Элементы вариационного исчисления. Молчанова Л.А. - 17 стр.

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z := x 0.2733 e
(1.0 x)
+ 0.69877 e
(1.0 x)
3
2
+
x
2
24 sin(
π (x+1)
2
)
π (π
2
+ 4)
3
2
+
x
2
24 sin(
π (x + 1)
2
)
π (π
2
+ 4)
+
sin(π (x + 1))
π (1 + π
2
)
3
2
+
x
2
24 sin(
π (x + 1)
2
)
π (π
2
+ 4)
+
sin(π (x + 1))
π (1 + π
2
)
8 sin(
3 π (x + 1)
2
)
π (9 π
2
+ 4)
N= 1
solve
Metod Ritcha
1
1.2
1.4
1.6
1.8
2
–0.8
–0.4
0
0.4
0.8
x
èñïîëüçîâàíèåì ñèíóñîâ
>N:=3;c1:='cross':c2:='circle':c3:='box':
                                  N:=3
>for j from 1 to N do
>a:=array(1..j):u:=Us(x,j):
>Ritz(F,u,1,j,a):
>pU_||j:=
>plot(Us(x,j),x=-1..1,color=black,style=point,
>symbol=c||j,title="Metod Pitcha",
>legend=cat("N=",convert(j,string))):
>end do:
>z:=(x)->0.2733*exp(-1.0*x)+0.69877*exp(1.0*x);
                z := x → 0.2733 e(−1.0 x) + 0.69877 e(1.0 x)
>pz:=plot(z(x),x=-1..1,legend="solve",color=black):
>Us(x,1);Us(x,2);Us(x,3);
                                       π (x+1)
                           3 x 24 sin( 2 )
                            + −
                           2 2   π (π 2 + 4)

                             π (x + 1)
                 3 x 24 sin(     2
                                       ) sin(π (x + 1))
                  + −                   +
                 2 2     π (π 2 + 4)       π (1 + π 2 )
                   π (x + 1)                         3 π (x + 1)
       3 x 24 sin(     2
                             ) sin(π (x + 1)) 8 sin(
                                                           2
                                                                 )
        + −                   +               −
       2 2     π (π 2 + 4)       π (1 + π 2 )     π (9 π 2 + 4)
>plots[display]({pUs_1,pz});

                                      Metod Ritcha
                                          2
                                        1.8
                                        1.6
                                        1.4
                                        1.2
                                          1

                        –0.8   –0.4           0           0.4 x   0.8

                                                   N= 1
                                                  solve




         Ðèñ. 4. Ãðàôèê ýêñòðåìàëåé ïðè èñïîëüçîâàíèè ñèíóñîâ


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