The het kernel on Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto My 28, 24 Nottion For f L (, we define ˆf : C by ˆf(ξ = (F f(ξ = f(xe 2πiξx dx, ξ. The sttement of the Riemnn-Lebesgue lemm is tht ˆf C (. We denote by S n the Fréchet spce of Schwrtz functions C. If α is multi-index, we define nd D α = i α D α = D α = D α Dαn n, α ( i D = D 2 + + D 2 n. ( i D n αn, 2 The het eqution Fix n, nd for t >, x, define k t (x = k(t, x = (4πt n/2 exp ( x 2. 4t We cll k the het kernel. It is strightforwrd to check for ny t > tht k t S n. The het kernel stisfies k t (x = (t /2 n k (t /2 x, t >, x. For > nd f(x = e π x 2, it is fct tht ˆf(ξ = n/2 e π ξ 2 /. Using this, for ny t > we get ˆk t (ξ = e 4π2 ξ 2t, ξ.
Thus for ny t >, k t (xdx = ˆk t ( =. Then the het kernel is n pproximte identity: if f L p (, p <, then f k t f p s t, nd if f is function on tht is bounded nd continuous, then for every x, f k t (x f(x s t. For ech t >, becuse k t S n we hve f k t C (, nd D α (f k t = f D α k t for ny multi-index α. 2 The het opertor is D t nd the het eqution is (D t u =. It is strightforwrd to check tht (D t k(t, x =, t >, x, tht is, the het kernel is solution of the het eqution. To get some prctice proving things bout solutions of the het eqution, we work out the following theorem from Follnd. 3 In Follnd s proof it is not pprent how the hypotheses on u nd D x re used, nd we mke this explicit. Theorem. Suppose tht u : [, C is continuous, tht u is C 2 on (,, tht (D t u(t, x =, t >, x, nd tht u(, x = for x. If for every ɛ > there is some C such tht then u =. u(t, x Ce ɛ x 2, D x u(t, x Ce ɛ x 2, t >, x, Proof. If f nd g re C 2 functions on some open set in R, such s (,, then g( t f f + f( t g + g = t (fg g = t (fg + = div t,x F, n j 2 f + f j= n j 2 g j= n j (f j g g j f j= where F = (fg, f g g f,..., f n g g n f. k, nd ny k t, belong merely to S n nd not to D(, which is demnded in the definition of n pproximte identity in Rudin s Functionl Anlysis, second ed. 2 Gerld B. Follnd, Introduction to Prtil Differentil Equtions, second ed., p., Theorem.4. 3 Gerld B. Follnd, Introduction to Prtil Differentil Equtions, second ed., p. 44, Theorem 4.4. 2
Tke t >, x, nd let f(t, x = u(t, x nd g(t, x = k(t t, x x for t >, x. Let < < b < t nd r >, nd define Ω = {(t, x : x < r, < t < b}. In Ω we check tht ( t f = nd ( t + g =, so by the divergence theorem, F ν = div t,x F = g( t f f + f( t g + g = g + f =. Ω Ω On the other hnd, s Ω Ω = {(b, x : x r} {(, x : x r} {(t, x : < t < b, x = r}, we hve F ν = Ω = = b + b + b + F (b, x (,,..., dx + F (t, x x x =r r dσ(xtn dt f(b, xg(b, xdx f(, xg(, xdx Ω F (, x (,,..., dx n (f j g g j f(t, x x j x =r r dσ(xtn dt j= u(b, xk(t b, x x dx u(, xk(t, x x dx n x =r j= k(t t, x x j u(t, x ( u(t, x j k(t t, x x xj r dσ(xtn dt, where σ is surfce mesure on { x = r} = rs n. As r, the first two terms tend to u(b, xk t b(x x dx = u(b, xk t b(x xdx = u(b, k t b(x nd u(, xk t (x x dx = u(, xk t (x xdx = u(, k t (x respectively. Let ɛ < 4(t, nd let C be s given in the sttement of the theorem. Using j k(t, x = xj 2t k(t, x, for ny r > the third term is bounded by n b x =r ( Ce ɛr2 x x 2t k(t t, x x + k(t t, x x Ce ɛr2 dσ(xt n dt, 3
which is bounded by b ( x + r n Ce ɛr2 2 x =r ( + (4π(t b n/2 exp r 2 4(t dσ(xt n dt, nd writing η = 4(t ɛ nd ω n = 2πn/2 Γ(n/2, the surfce re of the sphere of rdius in, this is equl to (b n r n ω n Ce ηr2 ( x + r 2 which tends to s r. Therefore, + (4π(t b n/2, u(b, k t b(x = u(, k t (x. One checks tht s b t, the left-hnd side tends to u(t, x, nd tht s, the right-hnd side tends to u(, x =. Therefore, u(t, x =. This is true for ny t >, x, nd s u : [, C is continuous, it follows tht u is identiclly. 3 Fundmentl solutions We extend k to R s { (4πt n/2 exp ( x 2 4t t >, x k(t, x = t, x. This function is loclly integrble in R, so it mkes sense to define Λ k D (R by Λ k φ = R φ(t, xk(t, xdxdt, φ D(R. Suppose tht P is polynomil in n vribles: P (ξ = c α ξ α = c α ξ α ξαn n. We sy tht E D ( is fundmentl solution of the differentil opertor P (D = c α D α = c α i α D α if P (DE = δ. If E = Λ f for some loclly integrble f, Λ f φ = φ(xf(xdx, we lso sy tht the function f is fundmentl solution of the differentil opertor P (D. We now prove tht the het kernel extended to R in the bove wy is fundmentl solution of the het opertor. 4 4 Gerld B. Follnd, Introduction to Prtil Differentil Equtions, second ed., p. 46, Theorem 4.6. 4
Theorem 2. Λ k is fundmentl solution of D t. Proof. For ɛ >, define K ɛ (t, x = k(t, x if t > ɛ nd K ɛ (t, x = otherwise. For ny φ D(R, ɛ (k(t, x K ɛ (t, xφ(t, xdxdt = k(t, xφ(t, xdxdt R R n ɛ φ k(t, xdxdt R n ɛ = φ dt = φ ɛ. This shows tht Λ Kɛ Λ k in D (R, with the wek-* topology. It is fct tht for ny multi-index, E D α E is continuous D (R D (R, nd hence (D t Λ Kɛ (D t Λ k in D (R. Therefore, to prove the theorem it suffices to prove tht (D t Λ Kɛ δ (becuse D (R with the wek-* topology is Husdorff. Let φ D(R. Doing integrtion by prts, (D t Λ Kɛ (φ = Λ Kɛ ((D t φ = K ɛ (t, x(d t φ(t, x φ(t, xdxtx R R n = k(t, xd t φ(t, x k(t, x φ(t, xdxtx ɛ R ( n = k(ɛ, xφ(ɛ, x φ(t, xd t k(t, xdt dx ɛ + φ(t, x k(t, xdxdt ɛ R n = k(ɛ, xφ(ɛ, xdx R n φ(t, x(d t k(t, xdtdx ɛ R n = k(ɛ, xφ(ɛ, xdx. So, using k t (x = k t ( x nd writing φ t (x = φ(t, x, (D t Λ Kɛ (φ = k ɛ ( xφ ɛ (xdx = k ɛ φ ɛ ( = k ɛ φ ( + k ɛ (φ ɛ φ (. Using the definition of convolution, the second term is bounded by sup φ ɛ (x φ (x k ɛ = sup φ ɛ (x φ (x, x x 5
which tends to s ɛ. Becuse k is n pproximte identity, k ɛ φ ( φ ( s ɛ. Tht is, (D t Λ Kɛ (φ φ ( = δ(φ s ɛ, showing tht (D t Λ Kɛ proof. δ in D (R nd completing the 4 Functions of the Lplcin This section is my working through of mteril in Follnd. 5 For f S n nd for ny nonnegtive integer k, doing integrtion by prts we get F (( k f(ξ = (( k f(xe 2πiξx dx = (4π 2 ξ 2 k (F f(ξ, ξ. Suppose tht P is polynomil in one vrible: P (x = c k x k. Then, writing P ( = c k ( k, we hve F (P ( f(ξ = c k F (( k f(ξ = c k (4π 2 ξ 2 k (F f(ξ = (F f(ξp (4π 2 ξ 2. We remind ourselves tht tempered distributions re elements of S n, i.e. continuous liner mps S n C. The Fourier trnsform of tempered distribution Λ is defined by Λf = (F Λf = Λ ˆf, f S n. It is fct tht the Fourier trnsform is n isomorphism of loclly convex spces S n S n. 6 Suppose tht ψ : (, C is function such tht Λf = f(ξψ(4π 2 ξ 2 dξ, f S n, is tempered distribution. We define ψ( : S n S n by ψ( f = F ( ˆfΛ, f S n. Define ˇf(x = f( x; this is not the inverse Fourier trnsform of f, which we denote by F. As well, write τ x f(y = f(y x. For u S n nd φ S n, we define the convolution u φ : C by One proves tht u φ C (, tht (u φ(x = u(τ x ˇφ, x. D α (u φ = (D α u φ = u (D α φ 5 Gerld B. Follnd, Introduction to Prtil Differentil Equtions, second ed., pp. 49 52, 4B. 6 Wlter Rudin, Functionl Anlysis, second ed., p. 92, Theorem 7.5. 6
for ny multi-index, tht u φ is tempered distribution, tht F (u φ = ˆφû, nd tht û ˆφ = F (φu. 7 We cn lso write ψ( in the following wy. There is unique κ ψ S n such tht F κ ψ = Λ. For f S n, we hve F (κ ψ f = ˆf ˆκ ψ = ˆfΛ, but, using the definition of ψ( we lso hve F (ψ( f = F F ( ˆfΛ = ˆfΛ, so κ ψ f = ψ( f. Moreover, κ ψ f C ( ; this shows tht ψ( f cn be interpreted s tempered distribution or s function. We cll κ ψ the convolution kernel of ψ(. For fixed t >, define ψ(s = e ts. Then Λ : S n C defined by Λf = f(ξψ(4π 2 ξ 2 dξ = f(ξ exp ( 4π 2 ξ 2 t dξ = f(ξˆk t (ξdξ is tempered distribution. Using the Plncherel theorem, we hve Λf = ˆf(ξkt (ξdξ. With κ ψ S n such tht F κ ψ = Λ, we hve Λf = (F κ ψ (f = κ ψ ( ˆf. Becuse f ˆf is bijection S n S n, this shows tht for ny f S n we hve κ ψ (f = f(ξk t (ξdξ. Hence, e t f = κ ψ f = k t f, t >, f S n. ( Suppose tht φ : (, C nd ω : (, (, re functions nd tht ψ(s = φ(τe sω(τ dτ, s >. Mnipulting symbols suggests tht it my be true tht nd then, for f S n, ψ( f = ψ( = φ(τe ω(τ fdτ = φ(τe ω(τ dτ, φ(τ(k ω(τ fdτ, 7 Wlter Rudin, Functionl Anlysis, second ed., p. 95, Theorem 7.9. 7
nd hence κ ψ (x = φ(τk ω(τ (xdτ, x. (2 Tke ψ(s = s β with < Re β < n 2. Becuse Re β < n 2, one checks tht Λf = f(ξ(4π 2 ξ 2 β dξ is tempered distribution. As Re β >, we hve nd writing φ(τ = τ β kernel of ( β is s β = τ β e sτ dτ nd ω(τ = τ, we suspect from (2 tht the convolution κ ψ (x = which one clcultes is equl to τ β k τ (xdτ, Γ ( n 2 β. (3 4 β π n/2 x n 2β Wht we hve written so fr does not prove tht this is the convolution kernel of ( β becuse it used (2, but it is strightforwrd to clculte tht indeed the convolution kernel of ( β is (3. This clcultion is explined in n exercise in Follnd. 8 Tking α = 2β nd defining Γ ( n α R α (x = Γ ( α 2 2 2α π n/2 x, < Re α < n, x n α Rn, we cll R α the Riesz potentil of order α. Tking s grnted tht (3 is the convolution kernel of ( β, we hve ( α/2 f = R α f, f S n. Then, if n > 2 nd α = 2 stisfies < Re α < n, we work out tht where ω n = 2πn/2 Γ(n/2, nd hence R 2 (x = (n 2ω n x n 2, ( f = R 2 f, f S n, 8 Gerld B. Follnd, Introduction to Prtil Differentil Equtions, second ed., p. 54, Exercise. 8
nd pplying we obtin f = (R 2 f = ( R 2 f, hence R 2 = δ. Tht is, R 2 is the fundmentl solution for. Suppose tht Re β >. Then, using the definition of s n integrl, with ψ(s = ( + s β, we hve ψ(s = Mnipulting symbols suggests tht ψ( = τ β e (+sτ dτ, s >. τ β e τ e τ dτ, nd using (, ssuming the bove is true we would hve for ll f S n, ψ( f = whose convolution kernel is τ β e τ e τ fdτ = We write α = 2β nd define, for Re α >, B α (x = Γ ( α 2 (4π n/2 τ β e τ k τ dτ. τ β e τ (k τ fdτ, τ α n 2 x 2 τ e 4τ dτ, x. We cll B α the Bessel potentil of order α. It is strightforwrd to show, nd shown in Follnd, tht B α <, so B α L (. Therefore we cn tke the Fourier trnsform of B α, nd one clcultes tht it is B α (ξ = ( + 4π 2 ξ 2 α/2, ξ, nd then ψ( = ( α/2 f = B α f, f S n. 5 Gussin mesure If µ is mesure on nd f : C is function such tht for every x the integrl f(x ydµ(y converges, we define the convolution µ f : C by (µ f(x = µ(τ x ˇf = (τ x ˇf(ydµ(y = ˇf(y xdµ(y = f(x ydµ(y. Let ν t be the mesure on with density k t. We cll ν t Gussin mesure. It stisfies ν t f(x = f(x ydν t (y = f(x yk t (ydy = f k t (x, x. 9