MATH529: L15 Fourier transform PDE solution

Overview

• Fourier transform solution of PDE BVPs

Fourier transform

• Fourier transform

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

• Inverse Fourier transform

  ${\displaystyle f\left(x\right)={ℱ}^{-1}\left\{F\left(\alpha \right)\right\}=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}F\left(\alpha \right)e^{-i\alpha x}d\alpha =\frac{1}{2\pi }(F,e^{-i\alpha x})}$

• Composition of direct and inverse

  ${\displaystyle {ℱ}^{-1}\left\{ℱ\left\{f\left(x\right)\right\}\right\}=f\left(x\right)}$

• Real, imaginary parts, $f:ℝ\to ℝ$

  ${\displaystyle \mathrm{Re}\left[F\left(\alpha \right)\right]=\mathrm{Re}[\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{i\alpha x}dx]=\mathrm{Re}[\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)(\cos \alpha x+i\sin \alpha x)dx]}$

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

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

Recall Fourier transform of derivatives

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

• 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 transforms 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

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

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

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

Solving PDE BVP by Fourier transforms

• Heat equation

  ${\displaystyle k\frac{{\partial }^{2}u}{\partial x^2}=\frac{\partial u}{\partial t},-\infty <x<\infty ,t>0,u(x,0)=f\left(x\right)=\{\begin{array}{ll}{u}_0& \left|x\right|<1\\ 0& \left|x\right|>1\end{array}.}$

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

  ${\displaystyle ℱ\{k\frac{{\partial }^{2}u}{\partial x^2}\}=ℱ\left\{\frac{\partial u}{\partial t}\right\}\Rightarrow ℱ\left\{\frac{\partial u}{\partial t}\right\}=\frac{\partial U(\alpha ,t)}{\partial t}=-k{\alpha }^{2}U\Rightarrow }$ ${\displaystyle U(\alpha ,t)=e^{-k{\alpha }^{2}t}U(\alpha ,0)=2\frac{u_0}{\alpha }\sin \alpha e^{-k{\alpha }^{2}t}=2u_0{e}^{-k{\alpha }^{2}t}\frac{\sin \alpha }{\alpha }}$ ${\displaystyle U(\alpha ,0)=\underset{-\infty }{\overset{\infty }{\int }}u(x,0)e^{i\alpha x}dx=\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{i\alpha x}dx=ℱ\left\{f\right\}=u_0\underset{-1}{\overset{1}{\int }}{e}^{i\alpha x}dx}$ ${\displaystyle U(\alpha ,0)=\frac{u_0}{i\alpha }{\left[e^{i\alpha x}\right]}_{x=-1}^{x=1}=\frac{u_0}{i\alpha }(e^{i\alpha }-e^{-i\alpha })=2\frac{u_0}{\alpha }\sin \alpha }$

Solving BVPs by Fourier transform (cont)

• Inverse Fourier transform

  ${\displaystyle u(x,t)={ℱ}^{-1}\left\{U(\alpha ,t)\right\}=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}U(\alpha ,t)e^{-i\alpha x}d\alpha \Rightarrow }$ ${\displaystyle u(x,t)=\frac{u_0}{\pi }\underset{-\infty }{\overset{\infty }{\int }}{e}^{-k{\alpha }^{2}t}\frac{\sin \alpha }{\alpha }{e}^{-i\alpha x}d\alpha =\frac{u_0}{\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{\sin \alpha }{\alpha }{e}^{-i\alpha x-k{\alpha }^{2}t}d\alpha }$ ${\displaystyle u(x,t)=\frac{u_0}{\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{\sin \alpha }{\alpha }{e}^{-{[\sqrt{kt}\alpha +\frac{ix}{2\sqrt{kt}}]}^2-\frac{x^2}{4kt}}d\alpha }$