Элементы вариационного исчисления. Молчанова Л.А. - 12 стр.

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eq := yn(x) − (
d
2
dx
2
yn(x)) = 0
bound2 := 2D(yn)(1) = 0
res := yn(x) =
e
3
e
(−x)
e
4
+1
+
ee
x
e
4
+1
yn := 0.3612626842e
(−1.x)
+ 0.04889158769e
x
var1
var2
primer 1
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
–0.8
–0.4
0
0.4
0.8
x
J[y, z] =
Z
2
−2
(y
02
+ z
02
+ 2yz)dx;
y(−2) = 1, y(2) = 0; z(−2) = 0, z(2) = 2.
                               q:=yn-yn2=0
>eq:=subs(yn=yn(x),yn2=diff(yn(x),x$2),lhs(eq))=0;
                                        d2
                       eq := yn(x) − ( dx 2 yn(x)) = 0

çàäàíèå åñòåñòâåííîãî ãðàíè÷íîãî óñëîâèÿ íà ïðàâîì êîíöå
>bound2:=subs(yn1=D(yn)(1),dfdy1)=0;
                         bound2 := 2D(yn)(1) = 0
ðåøåíèå óðàâíåíèÿ Ýéëåðà
>res:=dsolve({eq,yn(-1)=1} union{bound2},yn(x));
                                         3 (−x)      x
                       res := yn(x) = e ee4 +1 + eee
                                                   4 +1

>assign(res):yn:=evalf(yn(x));
               yn := 0.3612626842e(−1.x) + 0.04889158769ex
ïîñòðîåíèå ãðàôèêîâ ïåðâîãî è âòîðîãî âàðèàíòîâ çàäà÷è
>py:=plot(y,x=-1..1,title=primer 1",
>legend="accuracy solve",color=black):
> pyn:=plot(yn,x=-1..1,legend="natural condition",
"color=black,thickness=2):
>plots[display]({py,pyn});


                                         primer 1
                                           2
                                         1.8
                                         1.6
                                         1.4
                                         1.2
                                            1
                                         0.8
                                         0.6
                                         0.4
                       –0.8   –0.4           0          0.4 x   0.8

                                                 var1
                                                 var2




                 Ðèñ. 1. Ãðàôèê ýêñòðåìàëè ïðèìåðà 1
   Ïðèìåð 2. Íàéòè ýêñòðåìàëè ôóíêöèîíàëà
                                 Z   2
                     J[y, z] =           (y 02 + z 02 + 2yz)dx;
                                 −2

                 y(−2) = 1, y(2) = 0; z(−2) = 0, z(2) = 2.
Ïîñòðîèòü ãðàôèêè ýòèõ ýêñòðåìàëåé.
   Ðåøåíèå. Ïîäêëþ÷åíèå ïàêåòà
>restart:with(VariationalCalculus);

                                           12

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