方程组
f'''' - ff''' - 4gg'=0=F1
fg' - f'g - g''=0=F2
f,g分别为是z的函数,且 0<=z<=1
z=0 f(0)=A f'(0)=0 g(0)=0
z=1 f(1) f'(1) g(1)=B
----------------------------------
e.g. A=0 to +-2, B=100 to 1000
1<=i<=n+3, N=40 delta z=0.025
以F2为例,化简后
f(i)[g(i+1)- g(i-1)]/2delta z - g(i)[-f(i+2)+8f(i+1)-8f(i-1)+f(i-2)]/12delta z - [g(i+1)-2gi+g(i-1)]/delta z的平方=0
F1和F2排列分别是由f和g组成的4个矩阵
分别对F1和F2上的f(i),f(i+1),f(i+2),f(i-1),f(i-2)求导
以F2上的fi求导为例,结果是g(i+1)- g(i-1)]/2delta z
请问,这里的g(i+1)和g(i-1)怎么带入?
难道是 i
g(i=0)=0,
g(i=1)=g0.025=B*deltaz=100*0.025=2.5
g(i=2)=g0.05=B*deltaz=100*0.025=5
g(i=3)=g0.075=B*deltaz=100*0.025=7.5
....
?
请问这么算g(i)对吗?那么fi呢?
f'''' - ff''' - 4gg'=0=F1
fg' - f'g - g''=0=F2
f,g分别为是z的函数,且 0<=z<=1
z=0 f(0)=A f'(0)=0 g(0)=0
z=1 f(1) f'(1) g(1)=B
----------------------------------
e.g. A=0 to +-2, B=100 to 1000
1<=i<=n+3, N=40 delta z=0.025
以F2为例,化简后
f(i)[g(i+1)- g(i-1)]/2delta z - g(i)[-f(i+2)+8f(i+1)-8f(i-1)+f(i-2)]/12delta z - [g(i+1)-2gi+g(i-1)]/delta z的平方=0
F1和F2排列分别是由f和g组成的4个矩阵
分别对F1和F2上的f(i),f(i+1),f(i+2),f(i-1),f(i-2)求导
以F2上的fi求导为例,结果是g(i+1)- g(i-1)]/2delta z
请问,这里的g(i+1)和g(i-1)怎么带入?
难道是 i
g(i=0)=0,
g(i=1)=g0.025=B*deltaz=100*0.025=2.5
g(i=2)=g0.05=B*deltaz=100*0.025=5
g(i=3)=g0.075=B*deltaz=100*0.025=7.5
....
?
请问这么算g(i)对吗?那么fi呢?