常微分方程习题答案5.3
习题5.3
1、 假设A是nn矩阵,试证: ,
a) 对任意常数、都有 cc12
exp(A+A)=expA?expA cccc1122
b) 对任意整数k,都有
k(expA)=expkA
k,1,k (当k是负整数时,规定(expA),[(expA)])
证明:a) ?(A)?(A),(A)?(A) cccc1122
? exp(A+A)= expA?expA cccc1122
kb) k>0时,(expA),expA?expA……expA
,exp(A+A+……+A)
,expkA
k<0时,-k>0
k,1,k,k (expA),[(expA)]=[exp(-A)] = exp(-A)?exp(-A)……exp(-A)
,exp[(-A)(-k)]
,expkA
k 故k,都有(expA)=expkA ,
',(t)x2、 试证:如果是=Ax满足初始条件,的解,那么 ,,(t)0
,(t),[expA(t-t)] ,0
t-1-1,(t),(s)f(s)ds证明:由定理8可知,Ф(t)Ф(t) ,Ф(t) ,0,t0
-1-1又因为Ф(t)= expAt , Ф(t)=( expAt)= exp(-At), f(s)=0, 000
又因为矩阵 (At)?(- At)=(- At)?(At) 00
,(t)所以 ,[expA(t-t)] ,0
3、 试计算下面矩阵的特征值及对应的特征向量
i
2,33,,,,12,,,,a) b) 4,53,,,,43,,,,4,42,,
010121,,,,,,,,c) d) 0011,11,,,,
,,,,201,6,11,6,,,,
,,1,2,,,解:a)det(E,A)==(,5)(+1)=0 ,4,,3
?=5, =,1 ,,12
,,,,,0对应于,,=5的特征向量u=, () ,1,,2,,,
,,,,,0=,1的特征向量v=, () 对应于,,,2,,,,,,
,,,,b) det(E,A)=(+1)(+2)(,2),0 ?,,1,,2,,,2 ,,,312
1,,,,对应于,,1的特征向量u,, ( 0 ) 1,,,,1,,1
,,0,,
1,,,,,,0对应于,2的特征向量u,, ( ) ,1,2,,2
,,1,,
0,,,,,,0对应于,,2的特征向量u,, ( ) ,,13,,3
,,1,,
,,1,2,1
2,,1,1,1,,,c)det(E,A)=,(+1)(,3),0
,20,,1
?,,1(二重),,3 ,,12
ii
,1,,,,对应于,,1(二重)的特征向量u,, ( 0 ) ,2,,,,,1
,,,2,,
2,,,,,,0对应于,3的特征向量v,, ( ) ,1,,,2
,,2,,
,,10
,,,,,d) det(E,A)=0,1=(+3)(+1)(+2)=0
611,,6
?,,1,,,2,,,3 ,,,312
1,,,, 对应于,,1的特征向量u,, ( 0 ) ,,1,,,1,,1
,,1,,
1,,,,,,0 对应于,,2的特征向量u,, ( ) ,,2,2,,2
,,4,,
1,,,,,,0 对应于,,3的特征向量u,, ( ) ,,,33,,3
,,9,,
'x4、 试求方程组=Ax的一个基解矩阵,并计算expAt,其中A为:
,2112,,,,,,,,a) b) ,,,,43,12,,,,
2,33103,,,,,,,,c)4,53 d) 81,1,,,,,,,,51,14,42,,,,
,解:a)det(E,A)=0得,,,, 33,,12
1,,,,对应于的特征向量为u,, ( 0 ) ,,,,1,,2,3,,
iii
1,,,,0,,,对应于的特征向量为v,, ( ) ,2,,2,3,,
11,,,,,,,,?u,,v,是对应于,的两个线性无关的特征向量 ,,12,,,,2,32,3,,,,
3t,3t,,ee,,Ф(t)=是一个基解矩阵 3t,3t,,(23)e(23)e,,,,
3t,3t3t,3t,,1(23)e(23)eee,,,,,,, ExpAt= 3t,3t3t,3t,,23ee(23)e(23)e,,,,,,,
,b) 由det(E,A)=0得,5,,,1 ,,12
11,,,,,,解得u,,,,v,是对应于,的两个线性无关的特征向量 ,,12,,,,2,1,,,,
5t,t,,ee,, 则基解矩阵为Ф(t),5t,t,,2ee,,,
11,,,,11,,,133,,,,Ф(0), Ф(0), ,,212,1,,,,,,,33,,
5t,t5t,t,,e2eee,,1,1,,则expAt,Ф(t) Ф(0), 5t,t5t,t,,32e2e2ee,,,,
, c) 由det(E,A)=0得,2,,,2,,,1 ,,,312
2t,t,,e0e,,2t,2t,teee 解得基解矩阵Ф(t), ,,
,,2t,2tee0,,
1,11,,,,,1Ф,110(0), ,,
,,01,1,,
2t2t,t2t,t,,eeeee,,,,,,12t,2t2t,2t,t2t,teeeeeee,,,,, 则expAt,Ф(t) Ф(0), ,,
,,2t,2t2t,2t2teeeee,,,,,
iv
,d)由det(E,A)=0得,,3,,2,,,2, 77,,,312
,,,,,3t(2,7)t(2,7)t3eee,,,475475,,,,,,3t(2,7)t(2,7)t解得基解矩阵Ф(t),7 eee,,33,,1717,,,3t(2,7)t(2,7)t,,4eee33,,
,1则expAt,Ф(t) Ф(0),
,,87247247,,,,,3t(2,7)t(2,7)t,,,,eee333,,,,1567122287122287,,,,3t(2,7)t(2,7)t,,eee ,,99947,,32726272627,,,,,,3t(2,7)t(2,7)te,e,e,,999,,
'x5、试求方程组=Ax的基解矩阵,并求满足初始条件 ,(0),,的解,(t)123,,,,,,,,,a)A,,,,,,433,,,,
1030,,,,,,,,,b)A,81,1,,2 ,,,,
,,,,51,1,7,,,,
1211,,,,,,,,cA,),1,11,0,,,,
,,,,2010,,,,
5t,t,,ee,, 解:a)由第4题(b)知,基解矩阵为 5t,t,,2ee,,,
,,3,,,,,,,,,,,,, ,,,,,,,,,32,,,,,,,,,
所以 ,,2,,,1
5t,t,,2ee,,,(t), , 5t,t,,4ee,,,
v
b)由第4题(d)知,基解矩阵为
,,,,,3t(2,7)t(2,7)t3eee,,,475475,,,,,,3t(2,7)t(2,7)t 7Ф(t), eee,,33,,1717,,,3t(2,7)t(2,7)t,,4eee33,,所以
,,52742674267,,,,3t(2,7)t(2,7)t,,,,eee333,,
,,1364774814677481467,,,,,3t(2,7)t(2,7)t (),,,teee,,,99947,,2087178227178227,,,,,,,3t(2,7)t(2,7)te,e,e,,999,,
c) 由3(c)可知,矩阵A的特征值为,3,,,1(二重) ,,12
,,,,,,2,,,,,,,, 对应的特征向量为u,,u, ,12,,1,,,,4,2,,,,,,,,,,3,,
,,,,,,21,,,,,,,,,,,, ?,, 0,,,,,,,,4,2,,,,,,0,,,,,,,,3,,
111,,,,,,,,,,,,422,,,,,111,,,,,,,?,, 解得vv ,12,,,,244,,,,,111,,,,,,,,,,422,,,,,
3t,t ,(t),eEv,e[E,t(A,E)]v12
11,,3t,te,e,,22,,113t,t,,, e,e,,44,,113t,tee,,,22,,
vi
'x,(t)6、 求方程组=Ax,f(t)的解:
t112,,,e,,,,,,,,,,,a)(0),A,f(t),,,,,,,,,1431,,,,,,
0100,,,,,,,,, b)(0),0,A,001,f(t),0,,,,
,t,,,,,6,11,6e,,,,
,43sint,,,,,,,1,,,,,,,c)(0),,A,,f(t),,,,,,,,2,1,2cost2,,,,,,
'x解:a)令=Ax的基解矩阵为Ф(t)
,,,,()det()(5)(1)0p,E,A,,,, 所以,,5,,,,112
5t,t,t,t,,,,eeee,,1,1,,,,解得Ф(t),, 则Ф(t), 4t5t,t55tt,,,,,3e2ee2ee,,,,,,
,1,1,,1,1,,Ф(0), ,,,,213,,
312,,5t,tte,e,e,,,2045,,,(t)求得, 3115t,tt,,eee,,,,,1025,,
,b)由det(E,A)=0得,,1,,,2,,,3 ,,,312
设对应的特征向量为v,则 ,11
,,,,,, (E,A)v=0,得v, ,,,0,11,,1
,,,,,,
11,,,,,,,,,,,1,,32,,,,,, 取v,,同理可得v ,,v, 111123,,,,,,,,,,,,,1,3,2,,,,,,,,,,
vii
11,,,1,,,,23,, 则Ф(t),111 ,,
,,,1,2,3,,,,
131,,,2t,3t,t,te,e,e,te,,442,,351,2t,3t,t,t,,从而解得 ,(t),,2e,e,e,te,,442,,971,2t,3t,t,t4eeete,,,,,442,,
'xc)令=Ax的基解矩阵为Ф(t)
,由det(E,A)=0得,1,,2 ,,12
3,,t2tee,,解得对应的基解矩阵为Ф(t), 2,,t2t,,ee,,
3,,,t,t,23,,e,e,,,1,1,2,,?Ф(t), 从而Ф(0), 2,,,,,2t,2t2,2,,,,ee,,,
t,,11,,,,,,(t),(t)(0)(0),(t)(s)f(s)ds,0
t2t? ,,,,,,cost,2sint,e(,4,2,3),3e(1,,)1212,,,tt2,,2cost,2sint,e(,4,,2,,3),2e(1,,,,)1212,,
7、 假设m不是矩阵A的特征值。试证非齐线性方程组
mtx',Ax,ce
有一解形如
mt ,(t),pe
其中c,p是常数向量。
mt 证:要证是否为解,就是能否确定常数向量p ,(t),pe
mtmtmt pme,Ape,ce
则p(mE,A),c
由于m不是A的特征值
viii
mE,A,0故
mE,A存在逆矩阵
,mt1那么p,c(mE,A) 这样方程就有形如的解 ,(t),pe
8、 给定方程组
'',3',2,',,0xxxxx,11122 ,x',2x,x',x,01122,
a) 试证上面方程组等价于方程组u’=Au,其中
010ux,,,,,,11,,,,,,u,,A=,442 ,ux'21,,,,,,
,,,,,,2,1,1ux32,,,,,,b) 试求a)中的方程组的基解矩阵 c) 试求原方程组满足初始条件
x(0)=0, x’(0)=1, x(0)=0 112的解。
证:a)令 则方程组?化为 u,x,u,x',u,x112132
u'x'u,,,112,u'x''3u2uu'u,,,,, ,212133,u'x'u2uu,,,,,32213,
010,,,,即u’,u’=Au ? ,442u,,
,,2,1,1,,
反之,设x=u,x’=u,x=u则方程组?化为 111223
,,,,x''4x4x'2x,1112,x',2x,x',x,2112
,,,,x''2x'2xx'x,11122,,x'2xx'x,,,2112,
,b)由det(E,A)=0得,0,,1,,2 ,,,312
u,,01,,,2,,,uuu4,4,2,0由 得 u,,0,,0,,,1231
,,,uuu,2,,,02123,,,同理可求得u和u 23
ix
,,111,,,,,,,,,,,,取 ,0,,1,,2vvv,,,,123,,1,,,,20,,,,,,2,,
,,t2t1ee,,
t2t,,则是一个基解矩阵 (t),0e2e,,,1t,,2e02,,
c)令,则?化为等价的方程组?且初始条件变为u,x,u,x',u,x112132
而?满足此初始条件的解为: u(0),0,u(0),1,u(0),0.123
13,,t2t,2e,e,,0,,22,,,,2AtAttt, ? e,e1,,2e,3e,,,,
t,,,,01e,,,,,,,
于是根据等价性,?满足初始条件的解为?式
9、 试用拉普拉斯变换法解第5题和第6题。 证明:略。
10、 求下列初值问题的解:
x',x',0,,,12a)(0),1,(0),0,12x',x',1,12
x'',3x',2x,x',x,0,11122b),x',2x,x',x,0,1122
,,,(0),1,'(0),,1,(0),0 112
2,x'',mx,012c),2x'',mx,0,21
x(0),,,x'(0),,,x(0),,,x'(0),,11122324
11解:a)根据方程解得, , ,, x'x'1222
11?,t,,,,t, xcxc112222
? ,(0),11
x
11?0,,1 ?,1 ?,t,1 ccx,11122
?,(0),02
11?,0,,0 ?,0 ?,,t ccx,22222
1,t,1 综上:x12
1 ,,t x22
b)对方程两边取拉普拉斯变换,得
2,(),,1,3((),1),2(),(),(),0sXsssXsXssXsXs11122,sX(s),1,2X(s),sX(s),X(s),01122,
解得
2s,3211111X(s),,,,,,,1(s,1)(s,2)(s,2)3s,14s,212s,2
,s,21111X(s),,,,,2(s,1)(s,2)(s,2)3s,13s,2
211,t,2t2t,(t),e,e,e13412? 1,t2t,(t),(e,e)23
c)对方程两边取拉普拉斯变换,得
,,22,sX(s),s,,mX(s),01122,,,22sX(s),s,,mX(s),0,2342
22,,,sX(s),mX(s),s,1212即,22,,mX(s),sX(s),s,,1234
3222,,,,s,s,ms,m1234解得X(s),144s,m
3222,s,,s,m,s,m,3412X(s),244s,m
xi
m,t122m212m,,,,,,,2(t),[(,,)cost,(,,)sint],e112423424m4m4m24m22
mt122m212m,,,,,,2,[(,,)cost,(,,)sint],e12423424m4m4m24m22
m,t212m122m2,,,,,,,(t),[(,,,)cost,(,,)sint],e22341244m24m24m4m22
mt212m122m2,[(,,,,,)cost,(,,,,,,)sint],e2341244m24m24m4m22
,(x)11、 假设y,是二阶常系数线性微分方程初值问题
y'',ay',by,0, ,y(0),0,y'(0),1,
xy,,(x,t)f(t)dt 的解,试证是方程 ,0
y'',ay',by,f(x)
的解,这里f(x)为已知连续函数。
x,(x,t)f(t)dt 证明:y, ,0
xx,(0)f(x),,'(x,t)f(t)dt,,'(x,t)f(t)dt ?y’, ,,00
xxnny'',,(x,t)f(t)dt,,'(0)f(x),,(x,t)f(t)dt,f(x) ,,00
?
xxxy'',ay',by,,''(x,t)f(t)dt,f(x),a,'(x,t)f(t)dt,b,(x,t)f(t)dt ,,,000
x,[''(x,t),a'(x,t),b'(x,t),b(x,t)]f(t)dt,f(x),,,,,0
,f(x)
02412,04丁晶晶
02412,05徐雪輝
xii
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