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A solution <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503301.png" /> of a non-linear operator equation <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503302.png" />, in which <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503303.png" /> plays the role of parameter and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503304.png" /> that of the unknown. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503305.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503306.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503307.png" /> be Banach spaces and let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503308.png" /> be a non-linear operator that is continuous in a neighbourhood <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i0503309.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033010.png" /> and that maps <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033011.png" /> into a neighbourhood of zero in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033012.png" />. If the [[Fréchet derivative|Fréchet derivative]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033013.png" /> is continuous on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033014.png" />, if the operator <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033015.png" /> exists and is continuous and if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033016.png" />, then there are numbers <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033017.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033018.png" /> such that for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033019.png" /> the equation <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033020.png" /> has a unique solution <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033021.png" /> in the ball <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033022.png" />. Here if, additionally, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033023.png" /> is <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033024.png" /> times differentiable in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033025.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033026.png" /> is <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033027.png" /> times differentiable. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033028.png" /> is an [[Analytic operator|analytic operator]] in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033029.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050330/i05033030.png" /> is also analytic. These assertions generalize well-known propositions about implicit functions. For degenerate cases, see [[Branching of solutions|Branching of solutions]] of non-linear equations.
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A solution $y=f(x)$ of a non-linear operator equation $F(x,y)=0$, in which $x$ plays the role of parameter and $y$ that of the unknown. Let $X$, $Y$ and $Z$ be Banach spaces and let $F(x,y)$ be a non-linear operator that is continuous in a neighbourhood $\Omega$ of $(x_0,y_0)\in X\dotplus Y$ and that maps $\Omega$ into a neighbourhood of zero in $Z$. If the [[Fréchet derivative|Fréchet derivative]] $F_y(x,y)$ is continuous on $\Omega$, if the operator $[F_y(x_0,y_0)]^{-1}$ exists and is continuous and if $F(x_0,y_0)=0$, then there are numbers $\epsilon>0$ and $\delta>0$ such that for $||x-x_0||<\delta$ the equation $F(x,y)=0$ has a unique solution $y=f(x)$ in the ball $||y-y_0||<\epsilon$. Here if, additionally, $F(x,y)$ is $n$ times differentiable in $\Omega$, then $f(x)$ is $n$ times differentiable. If $F(x,y)$ is an [[Analytic operator|analytic operator]] in $\Omega$, then $f(x)$ is also analytic. These assertions generalize well-known propositions about implicit functions. For degenerate cases, see [[Branching of solutions|Branching of solutions]] of non-linear equations.
  
 
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Latest revision as of 14:17, 23 July 2014

A solution $y=f(x)$ of a non-linear operator equation $F(x,y)=0$, in which $x$ plays the role of parameter and $y$ that of the unknown. Let $X$, $Y$ and $Z$ be Banach spaces and let $F(x,y)$ be a non-linear operator that is continuous in a neighbourhood $\Omega$ of $(x_0,y_0)\in X\dotplus Y$ and that maps $\Omega$ into a neighbourhood of zero in $Z$. If the Fréchet derivative $F_y(x,y)$ is continuous on $\Omega$, if the operator $[F_y(x_0,y_0)]^{-1}$ exists and is continuous and if $F(x_0,y_0)=0$, then there are numbers $\epsilon>0$ and $\delta>0$ such that for $||x-x_0||<\delta$ the equation $F(x,y)=0$ has a unique solution $y=f(x)$ in the ball $||y-y_0||<\epsilon$. Here if, additionally, $F(x,y)$ is $n$ times differentiable in $\Omega$, then $f(x)$ is $n$ times differentiable. If $F(x,y)$ is an analytic operator in $\Omega$, then $f(x)$ is also analytic. These assertions generalize well-known propositions about implicit functions. For degenerate cases, see Branching of solutions of non-linear equations.

References

[1] T.H. Hildebrandt, L.M. Graves, "Implicit functions and their differences in general analysis" Trans. Amer. Math. Soc. , 29 (1927) pp. 127–153
[2] W.I. [V.I. Sobolev] Sobolew, "Elemente der Funktionalanalysis" , H. Deutsch , Frankfurt a.M. (1979) (Translated from Russian)
[3] M.M. Vainberg, V.A. Trenogin, "Theory of branching of solutions of non-linear equations" , Noordhoff (1974) (Translated from Russian)
[4] L. Nirenberg, "Topics in nonlinear functional analysis" , New York Univ. Inst. Math. Mech. (1974)


Comments

References

[a1] M.S. Berger, "Nonlinearity and functional analysis" , Acad. Press (1977)
How to Cite This Entry:
Implicit operator. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Implicit_operator&oldid=16566
This article was adapted from an original article by V.A. Trenogin (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article