> 
# Crank-Nicolson-esimerkki, luento ke 26.11.-96
# 
> eq1:=4*u[1,1]-u[2,1]=u[2,0];

                 eq1 := 4 u[1, 1] - u[2, 1] = u[2, 0]


                                u := u

> eq2:=4*u[2,1]-u[3,1]-u[1,1]=u[3,0]+u[1,0];

       eq2 := 4 u[2, 1] - u[3, 1] - u[1, 1] = u[3, 0] + u[1, 0]

> eq3:=4*u[3,1]-u[4,1]-u[2,1]=u[4,0]+u[2,0];

       eq3 := 4 u[3, 1] - u[4, 1] - u[2, 1] = u[4, 0] + u[2, 0]

> eq4:=4*u[4,1]-u[5,1]-u[3,1]=u[5,0]+u[3,0];

       eq4 := 4 u[4, 1] - u[5, 1] - u[3, 1] = u[5, 0] + u[3, 0]

> 
> for j from 0 to 5 do u[j,0]:=evalf(sin(j*0.2*Pi)) od;

                             u[0, 0] := 0


                        u[1, 0] := .5877852524


                        u[2, 0] := .9510565165


                        u[3, 0] := .9510565163


                        u[4, 0] := .5877852522


                             u[5, 0] := 0

> u[0,1]:=0;u[5,1]:=0;

                             u[0, 1] := 0


                             u[5, 1] := 0

> solve({eq1,eq2,eq3,eq4},{seq(u[j,1],j=1..4)});

{u[3, 1] = .6460385084, u[4, 1] = .3992737562, u[2, 1] = .6460385084,

    u[1, 1] = .3992737562}

> 
# Tästä on hyvä jatkaa ja lopulta hoitaa homma for-loopilla k-indeksin
# suhteen.
#