Math 220B - Summer 2003 Homework 1 Solutions

Transcription

1 Math 0B - Summer 003 Homework Soutions Consider the eigenvaue probem { X = λx 0 < x < X satisfies symmetric BCs x = 0, Suppose f(x)f (x) x=b x=a 0 for a rea-vaued functions f(x) which satisfy the boundary conditions Show there are no negative eigenvaues Answer: Let X be an eigenfunction with eigenvaue λ λ X, X = X, X using the assumption that X X x= x=0 0 = X, X X X x= x=0 0 Consider the eigenvaue probem, { X = λx a < x < b X satisfies certain BC s Suppose µ is an eigenvaue of mutipicity m > Let X,, X m denote m ineary independent eigenfunctions (which may or may not be orthogona) associated with the eigenvaue µ Use these eigenfunctions to construct m eigenfunctions Y,, Y m which are necessariy orthogona Answer: First, define Y = X X Next, we define a function orthogona to Y by subtracting the piece of X which is parae to Y Therefore, et Z = X X, Y Y We normaize, etting Y = Z / Z Continuing in this way, we et Z 3 = X 3 X 3, Y Y X 3, Y Y and then define Y 3 = Z 3 / Z 3 In genera, we define Z m = X m X m, Y Y X m, Y m Y m, and et Y m = Z m / Z m

2 3 Consider the eigenvaue probem X = λx X (0) + X(0) = 0 X() = 0 0 < x < (a) Find an equation for the positive eigenvaues Answer: If λ = β > 0, then X + β X = 0 = X(x) = A cos(βx) + B sin(βx) X (0) + X(0) = Bβ + A = 0 X() = 0 = A cos(β) + B sin(β) = 0 Combining these two equations, we concude that λ is a positive eigenvaue if and ony if β cos(β) + sin(β) = 0 or tan(β) = β (b) Show graphicay that there are an infinite number of positive eigenvaues Answer: Let f(β) = β g(β) = tan(β) We know that g(β) has vertica asymptotes at β = ( n + ) π/ and g(β) + as β ( n + ) π/ By graphing both functions, we see that f(β) and g(β) have an infinite number of intersections (c) Show that λ = 0 is an eigenvaue if and ony if = Find a corresponding eigenfunction in this case Answer: λ = 0 = X = 0 = X(x) = A + Bx X (0) + X(0) = 0 = B + A = 0 X() = 0 = A + B = 0

3 Combining these two equations impies B( ) = 0 Therefore, zero is an eigenvaue if and ony if = In this case, the corresponding eigenfunction is X 0 (x) = B 0 ( x) for any B 0 (d) Show that if, then there are no negative eigenvaues, but if >, then there is one negative eigenvaue Find the corresponding eigenfunction Answer: λ = γ < 0 = X(x) = A cosh(γx) + B sinh(γx) X (0) + X(0) = 0 = Bγ + A = 0 X() = 0 = A cosh(γ) + B sinh(γ) = 0 Combining these two equations, we concude that Bγ cosh(γ) + B sinh(γ) = 0 If B = 0, then X(x) 0 Therefore, we must have or γ cosh(γ) + sinh(γ) = 0, tanh(γ) = γ To ook for roots of this equation, we consider the graphs of the functions f(γ) = γ g(γ) = tanh(γ) We note that f (γ) = and g (γ) = sech (γ) Therefore, f (0) =, g (0) = Further, we note that g (γ) = sech (γ) tanh(γ) < 0 Aso, we note that g(γ) as γ + From this information, we concude that f(γ) and g(γ) ony intersect at γ = 0 if, and, consequenty, there are no negative eigenvaues if On the other hand, if >, f(γ) and g(γ) wi intersect once, impying there is exacty one negative eigenvaue if > 3

4 4 Use separation of variabes to sove u t ku xx = 0 0 < x <, t > 0 u(x, 0) = φ(x) u(0, t) = 0 u x (, t) = 0 Answer: We use separation of variabes, etting u(x, t) = X(x)T (t) Pugging this function into our PDE, we have kt = X X = λ We are ed to the eigenvaue probem { X = λx 0 < x < X(0) = 0 = X () Looking for positive eigenvaues, Therefore, T λ = β > 0 = X(x) = A cos(βx) + B sin(βx) X(0) = 0 = A = 0 X () = 0 = Bβ cos(β) = 0 = β n = ( n + ) π (( ) (( ) n + λ n = ) π n + X n (x) = B n sin ) π x By straightforward cacuation, we see these are a the eigenvaues Soving the equation for T n, Therefore, we et T = kλ n T = T n (t) = C n e kλnt u(x, t) = = T n (t) = C n e k (n+ (( ) n + C n sin ) π x e n= )π k (n+ t ) π! t In order for the initia condition u(x, 0) = φ(x) to be satisfied, we need ( ) (n+ sin )π x, φ(x) C n = ( ) ( ) (n+ (n+ sin x, sin x )π )π 4

5 5 Use separation of variabes to sove u t ku xx + u x = 0 0 < x <, t > 0 u(x, 0) = φ(x) u(0, t) = 0 u x (0, t) = 0 (Hint: Introduce a function f(x) such that v(x, t) = f(x)u(x, t) wi satisfy a PDE of the form v t kv xx + av = 0 with new initia and boundary conditions soution, sove for u) Answer: We wi give two methods Sove the equation for v, and from this Method We want to find a function f such that defining v = fu, then v satisfies a PDE of the form v t kv xx + av = 0 Let g = /f Therefore, u = gv Pugging this function into our PDE, we have gv t kgv xx + [ g + g]v x + [ kg + g ]v = 0 In order to eiminate the v x term, we want to choose g such that g + g = 0, which impies g(x) = e x/ Then et v = fu = (/g)u = e x/ u Soving for kg + g, we see that v wi satisfy the foowing initia/boundary-vaue probem v t kv xx + 4k v = 0 0 < x <, t > 0 v(x, 0) = e x/ φ(x) v(0, t) = 0 v x (, t) + v(, t) = 0 Now use separation of variabes on this initia/boundary-vaue probem Pugging in a function v(x, t) = X(x)T (t) impies We need to sove the ODE T kt = X X 4k = λ X X = λ + 4k Let µ = λ + We are ed to the eigenvaue probem 4k X = µx X(0) = 0 X () + X() = 0 0 < x < If µ = β > 0, then X(x) = A cos(βx) + B sin(βx) X(0) = 0 = A = 0 5

6 X () + X() = 0 = Bβ cos(β) + B sin(β) = 0 If B = 0, then X(x) 0 Therefore, we need or The corresponding eigenfunctions are β cos(β) + sin(β) = 0, tan(β) = β X n (x) = B n sin(β n x) If µ = 0, then X(x) = A + Bx X(0) = 0 = A = 0 X () + X() = 0 = B + B = 0 = B = 0 Therefore, zero is not an eigenvaue Using the resut from probem (), we can concude that there are no negative eigenvaues Recaing the definition of µ, we see that where Soving our equation for T n, we have λ n = β n + 4k tan(β n ) = β n T n (t) = C n e kλnt Therefore, we et v(x, t) = C n sin(β n x)e kλ nt n= where λ n and β n are defined above We want v(x, 0) = e x/ φ(x) Therefore, et C n = sin(β nx), e x/ φ(x) sin(β n x), sin(β n x) 6

7 Finay, we reca that u(x, t) = e x/ v(x, t) for v defined above Method Here is an aternate method, which aows one to use separation of variabes directy, but requires knowedge about Sturm-Liouvie probems (You are not responsibe for this materia, but I thought I d write up a soution using this method because some students used this technique) Let u = XT Pugging u into our equation, we have T kt = X k X X = λ We need to consider the eigenvaue probem { X + k X λx = 0 0 < x < X(0) = 0 = X () Mutipying by the integrating factor e x/k, we see this ODE can be rewritten as ( e x/k X ) λe x/k X = 0, an exampe of a Sturm-Liouvie probem We note that eigenfunctions of this Sturm- Liouvie probem are orthogona with respect to the weight function e x/k To find the soutions of our second-order ODE, X k X + λx = 0, we ook at the characteristic poynomia p(r) = r r + λ = 0 The roots of this k poynomia are given by ± 4λ k k Let µ = 4k λ Then r = The first case is µ > 0 In this case, = ± 4k λ r = ± µ r = ± i µ, which impies the associated eigenfunctions are given by X(x) = e x/ [A cos( µx) + B sin( µx)] 7

8 X(0) = 0 = A = 0 X () = 0 = [ sin( µ) + µ cos( µ)] = 0 = tan( µ) = µ Thus far, we have eigenvaues where and the associated eigenfunctions λ n = 4k + µ n tan( µ n ) = µ n X n (x) = e x/ sin( µ n x) Next, we consider µ = 0 In this case, r = which impies X(x) = Ae x/ + Bxe x/ X () = 0 = B X(0) = 0 = A = 0 [ e x/ + ] e/ = 0 But this can ony happen if B = 0 which impies X(x) 0 Therefore, µ 0 Next, we consider µ < 0 In this case, r = ± µ, which impies [ X(x) = e x/ Ae ] µx + Be µx X(0) = 0 = A + B = 0 X () = 0 = ( + ) µ e µ = µ or B = 0 Since both terms in the expression above are greater than, we must have B = 0 Therefore, µ 0 We concude that λ = 4k + µ 8

9 where µ > 0 such that and tan( µ) = µ X n (x) = e x/ sin( µ n x) The soution for our equation for T is given by T n (t) = C n e kλnt Therefore, et u(x, t) = C n e x/ sin( µ n x)e kλnt n= We want u(x, 0) = φ(x) That is, we want to choose C n such that C n e x/ sin( µ n x) = φ(x) n= Using the fact that eigenfunctions are orthogona with respect to the weight function e x/, we can mutipy both sides by sin( µ m x) and integrate over [0, ] with respect to the weight function e x/ In particuar, we concude that C n = 0 φ(x)ex/ sin( µ n x)e x/k dx 0 (ex/ sin( µ n x)) e x/k dx Simpifying, we have C n = 0 φ(x)e x/ sin( µ n x) dx 0 sin ( µ n x) dx 9