Chapter G Problem Sheet 7 (Lectures 13-15)
G.1 Fourier transform of exponential-like functions
Let \(a\in \mC \) with \(\re (a)>0\). Define \(u:\mR \to \mC \) by \(u(x)=\e ^{-a|x|}\).
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(i) Show that \(u\in L^1(\mR )\).
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(ii) Show that \(\hat {u}(\omega )=\sqrt {\frac {2}{\pi }}~\frac {a}{a^2+\omega ^2}\).
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Solution. (i) We have (using that the integrand is even)
\(\seteqnumber{0}{G.}{0}\)\begin{multline*} \int _{-\infty }^\infty |u(x)|\,dx =\int _{-\infty }^\infty \left |\e ^{-a|x|}\right |\,dx =\int _{-\infty }^\infty \e ^{-\re (a)|x|}\,dx =2\int _{0}^\infty \e ^{-\re (a)x}\,dx \\ =\lim _{R\to \infty }2\int _{0}^R \e ^{-\re (a)x}\,dx =\left [\frac {-2}{\re (a)}\e ^{-\re (a)x}\right ]_{x=0}^R =\lim _{R\to \infty }\frac {-2}{\re (a)}\e ^{-\re (a)R}+\frac {2}{\re (a)} =\frac {2}{\re (a)}. \end{multline*} (ii) We have
\(\seteqnumber{0}{G.}{0}\)\begin{multline*} \hat {u}(\omega ) =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega x}\e ^{-a|x|}\,dx =\frac {1}{\sqrt {2\pi }}\int _{0}^\infty \e ^{-i\omega x}\e ^{-ax}\,dx +\frac {1}{\sqrt {2\pi }}\int _{-\infty }^0 \e ^{-i\omega x}\e ^{ax}\,dx \\=\frac {1}{\sqrt {2\pi }}\int _{0}^\infty \e ^{(-i\omega -a) x}\,dx +\frac {1}{\sqrt {2\pi }}\int _{-\infty }^0 \e ^{(-i\omega +a) x}\,dx =\frac {1}{\sqrt {2\pi }}\left (\frac {1}{i\omega +a}+\frac {1}{-i\omega +a}\right ) \\ =\frac {1}{\sqrt {2\pi }}\left (\frac {-i\omega +a}{\omega ^2+a^2}+\frac {i\omega +a}{\omega ^2+a^2}\right ) =\frac {1}{\sqrt {2\pi }}~\frac {2a}{\omega ^2+a^2} =\sqrt {\frac {2}{\pi }}~\frac {a}{a^2+\omega ^2}. \end{multline*} □
G.2 The wave equation on the whole line
Let \(c>0\) and \(u_0:\mR \to \mR \) be given. Solve the wave equation
\[ \partial _{tt}u=c^2\partial _{xx}u,\qquad u(0,x)=u_0(x),\qquad \partial _t u(0,x)=0, \]
where \(x\in \mR \) and \(t>0\), by Fourier transforming in \(x\), solving the resulting ODE in \(t\) and inverse Fourier transforming the obtained solution of this ODE. Your answer should not contain any integrals (including convolutions).
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Solution. Taking the Fourier transform in \(x\) of the differential equation and using the relation between differentiation and Fourier transformation gives
\[ \partial _{tt} \hat {u}(t,\omega )=-c^2\omega ^2 \hat {u}(t,\omega ),\qquad \hat {u}(0,\omega )=\hat {u}_0(\omega ),\qquad \partial _t\hat {u}(0,\omega )=0. \]
Solving this initial value ODE gives
\[ \hat {u}(t,\omega )=\cos (c\omega t)\hat {u}_0(\omega ). \]
To inverse Fourier transform this, it is convenient to write the cosine in terms of complex exponentials:
\[ \hat {u}(t,\omega )=\frac {1}{2}\left [\e ^{ic\omega t}+\e ^{-ic\omega t}\right ]\hat {u}_0(\omega ). \]
By the space-shift formula we recognize \(\e ^{ic\omega t}\hat {u}_0(\omega )\) as the Fourier transform of \(x\mapsto u_0(x+ct)\) and \(\e ^{-ic\omega t}\hat {u}_0(\omega )\) as the Fourier transform of \(x\mapsto u_0(x-ct)\). Therefore
\[ u(t,x)=\frac {1}{2}\left [u_0(x+ct)+u_0(x-ct)\right ]. \]
This is (a special case of) d’Alembert’s formula which we will see again later in the module. □
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Definition G.1. For \(u\in L^1(0,\infty )\), the Fourier cosine transform, denoted \(\hat {u}_c\) or \(\mathcal {F}_c(u)\), is defined by
\[ \hat {u}_c(\omega ) =\sqrt {\frac {2}{\pi }}\int _0^\infty \cos (\omega x)u(x)\,dx, \]
and the Fourier sine transform, denoted \(\hat {u}_s\) or \(\mathcal {F}_s(u)\), is defined by
\[ \hat {u}_s(\omega ) =\sqrt {\frac {2}{\pi }}\int _0^\infty \sin (\omega x)u(x)\,dx. \]
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Remark G.3. Similarly as for the Laplace transform in MA20220 Ordinary Differential Equations and Control, for the sine and cosine transforms, the relation between transform and differentiation is a bit more complicated than it is for the Fourier transform (Corollary 17.13). We have (shown through integration by parts)
\[ \mathcal {F}_c(u'')(\omega )=-\omega ^2\hat {u}_c(\omega )-\sqrt {\frac {2}{\pi }}u'(0),\qquad \mathcal {F}_s(u'')(\omega )=-\omega ^2\hat {u}_s(\omega )+\omega \sqrt {\frac {2}{\pi }} u(0). \]
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Remark G.4. The convolution theorems for the sine and cosine transform are
\[ \mathcal {F}_s\left (t\mapsto \int _0^\infty f(\theta )\left [g(|t-\theta |)-g(t+\theta )\right ]\,d\theta \right ) =\sqrt {2\pi }~\mathcal {F}_s(f)\mathcal {F}_c(g), \]
and
\[ \mathcal {F}_c\left (t\mapsto \int _0^\infty f(\theta )\left [g(|t-\theta |)+g(t+\theta )\right ]\,d\theta \right ) =\sqrt {2\pi }~\mathcal {F}_c(f)\mathcal {F}_c(g). \]
G.3 The heat equation on a half-line
Let \(u_0:(0,\infty )\to \mR \) be given. Solve the heat equation
\[ \partial _t u=\partial _{xx} u,\qquad u(0,x)=u_0(x),\qquad u(t,0)=0, \]
where \(x>0\) and \(t>0\), by Fourier-Sine transforming in \(x\), solving the resulting ODE in \(t\) and inverse Fourier-Sine transforming the obtained solution of this ODE. Your answer will have the form of a convolution-like integral.
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Solution. Taking the Fourier-Sine transform in \(x\) of the differential equation and using the relation between differentiation and the Fourier-Sine transformation and using \(u(t,0)=0\) gives
\[ \partial _t \hat {u}_s(t,\omega )=-\omega ^2\hat {u}_s(t,\omega ). \]
The initial value transforms to
\[ \hat {u}_s(0,\omega )=\hat {u}_{0,s}(\omega ). \]
We therefore want to solve the ODE initial value problem for \(t\mapsto \hat {u}_s(t,\omega )\) (where \(\omega \in \mR \) is a parameter):
\[ \partial _t \hat {u}_s(t,\omega )=-\omega ^2\hat {u}_s(t,\omega ),\qquad \hat {u}_s(0,\omega )=\hat {u}_{0,s}(\omega ). \]
This gives
\[ \hat {u}_s(t,\omega )=\e ^{-\omega ^2t}\hat {u}_{0,s}(\omega ). \]
From our table of Fourier transforms, we have
\[ \mathcal {F}^{-1}\left (\omega \mapsto \e ^{-\omega ^2t}\right ) =\frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}. \]
Since the functions involved are even, we have that their Fourier transform and their Fourier-Cosine transform coincide. Hence
\[ \mathcal {F}_c\left (\frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}\right )=\e ^{-\omega ^2t}. \]
Using the convolution formula for the Fourier-Sine transform we then obtain
\[ u(t,x)=\frac {1}{\sqrt {4\pi t}} \int _{0}^\infty \left [\e ^{-(x-y)^2/(4t)}-\e ^{-(x+y)^2/(4t)}\right ]u_0(y)\,dy. \]
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G.4 Using Plancherel’s Theorem
Use Plancherel’s Theorem applied to the function
\[ u(x)=\begin {cases} x&x\in (-1,1)\\ 0&\text {otherwise}, \end {cases} \]
to determine
\[ \int _{-\infty }^\infty \frac {(\sin (y)-y\cos (y))^2}{y^4}\,dy. \]
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Solution. We determine the Fourier transform of the given function \(u\) (which we note is in \(L^1(\mR )\cap L^2(\mR )\)):
\(\seteqnumber{0}{G.}{0}\)\begin{align*} \hat {u}(\omega ) &=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega x}u(x)\,dx =\frac {1}{\sqrt {2\pi }}\int _{-1}^1 \e ^{-i\omega x}x\,dx \\ &=\frac {1}{\sqrt {2\pi }}\left [\frac {-1}{i\omega }\e ^{-i\omega x}x\right ]_{x=-1}^1 +\frac {1}{\sqrt {2\pi }}~\frac {1}{i\omega }\int _{-1}^1 \e ^{-i\omega x}\,dx \\ &=\frac {-1}{\sqrt {2\pi }}~\frac {2}{i\omega }~\frac {\e ^{i\omega }+\e ^{-i\omega }}{2} +\frac {1}{\sqrt {2\pi }}~\frac {1}{i\omega }\left [\frac {1}{-i\omega }\e ^{-i\omega x}\right ]_{x=-1}^1 \\ &=\frac {1}{\sqrt {2\pi }}~\frac {2i}{\omega }~\cos (\omega ) +\frac {1}{\sqrt {2\pi }}~\frac {2i}{\omega ^2}~\frac {\e ^{-i\omega }-\e ^{i\omega }}{2i} \\ &=\frac {1}{\sqrt {2\pi }}~\frac {2i}{\omega }~\cos (\omega ) -\frac {1}{\sqrt {2\pi }}~\frac {2i}{\omega ^2}~\sin (\omega ). \end{align*} It follows that
\[ \|\hat {u}\|^2_{L^2(\mR )} =\frac {2}{\pi }\int _{-\infty }^\infty \left (\frac {\omega \cos (\omega )-\sin (\omega )}{\omega ^2}\right )^2\,d\omega . \]
By Plancherel’s Theorem, this equals
\[ \|u\|^2_{L^2(\mR )}=\int _{-1}^1 x^2\,dx=\left [\frac {1}{3}x^3\right ]_{x=-1}^1=\frac {2}{3}. \]
It follows that
\[ \int _{-\infty }^\infty \left (\frac {\omega \cos (\omega )-\sin (\omega )}{\omega ^2}\right )^2\,d\omega =\frac {\pi }{2}~\frac {2}{3}=\frac {\pi }{3}. \]
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