MAGIC058 & MATH64062: Partial Differential Equations 1

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1 MAGIC58 & MATH6462: Partial Differential Equations Section 6 Fourier transforms 6. The Fourier integral formula We have seen from section 4 that, if a function f(x) satisfies the Dirichlet conditions, then it can be expressed as f(x) = 2 a + n= { ( nπ ) ( nπ )} a n cos l x + b n sin l x, l x l, (6.) where the Fourier coefficients are defined in the usual way. On using the integral forms for the Fourier coefficients, (6.) becomes That is, f(x) = 2l f(x) = 2l l l l l f(v) dv + f(v) dv + n= n= { l ( nπ ) ( nπ ) f(v) cos l l l v dv cos l x l ( nπ ) ( nπ ) } + f(v) sin l v dv sin l x. l { l ( nπ ) } f(v) cos l l l (v x) dv, l x l. (6.2) Now let us consider what happens as l. Provided f(v)dv <, i.e. the integral exists, then the first term on the right hand side of (6.2) tends to zero as l. However the sum on the right hand side is not so simple to deal with. Let k = π, where for large l, l k is very small, and note k is not an integer. Then n= { l ( nπ ) } f(v) cos l l l (v x) dv = π which, on writing we have n= F l (k, x) = l l n= { l ( nπ ) } f(v) cos l l l (v x) dv = π { l } k f(v) cos (n k(v x)) dv, l (6.3) f(v) cos (k(v x)) dv, (6.4) kf l (n k, x). (6.5) n=

2 MAGIC58 & MATH6462: Partial Differential Equations 2 Recall, from elementary calculus that b a g(k)dk g(k n ) k, (6.6) where k n = n k and k = b a N, N. Now consider the sum on the right hand side of (6.5). For fixed x and large l, k is small and positive. Thus, the points n k, n =, 2, 3,... are densely and evenly distributed along the whole of the positive k axis. Therefore we might presume that n= F l (k n, x) k n= F (k, x)dk as l. (6.7) However, (6.7) is only an assumption because: (i) the function F l (k, x) changes with k, l = π k, (ii) the forms of the sums in (6.6) and (6.7) are not quite the same. Nevertheless, (6.5) and (6.7) suggest that, as l, (6.2) becomes That is f(x) = π f(x) = π This is the Fourier integral formula for f(x). Theorem 6..: Fourier s integral theorem Let f(x) denote a function that satisfies: F (k, x)dk. f(v) cos (k(v x)) dvdk, < x <. (6.8) (i) f(x) and f (x) are piecewise continuous on every finite interval of the x-axis. (ii) f(x) is absolutely integrable on (, ). That is, f(x) dx converges. Then π f(v) cos (k(v x)) dvdk = { f(x) at a point of continuity, (f(x + a) + f(x a)) at a point of discontinuity. 2 No proof will be given here but we have sketched out a heuristic proof above. A different form of Fourier s integral formula can be obtained by using the exponential form of the cosine function. That is, f(x) = f(v)e ik(v x) dvdk + f(v)e ik(v x) dvdk.

3 MAGIC58 & MATH6462: Partial Differential Equations 3 In the second integral on the right hand side, by making the substitution k = t, we get f(x) = f(v)e ik(v x) dvdk which, on replacing the dummy variable name t by k, gives f(x) = f(v)e it(v x) dvdt, e ikx f(v)e ikv dvdk. (6.9) This is the exponential form of Fourier s integral formula, and can be written as f(x) = where F (k) = F (k)e ikx dk, (6.) f(v)e ikv dv. (6.) The function F (k) is the Fourier transform of f(x). It is sometimes written as F (k) = F {f(x)}. Equation (6.) is the inversion formula. This is often written as f(x) = F {F (k)}. Note, upper case letters denote Fourier transforms, so F (k) = Fourier transform of f(x) G(k) = Fourier transform of g(x) etc. The curly F symbol (F ) denotes the Fourier integral operator. Example Let g(x) = Then the Fourier transform of g(x) is: G(k) = = ] [ e ikx = ik {, x <,, x >. g(x)e ikx dx + = 2 sin(k) k Note that G(k) is a function of k only.. e ikx dx +

4 MAGIC58 & MATH6462: Partial Differential Equations Properties of Fourier transforms (see Handout 5) Example Find the Fourier transform of f(x) = e a x, a >. Hence, show that The Fourier transform is F (k) = Now we use the inversion formula: = cos(k) + k 2 dk = π 2e. e a x e ikx dx e (a+ik)x dx + [ ] e (a+ik)x [ e (a ik)x = + a + ik (a ik = a + ik + + a ik = F (k) = 2a a 2 + k. 2 e a x = = a π = 2a π 2ae ikx e (a ik)x dx ] a 2 + k dk 2 cos(kx) i sin(kx) dx a 2 + k 2 cos(kx) a 2 + k dx, 2 since cos(kx) is even and sin(kx) is odd. finally, letting a =, and x = ±, gives π 2e = cos(k) + k 2 dx. Note, in the last two examples, even functions of x give even functions of k. We can prove this for all f(x), and similarly that odd functions of x give odd functions of k: ( ) ( ) even even if f(x) is then F (k) is. (6.2) odd odd What do we do if we are given a F (k) (e.g. a 2 +k 2 ) by way of determining f(x)? Obviously we should imply the inverse transform (6.) but then how do we evaluate the integral? In general, we make recourse to contour integration for complex functions.

5 MAGIC58 & MATH6462: Partial Differential Equations 5 Example Find the function f(x), whose Fourier transform is F (k) = a 2 + k 2, a >. Now, we have f(x) = e ikx a 2 + k dk, 2 but how do we determine this integral? Well, let us consider the contour integral I = C e izk a 2 + z 2 dz where C = C + C 2 is the following closed contour I(z) C 2 R R C R R(z) for x <, and for x > : I(z) R C R R(z) R C 2 Take x < first. We know from Cauchy s theorem that e izx I = a 2 + z dz = i {residues of poles inside contour}. 2 C

6 MAGIC58 & MATH6462: Partial Differential Equations 6 In this case, there are poles of the integrand at z = ±ia, both of which are simple and so I = i {residue from pole at z = ia} ( ) (z ia)e izx = i lim z ia a 2 + z ( 2 = i lim ) (z ia)e izx z ia (z ia)(z + ia) = ie i(ia)x 2 ia = πeax a. Now take the radius of C 2, R, to become larger and larger. Then, if we write z = Re iθ, we have e izx = e i x Reiθ = e i x R cos(θ) e x R sin(θ), which, because sin(θ) > for < θ < π, e x R sin(θ) as R. Hence, the integrand on C 2 is exponentially small as R and so C 2. We therefore have I = lim + e izx lim R C R = C 2 a 2 + z dz 2, z=real which is the integral we want to determine (with z = k). Hence e ikx a 2 + k 2 dk = π a eax for x <. For x > the procedure is the same, except now we take the contour in the lower half z-plane. As R the C 2 vanishes again and we recover, from evaluating the residue at z = a, that e ikx a 2 + k dk = π 2 a e ax for x >. Hence f(x) = e ikx a 2 + k dk = 2 2a e a x if F (k) = which can be compared with a previous example. 6.3 Fourier cosine and sine transforms a 2 + k 2, We can reduce the Fourier transform pair to the range (, ) if we consider even or odd functions. Compare this with what we had for half-range Fourier series.

7 MAGIC58 & MATH6462: Partial Differential Equations 7 First suppose f(x) is even then Now call F (k) = = = 2 F (k) 2 f(x)e ikx dx (f(x) cos(kx) + if(x) sin(kx)) dx f(x) cos(kx)dx = F c (k) = f(x) cos(kx)dx (6.3) the Fourier cosine transform of f(x). The inversion formula is found from f(x) = but remember, F (k) is even if f(x) is even, so f(x) = = F (k)e ikx dk F (k) (cos(kx) i sin(kx)) dk 2F (k) cos(kx)dk, and F (k) = 2F c (k), so the inverse Fourier cosine transform is f(x) = 2 π F c (k) cos(kx)dk. (6.4) Note that this can be used on any function defined on (, ) (it just presumes an even extension of f(x) in (, )). Similarly if f(x) were odd then we get with the inversion formula Example F s (k) = f(x) = 2 π Find the Fourier sine transform of g(x) = f(x) sin(kx)dx (6.5) F s (k) sin(kx)dk. (6.6) {, x < b,, x > b.

8 MAGIC58 & MATH6462: Partial Differential Equations 8 So, G s (k) = = b g(x) sin(kx)dx sin(kx)dx + [ = ] b k cos(kx) = cos(kb). k Definition 6.2.: The Heaviside function The Heaviside function is defined as b dx H(x) = {, x >,, x <. (6.7) H(x) x It can be used to construct simple functions as follows: {, x > b, H(x b) =, x < b, and So, the top-hat function H(x + b) = g(x) = {, x > b,, x < b. {, x < b,, x > b, g(x) b b x can be written g(x) = H(x + b) H(x b). (6.8)

9 MAGIC58 & MATH6462: Partial Differential Equations 9 The Fourier transform of g(x) is so G(k) = b b e ikx dx = [ ] b ik eikx, b G(k) = 2 sin(kb). (6.9) k We will use (6.8) and (6.9) in the next section. Some special Fourier transforms: f(x) F (x) {, x < b x > b e a x x 2 + a 2 2 sin(kb) k 2a a 2 + k 2 π a e a k e bx2 π b e k2 /4b The Fourier transforms of many more functions can be obtained using the above together with properties (a)-(h) in the handout. Also, many Fourier sine and cosine transforms can be deduced from these and (a)-(h). 6.4 Boundary Value Problems Fourier transforms can be used for solving BVPs in which one or more variables have an infinite range. The effect of applying a Fourier transform to a PDE is to temporarily exclude one variable (e.g. x is turned into k, which is treated as a constant parameter until later). The resulting BVP is usually easier to solve; its solution being a function of the remaining variables plus the transform variable k. Once this solution has been found, the lost variable can be recovered by applying the inverse Fourier transform.

10 MAGIC58 & MATH6462: Partial Differential Equations To summarise IVBP PDE + BCs + ICs e.g. φ(x, t) apply F.T. w.r.t. x IVBP 2 ODE for Φ(k, t) + ICs + BCs solve IVBP 2 Φ(x, t) apply inverse F.T. φ(x, t) i.e. solution to IBVP Example Solve 2 y x 2 y 2 c 2 t =, < x <, 2 subject to the initial conditions: (i) y(x, ) = h(x), (ii) y (x, ) = v(x). t Remember we have already shown that the solution to this IBVP is D Alembert s solution: y(x, t) = 2 (h(x ct) + h(x + ct)) + 2c x+ct x ct v(z)dz, < x < t, (6.2) which we obtained in section 3.2 by (i) reducing the wave equation to canonical form, (ii) integrating to obtain y(x, t) = f(x ct) + g(x + ct) and (iii) applying the initial conditions. We shall now solve this IBVP by Fourier transforms and obtain (hopefully) the same result. Let Y (k, t) denote the Fourier transform of y(x, t) with respect to x, that is Y (k, t) = y(x, t)e ikx dx.

11 MAGIC58 & MATH6462: Partial Differential Equations Thus, the PDE becomes or, by property (e) ( 2 y x ) 2 y e ikx dx = 2 c 2 t 2 ( ik) 2 Y d 2 Y c 2 dt =. 2 So, we have d 2 Y dt + 2 c2 k 2 Y =, which we know has the general solution Y (k, t) = A(k)e ikct + B(k)e ikct, (6.2) a form more convenient than using sin(kct) and cos(kct). Note, now, that A and B although not functions of t, can clearly be functions of k. To find A and B we need to use the ICs. First, however, we need to Fourier transform them to remove x. Thus (i) = Y (k, ) = H(k), (ii) = dy (k, ) = V (k). dt where H(k) and V (k) are Fourier transforms of h(x) and v(x) respectively [Note, we write H(k) as the Fourier transform of h(x) so as not to confuse it with the Heaviside function]. Now, from (6.2), we have Y (k, ) = A(k) + B(k) = H(k), and, by differentiating with respect to t, we have which gives Hence, dy (k, ) = ikca(k) ikcb(k) = V (k), (6.22) dt A(k) = H(k) 2 Y (k, t) = ( H + V 2 ikc = 2 + V (k) 2ikc, The inverse Fourier transform therefore gives y(x, t) = H(k) B(k) = V (k) 2 2ikc. ) e ikct + ( H V ) e ikct 2 ikc ( H(k)e ikct + H(k)e ikct) + sin(kct) V (k)2. 2c k [ H(k)e ikct + 2 H(k)e ikct] e ikx dk + sin(kct) V (k)2 e ikx dk. 2c k

12 MAGIC58 & MATH6462: Partial Differential Equations 2 The first integral is easily dealt with. Property (d) tells us that so H(k)e ikct is the Fourier transform of h(x ct), and H(k)e ikct is the Fourier transform of h(x + ct), y(x, t) = (h(x ct) + h(x + ct)) + 2 2c sin(kct) V (k)2 e ikx dk. k We need to use convolution (property (g)) to deal with the remaining integral: V (k) is Fourier transform of v(x), 2 sin(kct) k is Fourier transform of (H(x + ct) H(x ct)), where H(x) is the Heaviside function (see (6.7), (6.9)). So, using the convolution property we have that so V (k)2 sin(kct) k is the Fourier transform of y(x, t) = 2 (h(x ct) + h(x + ct)) + 2c or finally = 2 (h(x ct) + h(x + ct)) + 2c y(x, t) = 2 (h(x ct) + h(x + ct)) + 2c as required. Example Use Fourier transforms to solve subject to x+ct v(z) (H(x + ct z) H(x ct z)) dz v(z) [H(x + ct z) H(x ct z)] dz x+ct x ct v(z)dz 2 φ x φ y 2 = in y, (i) φ(x, ) = f(x), (ii) φ(x, y) as y, (iii) φ(x, y) as x ±. x ct v(z)dz, v(z)dz, < x < t,

13 MAGIC58 & MATH6462: Partial Differential Equations 3 φ half of x-y plane φ φ φ(x, ) = f(x) Let Φ(k, y) denote the Fourier transform of φ(x, y) with respect to x. becomes Solutions are of the form or 2 Φ y + 2 ( ik)2 Φ = 2 Φ y 2 k2 Φ =. Then the PDE Φ(k, y) = A(k)e my + B(k)e my where m satisfies m 2 k 2 =. We have a choice of solutions. Either m = ±k or m = ± k. We want solutions which decay in y for all m values (or k values) in order to satisfy (ii). Hence, we choose Φ(k, y) = A(k)e k y + B(k)e k y and to ensure decay as y, we need B. Thus, Φ(k, y) = A(k)e k y (6.23) and in order to find A(k) we use the BC on y =. First, transform (i) to give and so Φ(k, ) = F (k), where F (k) = F (f(x)) (6.24) A(k) = F (k) from (6.23). Hence, using the inverse Fourier transform of (6.23), with the above result, we obtain φ(x, y) = F (k)e k y e ikx dk. Note, if the domain was finite for y, not infinite, then we would not need k, just k. We can improve on this solution a bit by again using the convolution. Note that F (k) is the Fourier transform of f(x) and e k y is the Fourier transform of y π(y 2 + x 2 )

14 MAGIC58 & MATH6462: Partial Differential Equations 4 from the example in section 6.2. Hence, the convolution theorem gives φ(x, y) = y f(v)dv, < x <, y. π(y 2 + (x v) 2 ) Note that the function f(x) given in BC (i) is arbitrary; we can therefore choose any function f(x) for which the Fourier transform exists. Suppose, for example, that f(x) = H(x + b) H(x b) f(x) b b x then φ(x, y) = y π = y π b b [H(v + b) H(v b)] (x v) 2 + y 2 dv (x v) 2 + y 2 dv. We can make the substitution v = x w, which gives or φ(x, y) = y π φ(x, y) = π x+b x b w 2 + y dw = [ ( )] x+b w tan 2 π y x b { ( ) ( )} x + b x b tan tan. y y So far we have considered two examples. In both cases we have used the Fourier transform to eliminate x and replace it with k. This was possible because the range of x was infinite. What happens if the range is semi-infinite? We can use the Fourier sine or cosine transform. But which one? Well, if we have a term in the governing PDE which is like 2 φ x 2 (6.25) and we have (i) the BC φ(x = ) = given (6.26)

15 MAGIC58 & MATH6462: Partial Differential Equations 5 then we use the Fourier sine transform. Whereas when we have (ii) the BC φ (x = ) = given (6.27) x then we use the Fourier cosine transform. Proof is given by the following examples. Note, if the PDE contained φ terms, then the sine or cosine transform are both useless! x Example - Diffusion equation Solve u t = 2 u x 2 in the region x, t, subject to the conditions Since u x (i) u(x, ) = f(x), (ii) u (, t) = b, b = constant, x u (iii) u, x as x. is known at x =, use the cosine transform with respect to x, i.e. U c = u(x, t) cos(kx)dx. Now, the following integral can be simplified, via integration by parts, to yield [ ] 2 u u x cos(kx)dx = 2 x cos(kx) u + k x sin(kx)dx, and integrating by parts again gives 2 u u cos(kx)dx = x2 x (, t) + k [u sin(kx)] k2 = b k 2 U c (k, t). u cos(kx)dx So, the PDE becomes (by multiplying by cos(kx) and performing the integral dx) U c t (k, t) = b k2 U c (k, t) or U c t (k, t) + k2 U c (k, t) = b. We can solve this by using the integrating factor e k2t : e k2 t U c t (k, t) + k2 e k2t U c (k, t) = d dt (U ce k2t ) = be k2 t

16 MAGIC58 & MATH6462: Partial Differential Equations 6 which integrates to give U c e k2t = b k 2 ek2t + A(k), where A(k) is the constant of integration that needs to be determined by condition (i). So and apply the cosine transform to (i) to give Hence or The solution is therefore U c (k, ) = U c (k, t) = b k 2 + A(k)e k2 t f(x) cos(kx)dx = F c (k). U c (k, ) = b k 2 + A(k) = F c(k) U c (k, t) = A(k) = F c (k) + b k 2. { F c (k) + b } e k2t b k 2 k, 2 which, using the inverse transform, gives u(x, t) = 2 {( F c (k) + b ) e k2t b } cos(kx)dk. π k 2 k 2 Suppose, for example, that b = and {, < x < d, f(x) =, x > d. Then and Example u(x, t) = 2 π F c (k) = e k2 t k d cos(kx)dx = sin(kd) k sin(kd) cos(kx)dk, x, t. A function φ(x, y) satisfies 2 φ x + 2 φ 2 y = 2 in the region x, y. The function is bounded in this region and (i) φ(, y) =, y >, (ii) φ(x, ) = f(x), x >.,

17 MAGIC58 & MATH6462: Partial Differential Equations 7 Show that φ(x, y) = y π { f(v) (v x) 2 + y 2 (v + x 2 ) + y 2 } dv. We know the value of φ at x =. Hence, use the sine transform with respect to x, i.e. Now Φ s (k, y) = φ(x, y) sin(kx)dx. ] [ 2 φ φ x sin(kx)dx = 2 x sin(kx) φ k x cos(kx)dx = k [φ cos(kx)] k2 φ sin(kx)dx. Hence, on applying the sine transform to the PDE, we obtain: with solution k 2 Φ s (k, y) + d2 Φ s (k, y) = dy2 Φ s (k, y) = A(k)e ky + B(k)e ky, y. The solution must be bounded as y, and note that k > for inverse sine transform. Hence B(k) = and Φ s (k, y) = A(k)e ky. (6.28) Now apply the sine transform to condition (ii) Φ s (k, ) = F s (k) = which, when compared with (6.28), yields The inversion formula gives, therefore, φ(x, y) = 2 π A(k) = F s (k). f(x) sin(kx)dx F s (k)e ky sin(kx)dk. We can, as in the previous examples, do better than this. We know so φ(x, y) = 2 π F s (k) = ( f(v) sin(kv)dv ) f(v) sin(kv)dv e ky sin(kx)dk

18 MAGIC58 & MATH6462: Partial Differential Equations 8 or, by interchanging orders of integration, φ(x, y) = 2 π = π f(v) sin(kv) sin(kx)e ky dk dv { } f(v) e ky [cos(k(v x)) cos(k(v + x))] dk dv. We can evaluate the inner integrals by comparing them with e bk cos(kx)dk = 2 e b k e ikx dk = b b 2 + x 2, where this is determined using the table of Fourier transforms in section 6.3. b = y, and x respectively is replaced by (v x) and (v + x), gives φ(x, y) = y { } f(v) π (v x) 2 + y dv. 2 (v + x) 2 + y 2 Letting I. D. Abrahams

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