heat_eq

heat_eq - P DE LECTURE NOTES, M ATH 237A-B 169 12. Heat...

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PDE LECTURE NOTES, MATH 237A-B 169 12. Heat Equation The heat equation for a function u : R + × R n C is the partial di f erential equation (12.1) μ t 1 2 u =0 with u (0 ,x )= f ( x ) , where f is a given function on R n . By Fourier transforming Eq. (12.1) in the x variables only, one f nds that (12.1) implies that (12.2) μ t + 1 2 | ξ | 2 ˆ u ( t, ξ )=0 with ˆ u (0 ˆ f ( ξ ) . and hence that ˆ u ( t, ξ e t | ξ | 2 / 2 ˆ f ( ξ ) . Inverting the Fourier transform then shows that u ( t, x F 1 ³ e t | ξ | 2 / 2 ˆ f ( ξ ) ´ ( x ³ F 1 ³ e t | ξ | 2 / 2 ´ F f ´ ( x : e t / 2 f ( x ) . From Example ?? , F 1 ³ e t | ξ | 2 / 2 ´ ( x p t ( x t n/ 2 e 1 2 t | x | 2 and therefore, u ( t, x Z R n p t ( x y ) f ( y ) d y. This suggests the following theorem. Theorem 12.1. Let (12.3) p t ( x y ):=(2 πt ) n/ 2 e | x y | 2 / 2 t be the heat kernel on R n . Then (12.4) μ t 1 2 x p t ( x y and lim t 0 p t ( x y δ x ( y ) , where δ x is the δ —functionat x in R n . More precisely, if f is a continuous bounded function on R n ,then u ( t, x Z R n p t ( x y ) f ( y ) dy is a solution to Eq. (12.1) where u (0 ) := lim t 0 u ( t, x ) . Proof. Direct computations show that ¡ t 1 2 x ¢ p t ( x y and an application of Theorem ?? shows lim t 0 p t ( x y δ x ( y ) or equivalently that lim t 0 R R 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 12.2 (Properties of e t / 2 ) . (1) For f L 2 ( R n ,dx ) , the function ³ e t / 2 f ´ ( x )=( P t f )( x Z R n f ( y ) e 1 2 t | x y | 2 (2 ) n/ 2 dy is smooth in ( t, x ) for t> 0 and x R n and is in fact real analytic. (2) e t / 2 acts as a contraction on L p ( R n ) for all p [0 , ] and 0 . Indeed, (3) Moreover, p t f f in L p as t 0 .
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170 BRUCE K. DRIVER Proof. Item 1. 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 di f erentiate under the integral. Item 2. | ( p t f )( x ) | Z R n | f ( y ) | p t ( x y ) dy and hence with the aid of Jensen’s inequality we have, k p t f k p L p Z R n Z R n | f ( y ) | p p t ( x y ) dydx = k f k p L p So P t is a contraction 0 . Item 3. It su ces 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 ) | Z R n | f ( y ) | P t ( x y ) dy k f k Z | y | R P t ( x y ) dy = k f k CR n e 1 2 t ( | x | R ) 2 and hence k p t f f k p L p = Z | y | R | p t f f | p dy + k f k n e 1 2 t ( | x | R ) 2 . Therefore p t f f in L p as t 0 f C c ( R n ) . Theorem 12.3 (Forced Heat Equation) . Suppose g C b ( R d ) and f C 1 , 2 b ([0 , ) × R d ) then u ( t, x ):= p t g ( x )+ Z t 0 p t τ f ( τ,x ) solves ∂u ∂t = 1 2 4 u + f with u (0 , · )= g.
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heat_eq - P DE LECTURE NOTES, M ATH 237A-B 169 12. Heat...

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