Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 7 docx

for later use.  2π 0 xy dt =  2π 0  a 0 2 + ∞  n=1  a n cos(nt) + b n sin(nt)   c 0 2 + ∞  n=1  c n cos(nt) + d n sin(nt)   dt =  2π 0  a 0 c 0 4 + ∞  n=1  a n c n cos 2 (nt) + b n d n sin 2 (nt)   dt = π  1 2 a 0 c 0 + ∞  n=1 (a n c n + b n d n )  In the arclength parametrization we have  dx ds  2 +  dy ds  2 = 1. In terms of t = 2πs/L this is  dx dt  2 +  dy dt  2 =  L 2π  2 . We integrate this identity on [0, 2π]. L 2 2π =  2π 0   dx dt  2 +  dy dt  2  dt = π  ∞  n=1  (nb n ) 2 + (−na n ) 2  + ∞  n=1  (nd n ) 2 + (−nc n ) 2   = π ∞  n=1 n 2 (a 2 n + b 2 n + c 2 n + d 2 n ) L 2 = 2π 2 ∞  n=1 n 2 (a 2 n + b 2 n + c 2 n + d 2 n ) 1414 We assume that the curve is parametrized so that the area is positive. (Reversing the orientation changes the sign of the area as defined above.) The area is A =  2π 0 x dy dt dt =  2π 0  a 0 2 + ∞  n=1  a n cos(nt) + b n sin(nt)   ∞  n=1  nd n cos(nt) − nc n sin(nt)   dt = π ∞  n=1 n(a n d n − b n c n ) Now we find an upper bound on the area. We will use the inequality |ab| ≤ 1 2 |a 2 + b 2 |, which follows from expanding (a − b) 2 ≥ 0. A ≤ π 2 ∞  n=1 n  a 2 n + b 2 n + c 2 n + d 2 n  ≤ π 2 ∞  n=1 n 2  a 2 n + b 2 n + c 2 n + d 2 n  We can express this in terms of the perimeter. = L 2 4π L 2 ≥ 4πA Now we determine the curves for which L 2 = 4πA. To do this we find conditions for which A is equal to the upper bound we obtained for it above. First note that ∞  n=1 n  a 2 n + b 2 n + c 2 n + d 2 n  = ∞  n=1 n 2  a 2 n + b 2 n + c 2 n + d 2 n  1415 implies that all the coefficients except a 0 , c 0 , a 1 , b 1 , c 1 and d 1 are zero. The constraint, π ∞  n=1 n(a n d n − b n c n ) = π 2 ∞  n=1 n  a 2 n + b 2 n + c 2 n + d 2 n  then becomes a 1 d 1 − b 1 c 1 = a 2 1 + b 2 1 + c 2 1 + d 2 1 . This implies that d 1 = a 1 and c 1 = −b 1 . a 0 and c 0 are arbitrary. Thus curves for which L 2 = 4πA have the parametrization x(t) = a 0 2 + a 1 cos t + b 1 sin t, y(t) = c 0 2 − b 1 cos t + a 1 sin t. Note that  x(t) − a 0 2  2 +  y(t) − c 0 2  2 = a 2 1 + b 2 1 . The curve is a circle of radius  a 2 1 + b 2 1 and center (a 0 /2, c 0 /2). Solution 28.20 1. The Fourier sine series has the form x(1 − x) = ∞  n=1 a n sin(nπx). The norm of the eigenfunctions is  1 0 sin 2 (nπx) dx = 1 2 . The coefficients in the expansion are a n = 2  1 0 x(1 − x) sin(nπx) dx = 2 π 3 n 3 (2 − 2 cos(nπ) − nπ sin(nπ)) = 4 π 3 n 3 (1 − (−1) n ). 1416 Thus the Fourier sine series is x(1 − x) = 8 π 3 ∞  n=1 odd n sin(nπx) n 3 = 8 π 3 ∞  n=1 sin((2n − 1)πx) (2n − 1) 3 . The Fourier cosine series has the form x(1 − x) = ∞  n=0 a n cos(nπx). The norm of the eigenfunctions is  1 0 1 2 dx = 1,  1 0 cos 2 (nπx) dx = 1 2 . The coefficients in the expansion are a 0 =  1 0 x(1 − x) dx = 1 6 , a n = 2  1 0 x(1 − x) cos(nπx) dx = − 2 π 2 n 2 + 4 sin(nπ) − nπ cos(nπ) π 3 n 3 = − 2 π 2 n 2 (1 + (−1) n ) Thus the Fourier cosine series is x(1 − x) = 1 6 − 4 π 2 ∞  n=1 even n cos(nπx) n 2 = 1 6 − 1 π 2 ∞  n=1 cos(2nπx) n 2 . 1417 -1 -0.5 0.5 1 -0.2 -0.1 0.1 0.2 -1 -0.5 0.5 1 -0.2 -0.1 0.1 0.2 Figure 28.13: The odd and even periodic extension of x(1 − x), 0 ≤ x ≤ 1. The Fourier sine series converges to the odd periodic extension of the function. Since this function is C 1 , continuously differentiable, we know that the Fourier coefficients must decay as 1/n 3 . The Fourier cosine series converges to the even periodic extension of the function. Since this function is only C 0 , continuous, the Fourier coefficients must decay as 1/n 2 . The odd and even periodic extensions are shown in Figure 28.13. The sine series is better because of the faster convergence of the series. 2. (a) We substitute x = 0 into the cosine series. 0 = 1 6 − 1 π 2 ∞  n=1 1 n 2 ∞  n=1 1 n 2 = π 2 6 1418 (b) We substitute x = 1/2 into the cosine series. 1 4 = 1 6 − 1 π 2 ∞  n=1 cos(nπ) n 2 ∞  n=1 (−1) n n 2 = − π 2 12 (c) We substitute x = 1/2 into the sine series. 1 4 = 8 π 3 ∞  n=1 sin((2n − 1)π/2) (2n − 1) 3 ∞  n=1 (−1) n (2n − 1) 3 = − π 3 32 1419 Chapter 29 Regular Sturm-Liouville Problems I learned there are troubles Of more than one kind. Some come from ahead And some come from behind. But I’ve bought a big bat. I’m all ready, you see. Now my troubles are going To have troubles with me! -I Had Trouble in Getting to Solla Sollew -Theodor S. Geisel, (Dr. Suess) 29.1 Derivation of the Sturm-Liouville Form Consider the eigenvalue problem on the finite interval [a . . . b], p 2 (x)y  + p 1 (x)y  + p 0 (x)y = µy, 1420 subject to the homogeneous unmixed boundary conditions α 1 y(a) + α 2 y  (a) = 0, β 1 y(b) + β 2 y  (b) = 0. Here the coefficient functions p j are real and continuous and p 2 > 0 on the interval [a . . . b]. (Note that if p 2 were negative we could multiply the equation by (−1) and replace µ by −µ.) The parameters α j and β j are real. We would like to write this problem in a form that can be used to obtain qualitative information about the problem. First we will write the operator in self-adjoint form. We divide by p 2 since it is non-vanishing. y  + p 1 p 2 y  + p 0 p 2 y = µ p 2 y. We multiply by an integrating factor. I = exp   p 1 p 2 dx  ≡ e P (x) e P (x)  y  + p 1 p 2 y  + p 0 p 2 y  = e P (x) µ p 2 y  e P (x) y    + e P (x) p 0 p 2 y = e P (x) µ p 2 y For notational convenience, we define new coefficient functions and parameters. p = e P (x) , q = e P (x) p 0 p 2 , σ = e P (x) 1 p 2 , λ = −µ. Since the p j are continuous and p 2 is positive, p, q, and σ are continuous. p and σ are positive functions. The problem now has the form, (py  )  + qy + λσy = 0, subject to the same boundary conditions, α 1 y(a) + α 2 y  (a) = 0, β 1 y(b) + β 2 y  (b) = 0. This is known as a Regular Sturm-Liouville problem. We will devote much of this chapter to studying the properties of this problem. We will encounter many results that are analogous to the properties of self-adjoint eigenvalue problems. 1421 Example 29.1.1 d dx  ln x dy dx  + λxy = 0, y(1) = y(2) = 0 is not a regular Sturm-Liouville problem since ln x vanishes at x = 1. Result 29.1.1 Any eigenvalue problem of the form p 2 y  + p 1 y  + p 0 y = µy, for a ≤ x ≤ b, α 1 y(a) + α 2 y  (a) = 0, β 1 y(b) + β 2 y  (b) = 0, where the p j are real and continuous and p 2 > 0 on [a, b], and the α j and β j are real can be written in the form of a regular Sturm-Liouville problem, (py  )  + qy + λσy = 0, on a ≤ x ≤ b, α 1 y(a) + α 2 y  (a) = 0, β 1 y(b) + β 2 y  (b) = 0. 29.2 Properties of Regular Sturm-Liouville Problems Self-Adjoint. Consider the Regular Sturm-Liouville equation. L[y] ≡ (py  )  + qy = −λσy. 1422 [...]... that for any smooth p(x), q(x), r(x) and s(x) the eigenfunctions belonging to distinct eigenvalues are orthogonal relative to the weight s(x) That is: b vm (x)vk (x)s(x) dx = 0 if λk = λm a 3 Find the eigenvalues and eigenfunctions for: d4 φ = λφ, dx4 φ(0) = φ (0) = 0, φ(1) = φ (1) = 0 Hint, Solution 144 2 29.5 Hints Hint 29.1 Hint 29.2 Hint 29.3 Hint 29 .4 Write the problem in Sturm-Liouville form... cos(nx)y(x) dx + αyn = fn π 2 ((−1)n y (π) − y (0)) − n2 yn + αyn = fn π Unfortunately we don’t know the values of y (0) and y (π) CONTINUE HERE 143 8 29 .4 Exercises Exercise 29.1 Find the eigenvalues and eigenfunctions of y + 2αy + λy = 0, y(a) = y(b) = 0, where a < b Write the problem in Sturm Liouville form Verify that the eigenvalues and eigenfunctions satisfy the properties of regular Sturm-Liouville... problem and thus the eigenvalues are real Use the Rayleigh quotient to show that there are only positive eigenvalues Informally show that there are an infinite number of eigenvalues with a graph Hint 29.6 Hint 29 .7 Find the solution for λ = 0, λ < 0 and λ > 0 A problem is self-adjoint if it satisfies Green’s identity Hint 29.8 Write the equation in self-adjoint form The Bessel equation of the first kind and. .. y(0) = 0 y(1) + y (1) = 0 Find the eigenfunctions for this problem and the equation which the eigenvalues satisfy Give the general solution in terms of these eigenfunctions Hint, Solution 144 0 Exercise 29 .7 Show that the eigenvalue problem, y + λy = 0, y(0) = 0, y (0) − y(1) = 0, (note the mixed boundary condition), has only one real eigenvalue Find it and the corresponding eigenfunction Show that this... l < z ≤ π This is an example that indicates that the results we obtained in class for eigenfunctions and eigenvalues with q(z) continuous and bounded also hold if q(z) is simply integrable; that is π |q(z)| dz 0 144 1 is finite Hint, Solution Exercise 29.10 1 Find conditions on the smooth real functions p(x), q(x), r(x) and s(x) so that the eigenvalues, λ, of: Lv ≡ (p(x)v (x)) − (q(x)v (x)) + r(x)v(x)... function defined on [a, b] can be expanded in a series of the eigenfunctions, ∞ f (x) ∼ cn φn (x), n=1 where the cn are the generalized Fourier coefficients, cn = φn |σ|f φn |σ|φn 1 Here the sum is convergent in the mean For any fixed x, the sum converges to 2 (f (x− )+f (x+ )) If f (x) is continuous and satisfies the boundary conditions, then the convergence is uniform 142 7 Result 29.2.1 Properties of regular... This form allows us to get upper and lower bounds on λ1 To derive this formula, we first write it in terms of the operator L λ1 = min u − u|L[u] u|σ|u Since u is continuous and satisfies the boundary conditions, we can expand u in a series of the eigenfunctions u|L[u] − =− u|σ|u =− ∞ ∞ n=1 cn φn L [ m=1 cm φm ] ∞ ∞ n=1 cn φn σ m=1 cm φm ∞ ∞ n=1 cn φn − m=1 cm λm σφm ∞ ∞ n=1 cn φn σ m=1 cm φm 142 6 We... ∈ Z+ We expand the solution in a cosine series ∞ y0 yn y(x) = √ + π n=1 2 cos(nx) π We also expand the inhomogeneous term ∞ f0 f (x) = √ + fn π n=1 2 cos(nx) π We multiply the differential equation by the orthonormal functions and integrate over the interval We neglect the 143 7 √ special case φ0 = 1/ π for now π 0 2 cos(nx)y dx + α π 2 cos(nx)y (x) π 2 ((−1)n y (π) − y (0)) + π π 0 π π + 0 0 2 cos(nx)y... quotient, we will find that there is a least eigenvalue Since the eigenvalues are distinct and have no finite cluster point, λn → ∞ as n → ∞ Thus the eigenvalues form an ordered sequence, λ1 < λ 2 < λ 3 < · · · Orthogonal Eigenfunctions Let λ and µ be two distinct eigenvalues with the eigenfunctions φ and ψ Green’s formula states ψ|L[φ] − L[ψ]|φ = 0 ψ| − λσφ − −µσψ|φ = 0 −λ ψ|σ|φ + µ ψ|σ|φ = 0 (µ − λ)... The associated eigenvalue problem is y + y = µy y(0) = 0 y(1) + y (1) = 0 Find the eigenfunctions for this problem and the equation which the eigenvalues must satisfy To do this, consider the eigenvalues and eigenfunctions for, y + λy = 0, y(0) = 0, y(1) + y (1) = 0 Show that the transcendental equation for λ has infinitely many roots λ1 < λ2 < λ3 < · · · Find the limit of λn as n → ∞ How is this limit . µy, for a ≤ x ≤ b, α 1 y(a) + α 2 y  (a) = 0, β 1 y(b) + β 2 y  (b) = 0, where the p j are real and continuous and p 2 > 0 on [a, b], and the α j and β j are real can be written in the form. convergent in the mean. For any fixed x, the sum converges to 1 2 (f(x − )+f(x + )). If f (x) is continuous and satisfies the boundary conditions, then the convergence is uniform. 142 7 Result 29.2.1 Properties. − 2 π 2 n 2 + 4 sin(nπ) − nπ cos(nπ) π 3 n 3 = − 2 π 2 n 2 (1 + (−1) n ) Thus the Fourier cosine series is x(1 − x) = 1 6 − 4 π 2 ∞  n=1 even n cos(nπx) n 2 = 1 6 − 1 π 2 ∞  n=1 cos(2nπx) n 2 . 141 7 -1 -0.5