Generalized Ordinary Differential Equations in Abstract Spaces and Applications. Группа авторов

Чтение книги онлайн.

Читать онлайн книгу Generalized Ordinary Differential Equations in Abstract Spaces and Applications - Группа авторов страница 23

Generalized Ordinary Differential Equations in Abstract Spaces and Applications - Группа авторов

Скачать книгу

alt="left-brace n Subscript k Baseline right-brace Subscript k element-of double-struck upper N Baseline subset-of double-struck upper N"/>, with n Subscript k plus 1 Baseline greater-than n Subscript k, for which left-brace f Subscript n Sub Subscript k Subscript Baseline left-parenthesis t Subscript i Baseline right-parenthesis right-brace Subscript k element-of double-struck upper N and left-brace f Subscript n Sub Subscript k Subscript Baseline left-parenthesis tau Subscript i Baseline right-parenthesis right-brace Subscript k element-of double-struck upper N are also relatively compact sets of upper X, for every i.

      This last statement implies that there exist left-brace y Subscript i Baseline colon i equals 0 comma 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue right-brace subset-of upper X and left-brace z Subscript i Baseline colon i equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue right-brace subset-of upper X satisfying

y Subscript i Baseline equals limit Underscript k right-arrow infinity Endscripts f Subscript n Sub Subscript k Subscript Baseline left-parenthesis t Subscript i Baseline right-parenthesis and z Subscript i Baseline equals limit Underscript k right-arrow infinity Endscripts f Subscript n Sub Subscript k Subscript Baseline left-parenthesis tau Subscript i Baseline right-parenthesis period

      Thus, there exists upper N element-of double-struck upper N such that

vertical-bar vertical-bar vertical-bar vertical-bar minus minus of ffnk left-parenthesis right-parenthesis tiyi less-than StartFraction epsilon Over 4 EndFraction and vertical-bar vertical-bar vertical-bar vertical-bar minus minus of ffnk left-parenthesis right-parenthesis tau izi less-than StartFraction epsilon Over 4 EndFraction comma parallel-to f Subscript n Sub Subscript q Subscript Baseline left-parenthesis t Subscript i Baseline right-parenthesis minus y Subscript i Baseline parallel-to less-than StartFraction epsilon Over 4 EndFraction and parallel-to f Subscript n Sub Subscript q Subscript Baseline left-parenthesis tau Subscript i Baseline right-parenthesis minus z Subscript i Baseline parallel-to less-than StartFraction epsilon Over 4 EndFraction comma

      for i equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue.

      Take t element-of left-bracket a comma b right-bracket and consider q element-of double-struck upper N such that q greater-than k. Therefore, either t equals t Subscript i, for some i element-of StartSet 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue EndSet, in which case, we have

parallel-to f Subscript n Sub Subscript k Subscript Baseline left-parenthesis t right-parenthesis minus f Subscript n Sub Subscript q Subscript Baseline left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f Subscript n Sub Subscript k Subscript Baseline left-parenthesis t Subscript i Baseline right-parenthesis minus y Subscript i Baseline parallel-to plus parallel-to f Subscript n Sub Subscript q Subscript Baseline left-parenthesis t Subscript i Baseline right-parenthesis minus y Subscript i Baseline parallel-to less-than StartFraction epsilon Over 2 EndFraction comma

      or t element-of left-parenthesis t Subscript i minus 1 Baseline comma t Subscript i Baseline right-parenthesis, for some i element-of StartSet 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue EndSet, in which case, we have

StartLayout 1st Row 1st Column parallel-to f Subscript n Sub Subscript k Baseline left-parenthesis t right-parenthesis minus f Subscript n Sub Subscript q Baseline left-parenthesis t right-parenthesis parallel-to 2nd Column less-than-or-slanted-equals parallel-to f Subscript n Sub Subscript k Baseline left-parenthesis t right-parenthesis minus f Subscript n Sub Subscript k Baseline left-parenthesis tau Subscript i Baseline right-parenthesis parallel-to plus parallel-to f Subscript n Sub Subscript k Baseline left-parenthesis tau Subscript i Baseline right-parenthesis minus f Subscript n Sub Subscript q Baseline left-parenthesis tau Subscript i Baseline right-parenthesis parallel-to plus parallel-to f Subscript n Sub Subscript q Baseline left-parenthesis t right-parenthesis minus f Subscript n Sub Subscript q Baseline left-parenthesis tau Subscript i Baseline right-parenthesis parallel-to 2nd Row 1st Column Blank 2nd Column less-than-or-slanted-equals parallel-to f Subscript n Sub Subscript k Baseline left-parenthesis t right-parenthesis minus f Subscript n Sub Subscript k Baseline left-parenthesis tau Subscript i Baseline right-parenthesis parallel-to plus parallel-to f Subscript n Sub Subscript q Baseline left-parenthesis t right-parenthesis minus f Subscript n Sub Subscript q Baseline left-parenthesis tau Subscript i Baseline right-parenthesis parallel-to plus parallel-to f Subscript n Sub Subscript k Baseline left-parenthesis tau Subscript i Baseline right-parenthesis minus z Subscript i Baseline parallel-to 3rd Row 1st Column Blank 2nd Column plus parallel-to f Subscript n Sub Subscript q Subscript Baseline left-parenthesis tau Subscript i Baseline right-parenthesis minus z Subscript i Baseline parallel-to less-than StartFraction epsilon Over 4 EndFraction plus StartFraction epsilon Over 4 EndFraction plus StartFraction epsilon Over 4 EndFraction plus StartFraction epsilon Over 4 EndFraction equals epsilon period EndLayout

      Hence, for every t element-of left-bracket a comma b right-bracket, left-brace f Subscript n Sub Subscript k Subscript Baseline left-parenthesis t right-parenthesis right-brace Subscript k element-of double-struck upper N Baseline subset-of upper X satisfies the Cauchy condition. Due to the fact that upper X is a complete space and because left-brace f Subscript n Sub Subscript k Subscript Baseline left-parenthesis t right-parenthesis right-brace Subscript k element-of double-struck upper N is a Cauchy sequence, the limit limit Underscript k right-arrow infinity Endscripts f Subscript n Sub Subscript k Baseline left-parenthesis t right-parenthesis exists.

      We conclude by considering

Скачать книгу