Show that (assuming sufficient smoothness of the domain and the data) \(u\) is a solution to the Dirichlet boundary value problem\[ -\Delta u=f \] In \(\Omega \) with B.C. \(u=g\) on \(\partial \Omega \) iff \(u\) is a minimizer of the energy functional, that is\[ E\left ( u\right ) =\min \left \{ E\left ( v\right ) :v\in C^{2}\left ( \bar{\Omega }\right ) \right \} \text{ such that }u=g\text{ on }\partial \Omega \] Here\[ E\left ( u\right ) =\int _{\Omega }\left ( \frac{1}{2}\left \vert \nabla u\right \vert ^{2}-fu\right ) dA \] (note, I will be using \(dA\) in the above integral assuming we are in \(\mathbb{R} ^{2}\). But the above can also be \(dV\) for \(\mathbb{R} ^{3}\) just as well and nothing will change in the derivation below. This is easier that writing \(dx\) and saying that \(x\) is a vector).
Solution
Since the proof is an iff, then we need to show both direction.
Forward direction Given that \(u\) solves \begin{equation} -\Delta u=f \tag{1} \end{equation} with \(\left . u\right \vert _{\partial \Omega }=g\). Then we need to show that \(E\left ( v\right ) \geq E\left ( u\right ) \) for all \(v\in C^{2}\left ( \bar{\Omega }\right ) \) that also satisfy same B.C.
Multiplying both sides of (1) by \(u-v\) and integrating over the domain gives\begin{equation} -\int _{\Omega }\left ( \Delta u\right ) \left ( u-v\right ) dA=\int _{\Omega }\left ( u-v\right ) fdA \tag{2} \end{equation} For the left integral \(\int _{\Omega }\left ( \Delta u\right ) \left ( u-v\right ) dA\), we will do integration by parts. Let \(\Delta u\equiv dV,u-v=U\), then \(\int _{\Omega }UdV=\int _{\partial \Omega }UV-\int _{\Omega }VdU\). Therefore \(dU=\nabla \left ( u-v\right ) \) and \(V=\nabla u\). After applying integration by parts the (2) now becomes\[ -\left ( \int _{\partial \Omega }\left ( u-v\right ) \frac{\partial u}{\partial \mathbf{n}}\ dL-\int _{\Omega }\nabla u\cdot \nabla \left ( u-v\right ) dA\right ) =\int _{\Omega }\left ( u-v\right ) fdA \] But \(\int _{\partial \Omega }\left ( u-v\right ) \frac{\partial u}{\partial \mathbf{n}}\ dL=0\) because \(u=v\) on the boundary \(\partial \Omega \,\) as both are \(g\). The above now simplifies to\begin{align*} \int _{\Omega }\nabla u\cdot \nabla \left ( u-v\right ) \ dA & =\int _{\Omega }\left ( uf-vf\right ) \ dA\\ \int _{\Omega }\nabla u\cdot \left ( \nabla u-\nabla v\right ) \ dA & =\int _{\Omega }\left ( uf-vf\right ) \ dA\\ \int _{\Omega }\left \vert \nabla u\right \vert ^{2}-\nabla u\cdot \nabla v\ dA & =\int _{\Omega }\left ( uf-vf\right ) \ dA\\ \int _{\Omega }\left \vert \nabla u\right \vert ^{2}-\int _{\Omega }fu\ dA & =\int _{\Omega }\left ( \nabla u\cdot \nabla v\right ) -vf\ dA \end{align*}
Now we use Schwarz triangle inequality and write \(\nabla u\cdot \nabla v\leq \frac{1}{2}\left ( \left \vert \nabla u\right \vert ^{2}+\left \vert \nabla v\right \vert ^{2}\right ) \). This comes from using \(ab\leq \frac{1}{2}\left ( a^{2}+b^{2}\right ) \). Using this in the RHS of the above gives
\begin{align*} \int _{\Omega }\left \vert \nabla u\right \vert ^{2}\ dA-\int _{\Omega }fu\ dA & \leq \int _{\Omega }\frac{1}{2}\left ( \left \vert \nabla u\right \vert ^{2}+\left \vert \nabla v\right \vert ^{2}\right ) -fv\ dA\\ \int _{\Omega }\left \vert \nabla u\right \vert ^{2}\ dA-\int _{\Omega }fu\ dA & \leq \int _{\Omega }\frac{1}{2}\left \vert \nabla u\right \vert ^{2}dA+\left ( \frac{1}{2}\int _{\Omega }\left \vert \nabla v\right \vert ^{2}-fv\ dA\right ) \\ \int _{\Omega }\frac{1}{2}\left \vert \nabla u\right \vert ^{2}dA-\int _{\Omega }fu\ dA & \leq \frac{1}{2}\int _{\Omega }\left \vert \nabla v\right \vert ^{2}-fv\ dA\\ \int _{\Omega }\frac{1}{2}\left \vert \nabla u\right \vert ^{2}-fu\ dA & \leq \frac{1}{2}\int _{\Omega }\left \vert \nabla v\right \vert ^{2}-fv\ dA \end{align*}
But by definition \(\int _{\Omega }\frac{1}{2}\left \vert \nabla u\right \vert ^{2}-fu\ dA=E\left ( u\right ) \) and \(\frac{1}{2}\int _{\Omega }\left \vert \nabla v\right \vert ^{2}-fv\ dA=E\left ( v\right ) \), therefore the above becomes\[ E\left ( u\right ) \leq E\left ( v\right ) \] Which is what we wanted to show. Now we will do the other direction.
Reverse direction Given that \(u\) minimizes energy among all test functions, i.e. given that \(E\left ( u\right ) =\min E\left ( w\right ) \), then need to show that \(-\Delta u=f\).
Consider \(w=u+\varepsilon v\) where \(v\) is any test function \(v\in C^{2}\left ( \bar{\Omega }\right ) \) and \(v=g\) at \(\partial \Omega \). Hence \[ \min \left ( E\left ( w\right ) \right ) =\min \left ( E\left ( u+\varepsilon v\right ) \right ) \] Therefore \(\min \left ( E\left ( u+\varepsilon v\right ) \right ) \) is achieved when \(\varepsilon =0\), since this then gives \(E\left ( u\right ) \) which by assumption is the minimum. Therefore\[ \frac{d}{d\varepsilon }E\left ( u+\varepsilon v\right ) =0 \] At \(\varepsilon =0\). But the above can be written as the following, using the definition of energy\begin{align} \frac{d}{d\varepsilon }\left ( \int _{\Omega }\frac{1}{2}\left \vert \nabla \left ( u+\varepsilon v\right ) \right \vert ^{2}-f\left ( u+\varepsilon v\right ) \ dA\right ) & =0\nonumber \\ \frac{d}{d\varepsilon }\left ( \int _{\Omega }\frac{1}{2}\left ( \nabla \left ( u+\varepsilon v\right ) \cdot \nabla \left ( u+\varepsilon v\right ) \right ) -f\left ( u+\varepsilon v\right ) \ dA\right ) & =0\tag{3} \end{align}
Expanding \(\nabla \left ( u+\varepsilon v\right ) \cdot \nabla \left ( u+\varepsilon v\right ) \) gives \begin{align} \nabla \left ( u+\varepsilon v\right ) \cdot \nabla \left ( u+\varepsilon v\right ) & =\left ( \nabla u+\varepsilon \nabla v\right ) \cdot \left ( \nabla u+\varepsilon \nabla v\right ) \nonumber \\ & =\left \vert \nabla u\right \vert ^{2}+2\varepsilon \nabla u\cdot \nabla v+\varepsilon ^{2}\left \vert \nabla v\right \vert ^{2}\tag{4} \end{align}
Substituting (4) into (3) gives\[ \frac{d}{d\varepsilon }\left ( \int _{\Omega }v\left ( \left \vert \nabla u\right \vert ^{2}+2\varepsilon \nabla u\cdot \nabla v+\varepsilon ^{2}\left \vert \nabla v\right \vert ^{2}\right ) -fu-\varepsilon fv\ dA\right ) =0 \] Now we move the derivative inside the take derivative w.r.t. \(\varepsilon \) giving\[ \left ( \int _{\Omega }\frac{1}{2}\left ( 2\nabla u\cdot \nabla v+2\varepsilon \left \vert \nabla v\right \vert ^{2}\right ) -fv\ dA\right ) =0 \] Evaluate at \(\varepsilon =0\) the above becomes\[ \int _{\Omega }\left ( \nabla u\cdot \nabla v\right ) dA-\int _{\Omega }fv\ dA=0 \] Integration by parts for the first integral. Let \(\nabla u=U,dV=\nabla v\), then \(\int _{\Omega }UdV=\int _{\partial \Omega }UV-\int _{\Omega }VdU\). Hence the above becomes\[ \left ( \int _{\partial \Omega }v\frac{\partial u}{\partial \mathbf{n}}\ dL-\int _{\Omega }v\Delta u\ dA\right ) -\int _{\Omega }fv\ dA=0 \] But \(v=0\) at boundary \(\partial \Omega \). The above simplifies to\begin{align*} -\int _{\Omega }v\Delta u\ dA-\int _{\Omega }fv\ dA & =0\\ \int _{\Omega }v\left ( -\Delta u\ -f\ \right ) dA & =0 \end{align*}
Since the above is true for all \(v\) test function then this implies that \(-\Delta u\ -f\ =0\) or \[ -\Delta u\ =f \] Which is what we wanted to show.
Find the Fourier transform of (f) \(f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}e^{-x}\sin x & & x>0\\ 0 & & x\leq 0 \end{array} \right . \)
Solution
\begin{align} \hat{f}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( x\right ) e^{-ikx}dx\nonumber \\ & =\frac{1}{\sqrt{2\pi }}\int _{0}^{\infty }e^{-x}\sin xe^{-ikx}dx\nonumber \\ & =\frac{1}{\sqrt{2\pi }}\int _{0}^{\infty }\sin xe^{-ikx-x}dx\nonumber \\ & =\frac{1}{\sqrt{2\pi }}\int _{0}^{\infty }\sin xe^{-x\left ( 1+ik\right ) }dx \tag{1} \end{align}
Integration by parts. \(\int udv=uv-\int vdu\). Let \(dv=e^{-x\left ( 1+ik\right ) },v=\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) },u=\sin x,du=\cos x\). Hence\begin{align*} I & =\int _{0}^{\infty }\sin xe^{-x\left ( 1+ik\right ) }dx\\ & =\left [ \sin x\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) }\right ] _{0}^{\infty }-\int _{0}^{\infty }\cos x\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) }dx\\ & =\frac{-1}{1+ik}\left [ \sin xe^{-x\left ( 1+ik\right ) }\right ] _{0}^{\infty }+\frac{1}{1+ik}\int _{0}^{\infty }\cos xe^{-x\left ( 1+ik\right ) }dx \end{align*}
But \(e^{-x\left ( 1+ik\right ) }=e^{-x}e^{-ikx}\) and this goes to zero as \(x\rightarrow \infty \) and since \(\sin x=0\) at \(x=0\) then the first term above is zero. The above reduces to\[ I=\frac{1}{1+ik}\int _{0}^{\infty }\cos xe^{-x\left ( 1+ik\right ) }dx \] Integration by parts. \(\int udv=uv-\int vdu\). Let \(dv=e^{-x\left ( 1+ik\right ) },v=\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) },u=\cos x,du=-\sin x\). The above becomes\begin{align*} I & =\frac{1}{1+ik}\left ( \left [ \cos x\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) }\right ] _{0}^{\infty }-\int _{0}^{\infty }\left ( -\sin x\right ) \frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) }dx\right ) \\ & =\frac{1}{1+ik}\left ( \left [ \cos x\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) }\right ] _{0}^{\infty }-\frac{1}{1+ik}\int _{0}^{\infty }\sin xe^{-x\left ( 1+ik\right ) }dx\right ) \end{align*}
But \(\int _{0}^{\infty }\sin xe^{-x\left ( 1+ik\right ) }dx=I\). The above becomes\begin{align*} I & =\frac{1}{1+ik}\left ( \left [ \cos x\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) }\right ] _{0}^{\infty }-\frac{1}{1+ik}I\right ) \\ & =\frac{1}{1+ik}\left [ \cos x\frac{e^{-x\left ( 1+ik\right ) }}{-\left ( 1+ik\right ) }\right ] _{0}^{\infty }-\left ( \frac{1}{1+ik}\right ) ^{2}I\\ I+\left ( \frac{1}{1+ik}\right ) ^{2}I & =\frac{-1}{\left ( 1+ik\right ) ^{2}}\left [ \cos xe^{-x\left ( 1+ik\right ) }\right ] _{0}^{\infty } \end{align*}
Now \(\left [ \cos xe^{-x\left ( 1+ik\right ) }\right ] _{0}^{\infty }=0-1=-1\). Hence the above reduces to\begin{align*} I\left ( 1+\left ( \frac{1}{1+ik}\right ) ^{2}\right ) & =\frac{1}{\left ( 1+ik\right ) ^{2}}\\ I & =\frac{\frac{1}{\left ( 1+ik\right ) ^{2}}}{1+\left ( \frac{1}{1+ik}\right ) ^{2}}\\ & =\frac{1}{1+\left ( 1+ik\right ) ^{2}}\\ & =\frac{1}{2-k^{2}+2ik} \end{align*}
Therefore \[ \int _{0}^{\infty }\sin xe^{-x\left ( 1+ik\right ) }dx=\frac{1}{2-k^{2}+2ik}\] Using (1) the Fourier transform becomes\[ \hat{f}\left ( k\right ) =\frac{1}{\sqrt{2\pi }}\frac{1}{2-k^{2}+2ik}\] This can be written as real and imaginary parts\begin{align*} \hat{f}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\frac{\left ( 2-k^{2}\right ) -2ik}{\left ( \left ( 2-k^{2}\right ) +2ik\right ) \left ( \left ( 2-k^{2}\right ) -2ik\right ) }\\ & =\frac{1}{\sqrt{2\pi }}\frac{\left ( 2-k^{2}\right ) -2ik}{\left ( 2-k^{2}\right ) ^{2}+4k^{2}}\\ & =\frac{1}{\sqrt{2\pi }}\left ( \frac{2-k^{2}}{k^{4}+4}-i\frac{2k}{k^{4}+4}\right ) \end{align*}
Find the inverse Fourier transform of the function \(\frac{1}{k+c}\) when (a) \(c=a\) is real (b) \(c=ib\) is pure imaginary.
Solution
Using shifting property where \(\mathcal{F}\left [ f\left ( x\right ) \right ] =\hat{f}\left ( k\right ) \) and let \(\hat{f}\left ( k\right ) =\frac{1}{k}\) then by shifting property \(\mathcal{F}\left [ e^{iax}f\left ( x\right ) \right ] =\hat{f}\left ( k-a\right ) \), (Theorem 7.4) therefore\begin{align} \mathcal{F}\left [ e^{-iax}f\left ( x\right ) \right ] & =\hat{f}\left ( k+a\right ) \nonumber \\ & =\frac{1}{k+a} \tag{1} \end{align}
We now just need to find \(f\left ( x\right ) \). From table of Fourier transforms on page 272, we see that \(\mathcal{F}\left [ \operatorname{sgn}\left ( x\right ) \right ] =\frac{1}{i}\sqrt{\frac{2}{\pi }}\frac{1}{k}\). Hence \[\mathcal{F}\left [ i\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \right ] =\frac{1}{k}\] Therefore \(f\left ( x\right ) =i\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \). Substituting this back into (1) gives\[\mathcal{F}\left [ ie^{-iax}\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \right ] =\frac{1}{k+a}\] Or\[\mathcal{F}^{-1}\left [ \frac{1}{k+a}\right ] =ie^{-iax}\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \]
Using shifting property, given that \(\mathcal{F}\left ( f\left ( x\right ) \right ) =\hat{f}\left ( k\right ) \), let \(\hat{f}\left ( k\right ) =\frac{1}{k}\) then by shifting property (Theorem 7.4) \(\mathcal{F}\left [ e^{i(ib)x}f\left ( x\right ) \right ] =\hat{f}\left ( k-ib\right ) \), then\begin{align} \mathcal{F}\left [ e^{bx}f\left ( x\right ) \right ] & =\hat{f}\left ( k+ib\right ) \nonumber \\ & =\frac{1}{k+ib} \tag{1} \end{align}
We now just need to find \(f\left ( x\right ) \). From table of Fourier transforms on page 272, we see that \(\mathcal{F}\left [ \operatorname{sgn}\left ( x\right ) \right ] =\frac{1}{i}\sqrt{\frac{2}{\pi }}\frac{1}{k}\). Hence \[\mathcal{F}\left [ i\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \right ] =\frac{1}{k}\] Therefore \(f\left ( x\right ) =i\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \). Substituting this back into (1) gives\[\mathcal{F}\left [ ie^{bx}\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \right ] =\frac{1}{k+ib}\] Or\[\mathcal{F}^{-1}\left [ \frac{1}{k+ib}\right ] =ie^{bx}\sqrt{\frac{\pi }{2}}\operatorname{sgn}\left ( x\right ) \]
Prove the Shift Theorem 7.4 which is
Theorem 7.4: if \(f\left ( x\right ) \) has Fourier transform \(\hat{f}\left ( k\right ) \), then the Fourier transform of the shifted function \(f\left ( x-\xi \right ) \) is \(e^{-ik\xi }\hat{f}\left ( k\right ) \). Similarly the transform of the product function \(e^{i\alpha x}f\left ( x\right ) \) for real \(\alpha \) is the shifted transform \(\hat{f}\left ( k-\alpha \right ) \) (note: using \(\alpha \) in place of the strange second \(k\) that the book uses)
Showing if \(f\left ( x\right ) \) has Fourier transform \(\hat{f}\left ( k\right ) \), then Fourier transform of the shifted function \(f\left ( x-\xi \right ) \) is \(e^{-ik\xi }\hat{f}\left ( k\right ) \). From definition, the Fourier transform of \(f\left ( x-\xi \right ) \) is given by \[\mathcal{F}\left [ f\left ( x-\xi \right ) \right ] =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( x-\xi \right ) e^{-ikx}dx \] Let \(x-\xi =u\). Then \(\frac{du}{dx}=1\). The above becomes (limits do not change)\begin{align*} \mathcal{F}\left [ f\left ( x-\xi \right ) \right ] & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( u\right ) e^{-ik\left ( u+\xi \right ) }du\\ & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( u\right ) e^{-iku}e^{-ik\xi }du\\ & =e^{-ik\xi }\overset{\hat{f}\left ( k\right ) }{\overbrace{\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( u\right ) e^{-iku}du}} \end{align*}
Therefore\[\mathcal{F}\left [ f\left ( x-\xi \right ) \right ] =e^{-ik\xi }\hat{f}\left ( k\right ) \] Which is what asked to show.
Showing that the Fourier transform of \(e^{i\alpha x}f\left ( x\right ) \) is \(\hat{f}\left ( k-\alpha \right ) \). From definition, the Fourier transform of \(e^{i\alpha x}f\left ( x\right ) \) is\begin{align*} \mathcal{F}\left [ e^{i\alpha x}f\left ( x\right ) \right ] & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{i\alpha x}f\left ( x\right ) e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( x\right ) e^{-ix\left ( k-\alpha \right ) }dx \end{align*}
But \(\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( x\right ) e^{-ix\left ( k-\alpha \right ) }dx\) is \(\hat{f}\left ( k-\alpha \right ) \) by replacing \(k\) with \(k-\alpha \) in the definition of Fourier transform. Hence\[\mathcal{F}\left [ e^{i\alpha x}f\left ( x\right ) \right ] =\hat{f}\left ( k-\alpha \right ) \] Which is what asked to show.
The two-dimensional Fourier transform of a function \(f\left ( x,y\right ) \) defined for \(\left ( x,y\right ) \in \mathbb{R} ^{2}\) is \begin{align*} \mathcal{F}\left [ f\left ( x,y\right ) \right ] & =\hat{f}\left ( k,l\right ) \\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f\left ( x,y\right ) e^{-i\left ( kx+ly\right ) }dxdy \end{align*}
(a) compute the Fourier transform of the following functions (i) \(e^{-\left \vert x\right \vert -\left \vert y\right \vert }\), (iii) The delta function \(\delta \left ( x-\xi \right ) \delta \left ( y-\eta \right ) \)
(b) Show that if \(f\left ( x,y\right ) =g\left ( x\right ) h\left ( y\right ) \) then \(\hat{f}\left ( k,l\right ) =\hat{g}\left ( k\right ) \hat{h}\left ( l\right ) \)
Solution
(i) The Fourier transform of \(e^{-\left \vert x\right \vert -\left \vert y\right \vert }\) is\begin{align} \hat{f}\left ( k,l\right ) & =\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-\left \vert x\right \vert -\left \vert y\right \vert }e^{-i\left ( kx+ly\right ) }dxdy\nonumber \\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-\left \vert x\right \vert }e^{-\left \vert y\right \vert }e^{-ikx}e^{-ily}dxdy\nonumber \\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }e^{-\left \vert y\right \vert }e^{-ily}\left ( \int _{-\infty }^{\infty }e^{-\left \vert x\right \vert }e^{-ikx}dx\right ) dy \tag{1} \end{align}
But \(\int _{-\infty }^{\infty }e^{-\left \vert x\right \vert }e^{-ikx}dx\) is the Fourier transform of \(f\left ( x\right ) =e^{-\left \vert x\right \vert }\) with \(\sqrt{2\pi }\) factor. In other words\[ \int _{-\infty }^{\infty }e^{-\left \vert x\right \vert }e^{-ikx}dx=\sqrt{2\pi }\hat{g}\left ( k\right ) \] Where \(\hat{g}\left ( k\right ) \) is used to indicate the Fourier transform of \(e^{-\left \vert x\right \vert }\). Hence (1) becomes\[ \hat{f}\left ( k,l\right ) =\frac{\sqrt{2\pi }}{2\pi }\hat{f}_{1}\left ( k\right ) \int _{-\infty }^{\infty }e^{-\left \vert y\right \vert }e^{-ily}dy \] But \(\int _{-\infty }^{\infty }e^{-\left \vert y\right \vert }e^{-ily}dy=\sqrt{2\pi }\hat{h}\left ( l\right ) \) Where \(\hat{h}\left ( l\right ) \) is used to indicate the Fourier transform of \(e^{-\left \vert y\right \vert }\). The above becomes\begin{align} \hat{f}\left ( k,l\right ) & =\frac{\sqrt{2\pi }}{2\pi }\hat{g}\left ( k\right ) \sqrt{2\pi }\hat{h}\left ( l\right ) \nonumber \\ & =\hat{g}\left ( k\right ) \hat{h}\left ( l\right ) \tag{2} \end{align}
So now we need to determine \(\hat{g}\left ( k\right ) \) and \(\hat{h}\left ( l\right ) \) and multiply the result. \begin{align*} \hat{g}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-\left \vert x\right \vert }e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi }}\left ( \int _{-\infty }^{0}e^{x}e^{-ikx}dx+\int _{0}^{\infty }e^{-x}e^{-ikx}dx\right ) \\ & =\frac{1}{\sqrt{2\pi }}\left ( \int _{-\infty }^{0}e^{-ikx+x}dx+\int _{0}^{\infty }e^{-ikx-x}dx\right ) \\ & =\frac{1}{\sqrt{2\pi }}\left ( \left [ \frac{e^{-ikx+x}}{1-ik}\right ] _{-\infty }^{0}+\left [ \frac{e^{-ikx-x}}{-1-ik}\right ] _{0}^{\infty }\right ) \\ & =\frac{1}{\sqrt{2\pi }}\left ( \frac{1}{1-ik}\left [ e^{-ikx}e^{x}\right ] _{-\infty }^{0}-\frac{1}{1+ik}\left [ e^{-ikx}e^{-x}\right ] _{0}^{\infty }\right ) \\ & =\frac{1}{\sqrt{2\pi }}\left ( \frac{1}{1-ik}\left ( 1-0\right ) -\frac{1}{1+ik}\left ( 0-1\right ) \right ) \\ & =\frac{1}{\sqrt{2\pi }}\left ( \frac{1}{1-ik}+\frac{1}{1+ik}\right ) \\ & =\frac{1}{\sqrt{2\pi }}\left ( \frac{\left ( 1+ik\right ) +\left ( 1-ik\right ) }{\left ( 1-ik\right ) \left ( 1+ik\right ) }\right ) \\ & =\frac{1}{\sqrt{2\pi }}\left ( \frac{2}{1+k^{2}}\right ) \\ & =\sqrt{\frac{2}{\pi }}\frac{1}{1+k^{2}} \end{align*}
Similarly\begin{align*} \hat{h}\left ( l\right ) & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-\left \vert y\right \vert }e^{-ily}dy\\ & =\sqrt{\frac{2}{\pi }}\frac{1}{1+l^{2}} \end{align*}
Hence from (2) the Fourier transform of \(e^{-\left \vert x\right \vert -\left \vert y\right \vert }\) is \begin{align*} \hat{f}\left ( k,l\right ) & =\hat{g}\left ( k\right ) \hat{h}\left ( l\right ) \\ & =\sqrt{\frac{2}{\pi }}\frac{1}{1+k^{2}}\sqrt{\frac{2}{\pi }}\frac{1}{1+l^{2}}\\ & =\frac{2}{\pi }\frac{1}{\left ( 1+k^{2}\right ) \left ( 1+l^{2}\right ) } \end{align*}
(ii) The Fourier transform of \(\delta \left ( x-\xi \right ) \delta \left ( y-\eta \right ) \). First we find the Fourier transform of \(\delta \left ( x-\xi \right ) \) and then the Fourier transform of \(\delta \left ( y-\eta \right ) \)\begin{align*} \hat{g}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }\delta \left ( x-\xi \right ) e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi }}e^{-ik\xi } \end{align*}
And\begin{align*} \hat{h}\left ( l\right ) & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }\delta \left ( y-\eta \right ) e^{-ily}dy\\ & =\frac{1}{\sqrt{2\pi }}e^{-il\eta } \end{align*}
Hence the Fourier transform of the product \(\delta \left ( x-\xi \right ) \delta \left ( y-\eta \right ) \) is (Using the product rule, which will be proofed in part b also).\begin{align*} \hat{f}\left ( k,l\right ) & =\hat{g}\left ( k\right ) \hat{h}\left ( l\right ) \\ & =\frac{1}{2\pi }e^{-ik\xi }e^{-il\eta } \end{align*}
The above could be rewritten in terms of trig functions using Euler relation if needed.
By definition, the Fourier transform of \(f\left ( x,y\right ) \) is\[ \hat{f}\left ( k,l\right ) =\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f\left ( x,y\right ) e^{-i\left ( kx+ly\right ) }dxdy \] But \(f\left ( x,y\right ) =g\left ( x\right ) h\left ( y\right ) \). Hence the above becomes\begin{align*} \hat{f}\left ( k,l\right ) & =\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g\left ( x\right ) h\left ( y\right ) e^{-i\left ( kx+ly\right ) }dxdy\\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g\left ( x\right ) h\left ( y\right ) e^{-ikx}e^{-ily}dxdy\\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }h\left ( y\right ) e^{-ily}\left ( \int _{-\infty }^{\infty }g\left ( x\right ) e^{-ikx}dx\right ) dy \end{align*}
But \(\int _{-\infty }^{\infty }g\left ( x\right ) e^{-ikx}dx=\sqrt{2\pi }\hat{g}\left ( k\right ) \). The above reduces to\[ \hat{f}\left ( k,l\right ) =\frac{1}{2\pi }\sqrt{2\pi }\hat{g}\left ( k\right ) \int _{-\infty }^{\infty }h\left ( y\right ) e^{-ily}dy \] But \(\int _{-\infty }^{\infty }h\left ( y\right ) e^{-ily}dy=\sqrt{2\pi }\hat{h}\left ( l\right ) \). Hence the above becomes\begin{align*} \hat{f}\left ( k,l\right ) & =\frac{1}{2\pi }\sqrt{2\pi }\hat{g}\left ( k\right ) \sqrt{2\pi }\hat{h}\left ( l\right ) \\ & =\hat{g}\left ( k\right ) \hat{h}\left ( l\right ) \end{align*}
Which is what asked to show.
Find the Fourier transform of (a) the error function \(\operatorname{erf}\left ( x\right ) =\frac{2}{\sqrt{\pi }}\int _{0}^{x}e^{-z^{2}}dz\)
Solution
Using\begin{equation} 1+\operatorname{erf}\left ( x\right ) =\frac{2}{\sqrt{\pi }}\int _{-\infty }^{x}e^{-z^{2}}dz \tag{1} \end{equation} Taking Fourier transform of both sides, and using the known relation from tables which says \[\mathcal{F}\left [ \int _{-\infty }^{x}f\left ( u\right ) du\right ] =\frac{1}{ik}\hat{f}\left ( k\right ) +\pi \hat{f}\left ( 0\right ) \delta \left ( k\right ) \] And using that Fourier transform of \(1\) is \(\sqrt{2\pi }\delta \left ( k\right ) \) then (1) becomes\[ \sqrt{2\pi }\delta \left ( k\right ) +\mathcal{F}\left [ \operatorname{erf}\left ( x\right ) \right ] =\frac{2}{\sqrt{\pi }}\left ( \frac{1}{ik}\hat{f}\left ( k\right ) +\pi \hat{f}\left ( 0\right ) \delta \left ( k\right ) \right ) \] Where \(\hat{f}\left ( k\right ) \) is the Fourier transform of \(e^{-u^{2}}\) (Gaussian) we derived in class as \(e^{-u^{2}}\Leftrightarrow \frac{1}{\sqrt{2}}e^{\frac{-k^{2}}{4}}\). The above becomes\begin{align*} \sqrt{2\pi }\delta \left ( k\right ) +\mathcal{F}\left [ \operatorname{erf}\left ( x\right ) \right ] & =\frac{2}{\sqrt{\pi }}\left ( \frac{1}{ik}\frac{1}{\sqrt{2}}e^{\frac{-k^{2}}{4}}+\pi \left [ \frac{1}{\sqrt{2}}e^{\frac{-k^{2}}{4}}\right ] _{k=0}\delta \left ( k\right ) \right ) \\ & =\frac{2}{\sqrt{\pi }}\left ( \frac{1}{ik}\frac{1}{\sqrt{2}}e^{\frac{-k^{2}}{4}}+\frac{\pi }{\sqrt{2}}\delta \left ( k\right ) \right ) \\ & =\frac{2}{\sqrt{\pi }}\frac{1}{ik}\frac{1}{\sqrt{2}}e^{\frac{-k^{2}}{4}}+\sqrt{2\pi }\delta \left ( k\right ) \end{align*}
Therefore the above simplifies to\begin{align*} \mathcal{F}\left [ \operatorname{erf}\left ( x\right ) \right ] & =\frac{2}{\sqrt{\pi }}\frac{1}{ik}\frac{1}{\sqrt{2}}e^{\frac{-k^{2}}{4}}\\ & =\sqrt{\frac{2}{\pi }}\frac{1}{ik}e^{\frac{-k^{2}}{4}}\\ & =-i\sqrt{\frac{2}{\pi }}\frac{1}{k}e^{\frac{-k^{2}}{4}} \end{align*}
Find the inverse Fourier transform of the following functions (d) \(\frac{k^{2}}{k-i}\)
Solution
Using property that \begin{align} \mathcal{F}\left [ f^{\prime }\left ( x\right ) \right ] & =ik\hat{f}\left ( k\right ) \nonumber \\\mathcal{F}\left [ f^{\prime \prime }\left ( x\right ) \right ] & =-k^{2}\hat{f}\left ( k\right ) \tag{1} \end{align}
Where in the above \(\mathcal{F}\left [ f\left ( x\right ) \right ] =\hat{f}\left ( k\right ) \). Comparing the above with \(\frac{k^{2}}{k-i}\), we see that \[ \hat{f}\left ( k\right ) =\frac{1}{k-i}\] Hence we need to find inverse Fourier transform of \(\frac{-1}{k-i}\) first in order to find \(f\left ( x\right ) \), and then take second derivative of the result. Writing \begin{align*} \frac{1}{k-i} & =\frac{1}{i\left ( \frac{k}{i}-1\right ) }\\ & =\frac{1}{i\left ( -ik-1\right ) }\\ & =\frac{-1}{i\left ( ik+1\right ) }\\ & =i\frac{1}{\left ( 1+ik\right ) } \end{align*}
From table (page 272 in textbook) we see that \[\mathcal{F}^{-1}\left [ \frac{1}{\left ( ik+1\right ) }\right ] =\sqrt{2\pi }e^{-x}\sigma \left ( x\right ) \] Using \(a=1\) in the table entry. Where \(\sigma \left ( x\right ) \) is the step function. Hence \[ i\mathcal{F}^{-1}\left [ \frac{1}{\left ( ik+1\right ) }\right ] =i\sqrt{2\pi }e^{-x}\sigma \left ( x\right ) \] Therefore \[ f\left ( x\right ) =i\sqrt{2\pi }e^{-x}\sigma \left ( x\right ) \] Now we take derivative of the above (using product rule)\[ f^{\prime }\left ( x\right ) =-i\sqrt{2\pi }e^{-x}\sigma \left ( x\right ) +i\sqrt{2\pi }e^{-x}\delta \left ( x\right ) \] Where \(\delta \left ( x\right ) \) is added since derivative of \(\sigma \left ( x\right ) \) has jump discontinuity at \(x=0\). Taking one more derivative gives\begin{align*} f^{\prime \prime }\left ( x\right ) & =i\sqrt{2\pi }e^{-x}\sigma \left ( x\right ) -i\sqrt{2\pi }e^{-x}\delta \left ( x\right ) -i\sqrt{2\pi }e^{-x}\delta \left ( x\right ) +i\sqrt{2\pi }e^{-x}\delta ^{\prime }\left ( x\right ) \\ & =i\sqrt{2\pi }e^{-x}\sigma \left ( x\right ) -2i\sqrt{2\pi }e^{-x}\delta \left ( x\right ) +i\sqrt{2\pi }e^{-x}\delta ^{\prime }\left ( x\right ) \end{align*}
Therefore\[\mathcal{F}^{-1}\left [ \frac{k^{2}}{k-i}\right ] =i\sqrt{2\pi }e^{-x}\sigma \left ( x\right ) -2i\sqrt{2\pi }e^{-x}\delta \left ( x\right ) +i\sqrt{2\pi }e^{-x}\delta ^{\prime }\left ( x\right ) \]
(a) Explain why the Fourier transform of a \(2\pi \) periodic function \(f\left ( x\right ) \) is a linear combinations of delta functions \(\hat{f}\left ( k\right ) =\sum _{n=-\infty }^{\infty }c_{n}\delta \left ( k-n\right ) \) where \(c_{n}\) are the complex Fourier series coefficients (3.65) of \(f\left ( x\right ) \) on \(\left [ -\pi ,\pi \right ] \)\begin{equation} c_{n}=\left \langle f,e^{inx}\right \rangle =\frac{1}{2\pi }\int _{-\pi }^{\pi }f\left ( x\right ) e^{-inx}dx \tag{3.65} \end{equation} (b) Find the Fourier transform of the following periodic functions (i) \(\sin 2x\) (ii) \(\cos ^{3}x\) (iii) The \(2\pi \) periodic extension of \(f\left ( x\right ) =x\) (iv) The sawtooth function \(h\left ( x\right ) =x\operatorname{mod}1\). i.e. the fractional part of \(x\)
Solution
Since \(f\left ( x\right ) \) is periodic, then its can be expressed as\[ f\left ( x\right ) =\sum _{n=-\infty }^{\infty }c_{n}e^{in\left ( \frac{2\pi }{T}\right ) x}\] But the period \(T=2\pi \) and the above simplifies to\begin{equation} f\left ( x\right ) =\sum _{n=-\infty }^{\infty }c_{n}e^{inx} \tag{1} \end{equation} Taking the Fourier transform of the above gives\begin{equation} \hat{f}\left ( k\right ) =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }f\left ( x\right ) e^{-ikx}dx \tag{2} \end{equation} Substituting (1) into (2) gives\begin{align*} \hat{f}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }\left ( \sum _{n=-\infty }^{\infty }c_{n}e^{inx}\right ) e^{-ikx}dx\\ & =\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }\left ( \sum _{n=-\infty }^{\infty }c_{n}e^{-ix\left ( k-n\right ) }\right ) dx \end{align*}
Changing the order of summation and integration\begin{align} \hat{f}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\sum _{n=-\infty }^{\infty }\left ( \int _{-\infty }^{\infty }c_{n}e^{-ix\left ( k-n\right ) }dx\right ) \nonumber \\ & =\frac{1}{\sqrt{2\pi }}\sum _{n=-\infty }^{\infty }c_{n}\left ( \int _{-\infty }^{\infty }e^{-ix\left ( k-n\right ) }dx\right ) \tag{3} \end{align}
But from tables we know that \(\mathcal{F}\left ( 1\right ) =\sqrt{2\pi }\delta \left ( k\right ) \). Which means that \[ \frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-ixk}dx=\sqrt{2\pi }\delta \left ( k\right ) \] Therefore, replacing \(k\) by \(k-n\) in the above gives\begin{align} \frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{-ix\left ( k-n\right ) }dx & =\sqrt{2\pi }\delta \left ( k-n\right ) \nonumber \\ \int _{-\infty }^{\infty }e^{-ix\left ( k-n\right ) }dx & =\left ( 2\pi \right ) \delta \left ( k-n\right ) \tag{4} \end{align}
Substituting (4) into (3) gives\begin{align*} \hat{f}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\sum _{n=-\infty }^{\infty }c_{n}\left ( 2\pi \right ) \delta \left ( k-n\right ) \\ & =\sqrt{2\pi }\sum _{n=-\infty }^{\infty }c_{n}\delta \left ( k-n\right ) \end{align*}
Note: The books seems to have a typo. It gives the above without the factor \(\sqrt{2\pi }\) at the front.
(i) \(\sin 2x\). Since this is periodic, then \(c_{n}=\frac{1}{2\pi }\int _{-\pi }^{\pi }\sin \left ( 2x\right ) e^{-inx}dx\). For \(n=2\) this gives \(c_{2}=-\frac{i}{2}\) and for \(n=-2\) it gives \(c_{-2}=\frac{i}{2}\) and it is zero for all other \(n\) values due to orthogonality of \(\sin \) functions. Using the above result obtained in part (a)\begin{align*} \hat{f}\left ( k\right ) & =\sqrt{2\pi }\sum _{n=-\infty }^{\infty }c_{n}\delta \left ( k-n\right ) \\ & =\sqrt{2\pi }c_{-2}\delta \left ( k+2\right ) +\sqrt{2\pi }c_{2}\delta \left ( k-2\right ) \\ & =\sqrt{2\pi }\frac{i}{2}\delta \left ( k+2\right ) -\sqrt{2\pi }\frac{i}{2}\delta \left ( k-2\right ) \\ & =i\sqrt{\frac{\pi }{2}}\delta \left ( k+2\right ) -i\sqrt{\frac{\pi }{2}}\delta \left ( k-2\right ) \end{align*}
(ii) \(\cos ^{3}x\). Since this is periodic, then \(c_{n}=\frac{1}{2\pi }\int _{-\pi }^{\pi }\cos ^{3}\left ( x\right ) e^{-inx}dx\). But \(\cos ^{3}\left ( x\right ) =\frac{1}{4}\cos \left ( 3x\right ) +\frac{3}{4}\cos \left ( x\right ) \). Hence only \(n=\pm 1,n=\pm 3\) will have coefficients and the rest are zero. \begin{align*} c_{-1} & =\frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{3}{4}\cos \left ( x\right ) e^{ix}dx=\frac{3}{8}\\ c_{1} & =\frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{3}{4}\cos \left ( x\right ) e^{-ix}dx=\frac{3}{8}\\ c_{-3} & =\frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{1}{4}\cos \left ( 3x\right ) e^{-3ix}dx=\frac{1}{8}\\ c_{3} & =\frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{1}{4}\cos \left ( 3x\right ) e^{-3ix}dx=\frac{1}{8} \end{align*}
Therefore, using result from part (a)\begin{align*} \hat{f}\left ( k\right ) & =\sqrt{2\pi }\sum _{n=-\infty }^{\infty }c_{n}\delta \left ( k-n\right ) \\ & =\sqrt{2\pi }\left ( \frac{1}{8}\delta \left ( k+3\right ) +\frac{3}{8}\delta \left ( k+1\right ) +\frac{3}{8}\delta \left ( k-1\right ) +\frac{1}{8}\delta \left ( k-3\right ) \right ) \\ & =\frac{1}{4}\sqrt{\frac{\pi }{2}}\left ( \delta \left ( k+3\right ) +3\delta \left ( k+1\right ) +3\delta \left ( k-1\right ) +\delta \left ( k-3\right ) \right ) \end{align*}
(iii) The \(2\pi \) periodic extension of \(f\left ( x\right ) =x\)
Since this is periodic, then \begin{align*} c_{n} & =\frac{1}{2\pi }\int _{-\pi }^{\pi }xe^{-inx}dx\\ & =\frac{2i}{n^{2}}\left ( n\pi \cos \left ( n\pi \right ) -\sin \left ( n\pi \right ) \right ) \\ & =\frac{2i}{n^{2}}\left ( n\pi \left ( -1\right ) ^{n}\right ) \\ & =\frac{2i}{n}\pi \left ( -1\right ) ^{n} \end{align*}
Therefore, using result from part (a)\begin{align*} \hat{f}\left ( k\right ) & =\sqrt{2\pi }\sum _{n=-\infty }^{\infty }c_{n}\delta \left ( k-n\right ) \\ & =\sqrt{2\pi }\sum _{n=-\infty }^{\infty }\frac{2i}{n}\pi \left ( -1\right ) ^{n}\delta \left ( k-n\right ) \\ & =2i\pi \sqrt{2\pi }\sum _{n=-\infty }^{\infty }\frac{\left ( -1\right ) ^{n}}{n}\delta \left ( k-n\right ) \qquad n\neq 0 \end{align*}
(iv) The sawtooth function
Find a solution to the differential equation \(-\frac{d^{2}u}{dx^{2}}+4u=\delta \left ( x\right ) \) by using the Fourier transform
Solution
Taking Fourier transform of both sides gives\begin{align*} -\left ( ik\right ) ^{2}\hat{u}\left ( k\right ) +4\hat{u}\left ( k\right ) & =\mathcal{F}\left [ \delta \left ( x\right ) \right ] \\ k^{2}\hat{u}\left ( k\right ) +4\hat{u}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }} \end{align*}
Solving for \(\hat{u}\left ( k\right ) \)\begin{align*} \hat{u}\left ( k\right ) \left ( k^{2}+4\right ) & =\frac{1}{\sqrt{2\pi }}\\ \hat{u}\left ( k\right ) & =\frac{1}{\sqrt{2\pi }}\frac{1}{k^{2}+4} \end{align*}
Finding inverse Fourier transform. From tables we see that \(\mathcal{F}\left ( e^{-a\left \vert x\right \vert }\right ) =\sqrt{\frac{2}{\pi }}\frac{a}{k^{2}+a^{2}}\). Using \(a=2\,\)\begin{align*} \mathcal{F}\left [ e^{-2\left \vert x\right \vert }\right ] & =\sqrt{\frac{2}{\pi }}\frac{2}{k^{2}+4}\\ \sqrt{\frac{\pi }{2}}\frac{1}{2}\mathcal{F}\left [ e^{-2\left \vert x\right \vert }\right ] & =\frac{1}{k^{2}+4}\\ \sqrt{\frac{\pi }{2}}\mathcal{F}\left [ \frac{1}{2}e^{-2\left \vert x\right \vert }\right ] & =\frac{1}{k^{2}+4}\\ \frac{1}{\sqrt{2\pi }}\sqrt{\frac{\pi }{2}}\mathcal{F}\left [ \frac{1}{2}e^{-2\left \vert x\right \vert }\right ] & =\frac{1}{\sqrt{2\pi }}\frac{1}{k^{2}+4}\\ \frac{1}{2}\mathcal{F}\left [ \frac{1}{2}e^{-2\left \vert x\right \vert }\right ] & =\frac{1}{\sqrt{2\pi }}\frac{1}{k^{2}+4}\\\mathcal{F}\left [ \frac{1}{4}e^{-2\left \vert x\right \vert }\right ] & =\frac{1}{\sqrt{2\pi }}\frac{1}{k^{2}+4} \end{align*}
Therefore \[ u\left ( x\right ) =\frac{1}{4}e^{-2\left \vert x\right \vert }\]