1. Trang chủ
  2. » Kỹ Thuật - Công Nghệ

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx

40 223 0

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 40
Dung lượng 309,96 KB

Nội dung

 x 2α+1 y    + λx 2α−1 y = 0, y(a) = y(b) = 0 Now we verify that the Sturm-Liouville properties are satisfied. • The eigenvalues λ n = α 2 +  nπ ln(b/a)  2 , n ∈ Z are real. • There are an infinite number of eigenvalues λ 1 < λ 2 < λ 3 < ··· , α 2 +  π ln(b/a)  2 < α 2 +  2π ln(b/a)  2 < α 2 +  3π ln(b/a)  2 < ··· There is a least eigenvalue λ 1 = α 2 +  π ln(b/a)  2 , but there is no greatest eigenvalue, (λ n → ∞ as n → ∞). • For each eigenvalue, we found one unique, (to within a multiplicative constant), eigenfunction φ n . We were able to choose the eigenfunctions to be real-valued. The eigenfunction φ n = x −α sin  nπ ln(x/a) ln(b/a)  . has exactly n − 1 zeros in the open interval a < x < b. 1454 • The eigenfunctions are orthogonal with respect to the weighting function σ(x) = x 2α−1 .  b a φ n (x)φ m (x)σ(x) dx =  b a x −α sin  nπ ln(x/a) ln(b/a)  x −α sin  mπ ln(x/a) ln(b/a)  x 2α−1 dx =  b a sin  nπ ln(x/a) ln(b/a)  sin  mπ ln(x/a) ln(b/a)  1 x dx = ln(b/a) π  π 0 sin(nx) sin(mx) dx = ln(b/a) 2π  π 0 (cos((n − m)x) − cos((n + m)x)) dx = 0 if n = m • The eigenfunctions are complete. Any piecewise continuous function f(x) defined on a ≤ x ≤ b can be expanded in a series of eigenfunctions f(x) ∼ ∞  n=1 c n φ n (x), where c n =  b a f(x)φ n (x)σ(x) dx  b a φ 2 n (x)σ(x) dx . The sum converges to 1 2 (f(x − ) + f(x + )). (We do not prove this property.) 1455 • The eigenvalues can be rel ated to the eigenfunctions with the Rayleigh quotient. λ n =  −pφ n dφ n dx  b a +  b a  p  dφ n dx  2 − qφ 2 n  dx  b a φ 2 n σ dx =  b a  x 2α+1  x −α−1  nπ ln(b/a) cos  nπ ln(x/a) ln(b/a)  − α sin  nπ ln(x/a) ln(b/a)  2  dx  b a  x −α sin  nπ ln(x/a) ln(b/a)  2 x 2α−1 dx =  b a   nπ ln(b/a)  2 cos 2 (·) − 2α nπ ln(b/a) cos (·) sin (·) + α 2 sin 2 (·)  x −1 dx  b a sin 2  nπ ln(x/a) ln(b/a)  x −1 dx =  π 0   nπ ln(b/a)  2 cos 2 (x) − 2α nπ ln(b/a) cos(x) sin(x) + α 2 sin 2 (x)  dx  π 0 sin 2 (x) dx = α 2 +  nπ ln(b/a)  2 Now we expand a function f(x) in a series of the eigenfunctions. f(x) ∼ ∞  n=1 c n x −α sin  nπ ln(x/a) ln(b/a)  , where c n =  b a f(x)φ n (x)σ(x) dx  b a φ 2 n (x)σ(x) dx = 2n ln(b/a)  b a f(x)x α−1 sin  nπ ln(x/a) ln(b/a)  dx 1456 Solution 29.4 y  − y  + λy = 0, y(0) = y(1) = 0. The factor that will p ut this equation in Sturm-Liouville form is F (x) = exp   x −1 dx  = e −x . The differential equation becomes d dx  e −x y   + λ e −x y = 0. Thus we see that the eigenfunctions will be orthogonal with respect to the weighting function σ = e −x . Substituting y = e αx into the differential equation yields α 2 − α + λ = 0 α = 1 ± √ 1 − 4λ 2 α = 1 2 ±  1/4 − λ. If λ < 1/4 then the solutions to the differential equation are exponential and only the trivial solution satisfi es the boundary conditions. If λ = 1/4 then the solution is y = c 1 e x/2 +c 2 x e x/2 and again only the trivial solution satisfies the boundary conditions. Now consider the case that λ > 1/4. α = 1 2 ± ı  λ − 1/4 The solutions are e x/2 cos(  λ − 1/4 x), e x/2 sin(  λ − 1/4 x). The left boundary condition gives us y = c e x/2 sin(  λ − 1/4 x). 1457 The right boundary condition demands that  λ − 1/4 = nπ, n = 1, 2, . . . Thus we see that the eigenvalues and eigenfunctions are λ n = 1 4 + (nπ) 2 , y n = e x/2 sin(nπx). If f(x) is a piecewise continuous function then we can expand it in a series of the eigenfunctions. f(x) = ∞  n=1 a n e x/2 sin(nπx) The coefficients are a n =  1 0 f(x) e −x e x/2 sin(nπx) dx  1 0 e −x ( e x/2 sin(nπx)) 2 dx =  1 0 f(x) e −x/2 sin(nπx) dx  1 0 sin 2 (nπx) dx = 2  1 0 f(x) e −x/2 sin(nπx) dx. Solution 29.5 Consider the eigenvalue problem y  + λy = 0 y(0) = 0 y(1) + y  (1) = 0. Since this is a Sturm-Liouville problem, there are only real eigenvalues. By the Rayleigh quotient, the eigenvalu es are λ = −φ dφ dx    1 0 +  1 0   dφ dx  2  dx  1 0 φ 2 dx , 1458 λ = φ 2 (1) +  1 0   dφ dx  2  dx  1 0 φ 2 dx . This demonstrates that there are only positive eigenvalues. The general solu tion of the differential equation for positive, real λ is y = c 1 cos  √ λx  + c 2 sin  √ λx  . The solution that satisfies the left boundary condition is y = c sin  √ λx  . For nontrivial solutions we must have sin  √ λ  + √ λ cos  √ λ  = 0 √ λ = −tan  √ λ  . The positive solutions of this equation are eigenvalues with corresponding eigenfunctions sin  √ λx  . In Figure 29.1 we plot the functions x and −tan(x) and draw vertical lines at x = (n − 1/2)π, n ∈ N. From this we see that there are an infinite number of eigenvalues, λ 1 < λ 2 < λ 3 < ···. In the limit as n → ∞, λ n → (n − 1/2)π. The limit is approached from above. Now consider the eigenvalue problem y  + y = µy y(0) = 0 y(1) + y  (1) = 0. From above we see that the eigenvalues satisfy  1 − µ = −tan   1 − µ  and that there are an infinite number of eigenvalu es. For large n, µ n ≈ 1 − (n − 1/2)π. The eigenfunctions are φ n = sin   1 − µ n x  . 1459 Figure 29.1: x and −tan(x). To solve the inhomogeneous problem, we expand the solution and the inhomogeneity in a series of the eigenfunctions. f = ∞  n=1 f n φ n , f n =  1 0 f(x)φ n (x) dx  1 0 φ 2 n (x) dx y = ∞  n=1 y n φ n We substitite the expansions into the diffe rential equation to determine the coefficients. y  + y = f ∞  n=1 µ n y n φ n = ∞  n=1 f n φ n y = ∞  n=1 f n µ n sin   1 − µ n x  1460 Solution 29.6 Consider the eigenvalue problem y  + y = µy y(0) = 0 y(1) + y  (1) = 0. From Exercise 29.5 we see that the eigenvalues satisfy  1 − µ = −tan   1 − µ  and that there are an infinite number of eigenvalu es. For large n, µ n ≈ 1 − (n − 1/2)π. The eigenfunctions are φ n = sin   1 − µ n x  . To solve the inhomogeneous problem, we expand the solution and the inhomogeneity in a series of the eigenfunctions. f = ∞  n=1 f n φ n , f n =  1 0 f(x)φ n (x) dx  1 0 φ 2 n (x) dx y = ∞  n=1 y n φ n We substitite the expansions into the diffe rential equation to determine the coefficients. y  + y = f ∞  n=1 µ n y n φ n = ∞  n=1 f n φ n y = ∞  n=1 f n µ n sin   1 − µ n x  Solution 29.7 First consider λ = 0. The general solution is y = c 1 + c 2 x. 1461 y = cx satisfies the boundary conditions. Thus λ = 0 is an eigenvalue . Now consider negative real λ. The general solution is y = c 1 cosh  √ −λx  + c 2 sinh  √ −λx  . The solution that satisfies the left boundary condition is y = c sinh  √ −λx  . For nontrivial solutions of the boundary value problem, there must be negative real solutions of √ −λ − sinh  √ −λ  = 0. Since x = sinh x has no nonzero real solutions, this equation has no solutions for negative real λ. There are no negative real eigenvalues. Finally consider positive real λ. The general solution is y = c 1 cos  √ λx  + c 2 sin  √ λx  . The solution that satisfies the left boundary condition is y = c sin  √ λx  . For nontrivial solutions of the boundary value problem, there must be positive real solutions of √ λ − sin  √ λ  = 0. Since x = sin x has no nonzero real solutions, this equation has no solutions for positive real λ. There are no positive real eigenvalues. There is only one real eigenvalue, λ = 0, with corresponding eigenfunction φ = x. 1462 [...]... cos(λ1 /4 x) + c2 cosh(λ1 /4 x) + c3 sin(λ1 /4 x) + c4 sinh(λ1 /4 x) Applying the condition φ(0) = 0 we obtain φ = c1 (cos(λ1 /4 x) − cosh(λ1 /4 x)) + c2 sin(λ1 /4 x) + c3 sinh(λ1 /4 x) The condition φ (0) = 0 reduces this to φ = c1 sin(λ1 /4 x) + c2 sinh(λ1 /4 x) We substitute the solution into the two right boundary conditions c1 sin(λ1 /4 ) + c2 sinh(λ1 /4 ) = 0 −c1 λ1/2 sin(λ1 /4 ) + c2 λ1/2 sinh(λ1 /4 ) = 0... log 2 = 147 4 1 dx + x 1 1 dx + x 2 1 dx x 2 1 dx x Chapter 31 The Laplace Transform 31.1 The Laplace Transform The Laplace transform of the function f (t) is defined ∞ e−st f (t) dt, L[f (t)] = 0 for all values of s for which the integral exists The Laplace transform of f (t) is a function of s which we will denote ˆ f (s) 1 A function f (t) is of exponential order α if there exist constants t0 and M such... S(x) < for all τ > T (x, ) c The sum is uniformly convergent if T is independent of x Similar to the Weierstrass M-test for infinite sums we have a uniform convergence test for integrals If there exists ∞ ∞ a continuous function M (t) such that |f (x, t)| ≤ M (t) and c M (t) dt is convergent, then c f (x, t) dt is uniformly convergent 147 0 If ∞ c f (x, t) dt is uniformly convergent, we have the following... sin(λ1 /4 ) = 0 The eigenvalues and eigenfunctions are λn = (nπ )4 , φn = sin(nπx), 146 9 n ∈ N Chapter 30 Integrals and Convergence Never try to teach a pig to sing It wastes your time and annoys the pig -? 30.1 Uniform Convergence of Integrals Consider the improper integral ∞ f (x, t) dt c The integral is convergent to S(x) if, given any > 0, there exists T (x, ) such that τ f (x, t) dt − S(x) < for all... |f (t)| < M eαt , for all t > t0 t ˆ If 0 0 f (t) dt exists and f (t) is of exponential order α then the Laplace transform f (s) exists for Here are a few examples of these concepts • sin t is of exponential order 0 1 Denoting the Laplace transform of f (t) as F (s) is also common 147 5 (s) > α • t e2t is of exponential order α for any α > 2 2 • et is not of exponential order α for any α • tn is of... Laplace transform of this function ∞ ˆ f (s) = e−st t et dt 0 ∞ t e(1−s)t dt = 0 ∞ ∞ 1 1 e(1−s)t dt t e(1−s)t − = 1−s 1−s 0 0 ∞ 1 e(1−s)t =− (1 − s)2 0 1 for (s) > 1 = (1 − s)2 147 6 Example 31.1.3 Consider the Laplace transform of the Heaviside function, 0 1 H(t − c) = for t < c for t > c, where c > 0 ∞ e−st H(t − c) dt L[H(t − c)] = 0 ∞ e−st dt = c e−st ∞ = −s c −cs e = for s (s) > 0 Example 31.1 .4 Next... as R → ∞ 1 48 2 ˆ Thus we have an expression for the inverse Laplace transform of f (s) 1 ı2π N α+ı∞ ˆ Res(est f (s), sn ) ˆ e f (s) ds = st α−ı∞ n=1 N ˆ Res(est f (s), sn ) ˆ L−1 [f (s)] = n=1 ˆ ˆ Result 31.2.1 If f (s) is analytic except for poles at s1 , s2 , , sN and f (s) → 0 as |s| → ∞ ˆ then the inverse Laplace transform of f (s) is N ˆ f (t) = L [f (s)] = ˆ Res(est f (s), sn ), −1 for t > 0... cut from s = 0 to s = −∞ and e−ıθ/2 1 √ = √ , s r 1 √ s √ s denotes the principal branch of s1/2 There is a for − π < θ < π 1 48 4 Let α be any positive number The inverse Laplace transform of α+ı∞ 1 f (t) = ı2π α−ı∞ 1 √ s is 1 est √ ds s We will evaluate the integral by deforming it to wrap around the branch cut Consider the integral on the contour + − shown in Figure 31.3 CR and CR are circular arcs... (s) The Inverse Laplace Transform The inverse Laplace transform in denoted ˆ f (t) = L−1 [f (s)] 147 7 We compute the inverse Laplace transform with the Mellin inversion formula α+ı∞ 1 ı2π f (t) = ˆ est f (s) ds α−ı∞ ˆ Here α is a real constant that is to the right of the singularities of f (s) To see why the Mellin inversion formula is correct, we take the Laplace transform of it Assume that f (t)... following a straight line to 1 + ıR; let CR be the contour starting at 1 + ıR and following a semicircular path down to 1 − ıR Let C be the combination of BR and CR This contour is shown in Figure 31.2 Im(s) α+iR CR BR Re(s) α-iR Figure 31.2: The Path of Integration for the Inverse Laplace Transform 1 48 0 Consider the line integral on C for R > 1 1 ı2π est C 1 1 ds = Res est 2 , 0 2 s s d st e = ds s=0 =t . s. 146 8 3. From part (a) we know that there are only positive eigenvalues. The general solution of the differential equation is φ = c 1 cos(λ 1 /4 x) + c 2 cosh(λ 1 /4 x) + c 3 sin(λ 1 /4 x) + c 4 sinh(λ 1 /4 x). Applying. ≤ √ 6 Thus the smallest zero of J 0 (x) is less than or equal to √ 6 ≈ 2 .44 94. (The smallest zero of J 0 (x) is approximately 2 .40 483 .) Solution 29.9 We assume that 0 < l < π. Recall that the. 1 /4 x). The left boundary condition gives us y = c e x/2 sin(  λ − 1 /4 x). 145 7 The right boundary condition demands that  λ − 1 /4 = nπ, n = 1, 2, . . . Thus we see that the eigenvalues and

Ngày đăng: 06/08/2014, 01:21