Problem set 5: Solutions Math 207B, Winter r(x)u(x)v(x) dx.

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1 Problem set 5: Soltions Mth 7B, Winter 6. Sppose tht p : [, b] R is continosly differentible fnction sch tht p >, nd q, r : [, b] R re continos fnctions sch tht r >, q. Define weighted inner prodct on L (, b) by, v r = r(x)(x)v(x) dx. Let A : D(A) L (, b) L (, b) be the opertor A = [ d r(x) dx p(x) d ] dx + q(x) with Dirichlet bondry conditions nd domin () Show tht D(A) = { H (, b) : () =, (b) = }., Av r = A, v r for ll, v D(A), mening tht A is self-djoint with respect to, r. (b) Show tht the eigenvles λ of the weighted Strm-Lioville eigenvle problem (p ) + q = λr, () =, (b) = re rel nd positive nd eigenfnctions ssocited with different eigenvles re orthogonl with respect to, r. Soltion. () Using integrtion by prts nd the bondry conditions stisfied by, v D(A), we hve, Av r = [ (pv ) + qv] dx = [p(ū v ūv )] b + [ (pū ) + qū] v dx = = A, v r. [ (p ) + q]v dx

2 (b) The relity of the eigenvles nd the orthogonlity of the eigenfnctions follows directly from the self-djointness of A. Moreover, if A = λ nd D(A) is normlized so tht r =, then n integrtion by prts gives λ =, A r = ū[ (p ) + q] dx = [ p + q ] dx. If λ =, then p dx =, so = nd = constnt. Then the bondry condition implies tht =, so λ = is not n eigenvle, nd λ >.

3 . A nonniform string of length one with wve speed c (x) = T/ρ (x) > is fixed t ech end, with zero initil displcement nd nonzero initil velocity. The trnsverse displcement y = (x, t) of the string stisfies the IBVP tt = c (x) xx < x <, t >, (, t) =, (, t) = t >, (x, ) = < x <, t (x, ) = g(x) < x <, Find the soltion in terms of the eigenvles λ n nd eigenfnctions φ n (x) of the weighted Strm-Lioville problem c φ n = λ n φ n, φ n () =, φ n () =, n =,, 3,... Soltion. Seprtion of vribles gives the soltions { φ n (x) cos(k n t), (x, t) = φ n (x) sin(k n t), where k n > with k n = λ n. From the previos qestion, λ n > nd the eigenfctions re orthogonl with respect to the weight r = c. Sperposing the seprted soltions tht re zero t t =, we get tht (x, t) = b n φ n (x) sin(k n t). The initil condition for t (x, ) is stisfied if g(x) = k nb n φ n (x). Using the orthogonlity of the eigenfnctions, we get tht or k n b n = g, φ n r, φ n r b n = c gφ n dx k n c φ n dx.

4 3. The Forier soltion of the initil vle problem is given by tt = xx < x <, t >, (, t) =, (, t) = t >, { x if x / (x, ) = ( x) if / < x <, t (x, ) = x, (x, t) = 8 π sin [πx] cos [πt] () Show tht the Forier series converges to continos fnction. Wht order of sptil (wek) L -derivtives does (x, t) hve? (b) Verify from the Forier soltion tht [ t (x, t) + x(x, t) ] dx = constnt for < t <. (c) Use mtlb (or nother progrm) to compte the prtil sm N (x, t) = 8 π N t t =.5 for N = 5 nd N = 5. sin [πx] cos [πt] (d) Use the ddition forml for sines to shows tht the Forier soltion cn be written in the form of the d Alembert soltion s (x, t) = F (x t) + F (x + t) for sitble fnction F : R R. Wht is F? Soltion. () We hve sin [πx] cos [πt],

5 nd <, so the Weierstrss M-test implies tht the series of continos fnctions converges niformly to continos fnction. We hve x (x, t) = 8 π cos [πx] cos [πt], so by Prsevl s theorem (nd the fct tht cos mπx L = /) x (, t) L = 3 π 3 π cos [πt], mening tht x (, t) L (, ) for ll t R. We hve the distribtionl derivtive xx (x, t) = 8 ( ) n sin [πx] cos [πt], which does not belong to L since cos [πt] diverges (except for specil vles of t, sch s t = /). It follows tht (, t) H (, ) hs one sptil L -derivtive. More generlly, if we lso consider frctionl derivtives of order s, then (, t) H s (, ) for s < 3/. (b) We hve t (x, t) = 8 π x (x, t) = 8 π sin [πx] sin [πt], cos [πx] cos [πt].

6 Prsevl s theorem implies tht so t (x, t) dx = 3 π x(x, t) dx = 3 π [ t (x, t) + x(x, t) ] dx = 3 π is constnt independent of time. At t = sin [πt], cos [πt], [ t (x, ) + x(x, ) ] dx = x(x, ) dx = 4, so we get the following sm from the Forier expnsion of the fnction x (x, ): = = π 8. (c) The prtil sms re shown below. Althogh the Forier series converges niformly, it isn t rpidly convergent since the soltion isn t smooth fnction of x. (d) Using the trigonometric identity we get tht sin A cos B = [sin(a B) + sin(a + B)], (x, t) = 4 π + 4 π sin [π(x t)] sin [π(x + t)] = F (x t) + F (x + t),

7 where F (x) = 4 π sin [πx]. The fnction F : R R is the odd, periodic extension of the initil dt, mening tht F ( x) = F (x), F (x + ) = F (x), nd { x if x / F (x) = x if / < x <,

8 4. Sppose tht (x, t) is smooth soltion of the wve eqtion tt = c, where x R n, nd the wve speed c > is constnt. () Show tht stisfies the energy eqtion ( t + c ) (c t t ) =. (b) For T >, let Ω T R n+ be the spce-time cone Ω T = { (x, t) R n+ : x < c (T t), < t < T }, nd for t T, let B(T t) be the sptil cross-section of Ω T t time t Define B(T t) = {x R n : x < c (T t)}. e T (t) = B(T t) ( t + c ) dx, nd show tht e T (t) e T (). (c) Sppose tht, re smooth soltion of the wve eqtion sch tht i (x, ) = f i (x), it (x, ) = g i (x) i =, where f = f, g = g in x c T, show tht = in Ω T. hint. For (b), pply the divergence theorem in spce-time to the eqtion in () over the trncted cone {(x, t ) Ω T : < t < t}, nd note tht the spce-time norml to the side of the cone Ω T is N = (ˆx, c )/ + c where ˆx = x/ x. For (c), consider =. Soltion. () Mltiplying the wve eqtion by t, we hve Using the identities ( ) t tt = t we get tht t tt c t =., t = ( t ) t ( t + c ) (c t t ) =. ( ), t

9 For < T < T, let Ω T,T denote the trncted cone Ω T,T = { (x, t) R n+ : x < c (T t) nd < t < T }. The bondry of Ω T,T consists of the bottom Ω T,T = Γ Σ Γ Γ = { (x, ) R n+ : x c T } with otwrd spce-time norml N = (, ), the side Σ = { (x, t) R n+ : x = c (T t) nd < t < T } with otwrd spce-time norml N = (ˆx, c )/ + c, nd the top Γ = { (x, T ) R n+ : x c (T T ) } with otwrd spce-time norml N = (, ). Integrting the energy eqtion over Ω T,T nd pplying the divergence theorem, we get tht { ( = Ω T,T t + c ) (c t t )} dxdt { ( = t + c ) } ν c t n ds Ω T,T where N = (n, ν) is the otwrd spce-time norml to Ω T,T. Splitting the bondry integrl into n integrl over the bottom, side, nd top, we get tht { e T (T c ( ) = e T () + c t + c ) } c t ˆx ds, where e T (T ( ) = Γ t + c ) dx, e T () = Σ Γ ( t + c ) dx.

10 Completing the sqre, nd sing the fct tht ˆx is nit vector, we get tht ( t + c ) c t ˆx = ( tˆx c ) ( tˆx c ). It follows tht e T (T ) e T () for < T < T. A physicl interprettion of this ineqlity is tht energy cn only propgte ot of the cone Ω T, not into it. (c) If, re two soltions with the sme initil dt for i nd it in x c T, then = hs zero initil dt, in x c T nd therefore e T () =. The previos ineqlity implies tht e T (t) e T () for < t < T, so e T (t) =. It follows tht t = nd = in Ω T, mening tht = constnt. Then the initil condition implies tht = nd = in Ω T. Remrk. As this reslt shows, the soltion of the wve eqtion t some point in spce-time cn only depend on, or inflence, the soltion t other points of spce-time tht cn be reched by trveling t speeds less thn or eql to c.