(Nonlinear after Linear)-Differential Equations
I know the theory needed to solve nonhomogeneous linear differential equations looking like
$$dot{x}(t) = A(t)x + b(t)tag{1}$$
in finite-dimensional spaces.
Let $A(t)$ be a linear form, i.e. mapping into $mathbb{R}$, and $b$ be real-valued. If I have a monotonous smooth diffeomorphism $varphiinoperatorname{Diff}^infty(mathbb{R},[0,1])$, how does solving $(1)$ relate to solving the possibly non-linear differential equation
$$dot{x}(t) = varphileft(A(t)x + b(t)right),tag{2}$$
in terms of solvability, stability etc. Do the solutions of $(1)$ and $(2)$ relate at all?
Many thanks in advance.
real-analysis differential-equations diffeomorphism
asked Nov 21 '18 at 16:11
Ramen
453412
add a comment |
I know the theory needed to solve nonhomogeneous linear differential equations looking like
$$dot{x}(t) = A(t)x + b(t)tag{1}$$
in finite-dimensional spaces.
Let $A(t)$ be a linear form, i.e. mapping into $mathbb{R}$, and $b$ be real-valued. If I have a monotonous smooth diffeomorphism $varphiinoperatorname{Diff}^infty(mathbb{R},[0,1])$, how does solving $(1)$ relate to solving the possibly non-linear differential equation
$$dot{x}(t) = varphileft(A(t)x + b(t)right),tag{2}$$
in terms of solvability, stability etc. Do the solutions of $(1)$ and $(2)$ relate at all?
Many thanks in advance.
real-analysis differential-equations diffeomorphism
asked Nov 21 '18 at 16:11
Ramen
453412
Even in the case where $varphi(X) = Phi X$ is linear, we can still get drastic changes in the behavior of solutions, since $Phi A$ can have very different eigenvalues/vectors than $A$. In the nonlinear case, solvability of $varphi(Ax+b)$ is more dependent on the solvability of $varphi(x)$.
– AlexanderJ93
Nov 21 '18 at 21:30
@AlexanderJ93 Thank you for your comment. Note that if $varphi$ was linear, it would essentially be just a number so eigenvalues/vectors would only be scaled, right? Also, $varphi$ wouldn't be bounded. I am talking specifically about the smooth, monotonous $mathbb{R}to [0,1]$ case, so I hoped one could say something, anything... Still, I expected the answer being along your lines to be the more probable case.
– Ramen
Nov 22 '18 at 19:06
add a comment |
I know the theory needed to solve nonhomogeneous linear differential equations looking like
$$dot{x}(t) = A(t)x + b(t)tag{1}$$
in finite-dimensional spaces.
Let $A(t)$ be a linear form, i.e. mapping into $mathbb{R}$, and $b$ be real-valued. If I have a monotonous smooth diffeomorphism $varphiinoperatorname{Diff}^infty(mathbb{R},[0,1])$, how does solving $(1)$ relate to solving the possibly non-linear differential equation
$$dot{x}(t) = varphileft(A(t)x + b(t)right),tag{2}$$
in terms of solvability, stability etc. Do the solutions of $(1)$ and $(2)$ relate at all?
Many thanks in advance.
real-analysis differential-equations diffeomorphism
asked Nov 21 '18 at 16:11
Ramen
453412
I know the theory needed to solve nonhomogeneous linear differential equations looking like
$$dot{x}(t) = A(t)x + b(t)tag{1}$$
in finite-dimensional spaces.
Let $A(t)$ be a linear form, i.e. mapping into $mathbb{R}$, and $b$ be real-valued. If I have a monotonous smooth diffeomorphism $varphiinoperatorname{Diff}^infty(mathbb{R},[0,1])$, how does solving $(1)$ relate to solving the possibly non-linear differential equation
$$dot{x}(t) = varphileft(A(t)x + b(t)right),tag{2}$$
in terms of solvability, stability etc. Do the solutions of $(1)$ and $(2)$ relate at all?
Many thanks in advance.
real-analysis differential-equations diffeomorphism
real-analysis differential-equations diffeomorphism
asked Nov 21 '18 at 16:11
Ramen
453412
asked Nov 21 '18 at 16:11
Ramen
453412
edited Nov 22 '18 at 19:06
asked Nov 21 '18 at 16:11
Ramen
453412
asked Nov 21 '18 at 16:11
Ramen
453412
asked Nov 21 '18 at 16:11
Ramen
453412
453412
Even in the case where $varphi(X) = Phi X$ is linear, we can still get drastic changes in the behavior of solutions, since $Phi A$ can have very different eigenvalues/vectors than $A$. In the nonlinear case, solvability of $varphi(Ax+b)$ is more dependent on the solvability of $varphi(x)$.
– AlexanderJ93
Nov 21 '18 at 21:30
@AlexanderJ93 Thank you for your comment. Note that if $varphi$ was linear, it would essentially be just a number so eigenvalues/vectors would only be scaled, right? Also, $varphi$ wouldn't be bounded. I am talking specifically about the smooth, monotonous $mathbb{R}to [0,1]$ case, so I hoped one could say something, anything... Still, I expected the answer being along your lines to be the more probable case.
– Ramen
Nov 22 '18 at 19:06
add a comment |
Even in the case where $varphi(X) = Phi X$ is linear, we can still get drastic changes in the behavior of solutions, since $Phi A$ can have very different eigenvalues/vectors than $A$. In the nonlinear case, solvability of $varphi(Ax+b)$ is more dependent on the solvability of $varphi(x)$.
– AlexanderJ93
Nov 21 '18 at 21:30
@AlexanderJ93 Thank you for your comment. Note that if $varphi$ was linear, it would essentially be just a number so eigenvalues/vectors would only be scaled, right? Also, $varphi$ wouldn't be bounded. I am talking specifically about the smooth, monotonous $mathbb{R}to [0,1]$ case, so I hoped one could say something, anything... Still, I expected the answer being along your lines to be the more probable case.
– Ramen
Nov 22 '18 at 19:06
Even in the case where $varphi(X) = Phi X$ is linear, we can still get drastic changes in the behavior of solutions, since $Phi A$ can have very different eigenvalues/vectors than $A$. In the nonlinear case, solvability of $varphi(Ax+b)$ is more dependent on the solvability of $varphi(x)$.
– AlexanderJ93
Nov 21 '18 at 21:30
Even in the case where $varphi(X) = Phi X$ is linear, we can still get drastic changes in the behavior of solutions, since $Phi A$ can have very different eigenvalues/vectors than $A$. In the nonlinear case, solvability of $varphi(Ax+b)$ is more dependent on the solvability of $varphi(x)$.
– AlexanderJ93
Nov 21 '18 at 21:30
@AlexanderJ93 Thank you for your comment. Note that if $varphi$ was linear, it would essentially be just a number so eigenvalues/vectors would only be scaled, right? Also, $varphi$ wouldn't be bounded. I am talking specifically about the smooth, monotonous $mathbb{R}to [0,1]$ case, so I hoped one could say something, anything... Still, I expected the answer being along your lines to be the more probable case.
– Ramen
Nov 22 '18 at 19:06
@AlexanderJ93 Thank you for your comment. Note that if $varphi$ was linear, it would essentially be just a number so eigenvalues/vectors would only be scaled, right? Also, $varphi$ wouldn't be bounded. I am talking specifically about the smooth, monotonous $mathbb{R}to [0,1]$ case, so I hoped one could say something, anything... Still, I expected the answer being along your lines to be the more probable case.
– Ramen
Nov 22 '18 at 19:06
add a comment |
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Even in the case where $varphi(X) = Phi X$ is linear, we can still get drastic changes in the behavior of solutions, since $Phi A$ can have very different eigenvalues/vectors than $A$. In the nonlinear case, solvability of $varphi(Ax+b)$ is more dependent on the solvability of $varphi(x)$.
– AlexanderJ93
Nov 21 '18 at 21:30
@AlexanderJ93 Thank you for your comment. Note that if $varphi$ was linear, it would essentially be just a number so eigenvalues/vectors would only be scaled, right? Also, $varphi$ wouldn't be bounded. I am talking specifically about the smooth, monotonous $mathbb{R}to [0,1]$ case, so I hoped one could say something, anything... Still, I expected the answer being along your lines to be the more probable case.
– Ramen
Nov 22 '18 at 19:06