The Fourier transform of a function \(f\) is denoted either by \(\mathcal{F}\{f(x)\}\) or \(\bar{f}(\xi)\) and is defined as \[\bar{f}(\xi)=\int\limits_{-\infty}^\infty f(x)e^{i\xi x}\mathrm{d}x.\]
The Fourier inversion theorem states that provided the inverse exists, it is given by \[f(x)=\frac{1}{2\pi}\int\limits_{-\infty}^\infty\bar{f}(\xi)e^{-i\xi x}\mathrm{d}\xi.\] In the lectures we have discussed different functional spaces and how this impacts notions like existence (i.e. does the integral converge?)
In the lectures we have met a number of important results that are useful when solving Fourier transform problems. These include the derivative theorem and the convolution theorem.
The derivative theorem states that provided \(f\in L^1(\mathbb{R})\), \[\mathcal{F}\{f'(x)\}=-i\xi\mathcal{F}\{f(x)\}=-i\xi\bar{f}(\xi).\]
The convolution theorem states that \[\mathcal{F}\{f*g\}=\mathcal{F}\{f\}\mathcal{F}\{g\},\] where \(f*g\) denotes the convolution of \(f\) and \(g\), which is defined by \begin{align}(f*g)(x)&=\int\limits_{-\infty}^\infty f(y)g(x-y)\mathrm{d}y\\&=\int\limits_{-\infty}^\infty f(x-y)g(y)\mathrm{d}y.\end{align}
The heat equation with initial conition \(u(x,0)=f(x)\) has solution given by \[u(x,t)=\frac{1}{2c\sqrt{\pi t}}\int\limits_{-\infty}^\infty {e^{-\frac{(x-y)^2}{4c^2t}}f(y)\mathrm{d}y}.\] You don't need to remember this formula for the exam, but you should know how to derive it by using Fourier transforms; the video below revises the derivation we saw in the lectures.
The Fourier transform of a function \(f\) is denoted either by \(\mathcal{F}\{f(x)\}\) or \(\bar{f}(\xi)\) and is defined as \[\bar{f}(\xi)=\int\limits_{-\infty}^\infty f(x)e^{i\xi x}\mathrm{d}x.\]
The Fourier inversion theorem states that provided the inverse exists, it is given by \[f(x)=\frac{1}{2\pi}\int\limits_{-\infty}^\infty\bar{f}(\xi)e^{-i\xi x}\mathrm{d}\xi.\] In the lectures we have discussed different functional spaces and how this impacts notions like existence (i.e. does the integral converge?)
In the lectures we have met a number of important results that are useful when solving Fourier transform problems. These include the derivative theorem and the convolution theorem.
The derivative theorem states that provided \(f\in L^1(\mathbb{R})\), \[\mathcal{F}\{f'(x)\}=-i\xi\mathcal{F}\{f(x)\}=-i\xi\bar{f}(\xi).\]
The convolution theorem states that \[\mathcal{F}\{f*g\}=\mathcal{F}\{f\}\mathcal{F}\{g\},\] where \(f*g\) denotes the convolution of \(f\) and \(g\), which is defined by \begin{align}(f*g)(x)&=\int\limits_{-\infty}^\infty f(y)g(x-y)\mathrm{d}y\\&=\int\limits_{-\infty}^\infty f(x-y)g(y)\mathrm{d}y.\end{align}
The heat equation with initial conition \(u(x,0)=f(x)\) has solution given by \[u(x,t)=\frac{1}{2c\sqrt{\pi t}}\int\limits_{-\infty}^\infty {e^{-\frac{(x-y)^2}{4c^2t}}f(y)\mathrm{d}y}.\] You don't need to remember this formula for the exam, but you should know how to derive it by using Fourier transforms; the video below revises the derivation we saw in the lectures.