Assignment 7 Due Tuessday, March 29, 2016

Transcription

1 Math 45 / AMCS 55 Dr. DeTurck Assignment 7 Due Tuessday, March 9, 6 Topics for this week Convergence of Fourier series; Lapace s equation and harmonic functions: basic properties, compuations on rectanges and cubes Fourier!), Poisson s formua for the disk Seventh Homework Assignment - due Tuesday, March 9 Reading: Read sections 5.4, 5.5, and 6. through 6.3 of the text Be prepared to discuss the foowing probems in cass: Page 34 page 9 in the first ed) probems, 3 Page 45 page 39 in the first ed) probems, 3 Page 6 page 54 in the first ed) probems, 5, 7 Page 64 page 58 in the first ed) probems, 6 *Page 7 page 63 in the first ed) probems, Page 34, probem. The Fourier sine series of fx) x on, ) is where x A n Integrate by parts with u x and dv sin nπx A n x nπx cos nπ cos nπx dx nπ Parseva s theorem says that x fx)) dx A n sin nπx x sin nπx dx dx, so du dx and v nπx cos nπ x nπx cos nπ ) n nπ A n sin nπx. sin nπx to get nπx n sin )n π nπ dx so we have x dx n π 3 n π

2 Mutipy both sides of the equation n π by π 3 and obtain π 6 n. Page 34, probem 3. We can obtain the Fourier cosine series for x by integrating the Fourier sine series of x, which we found in the preceding probem: x x t dt x ) n 4 n π cos nπx ) n 4 nπ sin nπt dt ) n 4 n π ) n 4 n π cos nπt We can evauate the constant sum on the right by integrating both sides from to a the cosine terms wi integrate to zero) and obtain so, dividing this by we can concude that This time, Parseva s theorem says x dx 3 3 x 3 ) n 4 n π ) n 4 3 n π, cos nπx x x 4 dx dx 6 4 n 4 π 4 cos nπx dx n 4 π 4 Now mutipy both sides of the equation n 4 π 4 by π4 8 5 and obtain π 4 4 π4 7 n 4 Rearrange this to get n 4 π4 4 π4 7 π4 9 Page 45, probem. This was probem 5b) on the first midterm. But et s try it using the hint. Let ϕt) f tg f tg, f tg f, f t f, g t g, g. We know that ϕt). And the minimum of ϕ happens where ϕ : ϕ t) f, g t g, g so ϕ for t f, g g, g.

3 3 And the minimum is ) f, g ϕ g, g f, g f, f g, g f, g ) f, g g, g g, g f, g f, g f, f g, g g, g f, f f, g g, g The minimum vaue of ϕ must be non-negative, so f, f f, g g, g in other words f, f g, g f, g. Taking the square root of both sides gives the Schwarz inequaity. Page 45, probem 3. A sight variation of this was aso on the midterm this is Poincaré s inequaity), but here s a proof of the version given: Use the square of the) Schwarz inequaity with f and g f, which says that, f ) f Now and So now Schwarz says ), f ) f x) dx f) f)) f dx f) f)) f x)) dx which is the version of Poincaré that we were to prove. f x)) dx f x)) dx Page 6, probem. If u is a soution of u k u that depends ony on r, then as we have shown previousy, or ese from the text) ur) satisfies Foowing the hint, et u v/r. Then The equation becomes: u r u k u. u v r v r and u v r v r v r 3. v r v r v r 3 v r r v ) r v r v k r

4 4 or in other words v k v. Since we re given that k > in particuar, k ), the soutions of this equation are v c e kr c e kr, which in turn yied ur) c e kr r c e kr r. Page 6, probem 5. To sove u on r < a in R with u when r a, it s enough to ook for radia soutions since we wi have shown that the soution of this Dirichet probem is unique). So we seek a soution ur) to the ordinary differentia equation: u r u that defines a smooth function at the origin and for which ua). The genera soution of u r u is u c c n r and a particuar soution of the inhomogeneous equation is so the genera soution is u p x 4 u g x 4 c c n r. We don t want a n r term since that wi become infinite as r, so we set c. In order for ua), we shoud set c 4 a, and so the soution of the probem is ur) 4 r a ) 4 x y a ). Page 6, probem 7. The genera soution of is u u rr r u r u c c r r 6. Now we need to arrange for u when r a and r b, in other words to sove the inear system a c 6 b a 6 b for c and c. The soution is c b a b a c c 6 a 6 b u aba b) b ab a ) r ) 6 r ab 6 b a a b ) a b 6 a b )

5 5 Page 64, probem. Since this is a pure Neumann probem, it is important to check that the integra of the norma derivative around the boundary of the region is zero, and it is, since the vaue of the outward-pointing norma is a on a segment of ength b where x ) and is equa to b on a segment of ength a where y ) and zero on the other two segments. So there is a soution to this probem, and it is unique ony up to adding a constant. We coud try writing the soution as the sum of two Fourier cosine series: ux, y) C A n cosha x) cos nπy b B n coshb y) cos nπx a but it is simper to note that since the boundary conditions are constants, we can seek the soution of the probem as ux, y) F x) Gy), where F x) is a quadratic poynomia with F ) a and F a), and where Gy) is aso a quadratic poynomia with G ) b and Gb). From the conditions on F, we have that F is inear in x with sope, so F x) x a, and ikewise G y) b y. So F x) x ax and Gx) by y up to an additive constant), so ux, y) F x) Gy) C x y ax by C. It s easy to check that this is harmonic and satisfies a the boundary conditions. Page 64, probem 6. We have to sove the Neumann probem on the cube, with u z x, y, ) gx, y) and the norma derivatives zero on the other five sides. So in our separated soutions Xx)Y y)zz) we wi have X ) X ), Y ) Y ) and Z ). So our separated soutions are XY Z cos mπx cos nπy cosh m n πz and the soution of the probem is Now ux, y, z) u z x, y, ) n m n m So, except when m n i.e., the constant term) m n π sinh m n πa mn A mn cos mπx cos nπy cosh m n πz. m n π sinh m n π cos mπx cos nπy gx, y), cos mπx cos nπy cos mπx cos nπy, cos mπx cos nπy gx, y) cos mπx cos nπy dx dy cos mπx cos nπy dx dy 4 gx, y) cos mπx cos nπy dx dy

6 6 and A is undetermined and we expect the soution of the Neumann to be unique ony up to the addition of an arbitrary constant). So the soution is ux, y) A n n n,m),) 4 gx, y) cos mπx cos nπy dx dy m n π sinh m n π cos mπx cos nπy cosh m n πz Page 7, probem. a) Since u 3 sin θ on the boundary of the disk, and the maximum of this function of θ is 4, we have that 4 is the maximum vaue of u throughout the disk by the maximum principe. b) The vaue of u at the origin is the average of u on the boundary of the disk, namey. Page 7, probem. We need the harmonic function u A r n A n cos nθ B n sin nθ) to equa 3 sin θ when r a. This wi be so if a the A n except n, for which A and a the B n except for n, for which B 3 a. So u 3 r sin θ. a Write up soutions of the foowing to hand in: Page 34 page 9 in the first ed) probems 4, 5 Page 45 page 39 in the first ed) probems 4, Page 6 page 54 in the first ed) probems 4, 6, 8 Page 64 page 58 in the first ed) probems 4, 7 *Page 7 page 63 in the first ed) probem 3 Page 34, probem 4. To find the sum of the series n 6 we find the Fourier series for x3, using the work from probems and 3 above. From probem 3, we know that the cosine series for x is x 3 ) n 4 n π cos nπx

7 7 so we integrate 3 times this to get x x x 3 3t dt ) n n π x ) n 3 n 3 π 3 sin nπx cos nπt dt x And this wi do the series on the right is the series for x 3 x: x 3 x and we can appy Parseva s theorem to this: ) n 3 x 3 x) dx n 3 π n 6 π 6 sin nπx sin nπx ) n 3 dx n 3 π 3 sin nπt Since x 3 x) x 6 x 4 4 x, the integra on the eft gives ) And as usua, a the integras on the right evauate to, so we obtain: n 6 π 6 x Mutipy both sides by π6 7 7 and obtain π n 6. Page 34, probem 5. a) To write we have B n, cosn )x cosn )x, cosn )x B n cosn )x n π cosn )x dx cos n )x dx n sinn )x π π )n 4 n )π b) The series wi converge for a x to the extension of the that is even across x, odd across xπ and π-periodic. So, on the interva π < x < π, the series converges to for π < x < π for x for < x < π for x π for π < x < π

8 8 c) The series is n for < x < π. And the Parseva s theorem says: or: dx Mutipy both sides by 8π/ to get n ) n 4 n )π cosn )x 6 n ) π π n n 8 n ) π n ) π 8 cos n )x dx Page 45, probem 4. Preiminaries: The boundary conditions for Xx) in the separated soutions are X ) X) X) and X ) X) X) If f and g are two functions that satisfy these boundary conditions, then f x)gx) fx)g x) f )g) f)g ) f )g) f)g ) ) f) f))g) f)g) g)) f) f))g) f)g) g)) ) f)g) f)g) f)g) f)g) f)g) f)g) f)g) f)g) and so by Theorems and of section 5.3, a the eigenvaues are rea and eigenfunctions corresponding to distinct eigenvaues are orthogona. We are going to use Poincaré s inequaity in part c) to show that there are no negative eigenvaues, but here is a more direct proof: Suppose λ β, then X c cosh βx c sinh βx and X βc sinh βx βc cosh βx. The boundary condition X ) X) X) says βc c c cosh β c sinh β and the boundary condition X ) X) X) says βc sinh β βc cosh β c c cosh β c sinh β We can simpify things a bit by subtracting the first equation from the second, dividing the resut by β, and repacing the second equation with c sinh β c cosh β c

9 9 Then the first boundary condition and this ast equation comprise the foowing system of two inear equations in the two unknowns c and c : cosh β β sinh β sinh β cosh β c c If β is to be an eigenvaue, the determinant of the matrix on the eft must be zero. The determinant is cosh β cosh β sinh β β sinh β cosh β β sinh β. We need to show there are no nonzero vaues of β for which this quantity is zero. So we study the function fz) cosh z z sinh z We have f) and f is an even function. We re going to write the Macaurin series for f and see that a the coefficients of even powers of z are negative, and a the coefficients of odd powers of z are zero, which wi show that fz) < for z. The series for cosh z and sinh z are. cosh z n z n n)! and sinh z x n n )! just ike the series for cosine and sine, but without the aternating signs). So the series for cosh z z n n)! since subtracting cances the n term, and the series for z sinh z z n n )! cosh z z sinh z ) n z n n )! n n n z n n )! which proves the caim that there are no vaues of z other than z for which this function is zero. Hence there are no negative eigenvaues. Now we can begin the probem. a) Zero is a doube eigenvaue of the probem, since any inear function Xx) A Bx satisfies the equation X together with the boundary conditions, which as they are printed in the book) say essentiay that the sope equas the sope equas the sope. Now we can concentrate on the positive eigenvaues and their eigenfunctions. The beginning of the anaysis is not so different from the negative case above, except we have trigonometric rather than hyperboic functions. So, suppose λ β, then X c cos βx c sin βx and X βc sin βx βc cos βx. The boundary condition X ) X) X) says βc c c cos β c sin β

10 and the boundary condition X ) X) X) says βc sin β βc cos β c c cos β c sin β We can simpify things a bit by subtracting the first equation from the second, dividing the resut by β, and repacing the second equation with c sin β c cos β c Then the first boundary condition and this ast equation comprise the foowing system of two inear equations in the two unknowns c and c : cos β β sin β c. sin β cos β c If β is to be an eigenvaue, the determinant of the matrix on the eft must be zero. The determinant is cos β cos β sin β β sin β cos β β sin β. Using the doube-ange formuas for sine and cosine, we can rewrite this as 4 sin β 4 β) sin β cos β. So the vaues of β that give eigenvaues are the roots of sin β β cos β sin β ) If the first factor is zero, i.e., sin πn β, then β and the system of equations for c and c becomes: πn c c so that c and the corresponding eigenfunction is X c cos πnx for n,, 3,...). If the second factor is zero, then β must be a root of the equation x tan x. By graphing y x and y tan x on the same graph and observing where the graphs cross, you can see that the vaues of β that yied eigenvaues are a near and a bit ess than and getting coser and coser to) n )π.

11 In this case we start by rewriting the system of equations for c and c in terms of the ange β and the doube-ange formuas, since we know that sin β β cos β: sin β ) β sin β cos β sin β cos β sin β c c The second equation after dividing by sin β) becomes cos β)c sin β)c Now repace sin β with β cos β and divide out the cosine factor to get c βc. So the eigenfunction in this case wi be a mutipe of) or, using that β λ, Xx) β cos βx sin βx Xx) λ cos λ x sin λ x. To summarize, we have three sets of eigenvaues: λ, with eigenfunction A Bx λ 4n π with eigenfunction cos nπx, for n,, 3,... λ n β n where β n are the positive roots of x tan x there are infinitey many of these) with eigenfunctions X n x) λ n cos λ n x sin λ n x. So the soution of the heat equation is ux, t) A Bx P n e 4kn π t/ cos nπx Q n e kλnt λn cos λ n x sin ) λ n x where the P n and Q n are the Fourier coefficients of the initia data ϕx). b) In the soution ux, t) given above, a the terms of the series have factors that are exponentias of negative numbers times t. So as t, a these terms go to zero rapidy!) and we are eft with im ux, t) A Bx. t c) Using integration by parts Green s first identity): Suppose Xx) is an eigenfunction with

12 eigenvaue λ; in the integration by parts, et u X and dv X dx so that du X dx and v X : λ X, X λx, X X, X X x)xx) dx Xx)X x) X) X) X) ) X) X x)) dx X) X)) X) x) X x)) dx ) X x)) dx by the resut of probem 3. But since λ X, X and of course X, X >, we must have λ. So there are no negative eigenvaues. d) To find A and B, we must be carefu to use orthogona eigenfunctions, and the functions and x are not orthogona on the interva x. However, the functions and x are orthogona:, x x dx x x A Bx ϕ, ϕ, x, x x, x ) Now,, and x, x x ) dx x 3 3 ) 3 ) 3 3, the components of ϕ in the directions of and of x are ϕ,, and ϕ, x x, x and we have, dx and x, x x ) dx x ) )3 3 4

13 3 ϕ, ϕ, x Ax B, x ϕx) dx 3 ϕx) dx, x 3 4 6x ) ϕx) dx x ) x ) ) ϕx) dx x x ) ) ϕx) dx ) x 3 6 ) ) ϕx) dx 3 x x ) ) ϕx) dx x So A 4 6x ) ϕx) dx and B x 3 6 ) ϕx) dx Page 45, probem. Since we know we can integrate a Fourier series term by term but not necessariy differentiate one, start with the series for f x): where A π f x) A f x) dx and A n π A n cos nx B n sin nx f x) cos nx dx, B n π Because fx) satisfies periodic boundary conditions, we have fπ) f) So A. Next, the Fourier series for fx) is where C π fx) C fx) dx and C n π We are given in the probem that and C n π D n π f x) dx πa C n cos nx D n sin nx fx) cos nx dx, D n π fx) cos nx dx fx) sin nx nπ f x) sin nx dx for n,,... fx) sin nx dx for n,,... fx) dx, so C. Aso, on integration by parts, π fx) sin nx dx fx) cos nx nπ π f x) sin nx dx B n nπ n f x) cos nx dx A n nπ n

14 4 So Parseva s theorem tes us that using that the integras of cos nx and sin nx from to π are equa to π for a n) ) A f x)) dx A n Bn)π n n B n n π Cn Dn)π fx)) dx. And from this we can aso see that equaity hods if and ony if A n B n C n D n for a n. Page 6, probem 4. We know from the very first assignment perhaps) that for functions u that depend ony on r that u u rr r u r The differentia equation r u ru is a Cauchy-Euer equation with genera soution u c c r. If ua) A and ub) B, then we have two equations in two unknowns: a c A b B the soution of which is c b a A c b a B so u aa bb a b c ab b A a B a b B A abb A) a b)r aa bb a b abb A) Page 6, probem 6. The genera soution of is u u rr r u r u c c n r r 4. Now we need to arrange for u when r a and r b, in other words to sove the inear system n a c 4 a n b 4 b c for c and c. The soution is c n b n a n b n a c 4 a 4 b 4 b n a a n b) n b n a 4 a b ) u ) b n a a n b a b ) n r r 4n b n a) 4

15 5 Page 6, probem 8. The genera soution of is u u rr r u r u c c r r 6. Now we need to arrange for u when r a and for u r when r b, in other words, to sove the inear system a c 6 b a 3 b c for c and c. The soution is c b b a c 6 a 3 b b 6 a b a b) 3 b u ) b 3 6 r a b3 a r As a, the hoe in the midde of the sphere coses up, and the probem woud seem to approach a Neumann probem on the soid ba. Unfortunatey, though, the necessary condition for a soution wi not be satisfied, since the norma derivative on the surface of the ba is zero, but u in the interior and the integras of these can t agree. That is why there is a singuarity forming with the a in the denominator of the constant term. The soution wi approach near the r b boundary. Page 64, probem 4. To find the harmonic function ux, y) with the given boundary conditions, we wi need to add together harmonic functions v and w that have inhomogeneous conditions on ony one side. So et v satisfy vx, ) x vx, ) v x, y) v x, y) and et w satisfy wx, ) wx, ) w x, y) w x, y) y. Since v x is zero for x and x, the series for v wi have cosines of x, and since v is zero when y but not when y, we have hyperboic sines of y, except when λ, where we have y) as foows: vx, y) a y) a n sinh nπ y) cos nπx So for y, we want, we need vx, ) a a n sinh nπ cos nπx x a x dx

16 6 and a n x cos nπx dx sinh nπ even n 4 n π odd n sinh nπ vx, y) y) k x sin nπx sinh nπ nπ ) cos nπx n π 4 sinhk )π y) k ) π cosk )πx sinhk )π Now w is zero for y and y so the series for w wi have sines of y, and since w x is zero when x but not when x, we have hyperboic cosines of x as foows: For x, we want so we need b n wx, y) So finay, nπ sinh nπ wx, y) w x, y) b n cosh nπx sin nπy nπb n sinh nπ sin nπy y y sin nπy dy y cos nπy nπ sinh nπ nπ ) ) n )n ) nπ sinh nπ nπ n 3 π 3 k ux, y) vx, y) wx, y) y sin nπy n π n π sinh nπ n π sinh nπ 8 n 4 π 4 sinh nπ k ) π sinhk )π 8 k ) 4 π 4 sinhk )π y) k k k k π cosh kπx sin kπy sinh kπ ) cos nπy n 3 π 3 even n odd n coshk )πx sink )πy 4 sinhk )π y) k ) π sinhk )π cosk )πx k π cosh kπx sin kπy sinh kπ k k ) π sinhk )π 8 k ) 4 π 4 sinhk )π coshk )πx sink )πy

17 7 Page 64, probem 7. Since u, y) uπ, y), we have X) Xπ) in the separated soutions, and so Xx) sin nx with eigenvaue n ). So Y y) ae ny be ny, but because we want the soution to decay to zero as y, we must have a. So the soution is ux, y) B n e ny sin nx where B n π hx) sin nx dx. b) If we didn t have the condition at infinity, then we woud have ux, y) A n e ny B n e ny ) sin nx and there woud be no way to determine the coefficients, since we woud know ony that A n B n π hx) sin nx dx. Page 7, probem 3. Let s start by proving the trig identity for sin 3θ: So we have We need the harmonic function sin 3θ sinθ θ) sin θ cos θ cos θ sin θ sin θcos θ sin θ) cos θsin θ cos θ) 3 sin θ cos θ sin 3 θ 3 sin θ sin θ) sin 3 θ 3 sin θ 4 sin 3 θ. u A sin 3 θ 3 4 sin θ 4 sin 3θ r n A n cos nθ B n sin nθ) to equa 3 4 sin θ 4 sin 3θ when r a. This wi be so if a the A n and a the B n except for n and n 3, for which So B 3 4a and B 3 4a 3. u 3 4a r sin θ 4a 3 r3 sin 3θ.