Q[a, b](t) = Combining these two assumptions we have, for any bounded interval [a, b]: u(x, t)dx = k(u x (b, t) u x (a, t)).

Transcription

1 THE ONE-DIMENSIONAL HEAT EQUATION. 1. Derivtion. Imgine dilute mteril species free to diffuse long one dimension; gs in cylindricl cvity, for exmple. To model this mthemticlly, we consider the concentrtion of the given species s function of the liner dimension nd time: u(x, t), defined so tht the totl mount Q of diffusing mteril contined in given intervl [, b] is equl to (or t lest proportionl to) the integrl of u: Q[, b](t) = u(x, t)dx. Then the conservtion lw of the species in question sys the rte of chnge of Q[, b] in time equls the net mount of the species flowing into [, b] through its boundry points; tht is, there exists function d(x, t), the diffusion rte from right to left, so tht: dq[, b] dt = d(b, t) d(, t). It is n experimentl fct bout diffusion phenomen (which cn be understood in terms of collisions of lrge numbers of prticles) tht the diffusion rte t ny point is proportionl to the grdient of concentrtion, with the right-to-left flow rte being positive if the concentrtion is incresing from left to right t (x, t), tht is: d(x, t) > when u x (x, t) >. So s first pproximtion, it is nturl to set: d(x, t) = ku x (x, t), k > ( Fick s lw of diffusion ). Combining these two ssumptions we hve, for ny bounded intervl [, b]: d dt u(x, t)dx = k(u x (b, t) u x (, t)). Since b is rbitrry, differentiting both sides s function of b we find: u t (x, t) = ku xx (x, t), the diffusion (or het) eqution in one dimension. (In the cse of het we tke u(x, t) to be the temperture, nd ssume there is function c(x) > throughout the conducting medium so tht the therml energy dq dded to the system by lump of mteril of mss 1

2 ρ(x)dx (where ρ(x) is the mss density) t temperture u(x, t) is dq = c(x)ρ(x)u(x, t)dx. Tking c, ρ to be constnt, we re bck in the erlier sitution.) We ll consider four types of problems for the het eqution: (i) The Cuchy problem for the eqution on the whole rel line, where the initil temperture (or concentrtion) u (x) is given nd we seek u(x, t), the solution giving its evolution in time; (ii) Boundry-vlue problems on the hlf-line x >, where we ssume either the temperture is held constnt t x = (so het flows in or out of the system t the origin), or tht there is no diffusion of het t x = (so u x = t the origin.) (iii) Boundry-vlue problems on bounded intervl [, L], or periodic boundry conditions on [ L, L]. (iv) Non-homogeneous problems, corresponding to het source inside the conducting mteril: u t ku xx = f(x, t), on the whole line or on n intervl. Bsic observtions: (i) The time-independent solutions of the het eqution re liner functions, u = Ax + B. By subtrcting n pproprite liner function, for the boundry condition where u is held constnt t the endpoints we cn lwys ssume the constnts re zero (Dirichlet BC). For Neumnn-type boundry conditions u x =, or periodic, the constnts re lwys solutions. (iii) In generl we expect the eqution to grdully smoothe out ny oscilltions in u nd drive it towrds the simplest time-independent solution consitent with the boundry conditions (liner, or constnt.) If the limit s t is constnt, it hs to be zero for Dirichlet BC, nd the verge vlue of the initil dt u, for Neumnn or periodic BC on bounded intervl. The reson is tht the integrl of u over the whole intervl (nd therefore its verge vlue) is constnt in time (under these BC): d dt u(x, t)dx = u t (x, t)dx = k under Neumnn BC, nd similrly under periodic BC. u xx (x, t)dx = k(u x (L, t) u x (, t)) =, (iv) Scling. The chnge of vribles t s = kt, x x chnges the eqution u t = ku xx to u s = u xx ; the chnge of vribles t t, x y = kx chnges it to u t = u yy. The simultneous chnge t s = λt, x y = λx (with λ > ) leds bck to the originl eqution: u s = ku yy (verify these 2

3 sttments). Thus we see tht one time dimension corresponds to two spce dimensions, in the sense tht ny scling of the vribles tht doesn t chnge the rtio x/ t lso leves the eqution unchnged. We lso see tht one cn lwys ssume the constnt k equls one, by rescling the time vrible. We will usully do tht, nd the correct expressions for the eqution with generl k > cn be recovered simply by mking the chnge t kt. 2. The het kernel on the rel line. 2.1 Derivtion. To look for exct solutions of u t = u xx on R (for t > ), we remember the scling fct just observed nd try to find solutions of the form: u(x, t) = p( x ), p = p(y). t The het eqution quickly leds to the ODE for p(y): p (y) = y 2 p (y), nd setting q(y) = p (y), we find first-order liner ODE with n esily derived generl solution: q (y) = y 2 q(y) q(y) = Ce y2 /4, C >, y R. Thus the function P (x, t) below is solution of the het eqution on the rel line: P (x, t) = x/ T e 2 /4 dy = 1 2 Now recll the well-known fct from Clculus: π e p2 dp = 2 x/ nd the relted definition of the Error Function : ERF (z) = π z This function hs the properties: e p2 dp. e p2 dp, z R. lim ERF (z) =, lim ERF (z) = 1, ERF () = 1 z z 2, ERF (z) 1 2 is n odd function of z. 3

4 And so insted of using P (x, t) we define: H(x, t) = ERF ( x ), solution of the het eqution u t u xx =. The normliztion constnts re chosen so tht the limit of this solution s t + is (pointwise in x) the (discontinuous!) function θ(x) defined by: θ(x) =, x < ; θ() = ERF () = 1, θ(x) = 1, x >. 2 (Engineers will recognize this s Heviside s unit step function t zero.) Unfortuntely we only found n exct solution of the het eqution given s n integrl, so just in cse we check whether u(x, t) = exp( x 2 /) is solution of u t = u xx nd find tht it is not. Computing its integrl over the whole rel line, we find (check!): This moves us to consider the function: e x2 dx = 4πt. h(x, t) = 1 4πt e x2, nd it turns out this one works (nd is clled the het kernel on the rel line): Exercise: Show tht h(x, t) is solution of u t = u xx for x R, t >, nd stisfies: h(x, t)dx = Reltion with the norml distribution. To understnd the behvior of h(x, t), we note tht its grph is n even bell-shped curve centered t nd with thickness of the pek pprently relted to t (shrper pek, the smller t > is.) Recll the stndrd norml probbility density function from Probbility: p(x;, 1) = 1 e x2 2, 2π s well s the more generl norml probbility density function with men vlue µ R nd stndrd devition σ > : p(x; µ, σ) = 1 σ (x µ) 2 2π e 2σ 2. 4

5 Its grph is the bell-shped curve with pek vlue t x = µ nd pek width proportionl to σ. Its integrl over the whole rel line is one, for ll µ R, σ >. If you ve studied some probbility (nd you should), you ll recll tht for normlly distributed rndom vrible X tking vlues on R, with expected vlue (men) µ nd stndrd devition σ (or vrince σ 2 ), we hve the probbilities: P r µ,σ [ X b] = p(x; µ, σ)dx. The cumultive distribution function for p(x;, 1) is exctly ERF! P r,1 [X z] = z nd with men µ nd stndrd devition σ: P r µ,σ [X x] = x p(x;, 1)dx = ERF (z), p(y; µ, σ)dy = ERF ( x µ σ 2 ). The connection with the het eqution then is: we hve the correspondence σ 2 2t (time corresponds to vrince). For the het kernel nd norml probbility density function.: p(x;, 1) = h(x, 1 2 ); σ2 p(x; µ, σ) = h(x µ, 2 ). For the integrted solution H(x, t) nd norml cumultive distribution function: ERF (z) = H(z, 1 µ ); ERF (x 2 σ σ2 ) = H(x µ, 2 2 ). 2.3 Solutions for step-function initil dt. It is esy to use linerity nd the trnsltion invrince (u(x, t) is solution of the het eqution if u(x, t) is) to write down the solutions of u t = u xx with for simple discontinuous initil dt. For exmple, consider the function: θ,b (x) =, x <, θ,b (x) = 1, < x < b, θ,b (x) =, x > b, θ,b () = θ,b (b) =

6 Then θ,b (x) = θ (x) θ b (x), where θ (x) = θ(x ) is the Heviside unit step function with jump discontinuity t x =. It is esy to see tht: H,b (x, t) = H(x, t) H(x b, t) = ERF ( x ) ERF ( x b ) is solution of u t u xx = on the rel line, converging pointwise to θ,b (x) s t +. This generlizes to rbitrry step functions : tke prtition N N+1 of the intervl [ 1, N+1 ] into N djcent sub-intervls, nd given rel constnts c 1,..., c N define: u (x) = N c i θ i, i+1 (x). i=1 This is piecewise-constnt function with jump discontinuities t the i, nd vnishing outside [ 1, N+1 ]. The solution of the het eqution with this initil function is simply: u(x, t) = N c i H i, i+1 (x, t) = i=1 N i=1 c i (ERF ( x i ) ERF ( x i+1 )). Exercise: Show tht for ech i = 2,..., N, we hve: lim t + u( i, t) = c i 1 +c i Solution of the Cuchy problem. We first recll some bsic definitions in Anlysis. Let f : I R be function, where I R is n open intervl. Definition. Let x I. f is continuous t x I if: ( ɛ > )( δ > )( x y < δ, y I f(x) f(y) < ɛ). Note tht, in generl, δ depends both on ɛ nd on x. f is uniformly continuous on I if: ( ɛ > )( δ > )( x, y I)( x y < δ f(x) f(y) < ɛ). In contrst, in this cse δ depends only on ɛ; the sme δ works for ll x nd y. It follows from the Men Vlue Theorem tht if f is continuous nd bounded in I, then f is uniformly continuous on I. (If you ve studied some Anlysis, you cn esily verify this.) 6

7 We sy f(x, t) f (x) pointwise on I s t + if, for ech x I, we hve lim t + f(x, t) = f (x). This mens: ( x I)( ɛ > )( τ > )( < t < τ f(x, t) f (x) < ɛ). Here τ in generl depends both on ɛ nd on x. We sy f(x, t) f (x) uniformly on I if lim t + f(x, t) f (x) =, where we use the sup norm for bounded functions on I (tht is, g is the smllest positive M so tht g(x) M for ll x I.) Equivlently, uniform convergence mens: ( ɛ > )( τ > )( < t < τ ( x I)( f(x, t) f (x) < ɛ)). Here τ depends on ɛ, but not on the point x in I. Theorem. Let u be continuous, bounded function on R. Denoting by h(x, t) the het kernel on R, consider the function u defined by n improper integrl: u(x, t) = h(x y, t)u (y)dy. We hve: (i) If u is continuous t x R, then u(x, t) u (x) s t + (pointwise convergence.) (ii) If u is uniformly continuous on R, then u(x, t) u (x) s t +, uniformly on R. (iii) We hve, for ech x R: u x (x, t) = h x (x y, t)u (y)dy. The sme holds for u xx nd u t. Since h t (x y, t) = h xx (x y, t) for ech y, t, it follows tht u t = u xx, so u is solution of the het eqution. (iv) The function (x, t) u(x, t) is smooth in R R + (tht is, hs continuous prtil drivtives of ll orders in x nd t, for ny (x, t) with t > ), even if u is just continuous. Importnt Remrk. Prt (iv) is sying tht even if the grph of u is jgged (non-differentible) in fct even if it hs jump discontinuities, though we won t prove this the grph of u(, t) is smooth for ny t >. This instntneous smoothing property is in strk contrst to wht we observed for the WE, which propgtes the singulrities of the initil dt for ll time (lng chrcteristic lines.) 7

8 Corollry. Suppose there exist constnts M, N R so tht, for ll x R, we hve N u (x) M. Then the sme inequlities hold t ny time t > : N u(x, t) M, for ll t >. Proof. Just note tht, since the het kernel integrtes to one over the whole rel line, we hve: u(x, t) N = h(x y, t)(u (y) N)dy, nd this must be greter thn or equl to zero, since h(x y, t) > nd u (y) N. the other inequlity is proved in the sme wy. In prticulr, we hve the stbility estimte for the sup norm : for ech t >. u(, t) u, Convergence of improper integrls. The theorem dels with improper integrls depending on prmeter, of the type: f(x, t, y)dy, f : R R R smooth, (x, t) R R R +, rectngle. We recll some definitions nd theorems used in the proof. An improper integrl of this type converges bsolutely for given (x, t) R if: lim [ A,B A B or equivlently if, t the given (x, t): ( ɛ > )( A > )( B > A) B f(x, t, y) dy + f(x, t, y) dy] =, A A B B f(x, t, y) dy + f(x, t, y) dy < ɛ. A Here A depends on ɛ, nd in generl lso on the point (x, t). (The djective bsolutely refers to the bsolute vlue, so it is utomtic if the integrnd is positive.) On the other hnd, we sy the integrl converges bsolutely nd uniformly in R if: ( ɛ > )( A > )( B > A)( (x, t) R) A B B f(x, t, y) dy+ f(x, t, y) dy < ɛ. A 8

9 Tht is, in this cse the sme A works for ll (x, t) R: how fr out we hve to tke the til prts of the grph of y f(x, t, y) to be ɛ-close to the vlue of the integrl depends only on ɛ, not on (x, t). It is theorem of Anlysis tht if f(x, t, y) nd f x (x, t, y) re continuous in R R, nd if the improper integrls f(x, t, y)dx nd f x(x, t, y)dx converges bsolutely nd uniformly in R, then the function in R tking (x, t) to f(x, t, y)dx is continuously differentible in R, nd: f(x, t, y)dx = f x (x, t, y)dx. x Proof of Theorem. For (i) nd (ii), given ɛ > (nd lso x R in cse (i)) we must find τ > so tht u(x, t) u (x) < ɛ if < t < τ, where τ my depend on x in cse (i), but depends only on ɛ in cse (ii). Choose δ > so tht u (x + z) u (x) < ɛ/2 if z < δ. Note δ depends on x nd ɛ in cse (i), but only on ɛ if u is uniformly contiuous (cse (ii)). Now split the integrl defining u(x, t) into two prts (with the chnge of vrible y z = y x, using lso tht h(x, t) is even in x): u(x, t) u (x) = = ( + z >δ h(x y, t)[u (y) u (x)]dy = δ δ )h(z, t)(u (x + z) u (x))dz = A + B, nd from the choice of δ the second integrl my be estimted by: B ɛ 2 δ δ h(z, t)dz < ɛ 2 for ny t >, h(z, t)(u (x+z) u (x))dz since h(z, t)dz = 1 (nd h(z, t) > for ll z, t). To estimte the integrl A, observe tht: h(z, t)dz = 1 e z2 / dz 4πt z >δ = 1 π e y2 y > δ z >δ dy = 2(1 ERF ( δ )) (the reder should verify the lst equlity), so tht: A 4 u (1 ERF ( δ )), 9

10 nd this will be smller thn ɛ/2 for ny < t < τ, provided we choose τ so tht: δ (1 ERF ( )) = ɛ 4τ 8 u. (This is possible since lim z ERF (z) = 1.) We see tht τ depends only on ɛ, u nd δ (hence on u nd ɛ in cse (ii), nd dditionlly on x in cse (i).) This concludes the proofs of prts (i) nd (ii) of the Theorem. To prove prt (iii) using the differentibility criterion reclled bove, we just hve to show tht the integrl: 1 h x (x y, t)u (y) dy = 2t 4πt x y e (x y)2 / u (y) dy is uniformly convergent for (x, t) R [τ, ), for ny τ >. Mking the chnge of vrible y z = (y x)/, this equls: 1 2t π z e z2 u (x + z ) dz < u 2τ π z e z2 dz for t > τ. Now, since z e z2 dz is convergent, given ɛ > we my choose A > (depending on τ nd u ) so tht, for ny B > A: ( A B + B A ) z e z2 dz < 2τ π u 1 ɛ, nd then the sum of the til ends of the previous integrl (defined by limits of integrtion z = A, B nd z = A, B) will be smller thn ɛ, for ny x R nd ny t > τ. This concludes the proof of uniform convergence of the integrl. The proofs for u xx nd u t follow the sme model, using the convergence of the integrl z 2 e z2 dz. This concludes the proof of of prt (iii) of the theorem. For prt (iv), the rgument showing existence nd continuity of higherorder prtil derivtives of u(x, t) in the open hlf-plne {(x, t); t > } is completely nlogous. For instnce, showing the existence nd continuity of the prtil derivtive of order k in x involves (in the sme wy s bove) the convergence of the improper integrl: z k e z2 dz. 1

11 4. The Mximum Principle. It is physiclly resonble tht, since temperture tends to become more uniform under the usul boundry conditions, the mximum nd minimum tempertures sould be ttined either t time zero or on the boundry. This is reflected in n importnt property of solutions of the het eqution, the mximum principle. At the most bsic level, it is consequence of the second derivtive test for functions of two vribles. In the following by upper hlf plne we men the set {(x, t); x R, t }. Clculus mximum principle. A solution of the het eqution cnnot hve nondegenerte locl mximum or minimum in ny open region of the upper hlf-plne. Proof. At nondegenerte locl mx (resp. min) we hve u t = nd u xx < (resp. u xx > ), nd this is incomptible with the eqution u t = ku xx. Let R = [, b] [, T ] be rectngle in the upper hlf-plne. The prbolic boundry of R is the usul boundry of R, except for the top edge: p R = [, b] {} {, b} [, T ]. Mximum principle. Let u(x, t) be solution of the het eqution u t = ku xx in rectngle [, L] [, T ] in the upper hlf-plne. Then the mximum nd the minimum vlues of u in R re ttined on the prbolic boundry p R. In other words: mx u(x, t) = mx u(x, t), min (x,t) R (x,t) pr u(x, t) = min u(x, t). (x,t) R (x,t) pr Proof. Let C > be rbitrry, nd define v(x, t) = u(x, t) + Cx 2. Then v stisfies in R: v t kv xx = u t ku xx 2kC = 2kC <. Thus the mximum vlue of v in R cnnot be ttined t n interior point of R (t such n interior mx, we d hve v t =, v xx, hence v t kv xx ). Note tht now we don t hve to ssume the mx is nondegenerte! If it is ttined t point (x, T ) of the top edge t = T, we still hve v xx (x, T ), 11

12 nd v(x, t) v(x, T ) for t < t, so v t (x, T ). But then v t kv xx t (x, T ), gin contrdiction. Thus mx R v = mx pr v mx pr u + CL 2. Since mx R u = mx R (v Cx 2 ) mx R v, we conclude mx R u mx pru + CL 2. Now let C to conclude mx R u mx pr u. The opposite inequlity is cler, since p R R. Hence they re equl. A similr proof pplies to the minimum. The mximum principle leves open the possibility tht the mximum (or minimum) vlues of u re ttined not just on the prbolic boundry, but lso elsewhere on R. This in fct cnnot hppen (unless u is constnt in R), but is hrder to prove. Strong mximum principle. If u is solution of the het eqution in rectngle R on the upper hlf-plne, the mximum nd minimum vlues of u in R re ttined only on p R, unless u is constnt on R. For proof, see [Protter-Weinberger]. Appliction. The mximum principle immeditely implies uniqueness of solutions of the non-homogeneous Dirichlet problem in [, L]: u t ku xx = f(x, t) in R = [, L] [, T ], u(, t) = g(t), u(l, t) = h(t), u(x, ) = u (x). Just observe tht if u 1 nd u 2 re two solutions of this problem in [, L], their difference w = u 1 u 2 is solution of w t = kw xx in the sme intervl, with zero boundry conditions nd zero initil condition. Thus its mximum nd minimum vlues on p R re both zero, so w on p R, nd by the mximum/minimum principle it follows tht w in R, so u 1 u 2 in R. Question: How would you use the Mximum Principle to show uniqueness for the non-homogeneous Neumnn problem in [, L]? (Tht is, u x (, t) = g(t), u x (L, t) = h(t)). Note tht u x is lso the solution of non-homogeneous het eqution, if u is. 12

13 PROBLEMS. 1. Solve the het eqution u t u xx = on the rel line, with initil exponentil initil dt u (x) = e x, where R is constnt. (Answer: e t2 +x ) Hint: In the integrl with the het kernel, complete squres to obtin: (x y) 2 y = ( x y + t) 2 t 2 x, then mke the chnge of vribles z = x y + t. 2. Solve the het eqution u t u xx =, for the following two initil conditions: (i) u (x) = 1 for x > nd u (x) = 2 for x <. (ii) u (x) = x, x < or x > 4; u (x) = x + 2, < x < 4. (Hint: write u (x) = x + 2(θ (x) θ b (x)) for suitble nd b, then use linerity.) 3. [Struss] Consider the solution 1 x 2 2t of u t u xx =. Find the loctions of its mximum nd minimum points in the closed rectngle { x 1, t T }. 4.[Struss] Consider the diffusion eqution u t = u xx on the intervl [, 1], with initil condition u (x) = 4x(1 x) nd boundry conditions u(, t) = u(1, t) =. (i) Show tht < u(x, t) < 1 for ll t > nd < x < 1. (ii) Show tht u(x, t) = u(1 x, t) for ll t nd x [, 1]. (Hint: uniqueness. (iii) Use the energy method to show tht 1 u2 dx is strictly decresing function of t. 5. [Struss] Solve the diffusion eqution u t = u xx on the rel line, with initil condition: u (x) = e x, x > ; u (x) =, x <. Hint: Use the sme trick s in problem 1. Answer:e t2 +x ERF ( x t). 6.[Struss] Solve the diffusion eqution u t u xx = on the rel line with initil condition: u (x) = x 2 3x + 7. Hint:Show tht u xxx solves the het eqution with initil dt zero, hence (by uniqueness) is the identiclly zero function for ech t >. This mens: u(x, t) = A(t)x 2 + B(t)x + C(t). Now substitute bck in the eqution to find the coefficients A(t), B(t), C(t). 13

14 Remrk. Clerly this method pplies to ny initil condition tht is polynomil function of x. In fct, you cn show tht if u (x) is polynomil of degree d, then so is u(x, t), for ech t >. 5. Stbility nd energy. 5.1 Norms in function spces. A norm in vector spce V is n ssignment v v of nonnegtive rel number to ech vector in v V, stisfying: (i) v = only when v = ; cv = c v, for ech v V nd ech c R; (iii) v + w v + w, for ech v, w V (tringle inequlity.) Exmples. In finite dimensions (tht is, in R n ), the best-known exmple is the eucliden norm 2 : if v = (x 1,..., x n ) R n : v 2 = (x x 2 n) 1/2. For ny p >, we cn define nlogously: v p = ( x 1 p x n p ) 1/p (why do we need the bsolute vlues?) It turns out the p re norms, but only for p 1 (the problem is the tringle inequlity.) The following is lso norm in R n : v = mx{ x 1,..., x n }. Exercise. For ny norm, we hve the unit bll : B = {v V ; v 1}. Sketch the unit blls for the norms in R 2 : 1, 2, 3 nd. Exercise. Show tht in R n : lim p v p = v. (This explins the nottion.) Remrk. It is fct of Anlysis tht, in finite-dimensionl vector spce V, ny two norms nd re equivlent, in the sense tht there exists constnt C > 1 so tht: v V : 1 C v v C v. (In prticulr, convergence of sequence to given limit is independent of the choice of norm.) This is definitely flse in infinite dimensions. (A simple exmple is given below.) Here we re primrily concerned with infinite-simensionl vector spces of functions, for the moment subspces of the spce of piecewise continuous, 14

15 bounded functions on the rel line or on n intervl. On C b (R) or C b ([, b]) (continuous nd bounded, or even piecewise continuous nd bounded) we hve the uniform norm : f = sup f(x), x I where I = R or I = [, b], nd sup ( supremum ) is the smllest M > so tht f(x) M for ll x I (often, but not lwys, this is the mximum of f(x) in I). We lso hve the L p norms on the spce of piecewise continuous functions on bounded intervl [, b]: f p = ( f p (x)dx) 1/p, p 1. Of these by fr the most importnt is the L 2 norm, which is ssocited to the L 2 inner product: f 2 = f, f 1/2, where f, g = f(x)g(x)dx. Remrk. (i) Something is being swept under the rug here. If piecewise continuous function hs zero L p norm, it is not quite true tht it is the zero function (it could be nonzero t finitely mny points in [, b]. If we consider only continuous functions (one-sided continuous t nd b), this problem disppers (but this solution is less stisfying thn it looks.) (ii) Note tht for ny continuous function f in [, b] we hve: f 2 = ( f 2 (x)dx) 1/2 (b ) 1/2 f. However, we don t hve n inequlity in the other direction. For the sequence f n (x) = x n in [, 1], we hve f n 2 = (2n + 1) 1/2 but f n = 1 for ll n. This shows these two norms in C[, 1] re not equivlent. 5.2 First energy rgument. We show the L 2 norm of solutions of the het eqution is decresing in time. Let u(x, t) be solution of u t ku xx = in the usul spces V D [, L], V N [, L], V per [ L, L]. Integrting by prts we find: d dt u 2 dx = 2 uu t dx = 2k 15 uu xx dx = u 2 xdx + uu x L,

16 nd the lst term is zero in ny of the usul spces. This shows u2 dx is decresing in time, unless u is constnt solution. This implies the following stbility estimte for solutions u 1, u 2 of the het eqution in V D, V N or V per with initil conditions u 1, u2 (resp.): u 1 (, t) u 2 (, t) 2 u 1 u 2 2. (Recll stbility estimte for PDE shows tht if the dt of the problem re close (in given norm), then the solutions re lso close, for ny given t >.) Exercise. Cn you use this method to show the L p norms of the solution (for p 2) re lso decresing in time? 5.3 Second energy rgument. One wy to define n energy for function on n intervl is to mesure the devition from being constnt. A simple wy to do this is to set: E[f] = 1 2 (f x ) 2 dx, f C 1 pw[, b]. (C 1 pw: continuous, with piecewise continuous derivtive). Thus E[f] = exctly if f is constnt. Let s compute the time derivtive of the energy for solution of the het eqution u t ku xx = in n intervl [, b], in one of the usul spces V D, V N, V per : de[u] dt = u x u xt dx = u xx u t dx + u x u t x=b x= = k (u xx ) 2 dx, since the boundry term u x u t b vnishes for u in V D, V N or V per. Thus the energy is decresing in time (unless u is constnt, in which cse its energy is zero). One might be tempted to use the energy to define norm, but tht doesn t work, since the constnt functions hve zero energy. It would work s norm in V D, since zero is the only constnt in V D. One option is to dd the L 2 integrl to the energy; this defines the H 1 norm : f H 1 = ( f 2 dx + f 2 xdx) 1/2. 16

17 This norm is ssocited to the H 1 inner product in C 1 pw[, b]: f, g H 1 = fgdx + f x g x dx. It then follows from the bove tht the H 1 norm is decresing in time for solutions of the het eqution, in prticulr: u(, t) H 1 u H 1. This is different stbility estimte for the het eqution. Importnt Remrk. There is deeper connection between the energy E[f] nd the het eqution. Consider one-prmeter fmily of functions in vector spce V [, b] in the intervl [, b], with initil velocity g V [, b]: Then: f(, s) V [, b], s I R, d ds f s s= = g V [, b]. d ds E[f s] s= = f x g x dx = f xx gdx + f x g b. If we tke V of Dirichlet, Neumnn or periodic type, the boundry term vnishes. Endowing V with the L 2 inner product, we my write this s: d ds E[f s] s= = f xx, g L 2 = L[f ], g. By nlogy with finite-dimensionl vector clculus, we my sy tht the left-hnd side is the directionl derivtive of E t f in the direction given by g. Since g is rbitrry in V, we could sy the grdient vector of E t f is f xx = L[f ]. Thus the het eqution u t = u xx corresponds to the grdient flow: u t = grd E(u) = L[u] Reclling the usul interprettion of grdient flows, this suggests the het eqution decreses the energy of function t the fstest possible rte. (All of this cn be mde rigorous.) 17

18 6. The het eqution with boundry conditions. We consider first the cse of Dirichlet, Neumnn or periodic boundry conditions on bounded intervl, corresponding to the function spces V D [, L], V N [, L], V per [ L, L]. Denote by V ny of these spces, endowed with the L 2 inner product. Given u V, we consider the problem: u t ku xx =, u(, t) V, u(x, ) = u (x). We suppose V dmits n orthonorml bsis ϕ n of eigenfunctions of the differentil opertor L[f] = f xx. As for the wve eqution, eigenfunctions evolve by multipliction by function of t: L[ϕ] = λϕ, u(x, ) = ϕ(x) u(x, t) = A(t)ϕ(x), A() = 1. nd the het eqution implies A(t) solves n ODE: A (t)ϕ(x) = u t = u xx = L[u] = A(t)L[ϕ] = λa(t)ϕ A (t) = λa(t). Thus A(t) = e λt nd u(x, t) = e λt ϕ(x). A generl initil condition u V dmits (forml) expnsion in terms of eigenfunctions: where: u (x) A n ϕ n (x), L[ϕ n ] = λ n ϕ n, A n = u, φ n. The corresponding solution of the het eqution then hs the (forml) expnsion: u(x, t) A n e λnt ϕ n (x). Using the definition of the L 2 inner product, we find (if the intervl is [, L], sy): u(x, t) e λnt u (y)ϕ n (y)ϕ n (x)dy, nd ssuming integrtion nd infinite sum my be interchnged, we write this in fmilir form: u(x, t) e λnt ϕ n (x)ϕ n (y)u (y)dy = h V (x, y, t)u (y)dy, 18

19 where we hve set: h V (x, y, t) = e λnt ϕ n (x)ϕ n (y), the forml het kernel in the function spce V. We ll develop this explicitly for V D [, L] nd V per [ L, L], leving the cse of V N for the exercises. In V D [, L], the eigenfunction expnsion corresponds to the Fourier sine series: u (x) b n sin( nπx L ), b n = 2 L Since the eigenvlues re λ n = (nπ/l) 2, the solution is: u (y) sin( nπy L )dy. u(x, t) b n e n2 π 2 L 2 t sin( nπx L ) 2 L = e n2 π 2 L 2 t sin( nπx L ) sin(nπy L )u (y)dy h D (x, y, t)u (y)dy, where h D (x, y, t), the forml het kernel in [, L], is given by: h D (x, y, t) = 2 L e n2 π 2 L 2 t sin( nπx L ) sin(nπy L ). In fct, using sin sin b = 1 2 [cos( b) cos( + b)] we hve: h D (x, y, t) = k D (x y, t) k D (x+y, t), k D (x, t) = 1 L e n2 π 2 L 2 t cos( nπx L ). (Compre with the Dirichlet kernel for the hlf-line, described below.) Observe the following: (i) h D (x, y, t) = h D (y, x, t); (ii)h D (x,, t) V D [, L] for ll (x, t) in the upper hlf-plne {t > }. (Also h D (, y, t) V D [, L], for ech (y, t).) 19

20 Turning to periodic boundry conditions, we consider the evolution of u V per [ L, L] under u t u xx =. The eigenfunction expnsion corresponds to the full Fourier series of u : where: n = 1 L L u (x) 2 + n cos( nπx L ) + b n sin( nπx L ), cos( nπy L )u (y)dy, n ; b n = 1 L L sin( nπy L )u (y)dy, n 1. (Note the constnt term /2 is the verge vlue of u over [ L, L].) The forml Fourier series of the solution is then: u(x, t) 2 + e n2 π 2 L 2 t [ n cos( nπx L ) + b n sin( nπx L )]. Substituting the expressions for the Fourier coefficients nd formly exchnging integrtion nd infinite sum, we find: u(x, t) 1 L or: L { e n2 π 2 L 2 t [cos( nπx L ) cos(nπy L )+sin(nπx L ) sin(nπy L )]}u (y)dy, u(x, t) L h per (x y, t)u (y)dy, where h per (x, t), the forml het kernel in V per [ L, L], is given by: h per (x, t) = 1 2L + 1 e n2 π 2 L L 2 t cos( nπx L ). Note h per (x, t) is even in x, nd in V per [ L, L] for ech t >. observe the reltion with the Dirichlet het kernel in [, L]: We lso h D (x, y, t) = h per (x y, t) h per (x + y, t), h per (x, t) = 1 2L + k D(x, t). Exercise. Write down the corresponding development in V N [, L], including the het kernel in V N [, L], nd its reltion with h per. Hint: recll the eigenfunction expnsion in V N [, L] corresponds to the Fourier cosine series: u (x) 2 + n cos( nπx L ), n = 2 u (y) cos( nπy )dy, n. L L 2

21 Problems on the hlf-line. As for the wve eqution, boundry vlue problems on the hlf-line {x > } cn be solved by extending the initil dt to the whole rel line. Consider the Dirichlet problem: u t ku xx = on {x >, t > }, u(x, ) = u (x), u(, t) = for ll t >. If we wnt continuous solutions up to t =, the comptibility condition u () = is required (we ssume u is defined, continuous nd bounded on {x }). Denote by u o the odd extension of u to the rel line (tht is, u o (x) = u ( x) if x < ). This extension is continuous, since u () =. We cn write down the solution using the het kernel: u(x, t) = h(x y, t)u o (y)dy, but it is useful to hve n expression using only the originl u. The integrl on the negtive hlf-line cn be rewrtitten using the chnge of vrible y z = y, to give: u(x, t) = [h(x y, t) h(x + y, t)]u (y)dy. 21

22 PROBLEMS. 1. Solve u t = u xx on the hlf-line {x > }, with Dirichlet boundry condition u(, t) = nd initil condition u (x) = sinh x = 1 2 (ex e x ). 2. (i) Find formul for the solution of the Neumnn problem on the hlf-line {x > }, with boundry condition u x (, t) = nd initil condition u (x), where u (x) =. Write down the nswer in the form: u(x, t) = h n (x, y, t)u (y)dy (tht is, find the Neumnn het kernel h N (x, y, t) for the hlf-line. (ii) Solve u t = u xx on {x > }, with Neumnn boundry conditions nd initil condition u (x) = cosh(x) = 1 2 (ex + e x ). 3. Let u be solution of the het eqution in V N [, L]. Use the energy method to show the integrl (u ū)2 dx is decresing in t (unless u u is constnt). Here ū is the verge vlue of the solution over the intervl [, L] (which is independent of t). 4. The function u (x) = sin 2 x is in ll three spces V D [, π], V N [, π], V per [ π, π]. Solve the het eqution u t = u xx with initil condition u, in ech of these spces. Hint: In V N nd V per, u is finite liner combintion of eigenfunctions; to solve the problem in V D, compute its Fourier sine series. 5. Suppose u(x, t) solves the het eqution with Neumnn boundry conditions in [, π], nd the initil condition u is finite liner combintion of eigenfunctions: N u (x) = n cos nx. n= Find constnts C > nd λ > so tht u(x, t) ū Ce λt. 6. Find n expression (s n infinite series of functions) for the het kernel in V N [, π], nd its reltion with the periodic het kernel in [ π, π], h per. Hint: recll the eigenfunction expnsion in V N [, L] corresponds to the Fourier cosine series: u (x) 2 + n cos nx n = 2 π π u (y) cos(ny)dy, n. 22