Задача Штурма-Лиувилля. Молчанова Л.А. - 12 стр.

UptoLike

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:= cos(
λ˜a
λ
˜
b)
λ :=
π
2
(1 + 2 Z1 ˜)
2
4 (
˜
b + ˜a)
2
λ :=
π
2
(1 + 2 k)
2
4 (
˜
b + ˜a)
2
C1 sin
4
s
π
2
(1 + 2
˜
k)
2
(
˜
b + ˜a)
2
x
4
+ C2 cos
4
s
π
2
(1 + 2
˜
k)
2
(
˜
b + ˜a)
2
x
4
C1 :=
C2 cos(
π (1+2
˜
k) ˜a
2 (
˜
ba)
)
sin(
π (1+2
˜
k) ˜a
2 (
˜
ba)
)
C2 sin(
π ˜aπ x+2 π ˜a
˜
k2 π x
˜
k
2
˜
b+2 ˜a
)
sin(
π ˜a+2 π ˜a
˜
k
2
˜
b+2 ˜a
)
Yn := (x,
˜
k) sin(
π ˜a π x + 2 π ˜a
˜
k 2 π x
˜
k
2
˜
b + 2 ˜a
)
Ïðèðàâíèâàåì íóëþ ýòîò îïðåäåëèòåëü è ðåøàåì ïîëó÷åííîå õàðàêòåðè-
ñòè÷åñêîå óðàâíåíèå:
>Delta:=select(has,Delta,[cos]);
                                √     √
                       ∆ := cos( λã − λb̃)

   >_EnvAllSolutions:=true:lambda:=solve(Delta,lambda);
                                  π 2 (1 + 2 _Z1 ˜)2
                           λ :=
                                      4 (−b̃ + ã)2

>lambda:=subs(_Z1='k',lambda);
                                     π 2 (1 + 2 k)2
                              λ :=
                                     4 (−b̃ + ã)2

Íàõîäèì ñîáñòâåííûå ôóíêöèè:
  >assume(k,posint):y(x);
             s                                  s                       
             √    π 2 (1 + 2 k̃)2                 √  π 2 (1 + 2 k̃)2
             4                     x            4                       x
                   (−b̃ + ã)2                      (−b̃ + ã)2          
    _C1 sin 
            
                                      + _C2 cos 
                                                
                                                                            
                                                                            
                      4                                4                 


>C1:=solve(eq1,_C1);

                                  _C2 cos( π2(1+2  k̃) ã
                                              (−b̃+ã)
                                                          )
                         C1 :=
                                     sin( π2(1+2  k̃) ã
                                             (−b̃+ã)
                                                         )

>simplify(subs(_C1=C1,y(x))):combine(%);

                       _C2 sin( π ã−π x+2  π ã k̃−2 π x k̃
                                        −2 b̃+2 ã
                                                             )
                                 sin( π−2
                                       ã+2 π ã k̃
                                          b̃+2 ã
                                                    )

>Yn:=unapply(select(has,%,[x]),x,k);

                                     π ã − π x + 2 π ã k̃ − 2 π x k̃
              Yn := (x, k̃) → sin(                                     )
                                               −2 b̃ + 2 ã
Ïðîâåðèì äèôôåðåíöèàëüíîå óðàâíåíèå:


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