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4.3.2: Famous Transformations

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    188860
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    Some differential equations are particularly difficult to answer with ordinary means. Linear algebra gives us help here too.

    In a finite dimensional vector space \(V\), we have learned that with any basis \(\mathcal{B}\) of \(V\), we can use a linear transformation to coordinatize each vector in \(V\) and "turn it into" a vector in \(\mathbb{R}^n\). This is done because we have powerful tools that make solving questions easier in \(\mathbb{R}^n\), and when we are done we un-coordinatize our answers to transform them back into \(V\).

    This is a common mathematical trick. Transform objects into another vector space (where calculations are easier), solve the problem, and then transform them back.

    Definition \(\PageIndex{2}\): Laplace Transformations

    The Laplace Transformation of a function \(f(t)\) is defined to be the function \(\mathcal{L}(f)=\displaystyle{\int_0^\infty f(t)e^{-st}dt}\)

    This is a linear transformation:

    \[ \mathcal{L}(f+g)=\displaystyle{\int_0^\infty [f(t)+g(t)]e^{-st}dt=\int_0^\infty f(t)e^{-st}dt+\int_0^\infty g(t)e^{-st}dt=\mathcal{L}(f)+\mathcal{L}(g)} \nonumber\]

    \[\mathcal{L}(cf)=\displaystyle{\int_0^\infty [cf(t)]e^{-st}dt=c\int_0^\infty f(t)e^{-st}dt=c\mathcal{L}(f)}\nonumber\]

    We restrict the domain of \(\mathcal{L}\) to be \(V=\left\{ f\in\mathbb{F}: |f(x)|\le Me^{cx} \text{ for some } M,c\in \mathbb{R} \right\} \) (so that the integral always converges), and then define \(W=\text{Range}(\mathcal{L})\) so that \(\mathcal{L}: V\to W\) is invertible (but the inverse is ugly to look at)

    It is (easily) verified that the following formulas (which we will use in the example below) hold for the Laplace transformation:

    \(\mathcal{L}(f'(t))=s\mathcal{L}(f)-f(0)\) and \(\mathcal{L}(e^{at})=\dfrac{1}{s-a} \longleftrightarrow \mathcal{L}^{-1}\left(\dfrac{1}{s-a}\right)=e^{at}\)

    Example \(\PageIndex{5}\): Eigenvalues of the Derivative

    Find all real eigenvalues and eigenvectors of the derivative transformation \(T:C'(\mathbb{R})\to C'(\mathbb{R})\) with \(T(f)=f'\).

    Solution

    We want to find the functions \(f\in C'(\mathbb{R})\) which satisfy \(T(f)=\lambda f\)

    Ultimately, we are looking to solve the differential equation \(y'=\lambda y\). We will do so by applying the Laplace Transformation to both sides:

    \[ \mathcal{L} (y')=\mathcal{L} (\lambda y) \implies s\mathcal{L}(y)-y(0)=\lambda \mathcal{L}(y)\nonumber\]

    We can solve for \(\mathcal{L}(y)\) to get \(\mathcal{L}(y)=\dfrac{y(0)}{s-\lambda}\)

    Now we apply the inverse transformation to both sides:

    \[ \mathcal{L}^{-1} \left( \mathcal{L}(y)\right)=\mathcal{L}^{-1} \left( \dfrac{y(0)}{s-\lambda}\right) \implies y=y(0)e^{\lambda t} \nonumber\]

    Therefore all real numbers are eigenvalues of \(T\), and each has a one dimensional eigenspace \(E_\lambda=\text{Span}\{e^{\lambda t}\}\)

    Definition \(\PageIndex{3}\): Fourier Transformations

    The Fourier Transformation of a function \(f(t)\) is defined to be the function \(\mathcal{F}(f)=\displaystyle{\int_{-\infty}^\infty f(t)e^{-2\pi i \, st}dt}\)

    This is also a linear transformation:

    \[ \mathcal{F}(f+g)=\displaystyle{\int_{-\infty}^\infty [f(t)+g(t)]e^{-2\pi i \, st}dt=\int_{-\infty}^\infty f(t)e^{-2\pi i\, st}dt+\int_{-\infty}^\infty g(t)e^{-2\pi i \, st}dt=\mathcal{F}(f)+\mathcal{F}(g)} \nonumber\]

    \[\mathcal{F}(cf)=\displaystyle{\int_{-\infty}^\infty [cf(t)]e^{-2\pi i \, st}dt=c\int_{-\infty}^\infty f(t)e^{-2\pi i \, st}dt=c\mathcal{F}(f)}\nonumber\]

    We restrict the domain of \(\mathcal{F}\) to be \(V=\left\{ f\in\mathbb{F}: \displaystyle{ \lim_{|x|\to \infty}|f^{(n)}(x)|=0} \text{ for all } n \right\} \) (so that the integral always converges), and then define \(W=\text{Range}(\mathcal{F})\) so that \(\mathcal{F}:V\to W\) is invertible (the inverse is again ugly to look at)

    Fourier claimed (in 1822) that almost every function can be decomposed into a (possibly infinite) linear combination of sine and/or cosine functions of varying frequencies (later dubbed a Fourier Series). So, instead of being used to solve differential equations, Fourier Transformations are used to describe a given function \(f\) in terms of a complex valued function of \(s\) which describes its frequency.


    4.3.2: Famous Transformations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.