Homework 6 Math 309 Spring 2016

Transcription

1 Homework 6 Math 309 Spring 2016 Due May 18th Name: Solution: KEY: Do not distribute! Directions: No late homework will be accepted. The homework can be turned in during class or in the math lounge in Pedelford C-wing. The math lounge is at the end of the hallway on the left. There are mailboxes on the wall and in the mailbox labelled Van Meter there will be an envelope labelled 309. Put your homework in the envelope. The grader will collect this envelope at 2:30pm. Staple your homework and make sure your homework is organized and clear. Points may be taken off for homework that is not stapled or unclear. Question Points Score Total: 34

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3 1. (8 points) Insulated Heat Equation Problem Consider a uniform rod of length L with an initial temperature given by u(x, 0) = sin(πx/l) with 0 x L. Assume that both ends of the bar are insulated (this is a homogeneous von Neumann boundary condition for t > 0). (a) (2 points) Find the temperature u(x, t). (Note: the initial condition u(x, 0) does not satisfy the boundary conditions, which is fine since we are only asking the boundary conditions to be satisfied for t > 0) Solution: We need to determine the temperature initially in terms of a cosine series. This means reflecting sin(πx/l) evenly and then extending periodically. In other words, we re really looking for the cosine series of sin(πx/l)). Using Euler-Fourier, we obtain a n = 1 L L L sin(πx/l) cos(nπx/l)dx = 2 L L 0 sin(πx/l) cos(nπx/l)dx. Now in order to complete the last integral on the right, we can adopt several strategies. The most obvious thing is to integrate by parts twice, and then compare sides however, that is a lot of work. A shorter strategy is to use the addition angle formulas for sine to write: sin(πx/l) cos(nπx/l) = 1 (sin((1 + n)πx/l) + sin((1 n)πx/l)). 2 With this in mind, the above integral becomes a n = 1 L L 0 (sin((1 + n)πx/l) + sin((1 n)πx/l))dx = 2 π ( 1 + () n 1 n 2 However, notice that in our derivation of this formula, we divided by 1 n, and therefore the expression we obtained for a n does not apply when n = 1. We must treat this case separately! We calculate using the double angle formula a 1 = 2 L L We conclude that 0 sin(πx/l) cos(πx/l)dx = 1 L L u(x, 0) = a a n cos(nπx/l) = 2 π + 2 π This tells us that n=1 u(x, t) = 2 π + n=2 2 π 0 ( 1 + () n 1 n 2 ). sin(2πx/l)dx = 1 2π cos(2πx/l) 1 0 = 0. n=2 ( 1 + () n 1 n 2 ) e n2 π 2 α 2 t/l 2 cos(nπx/l). ) cos(nπx/l). Page 2

4 (b) What is the steady state temperature as t? Solution: As t, the exponential terms die off, leaving only a 0 /2. Therefore the steady state temperature is 2/π. (c) Let α 2 = 1 and L = 40. Plot u vs. x for several values of t. Solution: 2. (8 points) Nonhomogeneous Boundary Conditions Let an aluminium rod of length 20 cm be initially at the uniform temperature of 25 degrees C. Suppose that at time t = 0, the end x = 0 is cooled to 0 degrees C while the other end x = 20 is heated to 60 degrees C, and both are thereafter maintained at those temperatures. (a) Find the temperature u(x, t). (Note: the initial condition u(x, 0) does not satisfy the boundary conditions, which is fine since we are only asking the boundary conditions to be satisfied for t > 0) Solution: We have Dirichlet boundary conditions, so this means that we should be thinking about a sine expansion. The first thing we should do is obtain the steady state solution u steady (x). Note that it is not time dependent (since it s steady state!!). Since it satisfies the heat equation, we know that u steady (x) = 0, and therefore u steady (x) = ax + b for some constants a and b. Now since u steady (0) = 0 and u steady (20) = 60, we can work out a and b, obtaining (a = 3, b = 0): u steady (x) = 3x. Next, we should solve for the transient solution, which has homogeneous Dirichlet boundary conditions and initially satisfies u trans (x, 0) = 25 u steady (x) = 25 3x. The sine expansion of this is given by u trans (x, 0) = n=1 10 nπ (5 + 7()n ) sin(nπx/20). Then since α 2 = 0.86 for aluminum, we find that the transient solution is: 10 u trans (x, t) = nπ (5 + 7()n )e n2π2 (0.86)t/400 sin(nπx/20). n=1 To get the solution to the inhomogeneous heat equation problem above, we need to add this transient solution to the steady state solution u steady (x). Therefore 10 u = u trans (x, t)+u steady (x, t) = 3x+ nπ (5+7()n )e n2π2 (0.86)t/400 sin(nπx/20). n=1 Page 3

5 (b) What is the steady state temperature as t? Solution: As t, the transient solution dies off, leaving behind only the steady state solution u steady = 3x. (c) Plot u vs. x for several values of t. Solution: (d) Determine how much time must elapse before the temperature at x = 5 comes within 1 degrees C of its steady state value. Solution: We are really asking how much time elapses until u trans (5, t) 1. We can determine this by plotting u trans (5, t) and determining what time it drops below 1 degree Celsius. Using partial sums, we approximate t = is about when u(5, t) is within its steady state value. 3. (8 points) Schrödinger Equation In quantum mechanics, the position of a point particle in space is not certain it s described by a probability distribution. The probability distribution of the position of the particle is ψ(x, t) 2, where ψ(x, t) is the wave function of the particle. (Note: the wave function ψ(x, t) can be complex-valued!!). The one-dimensional, time-dependent Schrödinger equation, describing the wave function ψ(x, t) of a particle of mass m interacting with a potential v(x) is given by i ψ t (x, t) = 2 2m ψ xx(x, t) + v(x)ψ(x, t) where is some universal constant. The potential v(x) can be imagined as a function describing the particles interaction with whatever stuff is in the space surrounding the particle, eg. walls, external forces, etc. (a) Use separation of variables to replace this partial differential equation with a pair of two ordinary differential equations Solution: We assume ψ(x, t) = F (x)g(t). Then inserting this into Schrödinger s equation, we obtain i F (x)g (t) = 2 2m F (x)g(t) + v(x)f (x)g(t). Now if we divide out by a G(t) and a F (x) we find i G (t)/g(t) = 2 2m F (x)/f (x) + v(x). Page 4

6 The function on the left hand side is a function of t only. The function on the right hand side is a function of x only. Therefore the only way that the above equality can work is if both sides are equal to some constant E. Therefore i G (t)/g(t) = E, 2 2m F (x)/f (x) + v(x) = E. Simplifying, this gives us the system of two ordinary differential equations i G (t) = EG(t). 2 2m F (x) + v(x)f (x) = EF (x). The latter equation of these two equations is known as the time-independent Schrödinger equation. (b) If v(x) is a potential corresponding to an infinite square well : { 0, < x < 1 v(x) =, x 1 Then ψ(x, t) must be zero whenever x 1 and therefore ψ(x, t) is the wave function of a particle trapped in a one-dimensional box! In other words, this potential describes a particle surrounded by impermeable walls. In this case, Schrödinger s equation reduces to i ψ t (x, t) = 2 2m ψ xx(x, t), < x < 1, t > 0 ψ(, t) = ψ(1, t) = 0, t > 0 Suppose that initially the wave function is known to be Determine ψ(x, t) for all t > 0. ψ(x, 0) = 3 5 sin(πx) sin(3πx). Solution: This is just like the heat equation, with α 2 = i and L = 1. Thus 2m given the initial condition, the solution that we are looking for is ψ(x, t) = 3 π2 e i 2m t sin(πx) π2 e i 2m t sin(3πx). 5 5 (c) Since ψ(x, t) 2 is the probability distribution of the particle s position at time t, the probability that the particle is somewhere in the box between l 1 and l 2 is given by P(l 1 pos l 2 ) = l2 l 1 ψ(x, t) 2 dx. Page 5

7 Show that the probability P( pos 1) that the particle is between and 1 is always 1 (in other words, the particle is always in the box!). Solution: Note that and therefore ψ(x, t) = 3 π2 ei 2m t sin(πx) π2 ei 2m t sin(3πx), 5 5 ψ(x, t) 2 = ψ(x, t)ψ(x, t) = 9 25 sin2 (πx) sin2 (3πx) (e i 8 π2 2m t + e i 8 π2 2m t ) sin(πx) sin(3πx). If we now integrate over the domain of the box (from to 1), orthogonality tells us the integral of sin(πx) sin(3πx) dies off! Therefore we obtain: 1 ψ(x, t) 2 dx = sin 2 (πx)dx sin 2 (3πx)dx = = 1. This shows that the probability that the particle is in the box at any time t is 1 e.g. it is a certainty. (d) What is the probability P( pos 0) that the particle is in the first half of the box at any given time? Solution: We can use the work from above to write 1 ψ(x, t) 2 dx = sin 2 (πx)dx sin 2 (3πx)dx+ 12 ) (e 0 i 8 π2 2m t + e i 8 π2 2m t sin(πx) sin( 25 However, since we re not integrating over the full period, we cannot appeal to orthogonality to say that the cross-term dies anymore. However, direct calculation shows that it does indeed die anyway. The sum of the first two integrals is easily calculated to be 1/2. Therefore the probability that the particle is in the first half of the box at any time t is exactly 1/2. In other words at any time the particle is equally likely to be in either side of the box. 4. (8 points) The Heat Equation in Two Dimensions We consider the two dimensional heat equation u t α 2 (u xx + u yy ) = 0. (a) Assume that u is of the form u(x, y, t) = F (x)g(y)t (t), and show that the heat equation reduces to the system of three ordinary differential equations T (t) + λt = 0 F (x) + λ µ F (x) = 0 α 2 G (y) + µ G(y) = 0 α 2 Page 6

8 for some constants λ and µ. Solution: Pluggin u(x, y, t) = F (x)g(y)t (t) into the two dimensional heat equation, we obtain F (x)g(y)t (t) = α 2 (F (x)g(y)t (t) + F (x)g (y)t (t)). Then dividing on both sides by F (x)g(y)t (t), we obtain: T (t)/t (t) = α 2 (F (x)/f (x) + G (y)/g(y)). The expression on the left hand side is a function of t only, while the expression on the right hand side is independent of t. Therefore both must be equal to a constant λ: Simplifying this, we get Therefore T (t)/t (t) = α 2 (F (x)/f (x) + G (y)/g(y)) = λ. T (t)/t (t) = λ α 2 F (x)/f (x) + α 2 G (y)/g(y) = λ. α 2 F (x)/f (x) = α 2 G (y)/g(y) λ. Again the expression on the left hand side is a function of x only, and on the right we have a function of y only, and therefore both are equal to a constant µ: It follows that α 2 F (x)/f (x) + λ = α 2 G (y)/g(y) = µ. α 2 F (x)/f (x) = µ λ α 2 G (y)/g(y) = µ. Simplyfying things further we obtain the system of three ordinary differential equations listed in (a) above. (b) Assume that u(x, y, t) = F (x)g(y)t (t) satisfies the heat equation above in the rectangular region [0, L] [0, M] and also satisfies the Dirichlet boundary conditions u(0, y, t) = 0, u(l, y, t) = 0, u(x, 0, t) = 0, u(x, M, t) = 0. Find all possible functions u(x, y, t) satisfying the above conditions. [Hint: they should be indexed by pairs of positive integers (m, n)] Solution: The Dirichlet boundary conditions result in boundary conditions on Page 7

9 our various ODE s. In particular F (x) + λ µ α 2 F (x) = 0, F (0) = 0, F (L) = 0 and also G (y) + µ G(y) = 0, G(0) = 0, G(M) = 0. α2 Then from our experience with boundary value problems, to get a nontrivial solution this says that µ = n 2 π 2 α 2 /M 2 and that λ µ = m 2 π 2 α 2 /L 2 for some integers m and n, and in this case the solutions we obtain are F (x) = A sin(mπx/l), G(x) = B sin(nπy/m) for some constants A and B. Then λ = m 2 π 2 α 2 /L 2 +n 2 π 2 α 2 /M 2, and therefore ( ) m 2 π 2 α 2 T = C exp t + n2 π 2 α 2 t. L 2 M 2 Thus we obtain the solution ( ) m 2 π 2 α 2 u(x, y, t) = F (x)g(y)t (t) = ABC exp t + n2 π 2 α 2 t sin(mπx/l) sin(nπx/m). L 2 M 2 This motivates us to set ( ) m 2 π 2 α 2 u mn (x, y, t) = exp t + n2 π 2 α 2 t sin(mπx/l) sin(nπx/m). L 2 M 2 Then by the superposition principle, any solution of the form u(x, y, t) = m=1 n=1 a mn u mn (x, y, t) is a solution, for constants a mn. In fact, all solutions may be written this way! (c) Use (b) to find a solution to the two dimensional heat equation with Dirichlet boundary conditions u t (u xx + u yy ) = 0, u(0, y, t) = 0, u(1, y, t) = 0, u(x, 0, t) = 0, u(x, 1, t) = 0, with the initial condition that u(x, y, 0) = sin(3πx) sin(2πy) + sin(2πx) sin(4πy). Create a surface plots of your solution for several values of t. Page 8

10 Solution: From part (b), we know to try to write u(x, y, t) = m=1 n=1 a mn u mn (x, t) for some constants a mn, where here α 2 = 1, L = 1, M = 1. we must choose the constants so that u(x, y, 0) = m=1 n=1 a mn u mn (x, y, 0) = m=1 n=1 a mn sin(mπx) sin(nπx). is equal to our initial condition. Looking at initial condition, this is easy! Just choose a 32 = 1, a 24 = 1 and a mn = 0 otherwise. Thus: u(x, y, t) = u 32 (x, y, t)+u 24 (x, y, t) = e 3π2t sin(3πx) sin(2πy)+e 20π2t sin(2πx) sin(4πy). Page 9