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Lecture 14: A gradient estimate for the heat equation on a ball. 1 Adapting the proof of the gradient estimate to Rn Last time we proved a gradient estimate for solutions of the heat equation on a torus. In this lecture we will adapt the argument to prove the same theorem on Rn . Theorem 1.1 If u : Rn × [0, ∞) → R is positive and satisfies the heat equation then |∇u|2 ut n t − ≤ . (1) 2 u u 2 We will sketch the proof. We work on an interval [0, T ], andnote that the result we 2 want follows immediately from this . Define f = log u and F = t |∇u| − uut . Let φ be a u2 cutoff function on the ball B1 (0) with 0 < φ < 1 on the interior and φ = 0 on the boundary. We can stretch this to get a cutoff function φr on Br (0) by taking φr (x) = φ(x/r) . If F is non positive the result is trivially true, so we can assume that φr F has an interior maximum without loss of generality. At this maximum d(φr F ) ≥ 0, and φr ∇F = −F ∇φr dt We’ll use these to get a bound on F . Calculate 4(φr F ) ≤ 0, 0 ≥ d 4− dt (2) (φr F ) ≥ φr 4F + 2∇F · ∇φr + F 4φr − φr (3) dF dt (4) We need to estimate some of these. The calculations are very similar to last time. We start with 1 4F = t4 |∇u|2 ut − u2 u (5) = t4(|∇f |2 − ft ) 2 2 ∂ f = 2t + 2t∇(4f ) · ∇f − t4ft ∂xi ∂xj by Bochner. Calculate 4f = ∂ ∂u/∂xi ∂xi u = 4u u − |∇u|2 u2 (6) (7) = − Ft to get 2 ∂2f 4F = 2t − 2∇F · ∇f − t4ft . (8) ∂xi ∂xj P P Recall the inequality ( Ai i)2 ≤ n (A2ii ) for all matrices A from last time, and apply it to the hessian of f to give 4F = ≥ 2t (4f )2 − 2∇F · ∇f − t4ft n 2F 2 − 2∇F · ∇f − 4ft . nt (9) (10) We also need an estimate on Ft . We have Ft = |∇f |2 − ft + t(2∇f · ∇ft ) − tftt , (11) Ft = |∇f |2 − ft + t(2∇f · ∇ft ) − t(4f + |∇f |2 )t F − t4ft . = t (12) and 4f + |∇f |2 = ft , so Putting 4, 10 and 13 together we get 2 2F F − 2∇F · ∇f − 0 ≥ φr + 2∇F · ∇φr + F 4φr . nt t (13) (14) Recall that φr ∇F = −F ∇φr , so F 2F 2F 2 2F + ∇φr · ∇f − − |∇φr | + F 4φr nt φr t φr 2F 2 1 |∇φr |2 4φr ≥ F φr + ∇φr · ∇f − − 2 + . nt φr t φ2r φr 0 ≥ φr 2 (15) (16) 2 2 ∂f 1 ∂φr r ∂f − for all > 0. to show ≥ − Now use an absorbing inequality ∂φ ∂xi ∂xi ∂xi ∂xi that 1 ∇φr · ∇f ≥ − |∇φr |2 − |∇f |2 (17) for all > 0. Consequently 2F 1 2 1 |∇φr |2 4φr 2 2 . (18) 0 ≥ F φr − |∇φr | + |∇f | − − 2 + nt φr t φ2r φr Let r → ∞ so that |∇φr | and 4φr tend to zero and φr → 1, and get 2F 1 2 0≥F . − 2|∇f | − nt t (19) Finally we let → 0 and recover F 0≥ t 2F −1 . n From this we getF ≤ n/2 as required. 3 (20)