A resolvent for an iteration method for nonlinear partial differential equations
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- by J. W. Neuberger
- Trans. Amer. Math. Soc. 226 (1977), 321-343
- DOI: https://doi.org/10.1090/S0002-9947-1977-0425705-X
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Abstract:
For each of m and n a positive integer denote by $S(m,i)$ the space of all real-valued symmetric i-linear functions on ${E_m},i = 1,2, \ldots ,n$. Denote by L a nonzero linear functional on $S(m,n)$, denote by f a real-valued analytic function on ${E_m} \times R \times S(m,1) \times \cdots \times S(m,[n/2])$ and denote by $\alpha$ a member of $D(f)$. Denote by H the space of all real-valued functions U, analytic at the origin of ${E_m}$, so that $\alpha = (0,U(0),U’(0), \ldots ,{U^{([n/2])}}(0))$. For $U \in H,{f_U}(x) \equiv f(x,U(x),U’(x), \ldots ,{U^{([n/2])}}(x))$ for all x for which this is defined. A one-parameter semigroup (nonlinear if $f \ne 0$) K on H is constructed so that if $U \in K$, then $K(\lambda )U$ converges, as $\lambda \to \infty$, to a solution Y to the partial differential equation $L{Y^{(n)}} = {f_Y}$. A resolvent j for this semigroup is determined so that $J(\lambda )U$ also converges to y as $\lambda \to \infty$ and so that $J{(\lambda /n)^n}U$ converges to $K(\lambda )U$ as $n \to \infty$. The solutions $Y \in H$ of $L{Y^{(n)}} = {f_Y}$ are precisely the fixed points of the semigroup K.References
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Bibliographic Information
- © Copyright 1977 American Mathematical Society
- Journal: Trans. Amer. Math. Soc. 226 (1977), 321-343
- MSC: Primary 47H99; Secondary 35A35, 35R20
- DOI: https://doi.org/10.1090/S0002-9947-1977-0425705-X
- MathSciNet review: 0425705