University of Toronto at Scarborough Department of Computer and Mathematical Sciences, Mathematics MAT C46S 2013/14.

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1 University of Toronto at Scarborough Department of Computer and Mathematical Sciences, Mathematics MAT C46S 213/14 Problem Set #4 Due date: Thursday, March 2, 214 at the beginning of class Part A: Let g(x) be a solution of the BVP, (p(x)y ) + q(x)y =, B a (y) =, B b (y) =. Show that if f(x) is a function for which the BVP (p(x)y ) +q(x)y = f(x), B a (y) =, B b (y) = has a solution then b a f(x)g(x)dx =. (Recall that B a (y) is the boundary condition at the point a, and similarly B b (y).) B a (y) = α 1 y(a)+α 2 y (a) Ly = (ay ) +by. L is self-adjoint on V := {h B a (h) = = B b (h)} in other words < Lu,v >=< u,lv > for all u,v V, where< r,s >= b a r(x)s(x)dx. By hypothesis g V and Lg =. Also there is z V s.t. Lz = f. So b a f(x)g(x)dx =< f,g >=< Lz,g >=< z,lg >=< z, >= An insulated copper wire 2 cm long is heated to 1 degrees and then one end is placed next to an ice bath at degrees while the other end is insulated. Determine the temperature at the centre of the bar after 4 seconds. Recall that the boundary condition at the end that is insulated is u (x = L) = 1

2 while the boundary condition at the end in the ice bath is. u(x = ) = (Note that according to Table in the text on p. 612, the value of the constant α 2 in the heat equation α 2 u xx = u t is 1.14.) u(,t) = and u x (2,t) = for all t. u(x,) = 1 for < x < 2. Set so so for some λ. Hence The boundary conditions give α 2 u xx = u t. u(x,t) = X(x)T(t) α 2 X T = XT α 2 X /X = λ = T /T α 2 X +λx = (1) T +λt = (2) u(,t) =, (3) u x (2,t) =, (4) u(x,) = 1, x 2 (5) From (4) we have X(x)T() = 1, so T(). From (3) we have X()T(t) = for all t, so X() =. From (4) we have X (2)T(t) = for all t, so X (2) =. Hence α 2 X +λx = and X() = and X (2) =. We need a nonzero solution since X = doesn t satisfy (5). The boundary conditions are X() =,X (2) =. Case 1: λ < Let λ = κ 2, κ >. 2

3 X 1 = e κx/2, X 2 = e κx/2. We need X(t) = c 1 eκx+c 2 e κx X 1 = κe κx/2, X 2 = κe κx/2 X() = tells us c 1 = c 2 so X(x) = csinhκx. So X (x) = cκcoshκx. So X (2) = cκcosh2κ = tells us cosh2κ = which is a contradiction (cosh(x) is never zero for any x). Case 2: λ = Here X(x) = a+bx,x (x) = b. X() = tells us a =. X (2) = tells us b =. So the only solution is X(x) =. Case 3: λ >. Let λ = k 2, k >. X 1 = coskx/α, X 2 = sinkx/α, B (X 1 ) = 1, B (X 2 ) = X 1 = k/αsinkx/α, X 2 = k/αcoskx/α, B 2 (X 1 ) = k/αsin2k/α, B 2 (X 2 ) = k/αcos2k/α So So 2k/α = (2n 1)/2π, n = 1,2,3,... λ n = α2 π ((2n 1)/2)2 Let X = c 1 X 1 +c 2 X 2. = X() = c 1 +, so C 1 =. Hence φ n = sin((2n 1)πx/4) and < φ n,φ n >= f(x) = 1, so 1 = < f,φ n >= 1 sin 2 (2n 1)πx/(4)dx = 2 sin 2 (2n 1)πv/2dv) 1 cos (2n 1)πv dv = 1(v sin(2n 1)v 2 (2n 1)π v=1 v== 1. 1sin (2n 1)πx 1 dx = 2 sin((2n 1)πv/2)dv 4 = 4 (2n 1)π ( cos((2n 1)πv/2) v=1 v= = 4 (1 cos(2n 1)π/2) = 4/((2n 1)π) (2n 1)π 3

4 So f = nb n φ n where b n = 4/((2n 1)π). Forφ n (x)thecorrespondingt satisfiest +λ n T =, λ n = α2 π (2n 1) 2 sot = Ae λnt. Let u n = e λnt sin(2n 1)πnx/4. Then with bn,λ n as above. x = 1,t = 1 : u(x,t) = b n e λnt sin (2n 1)πx 4 u(1,4) = b n e 4λn sin(2n 1)π/4 = a n b n e 4λn where a n = 2 1 when n = 1,2 mod 4 a n = 2 1 when n = 3,4 mod 4 Thus u(1,4) = 4 ( 1 π 2 /4 2 π e α2 + 1 π2 9/(4) 3π e α2 1 π2 25/(4) 5π e α2 1 ) π2 49/(4) 7π e α2 + = = 7.3 It will be approximately 7.3 degrees in the centre of the wire after 4 seconds. Question 3. Show that if f then the eigenvalue problem y + λf(x)y =, y() =, y(1) = cannot have any negative eigenvalues. We know y() =. Consider the smallest zero of y other than x =, call it b. So for < x < b, y(x) is not. We may assume y(x) > for that range (otherwise replace y by y). Now let z be the maximum of y(x) with < x < b Then y (z e) > and y (z+e) < for small enough e (because y (z) = ). Integrate y between z e and z+e: the integral is y (z+e) y (z e), so it is <. But if lambda <, because f(x) is greater than or equal to, y has the same sign as y. (in other words it is > ). This is a contradiction. Part B: Do the following problems from Boyce-Di Prima. Section1.6#8Findthesteady-statesolutionoftheheatconductionequationα 2 u xx = u t that satisfies the following set of boundary conditions: u(,t) = T,u x (L,t)+u(L,t) = 4

5 The steady state solution v(x) satisfies the differential equation v (x) = along with the boundary conditions v() = T, v (L)+v(L) = The general solution of this ODE is v(x) = Ax+B. The boundary condition v () = requires that B = T. It follows that v(x) = Ax+T, and v (L)+v(L) = A+AL+T. The second boundary condition requires that A = T/(1+L). Hence v(x) = Tx/(1+L)+T. Section 1.7#12 Dimensionless variables can be introduced into the wave equation a 2 u xx = u tt as follows: (a) Let s = x/l and show that the wave equation becomes a 2 u xx = L 2 u tt (b) Let τ = at/l and show that the wave equation becomes u xx = u tt The wave equation is given by a 2 2 u x = 2 u 2 t 2. Setting s = x/l, we have u x = u s s x = 1 u L s. It follows that 2 u x = 1 2 u 2 L s 2. Likewise, with τ = at/l, u t = a u L τ and 2 u t = a2 2 u 2 L 2 τ 2 Substitution into the original equation results in 2 u s = 2 u 2 τ 2. Section 1.8 #2: Find the solution u(x,t) of Laplace s equation in < x < a, < y < b with the boundary conditions u(,y) =,u(a,y) =, < y < b u(x,) = h(x), u(x,b) =, x a. 5

6 Using the method of separation of variables, write u(x,y) = X(x)Y(y). We arrive at the ordinary differential equation with X() = X(a) =, and X +λx =, Y λy = with Y(b) =. It follows that λ n = (nπ/a) 2 and X n (x) = sin(nπx/a). Write the solution of the other ODE as Y(y) = d 1 coshλ(b y)+d 2 sinhλ(b y). Imposing the boundary condition, we have d 1 =. Hence the fundamental solutions are given by u n (x,y) = sin(nπx/a)sinhλ n (b y). The general solution is Based on the boundary condition, u(x,y) = c n sin(nπx/a)sinh(λ n (b y)/a). h(x) = c n sin(nπx/a)sinh(nπb/a). The coefficients are calculated using the equation a c n sinh(nπb/a) = 2/a( h(x) sin(nπx/a)dx. Section 1.8 #9 Show that the equation (23) has periodic solutions only if λ is real. Consider the equation Θ +λθ =. Let λ = µ 2, where µ = ν +iσ. The ODE can then be written as Θ (ν +iσ) 2 Θ =. The general solution is Collecting the real and imaginary parts, Θ(θ) = c 1 e (ν+iσ)t +c 2 e (ν+iσ)t. Θ(θ) = (c 1 e νt +c 2 e νt )cos(σt)+i(c 1 e νt c 2 e νt )sin(σt). For ν, the functions e νt and e νt are linearly independent. If the coefficients are nonzero, then the real and imaginary parts of Θ(θ) are not periodic. So Equation (23) in Section 1.8 has periodic solutions only if λ = (iσ) 2 = σ 2, with σ >. 6