4.6 Applications of Transforms to Differential Equations Establish the following result by formally inverting the Laplace transform.

Transcription

1 4.6 Applications of Transforms to Differential Equations Establish the following result by formally inverting the Laplace transform ˆF(s) = 1 s sinh(sy) sinh(sl) l> f (x) = y l + 2( 1) n nπ n=1 ( nπy sin l ) ( ) nπ x cos l See the remark at the end of Problem 18, which explains how to show how the inverse Laplace transform can be proven to be valid in a situation such as this where there are an infinite number of poles. 4.6 Applications of Transforms to Differential Equations A particularly valuable technique available to solve differential equations in infinite and semiinfinite domains is the use of Fourier and Laplace transforms. In this section we describe some typical examples. The discussion is not intended to be complete. The aim of this section is to elucidate the transform technique, not to detail theoretical aspects regarding differential equations. The reader only needs basic training in the calculus of several variables to be able to follow the analysis. We shall use various classical partial differential equations (PDEs) as vehicles to illustrate methodology. Herein we will consider well-posed problems that will yield unique solutions. More general PDEs and the notion of well-posedness are investigated in considerable detail in courses on PDEs. Example Steady state heat flow in a semiinfinite domain obeys Laplace s equation. Solve for the bounded solution of Laplace s equation 2 φ(x, y) x φ(x, y) y 2 = (4.6.1) in the region < x <, y >, where on y = we are given φ(x, ) = h(x) with h(x) L 1 L 2 (i.e., h(x) dx < and h(x) 2 dx < ). This example will allow us to solve Laplace s Equation (4.6.1) by Fourier transforms. Denoting the Fourier transform in x of φ(x, y) as ˆΦ(k, y): ˆΦ(k, y) = e ikx φ(x, y) dx taking the Fourier transform of Eq. (4.6.1), and using the result from Section 4.5 for the Fourier transform of derivatives, Eqs. (4.5.16a,b) (assuming the validity

2 286 4 Residue alculus and Applications of ontour Integration of interchanging y-derivatives and integrating over k, which can be verified a posteriori), we have Hence 2 ˆΦ y 2 k2 ˆΦ = (4.6.2) ˆΦ(k, y) = A(k)e ky + B(k)e ky where A(k) and B(k) are arbitrary functions of k, to be specified by the boundary conditions. We require that ˆΦ(k, y) be bounded for all y >. In order that ˆΦ(k, y) yield a bounded function φ(x, y), we need ˆΦ(k, y) = (k)e k y (4.6.3) Denoting the Fourier transform of φ(x, ) = h(x) by Ĥ(k) fixes (k) = Ĥ(k), so that ˆΦ(k, y) = Ĥ(k)e k y (4.6.4) From Eq. (4.5.1) by direct integration (contour integration is not necessary) we find that ˆF(k, y) = e k y is the Fourier transform of f (x, y) = 1 y, thus π x 2 +y 2 from the convolution formula Eq. (4.5.17) the solution to Eq. (4.6.1) is given by φ(x, y) = 1 π yh(x ) (x x ) 2 + y 2 dx (4.6.5) If h(x) were taken to be a Dirac delta function concentrated at x = ζ, h(x) = h s (x ζ) = δ(x ζ), then Ĥ(k) = e ikζ, and from Eq. (4.6.4) directly (or Eq. (4.6.5)) a special solution to Eq. (4.6.1), φ s (x, y) is φ s (x, y) = G(x ζ, y) = 1 π ( y (x ζ) 2 + y 2 ) (4.6.6) Function G(x ζ, y) is called a Green s function; it is a fundamental solution to Laplace s equation in this region. Green s functions have the property of solving a given equation with delta function inhomogeneity. From the boundary values h s (x ζ,) = δ(x ζ)we may construct arbitrary initial values φ(x, ) = h(ζ ) δ(x ζ)dζ = h(x) (4.6.7a)

3 4.6 Applications of Transforms to Differential Equations 287 and, because Laplace s equation is linear, we find by superposition that the general solution satisfies φ(x, y) = h(ζ ) G(x ζ, y) dζ (4.6.7b) which is Eq. (4.6.5), noting that ζ or x are dummy integration variables. In many applications it is sufficient to obtain the Green s function of the underlying differential equation. The formula (4.6.5) is sometimes referred to as the Poisson formula for a half plane. Although we derived it via transform methods, it is worth noting that a pair of such formulae can be derived from auchy s integral formula. We describe this alternative method now. Let f (z) be analytic on the real axis and in the upper half plane and assume f (ζ ) uniformly as ζ. Using a large closed semicircular contour such as that depicted in Figure we have f (z) = 1 2πi = 1 2πi f (ζ ) ζ z dζ f (ζ ) ζ z dζ where Im z > (in the second formula there is no singularity because the contour closes in the upper half plane and ζ = z in the lower half plane). Adding and subtracting yields f (z) = 1 2πi ( 1 f (ζ ) ζ z ± 1 ) dζ ζ z The semicircular portion of the contour R vanishes as R, implying the following on Im ζ = for the plus and minus parts of the above integral, respectively: calling z = x + iy and ζ = x + iy, alling f (x, y) = 1 πi f (x, y) = 1 π ( f (x, y = ) ( f (x, y = ) x x (x x ) 2 + y 2 y (x x ) 2 + y 2 ) dx ) dx f (z) = f (x, y) = u(x, y) + i v(x, y), Re f (x, y = ) = u(x, ) = h(x)

4 288 4 Residue alculus and Applications of ontour Integration and taking the imaginary part of the first and the real part of the second of the above formulae, yields the conjugate Poisson formulae for a half plane: v(x, y) = 1 π u(x, y) = 1 π ( h(x ) ( h(x ) x x (x x ) 2 + y 2 y (x x ) 2 + y 2 ) dx ) dx Identifying u(x, y) as φ(x, y), we see that the harmonic function u(x, y) (because f (z) is analytic its real and imaginary parts satisfy Laplace s equation) is given by the same formula as Eq. (4.6.5). Moreover, we note that the imaginary part of f (z), v(x, y), is determined by the real part of f (z) on the boundary. We see that we cannot arbitrarily prescribe both the real and imaginary parts of f (z) on the boundary. These formulae are valid for a half plane. Similar formulae can be obtained by this method for a circle (see also Example 1, Section 2.6). Laplace s equation, (4.6.1), is typical of a steady state situation, for example, as mentioned earlier, steady state heat flow in a uniform metal plate. If we have time-dependent heat flow, the diffusion equation φ t = k 2 φ (4.6.8) is a relevant equation with k the diffusion coefficient. In Eq. (4.6.8), 2 is the Laplacian operator, which in two dimensions is given by 2 = In x 2 y 2 one dimension, taking k = 1 for convenience, we have the following initial value problem: φ(x, t) t = 2 φ(x, t) x 2 (4.6.9) The Green s function for the problem on the line < x < is obtained by solving Eq. (4.6.9) subject to φ(x, ) = δ(x ζ) Example Solve for the Green s function of Eq. (4.6.9). Define ˆΦ(k, t) = e ikx φ(x, t) dx

5 4.6 Applications of Transforms to Differential Equations 289 whereupon the Fourier transform of Eq. (4.6.9) satisfies hence ˆΦ(k, t) t = k 2 ˆΦ(k, t) (4.6.1) ˆΦ(k, t) = ˆΦ(k, )e k2 t = e ikζ k2 t (4.6.11) where ˆΦ(k, ) = e ikζ is the Fourier transform of φ(x, ) = δ(x ζ). Thus, by the inverse Fourier transform, and calling G(x ζ,t) the inverse transform of (4.6.11), G(x ζ,t) = 1 2π = e (x ζ)2 /4t e ik(x ζ) k2t dk 1 2π e x ζ (k i 2t ) 2t dk (x ζ) 2 = e 4t 2 πt (4.6.12) where we use e u2 du = π. Arbitrary initial values are included by again observing that which implies φ(x, ) = h(x) = h(ζ )δ(x ζ)dζ φ(x, t) = G(x ζ,t)h(ζ ) dζ = 1 2 πt h(ζ ) e (x ζ)2 4t dζ (4.6.13) The above solution to Eq. (4.6.9) could also be obtained by using Laplace transforms. It is instructive to show how the method proceeds in this case. We begin by introducing the Laplace transform of φ(x, t) with respect to t: ˆΦ(x, s) = e st φ(x, t) dt (4.6.14) Taking the Laplace transform in t of Eq. (4.6.8), with φ(x, ) = δ(x ζ), yields 2 ˆΦ x 2 (x, s) s ˆΦ(x, s) = δ(x ζ) (4.6.15)

6 29 4 Residue alculus and Applications of ontour Integration Hence the Laplace transform of the Green s function to Eq. (4.6.9) satisfies Eq. (4.6.15). We remark that generally speaking, any function G(x ζ)satisfying LG(x ζ) = δ(x ζ) where L is a linear differential operator, is referred to as a Green s function. The general solution corresponding to φ(x, ) = h(x) is obtained by superposition: φ(x, t) = G(x ζ)h(ζ ) dζ. Equation (4.6.15) is solved by first finding the bounded homogeneous solutions on < x <, for (x ζ) > and (x ζ) < : ˆΦ + (x ζ,s) = A(s)e s1/2 (x ζ) ˆΦ (x ζ,s) = B(s)e s1/2 (x ζ) for x ζ> for x ζ< (4.6.16) where we take s 1/2 to have a branch cut on the negative real axis; that is, s = re iθ, π θ<π. This will allow us to readily invert the Laplace transform (Re s > ). The coefficients A(s) and B(s) in Eq. (4.6.16) are found by (a) requiring continuity of ˆΦ(x ζ,s) at x = ζ and by (b) integrating Eq. (4.6.15) from x = ζ ϵ, tox = ζ + ϵ, and taking the limit as ϵ +. This yields a jump condition on ˆ (x ζ,s): x [ ] ˆΦ x ζ = + (x ζ,s) = 1 (4.6.17) x x ζ = ontinuity yields A(s) = B(s), and Eq. (4.6.17) gives hence s 1/2 A(s) s 1/2 B(s) = 1 A(s) = B(s) = 1 2s 1/2 Using Eq. (4.6.16), ˆΦ(x ζ,s) is written in the compact form: (4.6.18a) (4.6.18b) ˆΦ(x ζ,s) = e s1/2 x ζ 2s 1/2 (4.6.19) The solution φ(x, t) is found from the inverse Laplace transform: φ(x, t) = 1 2πi c+i c i e s1/2 x ζ e st 2s 1/2 ds (4.6.2)

7 4.6 Applications of Transforms to Differential Equations 291 for c >. To evaluate Eq. (4.6.2), we employ the same keyhole contour as in Example in Section 4.5 (see Figure 4.5.2). There are no singularities enclosed, and the contours R and ϵ at infinity and at the origin vanish in the limit R, ϵ, respectively. We only obtain contributions along the top and bottom of the branch cut to find φ(x, t) = 1 2πi e ir1/2 x ζ e rt 2r 1/2 e iπ/2 e iπ dr + 1 2πi e ir1/2 x ζ e rt 2r 1/2 e iπ/2 e iπ dr (4.6.21) In the second integral we put r 1/2 = u; in the first we put r 1/2 = u and then take u u, whereupon we find the same answer as before (see Eq. (4.6.12)): φ(x, t) = 1 2π e u2 t+iu x ζ du = 1 2π e x ζ (u i 2t ) 2t e (x ζ)2 4t du = e (x ζ)2 /4t 2 πt (4.6.22) The Laplace transform method can also be applied to problems in which the spatial variable is on the semiinfinite domain. However, rather than use Laplace transforms, for variety and illustration, we show below how the sine transform can be used on Eq. (4.6.9) with the following boundary conditions: φ(x, ) =, φ φ(x =, t) = h(t), lim (x, t) =, (4.6.23) x x Define, following Section 4.5 φ(x, t) = 2 π ˆΦ s (k, t) sin kx dk (4.6.24a) ˆΦ s (k, t) = φ(x, t) sin kx dx (4.6.24b)

8 292 4 Residue alculus and Applications of ontour Integration We now operate on Eq. (4.6.9) with the integral dx sin kx, and via integration by parts, find 2 [ ] φ φ φ sin kx dx = (x, t) sin kx k cos kx dx x 2 x x= x = kφ(, t) k 2 ˆΦ s (k, t) (4.6.25) whereupon the transformed version of Eq. (4.6.9) is ˆΦ s t (k, t) + k 2 ˆΦ s (k, t) = kh(t) (4.6.26) The solution of Eq. (4.6.26) with φ(x, ) = is given by ˆΦ s (k, t) = t h(t )ke k2 (t t ) dt (4.6.27) If φ(x, ) were nonzero, then Eq. (4.6.27) would have another term. For simplicity we only consider the case φ(x, ) =. Therefore φ(x, t) = 2 π { t } dk sin kx h(t )e k2 (t t ) kdt (4.6.28) By integration we can show that (use sin kx = (e ikx e ikx )/2 and integrate by parts to obtain integrals such as those in (4.6.12)) J(x, t t ) = ke k2 (t t ) sin kx dk = π xe x 2 /4(t t ) 4(t t ) 3/2 (4.6.29) hence by interchanging integrals in Eq. (4.6.28), we have φ(x, t) = 2 π t When h(t) = 1, if we call η = x h(t )J(x, t t ) dt 2(t t ) 1/2, then dη = φ(x, t) = 2 π x 2 t ( x erfc 2 t e η2 dη ) x 4(t t ) 3/2 dt, and we have (4.6.3)

9 4.6 Applications of Transforms to Differential Equations 293 We note that erfc(x) is a well-known function, called the complementary error function: erfc(x) 1 erf(x), where erf(x) 2 x π dy. e y2 It should be mentioned that the Fourier sine transform applies to problems such as Eq. (4.6.23) with fixed conditions on φ at the origin. Such solutions can be extended to the interval < x < where the initial values φ(x, ) are themselves extended as an odd function on (, ). However, if we should replace φ(x =, t) = h(t) by a derivative condition, at the origin, say, φ (x =, t) = h(t), then the appropriate transform to use is a cosine x transform. Another type of partial differential equation that is encountered frequently in applications is the wave equation 2 φ x 1 2 φ = F(x, t) (4.6.31) 2 c 2 t 2 where the constant c, c >, is the speed of propagation of the unforced wave. The wave equation governs vibrations of many types of continuous media with external forcing F(x, t). IfF(x, t) vibrates periodically in time with constant frequency ω>, say, F(x, t) = f (x)e iωt, then it is natural to look for special solutions to Eq. (4.6.31) of the form φ(x, t) = Φ(x)e iωt. Then Φ(x) satisfies 2 Φ x 2 + ( ) ω 2 Φ = f (x) (4.6.32) c A real solution to Eq. (4.6.32) is obtained by taking the real part; this would correspond to forcing of φ(x, t) = φ(x) cos ωt. If we simply look for a Fourier transform solution of Eq. (4.6.32) we arrive at Φ(x) = 1 2π ˆF(k) k 2 (ω/c) 2 eikx dk (4.6.33) where ˆF(k) is the Fourier transform of f (x). Unfortunately, for the standard contour, k real, < k <, Eq. (4.6.33) is not well defined because there are singularities in the denominator of the integrand in Eq. (4.6.33) when k =±ω/c. Without further specification the problem is not well posed. The standard acceptable solution is found by specifying a contour that is indented below k = ω/c and above k =+ω/c (see Figure 4.6.1); this removes the singularities in the denominator. This choice of contour turns out to yield solutions with outgoing waves at large distances from the source F(x, t). A choice of contour reflects an imposed boundary condition. In this problem it is well known and is referred to as the

10 294 4 Residue alculus and Applications of ontour Integration ω k= - c k= + ω c Fig Indented contour x Sommerfeld radiation condition. An outgoing wave has the form e iω(t x /c) (as t increases, x increases for a given choice of phase, i.e., on a fixed point on a wave crest). An incoming wave has the form e iω(t+ x /c). Using the Fourier representation ˆF(k) = f (ζ )e iζ k dζ in Eq. (4.6.33), we can write the function in the form Φ(x) = f (ζ )H(x ζ,ω/c) dζ (4.6.34a) where H(x ζ,ω/c) = 1 2π e ik(x ζ) k 2 (ω/c) 2 dk (4.6.34b) and the contour is specified as in Figure Eq. (4.6.34b) yields ontour integration of H(x ζ,ω/c) = ie i x ζ (ω/c) 2(ω/c) (4.6.34c) At large distances from the source, x, we have outgoing waves for the solution φ(x, t): { i } φ(x, t) = Re f (ζ )e iω(t x ζ /c) dζ. (4.6.34d) 2(ω/c) Thus, for example, if f (ζ ) is a point source: f (ζ ) = δ(ζ x ) where δ(ζ x ) is a Dirac delta function concentrated at x, then (4.6.34d) yields φ(x, t) = 1 2(ω/c) sin ω(t x x /c). (4.6.34e) An alternative method to find this result is to add a damping mechanism to the original equation. Namely, if we add the term ϵ φ to the left-hand side t of Eq. (4.6.31), then Eq. (4.6.33) is modified by adding the term iϵω to the denominator of the integrand. This has the desired effect of moving the poles off the real axis (k 1 = ω/c + iϵα, k 2 =+ω/c iϵα, where α = constant) in the same manner as indicated by Figure By using Fourier transforms,

11 4.6 Applications of Transforms to Differential Equations 295 and then taking the limit ϵ (small damping), the above results could have been obtained. In practice, wave propagation problems such as the following one u t + 3 u =, < x <, u(x, ) = f (x), (4.6.35) x 3 where again f (x) L 1 L 2, are solved by Fourier transforms. Function u(x, t) typically represents the small amplitude vibrations of a continuous medium such as water waves. One looks for a solution to Eq. (4.6.35) of the form u(x, t) = 1 2π b(k, t)e ikx dk (4.6.36) Taking the Fourier transform of (4.6.35) and using (4.5.16) or alternatively, substitution of Eq. (4.6.36) into Eq. (4.6.35) assuming interchanges of derivative and integrand are valid (a fact that can be shown to follow from rapid enough decay of f (x) at infinity, that is, f L 1 L 2 ) yields b t ik3 b = (4.6.37a) hence b(k, t) = b(k, )e ik3 t (4.6.37b) where b(k, ) = f (x)e ikx dx (4.6.37c) The solution (4.6.36) can be viewed as a superposition of waves of the form e ikx iω(k)t, ω(k) = k 3. Function ω(k) is referred to as the dispersion relation. The above integral representation, for general f (x), is the best one can do, because we cannot evaluate it in closed form. However, as t, the integral can be approximated by asymptotic methods, which will be discussed in hapter 6, that is, the methods of stationary phase and steepest descents. Suffice it to say that the solution u(x, t) ast (the initial values are said to disperse as t ) and the major contribution to the integral is found near the location where ω (k) = x/t; that is, x/t = 3k 2 in the integrand (where the phase = kx ω(k)t is stationary: k = ). The quantity ω (k) is called the group velocity, and it represents the speed of a packet of waves

12 296 4 Residue alculus and Applications of ontour Integration centered around wave number k. Using asymptotic methods for x/t <, as t, u(x, t) can be shown to have the following approximate form u(x, t) c t ( 2 i=1 ) b(k i ) ki ei(k i x+ki 3t+φ i ) (4.6.38) k 1 = x/3t, k 2 = x/3t, c,φ i constant When x/t > the solution decays exponentially. As x/t, Eq. (4.6.38) may be rearranged and put into the following self-similar form u(x, t) d (3t) 1/3 A(x/(3t)1/3 ) (4.6.39) where d is constant and A(η) satisfies (by substitution of (4.6.39) into (4.6.35)) or A ηηη ηa η A = A ηη ηa = (4.6.4a) with the boundary condition A asη. Equation (4.6.4a) is called Airy s equation. The integral representation of the solution to Airy s equation with A, η, is given by A(η) = 1 2π e i(kη+k3 /3) dk (4.6.4b) (See also the end of this section, Eq. (4.6.57).) Its wave form is depicted in Figure Function A(η) acts like a matching or turning of solutions from one type of behavior to another: i.e., from exponential decay as η + to oscillation as η (see also Section 6.7). Sometimes there is a need to use multiple transforms. For example, consider finding the solution to the following problem: 2 φ x φ y 2 m2 φ = f (x, y), φ(x, y) as x 2 + y 2 (4.6.41) A simple transform in x satisfies φ(x, y) = 1 2π Φ(k 1, y)e ik 1x dk 1

13 4.6 Applications of Transforms to Differential Equations 297 A( η) η Fig Airy function We can take another transform in y to obtain φ(x, y) = 1 (2π) 2 ˆΦ(k 1, k 2 )e ik 1x+ik 2 y dk 1 dk 2 (4.6.42) Using a similar formula for f (x, y) in terms of its transform ˆF(k 1, k 2 ), we find by substitution into Eq. (4.6.41) φ(x, y) = 1 (2π) 2 ˆF(k 1, k 2 )e ik 1x+ik 2 y k k2 2 + m2 dk 1 dk 2 (4.6.43) Rewriting Eq. (4.6.43) using ˆF(k 1, k 2 ) = f (x, y )e ik 1x ik 2 y dx dy and interchanging integrals yields φ(x, y) = f (x, y )G(x x, y y ) dx dy (4.6.44a) where G(x, y) = 1 (2π) 2 e ik 1x+ik 2 y k k2 2 + m2 dk 1 dk 2 (4.6.44b)

14 298 4 Residue alculus and Applications of ontour Integration By clever manipulation, one can evaluate Eq. (4.6.44b). Using the methods of Section 4.3, contour integration with respect to k 1 yields G(x, y) = 1 4π e ik 2 y k 2 2 +m2 x k m 2 dk 2 (4.6.45a) Thus, for x G sgn(x) (x, y) = e ik 2 y k2 2+m2 x dk 2 x 4π (4.6.45b) where sgnx = 1 for x >, and sgnx = 1 for x <. Equation (4.6.45b) takes on an elementary form for m = ( k 2 2 = k 2 ): and we have G x (x, y) = x 2π(x 2 + y 2 ) G(x, y) = 1 4π ln(x 2 + y 2 ) (4.6.46) The constant of integration is immaterial, because to have a vanishing solution φ(x, y) as x 2 + y 2, Eq. (4.6.44a) necessarily requires that f (x, y)dx dy =, which follows directly from Eq. (4.6.41) by integration with m =. Note that Eq. (4.6.43) implies that when m =, for the integral to be well defined, ˆF(k 1 =, k 2 = ) =, which in turn implies the need for the vanishing of the double integral of f (x, y). Finally, if m, we only remark that Eq. (4.6.44b) or (4.6.45a) is transformable to an integral representation of a modified Bessel function of order zero: G(x, y) = 1 2π K ( m x 2 + y 2) (4.6.47) Interested readers can find contour integral representations of Bessel functions in many books on special functions. Frequently, in the study of differential equations, integral representations can be found for the solution. Integral representations supplement series methods discussed in hapter 3 and provide an alternative representation of a class of solutions. We give one example in what follows. onsider Airy s equation in the form (see also Eq. (4.6.4a)) d 2 y zy = (4.6.48) dz2

15 4.6 Applications of Transforms to Differential Equations 299 and look for an integral representation of the form y(z) = e zζ v(ζ ) dζ (4.6.49) where the contour and the function v(ζ ) are to be determined. Formula (4.6.49) is frequently referred to as a generalized Laplace transform and the method as the generalized Laplace transform method. (Here is generally not the Bromwich contour.) Equation (4.6.49) is a special case of the more general integral representation K (z, ζ )v(ζ )dζ. Substitution of Eq. (4.6.49) into Eq. (4.6.48), and assuming the interchange of differentiation and integration, which is verified a posteriori gives (ζ 2 z)v(ζ )e zζ dζ = (4.6.5) Using rearranging and integrating yields ze zζ v = d dζ (ezζ zζ dv v) e dζ [ e zζ v(ζ ) ] + ( ζ 2 v + dv ) e zζ dζ = (4.6.51) dζ where the term in brackets, [ ], stands for evaluation at the endpoints of the contour. The essence of the method is to choose and v(ζ ) such that both terms in Eq. (4.6.51) vanish. Taking implies that dv dζ + ζ 2 v = (4.6.52) v(ζ ) = Ae ζ 3 /3, A = constant (4.6.53) Thinking of an infinite contour, and calling ζ = Re iθ, we find that the dominant term as R in [ ] is due to v(ζ ), which in magnitude is given by v(ζ ) = A e R3 (cos 3θ)/3 (4.6.54) Vanishing of this contribution for large values of ζ will occur for cos 3θ >, that is, for π 2 + 2nπ <3θ <π + 2nπ, n =, 1, 2 (4.6.55) 2

16 3 4 Residue alculus and Applications of ontour Integration π/2 5π/6 π/6 7π/6 2 II III I π/ π/2 Fig Three standard contours So we have three regions in which there is decay: π 6 <θ< π (I) 6 π 2 <θ< 5π (II) 6 7π 6 <θ< 3π (III) 2 (4.6.56) There are three standard contours i, i = 1, 2 and 3, in which the integrated term [ ] vanishes as R, depicted in Figure The shaded region refers to regions of decay of v(ζ ). The three solutions, of which only two are linearly independent (because the equation is of second order), are denoted by y i (z) = α i i e zζ ζ 3 /3 dζ (4.6.57) α i being a convenient normalizing factor, i = 1, 2, 3. In order not to have a trivial solution, one must take the contour between any two of the decaying regions. If we change variables ζ = ik then the solution corresponding to i = 2, y 2 (z), is proportional to the Airy function solution A(η) discussed earlier (see Eq. (4.6.4b)). Finally, we remark that this method applies to linear differential equations with coefficients depending linearly on the independent variable. Generalizations to other kernels K (z,ζ)(mentioned below Eq. (4.6.49)) can be made and is employed to solve other linear differential equations such as Bessel functions, Legendre functions, etc. The interested reader may wish to consult a reference such as Jeffreys and Jeffreys (1962).

17 4.6 Applications of Transforms to Differential Equations 31 Problems for Section Use Laplace transform methods to solve the ODE L dy dt + Ry = f (t), y() = y, constants L, R > (a) Let f (t) = sin ω t, ω >, so that the Laplace transform of f (t) is ˆF(s) = ω /(s 2 + ω 2 ). Find y(t) = y e R L t + ω L ( (R/L) 2 + ω 2 )e R L t + R L 2 sin ω t ( (R/L)2 + ω 2 ) ω L cos ω t ( (R/L)2 + ω 2 ) (b) Suppose f (t) is an arbitrary continuous function that possesses a Laplace transform. Use the convolution product for Laplace transforms (Section 4.5) to find y(t) = y e R L t + 1 L t f (t )e R L (t t ) dt (c) Let f (t) = sin ω t in (b) to obtain the result of part (a), and thereby verify your answer. This is an example of an L,R circuit with impressed voltage f (t) arising in basic electric circuit theory. 2. Use Laplace transform methods to solve the ODE d 2 y dt 2 k2 y = f (t), k >, y() = y, y () = y (a) Let f (t) = e k t, k k, k >, so that the Laplace transform of f (t) is ˆF(s) = 1/(s + k ), and find y(t) = y cosh kt + y sinh kt + e kt k k 2 k 2 cosh kt k 2 k + (k /k) 2 k 2 sinh kt k2

18 32 4 Residue alculus and Applications of ontour Integration (b) Suppose f (t) is an arbitrary continuous function that possesses a Laplace transform. Use the convolution product for Laplace transforms (Section 4.5) to find y(t) = y cosh kt + y k sinh kt + t f (t ) sinh k(t t ) k dt (c) Let f (t) = e k t in (b) to obtain the result in part (a). What happens when k = k? 3. onsider the differential equation d 3 y dt 3 + ω 3 y = f (t), ω >, y() = y () = y () = (a) Find that the Laplace transform of the solution, Ŷ (s), satisfies (assuming that f (t) has a Laplace transform ˆF(s)) Ŷ (s) = ˆF(s) s 3 + ω 3 (b) Deduce that the inverse Laplace transform of 1/(s 3 + ω 3 ) is given by and show that h(t) = e ω t 3ω 2 2 3ω 2 eω t/2 cos y(t) = t ( ω 2 h(t ) f (t t ) dt ) π 3t 3 by using the convolution product for Laplace transforms. 4. Let us consider Laplace s equation ( 2 φ)/( x 2 ) + ( 2 φ)/( y 2 ) =, for < x < and y >, with the boundary conditions ( φ/ y)(x, ) = h(x) and φ(x, y) asx 2 + y 2. Find the Fourier transform solution. Is there a constraint on the data h(x) for a solution to exist? If so, can this be explained another way? 5. Given the linear free Schrödinger equation (without a potential) i u t + 2 u x 2 =, with u(x, ) = f (x)

19 4.6 Applications of Transforms to Differential Equations 33 (a) solve this problem by Fourier transforms, by obtaining the Green s function in closed form, and using superposition. Recall that du = e iπ/4 π. eiu2 (b) Obtain the above solution by Laplace transforms. 6. Given the heat equation φ t = 2 φ x 2 with the following initial and boundary conditions φ(x, ) =, φ (x =, t) = g(t), x φ lim φ(x, t) = lim x x x = (a) solve this problem by Fourier cosine transforms. (b) Solve this problem by Laplace transforms. (c) Show that the representations of (a) and (b) are equivalent. 7. Given the wave equation (with wave speed being unity) and the boundary conditions 2 φ t 2 2 φ x 2 = φ(x, t = ) =, φ (x, t = ) =, t φ(x =, t) =, φ(x = l, t) = 1 (a) obtain the Laplace transform of the solution ˆΦ(x, s) ˆΦ(x, s) = sinh sx s sinh sl (b) Obtain the solution φ(x, t) by inverting the Laplace transform to find φ(x, t) = x l + 2( 1) n n=1 nπ ( ) ( ) nπ x nπt sin cos l l (see also Problem (19), Section 4.5).

20 34 4 Residue alculus and Applications of ontour Integration 8. Given the wave equation and the boundary conditions 2 φ t 2 2 φ x 2 = φ(x, t = ) =, φ(x =, t) =, φ (x, t = ) =, t φ(x = l, t) = f (t) (a) show that the Laplace transform of the solution is given by ˆΦ(x, s) = ˆF(s) sinh sx sinh sl where ˆF(s) is the Laplace transform of f (t). (b) all the solution of the problem when f (t) = 1 (so that ˆF(s) = 1/s) to be φ s (x, t). Show that the general solution is given by φ(x, t) = t φ s t (x, t ) f (t t ) dt 9. Use multiple Fourier transforms to solve φ t ( 2 ) φ x + 2 φ = 2 y 2 on the infinite domain < x <, < y <, t >, with φ(x, y) as x 2 + y 2, and φ(x, y, ) = f (x, y). How does the solution simplify if f (x, y) is a function of x 2 + y 2? What is the Green s function in this case? 1. Given the forced heat equation φ t 2 φ = f (x, t), x 2 φ(x, ) = g(x) on < x <, t >, with φ,g, f as x (a) use Fourier transforms to solve the equation. How does the solution compare with the case f =? (b) Use Laplace transforms to solve the equation. How does the method compare with that described in this section for the case f =?

21 4.6 Applications of Transforms to Differential Equations Given the ODE zy + (2r + 1)y + zy = look for a contour representation of the form y = e zζ v(ζ ) dζ. (a) Show that if is a closed contour and v(ζ ) is single valued on this contour, then it follows that v(ζ) = A(ζ 2 + 1) r 1/2. (b) Show that if y = z s w, then when s = r, w satisfies Bessel s equation z 2 w + zw + (z 2 r 2 )w =, and a contour representation of the solution is given by w = Az r e zζ (ζ 2 + 1) r 1/2 dζ Note that if r = n + 1/2 for integer n, then this representation yields the trivial solution. We take the branch cut to be inside the circle when (r 1/2) integer. 12. The hypergeometric equation zy + (a z)y by = has a contour integral representation of the form y = (a) Show that one solution is given by y = 1 e zζ ζ b 1 (1 ζ) a b 1 dζ e zζ v(ζ )dζ. where Re b > and Re (a b) >. (b) Let b = 1, and a = 2; show that this solution is y = (e z 1)/(z), and verify that it satisfies the equation. (c) Show that a second solution, y 2 = vy 1 (where the first solution is denoted as y 1 ) obeys zy 1 v + (2zy 1 + (a z)y 1 )v = Integrate this equation to find v, and thereby obtain a formal representation for y 2. What can be said about the analytic behavior of y 2 near z =? 13. Suppose we are given the following damped wave equation: 2 φ x 1 2 φ 2 c 2 t ϵ φ 2 t = e iωt δ(x ζ ), ω, ϵ >

22 36 4 Residue alculus and Applications of ontour Integration (a) Show that ψ(x) where φ(x, t) = e iωt ψ(x) satisfies ψ + (( ) ω 2 iωϵ) ψ = δ(x ζ) c (b) Show that ˆΨ(k), the Fourier transform of ψ(x),is given by ˆΨ(k) = e ikζ k 2 ( ω c ) 2 + iωϵ (c) Invert ˆΨ(k) to obtain ψ(x), and in particular show that as ϵ + we have ψ(x) = ie i ω c x ζ 2 ( ) ω c and that this has the effect of deforming the contour as described in Figure In this problem we obtain the Green s function of Laplace s equation in the upper half plane, < x <, < y <, by solving 2 G x + 2 G 2 y 2 = δ(x ζ )δ(y η), G(x, y = ) =, G(x, y) as r 2 = x 2 + y 2 (a) Take the Fourier transform of the equation with respect to x and find that the Fourier transform, Ĝ(k, y) = G(x, y)e ikx dk, satisfies 2 Ĝ y 2 k2 Ĝ = δ(y η)e ikζ with Ĝ(k, y = ) = (b) Take the Fourier sine transform of Ĝ(k, y) with respect to y and find, for that it satisfies Ĝ s (k, l) = Ĝ(k, y) sin lydy Ĝ s (k, l) = sin lηe ikζ l 2 + k 2

23 4.6 Applications of Transforms to Differential Equations 37 (c) Invert this expression with respect to k and find whereupon G(x, l) = e l x ζ sin lη 2l G(x, y) = 1 π (d) Evaluate G(x, y) to find G(x, y) = 1 4π log e l x ζ sin lη sin ly l dl ( (x ζ) 2 + (y η) 2 ) (x ζ) 2 + (y + η) 2 Hint: Note that taking the derivative of G(x, y) (of part (c) above) with respect to y yields an integral for ( G)/( y) that is elementary. Then one can integrate this result using G(x, y = ) = to obtain G(x, y).

24