Part II. Analysis of PDE

Transcription

1 Prt II. Anlysis of PDE Lecture notes for MA342H P. Krgeorgis 1/38

2 Second-order liner equtions Consider the liner opertor L(u) which is defined by L(u) = 1 u xx + 2 u yy +b 1 u x +b 2 u y for some constnts 1, 2,b 1,b 2 such tht one of 1, 2 is nonzero. This opertor is clled elliptic when 1 2 > 0, it is clled hyperbolic when 1 2 < 0, nd it is clled prbolic when 1 2 = 0. The Lplce eqution u xx +u yy = 0 is typicl exmple of n elliptic eqution. The solutions of this eqution re usully clled hrmonic. The wve eqution u tt c 2 u xx = 0 is hyperbolic for ny c > 0. It is used to describe the motion of wve whose speed c is constnt. The het eqution u t ku xx = 0 is prbolic for ny k > 0. In this cse, u(x, t) denotes the temperture of n one-dimensionl object. 2/38

3 Mximum principle Theorem 1. Mximum principle Consider the second-order liner opertor L(u) = 1 u xx + 2 u yy +b 1 u x +b 2 u y in the elliptic cse 1, 2 > 0. Suppose A R 2 is open nd bounded. 1 If L(u) 0 within A, then mx A A u = mx A u. 2 If L(u) 0 within A, then min A A u = min A u. Moreover, the sme result holds in the prbolic cse 1 > 0 = 2. The function u ttins min/mx over the sets A A nd A becuse these sets re closed nd bounded, hence lso compct. The second prt of the theorem follows trivilly from the first. 3/38

4 Mximum principle: Proof of prt 1 Define w(x,y) = u(x,y)+δe λx for some δ,λ > 0. Since L(w) = L(u)+δL(e λx ) = L(u)+δλe λx ( 1 λ+b 1 ), one hs L(w) > 0 within A for ll lrge enough λ. If the mximum of w is ttined t n interior point, then L(w) = 1 w xx + 2 w yy +b 1 w x +b 2 w y = 1 w xx + 2 w yy 0 t tht point nd this is contrdiction. In prticulr, one hs mx u mx w = mxw = mx A A A A A A (u+δeλx ). Once we now let δ 0, we my finlly conclude tht mx A A u mx A u mx A A u = mx A A u = mx A u. 4/38

5 Extended mximum principle Theorem 2. Extended mximum principle Consider the liner opertor L(u) in the cse tht 1 > 0 nd 2 0. Suppose A is bounded, open subset of R 2 nd suppose c(x,y) is function which is continuous on A with c(x,y) 0 t ll points. 1 If L(u)+cu 0 within A, then mx A A u mx A u +. 2 If L(u)+cu 0 within A, then min A A u min A u. 3 If L(u)+cu = 0 within A, then mx A A u = mx A u. Here, u + denotes the positive prt of u, nmely u + = mx(u,0). Similrly, u denotes the negtive prt of u, nmely u = min(u,0). This theorem does not necessrily hold when c(x, y) is positive. 5/38

6 Extended mximum principle: Counterexmple We show tht the extended mximum principle does not necessrily hold when the function c(x, y) is positive. Consider the cse u xx +u yy +2u = 0, x,y (0,2π). As one cn esily check, this eqution is stisfied by the function u(x,y) = (sinx) (siny). Now, the set A = (0,2π) (0,2π) is certinly open nd bounded, while u vnishes on A since either x = 0,2π or y = 0,2π on A. If the extended mximum principle did hold, then one would hve mx A A u = mx u = 0. A This is not the cse, however, s the left hnd side is equl to 1. 6/38

7 Extended mximum principle: Proof of prt 1 Let B = {(x,y) A : u(x,y) > 0}. If this set is empty, then u 0 in A = u 0 in A A by continuity = mx A A u 0 mx A u+ nd the result follows. If B is nonempty, then L(u) cu 0 in B = mx B B u = mx B u by the stndrd mximum principle. This ctully gives mx u = mxu mx B B B B u+ = mx A u+ nd one lso hs u(x,y) 0 mx A u + for ll (x,y) A B. 7/38

8 Extended mximum principle: Proof of prt 2 Since L(u)+cu 0 within A by ssumption, one hs L(u) cu 0 = L( u)+c( u) 0 within A. Using the first prt of the theorem, we conclude tht mx ( u) mx A A A ( u)+. Here, the rightmost function cn lso be expressed s ( u) + = mx( u,0) = min(u,0) = u. Thus, one my combine the lst two equtions to deduce tht mx ( u) mx A A A ( u ) = min u min A A A u. 8/38

9 Extended mximum principle: Proof of prt 3 First of ll, it is esy to see tht ( ) mx u = mx mx A A A A u+, mx A A ( u ). We now estimte u + nd u. In view of prt 1, one certinly hs u + (x,y) = mx(u(x,y),0) mx A u+ for ll (x,y) A A. In view of prt 2, one similrly hs u (x,y) = min(u(x,y),0) min A u = mx A ( u ). Inserting these estimtes in the first eqution, we conclude tht ( ) mx u mx mx A A A u+,mx A ( u ) = mx A u mx A A u. 9/38

10 Uniqueness of solutions Theorem 3. Uniqueness of solutions Let L,A,c be s before nd suppose f,g re continuous. Then the following boundry vlue problem dmits t most one solution. { } L(u)+cu = f(x,y) in A. u = g(x,y) on A Indeed, ssume u,v re solutions nd let w = u v. Since L(w)+cw = L(u) L(v)+cu cv = f f = 0 within A, the third prt of the extended mximum principle gives mx A A w = mx A w = mx g g = 0. A In prticulr, w = 0 t ll points nd thus u = v t ll points. 10/38

11 First comprison principle Theorem 4. Comprison principle in bounded domins Let L,A,c be s before nd suppose the functions u,v stisfy { } L(u)+cu L(v)+cv in A. u(x,y) v(x,y) on A Then one ctully hs u(x,y) v(x,y) for ll (x,y) A A. To prove this theorem, we note tht w = v u stisfies L(w)+cw = L(v) L(u)+cv cu 0 within A. According to the extended mximum principle, this implies tht mx (v u) = mx w mx A A A A A w+. Since w + = mx(v u,0) = 0 on A, the result now follows. 11/38

12 First comprison principle: Exmple Let A = (0,1) (0,1) nd consider the boundry vlue problem { uxx +u yy = 1 in A u(x,y) = 0 on A }. Tke L(u) = u xx +u yy nd let v denote the zero function. Since L(u) = 1 0 = L(v) within A nd u v on A, we conclude tht u v in A A. This proves the lower bound u 0 for the unknown function u. To obtin n upper bound, let w = +b(x 2 +y 2 ) nd suppose L(w) L(u) 4b 1. To ensure tht w u on A, we need to hve 2b. Once we now tke = 2b = 1/2, we find tht u w = (2 x 2 y 2 )/4. 12/38

13 A priori estimtes Theorem 5. A priori estimtes Let A be n open, bounded subset of R 2 nd suppose u stisfies { } L(u) = f in A. u = 0 on A Then there exists positive constnt C such tht mx A A u C sup f. A More generlly, suppose tht L(u 1 ) = f 1 nd L(u 2 ) = f 2 in A. If the functions u 1,u 2 re equl on A, then the theorem implies tht mx u 1 u 2 C sup f 1 f 2. A A In prticulr, the solution of L(u) = f depends continuously on f. A 13/38

14 A priori estimtes: Proof We my ssume tht x 0 x x 1 for ll points (x,y) A. Let w(x,y) = c(e λx 1 e λx ) for some constnts c,λ 0 to be chosen lter nd note tht L(w) = cl(e λx ) = cλ( 1 λ+b 1 )e λx ce λx 0 for ll lrge enough λ. Setting c = e λx 0 sup A f, we now get L(w) sup f L(u) = L(w) L(u) L(w). A Since w u w on A, this implies w u w in A, so u(x,y) w(x,y) ce λx 1 = e λ(x 1 x 0 ) sup f. A 14/38

15 Second comprison principle Theorem 6. Comprison principle for the het eqution Let k > 0 nd b R. Suppose tht u,v stisfy the eqution u t ku xx +bu x = 0 for ll x [x 0,x 1 ] nd ll t 0. If u v when x = x 0,x 1 nd when t = 0, then u v t ll points. This result compres u, v over n unbounded domin. Our previous comprison principle ws only vlid over bounded domin. It is esy to check tht the given eqution hs seprble solutions u(x,t) = e (b2 +4k 2 m 2 )t 4k u(x,t) = e (b2 +4k 2 m 2 )t 4k e bx 2k sin(mx),,m R, e bx 2k cos(mx),,m R. 15/38

16 Second comprison principle: Proof Define w(x,y) = v(x,y) u(x,y)+δe λx for some δ,λ > 0. Since L(u) = u t +ku xx bu x = L(w) = δλe λx (kλ b), it is cler tht L(w) > 0 for ll lrge enough λ. Consider the mximum vlue of w over the set [x 0,x 1 ] [0,T]. If it is ttined t either n interior point or point with t = T, then w x = 0, w xx 0, w t 0 t tht point nd this gives the contrdiction L(w) 0. Thus, the mximum is ttined on the remining prt of the boundry nd v u w mx x=x 0,x 1 t=0 (v u+δe λx ). Once we now let δ 0, we my finlly conclude tht v u mx x=x 0,x 1 t=0 (v u) 0 t ll points. 16/38

17 Second comprison principle: Exmple Consider the boundry vlue problem for the het eqution u t u xx = 0 for ll 0 x π nd t 0 subject to the boundry nd initil conditions u(0,t) = u(π,t) = 0, u(x,0) = f(x). We ssume tht there exist constnts α β such tht αsinx f(x) βsinx for ll 0 x π. In view of this ssumption, it is esy to see tht the inequlity αe t sinx u(x,t) βe t sinx holds when t = 0 nd lso when x = 0,π. Since ll three functions stisfy the het eqution, the inequlity holds for ll x,t. It esily follows tht u(x,t) tends to zero s t. 17/38

18 Wve eqution, pge 1 To solve the wve eqution u tt = c 2 u xx, we express it in the form ( ) ( 2 2 c2 t2 x 2 u = 0 = t +c )( x t c ) u = 0. x This fctoristion suggests the use of vribles α, β such tht α = t +c x, β = t c x. Using these new vribles, one my now integrte twice to get α β u = 0 = β u = f(β) = u = f(β)dβ +G(α) = F(β)+G(α) for some functions F,G. Thus, it remins to determine α nd β. 18/38

19 Wve eqution, pge 2 Bsed on our computtion bove, the vribles α, β should stisfy α = t +c x, β = t c x. According to the chin rule, however, they must lso stisfy α = t α t + x α x, β = t β t + x β x. This gives t α = 1 = t β nd x α = c = x β, so one my tke t = α+β, x = cα cβ. In prticulr, the vribles α,β tht we seek my be defined s α = ct+x 2c, β = ct x. 2c 19/38

20 D Alembert s formul Theorem 7. Generl solution Consider the wve eqution u tt c 2 u xx = 0, where c > 0 is fixed constnt. Every solution of this eqution must hve the form u(x,t) = F(x ct)+g(x+ct). Theorem 8. D Alembert s formul Consider the wve eqution u tt c 2 u xx = 0, where c > 0 is fixed constnt. There is unique solution tht stisfies u(x, 0) = ϕ(x) nd lso u t (x,0) = ψ(x) for ll x R. In fct, this unique solution is u(x,t) = ϕ(x ct)+ϕ(x+ct) c x+ct x ct ψ(s) ds. 20/38

21 D Alembert s formul: Proof Every solution of the wve eqution must hve the form u(x,t) = F(x ct)+g(x+ct), u t (x,t) = cf (x ct)+cg (x+ct). To ensure tht the initil conditions hold, we must lso hve ϕ(x) = F(x)+G(x), ψ(x) = cf (x)+cg (x). Eliminting F (x) gives cϕ (x)+ψ(x) = 2cG (x), hence lso G(x) = ϕ(x) 2 F(x) = ϕ(x) c + 1 2c x 0 0 x ψ(s)ds+d 1, ψ(s)ds d 1. Since u(x,t) = F(x ct)+g(x+ct), the result now follows. 21/38

22 Chnge of vribles, pge 1 We use the sme pproch s before in order to solve the eqution u tt 2u xx u xt = u t 2u x. Once gin, the key step is to write the given eqution s ( t2 x 2 x t t +2 ) u = 0 x nd then fctor the left hnd side. In this cse, we use the formul t 2 2x 2 xt t+2x = (t 2x)(t+x 1) tht one obtins by finding the roots of the qudrtic on the left. According to this formul, we my express the given eqution s ( t 2 )( x t + ) x 1 u = 0. 22/38

23 Chnge of vribles, pge 2 Our next step is to introduce vribles α,β such tht α = t 2 x, β = t + x. According to the chin rule, these vribles must stisfy α = t α t + x α x, β = t β t + x β x. This gives t α = t β = x β = 1 nd x α = 2, so one my tke t = β +α, x = β 2α. In prticulr, the desired vribles α,β my be defined s α = t x 3, β = 2t+x. 3 23/38

24 Chnge of vribles, pge 3 It remins to use the new vribles α,β in order to solve ( ) ( ) α β 1 u = 0 = β 1 u = f(β). This is relly first-order liner ODE which implies tht ( ) e β u = e β f(β) = e β u = e β f(β)dβ +G(α) β = u = F(β)+e β G(α). Reclling the definitions of α, β nd simplifying, we finlly get ( ) ( ) 2t+x u(x,t) = F 1 +e 2t+x t x 3 G = F 2 (2t+x)+e x G 2 (t x). 24/38

25 Energy functions An energy function for time-dependent problem is non-negtive function E(t) which is either constnt or else decresing in time. Energy functions re usully employed for proving the uniqueness of solutions nd lso for deriving priori estimtes. An energy function for the wve eqution u tt c 2 u xx = 0 is ( ) 1 E(t) = 2 u t(x,t) 2 + c2 2 u x(x,t) 2 dx. Suppose tht u(x,t) vnishes t the endpoints x =,b. Then its energy E(t) is conserved becuse n integrtion by prts gives E (t) = = (u t u tt +c 2 u x u xt )dx u t (u tt c 2 u xx )dx = 0. 25/38

26 Applictions of energy Theorem 9. Uniqueness of solutions Let f,g,h,ϕ,ψ be given functions nd consider the eqution u tt c 2 u xx = f(x,t), where x b nd t 0. This eqution ffords t most one solution u(x, t) which stisfies u(,t) = g(t), u(b,t) = h(t), u(x,0) = ϕ(x), u t (x,0) = ψ(x). Theorem 10. Energy estimte Suppose tht u(x,t) stisfies the bove equtions with g = h = 0. Letting E(t) denote the energy for the wve eqution, one hs t E(t) E(0)+ f(x,s) L 2 ds. 0 26/38

27 Uniqueness of solutions: Proof Suppose u,v re both solutions. Then w = u v stisfies w tt c 2 w xx = 0, where x b nd t 0 subject to the boundry nd initil conditions w(,t) = w(b,t) = w(x,0) = w t (x,0) = 0. Since the ssocited energy E(t) is conserved, one hs E(t) = = ( ) 1 2 w t(x,t) 2 + c2 2 w x(x,t) 2 dx ( ) 1 2 w t(x,0) 2 + c2 2 w x(x,0) 2 dx = 0. This implies tht w t = w x = 0 t ll points, so w is constnt. In prticulr, w = 0 t ll points nd thus u = v t ll points. 27/38

28 Energy estimte: Proof First of ll, we integrte by prts to find tht E (t) = = (u t u tt +c 2 u x u xt )dx u t (u tt c 2 u xx )dx = u t f dx. Using the Cuchy-Schwrz inequlity, we conclude tht E (t) u t L 2 f L 2 2E(t) f L 2. This lso implies the desired energy estimte becuse E(t) E(0) = t 0 E (s)ds 2 E(s) t 0 f(x,s) L 2 ds. 28/38

29 Energy functions: Exmple 1 To find n energy function for given eqution, one my multiply the eqution by u or u x or u t, for instnce, nd then look for terms which re either perfect derivtives or else non-negtive. In the cse of the wve eqution u tt c 2 u xx = 0, one finds tht 0 = u t u tt c 2 u t u xx = 1 ( ) u 2 2 t t c2 (u t u x ) x +c 2 u tx u x = 1 ( ) u 2 2 t t c2 (u t u x ) x + c2 ( ) u 2 2 x t. Collecting the terms which re perfect time derivtives, we now get E(t) = ( ) 1 2 u2 t + c2 [ ] b 2 u2 x dx = E (t) = c 2 u t u x. In prticulr, E(t) is conserved, s long s u t = 0 when x =,b. 29/38

30 Energy functions: Exmple 2 For the het eqution u t ku xx = 0, multipliction by u gives 0 = uu t kuu xx = 1 2 ( u 2 ) t k(uu x) x +ku 2 x. We my thus define n energy function by setting E(t) = 1 2 u 2 dx = E (t) = k [ ] b u 2 xdx+ kuu x. This energy function is decresing, s long s u = 0 when x =,b. Another energy function for the het eqution is given by I(t) = k 2 u 2 xdx. It rises in similr wy when one multiplies with u t insted of u. 30/38

31 Blow-up theorems Theorem 11. First blow-up theorem Let ε,t > 0 nd p > 1 be given. Suppose y(t) is positive function which is continuously differentible on the intervl [0, T) with y (t) εy(t) p for ll 0 t < T. Then its mximl existence time T is necessrily finite. Theorem 12. Second blow-up theorem Let ε,t > 0 be given. Suppose y(t) is positive function which is twice continuously differentible on the intervl [0, T) with y(t)y (t) (1+ε)y (t) 2 for ll 0 t < T. If y (0) > 0, then the mximl existence time T is necessrily finite. 31/38

32 First blow-up theorem: Proof First of ll, we seprte vribles nd integrte to get y (t) εy(t) p = t 0 y (s)ds y(s) p εt. Using the substitution z = y(s), we conclude tht εt y(t) y(0) dz z p dz y(0) z p = y(0)1 p p 1. This lredy shows tht the mximl existence time is finite. It lso shows tht y(t) becomes infinite in finite time. Nmely, εt my only rech the vlue on the right hnd side, if the lst two inequlities become equlities nd this is only the cse when y(t). 32/38

33 Second blow-up theorem: Proof The function z(t) = y(t) ε is esily seen to stisfy z (t) = εy(t) ε 1 y (t), z (t) = εy(t) ε 2[ (1+ε)y (t) 2 y(t)y (t) ]. Using the given inequlity nd integrting, we conclude tht z (t) 0 = z (t) z (0) = z(t) z(0)+z (0)t. Since z(t) is positive nd z (0) < 0, this lso implies tht 0 z(t) z(0)+z (0)t = t z(0)/z (0). In fct, t my only rech the vlue on the right hnd side, if z(t) becomes zero nd this is only the cse when y(t). 33/38

34 Conserved quntity Let c > 0 nd consider the nonliner wve eqution u tt c 2 u xx = f(u), where x b nd t 0. If F(u) is n ntiderivtive of f(u), then the expression C(t) = ( ) 1 2 u t(x,t) 2 + c2 2 u x(x,t) 2 F(u(x,t)) dx is conserved for ll solutions u(x,t) tht vnish when x =,b. To see this, one my simply integrte by prts to find tht C (t) = = ( ut u tt +c 2 u x u xt f(u)u t ) dx u t ( utt c 2 u xx f(u) ) dx = 0. 34/38

35 Blow-up of nonliner wves Theorem 13. Nonliner wves Let p > 1 nd consider the nonliner wve eqution u tt c 2 u xx = u p 1 u, where x b nd t 0 subject to the boundry nd initil conditions u(,t) = u(b,t) = 0, u(x,0) = ϕ(x), u t (x,0) = ψ(x). Suppose tht ϕ, ψ re positive nd the conserved quntity C(t) = ( ) 1 2 u t(x,t) 2 + c2 2 u x(x,t) 2 u(x,t) p+1 dx p+1 is negtive. Then the solution u(x,t) my only exist for finite time. 35/38

36 Nonliner wves: Proof, pge 1 We pply our second blow-up theorem to the function y(t) = 1 2 u(x,t) 2 dx = y (t) = uu t dx. Since u tt c 2 u xx = u p 1 u, n integrtion by prts gives y (t) = = = u 2 t dx+ uu tt dx u 2 t dx+c 2 uu xx dx+ u 2 t dx c 2 u 2 xdx+ u p+1 dx u p+1 dx. We now relte this expression to the conserved quntity C(t). 36/38

37 Nonliner wves: Proof, pge 2 According to our computtion bove, one hs y (t) = u 2 t dx c 2 u 2 xdx+ u p+1 dx. There is lso conserved quntity C(t), which is given by C(t) = 1 2 u 2 t dx+ c2 2 u 2 xdx 1 p+1 nd it is negtive by ssumption. It esily follows tht y (t) y (t)+(p+1)c(t) = p+3 2 p+3 2 u 2 t dx+ c2 (p 1) 2 u 2 t dx. u p+1 dx u 2 xdx 37/38

38 Nonliner wves: Proof, pge 3 In view of our definition of y(t), we hve ctully shown tht y(t)y (t) p+3 4 u 2 dx u 2 t dx. We my thus use the Cuchy-Schwrz inequlity to find tht y(t)y (t) p+3 4 ( ) 2 uu t dx = p+3 y (t) 2. 4 Since p > 1, the coefficient on the right hnd side is bigger thn 1 nd our second blow-up theorem is pplicble. Noting tht y (0) = ϕ(x)ψ(x)dx > 0, we conclude tht the mximl existence time is finite. 38/38