《数学物理方程》课程PPT教学课件（讲稿）第八章习题4

例辱8长为的杆:出于杆内放射性物的衰变而引起热量 的变化,设热量的变化正比于e*,初始温度为φ(x),杆端保 持为零度,求杆的温度分布,如9()=Wo,试算出相应的结果 解:热量的变化 △O=C△ 时间用t表示 △O,A oc e △t △t 因此,泛定方程为 u,-afu=be 非齐次扩散方程 边界条件(x2)x-0=0(x)==0 初始条件 =09)(x

解：热量的变化 Q = Cu 时间用 t 表示 kx e t u C t Q −    =   因此，泛定方程为 kx t xx u a u be− − = 2 边界条件 u(x,t) x=0 = 0 u(x,t) x=l = 0 ( ) 0 u x 初始条件 t= = 非齐次扩散方程

u=u +u be l(x,)x=0=0u(x,)x== n(x,)-0=0 (x)x=0 L 1|==0(x) a3x=0v(x)2==0v(x1)=0 be t=T u(x,t)=v(x,t; rdr 由边界条件: v(x,tr)=∑ Ck expl nIa(t-T nA sIn 26 1丌 e sin ds=2bnt (n)2+(k)2

I II u = u + u 0 2 − xx = I t I u a u ( , ) x=0 = 0 I u x t ( , ) x=l = 0 I u x t ( ) 0 u x t I = = ( , ) x=0 = 0 II u x t ( , ) x=l = 0 II u x t t=0 = 0 II u kx xx II t II u a u be− − = 2 0 2 vtt − a vxx = v(x,t) x=0 = 0 v(x,t) x=l = 0 kx v t be− = +0 =  = t II u x t v x t d 0 ( , ) ( , ; )  ]sin ; ( ) ( , ; ) exp[ 2 2 2 2 1 l n x l n a t v x t C n k     − = −  = 由边界条件：    d l n e l b C l kx k sin 2 0  − = 2 2 ( ) ( ) 1 ( 1) 2 n lk e bn kl n + − − = −  

v(x,t t)= ∑ 2bn丌 n-=1 (n)+(W)0x2a2(t-r)1nm n"(x)=∑2bmz e(-1) n2丌a(t-r sIn n=1 (n丌)+() ∑ 1-e(-1)nmx n丌at 2bn丌 SIr dt exp nta'r (nr)2+(k)2 0 (nz)+(k)公助以上 1-e“(-1)”1 nn a n丌at 2bn丌 ]-1} ∑ 26 { exp[-nTa't i (n)+(lk)nTa

]sin ; ( ) exp[ ( ) ( ) 1 ( 1) ( , ; ) 2 2 2 2 2 1 2 2 l n x l n a t n lk e v x t bn n kl n       − − + − − =  = − ]; ( ) sin exp[ ( ) ( ) 1 ( 1) ( , ) 2 2 2 2 2 1 0 2 2 l n a t dt l n x n lk e u x t bn t n kl n II      − − + − − =   = − sin exp[ ] exp[ ]; ( ) ( ) 1 ( 1) 2 2 2 2 2 0 2 2 2 2 1 2 2 l n a dt l n a t l n x n lk e bn t n kl n         − + − − =  = − sin exp[ ]{exp[ ] 1}; ( ) ( ) 1 ( 1) 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 − − + − − =  = − l n a t l n a t l n x n a l n lk e bn n kl n       sin {1 exp[ ]}; ( ) ( ) 1 ( 1) 2 2 2 2 2 1 2 2 2 2 l n a t l n x n a l n lk e b n kl n     − − + − − =  = −

杂题1 例50均细导线,每单位长的电阻为R,通以恒定的电流 ,导线表窗跟周閥温度为零度的介威进行热交换,设初始温渡 和两端温度都是零度,试求导线上的温度分布 解 温度改变的三个因素: 1电流升温:△O=12RM=cM △12R △tcp 2热传导,傅立叶定律: △uk △2 O△v 3.自由冷却,牛顿定律: 9=mn-0)=4g △a△Ohu=b △t△S △ t cpAt Cp 泛定方程: u -au-bu=g Xx

杂题1 l 解： 温度改变的三个因素： 1.电流升温： Q = l Rt = cu 2 c l R t u 2 =   1. 2.热传导，傅立叶定律： x u a x u c k t u x x   =   =   2  3.自由冷却，牛顿定律： 2. 3. t S Q q h u    = ( − 0) =  bu c hu c t Q t u = =   =     泛定方程： ut − a uxx − bu = G 2

v(x,2)x0=0v(xD)x=04 0 只求特解 t=0 傅立叶级数法 u(x)=2O)m"7 nZA 2G nZX 2G nAIl 4G cOS 0 (2k+1)丌 T2k+()+[ (2k+1) 4G )2-b72k+1(1)= 2k+1)丌 241(O)=072k+1(O)=0 拉普拉斯变换: 2k+(p){p2+[( (2k+1)ma 4G )2-b]}= p(2k+1)r

u(x,t) x=0 = 0 ux (x,t) x=l = 0 0 u t=0 = 只求特解 傅立叶级数法   = = 0 ( , ) ( )sin n n l n x u x t T t   = l n dx l n x l G G 0 sin 2    = = 0 sin n n l n x G G  l l n x n G 0 cos 2   = − (2 1) 4 + = k G   (2 1) 4 ) ] ( ) (2 1) ''( ) [( 2 1 2 2 1 + − = + + + + k G b T t l k a T t k k T2k+1 (0) = 0 T2k+1 '(0) = 0 拉普拉斯变换：   (2 1) 4 ) ]} (2 1) ( ){ [( ~ 2 2 2 1 + − = + + + p k G b l k a T k p p

4G k+1 (p) (2k+1)Pp2+ 2k+lTa [( b] 反演 (2k+1)ma )2>b1n2-(2 sm1(2k+1)m -bt +1)ma p2+(2)2-bl (2k+1)ma b (2k+1)ma )2<b L (2k+1)mu (2k+1)ma )-b]b-( 2k+l)ra

) ] (2 1) [( 1 1 (2 1) 4 ( ) ~ 2 2 2 1 b l k a p k p G T k p − + + + + =   反演： ) 1 1 ( 1 = − p L bt l k a b l k a b l k a p − + − + = − + + − 2 2 2 2 1 ) (2 1) sin ( ) (2 1) ( 1 ] ) ] (2 1) [( 1 [    L b l k a  + 2 ) (2 1) b l k a  + 2 ) (2 1) t l k a b l k a b b l k a p 2 2 2 2 1 ) (2 1) sinh ( ) (2 1) ( 1 ] ) ] (2 1) [( 1 [    + − + − = − + + − L

4G (2k+1)ma T2+1(t)= dt (2k+1)ma2-b(t-r b (2k+1)r (2k+1)mu2-b 4G (2k+1)z(2k+1)ma 6s,(2k+1)m2-b(-) (2k+1)z(2k+1)mu -cO,(2k+1)m bt] )2-b (2k+1)m)2<b 4G (2k+1) 7x+()= inh 6 )2(t-r) (2k+1)丌 2k+l)Ta 4G (2k+1)mu cosh b (2k+1) b 2k+1)ra 4G [cos b (2k+1)ma (2k+1)z b-( (2k +lTa

) ( ) (2 1) sin ( ) (2 1) ( 1 (2 1) 4 ( ) 2 0 2 2 1      − − + − + + = +  b t l k a b l k a d k G T t t k b l k a  + 2 ) (2 1) t b t l k a b l k k a G 0 2 2 ) ( ) (2 1) cos ( ) (2 1) ( 1 (2 1) 4     − − + − + + = ) ] (2 1) [1 cos ( ) (2 1) ( 1 (2 1) 4 2 2 bt l k a b l k k a G − + − − + + =    b l k a  + 2 ) (2 1) ) ( ) (2 1) sinh ( ) (2 1) ( 1 (2 1) 4 ( ) 2 0 2 2 1      − + − + − + = +  t l k a b l k a b d k G T t t k t t l k a b l k a b k G 0 2 2 ) ( ) (2 1) cosh ( ) (2 1) ( 1 (2 1) 4     − + − + − − + = ) 1] (2 1) [cos ( ) (2 1) ( 1 (2 1) 4 2 2 − + − + − + = t l k a b l k a b k G   