华中师范偏微分方程考试题2000级卷答案

任课教师朱长线线封y2ux+x2uxxx+(sinx+siny)uxyy+3x2+1=封(x2+y2)uxy+(sinx+cosy)uxxyy=2x4+3x2siny密密ut+uux=uxxuxxx+uyyy+2ux-3uy+1+u2=0期末考试试卷（B卷）答案 线线封y2uxx2uxxx(sinxsiny)uxyy3x21封(x2y2)uxy(sinxcosy)uxxyy2x43x2siny密密utuuxuxx​

uxxxuyyy2ux3uy

1u20 x2uxx-2xyuxy+y2uyy=解：由于D=x2y2-x2y2=0, 又特征方程为：x2dy2+2xydxdy+y2dx2=0，即xdy+ydx=0xy=c1 则ux=yux uxx=y2uxx，uxy=ux+yuxh+xyuxxuy=uxx+uh uyy=x2uxx+xuxh+xuxh+uhhx2y2uxx-2xyux-2xy2uxh-2x2y2uxx+x2y2uxx+2xy2uxh+y2uhh=即-2xyux+y2uhh=0即x2uxx2xyuxyy2uyy

x2y2x2y2

x2dy22xydxdyy2dx20即xdyydx0

xyc1

y则uxyu

uxxy2u

uxyuyuxyuuyuxu uyyx2uxuxuux2y2u2xyu2xy2u2x2y2ux2y2u2xy2uy2u即2xyuy

0即2u2

第1页(共 第1页(共

x2uxxy2uyy

x2dy2y2dx20,dy

xyc1

yc

yxu

yu

uxu1u 2y

x2y

x2 uxx

u

x2ux4ux3u

uyy

u2ux2u

4y2u2yux u1u0令

v,

v1

v

v(,a()2

vu1u(,)2

a(s)dsb(1 u(,)2F()G(

1u(x,y)(xy)2F

y)G(xy)x密封线密封线2.求方程x2uxxy2uyy0的通解x2dy2y2dx20,dy=–y xy=c y=c x

u=yu-yu u=xu+1u 2y

x2y

x2 uxx=yuxx-x2uxh+x4uhh+x3uh uyy=xuxx+2uxh+x2uhh

+2y hxh

-1u=令 =v, v-1v= x

1u(x,h)=x2a(s)ds+1 uttuxx

0x

tu

t

cos2

ut

t

cos32

0xuux

u

tu(xt)X(x)T(t),T(t)X(x)X(x)T(t)X(x)T(t)（常数X T X(x)X(x)T(t)T(t)

X(0)T(t)

X(0)X(1)

1X(x)c1

xc

x22

c1c2

X(x)c1x第2页(共3页密封线 y)+G(xy)密封线u(x,y)=(xy)2F(

X(x)c1

xc2

c2

c1

(k1)22

k0,1,

(x)

cos(k12

k0,1,2,T(t)bcos(k1)tcsin(k1) u(x,t)(Acos(k1)tBsin(k1)t)cos(k1)

k0,1,2,

u(x,t)

(Acos(k1)t

sin(k1)t)cos(k1) k

22 22 Acos(k1

xcos1 B(k1)cos(k1)xcos322kk022

kk0

A1

B0

B13

B2

u(x,t)cos1tcos1x2sin3tcos3密封线 密封线utt-uxx= 0<x< t>u

ut

0£x£2222ux|x=0= t‡u(xtX(x)T(t),X T

1X(x)=ce--lx+ce-lx12

u1(x,

a2 1

2a2 2

u1|t0则u(xyt)u1(xtu2yt是Cauchy

u2|t02(

a2

uyy

u|t01(x)2(

uxx

2u1(x,

u2(y,

uyyu1(x,

2u(y,2,2uu1(x,t)

(y,t)u(x,t)u2(y,t)

uta

(uxxuyy)

u1(x,

22u1(x,aa

)u2(y,t)u1(x,

u2(y,

22u2(y,aa

)

u|t0(u1(x,t)u2(y,t))|t0u1|t0u2|t01(x)2(因此u(xyt)u1(xtu2yt是Cauchy

a2

uyy

u|t01(x)2(第3页(共3页 X(x)=c1x+第2页(共3页3.l0,则（1）的通解为：X(x)=c1 lx+c2 代入边界条件的c2=

k=0,1,

(x)=

cos(k+12

k=0,1,2,T(t)=bcos(k+1)pt+csin(k+1)p u(x,t)=(Acos(k+1)pt+Bsin(k+1)pt)cos(k+1)p k=0,1,2,

u(x,t)=

(Acos(k+1)pt+Bsin(k+1)pt)cos(k+1)p

¥Acos(k

px=

1

B(k

pcos(k

px=

kk=0

22

2 A0=1,A1=A2== B0=0,B1=3p,B2=B3==密封线密封线

u(x,t)=cos1ptcos1px+2sin3ptcos3p ), 1=a 1 2=a 2 则u(xyt)u1xtu2yt是Cauchy

uxx

(x,

u2