Name: Math 4567. Homework Set # VI April 2, 21 Chapter 5, page 113, problem 1), (page 122, problem 1), (page 128, problem 2), (page 133, problem 4), (page 136, problem 1). (page 146, problem 1), Chapter 8 (page 29, problem 1) Chapter 5 page 113, Problem 1 The initial temperature of a slab x π is everywhere and the fae x = is kept at that temperature. Heat is supplied through the fae x = π at a onstant rate ku x (π, t) = A >. Write u(x, t) = U(x, t) + Φ(x) and use the solution to the problem ( ) U t = ku xx, < x < π, t >, and U(x, ) = F (x) where U(, t) =, U x (π, t) =, < x < π, F (x) = { f(x) when < x < π f(2π x) when π < x < 2π, whih is U(x, t) = B n exp( n2 k 4 B n = 1 ( 1)n π to derive the final solution u(x, t) t) sin nx 2, f(x) sin nx 2 dx. Solution: We first find a funtion u = Φ(x) that satisfies the nonhomogeneous ondition ku x (π, t) = A and the homogeneous ondition u(, t) =. The differential equation is Φ (x) =, so Φ(x) = Bx + C. The nonhomogeneous boundary ondition says KB = A and the homogeneous ondition says C = thus Φ(x) = Ax. Then, setting k 1
u(x, t) = U(x, t) + Φ(x) we see that u(x, t) will be a solution of our original problem, provided U(x, t) satisfies problem ( ) where U(x, ) = f(x) = Φ(x) = A x for < x < π. Thus k u(x, t) = A k x + B n exp( n2 k nx t) sin 4 2, where B n = A ( 1)n π (1 ) x sin nx k π 2 dx. Note that B n = unless n = 2m 1 is odd. Sine x sin nx 2 dx = 2 { x os nx n 2 π + os nx } 2 dx = 2 n [ π os nπ 2 + 2 n sin nπ 2 ] we get the solution 2 = ( 2m 1 )2 ( 1) m+1, u(x, t) = A k { x + 8 π ( 1) m [ (2m 1) exp (2m ] 1)2 k t sin 2 4 m=1 } (2m 1)x. 2 Chapter 5, page 122, Problem 1 The faes and edges x = and x = π, ( < y < π) of a square plate x π, y π are insulated. The edges y = and y = π, ( < x < π) are kept at temperatures and f(x), respetively. Let u(x, y) be the steady state temperature distribution in the plate. Show that A = 1 f(x)dx, A π 2 n = Find u(x, y) if f(x) = u. Solution: The problem is u(x, y) = A y + A n sinh ny os nx, 2 π f(x) os nx dx, n = 1, 2,. π sinh nπ u xx + u yy =, < x < π, < y < π, 2
u x (, y) =, u x (π, y) =, < y < π, u(x, ) =, < x < π u(x, π) = f(x), < x < π. We use separation of variables u = X(x)Y (y) to find solutions satisfying the homogeneous onditions. The Sturm-Liouville eigenvalue problem is X (x) + λx(x) =, X () = X (π) =. From past work we know that the eigenvalues are λ n = n 2, n = 1, 2, with eigenfuntions X n (x) = os nx, and λ = with eigenfuntion X (x) = 1. The orresponding equations for Y (y) are Y (y) λy (y) =, Y () =. As has been shown earlier, for λ n = n 2 we have Y n (y) = sinh ny and for λ = we have Y (y) = y. Thus we an write u(x, y) = A y + A n os nx sinh ny, where Thus u(x, π) = f(x) = A π + A n os nx sinh nπ. A n sinh nπ = 2 π A π = 1 π f(x) os nx dx, n = 1, 2,, f(x)dx. If f(x) = u then A n = A π = u, so the solution is u(x, y) = u π y. Chapter 5, page 128, Problem 2 Let the faes of a wedge shaped plate ρ a, φ α be insulated Find the steady temperature u(ρ, φ) in the plate when u = on the rays φ =, φ = α ( < ρ < α) and u = f(φ) on the ar ρ = a ( < φ < α). Assume f is pieewise smooth and u is bounded. 3
Solution: Our problem in polar oordinates is to find a funtion u(ρ, φ) for ρ 2 u ρρ + ρu ρ + u φφ =, < ρ < a, < φ < α, u(ρ, ) = u(ρ, α) =, < ρ < α, u(a, φ) = f(φ), < φ < α, where f is pieewise smooth and u < M, i.e., u is bounded. Separating variables, u = R(ρ)Φ(φ) satisfies the homogeneous onditions if Φ satisfies the Sturm-Liouville problem and R satisfies Φ + λφ =, Φ() = Φ(α) =, ρ 2 R + ρr λr = and R is bounded. It is straightforward to show that the eigenvalues are λ n = n2 π 2 with eigenfuntions Φ α 2 n (φ) = sin nπφ, n = 1, 2,.. The α hange of variable ρ = e s gives the orresponding equation for R as R ss λ n R =. The general solutions are R n (ρ) = Aρ nπ/α + Bρ nπ/α, and the boundedness requirement yields R n (ρ) = ρ nπ/α. Thus u(ρ, φ) = B n ρ nπ/α sin nπφ α, and where Thus u(a, φ) = f(φ) = B n a nπ/α = 2 α α B n a nπ/α sin nπφ α, f(ψ) sin nπψ α u(ρ, φ) = 2 ( ρ α a )nπ/α sin nπφ α α dψ. f(ψ) sin nπψ α dψ. 4
Chapter 5, page 133, Problem 4 A string is strethed between points and π on x-axis and, initially at rest, is released from the position y = f(x). The equation of motion is y tt = y xx 2βy t, < x < π, t >, where < β < 1 and β is onstant. Show that α n = y(x, t) = e βt B n (os α n t + β α n sin α n t ) sin nx, n 2 β 2, B n = 2 f(x) sin nx dx, n = 1, 2,. π Solution: Set z(x, t) = e βt y(x, t). Then z(x, t) satisfies z tt = z xx + β 2 z, < π, t >, z(, t) = z(π, t) =, t > z t (x, ) = βf(x), z(x, ) = f(x), < x < π,. where f() = f(π) =. We look for a solution of the form Then A n(t) = 2 π z(x, t) = A n (t) sin nx, A n (t) = 2 π = 2 π z tt (x, t) sin nx dx = 2 π z xx (x, t) sin nx dx + 2β2 π = 2 π [z x(x, t) sin nx π n z(x, t) sin nx dx. (z xx (x, t) + β 2 z(x, t)) sin nx dx z(x, t) sin nx dx. z x (x, t) os nx dx] + β 2 A n (t) = 2 π [ nz(x.t) os nx π n 2 z(x, t) sin nx dx] + β 2 A n (t) = ( n 2 + β 2 )A n (t). 5
Thus so A n(t) + (n 2 β 2 )A n (t) =, A n (t) = B n os α n t + C n sin α n t where α n = n 2 β 2, n = 1, 2,. The ondition gives The ondition B n = 2 π z(x, ) = f(x) = B n sin nx f(x) sin nx dx, n = 1, 2,. z t (x, ) = βf(x) = C n α n sin x gives C n = βb n /α n. (Here we are assuming that it is permissible to differentiate the sum term-by-term. This assumption ould be avoided by taking z t (x, t) = E(t) sin nx and obtaining E n (t) by integration by parts, just as we did for A n (t).) Thus we obtain the formal solution y(x, t) = e βt z(x, t) = e βt B n (os α n t + β α n sin α n t Chapter 5, page 136, Problem 1 Solve the problem y tt = a 2 y xx + Ax sin ωt, < x <, t >, y(, t) = y(, t) =, y(x, ) = y t (x, ) =. Show that resonane ours for ω = ω n, where ω n = nπa, n = 1, 2,. Solution: We look for a solution in the form y(x, t) = 6 B n (t) sin nπx, ) sin nx,
Then B n(t) = 2 B n (t) = 2 y tt (x, t) sin nπx y(x, t) sin nπx dx = 2 dx. [a 2 y xx (x, t)+ax sin ωt] sin nπx = 2A( 1)n+1 sin ωt a2 n 2 π 2 B nπ 2 n (t), Where we have integrated by parts several times and applied the boundary onditions. Thus ( ) B n(t) + a2 n 2 π 2 B 2 n (t) = 2A( 1)n+1 sin ωt. nπ This is a nonhomogeneous equation. We need only find one solution and then add to it the general solution H n os anπx + K n sin anπx of the homogeneous equation to get the general solution. We look for a solution of the form B n (t) = C n sin ωt. Substituting this into equation ( ) and setting ω n = anπ/ we find a solution if C n = 2A( 1) n+1 /(ωn 2 ω 2 ). Thus B n (t) = 2A( 1)n+1 ωn 2 ω 2 and the general solution is B n (t) = A n os anπt + C n sin anπt sin ωt, + 2A( 1)n+1 ω 2 n ω 2 sin ωt. The boundary onditions are B n () = B n() =, and these are satisfied for A n = and C n = 2( 1) n ω/[ω n (ωn 2 ω 2 )]. Thus the final solution is B n (t) = 2( 1)n ω anπt sin + 2A( 1)n+1 sin ωt, ω n (ωn 2 ω 2 ) ωn 2 ω 2 unless ω = ω n for some n. In that ase we have resonane and the solution beomes unbounded. To see this, we look for a partiular solution of ( ) in the ase ω = ω n. Take the trial solution B n (t) = D n t os ω n t. Then we find a solution provided D n = ( 1) n A/nπω n : ( ) B n (t) = ( 1)n A nπω n t os ω n t. 7 dx
To this solution we an add a general solution of the homogeneous equation, but the resonant solution quikly dominates the bounded solution of the wave equation as t gets large. Chapter 5, page 146, Problem 1 Write λ = α 2, α > and show that the Sturm-Liouville problem has no solutions. X + λx =, X( π) = X(π), X ( π) = X (π), Solution: The general solution of the differential equation is so X(x) = Ae αx + Be αx, X (x) = α(ae αx Be αx ). The onditions X( π) = X(π), X ( π) = X (π) an be written as A sinh απ = B sinh απ, A sinh απ = B sinh απ, respetively. Sine sinh απ for α, we have A = B = B, so A = B =. Thus there are no negative eigenvalues. Chapter 8, page 29, Problem 1 Find the eigenvalues and eigenfuntions: X + λx =, X() =, X (1) =. Solution: If λ = then X(x) = Ax + B and X (x) = A. Thus the boundary onditions are B =, A = and λ = is not an eigenvalue. If λ = α 2, α > then X(x) = Ae αx + Be αx, X (x) = α(ae αx Be αx ). Thus the boundary onditions are A + B = and Ae α e α, or B = A where A osh α =. Sine osh α we have A = B = and λ = α 2 is not an eigenvalue. If λ = α 2, α > then X(x) = A os αx+b sin αx, X (x) = α( A sin αx+ B os αx), and the boundary onditions an be read as A =, α(b os α) =, 8
or os α =, so λ n = α 2 n where α n = (2n 1) π 2, X n(x) = sin α n x n = 1, 2,. Sine 1 X2 n(x)dx = 1 1 2 (1 os π(2n 1)x)dx = 1 2 eigenfuntions are φ n (x) = 2 sin α n x. the normalized 9