Solutions to sample final questions

Transcription

1 Solutions to sample final questions To construct a Green function solution, consider finding a solution to the equation y + k y δx x where x is a parameter In the regions x < x and x > x y + k y By searching for solutions of the form yx e mx it can be seen that m + k and thus m ±ik Hence for x < x, the solution can be expressed as yx A cos kx + B sin kx for some constants A and B In the region x > x the solution will also have the same form, but potentially with different constants C and D: yx C cos kx + D sin kx Since y, it follows that A, and since y π k, then D To set the two remaining constants, consider x x The function must be continuous there, B sin kx C cos kx There must be jump in the derivative of at x x, Ck sin kx Bk cos kx See appendix A for a derivation of this condition By substituting Eq into Eq gives sin kx Ck sin kx Ck cos kx Ck From Eq it follows that Thus the Green function solution is Gx, x C B { cos kx sin kx sin kx k cos kx k k for x < x, sin kx cos kx k for x > x Plots of Gx, x are shown in Fig for the cases of kx π/, π/6, π/4, π/3, 5π/

2 - - kgx, x -3-4 kx π/ kx π/6-5 kx π/4 kx π/3-6 kx 5π/ π/8 π/4 3π/8 π/ kx Figure : Plots of several Green functions corresponding to different values of x a If z x + iy and r a + ib then rz r z a ibx + iy a + ibx iy ax + by + iay ibx ax + by iay + ibx iay bx and this will be satisfied if ay bx, which can be written as y b/ax This is a straight line passing through a, b b If z x + iy then z + z x + iy + x iy x y + ixy + x y ixy x y and z z x + iyx iy x + y Hence z + z b a + z zb + a x y b a + x + y b + a x b + a y x a y b +x b + a y + x a + y b 4x b + y a

3 5 5 y -5 - ay bx -5 x + y a - b x Figure : Plots of the loci of the two complex equations considered in question and therefore Dividing by 4a b gives Plots of the two loci are shown in Fig 3 The volume of the box is given by and the constraint is 4x b + y a 4a b x a + y b Vd, e, f 8de f d a + e b + f c The given constraint describes an ellipsoid If the maximum of V was in the interior of the ellipsoid, then V d 8e f, V e 8d f, V f 8de These equations are only satisfied if at least two of d, e, and f are zero, and the corresponding volume will be V, which is not a maximum Hence the maximum volume must correspond to d, e, and f being on the surface of the ellipsoid 3

4 To use the method of Lagrange multipliers, consider minimizing the augmented function Vd, e, f, λ 8de f + λ d a e b f c Then V d Hence and therefore 8e f λd a, V e λ 4 a e f d The constraint equation therefore gives and thus after which it follows from Eq 3 that The volume is 8d f λe b, b d f e c de f V f 8de λ f c a d b e c f 3 d a + e b + f c 3d a e d a 3 b 3, f c 3 Vd, e, f 8abc a The eigenvalues are solutions to det A λi 4 λ λ 4 λ λ 4 λλ 5 Hence and 5 are eigenvalues For λ, an eigenvector will satisfy 4 u v u, v, is a solution For λ 5, an eigenvector will satisfy u 4 v and thus a solution is u, v, It should be noted that the two eigenvectors are orthogonal, as would be expected for a symmetric matrix 4

5 b The product of A with itself is A A and thus β 5 For the given identity, A n 5 n A, it can be seen that the case of n is trivially satisfied Now suppose that the case of n is true, and consider the case of n + Then A n+ A A n 5AA n 5A n 55 n A 5 n+ A and the result is true for n + Hence, by mathematical induction, the result must be true for all n c The exponential is expλa λa n n n! I + I + n n I + A 5 I + A 5 λa n n! λ n 5 n A n! n 5 n λ n n! + n and thus f λ and gλ e 5λ /5 5 n λ n e I + 5λ A n! 5 5 a If f x log x, then f The first derivative of f x log x is f x x and thus f The successive derivatives are f x x, f x x 3, f 4 x 3 x 4 It can be seen that each successive derivative brings an additional power of x and an integer coefficient, and a therefore a general expression is f n n! x x n 5

6 for any positive integer n, f n n! The Taylor series is therefore f n x n n!x n n! n x n! n n n n A comparison between the function and the first few terms of its Taylor series is shown in Fig 3a If the series is written as a n x n, then the radius of convergence R of this power series is given by R lim n a n+ a n lim n n n + Hence R To find the exact interval of convergence, consider the end points at x ±R At x, the series is n which is the harmonic series and diverges At x, the series is n n, n, n which converges by the alternating series theorem Hence the exact interval of convergence is x < b The power series is given by log x x 3 x + x 3 x + x 3 x + x x x 3 x4 x5 x6 x6 3 x x 3 x4 x5 5x6 6 A comparison between the function and the Taylor series is shown in Fig 3b 6 The integral can be evaluated by considering a keyhole contour as shown in Fig 4, consisting of four components A, B, C, and D A branch cut can be introduced along the positive real axis, where the value of z /3 is taken to be positive and real just above the cut An integral around a circular contour given by z ρe iθ for θ < π is π ρ Iρ /3 e iθ/3 ρe iθ idθ ρe iθ + It can be seen that Iρ as ρ, because the ρ factor in the denominator will dominate In addition Iρ as ρ as the ρ 4/3 factor in the numerator will 6

7 5 a y -5 - y log x y x -5 y x x y x x x - b -4-6 y y log x x 3 Taylor series x Figure 3: a Comparison of log x to its first, second, and third order Taylor series b Comparison of log x x 3 to its Taylor series up to terms in x 6 7

8 tend to zero In the limit, only the integrals I A and I C along the contours A and C will matter It can be seen that z I C /3 dz z + r /3 e πi/3 dr re πi + r /3 e πi/3 dr r + e πi/3 I A C By the residue theorem, the integral around the entire contour will be determined by the residue at the enclosed pole of order at z The residue is given by z /3 Res z +, z d z /3 z + lim z dz z + Hence and therefore d lim z dz z/3 z /3 lim z 3 /3 e iπ/3 3 I A + I C πi e iπ/3 /3 3 I A e πi/3 πi e iπ/3 /3 3 By making use of e iπ, it can be seen that /3 3 3 e iπ/3 I A πi /3 e πi/3 3 πie iπ /3 3e πi/3 e πi/3 π /3 3 sin π 3 π /3 π 3 3/ 7 By using the chain rule f a x f a x + y f a y b f x + a f y 4 and f b x f b x + y f b y a f x b f y 5 8

9 z B Im D A C Re z R z r' Figure 4: Keyhole contour considered in question 4 The integrand has a pole of order at z, and a branch cut can be introduced along the positive real axis A closed contour can be constructed as A a section from r to R above the cut, B the circle z R, C a section from R to r below the cut, and D the circle z r Taking Eq 4 multiplied b, plus Eq 5 multiplied by a, gives b f a + a f b b + a f f + ab ab x y Similarly f x f y b f b + a a + a a + b f a a f b + a b b a + b f b 8 a The Laplace transform of f is given by L f p f te pt dt [ f te pt] pfp f f t pe pt dt By applying this result recursively, it can be seen that the Laplace transform of f is L f p pl f p f ppfp f f p Fp p f f 9

10 b Taking the Laplace transform of the equation gives p Fp p f f + 9pFp f + 8Fp which can be rearranged to give p + 9p + 8Fp p 9 Fp c The Bromwich inversion integral is p + p + 8p + f t c+i e pt p + dp πi c i p + 8p + where c > A closed contour can be made by making use of a large semicircle in the left half plane The integrand has two simple poles at p and p 8, both of which will be enclosed by the contour The residues at these poles are and Res e pt p + p + 8p +, p e pt p + lim p p + 8 e pt p + Res p + 8p +, p 8 e pt p + lim p 8 p + Hence the integral is f t πi πi 9e t 7 e 8t 7 9 a The Fourier transform of f x is given by π f xe ixα dx 9e t e 8t 7 and by using integration by parts this can be written as [ f x iαe ixα] π f x iαe ixα 9e t 7 e 8t 7 Since f x as x ± is a condition for using Fourier transforms, it follows that this is equal to iα f x iαe ixα, which is iα f α

11 b The Fourier transform of the delta function is δα π δxe iαx dx e iα π π Using this result, and the result from part a, the Fourier transform of the differential equation is c The solution is given by iα f α + iα f α f π f α πα iα + πα iα + i f x f αe iαx dα e iαx dα π α iα + i The integrand has simple poles at α i and α i The residues are e iαx Res α iα + i, α i e iαx α i lim α i α + iα i e iαx lim α i α + i e x 3i and e iαx Res α iα + i, α i e iαx α + i lim α i α + iα i e iαx lim α i α i ex 3i Now consider the integral in Eq 6 If x, a closed contour can be constructed by integrating from R to R and then around the semicircle Re iθ for θ π in the upper half plane As R the integrand will become small on the semicircle due to the presence of the e iαx term Hence the integral along the real line will be given in terms of the residue at α i, f x e iαx dα π α iα + i πi π e x 3i e x 3 If x <, the integral can be closed in the lower half plane, and thus will be given by the residue at α i, plus an additional minus sign due to the contour being in the reverse direction f x πi π ex ex 3i 3 6

12 The first Laplace transform can be written as F p the Bromwich inversion integral f t πi c+i c i p + b p + b + iap + b ia F pe pt dp πi c+i c i p + be pt dp p + b + iap + b ia has two simple poles at p b ± ia To evaluate this integral, the contour can be closed with a large semicircle in the left half plane, which will enclose both of the poles The residues are Hence Res F pe pt, p b ± ia p + be pt lim p b±ia p + b ± ia iae b±iat ia f t πi e bt+iat + e bt iat e bt eiat + e iat πi e bt±iat e bt cos at A plot of the function is shown in Fig 5a for the case of a π and b / The second Laplace transform can be written as F p and there Bromwich inversion integral f t πi ap p + ia p ia c+i c i ape pt dp p + ia p ia has two poles of order at p ±ia Again, the contour can be closed in the left half plane, enclosing both poles The residues are Res F pe pt, p ±ia lim p ±ia Hence d dp ape pt p ± ia p ± ia ae pt + apte pt ape pt p ± ia lim p ±ia p ± ia 4 a 8ae ±iat ± 8ia te ±iat iae ±iat ia 6a 4 8a3 8ia 4 t + 8a 3 e ±iat 6a 4 f t πi te iat πi i te iat t sin at i and the function is plotted in Fig 5b for the case of a π ±te±iat i

13 ft a x ft b x Figure 5: Plots of the two functions a f t e bt cos at and b f t t sin at considered in question, for the case of a π and b The integral can be written as cos x x + x + i dx e iz + e iz z + i z i dz The integrand has a simple pole at z i and a pole of order at z i The term e ix will become small in the upper half plane, and the term e ix will become small in the lower half plane To make use of residue calculus, the integral must be split into two components I e iz z + i z i dz, I e iz z + i z i dz, which must be considered separately For I a closed contour can be made by using a large semicircle in the upper half plane, which will enclose the simple pole at z i The residue at this pole is e iz Res z + i z i, z i lim z i e iz z + i 8e I πi 8e iπ 4e For I a closed contour can be made by using a large semicircle in the lower half 3

14 plane, which will enclose the pole at z i The residue at this pole is e iz Res z + i z i, z i d e iz lim z i dz z i z iie iz e iz lim z i z i i e e i 3 8e I πi 3 8e 3iπ 4e where an additional minus sign has been incorporated due to the contour being clockwise Hence cos x x + x + i dx I + I iπ e a First note that since f is periodic with period π, π π f xdx π π f x + zdx for any constant z, since the shifted integral will still cover the entire range of f Let f s x have complex Fourier series c s ne inx Then c s π 4πl π π 4πl l Similarly for n, π l l dz c s n 4πl 4πl l l c n l x+l dx x l l dx π π π π π l l l l [ e inz in l π π f ydy f x + zdz f x + zdx l x+l e inx dx e inx dx e inz dz e inz c n dz ] l l π x l l l π π f ydy l l c dz c f x + zdz e inx+z f x + zdx c ne inl e inl lin 4 c n sin nl nl

15 If the sinc function is defined as { sin x sinc x x for x, for x then it can be seen that in general c s n c n sinc nl b The a n coefficients of the Fourier series can be expressed in terms of the complex Fourier coefficients as a n π cos nx f xdx π e inx + e inx f xdx c n + c n π π π π Similarly, the b n coefficients can be expressed as b n π π π sin nx f xdx π e inx e inx f xdx c n c n πi π i Hence, by converting to the complex Fourier series coefficients and back again, it can be seen that the Fourier coefficients of the smoothed series are and a s n c s n + c s n c n sinc nl + c n sinc nl c n + c n sinc nl a n sinc nl b s n cs n c s n i c n sinc nl c n sinc nl i c n c n sinc nl i b n sinc nl 3 a For the first function, u x 3x + y, v y x 3y and since u x v y the Cauchy Riemann equations are not satisfied and this function is not analytic For the second function u x 3x 3y, v y 3x 3y and the first Cauchy Riemann equation u x v y is satisfied In addition, u y 6yx, v x 6yx and thus the second Cauchy Riemann equation u y v x is satisfied also, and hence the function is analytic It can be verified that this function is equal to z 3 5

16 b The partial derivatives of the components of f are u x ex y x cos xy y sin xy, v y ex y y sin xy + x cos xy, u y ex y y cos xy x sin xy, v x ex y x sin xy + y cos xy and it can be seen that u x v y and u y v x, so the Cauchy Riemann equations are satisfied f is analytic To determine f as a function of z, in can be seen that 4 The total cost of the tunnel is a Fx, y, y dx a f x, y e x y e ixy e x +ixy y e x+iy e z a a + yds a a + y + y dx Since the integrand has no explicit x dependence, the Beltrami identity can be used, and C F y F y + y + y y + y + y + y + y for some constant C Hence and therefore Hence whereupon integration gives C + C y + y Cy C + y Cdy C + y dx, C C + y x x for some constant x This can be rearranged to give C + y x x In general, if there was no straightforward way to convert x and y into z, then the direct substitutions x z + z/, y z z/i could be employed If f is analytic, the terms involving z should cancel 6 4C

17 y C + x x 4C Now consider the boundary conditions, y±a By symmetry, x To find C, note that and therefore ya C + a 4C 4C4 4C + a 4C C a 4C C ± a, giving two valid solutions for each value of a The corresponding functional form is yx C + x 4C x a 4C x a ± a 5 The two triangles together make up the square, x 3 and y 3 First consider any extrema in the interior of the square; they must satisfy f x x y 3, f y x + 4y From the first equation y x 3 Substituting into the second equation gives x + 4x 3 7x 4 and therefore x, from which it follows that y At this point, f x, y The boundaries could also contain possible extrema: f, y y y, f y, y 4y, y /; f x, x 3x, f x x, x 3, x 3/; f 3, y 5y + y, f y 3, y 5 + 4y, y 5/4; f x, 3 x 6x +, f x x, 3 x 6, x 3 The values of the function at these locations are f, / /, f 3/, 9/4, f 3, 5/4 5/8, f 3, 3 3 Note that one of these points coincides with a corner The central line x y could also hold minima and maxima for either T or T : d d f x, x x 5x 4x 5, x 5/4 dx dx 7

18 3 5 y x Figure 6: Contour plot of the function f x, y x xy + y 3x y, showing the region T in red and the region T in blue For each region, the minimum is shown by a circle and the maximum is shown by a square The contours are at intervals of / In addition to this point, the three remaining corners could also be extrema, and function values at these points are f 5/4, 5/4 5/8, f,, f 3,, f, 3 A contour plot of f is shown in Fig 6 By looking at the possibilities, it can be seen that the minimum and maximum of f in T are at f, 4 and f 3, 3 3 respectively The minimum and maximum of f in T are at f 5/4, 5/4 5/8 and f, 3 respectively 6 The integral C z + n dz z z has a pole of order n + at z To evaluate the integral, the residue of the integral at z must be found By using the binomial theorem, z + n z z z n k n n k z k k z 8 n k n z k n k

19 The constant term corresponding to /z in the Laurent expansion will occur when k n, which will be when k n The coefficient of this term is n n! n n!n n! n! n! C z + n dz z z πi n! n! By using cos t eit + e it and the substitution z e it, so that dz ie it dt, the integral can evaluated as π π cos n t dt π n π π n i πi e it + e it n dt z + n dz π z iz n! n! πn! n n! Solutions to the additional practice questions 7 The area and perimeter of the pentagon are given by A wh + wl, P h + w + w + l respectively To maximize the area subject to a fixed perimeter, the augmented function Aw, h, l, λ wh + wl + λp h w w + l can be considered where λ is a Lagrange multiplier An extremal point therefore satisfies A w A h A l h + l λ wλ w + l 7 w λ 8 w lλ w + l 9 From Eq 8 it follows that λ w Equation 6 then becomes w lw w + l 9

20 w + l l Therefore w 3l and w l 3 Substituting into Eq 7 then gives h + l l 3 6l 4l h 3 + l Substituting into the constraint equation gives It follows that and P 3 + l + 3l + 3l + l l w The corresponding area is l P P 6 h P P P 6 3 3P 6 3P A wh + l P 3P 4 The area evaluates to A 66987P Note that the area enclosed by a square of perimeter P is P A s P P As would be expected, due to the extra degree of freedom provided by l, which creates a slightly rounder shape, the pentagon encloses slightly more area 8 Plots of f and g are shown Fig 7a and Fig 7b respectively The convolution is given by f gx f x ygydy and since gy for y < and zero otherwise, this can be written as f gx f x ydy

21 The function f will be non-zero for the range < x y < which is equivalent to x < y < x + If x >, then x > and the integration will evaluate to zero If x, then the integration will be non-zero over the range from x to, and thus f gx 3 x y x dy [3y y x 3] x 3 x3 3x + x x x 3x + x 3 3x x3 3x x + x 3 Note that both f and g are even If x <, then x z for z >, and f gx f z ygydy f z + qg qdq f z qgqdq f gz f g x f g is also even Therefore x+x 3 for x <, f gx xx+ 3 for x <, otherwise Since x + x will equal zero when x, the convolution is continuous there; by symmetry this will also be true at x The function f g is plotted in Fig 7c 9 For the alternative form of g, where gx x for < x <, the convolution will

22 f x gx y a b c - - x - - x Components y f gx - - x Figure 7: Two functions a f and b g, and c their convolution f g

23 f x gx y a b c - - x - - x Components y f gx - - x Figure 8: Two functions a f and b g, and c their convolution f g 3

24 be described by Eq but with an additional factor of y: f gx x x y y x dy y y x 3 xy x dy [6y 3y x 4 4xy x 3] x 6 3 x4 4x x 3 6x + 3 4x x 8x + x 3 3x 4 4x +x x 3 + 4x 4 6x + x x x4 x + 6x xx + 4x For this case, f is even and g is odd If x <, then x z for z >, and f gx f z ygydy f z + qg qdq f z qgqdq f gz f g x and thus f g is odd Plots of f, g, and f g are shown in Fig 8 It is worth noting that the integral in Eq can be approximated as a sum, f gx f x ydy N N f x n N n N Each of the terms f x n N n N N corresponds to a copy of f translated by N Hence the convolution can be viewed as a sum over a number of translated components These are plotted in Fig 7c and it can be seen the convolution is approximately an average of these components For the alternative form of g, the convolution can be approximated as f gx y f x ydy N N f x n N n N n N N This case can be viewed similarly, although here the components are now weighted by the x factor These components are plotted in Fig 8c 4

25 Differentiating both sides of the geometric series formula gives na n d n da a a and multiplying both sides by a gives Hence na n n λe ix a a f x λe ix nλ n e inx n This is an expression for f as a complex Fourier series with coefficients { nλ n for n >, c n for n The smoothed Fourier series is therefore given by f s x n il il il il nλ n sin nl e inx nl n n λ n e inl e inl e inx λ n e inx+l e inx l λe ix+l λe ix l λe ix+l λe ix l λe ix+l e ix l il λe ix+l λe ix l λe ix sin l l λe ix cos l + λ e ix It can be seen that if l, then sin l l to f as expected and cos l, and the function will tend a The radial distance can be calculated as r a b + a b 4a b + a 4 a b + b 4 4 a + b a + b 5

26 b First note that ẋ ȧb + aḃ and ẏ ȧa ḃb, L mẋ + ẏ mȧ b + a ḃ + ȧ a + ḃ b mȧ + ḃ a + b m mȧb + aḃ + ȧa ḃb + m r a + b + m a + b + m a + b c The Euler Lagrange equation for bt can be calculated as d L L dt ḃ b d ma + b ḃ mȧ + dt ḃ b + 4mb a + b aȧ + bḃḃ + a + b b ȧ + ḃ b + which can be simplified to give ba + b + ḃaȧ + bḃ ȧ b + 4b a + b, 4b a + b Since the Lagrangian is invariant if a and b are switched, it follows that the Euler Lagrange equation for at is äa + b + ȧbḃ + aȧ ḃ a + If at C where C is a constant, then Eq 3 becomes ḃ 4a a + b 3 4 C + b 4 ḃ C + b, 5 where the positive sign for the square root has been chosen Differentiating gives b 4ḃb C + b 8b C + b 3 6 For at C to lead to a valid solution, Eq must also be satisfied It can be seen that the equation becomes bc + b 4b + bḃ C + b 7 6

27 and using Eqs 4 and 6 it can be seen that the left hand side is bc + b 8b + bḃ C + b 3 C + b 4b + C + b 8b C + b + 4b C + b 4b C + b Hence Eq 7 is satisfied and at C is a consistent solution To determine the evolution of b, note that Eq 5 can be rearranged to give C + b db dt C b + b3 3 t t for some constant t Using the given boundary condition of b, it follows that t This expression cannot easily be written as a function bt, since it would involve solving a cubic for b, but it is possible to write tb b3c + b 6 When b is large it can be seen that bt 3 6t Note that the trajectory at, bt is a parabola, which is physically reasonable for a mass moving in a gravitational field many comets follow near-parabolic trajectories as they orbit the Sun A Appendix: Derivation of the Green function condition In question, when solving the equation a solution of the form yx y + k y δx x 8 { A cos kx + B sin kx for x < x, C cos kx + D sin kx for x > x is constructed, and after the boundary conditions y y π k are taken into account, this becomes { B sin kx for x < x yx, C cos kx for x > x 7

28 To set the two remaining constants B and C, conditions to match the two sections of the solution at x x are needed Let the solutions for x < x and x > x be called y x and y + x respectively The solution y is taken to be continuous at x x and thus y x y + x, which gives B sin kx C cos kx 9 To get another condition, the derivatives of y and y + need to be considered Physically, if yx is viewed as the position of a mass as a function of time, the delta function term in Eq 8 corresponds to applying an impulsive force to the mass at the time x x In the same way that a ball hitting a wall experiences a short impulsive force over a small interval, which causes a rapid change in its velocity, it should be expected that δx x will cause a change in the derivative of y at x x To analyze this mathematically, integration must be used The total impulse of a delta function is given by δx and since this is localized at x, restricting to a small interval from ɛ to ɛ gives ɛ ɛ δx Now consider integrating Eq 8 over the interval from x ɛ to x + ɛ This gives x +ɛ x ɛ y + k ydx x +ɛ x ɛ δx x dx The integral of y will be y, and the integral of the delta function is a shifted version of Eq, and therefore [ y ] x +ɛ x x ɛ dx + +ɛ k y dx x ɛ As ɛ, the integral of k y will tend to zero, since y is continuous and the integration range gets smaller Since y x + ɛ y +x and y x ɛ y x, it follows that and therefore y +x y x Ck sin kx Bk cos kx Equations 9 and can then be solved to find B and C, from which a complete solution to Eq 8 can be constructed The condition in Eq is generally applicable For any differential equation like Eq 8 with the form y + f xy + gxy δx x for some functions f x and gx, the same derivation could be followed to obtain Eq 8