Homework 7 Math 309 Spring 2016 Due May 27th 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 1 8 2 8 3 8 4 8 Total: 32
1. (8 points) The Wave Equation Consider an elastic string of length L = 10 whose ends are held fixed. The string is set in motion from an initial position u(x, 0) = f(x) with initial velocity g(x), and the material properties of the string make u(x, t) satisfy the wave equation u tt c 2 u xx with c = 1. For each of the values of f(x) and g(x) below, determine (i) Determine the solution u(x, t) in terms of an infinite linear combination of the fundamental set of solutions u n (x, t) = sin(nπx/l) cos(cnπt/l) (ii) Plot u(x, t) vs. x for t = 0, 4, 8, 12, 16 (iii) Describe the motion of the string in a few sentences. You might want to plot some more values of t to see what is happening. (a) f(x) = 8x(L x) 2 /L 3 g(x) = 0. Solution: The coefficients of the sine expansion of f(x) are given by The solution is then given by u(x, n) = a n = 64 32 n 3 ( 1)n π3 n 3 π. 3 a n sin(nπx/l) cos(cnπt/l). n=1 Initially, the string has a smooth hump slightly skewed to the left. This hump moves off to the right, and when it reaches the end of the string, it is flipped upside down, and then moves to the left. When it next reaches the other end of the string it flips again, and moves back to the right, eventually returning to the initial state. (b) f(x) = { 1, x L/2 < 1 0, x L/2 1 g(x) = 0. Solution: The coefficients of the sine expansion of f(x) are given by The solution is then given by a n = 2 (cos(2nπ/5) cos(3nπ/5)). nπ u(x, n) = a n sin(nπx/l) cos(cnπt/l). n=1 Page 2
Initially, the string has a square bump in the center. This splits up into two square bumps, moving in opposite directions. When the bumps reach the ends, they are reflected, so that they are upside down and moving in the opposite directions they were before. As they pass over each other, they reform the original square bump, but upside down. This then splits up and the two squares move again to opposite ends of the string, where they are again reflected. These eventually meet again in the middle, returning the string to the original state. (c) f(x) = 0 g(x) = { 1, x L/2 < 1 0, x L/2 1 Solution: The coefficients of the sine expansion of f(x) are given by The solution is then given by a n = 2 (cos(2nπ/5) cos(3nπ/5)). nπ u(x, n) = n=1 a n L sin(nπx/l) sin(cnπt/l). cnπ 2. (8 points) Wave Equation with von Neummann Boundary Conditions Use separation of variables to find a solution to the wave equation u tt c 2 u xx = 0 with the homogeneous von Neumman boundary conditions u x (0, t) = 0, u x (L, t) = 0, and satisfying the initial condition u(x, 0) = cos(nπx/l), u t (x, 0) = 0, where here n is a nonnegative integer. Solution: For separation of variables, we propose a solution u(x, t) = F (x)t (t). Then the usual work reduces to the equations F (x) λ c 2 F (x) = 0, T (t) λt (t) = 0. Page 3
The interesting thing is how the von Neumann boundary conditions induce von Neumann boundary conditions on F (x). Since u x (0, t) = 0, we see that F (0)G(t) = 0. Since G(t) cannot be zero for a nontrivial solution, we must hvae F (0) = 0. Similarly, we see that F (L) = 0. Therefore we have the homogeneous von Neumann boundary value problem F (x) λ c 2 F (x) = 0, F (0) = 0, F (L) = 0. The usual considerations show that λ = n 2 π 2 c 2 /L 2, and that F (x) = C cos( λx/c) for some constant C. Furthermore, since u t (x, 0) = 0, we see that G (0) = 0, and therefore G(t) = B cos( λt). Therefore we see that u(x, t) = F (x)g(t) = BC cos( λx) cos( λt) = BC cos(nπx/l) cos(cnπt/l). To satisfy the initial condition, we take BC = 1, and therefore u(x, t) = cos(nπx/l) cos(cnπt/l). 3. (8 points) The Heat Equation in Polar Coordinates We consider the two dimensional heat equation u t α 2 (u xx u yy ) = 0. (a) Show that using polar coordinates, (r, θ), the heat equation becomes u t α (u 2 rr 1 r u r 1 ) r u 2 θθ = 0. Solution: This is the same calculation as done in the notes. Note that r 2 = x 2 y 2 and tan(θ) = y/x and therefore r x = cos(θ), r y = sin(θ), as well as θ x = sin(θ), θ y = cos(θ), r r Then from the chain rule we have x = r x θ x θ = cos(θ) 1 y = r y θ y θ = sin(θ) 1 r cos(θ) θ Page 4
Then we have that 2 x 2 = = ( ) 2 = x cos(θ) cos(θ) 1 cos(θ) 1 r sin(θ) ] cos(θ) ] θ and also cos(θ) = cos 2 (θ) 2 ] cos(θ) ] 1 1 Similarly, ] cos(θ) 1 ( 2 y = 2 y = = 1 r ] = 1 r ] 1 r sin(θ) ] θ 1 2. sin(θ) cos(θ) 2 θ 1 r 2 sin(θ) cos(θ) θ sin(θ) cos(θ) 2 θ 1 r sin2 (θ). = 1 r 2 sin2 (θ) 2 θ 2 1 r 2 sin(θ) cos(θ) θ. ) 2 = sin(θ) sin(θ) 1 r cos(θ) θ sin(θ) ] 1 1 r cos(θ) ] sin(θ) ] θ and also sin(θ) = sin 2 (θ) 2 sin(θ) ] 1 r cos(θ) ] 1r 2 = sin(θ) cos(θ) θ 1 r cos(θ) ] sin(θ) ] θ 1 r cos(θ) θ Adding all of this together, we obtain: ] r cos(θ) θ 1 r cos(θ) θ 2. θ 1 r 2 sin(θ) cos(θ) θ. = 1 r sin(θ) cos(θ) 1 r cos2 (θ). = 1 r 2 cos2 (θ) 2 θ 2 1 r 2 sin(θ) cos(θ) θ. 2 x 2 2 y = 2 2 1 2 r 1 2 r 2 θ. 2 Page 5
Applying this to the function u, we get u xx u yy = u rr 1 r u r 1 r u θθ. 2 Substiuting this into the heat equation leads to u t α (u 2 rr 1 r u r 1 ) r u 2 θθ = 0. (b) Assume that u is of the form u(r, θ, t) = R(r)S(θ)T (t), and show that the heat equation reduces to the system of three ordinary differential equations for some constants λ and µ. T (t) λt = 0 r 2 R (r) rr (r) 1 α 2 (r 2 λ µ)r = 0 S (θ) µ α 2 S(θ) = 0 Solution: Putting u(r, θ, t) = R(r)S(θ)T (t) into the above expression, we obtain: R(r)S(θ)T (t) = α 2 (R (r)s(θ)t (t) 1 r R (r)s(θ)t (t) 1 r 2 R(r)S (θ)t (t)). Dividing everything by R(r)S(s)T (t), we obtain T (t)/t (t) = α 2 (R (r)/r(r) 1 r R (r)/r(r) 1 r 2 S (θ)/s(θ)). The expression on the left hand side is a function of t only, while the right hand side is a function of r and θ only, and therefore both are equal to an arbitrary constant λ. This leads to and also T (t)/t (t) = λ α 2 (R (r)/r(r) 1 r R (r)/r(r) 1 r 2 S (θ)/s(θ)) = λ. Then after an algebraic manipulation, α 2 r 2 R (r)/r(r) α 2 rr (r)/r(r) r 2 λ = S (θ)/s(θ). Therefore again we have that both are equal to a constant µ: and also α 2 r 2 R (r)/r(r) α 2 rr (r)/r(r) r 2 λ = µ S (θ)/s(θ) = µ. After some simplification, this reduces to the expression in (b) above. Page 6
(c) Explain why µ = n 2 α 2 for some integer n. Hint: remember that θ is the angle counter-clockwise from the x-axis]. Solution: Since (r, θ) and (r, θ 2π) both refer to the same coordinate in polar coordinates, the function S should satisfy S(θ) = S(θ 2π). Moreover, we know that S satisfies S (θ) (µ/α 2 )S(θ) = 0. Depending on the value of µ/α 2, the solutions to this are either exponentials, polynomials or trig. functions. Since we require S to be periodic, we want them to be trig functions, and therefore we need µ/α 2 to be positive. In this case, the general solution to the differential equation is S(θ) = A cos( µθ/α) B sin( µθ/α). Now we need S to be 2π periodic, and therefore we need µ/α = n for some integer n. Therefore µ = n 2 α 2. (d) Find the general solution to the above system of equations in the case that λ = 0 and µ = α 2. Hint: to solve for R(r), propose a solution of the form R(r) = r b ] Solution: Since λ = 0, the solution for T is T = E for some constant E. Moreover, the general solution for S is S(θ) = A cos(θ) B sin(θ) Finally, we propose a solution R(r) = r b to the equation r 2 R (r) rr (r) R(r) = 0, and therefore b(b 1)r b 2 r b br b r b = 0. This leads to b 2 1 = 0, and therefore b = ±1. Thus we obtain two solutions: r and r 1. The general solution is therefore R(r) = Cr Dr 1. Putting this all together, we obtain u(r, θ, t) = R(r)S(θ)T (t) = (Cr Dr 1 )(A cos(θ) B sin(θ))e. Since A, B, C, D, E are all arbitrary constants, we can rewrite this as: u(r, θ, t) = C 1 r cos(θ) C 2 r sin(θ) C 3 r 1 cos(θ) C 4 r 1 sin(θ). 4. (8 points) Laplace Equation on a Semicircle Solve Laplaces equation inside a semicircle of radius a (0 < r < a, 0 < θ < π) subject to the boundary conditions: u = 0 on the diameter and u(a, θ) = g(θ). Page 7
Solution: Translating the Laplace equation to polor coordinates we get u rr 1 r u r 1 r u 2 θθ = 0 u(a, θ) = g(θ) u(r, 0) = 0 u(r, π) = 0 u(0, θ) = 0. Separation of variables gives us the differential equations r 2 R rr λr = 0 Θ λθ = 0 and initial conditions Θ(0) = Θ(π) = 0 R(0) = 0. We see that we can solve for Θ using this standard boundary value problem to get λ = n 2 and Θ(θ) = sin(nθ). We see that R = k 1 r n k 2 r n but since R(0) = 0 we must have k 2 = 0. Thus, R = r n is a solution for r. Putting this together and summing over all n we have u(r, θ) = c n r n sin(nθ). n=1 To satisfy the initial condition u(a, θ) = g(θ) we use a sine series and have a n c n = π 0 g(θ) sin(nθ)dθ. Page 8