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18.303 Problem Set 7 Solutions Problem 1 (10+10 points): g(x0 )d3 x0 , the second term is: Z Z · (x0 − x1 )d3 x0 = ∇0 g|x1 · x0 d3 x0 − x1 volume(V ) , (a) If we plug the Taylor expansion of g into ∇0 g|x1 R V V V 0 where we have pulled the constant (x -independent) terms out of the integral. Hence, this is zero (for all x0 and x) if Z 1 x 0 d3 x 0 , x1 = volume(V ) V which is just the centroid of the volume V . (b) From the notes, we can write g(x0 ) more explicitly as g(x0 ) = Hence, if we compute the Hessian Hij = H= 16π 2 |x1 (x0 − x0 ) (x0 − x) · . 4π|x0 − x|3 4π|x0 − x0 |3 ∂2g ∂x0i ∂x0j , evaluated at x0 = x1 , we get (after some algebra): 2I 9(x1 − x)(x1 − x)T (x1 − x0 )(x1 − x0 )T + 3 3 − x| |x1 − x0 | 16π 2 |x1 − x|5 |x1 − x0 |5 3(x1 − x)(x1 − x)T 3(x1 − x0 )(x1 − x0 )T − − , 2 5 3 16π |x1 − x| |x1 − x0 | 16π 2 |x1 − x|3 |x1 − x0 |5 where I is the 3 × 3 identity matrix and T is transposition of the column vectors. The first time comes when both derivatives fall on the numerator, the second term when both fall on the denominator, and the third and fourth terms are when one derivative acts on the numerator and one on the denominator. [Note that the (x0 − x0 )(x0 − x0 )T is a column vector times a row vector, an “outer product”, which by the rules of matrix multiplication produces a 3 × 3 matrix.] You may not have been able to write H in such a compact form. The resulting “quadrupole” term in B̂u0 is then Z Z X 0 T 0 3 0 ln(c2 /c1 ) (x − x1 ) H(x − x1 )d x = ln(c2 /c1 ) Hij (x0i − x1i )(x0j − x1j )d3 x0 . V i,j V R Note that the integrals V (x0i − x1i )(x0j − x1j )d3 x0 are properties of the geometric shape V only, independent of the source location x0 or the observer location x. From the formula for H, we can see by inspection that this term scales as ∼ 1/|x − x1 |3 and ∼ 1/|x0 − x1 |3 , which means that at large separations it is negligible compared to the dipole term from class. Problem 2 (5+5+(5+5) points): (a) We first need to compute u0.5 from the initial u0 . We can do that with second-order accuracy using, e.g. a Crank-Nicolson scheme: −1 A∆t A∆t 0.5 u = 1− 1− u0 . 4 4 Then, given u0 and u0.5 , or (by induction) from uk and uk+0.5 for k = 0, 1, 2, . . ., we can get uk+1 via the first leapfrog equation uk+1 = uk + Auk+0.5 ∆t and then we can get uk+1.5 from the second leapfrog equation uk+1.5 = uk+0.5 + Auk ∆t. In this way we can compute un and un+0.5 for as many n > 0 as desired. 1 (b) Applying the first leapfrog equation and then the second equation, we have un+1 − un un − un−1 − = A un+0.5 − un−0.5 = ∆tA2 un , ∆t ∆t or un+1 − 2un + un−1 = A2 un . ∆t2 2 But the left-hand side is just our familiar finite-difference approximation for ∂∂t2u evaluated at t = n∆t, and so 2 ∂ ∂u ∂ ∂u 2 this is an approximation for (deriving from the original PDE) ∂∂t2u = ∂t ∂t = ∂t Âu = Â ∂t = Â u. (c) If we plug in unm = eikm Gn and un+0.5 = eikm Gn+0.5 a into the first leapfrog equation, evaluated at the point m m in space, we obtain 2 ik −ik un+0.5 − 2un+0.5 + un+0.5 un+1 − unm G−1 m m−1 m ikm n+0.5 e − 2 + e ikm n+0.5 −4 sin (k/2) = ae G = ae G = eikm Gn = m+1 ∆t ∆t ∆x2 ∆x2 ∆x2 which simplifies to G0.5 − G−0.5 4∆t sin2 (k/2) . =− a ∆x2 If we plug into the second leapfrog equation, we similarly get 2 un − 2unm + unm−1 − un−0.5 un+0.5 1 − G−1 m m ikm n −4 sin (k/2) = aeikm Gn+0.5 = m+1 = e G , ∆t ∆t ∆x2 ∆x2 which simplifies to a(G0.5 − G−0.5 ) = − 4∆t sin2 (k/2) . ∆x2 (i) Combining the two equations, we have a = 1/a, or a2 = 1 =⇒ a = ±1. (ii) For a = +1, we get G0.5 − G−0.5 = − √ 4∆t sin2 (k/2) 1 ≤ 0 =⇒ G ≤ √ =⇒ G ≤ 1. 2 ∆x G 2