Heat Equation: Maximum Principles

Transcription

1 Heat Equatio: Maximum Priciple Nov. 9, 0 I thi lecture we will dicu the maximum priciple ad uiquee of olutio for the heat equatio.. Maximum priciple. The heat equatio alo ejoy maximum priciple a the Laplace equatio, but the detail are lightly differet. Recall that the domai uder coideratio i Ω T Ω [0, T); Ω T (Ω 0}) ( Ω [0, T]). () We have the followig trog maximum priciple. Theorem. (Maximum priciple of the heat equatio) Aume u C (Ω T ) C(Ω T) olve i Ω T. i. (Weak maximum priciple) The u t u=0 () maxu=maxu. (3) Ω T Ω T ii. (Strog maximum priciple) Furthermore, if Ω i coected ad there exit a poit (x 0,t 0 ) Ω T uch that the u i cotat i Ω t 0. u(x 0, t 0 ) = maxu, (4) Ω T Remark. Ituitively, the maximum priciple ca be explaied by the followig obervatio. Recall that u(x, t) = Φ(x y, t ) u(x, )dy (5) R for ay < t, ad the oegative fuctio Φ atifie R Φ(x y, t ) dy =, (6) it i clear that u(x, t) at time t i obtaied from averagig u(y, ) at a earlier time, ad therefore the maximum hould be decreaig. We preet two proof of thi theorem. The firt i through clever dicuio of ig, the ecod i through etablihig a mea value property. Note that the firt method oly yield the weak maximum priciple, that i the maximum iide i bouded by that o the boudary, itead of the trog maximum priciple, that i the maximum ca oly be attaied at the boudary, ule the fuctio i a cotat. Proof. Theorem 3. (Weak maximum priciple) Let Ω R be ope ad bouded. Let u C (Ω T ) C 0 (Ω T), ad atify i Ω T, we the have (If T <, the upreme ca be replaced by maximum). u t u 0 (7) upu= up u. (8) Ω T Ω T. Note that Ω t0 caot be replaced by Ω T, a ca be ee by the followig coideratio. Let u olve the heat equatio with iitial value M (cotat) ad (ide) boudary value g(x) M. Now if u t 0 a t ց0, we ca exted u to t<0 by ettig u M.

2 Heat Equatio: Maximum Priciple Remark 4. The coectio to the theory of Laplace equatio ca be ee by writig the iequality ito u u t. (9) However it i ot clear how to really ue thi fact a u t 0 oly hold at (x 0, t 0 ). Proof. If uffice to prove the cocluio for Ω T for all T < T. Fix oe uch T. By compacte of Ω T we kow that the maximum i attaied at ome poit (x 0, t 0 ).. Firt coider the cae Now if (x 0, t 0 ) Ω T, at (x 0, t 0 ), we have u > u t i Ω T. (0) a) u t (t 0 ) 0 becaue otherwie u(x 0, t 0 ε) >u(x 0, t 0 ) for ε mall eough; b) u(x 0, t 0 ) 0 becaue x 0 maximize u o Ω t 0 }. But the two cocluio cotradict each other.. For the geeral cae we coider a auxiliary fuctio We eaily check ad ca apply. Fially lettig ε ց 0 we obtai the cocluio. u u t, () v(x, t) =u(x, t) ε t. () v v t (3) From weak maximum priciple oe immediately obtai the uiquee for ohomogeeou iitial/boudary-value heat equatio whe the domai Ω i bouded. The ituatio i more complicated whe Ω i ot bouded. Theorem 5. (Weak maximum priciple for Ω = R ) Suppoe the u t u 0 i Ω T ; u(x, t) Me λ x i Ω T for M,λ>0; u(x, 0)= g(x), (4) upu upg. (5) Ω T R Remark 6. The reao why we eed a boud o the growth rate at ifiity i the followig. Recall that the effect of the heat operator i to obtai olutio at later time from averagig olutio at earlier time via a kerel of the form e x /4. Note that the kerel decay at ifiity a e c x. Therefore, if the olutio grow fater tha it ivere rate e λ x, the it i poible to iput extra tuff from ifiity ad thu violate uiquee. Here i a example of how oe ca tore eergy at ± ad the iput it through averagig by the heat kerel, ad thu obtai a ozero olutio with zero iitial data. Set u(x, t) g () (t) ()! x (6) 0 with g(t) e t k t > 0, k >. (7) 0 t = 0. Oe ca verify that u atifie the equatio by formally differetiatig iide the ummatio.. The reao we ue T < T here i to coclude u t 0 at the maximizer. Sice the equatio i olved o Ω T oly, u t at t=t i ot kow.

3 Nov. 9, 0 3 With ome calculatio 3, oe ca further how that the erie o the RHS i majorized by the Taylor expaio of [ ( x exp U(x, t) = t θ )] t k t > 0. (8) 0 t 0 for ome θ > 0. Note that U(x, t) caot be bouded by M e λx for ay cotat M,λ. Proof. Firt we divide (0, T) ito ubiterval with ize r <. It uffice to prove the claim o a ubiterval. From ow o we aume T < 4 λ 4 λ Coider the auxiliary fuctio ( x v(x, t) u(x, t) δ (4 π (T +ε t)) e 4(T +ε t) / ). (9) where ε i choe uch that T + ε < 4 λ. It i eay to verify that The trategy i the how firt v t v 0. (0) v(x, t) upg () R for ay (x, t) ad the lettig δ ց 0. We firt otice that, o the phere x y = R, we have v(x, t) Me λ( x +R) δ (4 π (T + ε t)) / e ( R 4(T +ε t) ) up R g () oce R i big eough. Now apply the weak maximum priciple o the domai B R (0, T), we obtai v(x, t) up g. Proof Mea value property. The mea value property for the heat equatio tur out to be much more complicated tha the oe for the Laplace equatio. We firt defie the heat ball : N Defiitio 7. (Heat ball) For fixed x R, t R, r > 0, we defie E(x, t; r) (y, ) R + t, Φ(x y, t ) r }. (3) Remark 8. We try to gai ome idea of what E(x, t; r) look like. It i defied by Thu i the time directio, we have (4π (t )) / e x y t t r 4 π where the lower boud i from the fact that e x y 4(t ). Furthermore 4(t ) r. (4) (5) E(x, t; r) = 0}=E(x, t; r) =t}=x}. (6) Next, to fid out the correct formula, we eed to fid a kerel K(x y, t ) uch that K(x y, t )dy d (7) E(x,t;r) 3. See pp. 3 i F. Joh Partial Differetial Equatio, 4ed. The cotat θ i determied i Problem 3 o p. 73 of the ame book.

4 4 Heat Equatio: Maximum Priciple i idepedet of r. Notice that where Thu we have Thi implie whe (4 π (t )) E(x,t;r) / e x y 4(t ) r Oe ca omehow 4 verify that 5 4 E(0,0;) Thu fially we have the correct formulatio. (4 π (t )) / e x y 4(t ) (8) t = t/r, =/r, x = x/r, y = y/r. (9) K(x y, t )dy d = K(x y, t )r + dy d. (30) E(x,t ;) K(x, t )=K(x, t) r + (3) x =x/r, t = t/r. (3) x dxdt =. (36) t Theorem 9. (Mea value property for the heat equatio) Let u C (Ω T ) olve the heat equatio, the for each E(x, t; r) Ω T. u(x, t)= x y 4 r E(x,t;r) u(y, ) dy d. (37) (t ) Proof. Without lo of geerality we ca et (x, t)=(0, 0) ad deote E(0, 0; r) by. Defie φ(r) r u(y, ) y dy d= u(ry, r ) y E dy d. (38) I the followig we will omit the prime ad imply ue y ad y. thu we compute φ (r) = d [ ] u(ry, r ) y dy d dr E y = ( yi u) y i + r ( u) y dy d i= = y r + ( yi u) y i + r + ( u) y i= A+B. (39) 4. I have o idea how to come up with thi particular fuctio. 5. A the computatio i tediou ad log, we put it i a a footote ad furthermore oly preet the mai tep. Firt otice x E(0,0;) e 4t / }= 0 t (4 π t) 4 π ; x ( t) log }. (33) 4 π t Thu the itegral become [ /4π t 0 x tlog 4πt x dx ] dt= π/ + (+)/ (+)/ Γ ( 4π + ) 0 ( ) t log 4 π t + dt. (34) where we have ued polar coordiate ad the formula α() = π/ Γ ( for the volume of -dimeio ball. Now ettig + ) =4πt ad uig the formula ( λ z Γ(z)= t λ log ) z dt; Γ(z + ) = z Γ(z), (35) 0 t we will ee after careful calculatio that mot term cacel out ad what remai i 4.

5 Nov. 9, 0 5 Now we itroduce ad ote that ψ = 0 o. Now we have ψ B = r + = r + log ( 4π)+ y 4 ( u) i= + log r, (40) y i ( yi ψ)dy d 4 u ψ +4 u yi y i ψ dy d. (4) i= Now itegrate by part w.r.t., we have B = r + 4u ψ + 4 u yi y i ψ dy d i= = r + 4u ψ + 4 ( u yi y i ) y 4 dy d = r + uyi y i dy d A. (4) Coequetly, uig the equatio, we have 4u ψ φ (r) = A+B = r + 4 uψ uyi y i dy d = r + 4 u yi ψ yi u y i y i dy d = 0. (43) Thu we have ( ) y φ(r)= limφ(t)=u(0, 0) lim t 0 t 0 t dy d = 4 u(0, 0). (44) E t From the mea value property we immediately have the trog maximum priciple: Proof. (of the trog maximum priciple) Suppoe there i (x 0, t 0 ) Ω T uch that u(x 0, t 0 ) = max Ω Tu, the by pickig r mall eough o that E(x 0, t 0 ; r) Ω T, ad uig the mea value property, we coclude that u i cotat iide E(x 0, t 0 ; r). Next for ay (y 0, 0 ) Ω T uch that the lie egmet coectig x 0, y 0 i i Ω T, we ca how that u(y 0, 0 )=u(x 0,t 0 ) wheever 0 <t 0 by coverig the lie egmet coectig (y 0, 0 ) ad (x 0,t 0 ) with the heat ball. Fially, ice Ω i coected, ay y 0 ca be coected from x 0 via fiitely may lie egmet. Ad therefore u(y, )=u(x 0, t 0 ) for all y Ω, < t 0. Remark 0. It tur out there i alo a verio of mea value property ivolvig oly the urface itegral over E(x, t; r). The idea i to compute the itegral of [Φ(x y, T t) r ] (u t u) over E(x, t; r) via itegratio by part. Thi will give a repreetatio formula for u(x, t), which tur out to be u(x, t)= x y r E(x,t;r) u(y, ) ds. (45) (t ) See J. Jot Partial Differetial Equatio, pp for detail. The trog maximum priciple ca alo be proved uig thi formula. See Theorem 4..3 o p.86 of J. Jot Partial Differetial Equatio.. Uiquee. A i the elliptic cae, uiquee i etablihed through maximum priciple.

6 6 Heat Equatio: Maximum Priciple Theorem. (Uiquee o bouded domai) Let g C( Ω T ), f C(Ω T ). The there exit at mot oe olutio u C (Ω T ) C(Ω T) of the iitial/boudary value problem I the cae Ω = R, we have to add extra coditio. u t u = f i Ω T ; u = g o Ω T. (46) Theorem. (Uiquee whe Ω = R ) Suppoe g C(R ), f C(R [0, T]). The there exit at mot oe olutio u C (R (0, T]) C(R [0, T]) of the iitial value problem atifyig the growth etimate for cotat M,λ>0. u t u= f i R (0, T); u= g o R t = 0} (47) u(x, t) M e λ x (48) Remark 3. There are i fact ifiitely may phyically icorrect olutio, which do ot atify the growth boud, to the zero iitial value problem. See Chapter 7 of F. Joh Partial Differetial Equatio. Exercie. Exercie. Let u C (Ω T ) C(Ω T) be a olutio to u t u=0 i Ω T ; u = 0 o Ω [0, T]; u = g o Ω t =0}, (49) where g 0. The u i poitive everywhere withi Ω T if g i poitive omewhere o Ω.