Уравнения математической физики (краевые задачи в пространствах Соболева). Салехов Л.Г - 14 стр.

                N
   a) pRIBLIVENNYE REENIQ.
   pUSTX m 2  POLOVIM
                                        X
                                        m
                          um (t x) =         gmi (t)i(x)
                                        i=0
GDE gmi DOLVNY BYTX PODOBRANY TAK, ^TOBY WYPOLNQLISX USLOWIQ:
  8 u (t x) = 0                                      ESLI t < 0 x 2 
  < m0
  : (um(t)jj ) + (grad um(t)jgrad j ) = (f (t)jj ) ESLI t > 0
     um (0) = u0m 
                                                                                       (I )
GDE um ESTX LINEJNAQ KOMBINACIQ IZ 1 2  : : :  m .
      0

   N.B. zDESX WWEDENY OBOZNA^ENIQ:
          um (t) = um (t x) u0m (t) = @um@t(t x)  f (t) = f (t x):
   |TI USLOWIQ BUDUT WYPOLNQTXSQ, ESLI
          8 g (t) = 0
          < mj0       P   m
                                                    ESLI t < 0
          : ggmj (t) + i=0 gmi (t)aij = fj (t) j = 0 1 : : :  m
              mj (0) = gmj
                         0 

GDE POLOVILI
 aij = (ijj )H () = (grad i jgrad j ) fj (t) = (f (t)jj ) gmi
                 1
                 0
                                                                   0
                                                                      = (u0m jj ):
iZWESTNO, ^TO \TA SISTEMA OBLADAET ODNIM I TOLXKO ODNIM REENIEM
W SMYSLE OBOB]ENNYH FUNKCIJ I ^TO gmj (t) 2 C (I ). dLQ DOKAZATELXST-
WA \TOGO MOVNO ISPOLXZOWATX ILI METOD \LEMENTARNYH REENIJ, ILI
METOD WARIACII POSTOQNNYH.
   b) aPRIORNAQ OCENKA.
   uMNOVAQ RAWENSTWO (I) NA gmj (t) I SUMMIRUQ PO j OT 0 DO m, POLU^AEM
SOOTNOENIE \NERGII:
                (u0m (t)jum(t)) + kum(t)k2H () = (f (t)jum(t)):
                                               1
                                               0

   wZQW WE]ESTWENNYE ^ASTI OT OBEIH ^ASTEJ RAWENSTWA I PROINTEGRI-
ROWAW PO t NA 0 t], IMEEM:
     1 (ku (t)k2 ; ku (0)k2) +
                                  Zt                               Zt
     2 m             m                 kum(s)kH    2
                                                       1
                                                       0 ()
                                                               ds = (f (s)jum(s))ds:
                                   0                                0

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