2l-2=0 即 n(0)=F limu=o u =c,ple a tc plea 由0)=F设:c(p)+c()=F()0 ~ ~ 2 2 2 2 - u = a p u dx d u ( ) F ~ 0 ~ = 0 ~ lim = ®¥ u x 即 ( ) ( ) x a p x a p u c p e c p e 1 2 ~ = + - u ( ) F c ( p) c (p) F(p) ~ : ~ 0 ~ 由 = 设 1 + 2 =