

Sine transform of derivatives

• $F_s\left(\alpha \right)={ℱ}_s\left\{f\left(x\right)\right\}={\int }_0^{\infty }f\left(x\right)\sin \alpha xdx$

• Evaluate ${ℱ}_s\left\{f^{\text{'}}\left(x\right)\right\}={\int }_0^{\infty }{f}^{\text{'}}\left(x\right)\sin \alpha xdx={\int }_0^{\infty }\sin \alpha xdf\left(x\right)$

  ${\displaystyle {ℱ}_s\left\{f^{\text{'}}\left(x\right)\right\}={[\sin \alpha xf\left(x\right)]}_{x=0}^{x\to \infty }-\alpha \underset{0}{\overset{\infty }{\int }}f\left(x\right)\cos \alpha xdx=-\alpha {ℱ}_c\left\{f\left(x\right)\right\}}$

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

  ${\displaystyle {ℱ}_s\left\{f^{\text{'}\text{'}}\right\}=-\alpha {ℱ}_c\left\{f^{\text{'}}\left(x\right)\right\}}$ ${\displaystyle {ℱ}_s\left\{f^{\text{'}\text{'}}\right\}=\alpha f\left(0\right)-{\alpha }^2{ℱ}_s\left\{f\right\}}$ ${\displaystyle }$

Heat equation BVP solved by Fourier transform

Heat equation

Fourier transform $ℱ\left\{u\right\}={\int }_{-\infty }^{\infty }u(x,t)e^{i\alpha x}dx=U(\alpha ,t)$,

Heat equation solution given by inverse Fourier transform

Inverse Fourier transform