Evans PDE Solutions, Chapter 2

Transcription

1 Auhors: Joe Benson, Denis Bashkirov, Minsu Kim, Helen Li, Ale Csar Evans PDE Soluions, Chaper Joe:,,; Denis: 4, 6, 4, 8; Minsu:,3, 5; Helen: 5,8,3,7. Ale:, 6 Problem. Wrie down an eplici formula for a funcion u solving he iniial-value problem { u + b Du + cu on R n (, ) u g on R n { } Here c R and b R n are consans. Sol: Fi and, and consider z(s) : u( + bs, + s) Then ż(s) b Du + u cu( + bs, + s) cz(s) Therefore, z(s) De cs, for some consan D. We can solve for D by leing s. Then, z( ) u( b, ) g( b) De c i.e. D g( b)e c Thus, u( + bs, + s) g( b)e c(+s) and so when s, we ge u(, ) g( b)e c. Problem. Prove ha Laplace s equaion u is roaion invarian; ha is, if O is an orhogonal n n mari and we define v() : u(o) ( R) hen v. Soluion: Le y : O, and wrie O (a i j ). Thus, v() u(o) u(y) where y j n i a ji i. This hen gives ha v i n j n j u y j y j i u y j a ji

2 Thus, v. v n a... a n.. a n... a nn O T u y. u y n D v O T D y u u y. u y n Now, v D v D v (O T D y u) (O T D y u) (O T D y u) T O T D y u (D y u) T (O T ) T O T D y u (D y u) T OO T D y u (D y u) T D y u because O is orhogonal (D y u) (D y u) u(y) Problem 3. Modify he proof of he mean value formulas o show for n 3 ha ( u() gds + nα(n)r n n(n )α(n) ) f d, n r n Soluion: Se and Then, φ () n provided φ(r) ( α(n) n (See he proof of Thm) φ() B(,r) u f in B (, r) u g on B(, r). B(,r) u(y)ds (y), < r, nα(n) n B(,) u(y)ds (y) nα(n)r n B(,r) B(,) u(y)dy ) n gds. nα(n)r n B(,r) ( f dy ) α(n) n B(,) f dy. α(n) n B(,)

3 3 Le ɛ > be given. () φ(ɛ) φ(r) r ɛ φ ()d Using inegraion by pars, we compue r r φ ()d f dyd ɛ ɛ nα(n) n B(,) r f dyd nα(n) ɛ n B(,) ([ f dy nα(n) n n B(,) ( r n(n )α(n) ɛ n ( : I n(n )α(n) r n r gds φ ()d. nα(n)r n B(,r) ɛ ] r B(,) B(,r) ɛ r ɛ n f ds d r n f dy + J ). n B(,r) B(,) f ds d ) f dy + ɛ n B(,ɛ) f dy ) Observe ha and J : ɛ n B(,ɛ) f dy C ɛ, for some consan C > B(,ɛ) f ()d n r d B(,) f ds. n As ɛ, I + J B(,ɛ) lim ɛ r ɛ f ()d. Thus, n φ ()d n(n )α(n)( f ()d B(,r) n ( n(n )α(n) ) f d. n r n Therefore, leing ɛ, we have from () ( u() φ() gds + nα(n)r n B(,r) n(n )α(n) B(,r) ) f d. n r n B(,r) r n B(,r) f dy ) Problem 4. We say v C (Ū) is subharmonic if (a) Prove for subharmonic v ha v() B(,r) v in U. v dy for all B(, r) U. (b) Prove ha herefore ma Ū v ma U v. (c) Le φ : R R be smooh and conve. Assume u is harmonic and v : φ(u). Prove v is subharmonic.

4 4 (d) Prove v : Du is subharmonic, whenever u is harmonic. Soluion. (a) As in he proof of Theorem, se φ(r) : v ds (y) and obain B(,r) φ (r) r v(y)dy. n For < ɛ < r, B(,r) r ɛ B(,r) φ (s)ds φ(r) φ(ɛ). Hence, φ(r) lim φ(ɛ) v(). Therefore, ɛ v dy v dy α(n)r n α(n)r n α(n)r n B(,r) r r ( ) v(z) ds (z) ds B(,s) nα(n)s n φ(s) ds r ns n v() ds v() r n (b) We assume ha U R n is open and bounded. For a momen, we assume also ha U is conneced. Suppose ha U is such a poin ha v( ) M : ma Ū v. Then for < r < dis(, U), M v( ) v dy M. B(,r) Due o coninuiy of v, an equaliy holds only if v M wihin B(, r). Therefore, he se u ({M}) U { U u() M} is boh open and relaively closed in U. By he connecedness of U, v is consan wihin he se U. Hence, i is consan wihin Ū and we conclude ha ma Ū v ma U v. Now le {U i i I} be he conneced componens of U. Pick any U and find j I such ha U j. We obain v() ma U j and conclude ha ma Ū v ma U v. (c) For (,..., n ) U and i, j n, v i j () v ma U j v ma v U φ(u()) φ (u()) u () u () + φ (u()) i j i j u i j (). Since φ is conve, hen φ () for any R. Recall ha u is harmonic and obain n ( ) u n ( ) u v φ (u) + u φ (u). i i (d) We se v : Du n k v i j () ( u k i ). k For (,..., n ) U and i, j n, n [ u u () () + u ] 3 u () (). i k i j k i j k i

5 5 Therefore, v n ( ) u + u ( u i k i k k k ), i ( ) u n u ( ) ( ) u v + u. i k k k i k i,k n k i,k n Problem 5: Prove ha here eiss a consan C, depending only on n, such ha ( ) ma u C ma g + ma f B(,) B(,) B(,) whenever u is a smooh soluion of u f in B (, ) u g on B(, ). Proof: Le M : ma B(,) f, hen we define v() u() + M n and w() u() + M n. We firs consider v(). Noe ha So, v() is a subharmonic funcion. From Problem 4 (b), we have v u M f M. Tha is Then, for w(), we have Again, we can ge i.e. ma B(,) ma v() ma v() ma g + M B(,) B(,) B(,) n. u() ma v() ma g + B(,) B(,) n ma f. B(,) w u M f M. ma w() ma w() ma g + M B(,) B(,) B(,) n. ma B(,) u() ma w() ma g + B(,) B(,) n ma f. B(,) Combining hese wo ogeher, we finally proved he problem. Problem 6. Use Poisson s formula for he ball o prove r n r (r + ) u() u() r + n rn u() (r ) n whenever u is posiive and harmonic in B (, r). This is an eplici form of Harnack s inequaliy.

6 6 Soluion. Since y B(, r), hen y + r. Therefore, u() r nα(n)r r nα(n)r B(,r) B(,r) r n r (r + ) n g(y) ds (y) y n g(y) (r + ) ds (y) r n rn (r + ) n g(y)ds (y) r n r u() (r + ) n B(,r) g(y)ds (y) nα(n)r n B(,r) r+ The inequaliy u() r n u() can be proven in a similar way. (r ) n Problem 7. Prove Poisson s formula for a ball: Assume g C( B(, r)) and le u() r g(y) nα(n)r B(,r) y ds (y) for n B (, r). Show ha Proof. Problem 8. Le u be he soluion of u in R n + u g on R n + given by Poisson s formula for he half-space. Assume g is bounded and g() for R n +, le. Show Du is no bounded near. (Hin: Esimae u(λe n) u() λ.) Proof: From formula (33) on page 37, we have u() n nα(n) and u() g(). Thus, using hin, we ge u(λe n ) u() g(y) λ nα(n) λe n y dy n nα(n) R n + y R n + Taking absolue value on boh sides, we have u(λe n) u() λ nα(n) I I. R n + g(y) y n dy, g(y) λe n y dy + n nα(n) y R n + y > R n + g(y) λe n y dy n nα(n) y > R n + g(y) λe n y n dy g(y) λe n y n dy

7 Since g is bounded, so i is obvious ha I is bounded and independen of λ. For I, in his case, g(y) y, so I y nα(n) y R n λe + n y dy n y (λ+ y ) n nα(n) B n (,r) y R n + y (λ + y ) n dy Noe ha for fied y, is increasing when λ is decreasing o, so by Monoone Convergence heorem, we have y lim λ nα(n) y R n (λ + y ) dy n + y y R n y dy n + y B n (,) y dy n dr ds (y) C y n r n rn dr. So, Du is unbounded near. 7 Problem. Suppose u is smooha nd solves u u in R n (, ). (i) Show u λ (, ) : u(λ, λ ) also solves he hea equaion for each λ R. (ii) Use (i) o show v(, ) : Du(, ) + u (, ) solves he hea equaion as well. (i) u λ (, ) λ u (λ, λ ) and u λi (, ) λu(λ, λ ) for each i. Then u λi i (, ) λ u i (λ, λ ). Consequenly, u λ λ u and u λ u λ λ (u u), so u λ solves he hea equaion for all λ R. (ii) We differeniae u(λ, λ ) u(λ,..., λ n, λ ) wih respec o λ we ge k u k (λ,..., λ k, λ ) + λu (λ,..., λ n, λ ) D(λ, λ ) + u (λ, λ ). k Taking λ, we hen have ha v(, ) Du(, ) + u (, ). u is smooh, so he second derivaives of u(λ, λ ) are coninuous, meaning he mied parials are equal. Therefore, v v u(λ, λ λ ) u(λ, λ λ ) u(λ, λ λ ) u(λ, λ λ ) (u λ λ u λ ), since u λ saisfies he hea equaion for all λ. Thus v does as well. Problem : Assume n and u(, ) v( ). a) Show u u if and only if () 4zv (z) + ( + z)v (z) (z > )

8 8 b) Show ha he general soluion of () is v(z) c z e s/4 s / ds + d c) Differeniae v( ) wih respec o and selec he consan c properly, so as o obain he fundamenal soluion Φ for n. Soluion: a) Assume ha u u. Then and So u u implies ha or If we le z, we ge 4z v u v ( ) ( ) ( ) u v + 4 v 4 v ( ) ( ) ( ) v + 4 v ( ) + ( + ) ( ) v ( v (z) + + z ) v (z) Muliplying his equaion by gives he desired equaliy. For he oher direcion, reverse he seps, and hence our proof is done. b) (by inegraing) 4zv + ( + z)v v v z 4 log(v ) log z z 4 + c v C v Cz / e z/4 z e s/4 s / ds + d

9 9 as is desired. c) v(z) c z e s/4 s / ds + d ( ) v c e s/4 s / ds + d ( ) v c ( ) e 4 / or ( ) v c e 4 Now we wan o inegrae over R and se he inegral equal o. Thus we ge Leing y or c e 4 d 4, we ge dy (4) / d and subsiuing, we ge c 4c 4e y dy e y dy Employing he ideniy e y dy π and solving for c, we ge Thus, c 4 π ( ) Φ(, ) : v c e 4 is easily shown o solve he equaion π e 4 Φ Φ Problem. Wrie down an eplici formula for a soluion of u u + cu f in R n (, ) u g on R n { }, where c R.

10 Soluion: Se v(, ) u(, )e C. Then, v u e C + Ce C u and v i i u i i e C. v v u e C + Ce C u e C u e C (u u + Cu) e C f. So, v is a soluion of v v e C f in R n (, ) v g on R n { }, By (7) (p.5), v(, ) Φ( y, )g(y)dy + R n Φ( y, s)e Cs f (y, s)dyds R n where Φ is he fundamenal soluion of he hear equaion. Since v(, ) u(, )e C, we have u(, ) e C( Φ( y, )g(y)dy + Φ( y, s)e Cs f (y, s)dyds ). R n R n Problem 3: Given g : [, ] R, wih g(), derive he formula u(, ) e 4( s) 4π ( s) 3/ g(s)ds, > for a soluion of he iniial/boundary-value problem u u inr + (, ) u onr + { }, u g on{ } [, ). Proof. We define So, we have and u(, ) g() >, v(, ) u(, ) + g(). u (, ) g () >, v (, ) u (, ) + g (), u (, ) >, v (, ) u (, ).

11 Hence, By formula (3) on page 49, we ge v(, ) Noe ha(page 46 Lemma) g () >, v (, ) v (, ) g (). v(, ), v(, ). { 4π( s) so when >, we le y z and obain Inegraing by pars, we ge u(, ) v(, ) + g() u(, ) v(, ) + e (y ) 4( s) g (s)dyds 4π( s) e (y ) 4( s) dy, g (s)ds 4π ( s) π ( s) π ( s) / I + I + } e (y ) 4( s) g (s)dyds 4π( s) e (y ) 4( s) dy e (y ) 4( s) dy g (s)ds e z 4( s) dz dg(s) e z 4( s) dz g(s) s s g(s) π ( s) 3/ ds + g(s) π ( s) / ds g(s) π ( s) 3/ ds g(s) π ( s) / ds g(s) g(s) π ( s) 3/ ds e z 4( s) dz e z 4( s) z 4( s) dz e z 4( s) dz z ( s) de z 4( s) e z 4( s) dz 4π ( s) 3/ ds ( z) e z 4( s) z z g(s) π ( s) 3/ ds I + e 4( s) 4π ( s) 3/ g(s)ds. e z 4( s) dz

12 Now, we focus on I and define w o be z 4ɛ, Thus, we proved Ne, we need o show ha Noe ha for any fied δ >. u(, ) I lim ɛ + g() lim ɛ + lim u(, ) lim + + ɛ / π π /4ɛ e z 4ɛ dz g( ɛ) e w dw. e 4( s) 4π ( s) 3/ g(s)ds, >. lim u(, ) g(). + + lim + g() lim + g() lim + 4π δ δ 4π 4π e 4( s) ( s) 3/ g(s)ds δ δ 4π e 4( s) ( s) 3/ g(s)ds e 4( s) ( s) 3/ ds e 4s s3/ ds For fied, we le s /w and ge Hence, we are done. lim u(, ) g() lim + + g() lim + g() π /δ π π /δ w 3 e w 3 4 e w 4 dw e w 4 dw g(). w dw 3 Problem 4. We say v C (U T) is a subsoluion of he hea equaion if v v in U T. (a) Prove for a subsoluion v ha v(, ) 4r n for all E(, ; r) U T. (b) Prove ha herefore ma ŪT v ma ΓT v E(,;r) y v(y, s) ( s) dyds

13 3 Soluion. (a) We may well assume upon ranslaing he space and ime coordinaes ha and. As in he proof of Theorem 3, se φ(r) : v(y, s) y r n s dyds, and derive For < ɛ < r, ψ(y, s) : n E(r) log( 4πs) + y 4s + n log r φ (r) 4n vψ n n v r n+ yi y i dyds E(r) s i n 4nv r n+ yi ψ yi n s v y i y i dyds. i r ɛ E(r) φ (z)dz φ(r) φ(ɛ). y Hence, φ(r) lim φ(ɛ) v(, ) lim ɛ ɛ ɛ E(ɛ) dyds 4v(, ), and he saemen follows. n s (b) Suppose here eiss a poin (, ) U T wih u(, ) M : ma ŪT u. Then for all sufficienly small r >, E(, ; r) U T. Using he resul proved above, we deduce M v(, ) y v(y, s) dyds M, 4r n ( s) since E(, ;r) y 4r n E(, ;r) ( s) dyds. Conclude ha u E(, ;r) M. The argumen used in he proof of Theorem 4 will finish he proof. Problem 5. (a) Show he general soluion of he PDE u y is u(, y) F() + G(y) for arbirary funcions F,G. (b) Using he change of variables ξ +, η, show u u if and only if u ξη. (c) Use (a),(b) o rederive d Alember s formula. Soluion: (a) u y u f () u(, y) f ()d + G(y) u y u y g(y) u(, y) g(y)dy + F()

14 4 This implies u(, y) F() + G(y). (b) ξ+η, y ξ η Define ũ : u ( ξ+η, ξ η ) ũ ξ u + u and ũ ξη 4 u 4 u + 4 u 4 u 4 (u u ) Hence, ũ ξη u u. (c) By (b), u u u ξη, and u(ξ, η) F(ξ) + G(η) by (a),i.e, u(, y) F( + ) + G( ). Since u(, ) g, u (, ) h, (3) u(, ) F() + G() g(), Inegraion (4) F() G() Thus, u (, ) F () G () h() h(y)dy + C, C:consan. () + (3); F() ( g() + h(y)dy + C ) () (3); G() ( g() h(y)dy C ) u(, y) F( + ) + G( ) ( + g( + ) + h(y)dy + C ) + ( g( ) h(y)dy C ) ( + g( + ) + h(y)dy + C + g( ) + h(y)dy C ) [ g( + ) + g( ) ] + + h(y)dy ( R, ). Problem 6. Assume E (E, E, E 3 ) and B (B, B, B 3 ) solve Mawell s equaions: E curl B B curl E div B div E Show ha u u where u B i or E i for i,, 3. Soluion.

15 5 curl(curl E) curl( B ) ) ( B 3 y + B z, B 3 + B z, B + B y curl B E E However, we also know ha curl(curl E) (div E) E E. Then E i saisfies u u for i,, 3. Similarly, curl(curl B) curl E B, and curl(curl B) (div B) B B, so B i saisfies u u for i,, 3. Problem 7.(Equipariion of energy) Le u C (R [, )) solve he iniial value problem for he wave equaion in one dimension: u u in R (, ) u g; u h on R { }. Suppose g, h have compac suppor. The kineic energy is k() : energy is p() : u (, )d. Prove (i) k() + p() is consan in. Proof. (i.) We define e() k() + p() we have d e() d (ii) k() p() for all large enough imes. u u d Hence, e() e(). (ii.)by d Alember s formula on page 68, we have So, and u(, ) ( u + u u u + u u d u (, )d and he poenial ) d. Since g, h have compac suppor, so u u d u (u u ) d. [ g( + ) + g( ) ] + + h(y)dy. u [ g ( + ) g ( ) ] + [h( + ) + h( )], u [ g ( + ) + g ( ) ] + [h( + ) h( )].

16 6 We assume ha here eiss a posiive consan M so ha [ M, M] supp(g ) and [ M, M] supp(h). Noe ha for a fied > M, M M < M + M and M + M M + M <. Thus, when > M : (a) < M + M. Then we have So, h( + ) g( + ). (b) M + M <. Then, u 4 g ( ) + 4 h( ) g ( )h( ) u. u 4 g ( + ) + 4 h( + ) + g ( + )h( + ) u. (c) Oherwise g ( + ) g ( ) h( + ) h( ). So, combining all he cases, i is obvious ha when > M, k() p(). Problem 8. Le u solve { u u in R 3 (, ) u g, u h on R 3 { }, where g, h are smooh and have compac suppor. Show here eiss a consan C such ha u(, ) C/ ( R 3, > ). Soluion. From he condiions i follows ha here eis R, M > such ha sp g, sp h B(, R) and g(y) M, Dg(y) M, h(y) M for any y R 3. Kirchhoff s formula gives he soluion of he iniial-value problem: u(, ) h(y) + g(y) + Dg(y) (y ) ds (y). B(,) Denoe by Σ he inersecion B(, ) B(, R). Observe ha he area of Σ is no greaer han he area of he sphere B(, R). Then, for >, we obain h(y) + Dg(y) (y ) ds (y) B(,) 4π h(y) + Dg(y) (y ) ds (y) B(,) B(,R) h(y) + Dg(y) y ds (y) 4π B(,) B(,R) 4π 4πR (M + M) R M.

17 For >, using he same argumen, we ge g(y) ds (y) 4π B(,) B(,) B(,R) g(y) ds (y) 4π 4πR M R M R M. Noice now ha he area Σ is no greaer han he area of he sphere B(, ). Then for <, g(y) ds (y) 4π g(y) ds (y) 4π 4π M M. B(,) B(,) B(,R) Wihou loss of generaliy, we can ake R >. Then, combining he esimaes obained above, we conclude u(, ) 3R M. 7 Evans PDE Soluions, Chaper 5 Ale: 4, Helen: 5, Rob H.: Problem. Suppose k {,,...}, < γ <. Prove C k,γ (Ū) is a Banach space. Soluion:. Firs we show ha C k,γ (Ū) is a norm, where we recall ha and u C k,γ (Ū) D α u C(Ū) + [D α u] C,γ (Ū), α k [u] C,γ (Ū) sup y U α k { u() u(y) y γ }. For he sake of opaqueness we now omi subscrips on all norms unless i is unclear from cone.. For any λ R we have firs λu() λu(y) u() u(y) [λu] sup λ sup λ [u],,y U y γ,y U y γ and cerainly So D α (λu) C(Ū) λd α u λ D α u. λu D α (λu) + [D α (λu)] α k α k λ D α u + λ [D α u] α k λ u. 3. If u i is obvious ha u. On he oher hand, u implies ha D α u C(Ū) α k

18 8 for every α k. In paricular his is rue for α so ha he supremum of D u u on U is, i.e. u. 4. Finally we mus prove he riangle inequaliy. We know he riangle inequaliy is rue for he sup norm C(Ū). We can also see ha for any α which makes sense [D α (u + v)] [D α u + D α v] [D α u] + [D α v]. Therefore we can easily conclude u + v D α (u + v) + [D α (u + v)] α k α k ( D α u + D α v ) + ([D α u] + [D α v]) α k u + v. 5. We need only show ha C k,γ (U) is complee. So le {u m } be a Cauchy sequence. Then {u m (){ is a Cauchy sequence for every, so define u o be he poinwise limi of he u m. Now if V is any bounded subse of U, hen V is compac, so ha u m u uniformly on any V. Since he u m are uniformly coninuous on V by assumpion, his implies ha u is uniformly coninuous on V as well (and so, a foriori u C(U)). Therefore u C(Ū). Wha we would really like would be o have u C k (Ū). Bu similar argumens show ha u has derivaives D α u for all α k on U by resricing firs o bounded subses of U o find he derivaives and hen using uniform convergence on hese subses o show he derivaives mus also be uniformly coninuous on bounded subses since he D α u m were. This leaves us wih only showing ha he norm of u is finie, so ha in fac u C k,γ (U). Bu for every n we have u n u sup D α u n () D α D α u n () D α u n (y) D α u() + D α u(y) u() + sup α k U α k,y U y γ lim m α k lim m u n u m. sup D α u n () D α u m () + U α k sup,y U α k D α u n () D α u n (y) D α u m () + D α u m (y) y γ In paricular, since {u m } is Cauchy here is some N so ha n, m N implies u n u m. Leing m approach, his implies ha u N u <. Now he riangle inequaliy applies o give Problem 4. u u N u + u N < + u N <. Assume U is bounded and U N i V i. Show here eis C funcions ζ i (i,..., N) such ha ζ, supp ζ i V i i,..., N N i ζ i on U. The funcions {ζ i } N for a pariion of uniy.

19 Soluion. Assume U is bounded and U N i V i. Wihou loss of generaliy, we may assume ha he V i are open, for if hey are no, we can replace V i by is inerior. We noe ha, since U is bounded, U is compac. Each U has a compac neighbourhood N conained in V i for some i. Then {N} is an open cover of U, which hen has a finie subcover N,..., N n. We now le F i be he union of he N k conained in V i. F i is he compac since i is he finie union of compac ses. The C version of Urysohn s Lemma (Folland, p.45) allows us o find smooh funcions ξ,..., ξ N such ha ξ i on F i and supp(ξ i ) V i. Since he F i cover U, U { : n ξ i () > } and we can use Urysohn again o find ζ C wih ζ on U and supp(ζ) { : n ξ i () > }. Now, we le ξ N ζ, so N+ ξ i > everywhere. We hen ake ξ i ζ i N+ ξ j as our pariion of uniy. Problem 5 (Helen) Prove ha if n and u W,p (, ) for some p <, hen u is equal a.e. o an absoluely coninuous funcion, and u which eiss a.e. belongs o L p (, ). Proof. Since u W,p (, ), so by definiion on page 4 and 44, we have some funcion v L p (, ) such ha u Dφd vφd, φ Cc ((, )). (,) (,) Noe ha v L p (, ), so by Hölder s inequaliy, we have v L v L p L q <, which means v L (, ). Thus, we can define funcion f () on (, ) by he following formula f () u( ) + v()d, (, ). According o he Fundamenal Theorem of Calcalus, f is absoluely coninuous. Now we will prove u f a.e. By he definiion of f, we have f v a.e. So for any φ Cc ((, )) we ge f Dφd f φd vφd. Therefore, (,) (,) (,) ( f u) Dφd φ C c (,) ((, )), which means u f + cons. And noe ha u( ) f ( ), hence u f a.e. So u eiss a.e. and saisfy u v a.e., so u L p (, ). 9