pde_notes03

pde_notes03 - x (0 , l ) and t (0 , T ] . Then v ( x , t )...

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Notes for TUT course MAT-51316 Robert Pich´e 4.9.2007 3 Maximum Principle for 1-D Diffusion Equation Theorem: If u t = ku xx in (0 , l ) × [0 , T ] then u achieves its maximum in R = [0 , l ] × [0 , T ] at t = 0 or at x = 0 or at x = l (and possibly elsewhere in R as well). Proof: Denote m = max R u, m init = max [0 ,l ] ×{ 0 } u, m bndry = max { 0 ,l [0 ,T ] u, and let M = max( m init , m bndry ) . We assume m > M and derive a contradiction. Let v ( x, t ) = u ( x, t ) + ± · ( T - t ) for all ( x, t ) R , with 0 < ± < ( m - M ) /T . Then v ( x, 0) = u ( x, 0) + ±T m init + ±T M + ±T for x [0 , l ] , and v (0 , t ) v (0 , t ) + ±t = u (0 , t ) + ±T m bndry + ±T M + ±T for t [0 , T ] , and similarly for v ( l, t ) , so M + ±T is an upper bound for v at the initial time ( t = 0 ) or on the boundary ( x = 0 or x = l ). By assumption, u achieves its maximum in R at some point other than initially or on the boundary, say at
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Unformatted text preview: x (0 , l ) and t (0 , T ] . Then v ( x , t ) = u ( x , t ) + ( T-t ) = m + ( T-t ) m &gt; M + T, so v achieves its maximum in R at some point other than initially or on the bound-ary, say at x 1 (0 , l ) and t 1 (0 , T ] . Then u t ( x 1 , t 1 ) = ku xx ( x 1 , t 1 ) = kv xx ( x 1 , t 1 ) , and so v t ( x 1 , t 1 ) = u t ( x 1 , t 1 )- &lt; . This inequality implies that v ( t, x 1 ) &gt; v ( t 1 , x 1 ) for some t (0 , t 1 ) , which con-tradicts the statement that v achieves its maximum at ( x 1 , t 1 ) . 1...
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