Math 220A - Fall 2002 Homework 8 Solutions

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1 Math A - Fall Homework 8 Solutions 1. Consider u tt u = x R 3, t > u(x, ) = φ(x) u t (x, ) = ψ(x). Suppose φ, ψ are supported in the annular region a < x < b. (a) Find the time T 1 > suh that u(x, t) is definitely zero for t > T 1 in the ase when i. x > b ii. a < x < b iii. x < a. T 1 = x + b. T 1 = x + b. T 1 = x + b. (b) Find the time T > suh that u(x, t) is definitely zero for < t < T in the ase when i. x > b ii. x < a. T = x b. T = a x. () Consider the same questions for n = dimensions. Sine u(x, t) depends on the values of the initial data in B(x, t), there is no time T 1 suh that we an guarantee that u(x, t) for all t > T 1. In answer to part (b), we again have T = ( x b)/ if x > b and T = (a x )/ if x < a. 1

2 . Solve u tt u = (x, y, z) R 3, t > u(x, y, z, ) = 1 u t (x, y, z, ) = x + y + z. Note: Below x, y and z represent vetors in R 3. Our solution is given by Kirhoff s formula as u(x, t) = φ(y) + φ(y) (y x) + tψ(y) ds(y) B(x,t) = 1 + t y ds(y). B(x,t) By making a hange of variables y = x + tz, we an rewrite this as u(x, t) = [1 + t x + tz ] ds(z) = [1 + t x + t x z + t 3 z ] ds(z). Now 1 ds(z) = 1 Therefore, we onlude that our solution is t x ds(z) = t x t x z ds(z) = t 3 z ds(z) = t 3. u(x, t) = 1 + t x + t Solve u tt u = (x, y) R, t > u(x, y, ) = u t (x, y, ) = x + y Note: Below, x, y and z represent vetors in R. The solution is given by the formula u(x, t) = 1 tφ(y) + t ψ(y) + t φ(y) (y x) dy π t ( t y x ) = 1 π t = 1 B(x,t) B(x,t) B(x,t) t y dy ( t y x ) t y dy ( t y x )

3 Now making the hange of variables y = x + tz, we have u(x, t) = 1 t x + tz dz B(,1) ( t tz ) = 1 t x + tz dz B(,1) t(1 z ) = t x + tx z + t z dz. (1 z ) B(,1) Now the first term an be evaluated as follows. t x 1 t x dz = π (1 z ) π B(,1) = t x. π 1 r dr dθ (1 r ) For the seond term, using the fat that z 1 /(1 z ) is odd with respet to the z axis (and similarly z /(1 z ) is odd with respet to the z 1 axis), we onlude that the seond term is zero. For the last term, we evaluate as follows, t 3 π B(,1) Therefore, our solution is given by z (1 z ) ds(z) = t 3 π 1 π = t 3 3. u(x, t) = t x + t 3 3. r 3 dr dθ (1 r ) 4. Solve { Ut + AU x = U(x, ) = Φ(x) where and 1 1 A = 1 1 sin(x) Φ(x) = 1 e. First, we diagonalize our matrix A by looking for our eigenvalues. We look at det(a λi) =. We see that our eigenvalues are given by λ =,. First, for 3

4 λ 1 =, we see that A λ 1 I = A = 1 1. Therefore, an eigenvetor assoiated with λ 1 = is given by v 1 = [ 1 1 ] T. Then for λ =, we have A λ I = 1 1. Therefore, we have two linearly independent eigenvetors, v = [1 1 ] T and v 3 = [ 1] T. Consequently, letting 1 1 Q = and we see that Λ =, Q 1 AQ = Λ. In partiular, plugging A = QΛQ 1 into our equation, we have Multiplying the equation by Q 1, we have U t + QΛQ 1 U x =. Q 1 U t + ΛQ 1 U x =. Then, letting V = Q 1 U, we have the deoupled initial-value problem { Vt + ΛV x = where V (x, ) = Ψ(x) sin(x) + 1 Ψ(x) = Q 1 1 Ψ(x) = sin(x) + 1. e We have three linear transport equations. First, (v 1 ) t = v 1 (x, ) = 1 ( sin(x) + 1) 4

5 implies Seond, implies v 1 (x, t) = 1 ( sin(x) + 1). (v ) t + (v ) x = v (x, ) = 1 (sin(x) + 1) v (x, t) = 1 (sin(x t) + 1). Last, { (v3 ) t + (v 3 ) x = implies Therefore, v 3 (x, ) = e v 3 (x, t) = e. V = 1 sin(x) + 1 sin(x t) + 1 e and U = QV implies our solution is given by U(x, t) = 1 sin(x) + sin(x t) sin(x) + sin(x t) +. e 5. Consider the symmetri hyperboli system [ ] [ ] [ ] u1 1 u1 + u t u x 1 + [ ] [ ] 1 u1 1 u x = [ ]. (a) Find the smallest ball in R in whih the domain of dependene of U(3, 4, 1) will lie. That is, find M suh that the value of U at the point (3, 4, 1) depends at most on the value of the initial data U(x 1, x, ) in the ball of radius 1M about (3, 4). For a symmetri hyperboli system, the domain of dependene of the solution at a point ( x, t ) is ontained within the ball of radius Mt where M = max λ i ( ξ). ξ =1, i=1,...,m where λ i (ξ), i = 1,..., m are the m eigenvalues of the matrix A(ξ) = m ξ i A i. i 5

6 Here, for ξ R, The eigenvalues of A(ξ) are given by Now A(ξ) = ξ 1 A 1 + ξ A [ ] ξ1 ξ =. ξ ξ 1 λ 1 = 3ξ 1 + ξ 1 + 4ξ λ = 3ξ 1 ξ 1 + 4ξ max λ i ( ξ) =, i = 1,. ξ =1 This an be found by maximizing λ 1, λ subjet to the onstraint ξ = 1. Therefore, M =, and onsequently, the domain of dependene for the point (3, 4, 1) is the ball {(x, y) R : x 3 + y 4 ( 1) }. (b) Show that the ball you found in part (a) is the smallest ball in whih you an guarantee the domain of dependene will lie, by showing there exists a diretion ξ = (ξ 1, ξ ), where ξ = 1 for whih there exists a plane wave solution U(x 1, x, t) = V (x ξ Mt); that is, a plane wave solution whih travels at speed M. You don t need to alulate the plane wave solution. From part (a), we see max ξ =1 λ 1 ( ξ) ours at ξ = (1, ), in whih ase λ 1 (1, ) =. Similarly, max ξ =1 λ ( ξ) ours at ξ = ( 1, ), in whih ase λ ( ξ) =. As this is a symmetri hyperboli equation, we know there are m plane wave solutions for every diretion ξ R. In partiular, for eah ξ R, the m plane wave solutions have speed λ i (ξ) for i = 1,..., m. To show this expliitly, in our ase above, we look for a plane wave solution v( ξ x σt) where ξ = (1, ). In other words, we are looking for a solution of the form v(x 1 σt) for some σ. Plugging this into our system, we have [ ] 1 σ v (x 1 σt) + v (x 1 σt) =. This means we need to look for a funtion v (x 1 σt) and a value σ suh that [ ] 1 v (x 1 σt) = σ v (x 1 σt). In other words, an eigenvetor v and a orresponding eigenvalue σ for [ ] 1. Clearly, is an eigenvalue of this matrix with orrsponding eigenvetor r. Letting v (x 1 σt) = r, we see we have found a plane wave solution whih travels in the diretion (1, ) with speed. 6

7 () Find two plane wave solutions whih propagate in the diretion (ξ 1, ξ ) = (3/5, 4/5); that is, find two general solutions of the form V 1 (ξ x σ 1 t), V (ξ x σ t). We look for a plane wave solution v( ξ x σt) whih travels in the diretion ξ = (3/5, 4/5). We plug into our system. Doing so, we have v( ξ x σt) = v(ξ 1 x 1 + ξ x σt) σ v + ξ 1 A 1 v + ξ A v =, where v = v (ξ 1 x 1 + ξ x σt) [ ] 1 A 1 = [ ] 1 A =. 1 Therefore, we need to look for an eigenvetor v and a orresponding eigenvalue σ of A( ξ) = ξ 1 A 1 + ξ A at ξ = (3/5, 4/5). Now The eigenvalues are given by [ ] A( ξ) ξ1 ξ = ξ ξ 1 ( ) [ ] = λ 1 = 1 1 (9 + 73) λ = 1 1 (9 73) with orresponding eigenvetors [ 1 r 1 = ( 3 + ] 73) 8 1 [ 1 r = ( 3 ] 73) 8 1 Therefore, any funtion v 1( ξ x λ 1 t) whih is a multiple of r 1 will be an eigenfuntion of A( ξ), and, therefore, a plane wave solution. (Similarly, any funtion v ( ξ x λ t) whih is a multiple of r.) Therefore, two general plane wave solutions in the diretion ξ = (3/5, 4/5) with speeds λ 1 and λ given above, are of the form v 1 ( ξ x λ 1 t) = f( ξ x λ 1 t) r 1, 7

8 and v ( ξ x λ t) = g( ξ x λ t) r for arbitrary funtions f and g, where r 1 and r are given above. 8