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 }