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Solve the nonlinear equation
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Suppose that $f:Eto F$(between Banach spaces), is of the form
$$f(x)=f(0)+D(x)+N(x).$$
Here $D$ is a linear term, whose kernel is of finite dimension, and admits a right inverse $G$, i.e. $D(G)(cdot)=Id_F$. The nonlinear term $N(x)$ satisfies that
$$|G(N(x))-G(N(y))|leq C(|x|+|y|)|x-y|$$
for some constant $C>0$ and $x,y$ in a small neighborhood $B_epsilon(C)(0)$.
Q: How to show that there is a unique zero point $x_0$ of $f$, i.e. $f(x_0)=0$, in $B_epsilon(0)cap G(F)$, with the initial condition $|G(f(0))|leqepsilon/2$?
functional-analysis differential-equations nonlinear-analysis
asked Aug 6 at 1:03
DLIN
376314
add a comment |Â
up vote
1
down vote
favorite
Suppose that $f:Eto F$(between Banach spaces), is of the form
$$f(x)=f(0)+D(x)+N(x).$$
Here $D$ is a linear term, whose kernel is of finite dimension, and admits a right inverse $G$, i.e. $D(G)(cdot)=Id_F$. The nonlinear term $N(x)$ satisfies that
$$|G(N(x))-G(N(y))|leq C(|x|+|y|)|x-y|$$
for some constant $C>0$ and $x,y$ in a small neighborhood $B_epsilon(C)(0)$.
Q: How to show that there is a unique zero point $x_0$ of $f$, i.e. $f(x_0)=0$, in $B_epsilon(0)cap G(F)$, with the initial condition $|G(f(0))|leqepsilon/2$?
functional-analysis differential-equations nonlinear-analysis
asked Aug 6 at 1:03
DLIN
376314
1
What do you mean by a "zero point" of $f$? A fixed point, that is, $x_0$ such that $f(x_0) = x_0$? No, that can't be right since $f: E to F$; so then what? Cheers!
– Robert Lewis
Aug 6 at 1:18
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that $f:Eto F$(between Banach spaces), is of the form
$$f(x)=f(0)+D(x)+N(x).$$
Here $D$ is a linear term, whose kernel is of finite dimension, and admits a right inverse $G$, i.e. $D(G)(cdot)=Id_F$. The nonlinear term $N(x)$ satisfies that
$$|G(N(x))-G(N(y))|leq C(|x|+|y|)|x-y|$$
for some constant $C>0$ and $x,y$ in a small neighborhood $B_epsilon(C)(0)$.
Q: How to show that there is a unique zero point $x_0$ of $f$, i.e. $f(x_0)=0$, in $B_epsilon(0)cap G(F)$, with the initial condition $|G(f(0))|leqepsilon/2$?
functional-analysis differential-equations nonlinear-analysis
asked Aug 6 at 1:03
DLIN
376314
Suppose that $f:Eto F$(between Banach spaces), is of the form
$$f(x)=f(0)+D(x)+N(x).$$
Here $D$ is a linear term, whose kernel is of finite dimension, and admits a right inverse $G$, i.e. $D(G)(cdot)=Id_F$. The nonlinear term $N(x)$ satisfies that
$$|G(N(x))-G(N(y))|leq C(|x|+|y|)|x-y|$$
for some constant $C>0$ and $x,y$ in a small neighborhood $B_epsilon(C)(0)$.
Q: How to show that there is a unique zero point $x_0$ of $f$, i.e. $f(x_0)=0$, in $B_epsilon(0)cap G(F)$, with the initial condition $|G(f(0))|leqepsilon/2$?
functional-analysis differential-equations nonlinear-analysis
asked Aug 6 at 1:03
DLIN
376314
edited Aug 7 at 13:47
asked Aug 6 at 1:03
DLIN
376314
asked Aug 6 at 1:03
DLIN
376314
asked Aug 6 at 1:03
DLIN
376314
376314
1
What do you mean by a "zero point" of $f$? A fixed point, that is, $x_0$ such that $f(x_0) = x_0$? No, that can't be right since $f: E to F$; so then what? Cheers!
– Robert Lewis
Aug 6 at 1:18
add a comment |Â
1
What do you mean by a "zero point" of $f$? A fixed point, that is, $x_0$ such that $f(x_0) = x_0$? No, that can't be right since $f: E to F$; so then what? Cheers!
– Robert Lewis
Aug 6 at 1:18
1
1
What do you mean by a "zero point" of $f$? A fixed point, that is, $x_0$ such that $f(x_0) = x_0$? No, that can't be right since $f: E to F$; so then what? Cheers!
– Robert Lewis
Aug 6 at 1:18
What do you mean by a "zero point" of $f$? A fixed point, that is, $x_0$ such that $f(x_0) = x_0$? No, that can't be right since $f: E to F$; so then what? Cheers!
– Robert Lewis
Aug 6 at 1:18
add a comment |Â
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1
What do you mean by a "zero point" of $f$? A fixed point, that is, $x_0$ such that $f(x_0) = x_0$? No, that can't be right since $f: E to F$; so then what? Cheers!
– Robert Lewis
Aug 6 at 1:18