Homework 3 solutions Math 136 Gyu Eun Lee 2016 April 15 A problem may have more than one valid method of solution. Here we present just one. Arbitrary functions are assumed to have whatever regularity properties are required, unless otherwise specified. 1.6.3 A rotation of the xy-plane is a linear transformation whose matrix is of the form ( ) a b R = b a where a 2 + b 2 = 1. 12 Each rotation defines a change of variable like so: ( ) ( ) ( ) ξ x ax by = R =. ν y bx + ay Given the equation a 11 u xx + 2a 12 u xy + a 22 u xx + b 1 u x + b 2 u y + cu = 0, (1) and a rotation R, replacing u(x,y) with u(ξ,η) = u(r(x,y)) gives us a new equation d 11 u ξ ξ + 2d 12 u ξ η + d 22 u ηη + e 1 u ξ + e 2 u η + f u = 0, (2) where the coefficients d i j, e i, and f depend on R. To say that the PDE is invariant under rotations means that the coefficients must actually be independent of R; 3 that is, d i j = a i j, b i = e i, and c = f for all choices of R (equivalently, for all choices of rotation angle θ), and our problem is to determine which conditions on a i j, b i, and c make this true. Since the second-order derivatives in Equation (1) correspond under the change of variable exactly to the second-order derivatives in Equation (2), and similarly for the first-order and zero-th order derivatives, it suffices to work with each order of derivatives separately. For the zero-th order term, clearly changing from xy-variables to ξ η-variables does not change the coefficient in front of u. Therefore we must have c = f for the zero-th order terms to be invariant. For the first-order term, a calculation with the chain rule shows that u x = a u ξ + b u η, c 2016, Gyu Eun Lee. Licensed under the Creative Commons Attribution-NonCommercial 4.0 International license (CC BY-NC 4.0), https://creativecommons.org/licenses/by-nc/4.0/. Last updated: April 20, 2016 1 Such a matrix is said to be a special orthogonal matrix, i.e. a matrix of determinant 1 with orthogonal rows and columns. The collection of all such matrices is called the special orthogonal group, denoted SO(2). 2 In particular, we can take a = cosθ, b = sinθ for some angle θ. However, we will see that expressing the matrix in this form only makes the notation cumbersome. 3 Equivalently, if u(x,y) is a solution to (1), then so is u(ξ,η) = u(r(x,y)) for all rotations R. 1
Therefore u u = b y ξ + a u η. Au x + Bu y = (aa bb)u ξ + (ba + ab)u η. If the equation is to be invariant under rotation, then we must have ( ) ( ) ( )( ) A aa bb a b A = = B ba + ab b a B for all choices of a,b with a 2 + b 2 = 1; that is, the 2-d vector (A,B) must be invariant under all rotations. But the only vector that is invariant under all rotations is the zero vector; therefore A = B = 0. For the second-order terms, we find the partial derivatives to be 2 u x 2 = 2 u a2 ξ 2 + 2ab 2 u ξ η + 2 u b2 η 2, Therefore 2 u x y = ab 2 u ξ 2 + (a2 b 2 ) 2 u ξ η + ab 2 u η 2, 2 u y 2 = 2 u b2 ξ 2 2ab 2 u η ξ + 2 u a2 η 2. Au xx + 2Bu xy +Cu yy = (a 2 A 2abB + b 2 C)u ξ ξ + (2abA + 2(a 2 b 2 )B 2abC)u ξ η + (b 2 C + 2abB + a 2 C)u ηη. If the second-order terms are to be invariant under all rotations, then A a 2 2ab b 2 A B = 2ab 2(a 2 b 2 ) 2ab B C b 2 2ab a 2 C for all choices of a,b with a 2 +b 2 = 1. Necessary conditions on A,B,C can now be obtained by taking particular choices of a and b. Choosing a = 1, b = 0 gives us A 1 0 0 A B = 0 2 0 B, C 0 0 1 C and in particular B = 2B, which tells us B = 0. Choosing a = 0, b = 1 now gives us A 0 0 1 A 0 = 0 2 0 0, C 1 0 0 C 2
or equivalently A = C. Therefore we conclude that the only choice of coefficients making Equation (1) invariant under all rotations is a(u xx + u yy ) + cu = 0. Remark: Another way one could prove this statement is to rephrase the invariance of coefficients in terms of matrices. For example, the statement that the second-order coefficients are invariant is equivalent to: if A = (a i j ) is the matrix of second-order coefficients, then BAB t = A for all rotation matrices B. Equivalently, since all rotation matrices satisfy B t = B 1, BA = AB; i.e. A commutes with all rotations. Then one needs to find necessary and sufficient conditions on A for this to be true. Several people tried this method, but failed in one key step: it is not true that for a general matrix A, if BAB t = A for all rotation matrices B then A is a multiple of the identity. In 2 dimensions, a counterexample is when A is a rotation matrix, because rotations about the same axis always commute. The key is to use the fact that A is also a symmetric matrix, and a result called the spectral theorem. A proof of this sort that did not invoke the symmetry of A did not receive full credit. 2.1.5 By d Alembert s formula, u(x,t) = 1 2 [φ(x + ct) + φ(x ct)] + 1 2c x+ct x ct ψ(s) ds = 1 2c x+ct x ct ψ(s) ds. Since ψ is zero outside the interval [ a,a], this integral is equal to the length of the intersection of [ a, a] and [x ct, x + ct]. When t = ka/2c for k some positive integer, Then we have the following cases: [x ct,x + ct] = [x ka 2,x + ka 2 ]. 3
(a) If x > a + ka 2, then [x ka 2,x + ka 2 ] and [ a,a] are disjoint intervals. So for x in this region, u(x, ka 2c ) = 0. (b) If a x ka 2 a x + ka 2, then (c) If then (d) If u(x,t) = a x + ka 2. x ka 2 a x + ka 2 a, u(x,t) = x + ka 2 + a. a x ka 2 < x + ka 2 a, then u(x,t) = ka. (e) If x ka 2 a < a x + ka 2, then u(x,t) = 2a. Whether or not there exist x that fall into any of these cases depends on the value of k. We look at just the example of k = 1, i.e. t = a/2c. In this case, [x a 2,x + a 2 ] is an interval of length a. Since [ a,a] has length 2a, case (e) cannot be valid for any x, but all other cases are valid, and the so the string profile for u(x, 2c a ) is a trapezoid: 0 x < 3a 2, 1 2c (x + 3a 2 ) 3a 2 x < a 2, u(x,t) = a 2c a 2 x < 2 a, 1 2c ( 3a 2 x) a 2 x < 3a 2, 0 x 3a 2. The string profiles should be: (a) t = 2c a : a trapezoid of height 2c a (b) t = a c : a triangle of height a c (c) t = 3a 2c : a trapezoid of height a c (d) t = 2a c : a wider trapezoid of height a c (e) t = 5a c : an even wider trapezoid of height a c 4
2.1.8 (a) Suppose u solves the spherical wave equation u tt = c 2 (u rr + 2 r u r ), and set v = ru. Then v tt = ru tt and v rr = ru rr + 2u r. Then v tt = ru tt = c 2 (ru rr + 2u r ) = c 2 v rr. (b) The general solution is where f and g are arbitrary functions. v(r,t) = f (r + ct) + g(r ct) (c) The initial conditions u(r,0) = φ(r), u t (r,0) = ψ(r) give us v(r,0) = rφ(r), v t (r,0) = rψ(r). Therefore by d Alembert s formula Then v(r,t) = 1 2 [(r + ct)φ(r + ct) + (r ct)φ(r ct)] + 1 2c r+ct r ct sψ(s) ds. r+ct u(r,t) = v(r,t) = 1 1 [(r + ct)φ(r + ct) + (r ct)φ(r ct)] + sψ(s) ds. r 2r 2cr r ct Remark: The reason we assume φ and ψ are even is because u is a solution of the spherical wave equation in 3 dimensions. In principle u(r,t) is then defined only for r 0, since the distance r to the origin cannot be negative. Then when we change to v = ru, we cannot solve the IVP as we have been doing in this chapter, because the domain of the IVP is (r,t) [0, ) (, ) instead of (, ) (, ). However, we can extend the initial conditions u(r,0) = φ(r) and u t (r,0) = ψ(r) to the negative r-axis by reflecting φ and ψ across the y-axis; the resulting reflections are even, and now the domain of the IVP is (, ) (, ), allowing us to use the techniques of section 2.1. Other than this, the assumption that φ and ψ are even does not add much to this problem. This technique is known as even reflection, and we will see it again in chapter 3. 2.2.1 For the sake of ease of notation, we will assume c = ρ = T = 1. The general case should be obvious from this case; in fact, by scaling the wave equation and making some linear changes of variable, it is not hard to see that the general case can be reduced to this case. (Exercise.) When c = ρ = T = 1, the wave equation is and the energy is E(t) = 1 2 u tt = u xx (u 2 t + u 2 x) dx. 5
In principle this is a function of t; however, by conservation of energy, de/dt = 0, and therefore E is a constant function of t. By the initial condition u(x,0) = φ(x) 0, we have u x (x,0) 0. We also know that u t (x,0) = ψ(x) 0. Therefore E(0) = 1 2 (u t (x,0) 2 + u x (x,0) 2 ) dx = 1 2 0 dx = 0. Since E is constant in t, this implies E(t) = 0 for all t. But since the integrand u t (x,t) 2 + u x (x,t) 2 is non-negative for all x and t, this implies that u t (x,t) 0 and u x (x,t) 0. The only continuously differentiable functions u that satisfy both these equations are the constant functions, so u must be constant. Since we know u(x,0) = 0, this constant must be 0; that is, u(x,t) 0. 2.2.4 Since u solves the wave equation, we know that for some functions f and g. Then u(x,t) = f (x +t) + g(x t) u(x + h,t + k) = f (x + h +t + k) + g(x + h t k), u(x h,t k) = f (x h +t k) + g(x h t + k), u(x + k,t + h) = f (x + k +t + h) + g(x + k t h), u(x k,t h) = f (x k +t h) + g(x k t + h). Comparing the sum of the first two lines, and the sum of the last two lines, we see that they are indeed equal. 6