Lecture 23: Fourier Transforms and Other Integral Transforms

LECTURE 23Fourier Transforms and Other Integral Transforms1. Integral TransformsSeveral times in this course we encountered formulas of the sort∫(1) g (s) = f (t) K (s, t) dtFor example, in the last lecture we utilized the the Laplace transform of a function isL [f] (x) :=∫ ∞0f (t) e −st dtand even earlier in our discussion of Fourier series we utilized Fourier transforms viaF (n) = 2 π∫ π0f (t) cos (nt) dtWe refer to a formula of the general form (1) as an integral transform. The function g (s) is called the(integral) transform of f (t) by the kernel K (s, t) .A particularly good way to think about such transforms is that they are analogous to a linear transformaionbetween two vector spaces: Suppose T : R n → R m is a linear transformation between the spaceof n-dimensional vectors and the space of m-dimensional vectors. Then the transformation is completelydetermined by its action on the standard basis (or any basis) of R n . Supposem∑(2) T (e i ) = A ji f jThinking of the “components” of a function f as being the set of its values f (t) at points t, and thinkingof replacing the discrete sum over components in (2) by a continuous integral over the components of f,the analogy should now be clear. It is worth remarking that the similarity between formulas (1) and (2)is mitigated by much more that simply reinterpreting a sum as an integral. The real point is that (1), like(2) is a linear transformation between two vector spaces.Also, just as in linear algebra where one uses linear transformations to transport a problem about anabstract vector space (e.g. a set of polynomial functions) to a concrete setting like R n in order to carry outcomputations, one can use integral transforms to transport problems, in particular differential equations,to a setting where they’re much easier to solve. The general scheme is something like thisoriginal problemsolution to original problemj=1integral transform−−−−−−−−−−−−−−−−→inverse integral transform←−−−−−−−−−−−−−−−−−−−−−−problem in transform space↓ relatively easy solutionsolution in transform spaceBy far the most useful integral transform is the Fourier transform of a functionF (ν) = √ 1 ∫ +∞f (t) e iνt dt2π−∞106

2. THE FOURIER TRANSFORM 107In many applications this is generally thought of as a way of identifying the frequency components of atime-varying function; but the utility of this transform really extends far beyond such applications. Twoother common integral transforms are(Fourier-Bessel) g (α) =(Mellin transform) g (α) =∫ ∞0∫ ∞0f (t) J n (αt) dtf (t) t α−1 dt2. The Fourier TransformThe big advantage of the Laplace transform is that it can be used to reduce a differential equation to analgebraic equation. However, in practice the Laplace transform also comes with a disadvantage, the inverseLaplace transform taking the solution in the transform space back to the original space is often difficult tocompute.Recall that when we were utilizing Fourier series, we were using formulasf (x) = a ∞ 02 + ∑ ( nπx) ∞∑ ( nπxa n cos + b n sinLLn=1n=1a n = 1 ∫ L( ) nπtf (t) cos dtL −LLb n = 1 ∫ L( ) nπtf (t) sin dtLL−Lto expand functions on the integral [−L, L] in terms of functions. Let’s consider what happens with theseformulas as L → ∞. We havef (x) = 1 ∫ Lf (t) dt + 1 ∞∑∫ L( ) nπt( nπx)f (t) cos cos dt2L −L Ln=1 −LL L+ 1 ∞∑∫ L( ) nπt( nπx)f (t) sin sin dtLn=1 −LL L= 1 ∫ L∞∑∫1 L ( nπ)f (t) dt +f (t) cos2L −LL −L L (t − x) dtn=1Now if f (t) is an integrable function on the real line; i.e.,∫ Lf (t) dt∣ ∣ < ∞ThenOn the other hand, if we setlimL→∞−L∫1 Lf (t) dt = 02L −Lω = nπ L)then∆w = π L∞∑n=1∫1 L ( nπ)f (t) cosL −L L (t − x) dt =∞∑n=1∆wπ∫ L−Lf (t) cos (w (t − x)) dt

2. THE FOURIER TRANSFORM 108and so when we let L → ∞, ∆w → 0 and the expression on the right, interpreted as a Riemann sum over∆w, becomes∫1 ∞ ∫ ∞dω f (t) cos (ω (t − x)) dxπ 0 −∞We can thus infer that, in the limit L → ∞, the Fourier expansion formula (1) becmesf (x) = 1 π∫ ∞0dω∫ ∞−∞f (t) cos (ω (t − x)) dxNow since f (t) cos (ω (t − x)) is a even function of ω we can rewrite (2) asf (x) = 12π∫ ∞−∞dω∫ ∞−∞f (t) cos (ω (t − x)) dxand because f (t) sin (ω (t − x)) is an odd function of ω, we also haveIf we now usewe can write0 = 12π∫ ∞−∞dω∫ ∞−∞f (t) sin (ω (t − x)) dxe iy = cos (y) + i sin (y)f (x) = 1 ∫ ∞ ∫ ∞dω f (t) e i(t−x)ω dx2π −∞ −∞because the integral over the odd part of e i(t−x)ω will not contribute to the total integral.Reversing the order of integration we obtain the formulaf (x) = 1 ∫ ∞(∫ ∞)f (t) e i(t−x)ω dω dt2π −∞−∞which suggests that12π∫ +∞−∞e i(t−x)ω dω = δ (t − x)On the other hand, suppose we define the Fourier Transform of f to beF [f] (ω) = √ 1 ∫ +∞f (x) e −ixω dx2π−∞and the Inverse Fourier Transform of a function g (ω) to beThenForF −1 [F [f]] (x) =F −1 [g] (x) = 1 √2π∫ ∞−∞F −1 [F [f]] (x) = f (x)= 12π=∫1 ∞√2π−∞∫ ∞−∞∫ ∞−∞= f (x)and so F and F −1 are truly inverses of each other.[∫ ∞f (t)g (w) e iωt dω]f (t) e −iωt dt e iωx dω−∞[∫ ∞]e iω(x−t) dω dt−∞f (t) δ (x − t) dt

3. THE WAVE AND EQUATION AND FOURIER TRANSFORMS 1093. The Wave and Equation and Fourier TransformsConsiderφ tt − c 2 φ xx = 0φ (x, 0) = f (x)φ t (x, 0) = 0Suppose we take the Fourier transforms of this PDE with respect to x :ThenΦ (ω, t) := F x [φ] =∫ +∞−∞φ (x, t) e −iωx dx[ ] ∫ ∂φ ∞∂ 2 φF x∂x 2 =−∞ ∂x 2 e−iωx dxIntegrating by parts twice and assuming lim x→±∞ φ (x, 0) = 0 (which is reasonable for any finite energysolution) so that the contributions from the boundaries vanish, we have[ ] ∫ ∂φ ∞∂ 2 ( )φ ∂2∫ ∞F x∂x 2 =−∞ ∂x 2 ∂x 2 e−iωx dx = −ω 2 φ (x, t) e −iωx dx = −ω 2 F x [φ]−∞and so the Fourier transforms of the differential equation and initial condiations are∂ 2∂t 2 Φ (ω, t) + c2 ω 2 Φ (ω, t) = 0The differential equation for Φ (ω, t) is easily solvedImposing the boundary conditions on Φ (ω, t) yieldsAnd soWe thus haveΦ (ω, 0) = F (ω) := F x [f] (ω)Φ t (ω, 0) = 0Φ (ω, t) = A (ω) e icωt + B (ω) e −icωtA (ω) + B (ω) = F (ω)icωA (ω) − icωB (ω) = 0A (ω) = B (ω) = 1 2 F (ω)Φ (ω, t) = 1 2 F (ω) eicωt + 1 F (ω) e−icωt2as the Fourier transform of the solution φ (x, t) of the original PDE/BVP. To recover φ (x, t) all we need tois apply the inverse Fourier transform∫1 ∞( 1φ (x, t) = √2π 2 F (ω) eicωt + 1 )2 F (ω) e−icωt e iωx dω=12 √ 2π−∞∫ ∞−∞= 1 2 f (x + ct) + 1 f (x − ct)2F (ω) e iω(x+ct) dω + 1 ∫ ∞2 √ F (ω) e iω(x−ct) dω2π −∞

4. DIFFUSION EQUATION WITH A INSTANTANEOUS SOURCE 1104. Diffusion Equation with a Instantaneous SourceConsider the PDE/BDEφ t − a 2 φ xx = 0φ (x, 0) = Sδ (x)which would govern the diffusion of S particles released at a point x = 0 at time t = 0 in a 1-dimensionalmedium. Taking the Fourier transform with respect to x of both the PDE and the initial condition yields∂∂t Φ (ω, t) + a2 ω 2 Φ = 0Φ (ω, 0) =∫1 ∞√2π−∞The differential equation for Φ (ω, t) has as its general solutionΦ (ω, t) = A (ω) e −a2 ω 2 tSe −iωx dx =To obtain the coefficient A (ω), we just apply the Fourier transform of the boundar condition. This gives usΦ (ω, t) =S √2πe −a2 ω 2 tS √2πNow it turms out thatF −1 ( e pω2) = 1 √ 2pe − x24p(This one can look up in a Table of Fourier Transforms). Finally, we recover the solution to the originalproblem by applying the inverse Laplace transform. We thus arrive atS 1φ (x, t) = √ √2π 2a2 t exp ( −x 2 /4a 2 t )=S√4πa2 t e−x2 /(4a 2 t)