 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