PROBLEMS AND SOLUTIONS 2

Transcription

1 PROBEMS AND SOUTIONS Problem 5.:1 Statemet. Fid the solutio of { u tt = a u xx, x, t R, u(x, ) = f(x), u t (x, ) = g(x), i the followig cases: (b) f(x) = e x, g(x) = axe x, (d) f(x) = 1, g(x) =, (f) f(x) =, g(x) = si x. Solutio. We ca directly apply D Alambert s formula (b) We have hece u(x, t) = f(x + at) + f(x at) + 1 a xe x dx = e x + C, g(s)ds. u(x, t) = e (x+at) + e () = e (x+at) + e () = e (). + 1 a ase s ds + 1 ( e (x+at) + e () ) (d) u(x, t) = 1. (f) We have u(x, t) = 1 a si s ds = 1 a (1 cos s)ds = 1 si (x at) si (x + at) t +. 8a Statemet. Solve Problem 5.: u tt = a u xx, x <, < t <, u x (, t) =, u(x, ) = x 3, u t (x, ) =. Date: Witer 16. 1

2 PROBEMS AND SOUTIONS Solutio. We eed first to exted the iitial data by eve reflectio to the etire real lie < c <, apply D Alambert s formula to solve the problem o the lie, ad fially restrict to the half lie x. The eve reflectio of the iitial datum x 3 is x 3. et us apply the D Alambert formula u(x, t) = x at 3 + x + at 3. Note that sice f(x) = x 3 is a C fuctio, u(x, t) satisfies the wave equatio for all x ad t. We ca calculate u x (x, t) = 3(x at) x at + 3(x + at) x + at, which implies that u x (, t) = for all t. To solve the origial problem, we just eed to restrict our attetio to the regio x. Problem 5.1:1 Statemet. Solve the problem u tt = a u xx, x, < t <, u(, t) = u(, t) =, u(x, ) = f(x), u t (x, ) = g(x), i the followig cases (a) f(x) = 3 si( x x ) si( ), g(x) = 1 x si( ), (b) f(x) = si 3 ( x ), g(x) =, (c) f(x) =, g(x) = si( x ) cos ( x ), (d) f(x) = si 3 ( x x ), g(x) = si( ) cos ( x ). Solutio. If the iitial data satisfy f(x) = N =1 α si( x ), g(x) = N =1 β si( x ), the the solutio is give by N ( u(x, t) = α at ) + β a si(at =1 Applyig this formula, we get the followig. ) ) si( x ). (a) u(x, t) = 3 at x at x ) si( ) ) si( ) + at x a si( ) si( ). (b) From the triple agle formula si 3θ = 3 si θ si 3 θ, we have si 3 θ = 3 si θ 1 si 3θ, hece u(x, t) = 3 at x ) si( ) 1 3at 3x ) si( ). (c) Agai usig the triple agle formula g(x) = si( x ) si3 ( x ) = 1 x si( ) + 1 3x si( ), which leads to u(x, t) = at x a si( ) si( ) + 3at 3x 1a si( ) si( ). (d) By combiig (b) ad (c) above, we ifer u(x, t) = ( 3 at ) + at a si( )) si( x ) + ( 1 3at ) + 1a 3at si( )) si( 3x ).

3 PROBEMS AND SOUTIONS 3 Problem 5.1:6 Statemet. Cosider the problem u tt = u xx, x, < t <, u(, t) =, u(, t) =, u(x, ) = x( x), u t (x, ) =. (a) Fid a fuctio that satisfies the equatio ad the boudary coditios exactly, ad the iitial coditio to withi a error of.1. (b) By computig u tt ad u xx at (x, t) = (, ), show that there is o C solutio of the problem. Solutio. (a) et us first compute the sie series coefficiets for f(x) = x( x), which are give by We eed the itegrals ad Hece b = x si x dx = f(x) si x dx, = 1,,.... x cos x x si x dx = x cos x = ( 1)+1 = ( 1) cos x dx = ( 1)+1, x cos x dx b = ( 1) +1 ( 1)+1 + (1 ( 1) ) ad the sie series for f is f(x) = 8 si(m + 1)x (m + 1). Obviously, for ay M >, the fuctio u M (x, t) = 8 M + x si x si x dx cos x + = ( 1)+1 + (( 1) 1). m= m= si(m + 1)x (m + 1) m + 1)t, = (1 ( 1) ), satisfies the wave equatio u tt = u xx ad the boudary coditios u(, t) = u(, t) =. We just eed to choose M so large that u M (x, ) f(x).1 for all x. We ca estimate this error as follows: f(x) u M (x, ) 8 m=m+1 M+1 si(m + 1)x (m + 1) 8 dθ θ = (M + 1). m=m+1 1 (m + 1) Therefore, to esure u M (x, ) f(x).1, it is sufficiet to take M 1, i.e., M 637.

4 PROBEMS AND SOUTIONS Problem 5.1:9 Statemet. Use separatio of variables to fid all product solutios of the problem { u tt = a u xx ku t, x, < t <, u(, t) =, u(, t) =, for the strig with air resistace ad fixed eds (assume k > ). Solutio. Puttig u(x, t) = X(x)T (t), we have ad divisio by XT gives XT = a X T kxt, T T + k T X = a T X = a α, where α is a costat. If α =, the the equatio for X becomes X =, meaig that X(x) = Ax + B. But the boudary coditios X() = X() = forces X(x). So this case is trivial. Now if α = λ > with λ >, we have X(x) = Ae λx + Be λx. The boudary coditios give A + B = ad Ae λ + Be λ =, implyig that A = B =. Fially, cosider the remaiig case α = λ < with λ >. The geeral solutio for X is X(x) = A λx) + B si(λx), ad from X() = we immediately get A =. The X() = B si(λ) = gives the coditio λ = for some positive iteger. To coclude the aalysis of the equatio for X, the oly solutios are X (x) = si( x ), = 1,,..., ad their liear combiatios. We shall cosider the equatio for T. With α = λ =, we have T + kt + ω T =, where ω = a. This (stadard equatio for damped oscillator) ca easily be solved by the asatz T (t) = e µt, which yields µ = k ± k If ω < k, we have two mootoe solutios ω. T (t) = Ae k+ t + Be k t, k ± = k ± k If ω > k, we have the oscillatig solutios T (t) = e kt/ (A cos ω t + B si ω t), ω = If it so happes that ω = k, we have T (t) = e kt/ (A + Bt). To coclude, all product solutios of the give problem are give by u(x, t) = T (x) si( x ), ω. ω k.

5 PROBEMS AND SOUTIONS 5 as rages over the positive itegers, where T is oe of the above three fuctios depedig o how ω = a compares with k. Note that give, T is oe ad oly oe of the above three choices.