Math 5440 Problem Set 7 Solutions

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1 Math 544 Math 544 Problem Set 7 Solutions Aaron Fogelson Fall, 13 1: (Logan, 3. # 1) Verify that the set of functions {1, cos(x), cos(x),...} form an orthogonal set on the interval [, π]. Next verify that the set of functions cos(nπx/l), n =, 1,,... form an orthogonal set on the interval [, L]. If f(x) = c n cos(nπx/l), n= in the mean-square sense on [, L], what are the formulas for the c n? Let m and n. Then π cos(mx) cos(nx)dx = if m =n = π 1 {cos((m+n)x)+cos((m n)x)} dx (1) 1 sin((m+n)x) π m+n + 1 sin((m n)x) π m n =. () To show{1, cos(πx/l), cos(πx/l),...} are orthogonal on[, L], note that with the change of variables y = πx/l, L cos ( nπx ) cos L ( mπx from the above provided m = n. If f(x) = c n cos(nπx/l), n= then L ) dx = L π cos(ny) cos(my)dy =, π in L [, L], ( f, cos(mπx/l)) = c m (cos(mπx/l), cos(mπx/l)). From (1), in case m = n, we see that (cos(mπx/l), cos(mπx/l)) = (L/π)(π/) = L/, so c n = L ( nπx ) f(x) cos dx. L L 1

2 : (Logan, 3. # 4) For which powers of r is the function f(x) = x r in L [, 1]? In L [, ]? 1 x r dx 1 = lim x r dx a a x r+1 = lim r+1 1 r = 1/ a a ln(x) 1 a r = 1/ { 1 = lim r+1 r+1 ar+1 r = 1/ a ln(a) r = 1/. The limit is finite for r > 1/ because then the exponent r+1 >, so x r L [, 1] for r > 1/. Only for r > 1/ could xr dx be finite because this integral is greater than the integral of x r over the smaller interval [, 1]. Assume r > 1/. x r dx x = r+1 lim b r+1 1 = lim b r+ 1 br+1. Only if r+1 < is this finite, but we ve already set that r > 1/, so there is no value of r for which x r L [, ]. b

3 3: (Logan, 3. # 5) Let f be defined and integrable on [, L]. The orthognal expansion b n sin n=1 ( nπx ), b n = L f(x) sin L L ( nπx ) dx, L is called the Fourier sine series for f on [, L]. Find the Fourier sine series for f(x) = cos(x) on [, π/]. What is the Fourier sine series for f(x) = sin x on [, π]? For f(x) = cos(x) on [, π/], π/ b n = cos(x) sin(nx)dx π/ = 4 1 π/ {sin(nx+x)+sin(nx x)} dx π = { cos((n+1)x) cos((n 1)x) } π/ π n+1 n 1 = { cos((n+1)π/) cos((n 1)π/) } + π n+1 n 1 π { } n 1+n+1 { 1 n } n 1 So = π (n+1)(n 1) = 8 n π 4n 1. cos(x) = n=1 8n π(4n 1) sin(nx). The Fourier Sine Series for sin(x) on [, π] is the single term sin(x). 3

4 4: (Logan, 3. # 8) For f, g L [a, b], prove the Cauchy-Schwarz inequality ( f, g) f g. Let q(t) = ( f + tg, f + tg) = ( f, f)+( f, g)t+(g, g)t. Because q(t), q(t) has 1 or no real roots. The roots t ± are t ± = ( f, g)± 4( f, g) 4(g, g)( f, f), (g, g) and for there to be at most one real root the discriminant must be nonpositive. Hence ( f, g) (g, g)( f, f), or ( f, g) (g, g) 1/ ( f, f) 1/, 4

5 5: (Logan, 3.3 # ) Find the Fourier series for f(x) = x on [ π, π]. Sketch a graph of the function defined on all of R to which the Fourier series converges. Is the convergence pointwise? Use the Fourier series to show Graph the frequency spectrum. π 1 = The Fourier coefficients for f(x) = x on [ π, π] are and a n = 1 π b n = 1 π π π π π x cos(nx)dx, x sin(nx)dx. Since x is even and sin(nx) is odd, their product is odd, and the integrals defining b n are all, so b n = for all n. The integral for a can be evaluated easily and give a = π /3. The integrals for a n for n 1 can be found using integration by parts twice as follows So a n { = 1 x π sin(nx) π n π = π x sin(nx)dx πn π = { x cos(nx) π πn n = { π( 1) n πn n π π = n (( 1)n ) = 4 n ( 1)n π x sin(nx) } dx n π π + π( 1)n n ( cos(nx) n + sin(nx) n x = π n n=1 ( 1)n cos(nx). ) } dx Dividing both sides of this equation by 4 and evaluating the result at x =, we get = π 1 +( ) which upon rearranging gives the desired sum. The function to which the Fourier series for x on[ π, π] is the extension of x from this interval to the entire real line as a π periodic function. The extended function is twice continuously differentiable except when x equals an integer multiple of π. The extended function is continuous for all x and is π periodic by construction. The first derivative of the extended function is continuous 5 π π }

6 except when x is an integer multiple of π. By inspection a n = O(1/n ) and b n =, so the series converges pointwise to the extended function at points at which that function is continuous, that is, for all x. Because the series with terms a n+ b n converges and the extended function is continuous and π-periodic, the convergence of the Fourier series is in fact uniform x x x x Figure.1: Upper left: The extended function. Upper Right: A three term approximation. Lower Left: A five term approximation. Lower Right: A seven term approximation. 6

7 Figure.: Frequency Spectrum. 7

8 6: (Logan, 3.3 # 5) Let f(x) = 1/ on π < x < and f(x) = 1/ on < x < π. Show that the Fourier series for f is n=1 (n 1)π sin(n 1)x. If s N (x) denotes the sum of the first N terms, sketch the graphs of s 1 (x), s 3 (x), s 7 (x), and s 1 (x) and compare with f(x). Observe that the approximations overshoot f(x) in the neighborhood of x =, and the overshoot is not improved regardless of how many terms are taken in the approximation. This overshoot behavior of Fourier series near a discontinuity is call the Gibbs phenomenon. Since f(x) is odd, a n = for all n. b n = π = 1 f(x) sin(nx)dx π π = 1 { π } sin(nx)dx sin(nx)dx π π { = 1 cos(nx) π π n cos(nx) } n π = 1 { ( 1)n + 1 π n n + 1 } n ( 1)n n = 1 { } π n ( 1)n n { n even nπ n odd. All positive odd integers can be written n 1 for n = 1,, 3,..., so b n 1 = (n 1)π. 8

9 x x x x Figure.3: Plot of partial sums s 1 (x), s 3 (x), s 7 (x), and s 1 (x). 9

10 7: (Logan, 3.4 # 1) Show that the substitution of u(x, t) = g(t)y(x) into the PDE leads to the pair of differential equations u t = (p(x)u x ) x q(x)u, a < x < b, t >, g = λg, (p(x)y ) + q(x)y = λy, where λ is some constant. Let u(x, t) = g(t)y(x). Then u t (x, t) = g (t)y(x) and u x (x, t) = g(t)y (x). Substituting these expressions into the PDE, we find g (t)y(x) = (p(x)g(t)y (x)) x q(x)g(t)y(x) = g(t)(py ) qgy. Rearranging this equation yields g g = (py ) qy. y Since the left side is a function of t and the right side is a function of x, both sides must equal the same constant which we call λ. Hence, g = λg, (py ) + qy = λy. 1

11 8: (Logan, 3.4 #) Show that the SLP y (x) = λy(x), < x < L, y() =, y(l) =, has eigenvalues λ n = n π /L and corresponding eigenfunctions y n (x) = sin(nπx/l), n = 1,,... We know from the energy argument presented in class that the eigenvalues for this problem (with p(x) = 1, q(x) =, and Dirichlet boundary conditions) are positive. So let λ = k for k >. Then the general solution of the ODE is y(x) = A cos(kx)+ B sin(kx) where A and B are arbitrary constants. The boundary conidtions y() = implies that A =, and the boundary condition y(l) = then implies that = B sin(kl). We cannot choose B = because this would give us an identically zero function, so we must have sin(kl) =. Therefore, kl is an integer multiple of π, i.e., kl = nπ for positive integer n. So λ n = n π /L and y n (x) = sin(kx) = sin(nπx/l). 11

12 9: (Logan, 3.4 #3) Show that the SLP y (x) = λy(x), < x < L, y () =, y(l) =, with mixed Dirichlet and Neumann boundary conditions has eigenvalues and corresponding eigenfunctions ( (1+n)π λ n = L ) for n =, 1,... y n (x) = cos (1+n)πx L Again, the conditions of this SLP are such that the energy argument works and we therefore conclude that all eigenvalues are positive. Let λ = k for k >. The general solution of the ODE is y(x) = A cos(kx)+ B sin(kx) where A and B are arbitrary constants. The derivative of y(x) is y (x) = Ak sin(kx)+ Bk cos(kx). The boundary conidtion y () = implies that B =, and y(x) = A cos(kx), and the boundary condition y(l) = then implies that = A cos(kl). Hence Lk must ( (n 1)π L ( ) be an odd multiple of π/. So k n = (n 1)π ) L and yn (x) = cos, for n = 1,, 3,.... 1

13 1: (Logan, 3.4 #6) Consider the SLP y (x) = λy(x), < x < 1; y()+y () =, y(l) =. Is λ = an eigenvalue? Are there any negative eigenvalues? Show that there are infinitely many positive eigenvalues by finding an equation whose roots are those eigenvalues, and show graphically that there are infinitely many roots. The energy argument would give us the relation y() (y ) dx = λ y dx, and the sign of the left hand sign is ambiguous, so we cannot use the energy argument to conclude anything about the eigenvalues. So we consider, in turn, the possibilities λ =, λ <, and λ >. If λ =, the general soluton of the ODE is y(x) = Ax+B for constants A and B. So y (x) = A. Both boundary conditions reduce to the condition A+B =, so B = A. Therefore, λ = is an eigenvalue with eigenfunction y (x) = x 1. If λ <, let λ = k. The general soluton of the ODE is y(x) = Ae kx + Be kx for constants A and B. Then y (x) = Ake kx Bke kx, and the boundary conditions imply that = A+ B+k(A B) and = Ae k + Be k. So we seek A and B that satisfy the linear system ( 1+k 1 k e k e k )( A B ) = ( ) There is a nontrivial solution only if the determinant e k (1+k) e k (1 k) =. This is equivalent to the equation e k = 1+k 1 k. The question then is whether this equation has a positive solution k. The functions on the two sides of this equation are equal at k =, and so are their first and second derivatives. However, the series expansions for these functions, convergent for k [, 1) are and e k = 1+k+ 1 (k) (k) k 1 k = (1+k)(1+k+k + k ) = 1+k+k + k and we see that 1 k 1+k > ek for any positive k in the interval [, 1), because the first three terms in the two series match and the remaining terms are larger for 1 k 1+k. For k = 1 one side of the equation is undefined. For k > 1 one side of the equation is negative, while the other is positive. Hence there are no positive solutions to the above equation and therefore there are no negative eigenvalues. So consider λ = k > for k >. The general solution of the ODE is y(x) = A cos(kx)+b sin(kx). The boundary conditions imply that A+Bk = and A cos(k)+ B sin(k) =. So A = kb, and = [ k cos(k) + sin(k)]b. Hence k must satisfy 13

14 k = tan(k). From the graphs of the functions z = k and z = tan(k) we can see that there are an infinite number of positive intersections of these curves and therefore an infinite number of positive eigenvalues. 14

15 11: (Logan, 3.4 #7) Show that the SLP y (x) = λy(x), < x < ; y()+y () =, 3y()+y () =, has exactly one negative eigenvalue. Is zero an eigenvalue? How many positive eigenvalues are there? We consider, in turn, the possilities, λ <, λ =, and λ >. Suppose λ = k for k >. The general solution of y = k y is y = Ae kx + Be kx. To satisfy the boundary conditions, we must find a nontrivial soluton to the linear system ( 1+k 1 k )( ) A (3+k)e k (3 k)e k B = ( ) Setting the determinant of the matrix equal to gives us an equation for k: 3+4k 4k 3 4k 4k = e4k. The graphs of the functions on the two sides of this equation shows there is one positive intersection (between and 1/). Thus there is exactly one negative eigenvalue y k.3.4 Consider λ =. The general solution of the ODE is y(x) = Ax+B. To satisfy the boundary conditions, we must find a nontrivial solution of the linear system ( )( ) ( ) 1 A =. 8 3 B Since the determinant of the matrix in not zero, there is no nontrivial solution of this system, so λ = is not an eigenvalue. We know that a regular SLP has an infinite number of real eigenvalues. We know there is one negative eigenvalue and that zero is not an eigenvalue. Therefore there are 15

16 an infinite number of positive eigenvalues. This may also be shown directly. For λ = k, the general solution to the ODE is y(x) = A cos(kx)+ B sin(kx). To satisfy the boundary conditions, we must find a nontrivial solution of the linear system ( )( ) 1 k A = 3 cos(k) k sin(k) 3 sin(k)+k cos(k) B The determinant of this matrix is if and only if k satisfies tan(k) = 4k 4k + 3. ( ). We see graphically that there are infinitely many positive intersections of the curves z = tan(k) and z = 4k, so there are infinitely many positive eigenvalues. 4k y k 15 16

17 1: (Logan, 3.4 #9) Consider the SLP (x y ) = λy, 1 < x < π, y(1) = y(π) =. Use an energy argument to show that any eigenvalue must be nonnegative. Find the eigenvalues and eigenfunctions. We try the energy argument to see if we can tell the sign of the eigenvalues. Multiply both sides of the ODE by y(x), integrate over the interval [1, π], and use integration by parts and the boundary conditions to find that π 1 π x (y ) dx = λ y dx. 1 So λ. The only way that λ = could occur is if y (x) = for all x [1, π]. Then y(x) would be constant and because of the boundary conditions, the constant would have to be. Therefore λ = is not an eigenvalue for this problem. So assume λ = k for k >. The ODE becomes x y (x)+xy (x)+k y(x) = This is a Cauchy-Euler equation (discussed in Appendix to Logan). We look for solutions of the form y(x) = x r. Substituting this into the ODE we find that r must satisfy This has two solutions r + r+k =. r ± = k There are three possible scenarios. Case 1: If r +, r are real and unequal, then the general solution to the ODE is y(x) = Ax r + + Bx r. Case : If r +, r are real and both equal to r, then the general solution is y(x) = Ax r + Bx r ln(x). Case 3: If r +, r are complex conjugates r ± = a±ib, then the general solution is y(x) = Ax a sin(b ln(x))+bx a cos(b ln(x)). Case 1: y(x) = Ax r + + Bx r with r ± real and unequal. The boundary conditions are satisfied if ( 1 1 π r + π r )( A B ) = The determinant of this system is π r + π r which is never zero for two different real numbers r ±, so there are no eigenvalues in case 1. Case : For r + = r = r to occur, 1 4k =, and r = 1/, so the solution in this case is ( ). y(x) = Ax 1/ + Bx 1/ ln(x). The boundary conditions are satisfied if ( )( ) 1 A π 1/ π 1/ = ln(π) B 17 ( ).

18 The determinant of this system is not zero, so k = 1/4 is not an eigenvalue. Case 3: For case 3 to occur, r ± = 1 ± i 4k 1 with k > 1/4. Then the general solution is ( ) ( ) y(x) = Ax 1/ 4k 1 cos ln(x) + Bx 1/ 4k 1 sin ln(x). To satisfy the boundary conditions, we need to find a nontrivial solution of the linear system ( 1 ) π 1/ cos( 4k 1 ln(π) π 1/ sin( 4k 1 ln(π) Setting the determinant of this system to zero gives us the equation ( ) π 1/ 4k 1 sin ln(π) =. This is satisfied if and only if 4k 1 ln(π) = nπ for some positive integer n. That is only if λ = k = 1 ( ) nπ 4 + n = 1,, 3,.... ln(π) (A ) )) = B ( ). These are the eigenvalues of the problem and the corresponding eigenfunctions are ( ) nπx y n (x) = x 1/ sin. ln(π) 18

Math 201 Assignment #11

Math 201 Assignment #11 Math 21 Assignment #11 Problem 1 (1.5 2) Find a formal solution to the given initial-boundary value problem. = 2 u x, < x < π, t > 2 u(, t) = u(π, t) =, t > u(x, ) = x 2, < x < π Problem 2 (1.5 5) Find