✔ 最佳答案
Q1:
by separation of variable
u(x,t)=Σ a(n)X(x)T(t),
其中 X(x),T(t) 滿足 T'/(4T)= X"/X = -(n/2)^2, n為整數1,2,3,4,...
u(x,t)=Σ exp(-n^2 t)*a(n)sin(nx/2) (週期 4pi)
又 u(x,0)=x(2π-x), 0<x<2π
故將 x(2π-x)作Fourier sine (半幅)展開, 得
a(n)= 32/(π n^3), n=1,3,5, ..., a(2)=a(4)=a(6)=...=0
故u(x,t)=(32/π)Σ[n=1~∞] exp[-(2n-1)t] sin[(2n-1)x/2]/(2n-1)^3
Q2:
u"(x)=
{ 2 , x in[-1,0)
{-2 , x in(0,1]
so,
u'(x)=
{ 2x+a , x in[-1,0)
{-2x+a , x in(0,1]
Note: 因 u'(x)在x=0時conti, so兩者常數項相同
u(x)=
{ x^2+ax +b, x in[-1,0)
{-x^2+ax +b, x in(0,1]
又u(-1)=u(1)=0, so, a= -1, b=0, 故
u(x)=
{ x^2-x, x in [-1,0)
{- x^2-x, x in(0, 1]
Note:u(x), u'(x)在 x=0處必conti, 否則u"(x)在x=0處不會是
Jump discontinuous.
2010-01-31 13:33:58 補充:
Q1:
題目原本就對t求偏導,算是邊界條件,為何要改為對x求偏導呢?
就因對t偏導,so,u(x,t)=Σ exp(-n^2 t)*a(n)sin(nx/2)
若對x偏導才是u(x,t)=a(0)+Σ exp(-n^2 t)*a(n)cos(nx/2)
個人覺得本題要注意的是:為何取u(x,0)的半幅展開,又是哪個半幅展開?