uv = y2 u/v = x 4 y = 1/x 2 u = 1 y = 2/x 2 u = 2

Transcription

1 Section A.. Variable change, divergence and curl (a) x..4 (b) [3 u = x v = /x } uv = u/v = x 4 } = uv x = (u/v) /4 = x v = = x v = = /x u = = /x u = We need to find the Jacobian: J = x u v x v u = u / 4 u 3/4 v /4 v / + u /4 v / 4 v 5/4 u / = = 8 u /4 v 3/4 + 8 u /4 v 3/4 = = 4 u /4 v 3/4 I = = R R uv dx d = J du dv = x R (u/v) /4 u /4 v 3/4 4 u /4 v 3/4 = 4

2 Some students calculated the inverse Jacobian, and directl substituted the values of u and v as a function of x and to obtain the same result (because it cancels out). Although acceptable, the solution above is the correct one. [5 (c)stokes B dr = B da But B is in x plane, so that B is in z direction: B da = [ B z dxd B z = B x B x = x Γ = x dx d = I = R R A majorit of students did use Stokes to obtain the integral above, and related it to the previousl calculated value of I. Minor issues appeared with sorting out the correct sign. [7

3 . Gradients and divergence (a)divergence u = g = 3z (x + )k + xz 3 i + z 3 j = z 3 (xi + j) + 3z (x + )k In polar coordinates, u = z 3 ρ ê ρ + 3z ρ ê k u = x (xz3 ) + (z3 ) + z (3z (x + )) = = z 3 + z 3 + 6z(x + ) = = 4z 3 + 6z(x + ) Cartesian = 4z 3 + 6zρ Polar (b) Gauss: In polar coordinates, u dv = V = u da = 4z 3 dz V u dv [4z 3 + 6zρ πρ dρ dz πρ dρ + 6z dz Both integrals in z are odd in z, and so integrate to zero: u da = πρ 3 dρ [5 Attempts to integrate the fluxes using the full area can succeed, if care is taken to consider the full area, but that approach produced errors, as some students did not consider the cancelation of top and bottom of the clinder. (c) u da = because u = ( g) = (cf. vector identit ( φ) = for an scalar function φ. Alternativel, one can evaluate u using the definition of curl, to find u =. (d)stokes From Stokes theorem, Γ = u dr = C u da S where S is the surface that spans C. But u =, so Γ =. [4 (e)gradient potential v = g g = g[ g x i + g j + g z k = = x ( g )i + ( g )j + z ( g )k = = (g ) [6 [4 3

4 he scalar potential of v is defined b v = φ (or it could be φ), so that φ = ± g We cannot introduce a vector potential for u because the field is not solenoidal, i.e. u. [6 4

5 3. Wave equation (a). We have for the homogeneous solution p, using the intermediate variable ζ = x ± ct: p x = df ζ dζ x + dg ζ dζ x p x = f + g p t = df dζ p t = cf cg ζ t + dg ζ dζ t f + g = c (f + g ) p x = d f ζ dζ x + d g ζ dζ x p x = f + g p t = d f ζ dζ x + d g ζ dζ t p t = c f + c g QED. (b). Using p = X(x) (t): [4 p x = X p t = X c X = X c X X = We then solve two differential equations to ield: = ω X = A cos kx + B sin kx = C cos ωt + D sin ωt where k = ω/c. (c) Since we are seeking a solution that vanishes at x = a, but not at x =, we choose to retain the cosine term (other combinations will give different intermediate solutions, with the same final solution), and incorporate the constant A into the other constants: [5 p(a, t) = cos(ka)(c cos ωt + D sin ωt) = k n a = (n + )π/ ω n = k n c = (c/a)(n + )(π/) p(x, t) = cos(ω n x/c)(c cos ω n t + D sin ω n t) Check: p(, t) = cos(ω n /a)(c cos ω n t + D sin ω n t) p(a, t) = cos ω n a/c(c cos ωt + D sin ωt) = (d). Now the solution at x = must match the given unstead function p sin Ωt. We write the generic solution as a sum of all possible solutions: [5 p(, t) = p sin ωt = cos(ω n x/a)(c cos ω n t + D sin ω n t) n= 5

6 which still obe the initial boundar conditions. We can multipl both sides of the generic solution at x = them b sines or cosines of ω m t and integrate over a time : p sin Ωt cos ω m t dt = p sin Ωt sin ω m t dt = cos(ω n /a) (C }{{} n cos ω n t cos ω m t + D n sin ω n t cos ω m t) dt n= cos(ω n /a) (C }{{} n cos ω n t sin ω m t + D n sin ω n t sin ω m t) dt n= It can be shown that (from Maths datebook, Fourier decomposition) n= n= cos πpt sin πpt πqt cos dt = δ pq πqt sin dt = δ pq c π so that if we make p = n +, q = m +, a = π, or = 4a/c, this corresponds to the Fourier decomposition of p (t), and C n = c a D n = c a 4a/c 4a/c p (t) cos(n + ) π p (t) sin(n + ) π c a t dt c a t dt [7 (e) For a step function at the origin, the step wave will propagate unchanged at the speed of sound through the tube, and reflect at the end, where there is a pressure node. he solution can be approximated b using the same method in (d), where the step function is represented as the sum of harmonics in the corresponding series solution. [4 6

7 exact solu+on t Fourier representa+on of p (t)=h(,t) 7

8 Section B 4. QR decomposition and Least-Squares Solution (a) Q is an othonormal matrix. hus [ c R d = Q v As R is upper-triangular it osthen straight-forward to solve b substituting. [3 (b) he first column is simpl the normalised version of the first column of A and r = 3 q = 3 Projecting the second column and subtracting ields q = 3 he final column is required to be othogonal to the other two. hus Q = 3 ; R = 3 (c) Solving ields (ignoring last column) [ 3 [ c d = 3 [ 8 [8 Solving ields c = 9/9; d = 4/3 (d) For the squared error to be zero, then the last row of (q ) 3 v =. For a solution [ x z = x + z = [6 hus an point ling on the plane will have zero squared error. [8 8

9 = exp ( s σ / sµ) ) [7 5. Sum of random variables and moment generating functions (a): differentiating the expression ields hus mean is λ. Differentiating again ields hus the variance is g (s) = λ exp( λ( s)) λ exp( λ( s)) σ = λ + λ λ = λ (b): the general form is to convolve the two distributions. In this case one is discrete the other continuous. he form must be p z (Z) = p x (X)p (Z X) X= hus the distribution is continuous [4 (c)(i): he MGF can be written g (s) = p (x) exp( sx)dx ( = exp ) πσ σ (x xµ + µ + sxσ ) dx ( = exp ) πσ σ ((x (µ sσ )) s σ 4 + sσ µ) dx [5 (c)(ii) Differentiating this expression ields g z(s) = (sσ µ λ exp( s)) exp ( s σ / sµ + λ(exp( s) ) ) hus the mean is µ z = g z() = µ + λ as expected. he second differential is g z (s) = (σ + λ exp( s)) exp ( s σ / sµ + λ(exp( s) ) ) +(sσ µ λ exp( s)) exp ( s σ / sµ + λ(exp( s) ) ) equating to zero ields and subtracting the µ z σ + λ + (µ + λ) (µ + λ) = σ + λ hese are the values of adding two independent variables. [4 (d): an error occurs when the magnitude of the noise for an integer is greater than.5. For a single integer this can be written as (using the smmetr) Φ(.5/σ) he exception to this is that zero integer value is alwa correctl recogised when the received signal is negative (it will alwas be the closest). hus the overall expression is P e = p x ()Φ(.5/σ) + p x (X)Φ(.5/σ) = ( p x ())Φ(.5/σ) x= = ( exp( λ))φ(.5/σ) 9 [5

10 6. LU Decomposition and Matrix Sub-Spaces (a) Performing LU decomposition ields z = here z 5 = 5 /5 (b) o find the general solution find the solution - solve 4 L = [8 hus = 4 hen solve Ux = = 4 hus setting x 3 = and x 4 = x = o compute the null-space Hence one axis is And the second basis [ [ x x = [ = [ Hence the general solution is x = + α + α 4

11 (c) he left null-space is perpendicular to the column-space - simpl cross-product two columns 7 5 (d) he solution would change as there would be no left-null-space, and there would onl be a single null-space. [3 [ [4