Fourier transform of a complex-valued function $f(t)$ is a another complex-valued function $\hat{f}(w)$ that codifies the same information that $f(z)$ in the following way: for every $w\in \mathbb{R}$, the value $\hat{f}(w)$ represent the weight of the function $e^{2\pi iwt}$ in the decomposition of $f(t)$ as a sum of exponential
$$ f(t)=\cdots+\hat{f}(1)e^{2\pi it}+\hat{f}(1.1)e^{2\pi i1.1t}+\hat{f}(2)e^{2\pi i2t}+\cdots $$ $$ \cdots+\hat{f}(w)e^{2\pi iwt}+\cdots= $$ $$ =\int_{-\infty}^{+\infty}\hat{f}(w)e^{2 \pi iwt} dw $$The transform can be obtained with $\hat{f}(w)=\int_{-\infty}^{+\infty}f(t)e^{-2 \pi iwt} dt$.
A great explanation of the intuition behind Fourier transform: this video. It also explains why a real function possesses a conjugate symmetric Fourier transform ($\hat{f}(-w)=\overline{\hat{f}(w)}$).
It is a particular case of the Laplace transform. Related Mellin transform. See also translation and momentum operators.
Given a linear PDE $u_t=u_{xx}$ with initial data $u(x,0) = f(x)$, the first step in the solution process is to decompose the initial data function $f(x)$ into its Fourier series representation. This is achieved by expressing $f(x)$ as an infinite sum of complex exponentials.
$$ f(x) = \sum_{n=-\infty}^{+\infty} c_n e^{2\pi inx} $$where the coefficients $c_n$ are given by:
$$ c_n = \int_{-\infty}^{+\infty} f(x) e^{-2\pi inx} dx $$Also, we decompose
$$ u(x,t)=\sum_{n=-\infty}^{+\infty} u_n(t) e^{2\pi inx} $$where $u_n(t)$ represents the time-dependent coefficients of the Fourier series expansion of $u(x,t)$.
The next step is to substitute the Fourier series representation of $u(x,t)$ into the given PDE. By doing so, we obtain a set of ordinary differential equations (ODEs) for the coefficients $u_n(t)$.
For the given PDE $u_t = u_{xx}$, the substitution yields:
$$ \frac{du_n(t)}{dt} = -4\pi^2 n^2 u_n(t) $$This is a simple first-order ODE which can be solved for $u_n(t)$. The general solution is:
$$ u_n(t) = u_n(0) e^{-4\pi^2 n^2 t} $$Where $u_n(0)$ is the initial value of the coefficient, which can be obtained from the Fourier series representation of the initial data $u(x,0) = f(x)$. Specifically, $u_n(0) = c_n$, the Fourier coefficient of $f(x)$.
Finally, substituting the solution for $u_n(t)$ into the Fourier series representation of $u(x,t)$, we obtain the solution of the PDE in terms of the initial data $f(x)$:
$$ u(x,t) = \sum_{n=-\infty}^{+\infty} c_n e^{-4\pi^2 n^2 t} e^{2\pi inx} $$For problems defined on an infinite domain, the Fourier series might not be the most suitable tool. Instead, the Fourier Transform is employed to handle such cases. Given a linear PDE $u_t = u_{xx}$ with initial data $u(x,0) = f(x)$, the solution process begins by taking the Fourier Transform of the initial data function $f(x)$. This transforms the function from the spatial domain to the frequency domain. The Fourier Transform of $f(x)$ is given by:
$$ f(x) = \int_{-\infty}^{+\infty} \hat{f}(\omega) e^{2\pi i\omega x} d\omega $$where $\hat{f}(\omega)$ represents the function in the frequency domain.
Similarly,
$$ u(x,t) = \int_{-\infty}^{+\infty} \hat{u}(\omega,t) e^{2\pi i\omega x} d\omega $$Substituting the Fourier Transformed functions into the PDE, we obtain a differential equation in the frequency domain. For the PDE $u_t = u_{xx}$, this yields:
$$ \int_{-\infty}^{+\infty} \frac{d\hat{u}(\omega,t)}{dt} e^{2\pi i\omega x} d\omega= \int_{-\infty}^{+\infty}-4\pi^2 \omega^2 \hat{u}(\omega,t) e^{2\pi i\omega x} d\omega $$so
$$ \frac{d\hat{u}(\omega,t)}{dt} = -4\pi^2 \omega^2 \hat{u}(\omega,t) $$This equation is a first-order ODE in the frequency domain, which can be solved for $\hat{u}(\omega,t)$. The general solution is:
$$ \hat{u}(\omega,t) = \hat{u}(\omega,0) e^{-4\pi^2 \omega^2 t} $$Where $\hat{u}(k,0)$ is the initial value in the frequency domain, which is simply $\hat{f}(\omega)$, the Fourier Transform of the initial data $f(x)$.
To obtain the solution in the spatial domain, we take the inverse Fourier Transform of $\hat{u}(\omega,t)$. This gives us the solution of the PDE in terms of the initial data $f(x)$.
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Author of the notes: Antonio J. Pan-Collantes
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