Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 10 potx

We take the Laplace transform of the differential equation. s 2 ˆy(s) − sy(0) − y  (0) − ˆy(s) = ˆ f(s) s 2 ˆy(s) − s − ˆy(s) = ˆ f(s) ˆy(s) = s s 2 − 1 + ˆ f(s) s 2 − 1 By inspection, (of a table of Laplace transforms), we see that L −1  s s 2 − 1  = cosh(t)H(t), L −1  1 s 2 − 1  = sinh(t)H(t). Now we use the convolution theorem. L −1  ˆ f(s) s 2 − 1  =  t 0 f(τ) sinh(t − τ) dτ The solution for positive t is y(t) = cosh(t) +  t 0 f(τ) sinh(t − τ) dτ. Clearly the solution is continuous because the integral of a bounded function is continuous. The first derivative of the solution is y  (t) = sinh t + f(t) sinh(0) +  t 0 f(τ) cosh(t − τ) dτ y  (t) = sinh t +  t 0 f(τ) cosh(t − τ) dτ We see that the first derivative is also continuous. 1534 Solution 31.21 We consider the problem dy dt +  t 0 y(τ) dτ = e −t , y(0) = 1. We take the Laplace transform of the equation and solve for ˆy. sˆy −y(0) + ˆy s = 1 s + 1 ˆy = s(s + 2) (s + 1)(s 2 + 1) We expand the right side in partial fractions. ˆy = − 1 2(s + 1) + 1 + 3s 2(s 2 + 1) We use a table of Laplace transforms to do the inversion. y = − 1 2 e −t + 1 2 (sin(t) + 3 cos(t)) Solution 31.22 We consider the problem L di 1 dt + Ri 1 + q/C = E 0 L di 2 dt + Ri 2 − q/C = 0 dq dt = i 1 − i 2 i 1 (0) = i 2 (0) = E 0 2R , q(0) = 0. 1535 We take the Laplace transform of the system of differential equations. L  s ˆ i 1 − E 0 2R  + R ˆ i 1 + ˆq C = E 0 s L  s ˆ i 2 − E 0 2R  + R ˆ i 2 − ˆq C = 0 sˆq = ˆ i 1 − ˆ i 2 We solve for ˆ i 1 , ˆ i 2 and ˆq. ˆ i 1 = E 0 2  1 Rs + 1/L s 2 + Rs/L + 2/(CL)  ˆ i 2 = E 0 2  1 Rs − 1/L s 2 + Rs/L + 2/(CL)  ˆq = CE 0 2  1 s − s + R/L s 2 + Rs/L + 2/(CL)  We factor the polynomials in the denominators. ˆ i 1 = E 0 2  1 Rs + 1/L (s + α − ıω)(s + α + ıω)  ˆ i 2 = E 0 2  1 Rs − 1/L (s + α − ıω)(s + α + ıω)  ˆq = CE 0 2  1 s − s + 2α (s + α − ıω)(s + α + ıω)  Here we have defined α = R 2L and ω 2 = 2 LC − α 2 . 1536 We expand the functions in partial fractions. ˆ i 1 = E 0 2  1 Rs + ı 2ωL  1 s + α + ıω − 1 s + α − ıω  ˆ i 2 = E 0 2  1 Rs − ı 2ωL  1 s + α + ıω − 1 s + α − ıω  ˆq = CE 0 2  1 s + ı 2ω  α + ıω s + α − ıω − α − ıω s + α + ıω  Now we can do the inversion with a table of Laplace transforms. i 1 = E 0 2  1 R + ı 2ωL  e (−α−ıω)t − e (−α+ıω)t   i 2 = E 0 2  1 R − ı 2ωL  e (−α−ıω)t − e (−α+ıω)t   q = CE 0 2  1 + ı 2ω  (α + ıω) e (−α+ıω)t −(α − ıω) e (−α−ıω)t   We simplify the expressions to obtain the solutions. i 1 = E 0 2  1 R + 1 ωL e −αt sin(ωt)  i 2 = E 0 2  1 R − 1 ωL e −αt sin(ωt)  q = CE 0 2  1 − e −αt  cos(ωt) + α ω sin(ωt)  Solution 31.23 We consider the problem y  + 4y  + 4y = 4 e −t , y(0) = 2, y  (0) = −3 1537 We take the Laplace transform of the differential equation and solve for ˆy(s). s 2 ˆy −sy(0) − y  (0) + 4sˆy −4y(0) + 4ˆy = 4 s + 1 s 2 ˆy −2s + 3 + 4sˆy −8 + 4ˆy = 4 s + 1 ˆy = 4 (s + 1)(s + 2) 2 + 2s + 5 (s + 2) 2 ˆy = 4 s + 1 − 2 s + 2 − 3 (s + 2) 2 We take the inverse Laplace transform to determine the solution. y = 4 e −t −(2 + 3t) e −2t 1538 Chapter 32 The Fourier Transform 32.1 Derivation from a Fourier Series Consider the eigenvalue problem y  + λy = 0, y(−L) = y(L), y  (−L) = y  (L). The eigenvalues and eigenfunctions are λ n =  nπ L  2 for n ∈ Z 0+ φ n = π L e ınπx/L , for n ∈ Z The eigenfunctions form an orthogonal set. A piecewise continuous function defined on [−L . . . L] can be expanded in a series of the eigenfunctions. f(x) ∼ ∞  n=−∞ c n π L e ınπx/L 1539 The Fourier coefficients are c n =  π L e ınπx/L    f(x)   π L e ınπx/L    π L e ınπx/L  = 1 2π  L −L e −ınπx/L f(x) dx. We substitute the expression for c n into the series for f(x). f(x) ∼ ∞  n=−∞  1 2L  L −L e −ınπξ/L f(ξ) dξ  e ınπx/L . We let ω n = nπ/L and ∆ω = π/L. f(x) ∼ ∞  ω n =−∞  1 2π  L −L e −ıω n ξ f(ξ) dξ  e ıω n x ∆ω. In the limit as L → ∞, (and thus ∆ω → 0), the sum becomes an integral. f(x) ∼  ∞ −∞  1 2π  ∞ −∞ e −ıωξ f(ξ) dξ  e ıωx dω. Thus the expansion of f(x) for finite L f(x) ∼ ∞  n=−∞ c n π L e ınπx/L c n = 1 2π  L −L e −ınπx/L f(x) dx 1540 in the limit as L → ∞ becomes f(x) ∼  ∞ −∞ ˆ f(ω) e ıωx dω ˆ f(ω) = 1 2π  ∞ −∞ f(x) e −ıωx dx. Of course this derivation is only heuristic. In the next section we will explore these formulas more carefully. 32.2 The Fourier Transform Let f(x) be piecewise continuous and let  ∞ −∞ |f(x)|dx exist. We define the function I(x, L). I(x, L) = 1 2π  L −L   ∞ −∞ f(ξ) e ıωξ dξ  e −ıωx dω. Since the integral in parentheses is uniformly convergent, we can interchange the order of integration. = 1 2π  ∞ −∞   L −L f(ξ) e ıω(ξ−x) dω  dξ = 1 2π  ∞ −∞  f(ξ) e ıω(ξ−x) ı(ξ −x)  L −L dξ = 1 2π  ∞ −∞ f(ξ) 1 ı(ξ −x)  e ıL(ξ−x) − e −ıL(ξ−x)  dξ = 1 π  ∞ −∞ f(ξ) sin(L(ξ −x)) ξ −x dξ = 1 π  ∞ −∞ f(ξ + x) sin(Lξ) ξ dξ. 1541 In Example 32.3.3 we will show that  ∞ 0 sin(Lξ) ξ dξ = π 2 . Continuous Functions. Suppose that f(x) is continuous. f(x) = 1 π  ∞ −∞ f(x) sin(Lξ) ξ dξ I(x, L) − f(x) = 1 π  ∞ −∞ f(x + ξ) − f(x) ξ sin(Lξ) dξ. If f (x) has a left and right derivative at x then f(x+ξ)−f(x) ξ is bounded and  ∞ −∞    f(x+ξ)−f(x) ξ    dξ < ∞. We u se the Riemann-Lebesgue lemma to show that the integral vanishes as L → ∞. 1 π  ∞ −∞ f(x + ξ) − f(x) ξ sin(Lξ) dξ → 0 as L → ∞. Now we have an identity for f(x). f(x) = 1 2π  ∞ −∞   ∞ −∞ f(ξ) e ıωξ dξ  e −ıωx dω. Piecewise Continuous Functions. Now consider the case that f(x) is only piecewise continuous. f(x + ) 2 = 1 π  ∞ 0 f(x + ) sin(Lξ) ξ dξ f(x − ) 2 = 1 π  0 −∞ f(x − ) sin(Lξ) ξ dξ 1542 [...]... collect the result for positive and negative ω ∞ e2ω−3η dη F[f (x)] = F[f (x)] = 0 1 −|ω| 1 −2|ω| e − e 6 12 A better way to find the Fourier transform of f (x) = x4 1 + 5x2 + 4 is to first expand the function in partial fractions f (x) = 1/3 1/3 − 2 +1 x +4 x2 1 2 1 4 F[f (x)] = F 2 − F 2 6 x +1 12 x +4 1 −|ω| 1 −2|ω| = e − e 6 12 32 .4. 4 Parseval’s Theorem Recall Parseval’s theorem for Fourier series... sine transform pair is defined: ∞ f (x) = −1 ˆ Fs [fs (ω)] ˆ fs (ω) sin(ωx) dω =2 1 ˆ fs (ω) = Fs [f (x)] = π 0 ∞ f (x) sin(ωx) dx 0 32.7 Properties of the Fourier Cosine and Sine Transform 32.7.1 Transforms of Derivatives Cosine Transform Using integration by parts we can find the Fourier cosine transform of derivatives Let y be a function for which the Fourier cosine transform of y and its first and second... the Fourier transform 32.3.3 Analytic Continuation Consider the Fourier transform of f (x) = 1 The Fourier integral is not convergent, and its principal value does not exist Thus we will have to be a little creative in order to define the Fourier transform Define the two functions   1 0 for x > 0 for x > 0   f+ (x) = 1/2 for x = 0 , f− (x) = 1/2 for x = 0     0 for x < 0 1 for x < 0 1550 Note... convolution theorem 1555 Example 32 .4. 2 Using the convolution theorem and the table of Fourier transform pairs in the appendix, we can find the Fourier transform of 1 f (x) = 4 x + 5x2 + 4 We factor the fraction f (x) = 1 (x2 + 1)(x2 + 4) From the table, we know that F x2 2c = e−c|ω| + c2 for c > 0 We apply the convolution theorem F[f (x)] = F 1 8 1 = 8 1 2 4 2 + 1 x2 + 4 8x ∞ e−|η| e−2|ω−η| dη = −∞ 0... continuation we can define the transform for all nonzero ω ı F[f− (x)] = 2πω = 1551 dx Now we are prepared to define the Fourier transform of f (x) = 1 F[1] = F[f− (x)] + F[f+ (x)] ı ı =− + 2πω 2πω = 0, for ω = 0 When ω = 0 the integral diverges When we consider the closure relation for the Fourier transform we will see that F[1] = δ(ω) 32 .4 Properties of the Fourier Transform In this section we will explore... Transform The Fourier transform is useful in solving some differential equations on the domain (−∞ ∞) with homogeneous boundary conditions at infinity We take the Fourier transform of the differential equation L[y] = f and solve for y We ˆ take the inverse transform to determine the solution y Note that this process is only applicable if the Fourier transform of y exists Hence the requirement for homogeneous... If ω < 0, we can close the contour in the upper half plane to obtain ı ˆ f (ω) = , for ω < 0 2 For ω = 0 the integral vanishes because 1 x is an odd function ∞ ˆ(0) = 1 = − 1 dx = 0 f 2π −∞ x 1 549 We collect the results in one formula ı ˆ f (ω) = − sign(ω) 2 We write the integrand for ω > 0 as the sum of an odd and and even function ∞ − −∞ ∞ ı 1 1 −ıωx e dx = − − 2π −∞ x 2 ∞ 1 −ı cos(ωx) dx + − sin(ωx)... 1 = f− (x) + f+ (x) The Fourier transform of f+ (x) converges for (ω) < 0 ∞ 1 e−ıωx dx F[f+ (x)] = 2π 0 ∞ 1 e(−ı (ω)+ (ω))x dx = 2π 0 ∞ 1 e−ıωx = 2π −ıω 0 ı for (ω) < 0 =− 2πω Using analytic continuation, we can define the Fourier transform of f+ (x) for all ω except the point ω = 0 ı F[f+ (x)] = − 2πω We follow the same procedure for f− (x) The integral converges for 1 2π 1 = 2π (ω) > 0 0 e−ıωx dx F[f−... (see Exercise 32.3), that the Fourier sine transform of the first and second derivatives are ˆ Fs [y ] = −ω fc (ω) Fs [y ] = −ω 2 yc (ω) + ˆ 1565 ω y(0) π 32.7.2 Convolution Theorems Cosine Transform of a Product Consider the Fourier cosine transform of a product of functions Let f (x) and ˆ g(x) be two functions defined for x ≥ 0 Let Fc [f (x)] = fc (ω), and Fc [g(x)] = gc (ω) ˆ 1 π 1 = π 2 = π ∞ Fc [f... = x − ıα 2π = ı eαω F e−ıωx , iα x − iα We combine the results for positive and negative values of ω F 32.3.2 0 ı eαω 1 = x − ıα for ω > 0, for ω < 0 Cauchy Principal Value and Integrals that are Not Absolutely Convergent That the integral of f (x) is absolutely convergent is a sufficient but not a necessary condition that the Fourier transform ∞ ∞ of f (x) exists The integral −∞ f (x) e−ıωx dx may converge . problem y  + 4y  + 4y = 4 e −t , y(0) = 2, y  (0) = −3 1537 We take the Laplace transform of the differential equation and solve for ˆy(s). s 2 ˆy −sy(0) − y  (0) + 4sˆy −4y(0) + 4 y = 4 s + 1 s 2 ˆy. −2s + 3 + 4sˆy −8 + 4 y = 4 s + 1 ˆy = 4 (s + 1)(s + 2) 2 + 2s + 5 (s + 2) 2 ˆy = 4 s + 1 − 2 s + 2 − 3 (s + 2) 2 We take the inverse Laplace transform to determine the solution. y = 4 e −t −(2. define the Fourier transform. Define the two functions f + (x) =      1 for x > 0 1/2 for x = 0 0 for x < 0 , f − (x) =      0 for x > 0 1/2 for x = 0 1 for x < 0 . 1550 Note