《数学物理方程》课程PPT教学课件（讲稿）8.2 非齐次方程

82非齐次方程 波动方程 un x t 振动方程 u-au=f(x, t) (1)傅立叶级数法 例1 u-a COS ISn ot u,(, r= O u,(,D)=0 l=0=0(x)u1|=0=v(x) 解:源项恰好满足边界条件,故可设解为 (x,1)=∑Tn()cos nZA 带入泛定方程 2IT()+()T, (D]"=Acos,sin ot n=0 cOS z同次项 T"(O)+(")7()= Asin ot T"na}T1(1)=0(≠1)

8.2 非齐次方程 ( , ) 2 u a u f x t tt − xx = ( , ) 2 u a u f x t t − xx = 波动方程 振动方程 (1) 傅立叶级数法 例 1 t l x utt a uxx A   cos sin 2 − = ux (x,t) x=0 = 0 ux (x,t) x=l = 0 ( ) 0 u x t= = ( ). 0 u x t t= = 解： l x 源项 cos 恰好满足边界条件，故可设解为   = = 0 ( , ) ( ) cos n n l n x u x t T t  带入泛定方程 t l x A l n x T t l n a T t n n n     [ ''( ) ( ) ( )]cos cos sin 0 2  + =  = l nx cos 同次项 T t A t l a T t   ''( ) ( ) 1 ( ) sin 2 1 + = ''( ) ( ) ( ) 0 ( 1) 2 + T t = n  l n a T t n n 

解方程r"()+(m)3()=0(n≠1)必须知道Tn的初始条件 带入初始条件x0=∑(0os=0()=29 n7 u4(x,1)=∑T() v(x)=∑vn,cos 7(O)=9=7o(5kl5 7o'(0)=vo=15 ∥(d5 T2(0)=q J(5)c0+n5 d T"(0)=vn=1 y()cos nrts 得解 70()=+v0t n刀t 1(O)=%0s=n inin

带入初始条件    =  = = = = 0 0 ( ,0) (0) cos ( ) cos n n n n l n x x l n x u x T        =  = = = = 0 0 ( , ) '( ) cos ( ) cos n n n t n l n x x l n x u x t T t        d l T l  = = 0 0 0 ( ) 1 (0) ''( ) ( ) ( ) 0 ( 1) 2 + T t = n  l n a T t n n 解方程  必须知道 Tn 的初始条件    d l T l  = = 0 0 0 ( ) 1 '(0)       d l n l T l n n ( )cos 2 (0) 0  = =       d l n l T l n n ( )cos 1 '(0) 0  = = 得解 T t t 0 0 0 ( ) = + l n at n a l l n at T t n n n     ( ) = cos + sin

r"()+(2)7(0)= AsIn ot的解是齐次方程的通解和非齐次方程度特解的和。 zat l 通解 q1cos-,+—v1Sm 刀tmu 特解: osin nao-(ra/n) 刀ut A @"sin at] sin iat 1)211 @sin at mo2-(m/) ) mt;m、2 sin ot T()+(1)7() ozd at / sm @t]+ losin at_()2 sin at (m/D)21 (na/n2 Lo sin ot-G)sin at]= Asin ot t 7()= aat aa t (osin -sin @)+o, cos

T t A t l a T t   ''( ) ( ) 1 ( ) sin 2 1 + = 的解是齐次方程的通解和非齐次方程度特解的和。 通解： l at a l l at     1 cos + 1 sin 特解： ( sin sin ) ( / ) 1 2 2 t l a l at a a l Al        − − [ ( ) sin sin ] ( / ) 1 ''( ) 2 2 1 2 2 t l a l at l a a a l Al T t          − − = − [ sin sin ] ( / ) 2 2 2 t l at l a a l A       − − = − [ sin ( ) sin ] ( / ) ( ) ( ) 2 1 2 2 2 t l a l at l a a l A T t l a         − − = t A t l a t a l A t l a l at l a a l A t l at l a a l A T t l a T t                      [ sin ( ) sin ] sin ( / ) [ sin ( ) sin ] ( / ) [ sin sin ] ( / ) ''( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 2 2 2 1 − = − = − − − + − + = − l at a l l at t l a l at a a l Al T t             ( sin sin ) cos sin ( / ) 1 ( ) 1 2 2 − + 1 + 1 − =

Al rat a u(x, 4= (osin sin t )cos +po +yot+ a -(ra/n) nat nrcat nzax (, cos+yu sin -)cos n=1 nzda 并不是普遍地方法 (2)冲量定理法(波动方程)1-abn=(xD)m0xD=00x0)-=0 =9(x)l-0=D(x) a.解的分解 u'n=aur= un-au xr=f(, t) u(x,t 0 n(x,D)x0=0a(x,1)==0 0=0(x)l40=v(x) 0 这是普通道具有齐次边界条件的 初始值零点非齐次方程,可用 齐次方程 冲量定理法

( cos sin ) cos . ( sin sin )cos ( / ) 1 ( , ) 1 2 2 0 0 l n ax l n at n a l l n at t l ax t l a l at a a l Al u x t n n n                   = + + − + + + − = 并不是普遍地方法。 (2) 冲量定理法（波动方程） ( , ) 2 u a u f x t tt − xx = u(x,t) x=0 = 0 u(x,t) x=l = 0 ( ) 0 u x t= = ( ). 0 u x t t= = a. 解的分解 I II u = u + u 0 2 − xx = I tt I u a u ( , ) x=0 = 0 I u x t ( , ) x=l = 0 I u x t ( ) 0 u x t I = = ( ). 0 u x t t I = = ( , ) 2 u a u xx f x t II tt II − = ( , ) x=0 = 0 II u x t ( , ) x=l = 0 II u x t 0 t=0 = II u t t=0 = 0 II u 这是普通道具有齐次边界条件的 齐次方程 初始值零点非齐次方程，可用 冲量定理法

b.物理思想 时刻t单位长弦受力F(x,D)=f(x,D)

b. 物理思想 时刻 t 单位长弦受力 F(x,t) = f (x,t)

C.解 v(xt x=0 v(x,)x==0 -=/(x0 u(x, t=v(x,t; r)dr 数学检验: 边界条件:70=0 u"(x, dl=vx, t; r),dr=0 初始条件 vdt=0 u'l=0=lim u".(x, ) =lim a v(, t; r)dr t

C. 解 0 2 vtt − a vxx = v(x,t) x=0 = 0 v(x,t) x=l = 0 0 v t= +0 = ( , ) 0 v f x t t t= + =  = t II u x t v x t d 0 ( , ) ( , ; )  数学检验： 边界条件： ( , ) ( , ; ) 0 0 0 = 0 = =  = t x x II u x t v x t  d ( , ) ( , ; ) 0 0 = = =  = t x l x l II u x t v x t  d 初始条件 0 0 0 0  u t= = vd = II lim ( , ) 0 0 u u t x t II t t t II → = =    = → t t v x t d t 0 0 lim ( , ; ) 

积分号下的求导公式 B() g(t; r)dr g(t; r)dr+g[t; B(tI aB(1) gt; a(t)) da(t) a(t dt=0 齐次方程a(c)=2cxkx+9(x)-= vu, (x t; r)dr t+(x v, (x, t; r)dr+f(r, t) f(x,) a Ddr+f(, t) Odt+ =f(x)

积分号下的求导公式： t t g t t t t g t d g t t t g t d dt d t t t t   −   +   =   ( ) [ ; ( )] ( ) ( ; ) ( ; ) [ ; ( )] ( ) ( ) ( ) ( )             ( , ) ( , ; ) ( , ; ) ( , ; ) 0 0 v x t d v x t d v x t t t u x t t t t t = +   =       ( , ; ) . 0  = t vt x t  d 0 0 0 0 =0 = = + =   u t t vt t  d II 齐次方程 ( , ) ( , ; ) ( , ; ) 0 v x t d v x t t t u x t t t tt t +   =    ( , ; ) ( , ; ) 0 v x t d v x t t t t = tt +    ( ) ( , ) 0 2 2 u a u v a v d f x t t xx II tt xx II tt II − = − +   ( , ; ) ( , ) 0 v x t d f x t t = tt +    0 ( , ) 0 d f x t t = +   = f (x,t) ( ( , )) 0 v f x t t t= + =

例1 u -afu=Acos-sin ot n(x.D=0=0n(x2)2==0 4=。=0(x)anl-o=v(x) 解v- -a2ν=0 v_(x,t v,(,tre=0 r+0 0 Ier+0=Acos sin uf(x,t) 由边界条件:(x;)=∑r0r)cos2n 带入泛定方程:∑n”+(D7女=0 n元 n T=0 nra(t-T nra(t 70(2r)=4+B(t-z) T,=A(T)cOS +B(T)sIn v(x, t; t)=Ao+Br+2(A, (T)cos a(t- +B(rsin

例 1 t l x utt a uxx A   cos sin 2 − = ux (x,t) x=0 = 0 ux (x,t) x=l = 0 ( ) 0 u x t= = ( ). 0 u x t t= = 解 0 2 vtt − a vxx = vx (x,t) x=0 = 0 vx (x,t) x=l = 0 v t= +0 = 0 cos sin ( , ) 0 f x t l x vt t A   = + = 由边界条件：   = = 0 ( , ; ) ( , ) cos n n l n x v x t T t    带入泛定方程： [ '' ( ) ]cos 0 0 2  + =  n= n l n x T l n a T   '' ( ) 0 2 + T = l n a Tn  ( ; ) ( ) 0 0 0 T t  = A + B t − l n a t B l n a t Tn An n ( ) ( )sin ( ) ( ) cos       − + − = l n x l n a t B l n a t v x t A B A n n n          ) cos ( ) ( )sin ( ) ( , ; ) ( ( ) cos 1 0 0   = − + − = + +

带入初始条件: =r+0=0 nJA A+2A,(r)cos 0 A=0 n=1 B,=Aa-sin OT =r+0 f(x, t)Bo+2nm B, (r)cos "nIm= cos i sinor B=Bn=0(m≠1) v(x, t; T)=A-sin atsina( cOS Al wa(t-T) (x,t)= cOS drain or sin Al t cosdrsin or[sin cOS zat cosdrSn[s(o+)z+so-)r]+cosa[eo(o+,)x-co(-,)]} 2 Al nat cossin cos(o+ coS(O Tat. 1 +cOS sin(@+)I

带入初始条件： 0 v t= +0 = ( ) cos 0 1 0 + =  = l n x A A n n        ( ) cos cos sin 1 0 l x A l n x B l n a B n + n =  = ( , ) 0 v f x t t t= + = An = 0   1 sin a l B = A 0 ( 1) B0 = Bn = n  l x l a t a l v x t A       cos ( ) ( , ; ) sin sin − = l a t d l x a Al u x t t II ( ) ( , ) cos sin sin 0       − =  cos sin [sin cos cos sin ] 0 l a l at l a l at d l x a Al t           = −  cos {sin [sin( ) sin( ) ] cos [cos( ) cos( ) ]} 2 0                  l a l a l at l a l a l at d l x a Al t = + + − + + − −  sin( ) } 1 sin( ) 1 cos [ cos( ) ] 1 cos( ) 1 cos {sin [ 2 0 0 0 0 t t t t l a l l a a l l a at l a l l a a l l a at l x a Al                         − − + − + + − − − + + + − =

rat. 1 COs--sin cos(o+) coS(@-)t-( 2 Tat. 1 + cOS sin( o )} Al rat { sin ot na sIn ot]+sin )} 0 0+ u( 1) a 7 op-r a losin u sin ot 由外力决定的 n(x,1)=∑(A1cos +B1Sm灯mt naat n7 由初始条件决定的 初值确定叠加系数:A nTs nIs P(ssin-ds: B y(ssin ds n u(x, t=u(x, t)+u(, t)

sin( ) } 1 sin( ) 1 cos [ )] 1 1 cos( ) ( 1 cos( ) 1 cos { sin [ 2 t l a l a t l a l l a at l a l a t l a l a t l a l l a at l x a Al                         − − + − + + − + + − − − + + + = − )} 1 1 sin ] sin ( 1 sin 1 cos {[ 2 l a l l a at t l a t l l a x a Al              − + + + − − + = ( , ) cos { sin sin } 2 2 2 2 2 t l a l at l a l l x a Al u x t II         − − = ( , ) ( cos sin )sin . 1 l n x l n at B l n at u x t An n n I    =  +  = 初值确定叠加系数： ( )sin ; 2 0      d l n l A l n  = ( )sin . 2 0       d l n n a B l n  = u(x,t) u (x,t) u (x,t) I II = + 由初始条件决定的 由外力决定的