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Generalized Ordinary Differential Equations in Abstract Spaces and Applications


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d EndAbsoluteValue"/>, we have sigma-summation Underscript i equals 1 Overscript StartAbsoluteValue d EndAbsoluteValue Endscripts vertical-bar vertical-bar vertical-bar vertical-bar minus minus of ff left-parenthesis right-parenthesis ti of ff left-parenthesis right-parenthesis t minus minus i 1 less-than epsilon period

      If we denote by upper A upper C left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis the space of all absolutely continuous functions from left-bracket a comma b right-bracket to upper X, then it is not difficult to prove that

upper A upper C left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis subset-of upper S upper L left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis subset-of upper C left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis period

      In upper S upper L left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis, we consider the usual supremum norm, parallel-to dot parallel-to Subscript infinity Baseline, induced by upper C left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis.

      The next two versions of the Fundamental Theorem of Calculus for Henstock vector integrals, as described in Definition 1.41, are borrowed from [70, Theorems 1 and 2]. We use the term almost everywhere in the sense of the Lebesgue measure m.

      Theorem 1.77: If and are both differentiable and is such that for almost every , then and , that is,

integral Subscript a Superscript t Baseline alpha left-parenthesis s right-parenthesis f prime left-parenthesis s right-parenthesis d s equals integral Subscript a Superscript t Baseline alpha left-parenthesis s right-parenthesis d f left-parenthesis s right-parenthesis comma t element-of left-bracket a comma b right-bracket period

      Theorem 1.78: If is differentiable and is bounded, then , and there exists the derivative for almost every , that is,

StartFraction d Over d t EndFraction left-bracket integral Subscript a Superscript t Baseline alpha left-parenthesis s right-parenthesis d f left-parenthesis s right-parenthesis right-bracket equals alpha left-parenthesis t right-parenthesis f prime left-parenthesis t right-parenthesis comma almost everywhere in left-bracket a comma b right-bracket period

      Corollary 1.79: Suppose is differentiable and nonconstant on any nondegenerate subinterval of and is bounded and such that . Then, almost everywhere in .

      From Corollary 1.50, we know that if f element-of upper C left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis and alpha element-of upper K Subscript f Baseline left-parenthesis left-bracket a comma b right-bracket comma upper L left-parenthesis upper X comma upper Y right-parenthesis right-parenthesis, then alpha overTilde Subscript f Baseline element-of upper C left-parenthesis left-bracket a comma b right-bracket comma upper Y right-parenthesis. For the Henstock vector integral, we have the following analogue whose proof can be found in [70, Theorem 7].

      Theorem 1.80: If and , then we have .

      The next result is borrowed from [70, Theorem 5]. We reproduce its proof here.

      Theorem 1.81: Suppose and is such that almost everywhere on . Then, and , that is,

ModifyingAbove alpha With tilde Subscript f Baseline left-parenthesis t right-parenthesis equals integral Subscript a Superscript t Baseline alpha left-parenthesis s right-parenthesis d f left-parenthesis s right-parenthesis equals 0 comma for every t element-of left-bracket a comma b right-bracket period

      Proof. Consider the sets

StartLayout 1st Row 1st Column Blank 2nd Column upper E equals StartSet t element-of double-struck upper R colon alpha left-parenthesis t right-parenthesis not-equals 0 EndSet and upper E Subscript n Baseline equals StartSet t element-of upper E colon n minus 1 less-than parallel-to alpha left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals n EndSet comma 2nd Row 1st Column Blank 2nd Column for each n element-of double-struck upper N period EndLayout

      By hypothesis, m left-parenthesis upper E right-parenthesis equals 0, where m denotes the Lebesgue measure. Hence, m left-parenthesis upper E Subscript n Baseline right-parenthesis equals 0 for every n element-of double-struck upper N. In addition, f element-of upper S upper L left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis. Then, for every n element-of double-struck upper N and every epsilon greater-than 0, there exists a gauge delta Subscript n on upper E Subscript n such that, for every delta Subscript n-fine tagged partial division d equals left-parenthesis xi Subscript n Sub Subscript i Subscript Baseline comma left-bracket t Subscript n Sub Subscript i Subscript minus 1 Baseline comma t Subscript n Sub Subscript i Subscript Baseline right-bracket right-parenthesis element-of italic upper T upper P upper D Subscript left-bracket a comma b right-bracket, with xi Subscript n Sub Subscript i Baseline element-of upper E Subscript n, for i equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d EndAbsoluteValue, we have

sigma-summation Underscript i equals 1 Overscript StartAbsoluteValue d EndAbsoluteValue Endscripts parallel-to f left-parenthesis t Subscript n Sub Subscript i Subscript Baseline right-parenthesis minus f left-parenthesis </p>
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