Løsningsførslag i 4M

Transcription

1 Norges tekisk aturviteskapelige uiversitet Istitutt for matematiske fag Side 1 av 6 Løsigsførslag i 4M Oppgave 1 a) A sketch of the graph of the give f o the iterval [ 3, 3) is as follows: The Fourier coefficiet A = 1 x dx = /4. The Fourier coefficiets A = 1 x cos(x) dx = 1 [ x si(x) 1 = cos(x) = ( 1) 1, si(x) dx

2 Side av 6 for = 1,, 3,... The Fourier coefficiets B = 1 x si(x) dx = 1 [ x cos(x) + 1 = ( 1)+1 + si(x) = ( 1)+1, for = 1,, 3,... Therefore, the Fourier series of f is 4 + [ ( 1) 1 =1 cos(x) + ( 1)+1 cos(x) dx si(x). (1) b) We otice that the series give ca be obtaied after some algebra if we take x = i (1). However, f has a jump-discotiuity at x =. Thus, f(+) + f( ) = 4 + =1 ( 1) 1 cos(). Observe that f( ) = ad f(+) =. Rearragig terms i the above equatio, we get 1 (k + 1) = 8. k= Oppgave f(w) = 1 [ Sice x = x whe x < ad x = x otherwise, we have xe (a iw)x dx + Usig itegratio by parts, we get: I 1 = = xe(a iw)x = lim t xe (a+iw)x dx 1 (I 1 + I ). xe(a iw)x a iw 1 e (a iw)x dx x= a iw a iw e(a iw)x x= (a iw) x= [ ( t)e at e iwt 1 a iw (a iw) + e at e iwt (a iw) Sice it is give that a >, ad e iwt is bouded for all t, the third term above coverges to. Furthermore, we kow that t p e at, for ay power p, as t. Hece, the first term above coverges to. Thus I 1 = 1/(a iw)..

3 Side 3 av 6 I a similar way, the calculatio of I proceeds as follows: I = xe (a+iw)x a + iw + 1 e (a+iw)x dx a + iw = xe (a+iw)x a + iw e (a+iw)x (a + iw) = lim [ te at e iwt t a + iw + 1 (a + iw) e at e iwt. (a + iw) By exactly the same argumets as above to justify the limits, we get I = 1/(a+iw). Combiig the values of I 1 ad I, we get f(w) = 1 1 [ (a iw) + 1 (a + iw) 4iaw = (a + w ). () To evaluate the give itegral, we use the Fourier-iversio formula. By (), have: f(x) = 1 = 1 4iaw (a + w ) eiwx dw 4iaw(cos(wx) + i si(wx)) dw. (a + w ) Sice f(x) is real-valued, the imagiary part o the right-had side of the above equatio must equal zero. Hece f(x) = 1 4aw si(wx) dw x R. (a + w ) The above holds true for each x R because f is a cotiuous fuctio ad, as a >, satisfies all the coditios of the theorem that gives us the Fourier-iversio formula. Takig x = 1 ad a = 1 i the last equatios gives us the aswer: w(si(w) (1 + w ) dw = e x x=1 = e 1. Oppgave 3 a) Assumig u to be of the form u(x, t) = F (x)g(t), we get u t = F Ġ, u xx = F G. Substitutig ito the give PDE, we have F Ġ = (F 5F )G. Sice we seek o-trivial solutios, F G caot be idetically zero. Therefore, we divide the last equatio by F G to get Ġ G (t) + 5 = F (x) F (x) t > ad x (, ).

4 Side 4 av 6 This ca oly be possible if both sides of the above equatio are costat. This gives us the followig ODEs for F ad G: where k is a yet-udetermied costat. As u, the boudary coditios imply that Ġ + (5 k)g =, (3) F kf =, (4) F () = = F (). The equatio (4) has three differet types of solutios depedig o the sig of k Case 1. Whe k >. I this situatio, F (x) = C 1 e kx + C e kx. If F has to satisfy the boudary coditios at x =,, the a routie argumet shows that C 1 = C =. Thus, the case k > does ot yield ay o-trivial solutios. Case. Whe k =. I this situatio, F (x) = C 1 x + C, whece F (x) = C 1. The boudary coditios imply that C 1 =, but the costat C ca be o-zero. Thus, we ow eed to cosider the equatio that determies G. As k =, (3) implies that G(t) = Ae 5t, where A is some costat. Thus u (x, t) = A e 5t (5) is a solutio of the PDE ( ) of the product form satisfyig the boudary coditios, where A is a udetermied costat. Case 3. Whe k <. I this situatio, it is otatioally simpler to write k = p. The, F (x) = C 1 cos(px) + C si(px). The boudary coditios give us the equatios pc =, pc 1 si(p) + pc cos(p) =. I the preset case, p. Hece, C =, ad we are faced with the coditio si(p) =. This implies p =, =, ±1, ±,... As p, ad as cos( px) = cos(px), it suffices to oly cosider p = 1,, 3,... Correspodig to each of these values of p, k =, ad solvig (3) yields G(t) = Ae ( +5)t. Thus, correspodig to each, we have the solutio u (x, t) = A cos(x)e (5+ )t, = 1,, 3,..., (6) where A is a udetermied costat. All possible solutios of ( ) of the product form that satisfy the give boudary coditios are give by (5) ad (6).

5 Side 5 av 6 b) For the give problem, we ca use superpositio to look for a solutio of the form u(x, t) = A e 5t + A cos(x)e (5+ )t. Imposig the iitial coditio gives us A + =1 ( x ) A cos(x) = cos cos(5x). (7) =1 We use a half-rage expasio, with period, for the fuctio o the righthad side, ad the above equatio suggests a Fourier cosie series. But, rather tha settig up the itegrals to determie the Fourier coefficiets, we recall the trigoometric idetity: cos ( x ) = 1 + cos(x). By (7) ad orthogoality, A = for all except =, 1 ad 5. I the latter case: A = 1/, A 1 = 1/ ad A 5 =. Hece is the desired solutio. u(x, t) = e 5t + cos(x)e 6t cos(5x)e 3t Oppgave 4 a) We use the followig approximatio of the Laplace operator: u(x, y) = 1 (u(x + h, y) + u(x h, y) + u(x, y + h) + u(x, y h) 4u(x, y)) h Combied with the cetral differece approximatio of u x give i the problem we obtai the discretized equatio 1 = 1 (u(x + h, y) + u(x h, y) + u(x, y + h) + u(x, y h) 4u(x, y)) h + 1 (u(x + h, y) u(x h, y)). h Thus, with the otatio u i,j = u(i, j) for i, j =, 1,, 3 this becomes after some rearragemets u i,j = 3 8 u i+1,j u i 1,j u i,j u i,j b) The ukows are the u 1,1, u,1, u,1 ad u, while the other u i,j s are give due to the prescribed boudary values. Oe iteratio with the Gauss-Seidel method with startig values 1 yields u 1,1 = 3 8 u, u, u 1, u 1, 1 4 = = 1,

6 Side 6 av 6 u,1 = 3 8 u 3, u 1, u, u, 1 4 = = 13 16, ad u 1, = 3 8 u, u, u 1, u 1,1 1 4 = = 3 8, u, = 3 8 u 3, u 1, u, u,1 1 4 = = 3 4. These are the seeked approximatios of u(1, 1), u(, 1), u(1, ) ad u(, ). Oppgave 5 a) With the otatio P 1 = (x 1 )(x 1)(x 3 )(x ), the iterpolatig polyomial is give by P = x(x 1)(x 3 )(x ), P 3 = x(x 1 )(x 3 )(x ), P 4 = x(x 1 )(x 1)(x ), P 5 = (x 1 )(x 1)(x 3 ), 11P 1 (x)/p 1 () 7P (x)/p ( 1 )+3P 3(x)/P 3 (1)+5P 4 (x)/p 4 ( 3 )+65P 5(x)/P 5 (). This simplifies to 8x 3 + 6x 11. b) Sice Simpso s method is exact for polyomials of degree three the error is zero. Oppgave 6 a) Heu s method for solvig a ODE. The output is a approximatio of y(b) with itial value y(a) = ya, where y is a solutio of y = e y 1 o (a, b). The output would be Q.53. Follows from k 1 = e 1 1, k = e e 1 1 ad Q = (k 1 + k )/. b) Sice h = 1/ ad the method is of secod order we have y y + C/. Hece we obtai y y = 1/3 (P Q) = 1/3 ( ).1.