Lecture 17 Phys 3750 D M Riffe -1- 2/25/2013 Fourier Transforms and the Wave Equation Overview and Motivation: We first discuss a few features of the Fourier transform (FT), and then we solve the initial-value problem for the wave equation using the Fourier transform. Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation. I. FT Change of Notation In the last lecture we introduced the FT of a function ()xf through the two equations () ()∫∞∞−= dkekfxfikxˆ21π, (1a) () ()∫∞∞−−= dxexfkfikxπ21ˆ. (1b) Note that we have changed notation compared to the last lecture. Hereafter we designate the FT of any function by the same symbol, but with an overhead caret included. That is, the FT of ()xf we now write as ()kfˆ. As we shall see, this is useful when dealing with equations that include FTs of several functions. 1 II. Some Properties of the Fourier Transform We now discuss several useful properties of the Fourier transform. A. Translation The first property has to do with translation of the function ()xf . Let's say we are interested in ()0xxf − , which corresponds to translation of ()xf by 0x . Then, using Eq. (1a) we can write () ()()()[]∫∫∞∞−−∞∞−−==−dkeekfdkekfxxfikxikxxxik00ˆ21ˆ210ππ (2) 1 This notation is fairly common practice. At some point you may even see the FT of ()xf written as ()kf . At least we won't be doing that here!Lecture 17 Phys 3750 D M Riffe -2- 2/25/2013 Thus, we see that the FT of ()0xxf− is ()0ˆikxekf−. In other words, translation of ()xf by 0x corresponds to multiplying the FT ()kfˆ by 0ikxe−. B. Differentiation The second property has to do with the FT of ()xf′, the derivative of ()xf . Again, using Eq. (1a) we have () ()[]∫∞∞−=′dkekfikxfikxˆ21π. (3) So we see that FT of ()xf′ is ()kfikˆ. That is, differentiation of ()xf corresponds to multiplying ()kfˆ by ik . C. Integration Let's consider the definite integral of ()xf , () ()∫∫∫=∞∞−2121ˆ21xxikxxxekfdkdxxfdxπ. (4) Switching the order of integration on the rhs produces () ()()()∫∫∫∫∞∞−∞∞−−==122121ˆ21ˆ21ikxikxxxikxxxeeikkfdkedxkfdkxfdxππ. (5) So if we define ()xIf to be the indefinite integral of ()xf , we can rewrite Eq. (5) as () ()()()∫∞∞−−=−12ˆ2112ikxikxeeikkfdkxIfxIfπ (6)Lecture 17 Phys 3750 D M Riffe -3- 2/25/2013 So integration of ()xf essentially corresponds to dividing the Fourier transform ()kfˆ by ik .2 D. Convolution The last property concerning the a function and its FT has to do with convolution. Because you may not be familiar with convolution, let's first define it. Simply put, the convolution of two functions ()xf and ()xg , which we denote ()( )xgf * , is defined as ()() ()()∫∞∞−′′′−=xdxgxxfxgf * (7) Perhaps the most common place that convolution arises is in spectroscopy, where ()xg is some intrinsic spectrum that is being measured, and ()xf is the resolution function of the spectrometer that is being used to measure the spectrum.3 The convolution ()()xgf * is the spectrum that is then measured. Note that ()()xgf * is indeed a function of x, and so we can calculate its FT, which we denote ()kgf )*ˆ(. Using Eq. (1b) we can write () ()()∫∫∞∞−−∞∞−′′−′=ikxexgxxfxddxkgfπ21)*ˆ(, (8) which can be rearranged as () ( ) ()∫∫∞∞−∞∞−−′−′′′=ikxexxfxdxgxdkgfπ21)*ˆ(. (9) Now the quantity in brackets is the FT of ()xf translated by x′. From Sec. II.B above we know that this is ()xikekf′−ˆ , and so Eq. (9) can be expressed as 2 You might think that Eq. (6) could be simplified to ()()∫∞∞−=ikxeikkfdkxIfˆ21π, but this cannot be done because indefinite integration produces an undetermined integration constant. The constant does not appear in Eq. (6) because it is an equation for the difference of ()2xIf and ()1xIf . 3 The resolution function is often quite close to a Gaussian of a particular, fixed width.Lecture 17 Phys 3750 D M Riffe -4- 2/25/2013 () () ( )∫∞∞−′−′′=xikexgxdkfkgfˆ)*ˆ( . (10) Recognizing the integral as ()kgˆ2π we finally have () ()()kgkfkgfˆˆ2)*ˆ(π= . (11) So we see that the FT of the convolution is the product of the FT's of the individual functions (along with a factor of π2 ). One way you may hear this result expressed is that convolution in real space (x) corresponds to multiplication in k space. Equation (11) is known as the convolution theorem. III. Solution to the Wave Equation Initial Value Problem Way back in Lecture 8 we discussed the initial value problem for the wave equation () ()22222,,xtxqcttxq∂∂=∂∂ (12) on the interval ∞<<∞−x. For the initial conditions () ()xaxq =0, , (13a) () ()xbxtq=∂∂0, , (13b) we found that the solution to Eq. (12) can be written as () ()() ()′′+−++=∫+−ctxctxxdxbcctxactxatxq121,. (14) With the help of the Fourier transform we are now going to rederive this solution, and along the way we will learn something very interesting about the FT of ()txq , . We start by defining the (spatial) FT of ()txq , as () ()∫∞∞−−= dxetxqtkqikx,21,ˆπ, (15a)Lecture 17 Phys 3750 D M Riffe -5- 2/25/2013 so that we also have () ()∫∞∞−= dketkqtxqikx,ˆ21,π. (15b) We also define the FT of Eq. (13), the initial conditions, () ()kakqˆ0,ˆ=, (16a) () ()kbktqˆ0,ˆ=∂∂. (16b) Now each side of the Eq. (12) is a function of x and t , so we can calculate the FT of both sides of Eq (12), () ()∫∫∞∞−−∞∞−−∂∂=∂∂dxextxqcdxettxqikxikx22222,, (17) On the lhs of this equation we can pull the time derivative outside the integral. The lhs is then just the second time derivative of ()tkq ,ˆ. The rhs can be simplified by remembering that the FT of the (x) derivative of a function is ik times the FT of the original function. Thus the FT of ()22, xtxq ∂∂ is just 2k− times ()tkq ,ˆ, the FT of ()txq , . Thus we can rewrite Eq. (17) as ()()tkqckttkq,ˆ,ˆ2222−=∂∂ (18) This equation should look very familiar to you. What equation is it? None other than the harmonic oscillator equation! What does this tell us about ()tkq ,ˆ? It tells us that ()tkq ,ˆ