u =0with u(0,x)=f(x), (x) =

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1 PDE LECTURE NOTES, MATH 37A-B 69. Heat Equation The heat equation for a function u : R + R n C is the partial differential equation (.) µ t u =0with u(0,x)=f(x), where f is a given function on R n. By Fourier transforming Eq. (.) in the x variables only, one finds that (.) implies that (.) µ t + ξ û(t, ξ) =0with û(0,ξ)= ˆf(ξ). and hence that û(t, ξ) =e t ξ / ˆf(ξ). Inverting the Fourier transform then shows that ³ u(t, x) =F e t ξ / ˆf(ξ) (x) = ³F ³ e / t ξ Ff (x) =:e t / f(x). From Example??, F ³ e t ξ / (x) =p t (x) =t n/ e t x and therefore, u(t, x) = p t (x y)f(y)dy. R n This suggests the following theorem. Theorem.. Let (.3) p t (x y) :=(πt) n/ e x y /t be the heat kernel on R n. Then µ (.4) t x p t (x y) =0and lim p t (x y) =δ x (y), t 0 where δ x is the δ functionatx in R n. More precisely, if f is a continuous bounded function on R n,then u(t, x) = p t (x y)f(y)dy R n is a solution to Eq. (.) where u(0,x) := lim t 0 u(t, x). Proof. Direct computations show that t x pt (x y) = 0 and an application of Theorem?? shows lim t 0 p t (x y) = δ x (y) or equivalently that lim t 0 RR n p t (x y)f(y)dy = f(x) uniformly on compact subsets of R n. This shows that lim t 0 u(t, x) =f(x) uniformly on compact subsets of R n. Proposition. (Properties of e t / ). () For f L (R n,dx), the function ³ Rn e t / f (x) =(P t f)(x) = f(y) e t x y (πt) dy n/ is smooth in (t, x) for t>0 and x R n and is in fact real analytic. () e t / acts as a contraction on L p (R n,dx) for all p [0, ] and t>0. Indeed, (3) Moreover, p t f f in L p as t 0.

2 70 BRUCE K. DRIVER Proof. Item. is fairly easy to check and is left the reader. One just notices that p t (x y) analytically continues to Re t>0 and x C n andthenshowsthat it is permissible to differentiate under the integral. Item. (p t f)(x) f(y) p t (x y)dy R n and hence with the aid of Jensen s inequality we have, kp t fk p L f(y) p p p t (x y)dydx = kfk p L p R n R n So P t is a contraction t >0. Item 3. It suffices to show, because of the contractive properties of p t, that p t f f as t 0 for f C c (R n ). Notice that if f has support in the ball of radius R centered at zero, then (p t f)(x) f(y) P t (x y)dy kfk P t (x y)dy R n y R = kfk CR n e t ( x R) and hence kp t f fk p L p = y R Therefore p t f f in L p as t 0 f C c (R n ). p t f f p dy + kfk CR n e t ( x R). Theorem.3 (Forced Heat Equation). Suppose g C b (R d ) and f ([0, ) R d ) then C, b u(t, x) :=p t g(x)+ t 0 p t τ f(τ,x)dτ solves u t = 4u + f with u(0, ) =g. Proof. Because of Theorem., we may with out loss of generality assume g =0in which case Therefore and u(t, x) = t u t (t, x) =p t f(0,x)+ 0 = p t f 0 (x) 4 u(t, x) = t 0 p t f(t τ,x)dτ. t 0 t 0 p τ f(t τ,x)dτ t p τ f(t τ,x)dτ τ p t 4 f(t τ,x)dτ.

3 PDE LECTURE NOTES, MATH 37A-B 7 Hence we find, using integration by parts and approximate δ function arguments, that µ t 4 t µ u(t, x) =p t f 0 (x)+ p τ τ 4 f(t τ,x)dτ = p t f 0 (x)+lim 0 = p t f 0 (x) lim 0 t t µ + lim 0 τ 4 p τ µ τ 4 p τ f(t τ,x) t 0 p τ f(t τ,x)dτ = p t f 0 (x) p t f 0 (x) + lim 0 f(t τ,x)dτ p f(t, x) =f(t, x)... Extensions of Theorem.. Proposition.4. Suppose f : R n R is a measurable function and there exists constants c, C < such that f(x) Ce c x. Then u(t, x) :=p t f(x) is smooth for (t, x) (0,c ) R n and for all k N and all multi-indices α, µ Ã k µ! k (.5) D t α u(t, x) = D α p t f(x). t In particular u satisfies the heat equation u t = u/ on (0,c ) R n. Proof. The reader may check that D α µ t k p t (x) =q(t,x)p t (x) where q is a polynomial in its variables. Let x 0 R n and >0 be small, then for x B(x 0, ) and any β>0, x y = x x y + y y + x ³β x + β y β y β ³ x 0 +. Hence µ ) k g(y) :=sup( Dα p t (x y)f(y) t : t c and x B(x 0, ) ) sup ( q(t,x y) e t x y y (πt) n/ Cec : t c and x B(x 0, ) C(β,x 0, )supn (πt) n/ q(t,x y)e [ t( β )+ ] y o c : t c and x B(x0, ). By choosing β close to 0, the reader should check using the above expression that for any 0 <δ<(/t c) / there is a C < such that g(y) Ce δ y. In particular g L (R n ). Hence one is justified in differentiating past the integrals in p t f and this proves Eq. (.5).

4 7 BRUCE K. DRIVER Lemma.5. There exists a polynomial q n (x) such that for any β>0 and δ>0, µ y δ e β y dy δ n q n R βδ e βδ n Proof. Making the change of variables y δy andthenpassingtopolar coordinates shows y δ e β y dy = δ n y e βδ y dy = σ S n δ n e βδ r r n dr. R n R n Letting λ = βδ and φ n (λ) := R r= e λr r n dr, integration by parts shows! Ãe φ n (λ) = r n λr d = r= λ λ e λ + e λr (n ) (n )r r= λ dr = λ e λ + n λ φ n (λ). Iterating this equation implies φ n (λ) = λ e λ + n µ λ λ e λ + n 3 λ φ n 4(λ) and continuing in this way shows φ n (λ) =e λ r n (λ (n )!! )+ δ λ δ φ i (λ) where δ is the integer part of n/, i=0if n is even and i =if n is odd and r n is a polynomial. Since it follows that φ 0 (λ) = for some polynomial q n. r= e λr dr φ (λ) = r= φ n (λ) e λ q n (λ ) re λr dr = e λ λ, Proposition.6. Suppose f C(R n, R) such that f(x) Ce c x then p t f f uniformly on compact subsets as t 0. In particular in view of Proposition.4, u(t, x) :=p t f(x) is a solution to the heat equation with u(0,x)=f(x). Proof. Let M > 0 be fixed and assume x M throughout. By uniform continuity of f on compact set, given >0 there exists δ = δ(t) > 0 such that f(x) f(y) if x y δ and x M. Therefore, choosing a>c/ sufficiently small, p t f(x) f(x) = p t (y)[f(x y) f(x)] dy p t (y) f(x y) f(x) dy p t (y)dy + C (πt) n/ [e c x y + e c x ]e t y dy y δ y δ + C (πt) n/ e ( t a) y dy. y δ So by Lemma.5, it follows that Ã! p t f(x) f(x) + C (πt) n/ δ n q n β t a e ( t a)δ

5 PDE LECTURE NOTES, MATH 37A-B 73 and therefore lim sup t 0 sup x M p t f(x) f(x) 0 as 0. Lemma.7. If q(x) is a polynomial on R n, then X t n n p t (x y)q(y)dy = R n! n q(x). n n=0 Proof. Since f(t, x) := p t (x y)q(y)dy = p t (y) X a αβ x α y β dy = X C α (t)x α, R n R n f(t, x) is a polynomial in x of degree no larger than that of q. Moreover f(t, x) solves the heat equation and f(t, x) q(x) as t 0. Since g(t, x) := P n=0 tn n n! q(x) has n thesamepropertiesoff and is a bounded operator when acting on polynomials of a fixeddegreeweconcludef(t, x) =g(t, x). Example.8. Suppose q(x) =x x + x 4 3, then e t / q(x) =x x + x t x x + x 4 t 3 +! 4 x x + x 4 3 = x x + x t x 3 + t! 4 4! = x x + x tx 3 +3t. Proposition.9. Suppose f C (R n ) and there exists a constant C< such that X D α f(x) Ce C x, then α =N+ (p t f)(x) = e t / f(x) = Proof. Fix x R n and let f N (y) := NX X α N+ t k k! k f(x)+o(t N+ ) as t 0 α! Dα f(x)y α. Then by Taylor s theorem with remainder f(x + y) f N (y) C y N+ sup e C x+ty C y N+ e C[ x + y ] C y N+ e C y t [0,] and thus p t (y)f(x + y)dy p t (y)f N (y)dy C p t (y) y N+ e C y dy R n R n R n = Ct N+ p (y) y N+ e t R n = O(t N+ ). C y dy

6 74 BRUCE K. DRIVER Since f(x + y) and f N (y) agree to order N +for y near zero, it follows that NX t k NX t k NX t k p t (y)f N (y)dy = R k! k f N (0) = k! k yf(x + y) y=0 = k! k f(x) n which completes the proof... Representation Theorem and Regularity. In this section, suppose that Ω is a bounded domain such that Ω is a C submanifold with C boundary and for T>0 let Ω T := (0,T) Ω, and Γ T := ([0,T] Ω) ({0} Ω) bd(ω T )=([0,T] Ω) ({0,T} Ω) as in Figure 36 below. Figure 36. A cylindrical region Ω T and the parabolic boundary Γ T. Theorem.0 (Representation Theorem). Suppose u C, ( Ω T ) ( Ω T = Ω T = [0,T] Ω) solves u t = 4u + f on Ω T. Then u(t,x)= p T (x y)u(0,y)dy + p T t (x y)f(t, y)dydt (.6) Ω + [0,T ] Ω [0,T ] Ω pt t (x y)u(t, y) p T t (x y) u n y n (y) dσ(y)dt Proof. For v C, ([0,T] Ω), integration by parts shows fvdydt = v(u t 4v)dydt Ω T Ω T = ( v t + v u)dydt + vu t=t dy + t=0 Ω T Ω = ( v t 4v)udydt + vu T dy + 0 Ω T Ω [0,T ] Ω [0,T ] Ω v v n dtdσ µ u n u v u n dσ dt.

7 PDE LECTURE NOTES, MATH 37A-B 75 Given >0, taking v(t, y) :=p T + t (x y) (note that v t + 4v =0and v C, ([0,T] Ω)) implies f(t, y)p T + t (x y)dydt =0+ p (x y)u(t, y)dy p T + (x y)u(t, y)dy [0,T ] Ω + Let 0 above to complete the proof. Ω [0,T ] Ω Ω pt + t (x y) u(t, y) p T + t (x y) u n y n (y) dσ(y)dt Corollary.. Suppose f := 0 so u t (t, x) = 4u(t, x). Then u C ((0,T) Ω). Proof. Extend p t (x) for t 0 by setting p t (x) :=0if t 0. It is not to hard to check that this extension is C on R R n \{0}. Using this notation we may write Eq. (.6) as u(t, x) = p t (x y)u(0,y)dy + Ω [0, ) Ω pt τ n y (x y)u(t, y) p T t (x y) u n (y) dσ(y)dτ. The result follows since now it permissible to differentiate under the integral to show u C ((0,T) Ω). Remark.. Since x p t (x) is analytic one may show that x u(t, x) is analytic for all x Ω..3. Weak Max Principles. Notation.3. Let a ij,b j C ΩT satisfy aij = a ji and for u C (Ω) let (.7) Lu(t, x) = nx a ij (t, x)u xix j (x)+ i,j= nx b i (t, x)u xi (x). We say L is elliptic if there exists θ>0 such that X aij (t, x)ξ i ξ j θ ξ for all ξ R n and (t, x) Ω T. i= Assumption 3. In this section we assume L is elliptic. is elliptic. As an example L = Lemma.4. Let L be an elliptic operator as above and suppose u C (Ω) and x 0 Ω is a point where u(x) has a local maximum. Then Lu(t, x 0 ) 0 for all t [0,T]. Proof. Fix t [0,T] and set B ij = u xix j (x 0 ),A ij := a ij (t, x 0 ) and let {e i } n i= be an orthonormal basis for R n such that Ae i = λ i e i. Notice that λ i θ>0 for

8 76 BRUCE K. DRIVER all i. By the first derivative test, u xi (x 0 )=0for all i and hence Lu(t, x 0 )= X A ij B ij = X A ji B ij =tr(ab) = X e i ABe i = X Ae i Be i = X i λ i e i Be i = X i λ i e i u(t, x 0 ) 0. The last inequality if a consequence of the second derivative test which asserts vu(t, x 0 ) 0 for all v R n. Theorem.5 (Elliptic weak maximum principle). Let Ω be a bounded domain and L be an elliptic operator as in Eq. (.7). We now assume that a ij and b j are functions of x alone. For each u C Ω C (Ω) such that Lu 0 on Ω (i.e. u is L subharmonic) we have (.8) max u max u. Ω bd(ω) Proof. Let us first assume Lu > 0 on Ω. If u and had an interior local maximum at x 0 Ω then by Lemma.4, Lu(x 0 ) 0 which contradicts the assumption that Lu(x 0 ) > 0. So if Lu > 0 on Ω we conclude that Eq. (.8) holds. Now suppose that Lu 0 on Ω. Let φ(x) :=e λx with λ>0, then Lφ(x) = λ a (x)+b (x)λ e λx λ (λθ + b (x)) e λx. By continuity of b(x) we may choose λ sufficiently large so that λθ +b (x) > 0 on Ω in which case Lφ > 0 on Ω. The results in the first paragraph may now be applied to u (x) :=u(x)+ φ(x) (for any >0) to learn u(x)+ φ(x) =u (x) max u max u + max φ for all x Ω. bd(ω) bd(ω) bd(ω) Letting 0 in this expression then implies u(x) max u for all x Ω bd(ω) which is equivalent to Eq. (.8). Theorem.6 (Parabolic weak maximum principle). Assume u C, (Ω T \Γ T ) C(Ω T ). () If u t Lu 0 in Ω T then (.9) max u =maxu. Ω T Γ T () If u t Lu 0 in Ω T then min u =minu. Ω T Γ T Proof. Item. follows from Item. by replacing u u, so it suffices to proveitem. Webeginbyassumingu t Lu < 0 on Ω T and suppose for the sake of contradiction that there exists a point (t 0,x 0 ) Ω T \Γ T such that u(t 0,x 0 )=maxu. Ω T () If (t 0,x 0 ) Ω T (i.e. 0 < t 0 < T) then by the first derivative test u t (t 0,x 0 )=0and by Lemma.4 Lu(t 0,x 0 ) 0. Therefore, (u t Lu)(t 0,x 0 )= Lu(t 0,x 0 ) 0 which contradicts the assumption that u t Lu < 0 in Ω T.

9 PDE LECTURE NOTES, MATH 37A-B 77 () If (t 0,x 0 ) Ω T \Γ T with t 0 = T, then by the first derivative test, u t (T,x 0) 0 and by Lemma.4 Lu(t 0,x 0 ) 0. So again (u t Lu)(t 0,x 0 ) 0 which contradicts the assumption that u t Lu < 0 in Ω T. Thus we have proved Eq. (.9) holds if u t Lu < 0 on Ω T. Finally if u t Lu 0 on Ω T and >0, the function u (t, x) :=u(t, x) t satisfies u t Lu <0. Thereforebywhatwehavejustproved u(t, x) t max Ω T u =maxu max u for all (t, x) Ω T. Γ T Γ T Letting 0 in the last equation shows that Eq. (.9) holds. Corollary.7. Thereisatmostonesolutionu C, (Ω T \Γ T ) C(Ω T ) to the partial differential equation u t = Lu with u = f on Γ T. Proof. If there were another solution v, then w := u v would solve w t = Lw with w =0on Γ T. So by the maximum principle in Theorem.6, w =0on Ω T. We now restrict back to L = 4 and we wish to see what can be said when Ω = R n an unbounded set. Theorem.8. Suppose u C([0,T] R n ) C, ((0,T) R n ), u t 4u 0 on [0,T] Rn and there exists constants A, a < such that u(t, x) Ae a x for (t, x) (0,T) R n. Then Proof. Recall that sup u(t, x) K := sup u(0,x). (t,x) [0,T ] R n x R n p t (x) = solves the heat equation µ n/ e t x = t (.0) t p t (x) = 4p t(x). µ n/ e t x x t Since both sides of Eq. (.0) are analytic as functions in x, so 7 p t t (ix) = (4p t)(ix) = 4 xp t (ix) and therefore for all τ>0and t<τ p τ t (ix) = ṗ τ t (ix) = t 4 xp τ t (ix). 7 Similarly since both sides of Eq. (.0) are analytic functions in t, it follows that t p t(x) = ṗ t(x) = 4p t.

10 78 BRUCE K. DRIVER That is to say the function ρ(t, x) :=p τ t (ix) = µ τ t n/ e (τ t) x for 0 t<τ solves the heat equation. (This can be checked directly as well.) Let, τ > 0 (to be chosen later) and set v(t, x) =u(t, x) ρ(t, x) for 0 t τ/. Since ρ(t, x) is increasing in t, Hence if we require τ v(t, x) Ae a x >aor τ< " lim x µ n/ e τ x for 0 t τ/. τ a it will follows that # =. sup v(t, x) 0 t τ/ Therefore we may choose M sufficiently large so that v(t, x) K := sup u(0,z) for all x M and 0 t τ/. z Since µ t 4 µ v = t 4 u 0 we may apply the maximum principle with Ω = B(0,M) and T = τ/ to conclude for (t, x) Ω T that u(t, x) ρ(t, x) =v(t, x) sup v(0,z) K if 0 t τ/. z Ω We may now let 0 in this equation to conclude that (.) u(t, x) K if 0 t τ/. By applying Eq. (.) to u(t + τ/,x) we may also conclude u(t, x) K if 0 t τ. Repeating this argument then enables us to show u(t, x) K for all 0 t T. Corollary.9. The heat equation u t 4u =0on [0,T] Rn with u(0, ) =f( ) C (R n ) has at most one solution in the class of functions u C([0,T] R n ) C, ((0,T) R n ) which satisfy for some constants A and a. u(t, x) Ae a x for (t, x) (0,T) R n Theorem.0 (Max Principle a la Hamilton). Suppose u C, [0,T] R d satisfies () u(t, x) Ae a x for some A, a ( for all t T ) () u(0,x) 0 for all x (3) u t 4u i.e. ( t 4)u 0. Then u(t, x) 0 for all (t, x) [0,T] R d.

11 PDE LECTURE NOTES, MATH 37A-B 79 Proof. Special Case. Assume u t < 4u on [0,T] Rd,u(0,x) < 0 for all x R d and there exists M>0such that u(t, x) < 0 if x M and t [0,T].For the sake of contradiction suppose there is some point (t, x) [0,T] R d such that u(t, x) > 0. By the intermediate value theorem there exists τ [0,t] such that u(τ,x)=0. In particular the set {u =0} is a non-empty closed compact subset of (0,T] B(0,M). Let π :(0,T] B(0,M) (0,T] be projection onto the first factor, since {u 6= 0} is a compact subset of (0,T] B(0,M) if follows that t 0 := min{t π ({u =0})} > 0. Choose a point x 0 B(0,M) such that (t 0,x 0 ) {u =0}, i.e. u(t 0,x 0 )=0, see Figure 37 below. Since u(t, x) < 0 for all 0 t<t 0 and x R d,u(t 0,x) 0 Figure 37. Findingapoint(t 0,x 0 ) such that t 0 is as small as possible and u(t 0,x 0 )=0. for all x R d with u(t 0,x 0 )=0. This information along with the first and second derivative tests allows us to conclude u(t 0,x 0 )=0, 4u(t 0,x 0 ) 0 and u t (t 0,x 0 ) 0. This then implies that 0 u t (t 0,x 0 ) < 4u(t 0,x 0 ) 0 which is absurd. Hence we conclude that u 0 on [0,T] R d. General Case: Let p t (x) = e t d/ 4t x be the fundamental solution to the heat equation t p t = 4p t. Let τ>0to be determined later. As in the proof of Theorem.8, the function µ d/ ρ(t, x) :=p τ t (ix) = e 4(τ t) x for 0 t<τ τ t

12 80 BRUCE K. DRIVER is still a solution to the heat equation. Given >0, define, for t τ/, u (t, x) =u(t, x) t ρ(t, x). Then ( t 4)u =( t 4)u <0, u (0,x)=u(0,x) 0 <0 and for t τ/ u (t, x) Ae a x τ e d/ 4τ x. Hence if we choose τ such that 4τ >a,we will have u (t, x) < 0 for x sufficiently large. Hence by the special case already proved, u (t, x) 0 for all 0 t τ and >0. Letting 0 implies that u(t, x) 0 for all 0 t τ/. As in the proof of Theorem.8 we may step our way up by applying the previous argument to u(t + τ/,x) andthentou(t + τ,x), etc. to learn u(t, x) 0 for all 0 t T..4. Non-Uniqueness of solutions to the Heat Equation. Theorem. (See Fritz John 7). For any α>, let ½ e t α t>0 (.) g(t) := 0 t 0 and define X g (k) (t)x k u(t, x) =. (k)! Then u C (R ) and (.3) u t = u xx and u(0,x):=0. In particular, the heat equation does not have unique solutions. Proof. We are going to look for a solution to Eq. (.3) of the form X u(t, x) = g n (t)x n n=0 in which case we have (formally) that X 0=u t u xx = (ġ n (t)x n g n (t)n(n )x n ) This implies = n=0 X [ġ n (t) (n +)(n +)g n+ (t)] x n. n=0 ġ n (.4) g n+ = (n +)(n +). To simplify the final answer, we will now assume u x (0,x)=0, i.e. g 0 in which case Eq. (.4) implies g n 0 for all n odd. We also have with g := g 0, g = ġ0 = ġ!,g 4 = ġ0 4 3 = g 4!,g 6 = g(3) 6!... g k = g(k) (k)!

13 PDE LECTURE NOTES, MATH 37A-B 8 and hence (.5) u(t, x) = X g (k) (t)x k. (k)! The function u(t, x) will solve u t = u xx for (t, x) R with u(0,x)=0provided the convergence in the sum is adequate to justify the above computations. Now let g(t) be given by Eq. (.) and extend g to C\(, 0] via g(z) =e z α where z α = e α log(z) = e α(ln r+iθ) for z = re iθ with π<θ<π. In order to estimate g (k) (t) we will use of the Cauchy estimates on the contour z t = γt where γ is going to be chosen sufficiently close to 0. Now Re(z α )=e α ln r cos(αθ) = z α cos(αθ) and hence g(z) = e Re(z α) = e z α cos(αθ). From Figure 38, we see Figure 38. Here is a picture of the maximum argument θ m that apointz on B(t, γt) may attain. Notice that sin θ m = γt/t = γ is independent of t and θ m 0 as γ 0. β(γ) :=min cos(αθ) : π <θ<πand re iθ t = γt ª is independent of t and β(γ) as γ 0. Therefore for z t = γt we have g(z) e z αβ(γ) e ([γ+]t) αβ(γ) = e β(γ) +γ t α e t α provided γ is chosen so small that β(γ) +γ. By for w B(t, tγ), the Cauchy integral formula and its derivative give g(w) = I g(z) z t =γt dz and πi z w g (k) (w) = k! I g(z) z t =γt dz πi k+ (z w) andinparticular (.6) g (k) (t) k! I g(z) k! z t =γt dz π k+ z w π e πγt k! t α = k+ γt γt k e t α.

14 8 BRUCE K. DRIVER We now use this to estimate the sum in Eq. (.5) as X u(t, x) g (k) (t)x k X (k)! e k! x k t α (γt) k (k)! X µ e x k x =expµ t α k! γt γt t α <. Therefore lim u(t, x) =0uniformly for x in compact subsets of R. Similarly one t 0 may use the estimate in Eq. (.6) to show u is smooth and X g (k) (t)(k)(k )x k X g (k) (t)x (k ) u xx = = (k)! ((k ))! k= X g (k+) (t)x k = = u t. (k)!.5. The Heat Equation on the Circle and R. In this subsection, let S L := {Lz : z S} be the circle of radius L. As usual we will identify functions on S L with πl periodic functions on R. GiventwoπL periodic functions f,g, let (f,g) L := πl f(x)ḡ(x)dx πl πl and denote H L := L πl to be the πl periodic functions f on R such that (f,f) L <. By Fourier s theorem we know that the functions χ L k (x) :=eikx/l with k form an orthonormal basis for H L and this basis satisfies d µ k dx χl k = χ L k. L Therefore the solution to the heat equation on S L, is given by u t = u xx with u(0, ) =f H L u(t, x) = X (f,χ L k )e ( L) k t e ikx/l k = X Ã! πl f(y)e iky/l dy e ( L) k t e ikx/l πl k πl = πl πl p L t (x y)f(y)dy where p L t (x) = X e ( L) k t e ikx/l. πl k If f is L periodic then it is nl periodic for all n N, so we also would learn u(t, x) = πnl πnl p nl t (x y)f(y)dy for all n N.

15 PDE LECTURE NOTES, MATH 37A-B 83 this suggest that we might pass to the limit as n in this equation to learn u(t, x) = p t (x y)f(y)dy where p t (x) := lim = π From this we conclude u(t, x) = R n pnl t R R (x) = lim L πl X e ( L) k t e i( L)x k k e ξt e iξx dξ = πt e x t. p t (x y)f(y)dy = πl πl n X p t (x y +πnl)f(y)dy and we arrive at the identity X e (x+πnl) t = X p t (x +πnl) = X e ( L) k t e ikx/l n πt πl n k which is a special case of Poisson s summation formula.