Assignment Number 3 Solutions

Transcription

1 Math 4354, Assigmet Number 3 Solutios 1. u t (x, t) = u xx (x, t), < x <, t > (1) u(, t) =, u(, t) = u(x, ) = x ( 1) +1 u(x, t) = e t si(x). () =1 Solutio: Look for simple solutios i the form u(x, t) = X(x)T (t). We obtai X λx =, The solutio of the secod equatio is T λt =. T (t) = Ce kλt (3) where C is a arbitrary costat. Furthermore, the boudary coditios give X()T (t) =, X()T (t) = for all t. Sice T (t) is ot idetically zero we obtai the desired eigevalue problem X (x) λx(x) =, X() =, X() =. (4) (a) If λ = the X(x) = ax + b so applyig the boudary coditios we get Zero is ot a eigevalue. (b) If λ = µ > the = X() = b, = X() = a a = b =. X(x) = a cosh(µx) + b sih(µx). Applyig the boudary coditios we have = X() = a a = = X() = b sih(µ) b =. Therefore, there are o positive eigevalues. Cosider the followig alterative argumet: If X (x) = λx(x) the multiplyig by X we have X(x)X (x) = λx(x). Itegrate this expressio from x = to x =. We have λ X(x) dx = X(x)X (x) dx = X (x) dx + X(x)X (x). 1

2 Sice X() = X() = we coclude λ = X (x) dx X(x) dx ad we see that λ must be less tha or equal to zero. (c) So, fially, cosider λ = µ so that X(x) = a cos(µx) + b si(µx). Applyig the boudary coditios we have = X() = a a = = X() = b si(µ). From this we coclude si(µ) = which implies ad therefore µ = λ = µ =, X (x) = si(x), = 1,,.. (5) From (14) we also have the associated fuctios T (t) = e t. The oly problem remaiig is to somehow pick the costats a so that the iitial coditio u(x, ) = ϕ(x) is satisfied. To do this we cosider what we leared from Fourier series. I particular we look for u as a ifiite sum u(x, t) = b e t si (x) ad we try to fid {b } satisfyig =1 ϕ(x) = u(x, ) = b si (x). But this othig more tha a Sie expasio of the fuctio ϕ o the iterval (, ). b = = = = [ x = ( 1)+1 ϕ(x) si (x) dx =1 x si (x) dx/ ( ) cos (x) x dx ( ) cos (x) u(x, t) = ( ) cos (x) x dx] ( 1) +1 e t si(x). (6) =1

3 . u t (x, t) = u xx (x, t), < x <, t > (7) u x (, t) =, u x (, t) = u(x, ) = x Solutio: Look for simple solutios i the form We obtai u(x, t) = X(x)T (t). X λx =, The solutio of the secod equatio is T λt =. T (t) = Ce λt (8) where C is a arbitrary costat. Furthermore, the boudary coditios give X ()T (t) =, X ()T (t) = for all t. Sice T (t) is ot idetically zero we obtai the desired eigevalue problem There are, i geeral, three cases: X (x) λx(x) =, X () =, X () =. (9) (a) If λ = the X(x) = ax + b so applyig the boudary coditios we get = X () = a, = X () = a a =. Notice that b is still a arbitrary costat. We coclude that λ = is a eigevalue with eigefuctio ϕ (x) = 1. (b) If λ = µ > the ad X(x) = a cosh(µx) + b sih(µx) X (x) = aµ sih(µx) + bµ cosh(µx). Applyig the boudary coditios we have = X () = bµ b = = X () = aµ sih(µ) a =. Therefore, there are o positive eigevalues. Cosider the followig alterative argumet: If X (x) = λx(x) the multiplyig by X we have X(x)X (x) = λx(x). Itegrate this expressio from x = to x =. We have λ X(x) dx = X(x)X (x) dx = X (x) dx + X(x)X (x). Sice X () = X () = we coclude λ = X (x) dx X(x) dx ad we see that λ must be less tha or equal to zero ( zero oly if X = ). 3

4 (c) So, fially, cosider λ = µ so that X(x) = a cos(µx) + b si(µx) ad X (x) = aµ si(µx) + bµ cos(µx). Applyig the boudary coditios we have = X () = bµ b = = X () = aµ si(µ). From this we coclude si(µ) = which implies µ = ad therefore λ = µ = (), X (x) = cos(x), = 1,,.. (1) From (8) we also have the associated fuctios T (t) = e t. Our work o Fourier series showed us that a = 1 ϕ(x) dx, a = ϕ(x) cos (x) dx. (11) As a explicit example for the iitial coditio cosider = 1 ad ϕ(x) = x(1 x). I this case (11) becomes We have ad a = 1 a = ϕ(x) dx, a = a = 1 ϕ(x) dx = 1 = 1 [ ] x =. ϕ(x) cos (x) dx = ϕ(x) cos (x) dx. x dx x cos (x) dx = = [ si (x) x = ( ) si (x) x dx ( ) cos (x) = (( 1) 1) = si (x) ] dx 4, odd., eve 4

5 I order to elimiate the odd terms i the expasio we itroduce a ew idex, k by = k where k = 1,,. So fially we arrive at the solutio u(x, t) = 4 k=1 1 (k 1) e (k 1)t cos((k 1)x). (1) 3. u t (x, t) = u xx (x, t), < x <, t > (13) u(, t) =, u x (, t) = u(x, ) = x Solutio: Look for simple solutios i the form u(x, t) = X(x)T (t). We obtai X λx =, T λt =. The solutio of the secod equatio is T (t) = Ce λt (14) where C is a arbitrary costat. Furthermore, the boudary coditios give X()T (t) =, X ()T (t) = for all t. Sice T (t) is ot idetically zero we obtai the desired eigevalue problem X (x) λx(x) =, X() =, X () =. (15) There are, i geeral, three cases: (a) If λ = the X(x) = ax + b so applyig the boudary coditios we get = X() = b, = X () = a a = b =. Zero is ot a eigevalue. (b) If λ = µ > the X(x) = a cosh(µx) + b sih(µx) ad X (x) = aµ sih(µx) + bµ cosh(µx). Applyig the boudary coditios we have = X () = aµ a = = X () = bµ cosh(µ) b =. Therefore, there are o positive eigevalues. 5

6 Cosider the followig alterative argumet: If X (x) = λx(x) the multiplyig by X we have X(x)X (x) = λx(x). Itegrate this expressio from x = to x =. We have λ X(x) dx = X(x)X (x) dx = X (x) dx + X(x)X (x). Sice X() = X () = we coclude λ = X (x) dx X(x) dx ad we see that λ must be less tha or equal to zero. (c) So, fially, cosider λ = µ so that ad X(x) = a cos(µx) + b si(µx) X (x) = aµ si(µx) + bµ cos(µx). Applyig the boudary coditios we have = X() = aµ a = = X () = bµ cos(µ). From this we coclude cos(µ) = which implies ad therefore µ = ( 1) ( ) ( 1) λ = µ =, X (x) = si(µ x), = 1,,.. (16) From (14) we also have the associated fuctios T (t) = e λt. We look for u as a ifiite sum u(x, t) = ad we try to fid {b } satisfyig ( ) ( 1)x b e λt si =1 ϕ(x) = u(x, ) = ( ) ( 1)x b si. =1 But this othig more tha a Sie type expasio of the fuctio ϕ o the iterval (, ). Usig ( ) ( 1)x X (x) = si 6

7 we have ϕ(x) = b k X k (x). k=1 We proceed as usual by multiplyig both sides by X (x) ad itegratig from to ad usig the orthogoality (described below i (18), () ). X (x)ϕ(x) dx = b k X (x)x k (x) dx k=1 which implies orthogoality: To see this recall that X j λ λ k ad therefore λ X (x)x k (x) dx = Therefore For = k we have b = ϕ(x)x (x) dx. (17), = k X (x)x k (x) dx =., k = λ j X j ad X j () =, X j() =. First cosider k so = = X (x)x k (x) dx (18) X (x)x k(x) dx + [X (x)x k (x)] X (x)x k (x) dx + [X (x)x k (x) X (x)x k(x)] = λ k X (x)x k (x) dx. (19) (λ λ k ) X (x)x k (x) dx = X (x) dx = X (x)x k (x) dx =. ( ) ( 1)x si dx () = 1 (1 cos (( 1)x)) dx = 1 ( ) 1 si (( 1)x) 1 =. (1) 7

8 ( ) 1 For the iitial coditio ϕ(x) = x. Recall that µ = b = = = [ ϕ(x)x (x) dx = ( x cos (µ ) x) dx µ x cos(µ x) µ + = [ cos(µ ) + si(µ ] x) µ = µ [ cos(( 1)/) + µ = 8( 1)+1 ( 1) x si (µ x) dx ] cos(µ x) dx µ si(( 1)/) (( 1)/) ] u(x, t) = 8 =1 ( ) ( 1) +1 ( 1) ( 1) e [( 1)/]t si x. () 8